Introduction to
Tensor Calculus
and
Continuum Mechanics
by J.H. Heinbockel
Department of Mathematics and Statistics
Old Dominion University
PREFACE
This is an introductory text which presents fundamental concepts from the subject
areas of tensor calculus, differential geometry and continuum mechanics. The material
presented is suitable for a two semester course in applied mathematics and is flexible
enough to be presented to either upper level undergraduate or beginning graduate students
majoring in applied mathematics, engineering or physics. The presentation assumes the
students have some knowledge from the areas of matrix theory, linear algebra and advanced
calculus. Each section includes many illustrative worked examples. At the end of each
section there is a large collection of exercises which range in difficulty. Many new ideas
are presented in the exercises and so the students should be encouraged to read all the
exercises.
The purpose of preparing these notes is to condense into an introductory text the basic
definitions and techniques arising in tensor calculus, differential geometry and continuum
mechanics. In particular, the material is presented to (i) develop a physical understanding
of the mathematical concepts associated with tensor calculus and (ii) develop the basic
equations of tensor calculus, differential geometry and continuum mechanics which arise
in engineering applications. From these basic equations one can go on to develop more
sophisticated models of applied mathematics. The material is presented in an informal
manner and uses mathematics which minimizes excessive formalism.
The material has been divided into two parts. The first part deals with an introduction to tensor calculus and differential geometry which covers such things as the indicial
notation, tensor algebra, covariant differentiation, dual tensors, bilinear and multilinear
forms, special tensors, the Riemann Christoffel tensor, space curves, surface curves, curvature and fundamental quadratic forms. The second part emphasizes the application of
tensor algebra and calculus to a wide variety of applied areas from engineering and physics.
The selected applications are from the areas of dynamics, elasticity, fluids and electromagnetic theory. The continuum mechanics portion focuses on an introduction of the basic
concepts from linear elasticity and fluids. The Appendix A contains units of measurements
from the Système International d’Unitès along with some selected physical constants. The
Appendix B contains a listing of Christoffel symbols of the second kind associated with
various coordinate systems. The Appendix C is a summary of useful vector identities.
J.H. Heinbockel, 1996
Copyright c 1996 by J.H. Heinbockel. All rights reserved.
Reproduction and distribution of these notes is allowable provided it is for non-profit
purposes only.
INTRODUCTION TO
TENSOR CALCULUS
AND
CONTINUUM MECHANICS
PART 1: INTRODUCTION TO TENSOR CALCULUS
§1.1 INDEX NOTATION
. . . . . . . . . . . . . .
Exercise 1.1
. . . . . . . . . . . . . . . . . . . . .
§1.2 TENSOR CONCEPTS AND TRANSFORMATIONS
Exercise 1.2 . . . . . . . . . . . . . . . . . . . . . .
§1.3 SPECIAL TENSORS . . . . . . . . . . . . . .
Exercise 1.3 . . . . . . . . . . . . . . . . . . . . . .
§1.4 DERIVATIVE OF A TENSOR . . . . . . . . . .
Exercise 1.4 . . . . . . . . . . . . . . . . . . . . . .
§1.5 DIFFERENTIAL GEOMETRY AND RELATIVITY
Exercise 1.5 . . . . . . . . . . . . . . . . . . . . . .
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1
28
35
54
65
101
108
123
129
162
PART 2: INTRODUCTION TO CONTINUUM MECHANICS
§2.1 TENSOR NOTATION FOR VECTOR QUANTITIES . . . .
Exercise 2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . .
§2.2 DYNAMICS . . . . . . . . . . . . . . . . . . . . . .
Exercise 2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . .
§2.3 BASIC EQUATIONS OF CONTINUUM MECHANICS . . .
Exercise 2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . .
§2.4 CONTINUUM MECHANICS (SOLIDS)
. . . . . . . . .
Exercise 2.4 . . . . . . . . . . . . . . . . . . . . . . . . . . .
§2.5 CONTINUUM MECHANICS (FLUIDS)
. . . . . . . . .
Exercise 2.5 . . . . . . . . . . . . . . . . . . . . . . . . . . .
§2.6 ELECTRIC AND MAGNETIC FIELDS . . . . . . . . . .
Exercise 2.6 . . . . . . . . . . . . . . . . . . . . . . . . . . .
BIBLIOGRAPHY
. . . . . . . . . . . . . . . . . . . . .
APPENDIX A
UNITS OF MEASUREMENT . . . . . . .
APPENDIX B
CHRISTOFFEL SYMBOLS OF SECOND KIND
APPENDIX C
VECTOR IDENTITIES . . . . . . . . . .
INDEX . . . . . . . . . . . . . . . . . . . . . . . . . .
171
182
187
206
211
238
243
272
282
317
325
347
352
353
355
362
363
1
PART 1: INTRODUCTION TO TENSOR CALCULUS
A scalar field describes a one-to-one correspondence between a single scalar number and a point. An ndimensional vector field is described by a one-to-one correspondence between n-numbers and a point. Let us
generalize these concepts by assigning n-squared numbers to a single point or n-cubed numbers to a single
point. When these numbers obey certain transformation laws they become examples of tensor fields. In
general, scalar fields are referred to as tensor fields of rank or order zero whereas vector fields are called
tensor fields of rank or order one.
Closely associated with tensor calculus is the indicial or index notation. In section 1 the indicial
notation is defined and illustrated. We also define and investigate scalar, vector and tensor fields when they
are subjected to various coordinate transformations. It turns out that tensors have certain properties which
are independent of the coordinate system used to describe the tensor. Because of these useful properties,
we can use tensors to represent various fundamental laws occurring in physics, engineering, science and
mathematics. These representations are extremely useful as they are independent of the coordinate systems
considered.
§1.1 INDEX NOTATION
~ and B
~ can be expressed in the component form
Two vectors A
~ = A1 b
e1 + A2 b
e2 + A3 b
e3
A
and
~ = B1 b
e1 + B2 b
e2 + B3 b
e3 ,
B
~ and
e2 and b
e3 are orthogonal unit basis vectors. Often when no confusion arises, the vectors A
where b
e1 , b
~ are expressed for brevity sake as number triples. For example, we can write
B
~ = (A1 , A2 , A3 )
A
and
~ = (B1 , B2 , B3 )
B
~ and B
~ are given. The unit vectors would
where it is understood that only the components of the vectors A
be represented
b
e1 = (1, 0, 0),
b
e2 = (0, 1, 0),
b
e3 = (0, 0, 1).
~ and B
~ is the index or indicial notation. In the index notation,
A still shorter notation, depicting the vectors A
the quantities
Ai ,
i = 1, 2, 3
and
Bp ,
p = 1, 2, 3
~ and B.
~ This notation focuses attention only on the components of
represent the components of the vectors A
the vectors and employs a dummy subscript whose range over the integers is specified. The symbol Ai refers
~ simultaneously. The dummy subscript i can have any of the integer
to all of the components of the vector A
~ Setting i = 2 focuses
values 1, 2 or 3. For i = 1 we focus attention on the A1 component of the vector A.
~ and similarly when i = 3 we can focus attention on
attention on the second component A2 of the vector A
~ The subscript i is a dummy subscript and may be replaced by another letter, say
the third component of A.
p, so long as one specifies the integer values that this dummy subscript can have.
2
It is also convenient at this time to mention that higher dimensional vectors may be defined as ordered
n−tuples. For example, the vector
~ = (X1 , X2 , . . . , XN )
X
with components Xi , i = 1, 2, . . . , N is called a N −dimensional vector. Another notation used to represent
this vector is
~ = X1 b
e1 + X2 b
e2 + · · · + XN b
eN
X
where
b
e1 , b
e2 , . . . , b
eN
are linearly independent unit base vectors. Note that many of the operations that occur in the use of the
index notation apply not only for three dimensional vectors, but also for N −dimensional vectors.
In future sections it is necessary to define quantities which can be represented by a letter with subscripts
or superscripts attached. Such quantities are referred to as systems. When these quantities obey certain
transformation laws they are referred to as tensor systems. For example, quantities like
Akij
eijk
δij
δij
Ai
Bj
aij .
The subscripts or superscripts are referred to as indices or suffixes. When such quantities arise, the indices
must conform to the following rules:
1. They are lower case Latin or Greek letters.
2. The letters at the end of the alphabet (u, v, w, x, y, z) are never employed as indices.
The number of subscripts and superscripts determines the order of the system. A system with one index
is a first order system. A system with two indices is called a second order system. In general, a system with
N indices is called a N th order system. A system with no indices is called a scalar or zeroth order system.
The type of system depends upon the number of subscripts or superscripts occurring in an expression.
m
, (all indices range 1 to N), are of the same type because they have the same
For example, Aijk and Bst
number of subscripts and superscripts. In contrast, the systems Aijk and Cpmn are not of the same type
because one system has two superscripts and the other system has only one superscript. For certain systems
the number of subscripts and superscripts is important. In other systems it is not of importance. The
meaning and importance attached to sub- and superscripts will be addressed later in this section.
In the use of superscripts one must not confuse “powers ”of a quantity with the superscripts. For
example, if we replace the independent variables (x, y, z) by the symbols (x1 , x2 , x3 ), then we are letting
y = x2 where x2 is a variable and not x raised to a power. Similarly, the substitution z = x3 is the
replacement of z by the variable x3 and this should not be confused with x raised to a power. In order to
write a superscript quantity to a power, use parentheses. For example, (x2 )3 is the variable x2 cubed. One
of the reasons for introducing the superscript variables is that many equations of mathematics and physics
can be made to take on a concise and compact form.
There is a range convention associated with the indices. This convention states that whenever there
is an expression where the indices occur unrepeated it is to be understood that each of the subscripts or
superscripts can take on any of the integer values 1, 2, . . . , N where N is a specified integer. For example,
3
the Kronecker delta symbol δij , defined by δij = 1 if i = j and δij = 0 for i 6= j, with i, j ranging over the
values 1,2,3, represents the 9 quantities
δ11 = 1
δ12 = 0
δ13 = 0
δ21 = 0
δ22 = 1
δ23 = 0
δ31 = 0
δ32 = 0
δ33 = 1.
The symbol δij refers to all of the components of the system simultaneously. As another example, consider
the equation
b
em · b
en = δmn
m, n = 1, 2, 3
(1.1.1)
the subscripts m, n occur unrepeated on the left side of the equation and hence must also occur on the right
hand side of the equation. These indices are called “free ”indices and can take on any of the values 1, 2 or 3
as specified by the range. Since there are three choices for the value for m and three choices for a value of
n we find that equation (1.1.1) represents nine equations simultaneously. These nine equations are
b
e1 = 1
e1 · b
b
e1 · b
e2 = 0
b
e1 · b
e3 = 0
b
e1 = 0
e2 · b
b
e2 = 1
e2 · b
b
e3 = 0
e2 · b
b
e1 = 0
e3 · b
b
e2 = 0
e3 · b
b
e3 = 1.
e3 · b
Symmetric and Skew-Symmetric Systems
A system defined by subscripts and superscripts ranging over a set of values is said to be symmetric
in two of its indices if the components are unchanged when the indices are interchanged. For example, the
third order system Tijk is symmetric in the indices i and k if
Tijk = Tkji
for all values of i, j and k.
A system defined by subscripts and superscripts is said to be skew-symmetric in two of its indices if the
components change sign when the indices are interchanged. For example, the fourth order system Tijkl is
skew-symmetric in the indices i and l if
Tijkl = −Tljki
for all values of ijk and l.
As another example, consider the third order system aprs , p, r, s = 1, 2, 3 which is completely skewsymmetric in all of its indices. We would then have
aprs = −apsr = aspr = −asrp = arsp = −arps .
It is left as an exercise to show this completely skew- symmetric systems has 27 elements, 21 of which are
zero. The 6 nonzero elements are all related to one another thru the above equations when (p, r, s) = (1, 2, 3).
This is expressed as saying that the above system has only one independent component.
4
Summation Convention
The summation convention states that whenever there arises an expression where there is an index which
occurs twice on the same side of any equation, or term within an equation, it is understood to represent a
summation on these repeated indices. The summation being over the integer values specified by the range. A
repeated index is called a summation index, while an unrepeated index is called a free index. The summation
convention requires that one must never allow a summation index to appear more than twice in any given
expression. Because of this rule it is sometimes necessary to replace one dummy summation symbol by
some other dummy symbol in order to avoid having three or more indices occurring on the same side of
the equation. The index notation is a very powerful notation and can be used to concisely represent many
complex equations. For the remainder of this section there is presented additional definitions and examples
to illustrated the power of the indicial notation. This notation is then employed to define tensor components
and associated operations with tensors.
EXAMPLE 1.1-1 The two equations
y1 = a11 x1 + a12 x2
y2 = a21 x1 + a22 x2
can be represented as one equation by introducing a dummy index, say k, and expressing the above equations
as
yk = ak1 x1 + ak2 x2 ,
k = 1, 2.
The range convention states that k is free to have any one of the values 1 or 2, (k is a free index). This
equation can now be written in the form
yk =
2
X
aki xi = ak1 x1 + ak2 x2
i=1
where i is the dummy summation index. When the summation sign is removed and the summation convention
is adopted we have
yk = aki xi
i, k = 1, 2.
Since the subscript i repeats itself, the summation convention requires that a summation be performed by
letting the summation subscript take on the values specified by the range and then summing the results.
The index k which appears only once on the left and only once on the right hand side of the equation is
called a free index. It should be noted that both k and i are dummy subscripts and can be replaced by other
letters. For example, we can write
yn = anm xm
n, m = 1, 2
where m is the summation index and n is the free index. Summing on m produces
yn = an1 x1 + an2 x2
and letting the free index n take on the values of 1 and 2 we produce the original two equations.
5
EXAMPLE 1.1-2. For yi = aij xj , i, j = 1, 2, 3 and xi = bij zj , i, j = 1, 2, 3 solve for the y variables in
terms of the z variables.
Solution: In matrix form the given equations can be expressed:
 
a11
y1
 y2  =  a21
y3
a31

a12
a22
a32
 
a13
x1
a23   x2 
a33
x3

and
 
x1
b11
 x2  =  b21
x3
b31
b12
b22
b32
 
b13
z1
b23   z2  .
b33
z3
Now solve for the y variables in terms of the z variables and obtain
 
a11
y1
 y2  =  a21
y3
a31

a12
a22
a32

a13
b11
a23   b21
a33
b31
b12
b22
b32
 
b13
z1
b23   z2  .
b33
z3
The index notation employs indices that are dummy indices and so we can write
yn = anm xm ,
n, m = 1, 2, 3 and xm = bmj zj ,
m, j = 1, 2, 3.
Here we have purposely changed the indices so that when we substitute for xm , from one equation into the
other, a summation index does not repeat itself more than twice. Substituting we find the indicial form of
the above matrix equation as
yn = anm bmj zj ,
m, n, j = 1, 2, 3
where n is the free index and m, j are the dummy summation indices. It is left as an exercise to expand
both the matrix equation and the indicial equation and verify that they are different ways of representing
the same thing.
EXAMPLE 1.1-3.
The dot product of two vectors Aq , q = 1, 2, 3 and Bj , j = 1, 2, 3 can be represented
~
~ Since the
B = |B|.
with the index notation by the product Ai Bi = AB cos θ i = 1, 2, 3, A = |A|,
subscript i is repeated it is understood to represent a summation index. Summing on i over the range
specified, there results
A1 B1 + A2 B2 + A3 B3 = AB cos θ.
Observe that the index notation employs dummy indices. At times these indices are altered in order to
conform to the above summation rules, without attention being brought to the change. As in this example,
the indices q and j are dummy indices and can be changed to other letters if one desires. Also, in the future,
if the range of the indices is not stated it is assumed that the range is over the integer values 1, 2 and 3.
To systems containing subscripts and superscripts one can apply certain algebraic operations. We
present in an informal way the operations of addition, multiplication and contraction.
6
Addition, Multiplication and Contraction
The algebraic operation of addition or subtraction applies to systems of the same type and order. That
i
is again a
is we can add or subtract like components in systems. For example, the sum of Aijk and Bjk
i
i
= Aijk + Bjk
, where like components are added.
system of the same type and is denoted by Cjk
The product of two systems is obtained by multiplying each component of the first system with each
component of the second system. Such a product is called an outer product. The order of the resulting
product system is the sum of the orders of the two systems involved in forming the product. For example,
if Aij is a second order system and B mnl is a third order system, with all indices having the range 1 to N,
then the product system is fifth order and is denoted Cjimnl = Aij B mnl . The product system represents N 5
terms constructed from all possible products of the components from Aij with the components from B mnl .
The operation of contraction occurs when a lower index is set equal to an upper index and the summation
convention is invoked. For example, if we have a fifth order system Cjimnl and we set i = j and sum, then
we form the system
N mnl
.
C mnl = Cjjmnl = C11mnl + C22mnl + · · · + CN
Here the symbol C mnl is used to represent the third order system that results when the contraction is
performed. Whenever a contraction is performed, the resulting system is always of order 2 less than the
original system. Under certain special conditions it is permissible to perform a contraction on two lower case
indices. These special conditions will be considered later in the section.
The above operations will be more formally defined after we have explained what tensors are.
The e-permutation symbol and Kronecker delta
Two symbols that are used quite frequently with the indicial notation are the e-permutation symbol
and the Kronecker delta. The e-permutation symbol is sometimes referred to as the alternating tensor. The
e-permutation symbol, as the name suggests, deals with permutations. A permutation is an arrangement of
things. When the order of the arrangement is changed, a new permutation results. A transposition is an
interchange of two consecutive terms in an arrangement. As an example, let us change the digits 1 2 3 to
3 2 1 by making a sequence of transpositions. Starting with the digits in the order 1 2 3 we interchange 2 and
3 (first transposition) to obtain 1 3 2. Next, interchange the digits 1 and 3 ( second transposition) to obtain
3 1 2. Finally, interchange the digits 1 and 2 (third transposition) to achieve 3 2 1. Here the total number
of transpositions of 1 2 3 to 3 2 1 is three, an odd number. Other transpositions of 1 2 3 to 3 2 1 can also be
written. However, these are also an odd number of transpositions.
7
EXAMPLE 1.1-4.
The total number of possible ways of arranging the digits 1 2 3 is six. We have
three choices for the first digit. Having chosen the first digit, there are only two choices left for the second
digit. Hence the remaining number is for the last digit. The product (3)(2)(1) = 3! = 6 is the number of
permutations of the digits 1, 2 and 3. These six permutations are
1 2 3 even permutation
1 3 2 odd permutation
3 1 2 even permutation
3 2 1 odd permutation
2 3 1 even permutation
2 1 3 odd permutation.
Here a permutation of 1 2 3 is called even or odd depending upon whether there is an even or odd number
of transpositions of the digits. A mnemonic device to remember the even and odd permutations of 123
is illustrated in the figure 1.1-1. Note that even permutations of 123 are obtained by selecting any three
consecutive numbers from the sequence 123123 and the odd permutations result by selecting any three
consecutive numbers from the sequence 321321.
Figure 1.1-1. Permutations of 123.
In general, the number of permutations of n things taken m at a time is given by the relation
P (n, m) = n(n − 1)(n − 2) · · · (n − m + 1).
By selecting a subset of m objects from a collection of n objects, m ≤ n, without regard to the ordering is
called a combination of n objects taken m at a time. For example, combinations of 3 numbers taken from
the set {1, 2, 3, 4} are (123), (124), (134), (234). Note that ordering of a combination is not considered. That
is, the permutations (123), (132), (231), (213), (312), (321) are considered equal. In general, the number of
n
n!
n
where m
are the
combinations of n objects taken m at a time is given by C(n, m) =
=
m!(n − m)!
m
binomial coefficients which occur in the expansion
(a + b)n =
n X
n n−m m
b .
a
m
m=0
8
The definition of permutations can be used to define the e-permutation symbol.
Definition: (e-Permutation symbol or alternating tensor)
The e-permutation symbol is defined

if ijk . . . l is an even permutation of the integers 123 . . . n

1
ijk...l
= eijk...l = −1
if ijk . . . l is an odd permutation of the integers 123 . . . n
e


0
in all other cases
EXAMPLE 1.1-5.
Find e612453 .
Solution: To determine whether 612453 is an even or odd permutation of 123456 we write down the given
numbers and below them we write the integers 1 through 6. Like numbers are then connected by a line and
we obtain figure 1.1-2.
Figure 1.1-2. Permutations of 123456.
In figure 1.1-2, there are seven intersections of the lines connecting like numbers. The number of
intersections is an odd number and shows that an odd number of transpositions must be performed. These
results imply e612453 = −1.
Another definition used quite frequently in the representation of mathematical and engineering quantities
is the Kronecker delta which we now define in terms of both subscripts and superscripts.
Definition: (Kronecker delta) The Kronecker delta is defined:
δij = δij =
1
0
if i equals j
if i is different from j
9
EXAMPLE 1.1-6. Some examples of the e−permutation symbol and Kronecker delta are:
e123 = e123 = +1
δ11 = 1
δ12 = 0
e213 = e213 = −1
δ21 = 0
δ22 = 1
δ31
δ32 = 0.
e112 = e
EXAMPLE 1.1-7.
112
=0
=0
When an index of the Kronecker delta δij is involved in the summation convention,
the effect is that of replacing one index with a different index. For example, let aij denote the elements of an
N × N matrix. Here i and j are allowed to range over the integer values 1, 2, . . . , N. Consider the product
aij δik
where the range of i, j, k is 1, 2, . . . , N. The index i is repeated and therefore it is understood to represent
a summation over the range. The index i is called a summation index. The other indices j and k are free
indices. They are free to be assigned any values from the range of the indices. They are not involved in any
summations and their values, whatever you choose to assign them, are fixed. Let us assign a value of j and
k to the values of j and k. The underscore is to remind you that these values for j and k are fixed and not
to be summed. When we perform the summation over the summation index i we assign values to i from the
range and then sum over these values. Performing the indicated summation we obtain
aij δik = a1j δ1k + a2j δ2k + · · · + akj δkk + · · · + aN j δN k .
In this summation the Kronecker delta is zero everywhere the subscripts are different and equals one where
the subscripts are the same. There is only one term in this summation which is nonzero. It is that term
where the summation index i was equal to the fixed value k This gives the result
akj δkk = akj
where the underscore is to remind you that the quantities have fixed values and are not to be summed.
Dropping the underscores we write
aij δik = akj
Here we have substituted the index i by k and so when the Kronecker delta is used in a summation process
it is known as a substitution operator. This substitution property of the Kronecker delta can be used to
simplify a variety of expressions involving the index notation. Some examples are:
Bij δjs = Bis
δjk δkm = δjm
eijk δim δjn δkp = emnp .
Some texts adopt the notation that if indices are capital letters, then no summation is to be performed.
For example,
aKJ δKK = aKJ
10
as δKK represents a single term because of the capital letters. Another notation which is used to denote no
summation of the indices is to put parenthesis about the indices which are not to be summed. For example,
a(k)j δ(k)(k) = akj ,
since δ(k)(k) represents a single term and the parentheses indicate that no summation is to be performed.
At any time we may employ either the underscore notation, the capital letter notation or the parenthesis
notation to denote that no summation of the indices is to be performed. To avoid confusion altogether, one
can write out parenthetical expressions such as “(no summation on k)”.
EXAMPLE 1.1-8. In the Kronecker delta symbol δji we set j equal to i and perform a summation. This
operation is called a contraction. There results δii , which is to be summed over the range of the index i.
Utilizing the range 1, 2, . . . , N we have
N
δii = δ11 + δ22 + · · · + δN
δii = 1 + 1 + · · · + 1
δii = N.
In three dimension we have δji , i, j = 1, 2, 3 and
δkk = δ11 + δ22 + δ33 = 3.
In certain circumstances the Kronecker delta can be written with only subscripts.
δij ,
For example,
i, j = 1, 2, 3. We shall find that these circumstances allow us to perform a contraction on the lower
indices so that δii = 3.
EXAMPLE 1.1-9.
The determinant of a matrix A = (aij ) can be represented in the indicial notation.
Employing the e-permutation symbol the determinant of an N × N matrix is expressed
|A| = eij...k a1i a2j · · · aN k
where eij...k is an N th order system. In the special case of a 2 × 2 matrix we write
|A| = eij a1i a2j
where the summation is over the range 1,2 and the e-permutation symbol is of order 2. In the special case
of a 3 × 3 matrix we have
a11
|A| = a21
a31
a12
a22
a32
a13
a23 = eijk ai1 aj2 ak3 = eijk a1i a2j a3k
a33
where i, j, k are the summation indices and the summation is over the range 1,2,3. Here eijk denotes the
e-permutation symbol of order 3. Note that by interchanging the rows of the 3 × 3 matrix we can obtain
11
more general results. Consider (p, q, r) as some permutation of the integers (1, 2, 3), and observe that the
determinant can be expressed
ap1
∆ = aq1
ar1
ap2
aq2
ar2
ap3
aq3 = eijk api aqj ark .
ar3
If (p, q, r)
is an even permutation of (1, 2, 3) then
∆ = |A|
If (p, q, r)
is an odd permutation of (1, 2, 3) then
∆ = −|A|
If (p, q, r)
is not a permutation of (1, 2, 3) then
∆ = 0.
We can then write
eijk api aqj ark = epqr |A|.
Each of the above results can be verified by performing the indicated summations. A more formal proof of
the above result is given in EXAMPLE 1.1-25, later in this section.
EXAMPLE 1.1-10.
The expression eijk Bij Ci is meaningless since the index i repeats itself more than
twice and the summation convention does not allow this. If you really did want to sum over an index which
occurs more than twice, then one must use a summation sign. For example the above expression would be
n
X
eijk Bij Ci .
written
i=1
EXAMPLE 1.1-11.
The cross product of the unit vectors

ek

b
b
ej = − b
ei × b
ek


0
b
e2 , b
e3 can be represented in the index notation by
e1 , b
if (i, j, k) is an even permutation of (1, 2, 3)
if (i, j, k) is an odd permutation of (1, 2, 3)
in all other cases
ek . This later result can be verified by summing on the
ej = ekij b
This result can be written in the form b
ei × b
index k and writing out all 9 possible combinations for i and j.
EXAMPLE 1.1-12.
Given the vectors Ap , p = 1, 2, 3 and Bp , p = 1, 2, 3 the cross product of these two
vectors is a vector Cp , p = 1, 2, 3 with components
Ci = eijk Aj Bk ,
i, j, k = 1, 2, 3.
(1.1.2)
The quantities Ci represent the components of the cross product vector
~ =A
~×B
~ = C1 b
e1 + C2 b
e2 + C3 b
e3 .
C
~ is to be summed over each of the indices which
The equation (1.1.2), which defines the components of C,
repeats itself. We have summing on the index k
Ci = eij1 Aj B1 + eij2 Aj B2 + eij3 Aj B3 .
(1.1.3)
12
We next sum on the index j which repeats itself in each term of equation (1.1.3). This gives
Ci = ei11 A1 B1 + ei21 A2 B1 + ei31 A3 B1
+ ei12 A1 B2 + ei22 A2 B2 + ei32 A3 B2
(1.1.4)
+ ei13 A1 B3 + ei23 A2 B3 + ei33 A3 B3 .
Now we are left with i being a free index which can have any of the values of 1, 2 or 3. Letting i = 1, then
letting i = 2, and finally letting i = 3 produces the cross product components
C1 = A2 B3 − A3 B2
C2 = A3 B1 − A1 B3
C3 = A1 B2 − A2 B1 .
~×B
~ = eijk Aj Bk b
ei . This result can be verified by
The cross product can also be expressed in the form A
summing over the indices i,j and k.
EXAMPLE 1.1-13.
Show
eijk = −eikj = ejki
for
i, j, k = 1, 2, 3
Solution: The array i k j represents an odd number of transpositions of the indices i j k and to each
transposition there is a sign change of the e-permutation symbol. Similarly, j k i is an even transposition
of i j k and so there is no sign change of the e-permutation symbol. The above holds regardless of the
numerical values assigned to the indices i, j, k.
The e-δ Identity
An identity relating the e-permutation symbol and the Kronecker delta, which is useful in the simplification of tensor expressions, is the e-δ identity. This identity can be expressed in different forms. The
subscript form for this identity is
eijk eimn = δjm δkn − δjn δkm ,
i, j, k, m, n = 1, 2, 3
where i is the summation index and j, k, m, n are free indices. A device used to remember the positions of
the subscripts is given in the figure 1.1-3.
The subscripts on the four Kronecker delta’s on the right-hand side of the e-δ identity then are read
(first)(second)-(outer)(inner).
This refers to the positions following the summation index. Thus, j, m are the first indices after the summation index and k, n are the second indices after the summation index. The indices j, n are outer indices
when compared to the inner indices k, m as the indices are viewed as written on the left-hand side of the
identity.
13
Figure 1.1-3. Mnemonic device for position of subscripts.
Another form of this identity employs both subscripts and superscripts and has the form
j k
k
δn − δnj δm
.
eijk eimn = δm
(1.1.5)
One way of proving this identity is to observe the equation (1.1.5) has the free indices j, k, m, n. Each
of these indices can have any of the values of 1, 2 or 3. There are 3 choices we can assign to each of j, k, m
or n and this gives a total of 34 = 81 possible equations represented by the identity from equation (1.1.5).
By writing out all 81 of these equations we can verify that the identity is true for all possible combinations
that can be assigned to the free indices.
An alternate proof of the e − δ identity is to consider the determinant
δ11
δ12
δ13
δ21
δ22
δ23
1 0
δ31
δ32 = 0 1
0 0
δ33
0
0 = 1.
1
By performing a permutation of the rows of this matrix we can use the permutation symbol and write
δ1i
δ1j
δ1k
δ2i
δ2j
δ2k
δ3i
δ3j = eijk .
δ3k
By performing a permutation of the columns, we can write
δri
δrj
δrk
δsi
δsj
δsk
δti
δtj = eijk erst .
δtk
Now perform a contraction on the indices i and r to obtain
δii
δij
δik
δsi
δsj
δsk
δti
δtj = eijk eist .
δtk
Summing on i we have δii = δ11 + δ22 + δ33 = 3 and expand the determinant to obtain the desired result
δsj δtk − δtj δsk = eijk eist .
14
Generalized Kronecker delta
The generalized Kronecker delta is defined by the (n × n) determinant
ij...k
δmn...p
i
δm
j
δm
= .
..
k
δm
δni
δnj
..
.
δnk
· · · δpi
· · · δpj
. .
..
. ..
· · · δpk
For example, in three dimensions we can write
ijk
δmnp
i
δm
j
= δm
k
δm
δni
δnj
δnk
δpi
δpj = eijk emnp .
δpk
Performing a contraction on the indices k and p we obtain the fourth order system
rs
rsp
r s
s
= δmnp
= ersp emnp = eprs epmn = δm
δn − δnr δm
.
δmn
As an exercise one can verify that the definition of the e-permutation symbol can also be defined in terms
of the generalized Kronecker delta as
··· N
.
ej1 j2 j3 ···jN = δj11 j22 j33 ···j
N
Additional definitions and results employing the generalized Kronecker delta are found in the exercises.
In section 1.3 we shall show that the Kronecker delta and epsilon permutation symbol are numerical tensors
which have fixed components in every coordinate system.
Additional Applications of the Indicial Notation
The indicial notation, together with the e − δ identity, can be used to prove various vector identities.
EXAMPLE 1.1-14.
Solution: Let
~×B
~ = −B
~ ×A
~
Show, using the index notation, that A
~ =A
~×B
~ = C1 b
e1 + C2 b
e2 + C3 b
e3 = Ci b
ei
C
and let
~ =B
~ ×A
~ = D1 b
e1 + D2 b
e2 + D3 b
e3 = Di b
ei .
D
We have shown that the components of the cross products can be represented in the index notation by
Ci = eijk Aj Bk
and Di = eijk Bj Ak .
We desire to show that Di = −Ci for all values of i. Consider the following manipulations: Let Bj = Bs δsj
and Ak = Am δmk and write
Di = eijk Bj Ak = eijk Bs δsj Am δmk
(1.1.6)
where all indices have the range 1, 2, 3. In the expression (1.1.6) note that no summation index appears
more than twice because if an index appeared more than twice the summation convention would become
meaningless. By rearranging terms in equation (1.1.6) we have
Di = eijk δsj δmk Bs Am = eism Bs Am .
15
In this expression the indices s and m are dummy summation indices and can be replaced by any other
letters. We replace s by k and m by j to obtain
Di = eikj Aj Bk = −eijk Aj Bk = −Ci .
~
~ = −C
~ or B
~ ×A
~ = −A
~ × B.
~ That is, D
~ = Di b
ei = −Ci b
ei = −C.
Consequently, we find that D
Note 1. The expressions
Ci = eijk Aj Bk
and
Cm = emnp An Bp
with all indices having the range 1, 2, 3, appear to be different because different letters are used as subscripts. It must be remembered that certain indices are summed according to the summation convention
and the other indices are free indices and can take on any values from the assigned range. Thus, after
summation, when numerical values are substituted for the indices involved, none of the dummy letters
used to represent the components appear in the answer.
Note 2. A second important point is that when one is working with expressions involving the index notation,
the indices can be changed directly. For example, in the above expression for Di we could have replaced
j by k and k by j simultaneously (so that no index repeats itself more than twice) to obtain
Di = eijk Bj Ak = eikj Bk Aj = −eijk Aj Bk = −Ci .
Note 3. Be careful in switching back and forth between the vector notation and index notation. Observe that a
~ can be represented
vector A
~ = Ai b
ei
A
or its components can be represented
~· b
A
ei = Ai ,
i = 1, 2, 3.
~ = Ai as this is a
Do not set a vector equal to a scalar. That is, do not make the mistake of writing A
misuse of the equal sign. It is not possible for a vector to equal a scalar because they are two entirely
different quantities. A vector has both magnitude and direction while a scalar has only magnitude.
EXAMPLE 1.1-15.
Verify the vector identity
~ · (B
~ × C)
~ =B
~ · (C
~ × A)
~
A
Solution: Let
~ ×C
~ =D
~ = Di b
ei
B
where
Di = eijk Bj Ck
~ ×A
~ = F~ = Fi b
ei
C
where
Fi = eijk Cj Ak
where all indices have the range 1, 2, 3. To prove the above identity, we have
~ · (B
~ × C)
~ =A
~ ·D
~ = Ai Di = Ai eijk Bj Ck
A
= Bj (eijk Ai Ck )
= Bj (ejki Ck Ai )
and let
16
since eijk = ejki . We also observe from the expression
Fi = eijk Cj Ak
that we may obtain, by permuting the symbols, the equivalent expression
Fj = ejki Ck Ai .
This allows us to write
~ · F~ = B
~ · (C
~ × A)
~
~ · (B
~ × C)
~ = Bj Fj = B
A
which was to be shown.
~ · (B
~ × C)
~ is called a triple scalar product. The above index representation of the triple
The quantity A
scalar product implies that it can be represented as a determinant (See example 1.1-9). We can write
A1
~ · (B
~ × C)
~ = B1
A
C1
A2
B2
C2
A3
B3 = eijk Ai Bj Ck
C3
A physical interpretation that can be assigned to this triple scalar product is that its absolute value represents
~ B,
~ C.
~ The absolute value is
the volume of the parallelepiped formed by the three noncoplaner vectors A,
needed because sometimes the triple scalar product is negative. This physical interpretation can be obtained
from an analysis of the figure 1.1-4.
Figure 1.1-4. Triple scalar product and volume
17
~ × C|
~ is the area of the parallelogram P QRS. (ii) the unit vector
In figure 1.1-4 observe that: (i) |B
b
en =
~ ×C
~
B
~
~
|B × C|
~ and C.
~ (iii) The dot product
is normal to the plane containing the vectors B
~
~
~· B×C =h
~· b
A
en = A
~ × C|
~
|B
~ on b
equals the projection of A
en which represents the height of the parallelepiped. These results demonstrate
that
~ × C|
~ h = (area of base)(height) = volume.
~ · (B
~ × C)
~ = |B
A
EXAMPLE 1.1-16.
Verify the vector identity
~ × B)
~ × (C
~ × D)
~ = C(
~ D
~ ·A
~ × B)
~ − D(
~ C
~ ·A
~ × B)
~
(A
~ =C
~ ×D
~ = Ei b
~×B
~ = Fi b
ei and E
ei . These vectors have the components
Solution: Let F~ = A
Fi = eijk Aj Bk
and
Em = emnp Cn Dp
~ = F~ × E
~ = Gi b
ei has the components
where all indices have the range 1, 2, 3. The vector G
Gq = eqim Fi Em = eqim eijk emnp Aj Bk Cn Dp .
From the identity eqim = emqi this can be expressed
Gq = (emqi emnp )eijk Aj Bk Cn Dp
which is now in a form where we can use the e − δ identity applied to the term in parentheses to produce
Gq = (δqn δip − δqp δin )eijk Aj Bk Cn Dp .
Simplifying this expression we have:
Gq = eijk [(Dp δip )(Cn δqn )Aj Bk − (Dp δqp )(Cn δin )Aj Bk ]
= eijk [Di Cq Aj Bk − Dq Ci Aj Bk ]
= Cq [Di eijk Aj Bk ] − Dq [Ci eijk Aj Bk ]
which are the vector components of the vector
~ D
~ ·A
~ × B)
~ − D(
~ C
~ ·A
~ × B).
~
C(
18
Transformation Equations
Consider two sets of N independent variables which are denoted by the barred and unbarred symbols
xi and xi with i = 1, . . . , N. The independent variables xi , i = 1, . . . , N can be thought of as defining
the coordinates of a point in a N −dimensional space. Similarly, the independent barred variables define a
point in some other N −dimensional space. These coordinates are assumed to be real quantities and are not
complex quantities. Further, we assume that these variables are related by a set of transformation equations.
xi = xi (x1 , x2 , . . . , xN )
i = 1, . . . , N.
(1.1.7)
It is assumed that these transformation equations are independent. A necessary and sufficient condition that
these transformation equations be independent is that the Jacobian determinant be different from zero, that
is
∂xi
x
=
J( ) =
x
∂ x̄j
∂x1
∂x1
∂x2
∂x1
∂x1
∂x2
∂x2
∂x2
∂xN
∂x1
∂xN
∂x2
..
.
···
···
..
.
···
..
.
∂x1
∂xN
∂x2
∂xN
..
.
6= 0.
∂xN
∂xN
This assumption allows us to obtain a set of inverse relations
xi = xi (x1 , x2 , . . . , xN )
i = 1, . . . , N,
(1.1.8)
where the x0 s are determined in terms of the x0 s. Throughout our discussions it is to be understood that the
given transformation equations are real and continuous. Further all derivatives that appear in our discussions
are assumed to exist and be continuous in the domain of the variables considered.
EXAMPLE 1.1-17.
The following is an example of a set of transformation equations of the form
defined by equations (1.1.7) and (1.1.8) in the case N = 3. Consider the transformation from cylindrical
coordinates (r, α, z) to spherical coordinates (ρ, β, α). From the geometry of the figure 1.1-5 we can find the
transformation equations
r = ρ sin β
α=α
0 < α < 2π
z = ρ cos β
with inverse transformation
ρ=
0<β<π
p
r2 + z 2
α=α
r
β = arctan( )
z
Now make the substitutions
(x1 , x2 , x3 ) = (r, α, z)
and
(x1 , x2 , x3 ) = (ρ, β, α).
19
Figure 1.1-5. Cylindrical and Spherical Coordinates
The resulting transformations then have the forms of the equations (1.1.7) and (1.1.8).
Calculation of Derivatives
We now consider the chain rule applied to the differentiation of a function of the bar variables. We
represent this differentiation in the indicial notation. Let Φ = Φ(x1 , x2 , . . . , xn ) be a scalar function of the
variables xi ,
i = 1, . . . , N and let these variables be related to the set of variables xi , with i = 1, . . . , N by
the transformation equations (1.1.7) and (1.1.8). The partial derivatives of Φ with respect to the variables
xi can be expressed in the indicial notation as
∂Φ ∂xj
∂Φ ∂x1
∂Φ ∂x2
∂Φ ∂xN
∂Φ
=
=
+
+
·
·
·
+
∂xi
∂x1 ∂xi
∂x2 ∂xi
∂xj ∂xi
∂xN ∂xi
for any fixed value of i satisfying 1 ≤ i ≤ N.
(1.1.9)
The second partial derivatives of Φ can also be expressed in the index notation. Differentiation of
equation (1.1.9) partially with respect to xm produces
∂Φ ∂xj
∂Φ ∂ 2 xj
∂
∂2Φ
=
+
.
∂xi ∂xm
∂xm ∂xj ∂xi
∂xj ∂xi ∂xm
(1.1.10)
This result is nothing more than an application of the general rule for differentiating a product of two
quantities. To evaluate the derivative of the bracketed term in equation (1.1.10) it must be remembered that
the quantity inside the brackets is a function of the bar variables. Let
∂Φ
= G(x1 , x2 , . . . , xN )
G=
∂xj
to emphasize this dependence upon the bar variables, then the derivative of G is
∂G ∂xk
∂ 2 Φ ∂xk
∂G
=
=
.
(1.1.11)
∂xm
∂xk ∂xm
∂xj ∂xk ∂xm
This is just an application of the basic rule from equation (1.1.9) with Φ replaced by G. Hence the derivative
from equation (1.1.10) can be expressed
∂Φ ∂ 2 xj
∂ 2 Φ ∂xj ∂xk
∂2Φ
=
+
j
i
m
i
m
∂x ∂x
∂x ∂x ∂x
∂xj ∂xk ∂xi ∂xm
where i, m are free indices and j, k are dummy summation indices.
(1.1.12)
20
EXAMPLE 1.1-18.
Let Φ = Φ(r, θ) where r, θ are polar coordinates related to the Cartesian coordinates
∂2Φ
∂Φ
and
(x, y) by the transformation equations x = r cos θ
y = r sin θ. Find the partial derivatives
∂x
∂x2
Solution: The partial derivative of Φ with respect to x is found from the relation (1.1.9) and can be written
∂Φ ∂r
∂Φ ∂θ
∂Φ
=
+
.
∂x
∂r ∂x
∂θ ∂x
(1.1.13)
The second partial derivative is obtained by differentiating the first partial derivative. From the product
rule for differentiation we can write
∂Φ ∂ 2 r
∂r ∂ ∂Φ
∂θ ∂ ∂Φ
∂Φ ∂ 2 θ
∂2Φ
=
+
+
+
.
∂x2
∂r ∂x2
∂x ∂x ∂r
∂θ ∂x2
∂x ∂x ∂θ
(1.1.14)
To further simplify (1.1.14) it must be remembered that the terms inside the brackets are to be treated as
functions of the variables r and θ and that the derivative of these terms can be evaluated by reapplying the
basic rule from equation (1.1.13) with Φ replaced by
∂Φ
∂r
and then Φ replaced by
∂Φ
∂θ .
This gives
∂ 2 Φ ∂θ
∂Φ ∂ 2 r
∂r ∂ 2 Φ ∂r
∂ 2Φ
+
=
+
∂x2
∂r ∂x2
∂x ∂r2 ∂x ∂r∂θ ∂x
∂ 2 Φ ∂θ
∂θ ∂ 2 Φ ∂r
∂Φ ∂ 2 θ
+
+
.
+
∂θ ∂x2
∂x ∂θ∂r ∂x
∂θ2 ∂x
(1.1.15)
y
and from
x
these relations we can calculate all the necessary derivatives needed for the simplification of the equations
From the transformation equations we obtain the relations r2 = x2 + y 2
and
tan θ =
(1.1.13) and (1.1.15). These derivatives are:
∂r
= 2x or
∂x
y
∂θ
= − 2 or
sec2 θ
∂x
x
sin2 θ
∂θ
∂2r
=
= − sin θ
∂x2
∂x
r
2r
∂r
x
= = cos θ
∂x
r
y
∂θ
sin θ
=− 2 =−
∂x
r
r
∂θ
∂r
+ sin θ ∂x
−r cos θ ∂x
∂2θ
2 sin θ cos θ
=
=
.
∂x2
r2
r2
Therefore, the derivatives from equations (1.1.13) and (1.1.15) can be expressed in the form
∂Φ
∂Φ sin θ
∂Φ
=
cos θ −
∂x
∂r
∂θ r
2
2
∂Φ sin θ cos θ ∂ 2 Φ
∂Φ sin θ
∂ 2 Φ cos θ sin θ ∂ 2 Φ sin2 θ
∂ Φ
2
+
2
+
=
+
cos
θ
−
2
.
∂x2
∂r r
∂θ
r2
∂r2
∂r∂θ
r
∂θ2 r2
By letting x1 = r, x2 = θ, x1 = x, x2 = y and performing the indicated summations in the equations (1.1.9)
and (1.1.12) there is produced the same results as above.
Vector Identities in Cartesian Coordinates
Employing the substitutions x1 = x, x2 = y, x3 = z, where superscript variables are employed and
e2 , b
e3 , we illustrated how various vector operations
denoting the unit vectors in Cartesian coordinates by b
e1 , b
are written by using the index notation.
21
Gradient.
In Cartesian coordinates the gradient of a scalar field is
grad φ =
∂φ
∂φ
∂φ
b
b
b
e1 +
e2 +
e3 .
∂x
∂y
∂z
The index notation focuses attention only on the components of the gradient. In Cartesian coordinates these
components are represented using a comma subscript to denote the derivative
b
ej · grad φ = φ,j =
∂φ
,
∂xj
j = 1, 2, 3.
The comma notation will be discussed in section 4. For now we use it to denote derivatives. For example
∂φ
∂2φ
, φ ,jk =
, etc.
φ ,j =
j
∂x
∂xj ∂xk
Divergence.
~ is a scalar field and can be
In Cartesian coordinates the divergence of a vector field A
represented
~ = div A
~ = ∂A1 + ∂A2 + ∂A3 .
∇·A
∂x
∂y
∂z
Employing the summation convention and index notation, the divergence in Cartesian coordinates can be
represented
~ = div A
~ = Ai,i =
∇·A
∂Ai
∂A1
∂A2
∂A3
=
+
+
∂xi
∂x1
∂x2
∂x3
where i is the dummy summation index.
~ = curl A
~ = ∇×A
~ in Cartesian coordinates, we note that the index
Curl. To represent the vector B
~ can
notation focuses attention only on the components of this vector. The components Bi , i = 1, 2, 3 of B
be represented
~ = eijk Ak,j ,
ei · curl A
Bi = b
for
i, j, k = 1, 2, 3
k
where eijk is the permutation symbol introduced earlier and Ak,j = ∂A
∂xj . To verify this representation of the
~ we need only perform the summations indicated by the repeated indices. We have summing on j that
curl A
Bi = ei1k Ak,1 + ei2k Ak,2 + ei3k Ak,3 .
Now summing each term on the repeated index k gives us
Bi = ei12 A2,1 + ei13 A3,1 + ei21 A1,2 + ei23 A3,2 + ei31 A1,3 + ei32 A2,3
Here i is a free index which can take on any of the values 1, 2 or 3. Consequently, we have
∂A3
∂A2
−
∂x2
∂x3
∂A1
∂A3
=
−
∂x3
∂x1
∂A2
∂A1
=
−
∂x1
∂x2
For
i = 1,
B1 = A3,2 − A2,3 =
For
i = 2,
B2 = A1,3 − A3,1
For
i = 3,
B3 = A2,1 − A1,2
~ in Cartesian coordinates.
which verifies the index notation representation of curl A
22
Other Operations. The following examples illustrate how the index notation can be used to represent
additional vector operators in Cartesian coordinates.
~ · ∇)A
~ are
1. In index notation the components of the vector (B
~ · ∇)A}
~ ·b
{(B
ep = Ap,q Bq
p, q = 1, 2, 3
This can be verified by performing the indicated summations. We have by summing on the repeated
index q
Ap,q Bq = Ap,1 B1 + Ap,2 B2 + Ap,3 B3 .
The index p is now a free index which can have any of the values 1, 2 or 3. We have:
for
p = 1,
for
p = 2,
for
p = 3,
A1,q Bq = A1,1 B1 + A1,2 B2 + A1,3 B3
∂A1
∂A1
∂A1
=
B1 +
B2 +
B3
∂x1
∂x2
∂x3
A2,q Bq = A2,1 B1 + A2,2 B2 + A2,3 B3
∂A2
∂A2
∂A2
=
B1 +
B2 +
B3
∂x1
∂x2
∂x3
A3,q Bq = A3,1 B1 + A3,2 B2 + A3,3 B3
∂A3
∂A3
∂A3
=
B1 +
B2 +
B3
∂x1
∂x2
∂x3
~ · ∇)φ has the following form when expressed in the index notation:
2. The scalar (B
~ · ∇)φ = Bi φ,i = B1 φ,1 + B2 φ,2 + B3 φ,3
(B
∂φ
∂φ
∂φ
= B1 1 + B2 2 + B3 3 .
∂x
∂x
∂x
~ × ∇)φ is expressed in the index notation by
3. The components of the vector (B
h
i
~ × ∇)φ = eijk Bj φ,k .
b
ei · (B
This can be verified by performing the indicated summations and is left as an exercise.
~ × ∇) · A
~ may be expressed in the index notation. It has the form
4. The scalar (B
~ × ∇) · A
~ = eijk Bj Ai,k .
(B
This can also be verified by performing the indicated summations and is left as an exercise.
~ in the index notation are represented
5. The vector components of ∇2 A
~ = Ap,qq .
b
ep · ∇2 A
The proof of this is left as an exercise.
23
EXAMPLE 1.1-19. In Cartesian coordinates prove the vector identity
~ = ∇ × (f A)
~ = (∇f ) × A
~ + f (∇ × A).
~
curl (f A)
~ = curl (f A)
~ and write the components as
Solution: Let B
Bi = eijk (f Ak ),j
= eijk [f Ak,j + f,j Ak ]
= f eijk Ak,j + eijk f,j Ak .
This index form can now be expressed in the vector form
~ = curl (f A)
~ = f (∇ × A)
~ + (∇f ) × A
~
B
~ + B)
~ = ∇·A
~ +∇·B
~
EXAMPLE 1.1-20. Prove the vector identity ∇ · (A
~+B
~ =C
~ and write this vector equation in the index notation as Ai + Bi = Ci . We then
Solution: Let A
have
~ + ∇ · B.
~
~ = Ci,i = (Ai + Bi ),i = Ai,i + Bi,i = ∇ · A
∇·C
~ · ∇)f = A
~ · ∇f
EXAMPLE 1.1-21. In Cartesian coordinates prove the vector identity (A
Solution: In the index notation we write
~ · ∇)f = Ai f,i = A1 f,1 + A2 f,2 + A3 f,3
(A
∂f
∂f
∂f
~ · ∇f.
= A1 1 + A2 2 + A3 3 = A
∂x
∂x
∂x
EXAMPLE 1.1-22. In Cartesian coordinates prove the vector identity
~ × B)
~ = A(∇
~
~ − B(∇
~
~ + (B
~ · ∇)A
~ − (A
~ · ∇)B
~
∇ × (A
· B)
· A)
~ × B)
~ is
Solution: The pth component of the vector ∇ × (A
~ × B)]
~ = epqk [ekji Aj Bi ],q
b
ep · [∇ × (A
= epqk ekji Aj Bi,q + epqk ekji Aj,q Bi
By applying the e − δ identity, the above expression simplifies to the desired result. That is,
~ × B)]
~ = (δpj δqi − δpi δqj )Aj Bi,q + (δpj δqi − δpi δqj )Aj,q Bi
b
ep · [∇ × (A
= Ap Bi,i − Aq Bp,q + Ap,q Bq − Aq,q Bp
In vector form this is expressed
~ × B)
~ = A(∇
~
~ − (A
~ · ∇)B
~ + (B
~ · ∇)A
~ − B(∇
~
~
∇ × (A
· B)
· A)
24
~
~ = ∇(∇ · A)
~ − ∇2 A
EXAMPLE 1.1-23. In Cartesian coordinates prove the vector identity ∇ × (∇ × A)
~ = eijk Ak,j and consequently the
~ is given by b
Solution: We have for the ith component of ∇ × A
ei · [∇ × A]
~ is
pth component of ∇ × (∇ × A)
~ = epqr [erjk Ak,j ],q
b
ep · [∇ × (∇ × A)]
= epqr erjk Ak,jq .
The e − δ identity produces
~ = (δpj δqk − δpk δqj )Ak,jq
b
ep · [∇ × (∇ × A)]
= Ak,pk − Ap,qq .
~
~ = ∇(∇ · A)
~ − ∇2 A.
Expressing this result in vector form we have ∇ × (∇ × A)
Indicial Form of Integral Theorems
The divergence theorem, in both vector and indicial notation, can be written
ZZ
Z
Z
ZZZ
b dσ
div · F~ dτ =
Fi,i dτ =
Fi ni dσ
i = 1, 2, 3
F~ · n
V
S
V
(1.1.16)
S
where ni are the direction cosines of the unit exterior normal to the surface, dτ is a volume element and dσ
is an element of surface area. Note that in using the indicial notation the volume and surface integrals are
to be extended over the range specified by the indices. This suggests that the divergence theorem can be
applied to vectors in n−dimensional spaces.
The vector form and indicial notation for the Stokes theorem are
Z
Z
Z
ZZ
b dσ =
F~ · d~r
(∇ × F~ ) · n
eijk Fk,j ni dσ =
Fi dxi
S
C
S
i, j, k = 1, 2, 3
(1.1.17)
C
and the Green’s theorem in the plane, which is a special case of the Stoke’s theorem, can be expressed
ZZ ∂F1
∂F2
−
∂x
∂y
Z
Z
Z
F1 dx + F2 dy
dxdy =
C
Fi dxi
e3jk Fk,j dS =
S
i, j, k = 1, 2
(1.1.18)
C
Other forms of the above integral theorems are
ZZ
ZZZ
b dσ
∇φ dτ =
φn
V
S
~ where C
~ is a constant vector. By replacing F~ by
obtained from the divergence theorem by letting F~ = φC
~ in the divergence theorem one can derive
F~ × C
ZZ
ZZZ ~
F~ × ~n dσ.
∇ × F dτ = −
V
S
In the divergence theorem make the substitution F~ = φ∇ψ to obtain
ZZ
ZZZ
b dσ.
(φ∇ψ) · n
(φ∇2 ψ + (∇φ) · (∇ψ) dτ =
V
S
25
The Green’s identity
ZZZ
φ∇2 ψ − ψ∇2 φ dτ =
V
ZZ
b dσ
(φ∇ψ − ψ∇φ) · n
S
is obtained by first letting F~ = φ∇ψ in the divergence theorem and then letting F~ = ψ∇φ in the divergence
theorem and then subtracting the results.
Determinants, Cofactors
For A = (aij ), i, j = 1, . . . , n an n × n matrix, the determinant of A can be written as
det A = |A| = ei1 i2 i3 ...in a1i1 a2i2 a3i3 . . . anin .
This gives a summation of the n! permutations of products formed from the elements of the matrix A. The
result is a single number called the determinant of A.
EXAMPLE 1.1-24.
In the case n = 2 we have
|A| =
a11
a21
a12
= enm a1n a2m
a22
= e1m a11 a2m + e2m a12 a2m
= e12 a11 a22 + e21 a12 a21
= a11 a22 − a12 a21
EXAMPLE 1.1-25.
In the case

a11
A =  a21
a31
n = 3 we can use either of the notations


 1
a12 a13
a1 a12 a13
a22 a23 
or
A =  a21 a22 a23 
a32 a33
a31 a32 a33
and represent the determinant of A in any of the forms
det A = eijk a1i a2j a3k
det A = eijk ai1 aj2 ak3
det A = eijk ai1 aj2 ak3
det A = eijk a1i a2j a3k .
These represent row and column expansions of the determinant.
An important identity results if we examine the quantity Brst = eijk air ajs akt . It is an easy exercise to
change the dummy summation indices and rearrange terms in this expression. For example,
Brst = eijk air ajs akt = ekji akr ajs ait = ekji ait ajs akr = −eijk ait ajs akr = −Btsr ,
and by considering other permutations of the indices, one can establish that Brst is completely skewsymmetric. In the exercises it is shown that any third order completely skew-symmetric system satisfies
Brst = B123 erst . But B123 = det A and so we arrive at the identity
Brst = eijk air ajs akt = |A|erst .
26
Other forms of this identity are
eijk ari asj atk = |A|erst
and eijk air ajs akt = |A|erst .
(1.1.19)
Consider the representation of the determinant
a11
|A| = a21
a31
a12
a22
a32
a13
a23
a33
by use of the indicial notation. By column expansions, this determinant can be represented
|A| = erst ar1 as2 at3
(1.1.20)
and if one uses row expansions the determinant can be expressed as
|A| = eijk a1i a2j a3k .
(1.1.21)
Define Aim as the cofactor of the element am
i in the determinant |A|. From the equation (1.1.20) the cofactor
of ar1 is obtained by deleting this element and we find
A1r = erst as2 at3 .
(1.1.22)
The result (1.1.20) can then be expressed in the form
|A| = ar1 A1r = a11 A11 + a21 A12 + a31 A13 .
(1.1.23)
That is, the determinant |A| is obtained by multiplying each element in the first column by its corresponding
cofactor and summing the result. Observe also that from the equation (1.1.20) we find the additional
cofactors
A2s = erst ar1 at3
and
A3t = erst ar1 as2 .
(1.1.24)
Hence, the equation (1.1.20) can also be expressed in one of the forms
|A| = as2 A2s = a12 A21 + a22 A22 + a32 A23
|A| = at3 A3t = a13 A31 + a23 A32 + a33 A33
The results from equations (1.1.22) and (1.1.24) can be written in a slightly different form with the indicial
notation. From the notation for a generalized Kronecker delta defined by
ijk
,
eijk elmn = δlmn
the above cofactors can be written in the form
1 1jk
1 1jk s t
e erst asj atk = δrst
aj ak
2!
2!
1
1 2jk s t
A2r = e123 esrt as1 at3 = e2jk erst asj atk = δrst
aj ak
2!
2!
1
1 3jk s t
A3r = e123 etsr at1 as2 = e3jk erst asj atk = δrst
aj ak .
2!
2!
A1r = e123 erst as2 at3 =
27
These cofactors are then combined into the single equation
Air =
1 ijk s t
δ a a
2! rst j k
(1.1.25)
which represents the cofactor of ari . When the elements from any row (or column) are multiplied by their
corresponding cofactors, and the results summed, we obtain the value of the determinant. Whenever the
elements from any row (or column) are multiplied by the cofactor elements from a different row (or column),
and the results summed, we get zero. This can be illustrated by considering the summation
1 ijk s t m
1
s t
δmst aj ak ar = eijk emst am
r aj ak
2!
2!
1
1 ijk
= eijk erjk |A| = δrjk
|A| = δri |A|
2!
2!
i
am
r Am =
Here we have used the e − δ identity to obtain
ijk
= eijk erjk = ejik ejrk = δri δkk − δki δrk = 3δri − δri = 2δri
δrjk
which was used to simplify the above result.
As an exercise one can show that an alternate form of the above summation of elements by its cofactors
is
r
arm Am
i = |A|δi .
EXAMPLE 1.1-26.
N
In N-dimensions the quantity δkj11jk22...j
...kN is called a generalized Kronecker delta. It
can be defined in terms of permutation symbols as
N
ej1 j2 ...jN ek1 k2 ...kN = δkj11jk22...j
...kN
(1.1.26)
Observe that
k1 k2 ...kN
N
= (N !) ej1 j2 ...jN
δkj11jk22...j
...kN e
This follows because ek1 k2 ...kN is skew-symmetric in all pairs of its superscripts. The left-hand side denotes
a summation of N ! terms. The first term in the summation has superscripts j1 j2 . . . jN and all other terms
have superscripts which are some permutation of this ordering with minus signs associated with those terms
having an odd permutation. Because ej1 j2 ...jN is completely skew-symmetric we find that all terms in the
summation have the value +ej1 j2 ...jN . We thus obtain N ! of these terms.
28
EXERCISE 1.1
I 1.
Simplify each of the following by employing the summation property of the Kronecker delta. Perform
sums on the summation indices only if your are unsure of the result.
I 2.
(a) eijk δkn
(c) eijk δis δjm δkn
(e) δij δjn
(b) eijk δis δjm
(d)
(f ) δij δjn δni
aij δin
Simplify and perform the indicated summations over the range 1, 2, 3
(a)
δii
(b) δij δij
(c) eijk Ai Aj Ak
(e) eijk δjk
(d)
(f ) Ai Bj δji − Bm An δmn
eijk eijk
~ = Ai
Express each of the following in index notation. Be careful of the notation you use. Note that A
~· b
is an incorrect notation because a vector can not equal a scalar. The notation A
ei = Ai should be used to
I 3.
express the ith component of a vector.
~ · (B
~ × C)
~
(a) A
~ A
~ · C)
~
(c) B(
~ × (B
~ × C)
~
(b) A
(d)
~ A
~ · C)
~ − C(
~ A
~ · B)
~
B(
(b) eijk = −ejik = −eikj = −ekji
I 4.
Show the e permutation symbol satisfies: (a)
I 5.
~ × (B
~ × C)
~ = B(
~ A
~ · C)
~ − C(
~ A
~ · B)
~
Use index notation to verify the vector identity A
I 6.
Let yi = aij xj and xm = aim zi where the range of the indices is 1, 2
eijk = ejki = ekij
(a) Solve for yi in terms of zi using the indicial notation and check your result
to be sure that no index repeats itself more than twice.
(b) Perform the indicated summations and write out expressions
for y1 , y2 in terms of z1 , z2
(c)
Express the above equations in matrix form. Expand the matrix
equations and check the solution obtained in part (b).
I 7.
Use the e − δ identity to simplify (a)
I 8.
Prove the following vector identities:
(a)
eijk ejik
(b)
eijk ejki
~ · (B
~ × C)
~ =B
~ · (C
~ × A)
~ =C
~ · (A
~ × B)
~
A
triple scalar product
~ × B)
~ ×C
~ = B(
~ A
~ · C)
~ − A(
~ B
~ · C)
~
(b) (A
I 9.
Prove the following vector identities:
~ × B)
~ · (C
~ × D)
~ = (A
~ · C)(
~ B
~ · D)
~ − (A
~ · D)(
~ B
~ · C)
~
(a) (A
~ × (B
~ × C)
~ +B
~ × (C
~ × A)
~ +C
~ × (A
~ × B)
~ = ~0
(b) A
(c)
~ × B)
~ × (C
~ × D)
~ = B(
~ A
~·C
~ × D)
~ − A(
~ B
~ ·C
~ × D)
~
(A
29
I 10.
~ = (1, −1, 0) and B
~ = (4, −3, 2) find using the index notation,
For A
(a)
Ci = eijk Aj Bk ,
i = 1, 2, 3
(b) Ai Bi
(c)
I 11.
What do the results in (a) and (b) represent?
Represent the differential equations
dy1
= a11 y1 + a12 y2
dt
dy2
= a21 y1 + a22 y2
dt
and
using the index notation.
I 12.
Let Φ = Φ(r, θ) where r, θ are polar coordinates related to Cartesian coordinates (x, y) by the transformation equations
x = r cos θ
and
y = r sin θ.
∂2Φ
∂Φ
,
and
(a) Find the partial derivatives
∂y
∂y 2
(b) Combine the result in part (a) with the result from EXAMPLE 1.1-18 to calculate the Laplacian
∇2 Φ =
∂2Φ ∂2Φ
+
∂x2
∂y 2
in polar coordinates.
I 13.
(Index notation) Let a11 = 3,
a12 = 4,
a21 = 5,
a22 = 6.
Calculate the quantity C = aij aij , i, j = 1, 2.
I 14.
Show the moments of inertia Iij defined by
ZZZ
(y 2 + z 2 )ρ(x, y, z) dτ
I11 =
ZR
ZZ
I22 =
2
2
(x + z )ρ(x, y, z) dτ
ZZZ
I23 = I32 = −
ZR
ZZ
I12 = I21 = −
R
ZZZ
I33 =
yzρ(x, y, z) dτ
xyρ(x, y, z) dτ
R
(x2 + y 2 )ρ(x, y, z) dτ
ZZZ
I13 = I31 = −
R
xzρ(x, y, z) dτ,
R
ZZZ
can be represented in the index notation as Iij =
xm xm δij − xi xj ρ dτ, where ρ is the density,
R
x1 = x, x2 = y, x3 = z and dτ = dxdydz is an element of volume.
I 15.
Determine if the following relation is true or false. Justify your answer.
b
ej × b
ek ) = ( b
ei × b
ej ) · b
ek = eijk ,
ei · ( b
i, j, k = 1, 2, 3.
Hint: Let b
em = (δ1m , δ2m , δ3m ).
I 16.
Without substituting values for i, l = 1, 2, 3 calculate all nine terms of the given quantities
(a)
I 17.
B il = (δji Ak + δki Aj )ejkl
(b)
Ail = (δim B k + δik B m )emlk
Let Amn xm y n = 0 for arbitrary xi and y i , i = 1, 2, 3, and show that Aij = 0 for all values of i, j.
30
I 18.
(a) For amn , m, n = 1, 2, 3 skew-symmetric, show that amn xm xn = 0.
(b) Let amn xm xn = 0,
m, n = 1, 2, 3 for all values of xi , i = 1, 2, 3 and show that amn must be skew-
symmetric.
I 19.
Let A and B denote 3 × 3 matrices with elements aij and bij respectively. Show that if C = AB is a
matrix product, then det(C) = det(A) · det(B).
Hint: Use the result from example 1.1-9.
I 20.
(a) Let u1 , u2 , u3 be functions of the variables s1 , s2 , s3 . Further, assume that s1 , s2 , s3 are in turn each
∂um
∂(u1 , u2 , u3 )
denote the Jacobian of the u0 s with
=
functions of the variables x1 , x2 , x3 . Let
∂xn
∂(x1 , x2 , x3 )
respect to the x0 s. Show that
∂ui
∂ui ∂sj
∂ui
∂sj
=
=
·
.
∂xm
∂sj ∂xm
∂sj
∂xm
∂xi
∂xi ∂ x̄j
i
=
= δm
and show that J( xx̄ )·J( xx̄ ) = 1, where J( xx̄ ) is the Jacobian determinant
j
m
∂ x̄ ∂x
∂xm
of the transformation (1.1.7).
(b) Note that
I 21.
A third order system a`mn with `, m, n = 1, 2, 3 is said to be symmetric in two of its subscripts if the
components are unaltered when these subscripts are interchanged. When a`mn is completely symmetric then
a`mn = am`n = a`nm = amn` = anm` = an`m . Whenever this third order system is completely symmetric,
then: (i) How many components are there? (ii) How many of these components are distinct?
Hint: Consider the three cases (i) ` = m = n
I 22.
(ii) ` = m 6= n
(iii) ` 6= m 6= n.
A third order system b`mn with `, m, n = 1, 2, 3 is said to be skew-symmetric in two of its subscripts
if the components change sign when the subscripts are interchanged. A completely skew-symmetric third
order system satisfies b`mn = −bm`n = bmn` = −bnm` = bn`m = −b`nm . (i) How many components does
a completely skew-symmetric system have? (ii) How many of these components are zero? (iii) How many
components can be different from zero? (iv) Show that there is one distinct component b123 and that
b`mn = e`mn b123 .
Hint: Consider the three cases (i) ` = m = n
I 23.
(ii) ` = m 6= n
(iii) ` 6= m 6= n.
Let i, j, k = 1, 2, 3 and assume that eijk σjk = 0 for all values of i. What does this equation tell you
about the values σij , i, j = 1, 2, 3?
I 24.
Assume that Amn and Bmn are symmetric for m, n = 1, 2, 3. Let Amn xm xn = Bmn xm xn for arbitrary
values of xi , i = 1, 2, 3, and show that Aij = Bij for all values of i and j.
I 25.
Assume Bmn is symmetric and Bmn xm xn = 0 for arbitrary values of xi , i = 1, 2, 3, show that Bij = 0.
31
I 26.
(Generalized Kronecker delta)
ij...k
δmn...p
(a)
i
δm
j
δm
= .
..
k
δm
δni
δnj
..
.
δnk
Define the generalized Kronecker delta as the n × n determinant
· · · δpi
· · · δpj
.
..
. ..
· · · δpk
where δsr is the Kronecker delta.
123
eijk = δijk
Show
(b) Show
ijk
eijk = δ123
(c)
Show
ij
δmn
= eij emn
(d)
Define
rs
rsp
δmn
= δmnp
and show
(summation on p)
rs
r s
s
= δm
δn − δnr δm
δmn
Note that by combining the above result with the result from part (c)
we obtain the two dimensional form of the e − δ identity
r
rn
= 12 δmn
(e) Define δm
I 27.
Let
denote the cofactor of
ari
a11
in the determinant a21
a31
(a) Show erst Air = eijk asj atk
I 28.
rst
δpst
= 2δpr
(summation on n) and show
rst
δrst
= 3!
(f ) Show
Air
r s
s
δn − δnr δm
.
ers emn = δm
a12
a22
a32
a13
a23 as given by equation (1.1.25).
a33
(b) Show erst Ari = eijk ajs akt
(a) Show that if Aijk = Ajik , i, j, k = 1, 2, 3 there is a total of 27 elements, but only 18 are distinct.
(b) Show that for i, j, k = 1, 2, . . . , N there are N 3 elements, but only N 2 (N + 1)/2 are distinct.
I 29.
I 30.
I 31.
Let aij = Bi Bj for i, j = 1, 2, 3 where B1 , B2 , B3 are arbitrary constants. Calculate det(aij ) = |A|.
(a)
For
A = (aij ), i, j = 1, 2, 3,
show
|A| = eijk ai1 aj2 ak3 .
(b)
For
A = (aij ), i, j = 1, 2, 3,
show
|A| = eijk ai1 aj2 ak3 .
(c)
For
A = (aij ), i, j = 1, 2, 3,
show
|A| = eijk a1i a2j a3k .
(d)
For
I = (δji ), i, j = 1, 2, 3,
show
|I| = 1.
Let |A| = eijk ai1 aj2 ak3 and define Aim as the cofactor of aim . Show the determinant can be
expressed in any of the forms:
(a) |A| = Ai1 ai1
where Ai1 = eijk aj2 ak3
(b) |A| = Aj2 aj2
where Ai2 = ejik aj1 ak3
(c) |A| = Ak3 ak3
where
Ai3 = ejki aj1 ak2
32
I 32.
Show the results in problem 31 can be written in the forms:
Ai1 =
I 33.
1
e1st eijk ajs akt ,
2!
Ai2 =
1
e2st eijk ajs akt ,
2!
Ai3 =
1
e3st eijk ajs akt ,
2!
or Aim =
1
emst eijk ajs akt
2!
Use the results in problems 31 and 32 to prove that apm Aim = |A|δip .

I 34.
1
Let (aij ) =  1
2
I 35.
Let

2 1
0 3  and calculate C = aij aij , i, j = 1, 2, 3.
3 2
a111 = −1,
a211 = 1,
a112 = 3,
a212 = 5,
a121 = 4,
a122 = 2
a221 = 2,
a222 = −2
a1121 = 3,
a1122 = 1
and calculate the quantity C = aijk aijk , i, j, k = 1, 2.
I 36.
Let
a1111 = 2,
a1112 = 1,
a1211 = 5,
a1212 = −2,
a2111 = 1,
a2112 = 0,
a2211 = −2,
a2212 = 1,
a1221 = 4,
a1222 = −2
a2121 = −2,
a2122 = −1
a2221 = 2,
a2222 = 2
and calculate the quantity C = aijkl aijkl , i, j, k, l = 1, 2.
I 37.
I 38.
Simplify the expressions:
(a) (Aijkl + Ajkli + Aklij + Alijk )xi xj xk xl
(c)
∂xi
∂xj
(b) (Pijk + Pjki + Pkij )xi xj xk
(d)
aij
∂ 2 xi ∂xj
∂ 2 xm ∂xi
r − ami
t
s
∂x ∂x ∂x
∂xs ∂xt ∂xr
Let g denote the determinant of the matrix having the components gij , i, j = 1, 2, 3. Show that
g1r
(a) g erst = g2r
g3r
g1s
g2s
g3s
g1t
g2t
g3t
i
δm
j
= δm
k
δm
δni
δnj
δnk
gir
(b) g erst eijk = gjr
gkr
Show that e
I 40.
Show that eijk emnp Amnp = Aijk − Aikj + Akij − Ajik + Ajki − Akji
emnp =
ijk
δmnp
git
gjt
gkt
δpi
δpj
δpk
I 39.
ijk
gis
gjs
gks
Hint: Use the results from problem 39.
I 41.
Show that
(a)
eij eij = 2!
(c)
eijkl eijkl = 4!
(b)
eijk eijk = 3!
(d)
Guess at the result
ei1 i2 ...in ei1 i2 ...in
33
I 42.
Determine if the following statement is true or false. Justify your answer. eijk Ai Bj Ck = eijk Aj Bk Ci .
Let aij , i, j = 1, 2 denote the components of a 2 × 2 matrix A, which are functions of time t.
a
a12
to verify that these representations are the same.
(a) Expand both |A| = eij ai1 aj2 and |A| = 11
a21 a22
(b) Verify the equivalence of the derivative relations
I 43.
dai1
daj2
d|A|
= eij
aj2 + eij ai1
dt
dt
dt
and
d|A|
=
dt
da11
dt
a21
da12
dt
a22
+
a11
da21
dt
a12
da22
dt
(c) Let aij , i, j = 1, 2, 3 denote the components of a 3 × 3 matrix A, which are functions of time t. Develop
appropriate relations, expand them and verify, similar to parts (a) and (b) above, the representation of
a determinant and its derivative.
I 44.
For f = f (x1 , x2 , x3 ) and φ = φ(f ) differentiable scalar functions, use the indicial notation to find a
formula to calculate grad φ .
~=0
(b) ∇ · ∇ × A
I 45.
Use the indicial notation to prove (a) ∇ × ∇φ = ~0
I 46.
If Aij is symmetric and Bij is skew-symmetric, i, j = 1, 2, 3, then calculate C = Aij Bij .
I 47.
Amn
I 48.
Assume Aij = Aij (x1 , x2 , x3 ) and Aij = Aij (x1 , x2 , x3 ) for i, j = 1, 2, 3 are related by the expression
∂Amn
∂xi ∂xj
= Aij m n . Calculate the derivative
.
∂x ∂x
∂xk
Prove that if any two rows (or two columns) of a matrix are interchanged, then the value of the
determinant of the matrix is multiplied by minus one. Construct your proof using 3 × 3 matrices.
I 49.
Prove that if two rows (or columns) of a matrix are proportional, then the value of the determinant
of the matrix is zero. Construct your proof using 3 × 3 matrices.
I 50.
Prove that if a row (or column) of a matrix is altered by adding some constant multiple of some other
row (or column), then the value of the determinant of the matrix remains unchanged. Construct your proof
using 3 × 3 matrices.
I 51.
Simplify the expression φ = eijk e`mn Ai` Ajm Akn .
I 52.
Let Aijk denote a third order system where i, j, k = 1, 2. (a) How many components does this system
have? (b) Let Aijk be skew-symmetric in the last pair of indices, how many independent components does
the system have?
I 53.
Let Aijk denote a third order system where i, j, k = 1, 2, 3. (a) How many components does this
system have? (b) In addition let Aijk = Ajik and Aikj = −Aijk and determine the number of distinct
nonzero components for Aijk .
34
I 54.
Show that every second order system Tij can be expressed as the sum of a symmetric system Aij and
skew-symmetric system Bij . Find Aij and Bij in terms of the components of Tij .
I 55.
Consider the system Aijk ,
i, j, k = 1, 2, 3, 4.
(a) How many components does this system have?
(b) Assume Aijk is skew-symmetric in the last pair of indices, how many independent components does this
system have?
(c) Assume that in addition to being skew-symmetric in the last pair of indices, Aijk + Ajki + Akij = 0 is
satisfied for all values of i, j, and k, then how many independent components does the system have?
I 56.
~ in indicial form. (b) Write the equation of the plane
(a) Write the equation of a line ~r = ~r0 + t A
~n · (~r − ~r0 ) = 0 in indicial form. (c) Write the equation of a general line in scalar form. (d) Write the
equation of a plane in scalar form. (e) Find the equation of the line defined by the intersection of the
planes 2x + 3y + 6z = 12 and 6x + 3y + z = 6. (f) Find the equation of the plane through the points
(5, 3, 2), (3, 1, 5), (1, 3, 3). Find also the normal to this plane.
I 57.
The angle 0 ≤ θ ≤ π between two skew lines in space is defined as the angle between their direction
vectors when these vectors are placed at the origin. Show that for two lines with direction numbers ai and
bi i = 1, 2, 3, the cosine of the angle between these lines satisfies
ai b i
√
cos θ = √
ai ai b i b i
I 58.
I 59.
I 60.
Let aij = −aji for i, j = 1, 2, . . . , N and prove that for N odd det(aij ) = 0.
∂λ
∂2λ
(b)
Let λ = Aij xi xj where Aij = Aji and calculate (a)
∂xm
∂xm ∂xk
Given an arbitrary nonzero vector Uk , k = 1, 2, 3, define the matrix elements aij = eijk Uk , where eijk
is the e-permutation symbol. Determine if aij is symmetric or skew-symmetric. Suppose Uk is defined by
the above equation for arbitrary nonzero aij , then solve for Uk in terms of the aij .
I 61.
If Aij = Ai Bj 6= 0 for all i, j values and Aij = Aji for i, j = 1, 2, . . . , N , show that Aij = λBi Bj
where λ is a constant. State what λ is.
I 62.
Assume that Aijkm , with i, j, k, m = 1, 2, 3, is completely skew-symmetric. How many independent
components does this quantity have?
I 63.
Consider Rijkm , i, j, k, m = 1, 2, 3, 4. (a) How many components does this quantity have? (b) If
Rijkm = −Rijmk = −Rjikm then how many independent components does Rijkm have? (c) If in addition
Rijkm = Rkmij determine the number of independent components.
I 64.
Let xi = aij x̄j , i, j = 1, 2, 3 denote a change of variables from a barred system of coordinates to an
unbarred system of coordinates and assume that Āi = aij Aj where aij are constants, Āi is a function of the
∂ Āi
.
x̄j variables and Aj is a function of the xj variables. Calculate
∂ x̄m
35
§1.2 TENSOR CONCEPTS AND TRANSFORMATIONS
~ as
e2 , b
e3 independent orthogonal unit vectors (base vectors), we may write any vector A
For b
e1 , b
~ = A1 b
e1 + A2 b
e2 + A3 b
e3
A
~ relative to the base vectors chosen. These components are the
where (A1 , A2 , A3 ) are the coordinates of A
~ onto the base vectors and
projection of A
~· b
~·b
~ = (A
~· b
e1 + (A
e2 ) b
e2 + (A
e3 ) b
e3 .
A
e1 ) b
~ 2, E
~ 3 ), not necessarily of unit length, we can then
~ 1, E
Select any three independent orthogonal vectors, (E
write
b
e1 =
~1
E
,
~ 1|
|E
b
e2 =
~2
E
,
~ 2|
|E
b
e3 =
~ can be expressed as
and consequently, the vector A
!
!
~·E
~2
~ ·E
~1
A
A
~1 +
~2 +
~=
E
E
A
~1 · E
~2 · E
~1
~2
E
E
~3
E
,
~ 3|
|E
~·E
~3
A
~3 · E
~3
E
!
~ 3.
E
Here we say that
~·E
~ (i)
A
,
~ (i) · E
~ (i)
E
i = 1, 2, 3
~ 2, E
~ 3 . Recall that the parenthesis about
~ relative to the chosen base vectors E
~ 1, E
are the components of A
the subscript i denotes that there is no summation on this subscript. It is then treated as a free subscript
which can have any of the values 1, 2 or 3.
Reciprocal Basis
~ 1, E
~ 2, E
~ 3 ) which are not necessarily orthogonal, nor of
Consider a set of any three independent vectors (E
~ in terms of these vectors we must find components (A1 , A2 , A3 )
unit length. In order to represent the vector A
such that
~ 1 + A2 E
~ 2 + A3 E
~ 3.
~ = A1 E
A
This can be done by taking appropriate projections and obtaining three equations and three unknowns from
which the components are determined. A much easier way to find the components (A1 , A2 , A3 ) is to construct
~ 2, E
~ 3 ). Recall that two bases (E
~ 1, E
~ 2, E
~ 3 ) and (E
~ 1, E
~ 2, E
~ 3 ) are said to be reciprocal
~ 1, E
a reciprocal basis (E
if they satisfy the condition
~ j = δj =
~i · E
E
i
1
0
if i = j
.
if i =
6 j
~ 1 = δ21 = 0 and E
~3 · E
~ 1 = δ31 = 0 so that the vector E
~ 1 is perpendicular to both the
~2 · E
Note that E
~ 3 . (i.e. A vector from one basis is orthogonal to two of the vectors from the other basis.)
~ 2 and E
vectors E
~2 × E
~ 3 where V is a constant to be determined. By taking the dot
~ 1 = V −1 E
We can therefore write E
~ 1 · (E
~2 × E
~ 3 ) is the volume
~ 1 we find that V = E
product of both sides of this equation with the vector E
~ 2, E
~ 3 when their origins are made to coincide. In a
~ 1, E
of the parallelepiped formed by the three vectors E
36
~ 1, E
~ 2, E
~ 3 ) a given set of basis vectors, then the reciprocal
similar manner it can be demonstrated that for (E
basis vectors are determined from the relations
~2 × E
~3 × E
~1 × E
~ 3,
~2 = 1 E
~ 1,
~3 = 1 E
~ 2,
~1 = 1 E
E
E
E
V
V
V
~2 × E
~ 3 ) 6= 0 is a triple scalar product and represents the volume of the parallelepiped
~ 1 · (E
where V = E
having the basis vectors for its sides.
~ 2, E
~ 3 ) and (E
~ 1, E
~ 2, E
~ 3 ) denote a system of reciprocal bases. We can represent any vector A
~
~ 1, E
Let (E
~ 2, E
~ 3 ) and represent A
~ in the form
~ 1, E
with respect to either of these bases. If we select the basis (E
~ 1 + A2 E
~ 2 + A3 E
~ 3,
~ = A1 E
A
(1.2.1)
~ relative to the basis vectors (E
~ 1, E
~ 2, E
~ 3 ) are called the contravariant
then the components (A1 , A2 , A3 ) of A
~ These components can be determined from the equations
components of A.
~ ·E
~ 1 = A1 ,
A
~·E
~ 2 = A2 ,
A
~·E
~ 3 = A3 .
A
~ 2, E
~ 3 ) and represent A
~ in the form
~ 1, E
Similarly, if we choose the reciprocal basis (E
~ 1 + A2 E
~ 2 + A3 E
~ 3,
~ = A1 E
A
(1.2.2)
~ 1, E
~ 2, E
~ 3 ) are called the covariant components of
then the components (A1 , A2 , A3 ) relative to the basis (E
~ These components can be determined from the relations
A.
~ ·E
~ 1 = A1 ,
A
~ ·E
~ 2 = A2 ,
A
~ ·E
~ 3 = A3 .
A
The contravariant and covariant components are different ways of representing the same vector with respect
to a set of reciprocal basis vectors. There is a simple relationship between these components which we now
develop. We introduce the notation
~ j = gij = gji ,
~i · E
E
~i · E
~ j = g ij = g ji
E
and
(1.2.3)
where gij are called the metric components of the space and g ij are called the conjugate metric components
of the space. We can then write
~1 · E
~ 1 ) + A2 (E
~2 · E
~ 1 ) + A3 (E
~3 · E
~ 1 ) = A1
~ ·E
~ 1 = A1 (E
A
~ ·E
~ 1 = A1 (E
~1 · E
~ 1 ) + A2 (E
~2 · E
~ 1 ) + A3 (E
~3 · E
~ 1 ) = A1
A
or
A1 = A1 g11 + A2 g12 + A3 g13 .
(1.2.4)
~ ·E
~ 3 one can establish the results
~·E
~ 2 and A
In a similar manner, by considering the dot products A
A2 = A1 g21 + A2 g22 + A3 g23
A3 = A1 g31 + A2 g32 + A3 g33 .
These results can be expressed with the index notation as
Ai = gik Ak .
~ ·E
~ 1,
Forming the dot products A
~ ·E
~ 2,
A
(1.2.6)
~·E
~ 3 it can be verified that
A
Ai = g ik Ak .
(1.2.7)
The equations (1.2.6) and (1.2.7) are relations which exist between the contravariant and covariant compo~1 + β E
~2 + γ E
~ 3 , then one can show
~ Similarly, if for some value j we have E
~j = αE
nents of the vector A.
j
ij
~ i . This is left as an exercise.
~ =g E
that E
37
Coordinate Transformations
Consider a coordinate transformation from a set of coordinates (x, y, z) to (u, v, w) defined by a set of
transformation equations
x = x(u, v, w)
y = y(u, v, w)
(1.2.8)
z = z(u, v, w)
It is assumed that these transformations are single valued, continuous and possess the inverse transformation
u = u(x, y, z)
v = v(x, y, z)
(1.2.9)
w = w(x, y, z).
These transformation equations define a set of coordinate surfaces and coordinate curves. The coordinate
surfaces are defined by the equations
u(x, y, z) = c1
v(x, y, z) = c2
(1.2.10)
w(x, y, z) = c3
where c1 , c2 , c3 are constants. These surfaces intersect in the coordinate curves
~r(u, c2 , c3 ),
~r(c1 , v, c3 ),
~r(c1 , c2 , w),
(1.2.11)
where
e2 + z(u, v, w) b
e3 .
~r(u, v, w) = x(u, v, w) b
e1 + y(u, v, w) b
The general situation is illustrated in the figure 1.2-1.
Consider the vectors
~ 1 = grad u = ∇u,
E
~ 2 = grad v = ∇v,
E
~ 3 = grad w = ∇w
E
(1.2.12)
evaluated at the common point of intersection (c1 , c2 , c3 ) of the coordinate surfaces. The system of vectors
~ 2, E
~ 3 ) can be selected as a system of basis vectors which are normal to the coordinate surfaces.
~ 1, E
(E
Similarly, the vectors
~ 1 = ∂~r ,
E
∂u
~ 2 = ∂~r ,
E
∂v
~ 3 = ∂~r
E
∂w
(1.2.13)
~ 1, E
~ 2, E
~ 3 ) which
when evaluated at the common point of intersection (c1 , c2 , c3 ) forms a system of vectors (E
we can select as a basis. This basis is a set of tangent vectors to the coordinate curves. It is now demonstrated
~ 2, E
~ 3 ) and the tangential basis (E
~ 1, E
~ 2, E
~ 3 ) are a set of reciprocal bases.
~ 1, E
that the normal basis (E
e2 + z b
e3 denotes the position vector of a variable point. By substitution for
Recall that ~r = x b
e1 + y b
x, y, z from (1.2.8) there results
e2 + z(u, v, w) b
e3 .
~r = ~r(u, v, w) = x(u, v, w) b
e1 + y(u, v, w) b
(1.2.14)
38
Figure 1.2-1. Coordinate curves and coordinate surfaces.
A small change in ~r is denoted
e2 + dz b
e3 =
d~r = dx b
e1 + dy b
∂~r
∂~r
∂~r
du +
dv +
dw
∂u
∂v
∂w
(1.2.15)
where
∂x
∂y
∂z
∂~r
b
b
b
=
e1 +
e2 +
e3
∂u
∂u
∂u
∂u
∂x
∂y
∂z
∂~r
(1.2.16)
b
b
b
=
e1 +
e2 +
e3
∂v
∂v
∂v
∂v
∂x
∂y
∂z
∂~r
b
b
b
=
e1 +
e2 +
e3 .
∂w
∂w
∂w
∂w
In terms of the u, v, w coordinates, this change can be thought of as moving along the diagonal of a paral∂~r
∂~r
∂~r
du,
dv, and
dw.
lelepiped having the vector sides
∂u
∂v
∂w
Assume u = u(x, y, z) is defined by equation (1.2.9) and differentiate this relation to obtain
du =
∂u
∂u
∂u
dx +
dy +
dz.
∂x
∂y
∂z
(1.2.17)
The equation (1.2.15) enables us to represent this differential in the form:
du = grad u · d~r
∂~r
∂~r
∂~r
du +
dv +
dw
du = grad u ·
∂u
∂v
∂w
∂~r
∂~r
∂~r
du = grad u ·
du + grad u ·
dv + grad u ·
dw.
∂u
∂v
∂w
(1.2.18)
By comparing like terms in this last equation we find that
~ 1 = 1,
~1 · E
E
~1 · E
~ 2 = 0,
E
~1 · E
~ 3 = 0.
E
Similarly, from the other equations in equation (1.2.9) which define v = v(x, y, z),
can be demonstrated that
dv =
(1.2.19)
and w = w(x, y, z) it
∂~r
∂~r
∂~r
du + grad v ·
dv + grad v ·
dw
grad v ·
∂u
∂v
∂w
(1.2.20)
39
∂~r
∂~r
∂~r
du + grad w ·
dv + grad w ·
dw.
dw = grad w ·
∂u
∂v
∂w
and
(1.2.21)
By comparing like terms in equations (1.2.20) and (1.2.21) we find
~ 1 = 0,
~2 · E
E
~2 · E
~ 2 = 1,
E
~2 · E
~3 = 0
E
~ 1 = 0,
~3 · E
E
~3 · E
~ 2 = 0,
E
~3 · E
~ 3 = 1.
E
(1.2.22)
The equations (1.2.22) and (1.2.19) show us that the basis vectors defined by equations (1.2.12) and (1.2.13)
are reciprocal.
Introducing the notation
(x1 , x2 , x3 ) = (u, v, w)
(y 1 , y 2 , y 3 ) = (x, y, z)
(1.2.23)
where the x0 s denote the generalized coordinates and the y 0 s denote the rectangular Cartesian coordinates,
the above equations can be expressed in a more concise form with the index notation. For example, if
xi = xi (x, y, z) = xi (y 1 , y 2 , y 3 ),
and y i = y i (u, v, w) = y i (x1 , x2 , x3 ),
i = 1, 2, 3
(1.2.24)
then the reciprocal basis vectors can be represented
~ i = grad xi ,
E
i = 1, 2, 3
(1.2.25)
and
~ i = ∂~r ,
E
∂xi
i = 1, 2, 3.
(1.2.26)
We now show that these basis vectors are reciprocal. Observe that ~r = ~r(x1 , x2 , x3 ) with
d~r =
∂~r
dxm
∂xm
(1.2.27)
and consequently
dxi = grad xi · d~r = grad xi ·
∂~r
m
i
~i · E
~ m dxm = δm
E
dx
=
dxm ,
∂xm
i = 1, 2, 3
(1.2.28)
Comparing like terms in this last equation establishes the result that
~ m = δi ,
~i · E
E
m
i, m = 1, 2, 3
which demonstrates that the basis vectors are reciprocal.
(1.2.29)
40
Scalars, Vectors and Tensors
Tensors are quantities which obey certain transformation laws. That is, scalars, vectors, matrices
and higher order arrays can be thought of as components of a tensor quantity. We shall be interested in
finding how these components are represented in various coordinate systems. We desire knowledge of these
transformation laws in order that we can represent various physical laws in a form which is independent of
the coordinate system chosen. Before defining different types of tensors let us examine what we mean by a
coordinate transformation.
Coordinate transformations of the type found in equations (1.2.8) and (1.2.9) can be generalized to
higher dimensions. Let xi , i = 1, 2, . . . , N denote N variables. These quantities can be thought of as
representing a variable point (x1 , x2 , . . . , xN ) in an N dimensional space VN . Another set of N quantities,
call them barred quantities, xi , i = 1, 2, . . . , N, can be used to represent a variable point (x1 , x2 , . . . , xN ) in
an N dimensional space V N . When the x0 s are related to the x0 s by equations of the form
xi = xi (x1 , x2 , . . . , xN ),
i = 1, 2, . . . , N
(1.2.30)
then a transformation is said to exist between the coordinates xi and xi , i = 1, 2, . . . , N. Whenever the
relations (1.2.30) are functionally independent, single valued and possess partial derivatives such that the
Jacobian of the transformation
J
x
x
=J
x1 , x2 , . . . , xN
x1 , x2 , . . . , xN
∂x1
∂x1
∂x1
∂x2
..
.
=
..
.
∂xN
∂x1
∂xN
∂x2
...
...
...
∂x1
∂xN
..
.
(1.2.31)
∂xN
∂xN
is different from zero, then there exists an inverse transformation
xi = xi (x1 , x2 , . . . , xN ),
i = 1, 2, . . . , N.
(1.2.32)
For brevity the transformation equations (1.2.30) and (1.2.32) are sometimes expressed by the notation
xi = xi (x), i = 1, . . . , N
and
xi = xi (x), i = 1, . . . , N.
(1.2.33)
¯ coordinates. For simplicity
Consider a sequence of transformations from x to x̄ and then from x̄ to x̄
¯ = z. If we denote by T1 , T2 and T3 the transformations
let x̄ = y and x̄
T1 :
T2 :
y i = y i (x1 , . . . , xN ) i = 1, . . . , N
i
i
1
N
z = z (y , . . . , y ) i = 1, . . . , N
or
T1 x = y
or T2 y = z
Then the transformation T3 obtained by substituting T1 into T2 is called the product of two successive
transformations and is written
T3 :
z i = z i (y 1 (x1 , . . . , xN ), . . . , y N (x1 , . . . , xN ))
i = 1, . . . , N
or T3 x = T2 T1 x = z.
This product transformation is denoted symbolically by T3 = T2 T1 .
The Jacobian of the product transformation is equal to the product of Jacobians associated with the
product transformation and J3 = J2 J1 .
41
Transformations Form a Group
A group G is a nonempty set of elements together with a law, for combining the elements. The combined
elements are denoted by a product. Thus, if a and b are elements in G then no matter how you define the
law for combining elements, the product combination is denoted ab. The set G and combining law forms a
group if the following properties are satisfied:
(i) For all a, b ∈ G, then ab ∈ G. This is called the closure property.
(ii) There exists an identity element I such that for all a ∈ G we have Ia = aI = a.
(iii) There exists an inverse element. That is, for all a ∈ G there exists an inverse element a−1 such that
a a−1 = a−1 a = I.
(iv) The associative law holds under the combining law and a(bc) = (ab)c for all a, b, c ∈ G.
For example, the set of elements G = {1, −1, i, −i}, where i2 = −1 together with the combining law of
ordinary multiplication, forms a group. This can be seen from the multiplication table.
×
1
-1
-i
i
1
1
-1
-i
i
-1
-1
1
i
-i
i
i
-i
1
-1
-i
-i
i
-1
1
The set of all coordinate transformations of the form found in equation (1.2.30), with Jacobian different
from zero, forms a group because:
(i) The product transformation, which consists of two successive transformations, belongs to the set of
transformations. (closure)
(ii) The identity transformation exists in the special case that x and x are the same coordinates.
(iii) The inverse transformation exists because the Jacobian of each individual transformation is different
from zero.
(iv) The associative law is satisfied in that the transformations satisfy the property T3 (T2 T1 ) = (T3 T2 )T1 .
When the given transformation equations contain a parameter the combining law is often times represented as a product of symbolic operators. For example, we denote by Tα a transformation of coordinates
having a parameter α. The inverse transformation can be denoted by Tα−1 and one can write Tα x = x or
x = Tα−1 x. We let Tβ denote the same transformation, but with a parameter β, then the transitive property
is expressed symbolically by Tα Tβ = Tγ where the product Tα Tβ represents the result of performing two
successive transformations. The first coordinate transformation uses the given transformation equations and
uses the parameter α in these equations. This transformation is then followed by another coordinate transformation using the same set of transformation equations, but this time the parameter value is β. The above
symbolic product is used to demonstrate that the result of applying two successive transformations produces
a result which is equivalent to performing a single transformation of coordinates having the parameter value
γ. Usually some relationship can then be established between the parameter values α, β and γ.
42
Figure 1.2-2. Cylindrical coordinates.
In this symbolic notation, we let Tθ denote the identity transformation. That is, using the parameter
value of θ in the given set of transformation equations produces the identity transformation. The inverse
transformation can then be expressed in the form of finding the parameter value β such that Tα Tβ = Tθ .
Cartesian Coordinates
At times it is convenient to introduce an orthogonal Cartesian coordinate system having coordinates
i
y,
i = 1, 2, . . . , N. This space is denoted EN and represents an N-dimensional Euclidean space. Whenever
the generalized independent coordinates xi , i = 1, . . . , N are functions of the y 0 s, and these equations are
functionally independent, then there exists independent transformation equations
y i = y i (x1 , x2 , . . . , xN ),
i = 1, 2, . . . , N,
(1.2.34)
with Jacobian different from zero. Similarly, if there is some other set of generalized coordinates, say a barred
system xi , i = 1, . . . , N where the x0 s are independent functions of the y 0 s, then there will exist another set
of independent transformation equations
y i = y i (x1 , x2 , . . . , xN ),
i = 1, 2, . . . , N,
(1.2.35)
with Jacobian different from zero. The transformations found in the equations (1.2.34) and (1.2.35) imply
that there exists relations between the x0 s and x0 s of the form (1.2.30) with inverse transformations of the
form (1.2.32). It should be remembered that the concepts and ideas developed in this section can be applied
to a space VN of any finite dimension. Two dimensional surfaces (N = 2) and three dimensional spaces
(N = 3) will occupy most of our applications. In relativity, one must consider spaces where N = 4.
EXAMPLE 1.2-1. (cylindrical coordinates (r, θ, z)) Consider the transformation
x = x(r, θ, z) = r cos θ
y = y(r, θ, z) = r sin θ
z = z(r, θ, z) = z
from rectangular coordinates (x, y, z) to cylindrical coordinates (r, θ, z), illustrated in the figure 1.2-2. By
letting
y 1 = x,
y 2 = y,
y3 = z
x1 = r,
x2 = θ,
x3 = z
the above set of equations are examples of the transformation equations (1.2.8) with u = r, v = θ, w = z as
the generalized coordinates.
43
EXAMPLE 1.2.2. (Spherical Coordinates) (ρ, θ, φ)
Consider the transformation
x = x(ρ, θ, φ) = ρ sin θ cos φ
y = y(ρ, θ, φ) = ρ sin θ sin φ
z = z(ρ, θ, φ) = ρ cos θ
from rectangular coordinates (x, y, z) to spherical coordinates (ρ, θ, φ). By letting
y 1 = x, y 2 = y, y 3 = z
x1 = ρ, x2 = θ , x3 = φ
the above set of equations has the form found in equation (1.2.8) with u = ρ, v = θ, w = φ the generalized
coordinates. One could place bars over the x0 s in this example in order to distinguish these coordinates from
the x0 s of the previous example. The spherical coordinates (ρ, θ, φ) are illustrated in the figure 1.2-3.
Figure 1.2-3. Spherical coordinates.
Scalar Functions and Invariance
We are now at a point where we can begin to define what tensor quantities are. The first definition is
for a scalar invariant or tensor of order zero.
44
Definition: ( Absolute scalar field) Assume there exists a coordinate
transformation of the type (1.2.30) with Jacobian J different from zero. Let
the scalar function
f = f (x1 , x2 , . . . , xN )
(1.2.36)
be a function of the coordinates xi , i = 1, . . . , N in a space VN . Whenever
there exists a function
f = f (x1 , x2 , . . . , xN )
(1.2.37)
which is a function of the coordinates xi , i = 1, . . . , N such that f = J W f,
then f is called a tensor of rank or order zero of weight W in the space VN .
Whenever W = 0, the scalar f is called the component of an absolute scalar
field and is referred to as an absolute tensor of rank or order zero.
That is, an absolute scalar field is an invariant object in the space VN with respect to the group of
coordinate transformations. It has a single component in each coordinate system. For any scalar function
of the type defined by equation (1.2.36), we can substitute the transformation equations (1.2.30) and obtain
f = f (x1 , . . . , xN ) = f (x1 (x), . . . , xN (x)) = f (x1 , . . . , xN ).
(1.2.38)
Vector Transformation, Contravariant Components
In VN consider a curve C defined by the set of parametric equations
C:
xi = xi (t),
i = 1, . . . , N
where t is a parameter. The tangent vector to the curve C is the vector
T~ =
dxN
dx1 dx2
,
,...,
dt dt
dt
.
In index notation, which focuses attention on the components, this tangent vector is denoted
Ti =
dxi
,
dt
i = 1, . . . , N.
For a coordinate transformation of the type defined by equation (1.2.30) with its inverse transformation
defined by equation (1.2.32), the curve C is represented in the barred space by
xi = xi (x1 (t), x2 (t), . . . , xN (t)) = xi (t),
i = 1, . . . , N,
with t unchanged. The tangent to the curve in the barred system of coordinates is represented by
∂xi dxj
dxi
=
,
dt
∂xj dt
i = 1, . . . , N.
(1.2.39)
45
i
Letting T , i = 1, . . . , N denote the components of this tangent vector in the barred system of coordinates,
the equation (1.2.39) can then be expressed in the form
i
T =
∂xi j
T ,
∂xj
i, j = 1, . . . , N.
(1.2.40)
This equation is said to define the transformation law associated with an absolute contravariant tensor of
rank or order one. In the case N = 3 the matrix form of this transformation is represented
 1
∂x
1
T
∂x1

2
 T  =  ∂x21
∂x
3
∂x3
T
1

∂x
∂x1
∂x2
∂x2
∂x2
∂x3
∂x2
∂x1
∂x3
∂x2
∂x3
∂x3
∂x3


T1
 2
 T
T3
(1.2.41)
A more general definition is
Definition: (Contravariant tensor) Whenever N quantities Ai in
i
a coordinate system (x1 , . . . , xN ) are related to N quantities A in a
coordinate system (x1 , . . . , xN ) such that the Jacobian J is different
from zero, then if the transformation law
i
A = JW
∂xi j
A
∂xj
is satisfied, these quantities are called the components of a relative tensor
of rank or order one with weight W . Whenever W = 0 these quantities
are called the components of an absolute tensor of rank or order one.
We see that the above transformation law satisfies the group properties.
EXAMPLE 1.2-3. (Transitive Property of Contravariant Transformation)
Show that successive contravariant transformations is also a contravariant transformation.
Solution: Consider the transformation of a vector from an unbarred to a barred system of coordinates. A
vector or absolute tensor of rank one Ai = Ai (x), i = 1, . . . , N will transform like the equation (1.2.40) and
i
A (x) =
∂xi j
A (x).
∂xj
(1.2.42)
Another transformation from x → x coordinates will produce the components
i
∂x j
A (x) =
A (x)
∂xj
i
(1.2.43)
Here we have used the notation Aj (x) to emphasize the dependence of the components Aj upon the x
coordinates. Changing indices and substituting equation (1.2.42) into (1.2.43) we find
i
A (x) =
i
∂x ∂xj m
A (x).
∂xj ∂xm
(1.2.44)
46
From the fact that
i
i
∂x
∂x ∂xj
=
,
∂xm
∂xj ∂xm
the equation (1.2.44) simplifies to
i
i
A (x) =
∂x m
A (x)
∂xm
(1.2.45)
and hence this transformation is also contravariant. We express this by saying that the above are transitive
with respect to the group of coordinate transformations.
Note that from the chain rule one can write
∂xm ∂x1
∂xm ∂x2
∂xm ∂x3
∂xm
∂xm ∂xj
m
=
+
+
1 ∂xn
2 ∂xn
3 ∂xn = ∂xn = δn .
j ∂xn
∂x
∂x
∂x
∂x
Do not make the mistake of writing
∂xm
∂xm ∂x2
2 ∂xn = ∂xn
∂x
∂xm ∂x3
∂xm
3 ∂xn = ∂xn
∂x
or
as these expressions are incorrect. Note that there are no summations in these terms, whereas there is a
summation index in the representation of the chain rule.
Vector Transformation, Covariant Components
Consider a scalar invariant A(x) = A(x) which is a shorthand notation for the equation
A(x1 , x2 , . . . , xn ) = A(x1 , x2 , . . . , xn )
involving the coordinate transformation of equation (1.2.30). By the chain rule we differentiate this invariant
and find that the components of the gradient must satisfy
∂A ∂xj
∂A
.
i = ∂xj
∂x
∂xi
(1.2.46)
Let
Aj =
∂A
∂xj
and
Ai =
∂A
,
∂xi
then equation (1.2.46) can be expressed as the transformation law
Ai = Aj
∂xj
.
∂xi
(1.2.47)
This is the transformation law for an absolute covariant tensor of rank or order one. A more general definition
is
47
Definition: (Covariant tensor)
1
Whenever N quantities Ai in a
N
coordinate system (x , . . . , x ) are related to N quantities Ai in a coordinate system (x1 , . . . , xN ), with Jacobian J different from zero, such
that the transformation law
Ai = J W
∂xj
Aj
∂xi
(1.2.48)
is satisfied, then these quantities are called the components of a relative
covariant tensor of rank or order one having a weight of W . Whenever W = 0, these quantities are called the components of an absolute
covariant tensor of rank or order one.
Again we note that the above transformation satisfies the group properties. Absolute tensors of rank or
order one are referred to as vectors while absolute tensors of rank or order zero are referred to as scalars.
EXAMPLE 1.2-4. (Transitive Property of Covariant Transformation)
Consider a sequence of transformation laws of the type defined by the equation (1.2.47)
x→x
x→x
∂xj
∂xi
∂xm
Ak (x) = Am (x) k
∂x
Ai (x) = Aj (x)
We can therefore express the transformation of the components associated with the coordinate transformation
x → x and
∂xj ∂xm
∂xj
Ak (x) = Aj (x) m
=
A
(x)
,
j
k
k
∂x
∂x
∂x
which demonstrates the transitive property of a covariant transformation.
Higher Order Tensors
We have shown that first order tensors are quantities which obey certain transformation laws. Higher
order tensors are defined in a similar manner and also satisfy the group properties. We assume that we are
given transformations of the type illustrated in equations (1.2.30) and (1.2.32) which are single valued and
continuous with Jacobian J different from zero. Further, the quantities xi and xi , i = 1, . . . , n represent the
coordinates in any two coordinate systems. The following transformation laws define second order and third
order tensors.
48
Definition: (Second order contravariant tensor) Whenever N-squared quantities Aij
mn
in a coordinate system (x1 , . . . , xN ) are related to N-squared quantities A
1
in a coordinate
N
system (x , . . . , x ) such that the transformation law
mn
A
(x) = Aij (x)J W
∂xm ∂xn
∂xi ∂xj
(1.2.49)
is satisfied, then these quantities are called components of a relative contravariant tensor of
rank or order two with weight W . Whenever W = 0 these quantities are called the components
of an absolute contravariant tensor of rank or order two.
Definition: (Second order covariant tensor) Whenever N-squared quantities
Aij in a coordinate system (x1 , . . . , xN ) are related to N-squared quantities Amn
in a coordinate system (x1 , . . . , xN ) such that the transformation law
Amn (x) = Aij (x)J W
∂xi ∂xj
∂xm ∂xn
(1.2.50)
is satisfied, then these quantities are called components of a relative covariant tensor
of rank or order two with weight W . Whenever W = 0 these quantities are called
the components of an absolute covariant tensor of rank or order two.
Definition: (Second order mixed tensor)
Aij
1
Whenever N-squared quantities
m
N
in a coordinate system (x , . . . , x ) are related to N-squared quantities An in
a coordinate system (x1 , . . . , xN ) such that the transformation law
m
An (x) = Aij (x)J W
∂xm ∂xj
∂xi ∂xn
(1.2.51)
is satisfied, then these quantities are called components of a relative mixed tensor of
rank or order two with weight W . Whenever W = 0 these quantities are called the
components of an absolute mixed tensor of rank or order two. It is contravariant
of order one and covariant of order one.
Higher order tensors are defined in a similar manner. For example, if we can find N-cubed quantities
Am
np
such that
∂xi ∂xα ∂xβ
(1.2.52)
∂xγ ∂xj ∂xk
then this is a relative mixed tensor of order three with weight W . It is contravariant of order one and
i
Ajk (x) = Aγαβ (x)J W
covariant of order two.
49
General Definition
In general a mixed tensor of rank or order (m + n)
...im
Tji11ji22...j
n
(1.2.53)
is contravariant of order m and covariant of order n if it obeys the transformation law
h x iW
i1
∂xim ∂xb1 ∂xb2
∂xbn
∂xi2
i1 i2 ...im
...am ∂x
T j1 j2 ...jn = J
Tba11ba22...b
·
·
·
·
·
·
·
n
x
∂xa1 ∂xa2
∂xam ∂xj1 ∂xj2
∂xjn
where
J
x
x
=
(1.2.54)
∂x
∂(x1 , x2 , . . . , xN )
=
∂x
∂(x1 , x2 , . . . , xN )
is the Jacobian of the transformation. When W = 0 the tensor is called an absolute tensor, otherwise it is
called a relative tensor of weight W.
Here superscripts are used to denote contravariant components and subscripts are used to denote covariant components. Thus, if we are given the tensor components in one coordinate system, then the components
in any other coordinate system are determined by the transformation law of equation (1.2.54). Throughout
the remainder of this text one should treat all tensors as absolute tensors unless specified otherwise.
Dyads and Polyads
Note that vectors can be represented in bold face type with the notation
A = Ai Ei
This notation can also be generalized to tensor quantities. Higher order tensors can also be denoted by bold
face type. For example the tensor components Tij and Bijk can be represented in terms of the basis vectors
Ei , i = 1, . . . , N by using a notation which is similar to that for the representation of vectors. For example,
T = Tij Ei Ej
B = Bijk Ei Ej Ek .
Here T denotes a tensor with components Tij and B denotes a tensor with components Bijk . The quantities
Ei Ej are called unit dyads and Ei Ej Ek are called unit triads. There is no multiplication sign between the
basis vectors. This notation is called a polyad notation. A further generalization of this notation is the
representation of an arbitrary tensor using the basis and reciprocal basis vectors in bold type. For example,
a mixed tensor would have the polyadic representation
ij...k
Ei Ej . . . Ek El Em . . . En .
T = Tlm...n
A dyadic is formed by the outer or direct product of two vectors. For example, the outer product of the
vectors
a = a 1 E 1 + a2 E 2 + a3 E 3
and b = b1 E1 + b2 E2 + b3 E3
50
gives the dyad
ab =a1 b1 E1 E1 + a1 b2 E1 E2 + a1 b3 E1 E3
a2 b 1 E 2 E 1 + a2 b 2 E 2 E 2 + a2 b 3 E 2 E 3
a3 b 1 E 3 E 1 + a3 b 2 E 3 E 2 + a3 b 3 E 3 E 3 .
In general, a dyad can be represented
A = Aij Ei Ej
i, j = 1, . . . , N
where the summation convention is in effect for the repeated indices. The coefficients Aij are called the
coefficients of the dyad. When the coefficients are written as an N × N array it is called a matrix. Every
second order tensor can be written as a linear combination of dyads. The dyads form a basis for the second
order tensors. As the example above illustrates, the nine dyads {E1 E1 , E1 E2 , . . . , E3 E3 }, associated with
the outer products of three dimensional base vectors, constitute a basis for the second order tensor A = ab
having the components Aij = ai bj with i, j = 1, 2, 3. Similarly, a triad has the form
T = Tijk Ei Ej Ek
Sum on repeated indices
where i, j, k have the range 1, 2, . . . , N. The set of outer or direct products { Ei Ej Ek }, with i, j, k = 1, . . . , N
i
are associated
constitutes a basis for all third order tensors. Tensor components with mixed suffixes like Cjk
with triad basis of the form
i
Ei Ej Ek
C = Cjk
where i, j, k have the range 1, 2, . . . N. Dyads are associated with the outer product of two vectors, while triads,
tetrads,... are associated with higher-order outer products. These higher-order outer or direct products are
referred to as polyads.
The polyad notation is a generalization of the vector notation. The subject of how polyad components
transform between coordinate systems is the subject of tensor calculus.
ei and a dyadic with components called dyads is written
In Cartesian coordinates we have Ei = Ei = b
ei b
ej or
A = Aij b
e1 b
e1 + A12 b
e1 b
e2 + A13 b
e1 b
e3
A =A11 b
e2 b
e1 + A22 b
e2 b
e2 + A23 b
e2 b
e3
A21 b
e3 b
e1 + A32 b
e3 b
e2 + A33 b
e3 b
e3
A31 b
ej are called unit dyads. Note that a dyadic has nine components as compared with a
where the terms b
ei b
vector which has only three components. The conjugate dyadic Ac is defined by a transposition of the unit
vectors in A, to obtain
e1 b
e1 + A12 b
e2 b
e1 + A13 b
e3 b
e1
Ac =A11 b
e1 b
e2 + A22 b
e2 b
e2 + A23 b
e3 b
e2
A21 b
e1 b
e3 + A32 b
e2 b
e3 + A33 b
e3 b
e3
A31 b
51
If a dyadic equals its conjugate A = Ac , then Aij = Aji and the dyadic is called symmetric. If a dyadic
equals the negative of its conjugate A = −Ac , then Aij = −Aji and the dyadic is called skew-symmetric. A
special dyadic called the identical dyadic or idemfactor is defined by
e1 + b
e2 + b
e3 .
e2 b
e3 b
J= b
e1 b
~ produces the
This dyadic has the property that pre or post dot product multiplication of J with a vector V
same vector V~ . For example,
~ · J = (V1 b
e1 + V2 b
e2 + V3 b
e3 ) · J
V
~
e1 · b
e1 + V2 b
e2 · b
e2 + V3 b
e3 · b
e3 = V
e1 b
e2 b
e3 b
= V1 b
~ = J · (V1 b
e1 + V2 b
e2 + V3 b
e3 )
and J · V
~
e1 + V2 b
e2 + V3 b
e3 = V
e1 b
e1 · b
e2 b
e2 · b
e3 b
e3 · b
= V1 b
A dyadic operation often used in physics and chemistry is the double dot product A : B where A and
B are both dyadics. Here both dyadics are expanded using the distributive law of multiplication, and then
em b
ej : b
en are combined according to the rule
each unit dyad pair b
ei b
b
ej : b
en = ( b
em b
ei · b
em )( b
ej · b
en ).
ei b
ei b
ej and B = Bij b
ei b
ej , then the double dot product A : B is calculated as follows.
For example, if A = Aij b
ei b
ej ) : (Bmn b
em b
en ) = Aij Bmn ( b
ej : b
en ) = Aij Bmn ( b
ei b
em b
ei · b
em )( b
ej · b
en )
A : B = (Aij b
= Aij Bmn δim δjn = Amj Bmj
= A11 B11 + A12 B12 + A13 B13
+ A21 B21 + A22 B22 + A23 B23
+ A31 B31 + A32 B32 + A33 B33
When operating with dyads, triads and polyads, there is a definite order to the way vectors and polyad
~ = Bi b
~ = Ai b
ei and B
ei vectors with outer product
components are represented. For example, for A
~B
~ = Am Bn b
em b
en = φ
A
there is produced the dyadic φ with components Am Bn . In comparison, the outer product
~A
~ = Bm An b
em b
en = ψ
B
produces the dyadic ψ with components Bm An . That is
~B
~ =A1 B1 b
e1 b
e1 + A1 B2 b
e1 b
e2 + A1 B3 b
e1 b
e3
φ=A
e2 b
e1 + A2 B2 b
e2 b
e2 + A2 B3 b
e2 b
e3
A2 B1 b
e3 b
e1 + A3 B2 b
e3 b
e2 + A3 B3 b
e3 b
e3
A3 B1 b
~A
~ =B1 A1 b
e1 b
e1 + B1 A2 b
e1 b
e2 + B1 A3 b
e1 b
e3
and ψ = B
e2 b
e1 + B2 A2 b
e2 b
e2 + B2 A3 b
e2 b
e3
B2 A1 b
e3 b
e1 + B3 A2 b
e3 b
e2 + B3 A3 b
e3 b
e3
B3 A1 b
are different dyadics.
~ is defined for both pre and post multiplication as
The scalar dot product of a dyad with a vector C
~ =A
~B
~ ·C
~ =A(
~ B
~ · C)
~
φ·C
~ ·φ=C
~ ·A
~B
~ =(C
~ · A)
~ B
~
C
These products are, in general, not equal.
52
Operations Using Tensors
The following are some important tensor operations which are used to derive special equations and to
prove various identities.
Addition and Subtraction
Tensors of the same type and weight can be added or subtracted. For example, two third order mixed
i
denote two third order
tensors, when added, produce another third order mixed tensor. Let Aijk and Bjk
mixed tensors. Their sum is denoted
i
i
= Aijk + Bjk
.
Cjk
That is, like components are added. The sum is also a mixed tensor as we now verify. By hypothesis Aijk
i
and Bjk
are third order mixed tensors and hence must obey the transformation laws
i
∂xi ∂xn ∂xp
∂xm ∂xj ∂xk
i
n
p
m ∂x ∂x ∂x
= Bnp
.
j
∂xm ∂x ∂xk
Ajk = Am
np
i
B jk
i
i
i
We let C jk = Ajk + B jk denote the sum in the transformed coordinates. Then the addition of the above
transformation equations produces
i
i
n
p
∂xi ∂xn ∂xp
i
i
m
m ∂x ∂x ∂x
C jk = Ajk + B jk = Am
+
B
=
C
.
np
np
np
∂xm ∂xj ∂xk
∂xm ∂xj ∂xk
Consequently, the sum transforms as a mixed third order tensor.
Multiplication (Outer Product)
The product of two tensors is also a tensor. The rank or order of the resulting tensor is the sum of
the ranks of the tensors occurring in the multiplication. As an example, let Aijk denote a mixed third order
l
denote a mixed second order tensor. The outer product of these two tensors is the fifth
tensor and let Bm
order tensor
il
l
= Aijk Bm
, i, j, k, l, m = 1, 2, . . . , N.
Cjkm
i
l
Here all indices are free indices as i, j, k, l, m take on any of the integer values 1, 2, . . . , N. Let Ajk and B m
il
denote the components of the given tensors in the barred system of coordinates. We define C jkm as the
il
l
is a tensor for by hypothesis Aijk and Bm
are tensors
outer product of these components. Observe that Cjkm
and hence obey the transformation laws
∂xα ∂xj ∂xk
∂xi ∂xβ ∂xγ
δ
m
δ
l ∂x ∂x
B = Bm
.
∂xl ∂x
The outer product of these components produces
α
Aβγ = Aijk
∂xα ∂xj ∂xk ∂xδ ∂xm
∂xi ∂xβ ∂xγ ∂xl ∂x
(1.2.56)
∂xα ∂xj ∂xk ∂xδ ∂xm
il
= Cjkm i
∂x ∂xβ ∂xγ ∂xl ∂x
transforms as a mixed fifth order absolute tensor. Other outer products are
αδ
α
δ
l
C βγ = Aβγ B = Aijk Bm
il
which demonstrates that Cjkm
analyzed in a similar way.
(1.2.55)
53
Contraction
The operation of contraction on any mixed tensor of rank m is performed when an upper index is
set equal to a lower index and the summation convention is invoked. When the summation is performed
over the repeated indices the resulting quantity is also a tensor of rank or order (m − 2). For example, let
Aijk , i, j, k = 1, 2, . . . , N denote a mixed tensor and perform a contraction by setting j equal to i. We obtain
Aiik = A11k + A22k + · · · + AN
N k = Ak
where k is a free index. To show that Ak is a tensor, we let
transformed components of
Aijk .
By hypothesis
Aijk
i
Aik
(1.2.57)
= Ak denote the contraction on the
is a mixed tensor and hence the components must
satisfy the transformation law
∂xi ∂xn ∂xp
.
∂xm ∂xj ∂xk
Now execute a contraction by setting j equal to i and perform a summation over the repeated index. We
i
Ajk = Am
np
find
∂xi ∂xn ∂xp
∂xn ∂xp
= Am
np
i
k
m
∂x ∂x ∂x
∂xm ∂xk
(1.2.58)
p
p
∂xp
m n ∂x
n ∂x
= Anp δm k = Anp k = Ap k .
∂x
∂x
∂x
Hence, the contraction produces a tensor of rank two less than the original tensor. Contractions on other
i
Aik = Ak = Am
np
mixed tensors can be analyzed in a similar manner.
New tensors can be constructed from old tensors by performing a contraction on an upper and lower
index. This process can be repeated as long as there is an upper and lower index upon which to perform the
contraction. Each time a contraction is performed the rank of the resulting tensor is two less than the rank
of the original tensor.
Multiplication (Inner Product)
The inner product of two tensors is obtained by:
(i) first taking the outer product of the given tensors and
(ii) performing a contraction on two of the indices.
EXAMPLE 1.2-5. (Inner product)
Let Ai and Bj denote the components of two first order tensors (vectors). The outer product of these
tensors is
Cji = Ai Bj , i, j = 1, 2, . . . , N.
The inner product of these tensors is the scalar
C = Ai Bi = A1 B1 + A2 B2 + · · · + AN BN .
Note that in some situations the inner product is performed by employing only subscript indices. For
example, the above inner product is sometimes expressed as
C = Ai Bi = A1 B1 + A2 B2 + · · · AN BN .
This notation is discussed later when Cartesian tensors are considered.
54
Quotient Law
Assume Brqs and Cps are arbitrary absolute tensors. Further assume we have a quantity A(ijk) which
we think might be a third order mixed tensor Aijk . By showing that the equation
Arqp Brqs = Cps
is satisfied, then it follows that Arqp must be a tensor. This is an example of the quotient law. Obviously,
this result can be generalized to apply to tensors of any order or rank. To prove the above assertion we shall
show from the above equation that Aijk is a tensor. Let xi and xi denote a barred and unbarred system of
coordinates which are related by transformations of the form defined by equation (1.2.30). In the barred
system, we assume that
r
qs
s
Aqp B r = C p
(1.2.59)
l
are arbitrary absolute tensors and therefore must satisfy the transformation
where by hypothesis Bkij and Cm
equations
∂xq ∂xs ∂xk
∂xi ∂xj ∂xr
s
∂x
∂xm
s
l
C p = Cm
.
l
∂x ∂xp
qs
B r = Bkij
qs
s
We substitute for B r and C p in the equation (1.2.59) and obtain the equation
q
s
s
k
m
r
ij ∂x ∂x ∂x
l ∂x ∂x
Aqp Bk
= Cm l
∂xi ∂xj ∂xr
∂x ∂xp
∂xs ∂xm
.
= Arqm Brql l
∂x ∂xp
Since the summation indices are dummy indices they can be replaced by other symbols. We change l to j,
q to i and r to k and write the above equation as
q
k
m
∂xs
r ∂x ∂x
k ∂x
Aqp i
− Aim p Bkij = 0.
∂xj
∂x ∂xr
∂x
Use inner multiplication by
∂xn
∂xs
and simplify this equation to the form
q
k
m
r ∂x ∂x
n
k ∂x
− Aim p Bkij = 0
δj Aqp i
∂x ∂xr
∂x
q
k
m
r ∂x ∂x
k ∂x
Aqp i
− Aim p Bkin = 0.
∂x ∂xr
∂x
or
Because Bkin is an arbitrary tensor, the quantity inside the brackets is zero and therefore
r
Aqp
m
∂xq ∂xk
k ∂x
= 0.
r − Aim
i
∂x ∂x
∂xp
This equation is simplified by inner multiplication by
r
∂xi ∂xl
∂xj ∂xk
to obtain
∂xm ∂xi ∂xl
=0
∂xp ∂xj ∂xk
∂xm ∂xi ∂xl
= Akim p
∂x ∂xj ∂xk
δjq δrl Aqp − Akim
l
Ajp
which is the transformation law for a third order mixed tensor.
or
55
EXERCISE 1.2
I 1.
Consider the transformation equations representing a rotation of axes through an angle α.
Tα :
x1
= x1 cos α − x2 sin α
x2
= x1 sin α + x2 cos α
Treat α as a parameter and show this set of transformations constitutes a group by finding the value of α
which:
(i) gives the identity transformation.
(ii) gives the inverse transformation.
(iii) show the transformation is transitive in that a transformation with α = θ1 followed by a transformation
with α = θ2 is equivalent to the transformation using α = θ1 + θ2 .
I 2.
Show the transformation
Tα :
x1
x2
= αx1
= α1 x2
forms a group with α as a parameter. Find the value of α such that:
(i) the identity transformation exists.
(ii) the inverse transformation exists.
(iii) the transitive property is satisfied.
I 3.
Show the given transformation forms a group with parameter α.
(
Tα :
I 4.
x1
=
x1
1−αx1
x2
=
x2
1−αx1
Consider the Lorentz transformation from relativity theory having the velocity parameter V, c is the
speed of light and x4 = t is time.
 1
x






 x2

x3




 x4

TV :
1
4
x −V x
= p
V2
1−
c2
= x2
= x3
1
x4 − Vcx2
= p
2
1− V2
c
Show this set of transformations constitutes a group, by establishing:
(i) V = 0 gives the identity transformation T0 .
(ii) TV2 · TV1 = T0 requires that V2 = −V1 .
(iii) TV2 · TV1 = TV3 requires that
V3 =
I 5.
V1 + V2
.
1 + V1c2V2
~ 2, E
~ 3 ) an arbitrary independent basis, (a) Verify that
~ 1, E
For (E
~2 × E
~ 3,
~1 = 1 E
E
V
~3 × E
~2 = 1 E
~ 1,
E
V
~ 1 · (E
~2 × E
~ 3)
is a reciprocal basis, where V = E
~1 × E
~3 = 1 E
~2
E
V
~ i.
~ j = g ij E
(b) Show that E
56
Figure 1.2-4. Cylindrical coordinates (r, β, z).
I 6.
For the cylindrical coordinates (r, β, z) illustrated in the figure 1.2-4.
(a) Write out the transformation equations from rectangular (x, y, z) coordinates to cylindrical (r, β, z)
coordinates. Also write out the inverse transformation.
(b) Determine the following basis vectors in cylindrical coordinates and represent your results in terms of
cylindrical coordinates.
~ 2, E
~ 3 . (ii)The normal basis E
~ 1, E
~ 2, E
~ 3 . (iii) êr , êβ , êz
~ 1, E
(i) The tangential basis E
where êr , êβ , êz are normalized vectors in the directions of the tangential basis.
~ = Ax b
e1 + Ay b
e2 + Az b
e3 can be represented in any of the forms:
(c) A vector A
~ 1 + A2 E
~ 2 + A3 E
~3
~ = A1 E
A
~ = A1 E
~ 1 + A2 E
~ 2 + A3 E
~3
A
~ = Ar êr + Aβ êβ + Az êz
A
depending upon the basis vectors selected . In terms of the components Ax , Ay , Az
(i) Solve for the contravariant components A1 , A2 , A3 .
(ii) Solve for the covariant components A1 , A2 , A3 .
(iii) Solve for the components Ar , Aβ , Az . Express all results in cylindrical coordinates. (Note the
components Ar , Aβ , Az are referred to as physical components. Physical components are considered in
more detail in a later section.)
57
Figure 1.2-5. Spherical coordinates (ρ, α, β).
I 7.
For the spherical coordinates (ρ, α, β) illustrated in the figure 1.2-5.
(a) Write out the transformation equations from rectangular (x, y, z) coordinates to spherical (ρ, α, β) coordinates. Also write out the equations which describe the inverse transformation.
(b) Determine the following basis vectors in spherical coordinates
~ 2, E
~ 3.
~ 1, E
(i) The tangential basis E
~ 2, E
~ 3.
~ 1, E
(ii) The normal basis E
(iii) êρ , êα , êβ which are normalized vectors in the directions of the tangential basis. Express all results
in terms of spherical coordinates.
~ = Ax b
e1 + Ay b
e2 + Az b
e3 can be represented in any of the forms:
(c) A vector A
~ 1 + A2 E
~ 2 + A3 E
~3
~ = A1 E
A
~ = A1 E
~ 1 + A2 E
~ 2 + A3 E
~3
A
~ = Aρ êρ + Aα êα + Aβ êβ
A
depending upon the basis vectors selected . Calculate, in terms of the coordinates (ρ, α, β) and the
components Ax , Ay , Az
(i) The contravariant components A1 , A2 , A3 .
(ii) The covariant components A1 , A2 , A3 .
(iii) The components Aρ , Aα , Aβ which are called physical components.
I 8.
Work the problems 6,7 and then let (x1 , x2 , x3 ) = (r, β, z) denote the coordinates in the cylindrical
system and let (x1 , x2 , x3 ) = (ρ, α, β) denote the coordinates in the spherical system.
(a) Write the transformation equations x → x from cylindrical to spherical coordinates. Also find the
inverse transformations.
( Hint: See the figures 1.2-4 and 1.2-5.)
(b) Use the results from part (a) and the results from problems 6,7 to verify that
Ai = Aj
∂xj
∂xi
for
i = 1, 2, 3.
(i.e. Substitute Aj from problem 6 to get Āi given in problem 7.)
58
(c) Use the results from part (a) and the results from problems 6,7 to verify that
i
A = Aj
∂xi
∂xj
for
i = 1, 2, 3.
(i.e. Substitute Aj from problem 6 to get Āi given by problem 7.)
I 9.
Pick two arbitrary noncolinear vectors in the x, y plane, say
~1 = 5 b
e1 + b
e2
V
~2 = b
and V
e1 + 5 b
e2
~3 = b
~2 . The vectors V
~1 and V~2 can be thought of
and let V
e3 be a unit vector perpendicular to both V~1 and V
as defining an oblique coordinate system, as illustrated in the figure 1.2-6.
~ 1 , V~ 2 , V~ 3 ).
(a) Find the reciprocal basis (V
(b) Let
~3
e2 + z b
e3 = αV~1 + β V~2 + γ V
~r = x b
e1 + y b
and show that
y
5x
−
24 24
5y
x
β=− +
24 24
γ=z
α=
(c) Show
x = 5α + β
y = α + 5β
z=γ
(d) For γ = γ0 constant, show the coordinate lines are described by α = constant
and
and sketch some of these coordinate lines. (See figure 1.2-6.)
(e) Find the metrics gij and conjugate metrices g ij associated with the (α, β, γ) space.
Figure 1.2-6. Oblique coordinates.
β = constant,
59
I 10.
Consider the transformation equations
x = x(u, v, w)
y = y(u, v, w)
z = z(u, v, w)
substituted into the position vector
e2 + z b
e3 .
~r = x b
e1 + y b
Define the basis vectors
~ 2, E
~ 3) =
~ 1, E
(E
∂~r ∂~r ∂~r
,
,
∂u ∂v ∂w
with the reciprocal basis
~2 × E
~ 3,
~1 = 1 E
E
V
~3 × E
~2 = 1 E
~ 1,
E
V
~1 × E
~3 = 1 E
~ 2.
E
V
where
~2 × E
~ 3 ).
~ 1 · (E
V =E
~2 × E
~ 3 ) and show that v · V = 1.
~ 1 · (E
Let v = E
I 11.
Given the coordinate transformation
x = −u − 2v
y = −u − v
z=z
(a) Find and illustrate graphically some of the coordinate curves.
(b) For ~r = ~r(u, v, z) a position vector, define the basis vectors
~ 1 = ∂~r ,
E
∂u
~ 2 = ∂~r ,
E
∂v
~ 3 = ∂~r .
E
∂z
~ 2, E
~ 3.
~ 1, E
Calculate these vectors and then calculate the reciprocal basis E
(c) With respect to the basis vectors in (b) find the contravariant components Ai associated with the vector
~ = α1 b
e1 + α2 b
e2 + α3 b
e3
A
where (α1 , α2 , α3 ) are constants.
~ given in part (c).
(d) Find the covariant components Ai associated with the vector A
(e) Calculate the metric tensor gij and conjugate metric tensor g ij .
(f) From the results (e), verify that gij g jk = δik
(g) Use the results from (c)(d) and (e) to verify that Ai = gik Ak
(h) Use the results from (c)(d) and (e) to verify that Ai = g ik Ak
~ on unit vectors in the directions E
~ 1, E
~ 2, E
~ 3.
(i) Find the projection of the vector A
~ 2, E
~ 3.
~ on unit vectors the directions E
~ 1, E
(j) Find the projection of the vector A
60
ei where y i = y i (x1 , x2 , x3 ), i = 1, 2, 3 we have by definition
For ~r = y i b
I 12.
i
~ j = ∂~r = ∂y b
ei . From this relation show that
E
∂xj
∂xj
m
~ m = ∂x b
ej
E
∂y j
and consequently
m
m
~i · E
~ j = ∂y ∂y ,
gij = E
∂xi ∂xj
I 13.
i
j
~i · E
~ j = ∂x ∂x ,
and g ij = E
∂y m ∂y m
i, j, m = 1, . . . , 3
Consider the set of all coordinate transformations of the form
y i = aij xj + bi
where aij and bi are constants and the determinant of aij is different from zero. Show this set of transformations forms a group.
I 14.
For αi , βi constants and t a parameter, xi = αi + t βi ,i = 1, 2, 3 is the parametric representation of
a straight line. Find the parametric equation of the line which passes through the two points (1, 2, 3) and
(14, 7, −3). What does the vector
I 15.
d~
r
dt
represent?
A surface can be represented using two parameters u, v by introducing the parametric equations
xi = xi (u, v),
i = 1, 2, 3,
a < u < b and c < v < d.
The parameters u, v are called the curvilinear coordinates of a point on the surface. A point on the surface
e1 + x2 (u, v) b
e2 + x3 (u, v) b
e3 . The vectors
can be represented by the position vector ~r = ~r(u, v) = x1 (u, v) b
and
∂~
r
∂v
∂~
r
∂u
are tangent vectors to the coordinate surface curves ~r(u, c2 ) and ~r(c1 , v) respectively. An element of
surface area dS on the surface is defined as the area of the elemental parallelogram having the vector sides
∂~
r
∂u du
and
∂~
r
∂v dv.
Show that
dS = |
p
∂~r
∂~r
×
| dudv = g11 g22 − (g12 )2 dudv
∂u ∂v
where
g11 =
∂~r ∂~r
·
∂u ∂u
g12 =
∂~r ∂~r
·
∂u ∂v
g22 =
∂~r ∂~r
·
.
∂v ∂v
~ × B)
~ · (A
~ × B)
~ = |A
~ × B|
~ 2 See Exercise 1.1, problem 9(c).
Hint: (A
I 16.
(a) Use the results from problem 15 and find the element of surface area of the circular cone
x = u sin α cos v
α a constant
(b) Find the surface area of the above cone.
y = u sin α sin v
0≤u≤b
z = u cos α
0 ≤ v ≤ 2π
61
I 17.
The equation of a plane is defined in terms of two parameters u and v and has the form
xi = αi u + βi v + γi
i = 1, 2, 3,
where αi βi and γi are constants. Find the equation of the plane which passes through the points (1, 2, 3),
(14, 7, −3) and (5, 5, 5). What does this problem have to do with the position vector ~r(u, v), the vectors
∂~
r ∂~
r
∂u , ∂v
I 18.
and ~r(0, 0)? Hint: See problem 15.
Determine the points of intersection of the curve x1 = t, x2 = (t)2 , x3 = (t)3 with the plane
8 x1 − 5 x2 + x3 − 4 = 0.
I 19.
~k = E
~i × E
~ j where v = E
~ 1 · (E
~2 × E
~ 3 ) and
and v −1 eijk E
I 20.
Let x̄i and xi , i = 1, 2, 3 be related by the linear transformation x̄i = cij xj , where cij are constants
~k = E
~i × E
~j
Verify the relations V eijk E
~2 × E
~ 3 )..
~ 1 · (E
V =E
n
denote the cofactor of cm
such that the determinant c = det(cij ) is different from zero. Let γm
n divided by
the determinant c.
(a) Show that cij γkj = γji cjk = δki .
(b) Show the inverse transformation can be expressed xi = γji x̄j .
(c) Show that if Ai is a contravariant vector, then its transformed components are Āp = cpq Aq .
(d) Show that if Ai is a covariant vector, then its transformed components are Āi = γip Ap .
I 21.
Show that the outer product of two contravariant vectors Ai and B i , i = 1, 2, 3 results in a second
order contravariant tensor.
ei the element of arc length squared is
Show that for the position vector ~r = y i (x1 , x2 , x3 ) b
m
m
∂y
∂y
2
i
j
~i · E
~j =
.
ds = d~r · d~r = gij dx dx where gij = E
∂xi ∂xj
I 22.
I 23.
i
k
i
p
i
absolute tensors, show that if Aijk Bnk = Cjn
then Ajk B n = C jn .
For Aijk , Bnm and Ctq
I 24.
Let Aij denote an absolute covariant tensor of order 2. Show that the determinant A = det(Aij ) is
p
an invariant of weight 2 and (A) is an invariant of weight 1.
I 25.
Let B ij denote an absolute contravariant tensor of order 2. Show that the determinant B = det(B ij )
√
is an invariant of weight −2 and B is an invariant of weight −1.
I 26.
(a) Write out the contravariant components of the following vectors
~1
(i) E
~2
(ii) E
~3
(iii) E
where
~ i = ∂~r
E
∂xi
for i = 1, 2, 3.
(b) Write out the covariant components of the following vectors
~1
(i) E
~2
(ii) E
~3
(ii) E
~ i = grad xi ,
where E
for i = 1, 2, 3.
62
I 27.
Let Aij and Aij denote absolute second order tensors. Show that λ = Aij Aij is a scalar invariant.
I 28.
Assume that aij , i, j = 1, 2, 3, 4 is a skew-symmetric second order absolute tensor. (a) Show that
bijk =
∂ajk
∂aki
∂aij
+
+
i
j
∂x
∂x
∂xk
is a third order tensor. (b) Show bijk is skew-symmetric in all pairs of indices and (c) determine the number
of independent components this tensor has.
I 29.
Show the linear forms A1 x + B1 y + C1 and A2 x + B2 y + C2 , with respect to the group of rotations
and translations x = x cos θ − y sin θ + h and y = x sin θ + y cos θ + k, have the forms A1 x + B 1 y + C 1 and
A2 x + B 2 y + C 2 . Also show that the quantities A1 B2 − A2 B1 and A1 A2 + B1 B2 are invariants.
I 30.
Show that the curvature of a curve y = f (x) is κ = ± y 00 (1 + y 02 )−3/2 and that this curvature remains
invariant under the group of rotations given in the problem 1. Hint: Calculate
dy
dx
=
dy dx
dx dx .
I 31.
Show that when the equation of a curve is given in the parametric form x = x(t), y = y(t), then
ẋÿ − ẏẍ
and remains invariant under the change of parameter t = t(t), where
the curvature is κ = ± 2
(ẋ + ẏ 2 )3/2
ẋ = dx
dt , etc.
I 32.
ij
Let Aij
k denote a third order mixed tensor. (a) Show that the contraction Ai is a first order
contravariant tensor. (b) Show that contraction of i and j produces Aii
k which is not a tensor. This shows
that in general, the process of contraction does not always apply to indices at the same level.
I 33.
Let φ = φ(x1 , x2 , . . . , xN ) denote an absolute scalar invariant. (a) Is the quantity
Is the quantity
I 34.
2
∂ φ
∂xi ∂xj
∂φ
∂xi
a tensor? (b)
a tensor?
Consider the second order absolute tensor aij , i, j = 1, 2 where a11 = 1, a12 = 2, a21 = 3 and a22 = 4.
Find the components of aij under the transformation of coordinates x1 = x1 + x2 and x2 = x1 − x2 .
I 35.
Let Ai , Bi denote the components of two covariant absolute tensors of order one. Show that
Cij = Ai Bj is an absolute second order covariant tensor.
I 36.
Let Ai denote the components of an absolute contravariant tensor of order one and let Bi denote the
components of an absolute covariant tensor of order one, show that Cji = Ai Bj transforms as an absolute
mixed tensor of order two.
I 37.
(a) Show the sum and difference of two tensors of the same kind is also a tensor of this kind. (b) Show
that the outer product of two tensors is a tensor. Do parts (a) (b) in the special case where one tensor Ai
is a relative tensor of weight 4 and the other tensor Bkj is a relative tensor of weight 3. What is the weight
of the outer product tensor Tkij = Ai Bkj in this special case?
I 38.
ij
j
Let Aij
km denote the components of a mixed tensor of weight M . Form the contraction Bm = Aim
j
and determine how Bm
transforms. What is its weight?
I 39.
Let Aij denote the components of an absolute mixed tensor of order two. Show that the scalar
contraction S = Aii is an invariant.
63
I 40.
Let Ai = Ai (x1 , x2 , . . . , xN ) denote the components of an absolute contravariant tensor. Form the
quantity Bji =
∂Ai
∂xj
and determine if Bji transforms like a tensor.
∂Ai
∂Aj
−
are the
Let Ai denote the components of a covariant vector. (a) Show that aij =
j
∂x
∂xi
∂ajk
∂aki
∂aij
+
+
= 0.
components of a second order tensor. (b) Show that
∂xk
∂xi
∂xj
I 42. Show that xi = K eijk Aj Bk , with K 6= 0 and arbitrary, is a general solution of the system of equations
I 41.
Ai xi = 0, Bi xi = 0, i = 1, 2, 3. Give a geometric interpretation of this result in terms of vectors.
~ = yb
e2 + x b
e3 where b
e1 , b
e2 , b
e3 denote a set of unit basis vectors which
Given the vector A
e1 + z b
~2 = 4 b
~3 = b
~1 = 3 b
e1 + 4 b
e2 , E
e1 + 7 b
e2 and E
e3 denote a set of
define a set of orthogonal x, y, z axes. Let E
I 43.
basis vectors which define a set of u, v, w axes. (a) Find the coordinate transformation between these two
~ 3, E
~ 3 . (c) Calculate the covariant components of A.
~
~ 1, E
sets of axes. (b) Find a set of reciprocal vectors E
~
(d) Calculate the contravariant components of A.
I 44.
ei b
ej denote a dyadic. Show that
Let A = Aij b
A : Ac = A11 A11 + A12 A21 + A13 A31 + A21 A12 + A22 A22 + A23 A32 + A31 A13 + A32 A23 + A23 A33
I 45.
~ = Ai b
~ = Bi b
~ = Ci b
~ = Di b
~ B,
~ ψ =C
~D
~ denote
ei , B
ei , C
ei , D
ei denote vectors and let φ = A
Let A
dyadics which are the outer products involving the above vectors. Show that the double dot product satisfies
~B
~ :C
~D
~ = (A
~ · C)(
~ B
~ · D)
~
φ:ψ=A
I 46.
Show that if aij is a symmetric tensor in one coordinate system, then it is symmetric in all coordinate
systems.
I 47.
Write the transformation laws for the given tensors. (a)
I 48.
Show that if Ai = Aj
and unbarred systems.
Akij
(b)
Aij
k
(c)
Aijk
m
∂xj
∂xj
i , then Ai = Aj ∂xi . Note that this is equivalent to interchanging the bar
∂x
I 49.
(a) Show that under the linear homogeneous transformation
x1 =a11 x1 + a21 x2
x2 =a12 x1 + a22 x2
the quadratic form
Q(x1 , x2 ) = g11 (x1 )2 + 2g12 x1 x2 + g22 (x2 )2
becomes
Q(x1 , x2 ) = g11 (x1 )2 + 2g12 x1 x2 + g 22 (x2 )2
where g ij = g11 aj1 ai1 + g12 (ai1 aj2 + aj1 ai2 ) + g22 ai2 aj2 .
(b) Show F = g11 g22 − (g12 )2 is a relative invariant of weight 2 of the quadratic form Q(x1 , x2 ) with respect
to the group of linear homogeneous transformations. i.e. Show that F = ∆2 F where F = g 11 g22 −(g12 )2
and ∆ = (a11 a22 − a21 a12 ).
64
I 50.
Let ai and bi for i = 1, . . . , n denote arbitrary vectors and form the dyadic
Φ = a1 b1 + a2 b2 + · · · + an bn .
By definition the first scalar invariant of Φ is
φ1 = a1 · b1 + a2 · b2 + · · · + an · bn
where a dot product operator has been placed between the vectors. The first vector invariant of Φ is defined
~ = a1 × b1 + a2 × b2 + · · · + an × bn
φ
where a vector cross product operator has been placed between the vectors.
(a) Show that the first scalar and vector invariant of
e2 + b
e3 + b
e3
e2 b
e3 b
Φ= b
e1 b
are respectively 1 and b
e1 + b
e3 .
e1 + f2 b
e2 + f3 b
e3 one can form the dyadic ∇f having the matrix components
(b) From the vector f = f1 b
 ∂f1 ∂f2 ∂f3 
∇f = 
∂x
∂f1
∂y
∂f1
∂z
∂x
∂f2
∂y
∂f2
∂z
∂x
∂f3
∂y
∂f3
∂z
.
Show that this dyadic has the first scalar and vector invariants given by
∂f2
∂f3
∂f1
+
+
∇·f =
∂x
∂y
∂z
∂f1
∂f2
∂f3
∂f2
∂f3
∂f1
b
b
b
−
e1 +
−
e2 +
−
e3
∇×f =
∂y
∂z
∂z
∂x
∂x
∂y
I 51.
Let Φ denote the dyadic given in problem 50. The dyadic Φ2 defined by
1X
ai × aj bi × bj
Φ2 =
2 i,j
is called the Gibbs second dyadic of Φ, where the summation is taken over all permutations of i and j. When
i = j the dyad vanishes. Note that the permutations i, j and j, i give the same dyad and so occurs twice
in the final sum. The factor 1/2 removes this doubling. Associated with the Gibbs dyad Φ2 are the scalar
invariants
1X
(ai × aj ) · (bi × bj )
2 i,j
1X
(ai × aj · ak )(bi × bj · bk )
φ3 =
6
φ2 =
i,j,k
Show that the dyad
Φ = as + tq + cu
has
the first scalar invariant φ1 = a · s + b · t + c · u
~ = a×s+b×t+c×u
the first vector invariant φ
Gibbs second dyad
Φ2 = b × ct × u + c × au × s + a × bs × t
second scalar of Φ φ2 = (b × c) · (t · u) + (c × a) · (u × s) + (a × b) · (s × t)
third scalar of Φ φ3 = (a × b · c)(s × t · u)
65
I 52. (Spherical Trigonometry) Construct a spherical triangle ABC on the surface of a unit sphere with
sides and angles less than 180 degrees. Denote by a,b c the unit vectors from the origin of the sphere to the
vertices A,B and C. Make the construction such that a · (b × c) is positive with a, b, c forming a right-handed
system. Let α, β, γ denote the angles between these unit vectors such that
a · b = cos γ
c · a = cos β
b · c = cos α.
(1)
The great circles through the vertices A,B,C then make up the sides of the spherical triangle where side α
is opposite vertex A, side β is opposite vertex B and side γ is opposite the vertex C. The angles A,B and C
between the various planes formed by the vectors a, b and c are called the interior dihedral angles of the
spherical triangle. Note that the cross products
a × b = sin γ c
b × c = sin α a
c × a = sin β b
(2)
define unit vectors a, b and c perpendicular to the planes determined by the unit vectors a, b and c. The
dot products
a · b = cos γ
b · c = cos α
c · a = cos β
(3)
define the angles α,β and γ which are called the exterior dihedral angles at the vertices A,B and C and are
such that
α=π−A
β =π−B
γ = π − C.
(4)
(a) Using appropriate scaling, show that the vectors a, b, c and a, b, c form a reciprocal set.
(b) Show that a · (b × c) = sin α a · a = sin β b · b = sin γ c · c
(c) Show that a · (b × c) = sin α a · a = sin β b · b = sin γ c · c
(d) Using parts (b) and (c) show that
sin β
sin γ
sin α
=
=
sin α
sin γ
sin β
(e) Use the results from equation (4) to derive the law of sines for spherical triangles
sin β
sin γ
sin α
=
=
sin A
sin B
sin C
(f) Using the equations (2) show that
sin β sin γb · c = (c × a) · (a × b) = (c · a)(a · b) − b · c
and hence show that
cos α = cos β cos γ − sin β sin γ cos α.
In a similar manner show also that
cos α = cos β cos γ − sin β sin γ cos α.
(g) Using part (f) derive the law of cosines for spherical triangles
cos α = cos β cos γ + sin β sin γ cos A
cos A = − cos B cos C + sin B sin C cos α
A cyclic permutation of the symbols produces similar results involving the other angles and sides of the
spherical triangle.
65
§1.3 SPECIAL TENSORS
Knowing how tensors are defined and recognizing a tensor when it pops up in front of you are two
different things. Some quantities, which are tensors, frequently arise in applied problems and you should
learn to recognize these special tensors when they occur. In this section some important tensor quantities
are defined. We also consider how these special tensors can in turn be used to define other tensors.
Metric Tensor
Define y i , i = 1, . . . , N as independent coordinates in an N dimensional orthogonal Cartesian coordinate
system. The distance squared between two points y i
and y i + dy i ,
i = 1, . . . , N is defined by the
expression
ds2 = dy m dy m = (dy 1 )2 + (dy 2 )2 + · · · + (dy N )2 .
(1.3.1)
Assume that the coordinates y i are related to a set of independent generalized coordinates xi , i = 1, . . . , N
by a set of transformation equations
y i = y i (x1 , x2 , . . . , xN ),
i = 1, . . . , N.
(1.3.2)
To emphasize that each y i depends upon the x coordinates we sometimes use the notation y i = y i (x), for
i = 1, . . . , N. The differential of each coordinate can be written as
dy m =
∂y m j
dx ,
∂xj
m = 1, . . . , N,
(1.3.3)
and consequently in the x-generalized coordinates the distance squared, found from the equation (1.3.1),
becomes a quadratic form. Substituting equation (1.3.3) into equation (1.3.1) we find
ds2 =
∂y m ∂y m i j
dx dx = gij dxi dxj
∂xi ∂xj
where
gij =
∂y m ∂y m
,
∂xi ∂xj
i, j = 1, . . . , N
(1.3.4)
(1.3.5)
are called the metrices of the space defined by the coordinates xi , i = 1, . . . , N. Here the gij are functions of
the x coordinates and is sometimes written as gij = gij (x). Further, the metrices gij are symmetric in the
indices i and j so that gij = gji for all values of i and j over the range of the indices. If we transform to
another coordinate system, say xi , i = 1, . . . , N , then the element of arc length squared is expressed in terms
of the barred coordinates and ds2 = g ij dxi dxj , where gij = g ij (x) is a function of the barred coordinates.
The following example demonstrates that these metrices are second order covariant tensors.
66
EXAMPLE 1.3-1.
Show the metric components gij are covariant tensors of the second order.
Solution: In a coordinate system xi , i = 1, . . . , N the element of arc length squared is
ds2 = gij dxi dxj
(1.3.6)
while in a coordinate system xi , i = 1, . . . , N the element of arc length squared is represented in the form
ds2 = g mn dxm dxn .
(1.3.7)
The element of arc length squared is to be an invariant and so we require that
gmn dxm dxn = gij dxi dxj
(1.3.8)
Here it is assumed that there exists a coordinate transformation of the form defined by equation (1.2.30)
together with an inverse transformation, as in equation (1.2.32), which relates the barred and unbarred
coordinates. In general, if xi = xi (x), then for i = 1, . . . , N we have
dxi =
∂xi
dxm
∂xm
and dxj =
∂xj
dxn
∂xn
(1.3.9)
Substituting these differentials in equation (1.3.8) gives us the result
∂xi ∂xj
g mn dx dx = gij m n dxm dxn
∂x ∂x
m
n
or
∂xi ∂xj
g mn − gij m n dxm dxn = 0
∂x ∂x
For arbitrary changes in dxm this equation implies that g mn = gij
∂xi ∂xj
and consequently gij transforms
∂xm ∂xn
as a second order absolute covariant tensor.
EXAMPLE 1.3-2. (Curvilinear coordinates) Consider a set of general transformation equations from
rectangular coordinates (x, y, z) to curvilinear coordinates (u, v, w). These transformation equations and the
corresponding inverse transformations are represented
Here y 1 = x, y 2 = y, y 3 = z
x = x(u, v, w)
u = u(x, y, z)
y = y(u, v, w)
v = v(x, y, z)
z = z(u, v, w).
w = w(x, y, z)
(1.3.10)
and x1 = u, x2 = v, x3 = w are the Cartesian and generalized coordinates
and N = 3. The intersection of the coordinate surfaces u = c1 ,v = c2 and w = c3 define coordinate curves
of the curvilinear coordinate system. The substitution of the given transformation equations (1.3.10) into
e2 + z b
e3 produces the position vector which is a function of the generalized
the position vector ~r = x b
e1 + y b
coordinates and
e2 + z(u, v, w) b
e3
~r = ~r(u, v, w) = x(u, v, w) b
e1 + y(u, v, w) b
67
and consequently d~r =
∂~r
∂~r
∂~r
du +
dv +
dw, where
∂u
∂v
∂w
~ 1 = ∂~r =
E
∂u
~ 2 = ∂~r =
E
∂v
∂~
~3 = r =
E
∂w
∂x
∂y
∂z
b
b
b
e1 +
e2 +
e3
∂u
∂u
∂u
∂x
∂y
∂z
b
b
b
e1 +
e2 +
e3
∂v
∂v
∂v
∂x
∂y
∂z
b
b
b
e1 +
e2 +
e3 .
∂w
∂w
∂w
(1.3.11)
are tangent vectors to the coordinate curves. The element of arc length in the curvilinear coordinates is
∂~r ∂~r
∂~r ∂~r
∂~r ∂~r
·
dudu +
·
dudv +
·
dudw
∂u ∂u
∂u ∂v
∂u ∂w
∂~r ∂~r
∂~r ∂~r
∂~r ∂~r
·
dvdu +
·
dvdv +
·
dvdw
+
∂v ∂u
∂v ∂v
∂v ∂w
∂~r ∂~r
∂~r ∂~r
∂~r ∂~r
·
dwdu +
·
dwdv +
·
dwdw.
+
∂w ∂u
∂w ∂v
∂w ∂w
ds2 = d~r · d~r =
(1.3.12)
Utilizing the summation convention, the above can be expressed in the index notation. Define the
quantities
x2 = v,
∂~r ∂~r
·
∂u ∂v
∂~r ∂~r
·
=
∂v ∂v
∂~r ∂~r
·
=
∂w ∂v
g13 =
g21
g22
g23
g32
g33
x3 = w. Then the above element of arc length can be expressed as
~i · E
~ j dxi dxj = gij dxi dxj ,
ds2 = E
where
∂~r ∂~r
·
∂u ∂w
∂~r ∂~r
·
=
∂v ∂w
∂~r ∂~r
·
=
∂w ∂w
g12 =
g31
and let x1 = u,
∂~r ∂~r
·
∂u ∂u
∂~r ∂~r
·
=
∂v ∂u
∂~r ∂~r
·
=
∂w ∂u
g11 =
m
m
~i · E
~ j = ∂~r · ∂~r = ∂y ∂y ,
gij = E
∂xi ∂xj
∂xi ∂xj
i, j = 1, 2, 3
i, j free indices
(1.3.13)
are called the metric components of the curvilinear coordinate system. The metric components may be
thought of as the elements of a symmetric matrix, since gij = gji . In the rectangular coordinate system
x, y, z, the element of arc length squared is ds2 = dx2 + dy 2 + dz 2 . In this space the metric components are

1 0
gij =  0 1
0 0

0
0.
1
68
EXAMPLE 1.3-3. (Cylindrical coordinates (r, θ, z))
The transformation equations from rectangular coordinates to cylindrical coordinates can be expressed
as x = r cos θ,
y = r sin θ,
z = z. Here y 1 = x, y 2 = y, y 3 = z
and x1 = r, x2 = θ, x3 = z, and the
e2 + z b
e3 . The derivatives of this position
position vector can be expressed ~r = ~r(r, θ, z) = r cos θ b
e1 + r sin θ b
vector are calculated and we find
~ 1 = ∂~r = cos θ b
e1 + sin θ b
e2 ,
E
∂r
~ 2 = ∂~r = −r sin θ b
e1 + r cos θ b
E
e2 ,
∂θ
~ 3 = ∂~r = b
e3 .
E
∂z
From the results in equation (1.3.13), the metric components of this space are

1 0
gij =  0 r2
0 0

0
0.
1
We note that since gij = 0 when i 6= j, the coordinate system is orthogonal.
Given a set of transformations of the form found in equation (1.3.10), one can readily determine the
metric components associated with the generalized coordinates. For future reference we list several different coordinate systems together with their metric components. Each of the listed coordinate systems are
orthogonal and so gij = 0 for i 6= j. The metric components of these orthogonal systems have the form

h21

0
gij =
0
0
h22
0

0
0 
h23
and the element of arc length squared is
ds2 = h21 (dx1 )2 + h22 (dx2 )2 + h23 (dx3 )2 .
1. Cartesian coordinates (x, y, z)
x=x
h1 = 1
y=y
h2 = 1
z=z
h3 = 1
The coordinate curves are formed by the intersection of the coordinate surfaces
x =Constant, y =Constant and z =Constant.
69
Figure 1.3-1. Cylindrical coordinates.
2. Cylindrical coordinates (r, θ, z)
x = r cos θ
r≥0
h1 = 1
y = r sin θ
0 ≤ θ ≤ 2π
h2 = r
z=z
−∞<z <∞
h3 = 1
The coordinate curves, illustrated in the figure 1.3-1, are formed by the intersection of the coordinate
surfaces
x2 + y 2 = r2 ,
y/x = tan θ
Cylinders
Planes
z = Constant
Planes.
3. Spherical coordinates (ρ, θ, φ)
x = ρ sin θ cos φ
ρ≥0
h1 = 1
y = ρ sin θ sin φ
0≤θ≤π
h2 = ρ
z = ρ cos θ
0 ≤ φ ≤ 2π
h3 = ρ sin θ
The coordinate curves, illustrated in the figure 1.3-2, are formed by the intersection of the coordinate
surfaces
x2 + y 2 + z 2 = ρ2
Spheres
x2 + y 2 = tan2 θ z 2
Cones
y = x tan φ Planes.
4. Parabolic cylindrical coordinates (ξ, η, z)
x = ξη
1
y = (ξ 2 − η 2 )
2
z=z
−∞<z <∞
p
ξ 2 + η2
p
h2 = ξ 2 + η 2
η≥0
h3 = 1
−∞<ξ <∞
h1 =
70
Figure 1.3-2. Spherical coordinates.
The coordinate curves, illustrated in the figure 1.3-3, are formed by the intersection of the coordinate
surfaces
x2 = −2ξ 2 (y −
ξ2
)
2
Parabolic cylinders
η2
)
Parabolic cylinders
2
z = Constant
Planes.
x2 = 2η 2 (y +
Figure 1.3-3. Parabolic cylindrical coordinates in plane z = 0.
5. Parabolic coordinates (ξ, η, φ)
x = ξη cos φ
ξ≥0
y = ξη sin φ
1
z = (ξ 2 − η 2 )
2
η≥0
p
ξ 2 + η2
p
h2 = ξ 2 + η 2
0 < φ < 2π
h3 = ξη
h1 =
71
The coordinate curves, illustrated in the figure 1.3-4, are formed by the intersection of the coordinate
surfaces
x2 + y 2 = −2ξ 2 (z −
ξ2
)
2
Paraboloids
η2
)
Paraboloids
2
y = x tan φ
Planes.
x2 + y 2 = 2η 2 (z +
Figure 1.3-4. Parabolic coordinates, φ = π/4.
6. Elliptic cylindrical coordinates (ξ, η, z)
x = cosh ξ cos η
ξ≥0
y = sinh ξ sin η
0 ≤ η ≤ 2π
q
h1 = sinh2 ξ + sin2 η
q
h2 = sinh2 ξ + sin2 η
z=z
−∞<z <∞
h3 = 1
The coordinate curves, illustrated in the figure 1.3-5, are formed by the intersection of the coordinate
surfaces
y2
x2
+
=1
cosh2 ξ
sinh2 ξ
y2
x2
−
=1
cos2 η sin2 η
Elliptic cylinders
Hyperbolic cylinders
z = Constant
Planes.
72
Figure 1.3-5. Elliptic cylindrical coordinates in the plane z = 0.
7. Elliptic coordinates (ξ, η, φ)
s
p
x = (1 − η 2 )(ξ 2 − 1) cos φ
p
y = (1 − η 2 )(ξ 2 − 1) sin φ
1≤ξ<∞
−1≤η ≤1
0 ≤ φ < 2π
z = ξη
h1 =
s
h2 =
h3 =
ξ 2 − η2
ξ2 − 1
ξ 2 − η2
1 − η2
p
(1 − η 2 )(ξ 2 − 1)
The coordinate curves, illustrated in the figure 1.3-6, are formed by the intersection of the coordinate
surfaces
y2
z2
x2
+
+
=1
Prolate ellipsoid
ξ2 − 1 ξ2 − 1 ξ2
x2
y2
z2
−
−
=1
Two-sheeted hyperboloid
η2
1 − η2
1 − η2
y = x tan φ Planes
8. Bipolar coordinates (u, v, z)
a sinh v
,
0 ≤ u < 2π
cosh v − cos u
a sin u
,
−∞ < v < ∞
y=
cosh v − cos u
z=z
−∞<z <∞
x=
h21 = h22
h22 =
a2
(cosh v − cos u)2
h23 = 1
73
Figure 1.3-6. Elliptic coordinates φ = π/4.
Figure 1.3-7. Bipolar coordinates.
The coordinate curves, illustrated in the figure 1.3-7, are formed by the intersection of the coordinate
surfaces
a2
Cylinders
sinh2 v
a2
x2 + (y − a cot u)2 =
Cylinders
sin2 u
z = Constant
Planes.
(x − a coth v)2 + y 2 =
74
9. Conical coordinates (u, v, w)
uvw
,
b 2 > v 2 > a2 > w 2 ,
ab
r
u (v 2 − a2 )(w2 − a2 )
y=
a
a2 − b 2
r
u (v 2 − b2 )(w2 − b2 )
z=
b
b 2 − a2
x=
u≥0
h21 = 1
u2 (v 2 − w2 )
− a2 )(b2 − v 2 )
u2 (v 2 − w2 )
h23 =
2
(w − a2 )(w2 − b2 )
h22 =
(v 2
The coordinate curves, illustrated in the figure 1.3-8, are formed by the intersection of the coordinate
surfaces
x2 + y 2 + z 2 = u2
2
2
Spheres
2
y
z
x
+ 2
+ 2
=0,
Cones
v2
v − a2
v − b2
y2
z2
x2
+
+
= 0,
Cones.
w2
w 2 − a2
w 2 − b2
Figure 1.3-8. Conical coordinates.
10. Prolate spheroidal coordinates (u, v, φ)
x = a sinh u sin v cos φ,
u≥0
h21 = h22
y = a sinh u sin v sin φ,
0≤v≤π
h22 = a2 (sinh2 u + sin2 v)
z = a cosh u cos v,
h23 = a2 sinh2 u sin2 v
0 ≤ φ < 2π
The coordinate curves, illustrated in the figure 1.3-9, are formed by the intersection of the coordinate
surfaces
y2
z2
x2
+
+
= 1,
2
2
(a sinh u)
(a sinh u)
(a cosh u)2
x2
y2
z2
−
−
= 1,
2
2
(a cos v)
(a sin v)
(a sin v)2
Prolate ellipsoids
Two-sheeted hyperboloid
y = x tan φ,
Planes.
75
Figure 1.3-9. Prolate spheroidal coordinates
11. Oblate spheroidal coordinates (ξ, η, φ)
ξ≥0
π
π
y = a cosh ξ cos η sin φ,
− ≤η≤
2
2
z = a sinh ξ sin η, 0 ≤ φ ≤ 2π
x = a cosh ξ cos η cos φ,
h21 = h22
h22 = a2 (sinh2 ξ + sin2 η)
h23 = a2 cosh2 ξ cos2 η
The coordinate curves, illustrated in the figure 1.3-10, are formed by the intersection of the coordinate
surfaces
y2
z2
x2
+
+
= 1,
2
2
(a cosh ξ)
(a cosh ξ)
(a sinh ξ)2
y2
z2
x2
+
−
= 1,
2
2
(a cos η)
(a cos η)
(a sin η)2
Oblate ellipsoids
One-sheet hyperboloids
y = x tan φ,
Planes.
12. Toroidal coordinates (u, v, φ)
a sinh v cos φ
,
cosh v − cos u
a sinh v sin φ
,
y=
cosh v − cos u
a sin u
,
z=
cosh v − cos u
x=
0 ≤ u < 2π
−∞ < v < ∞
0 ≤ φ < 2π
h21 = h22
h22 =
a2
(cosh v − cos u)2
h23 =
a2 sinh2 v
(cosh v − cos u)2
The coordinate curves, illustrated in the figure 1.3-11, are formed by the intersection of the coordinate
surfaces
a cos u 2
a2
,
=
x2 + y 2 + z −
sin u
sin2 u
2
p
cosh v
a2
,
x2 + y 2 − a
+ z2 =
sinh v
sinh2 v
y = x tan φ,
Spheres
Torus
planes
76
Figure 1.3-10. Oblate spheroidal coordinates
Figure 1.3-11. Toroidal coordinates
EXAMPLE 1.3-4. Show the Kronecker delta δji is a mixed second order tensor.
Solution: Assume we have a coordinate transformation xi = xi (x), i = 1, . . . , N of the form (1.2.30) and
i
possessing an inverse transformation of the form (1.2.32). Let δ j and δji denote the Kronecker delta in the
barred and unbarred system of coordinates. By definition the Kronecker delta is defined
i
δj
=
δji
=
0,
if
i 6= j
1,
if
i=j
.
77
Employing the chain rule we write
∂xm ∂xi
∂xm ∂xk i
∂xm
=
δ
n =
n
∂x
∂xi ∂x
∂xi ∂xn k
By hypothesis, the xi , i = 1, . . . , N are independent coordinates and therefore we have
(1.3.14)
∂xm
∂xn
m
= δ n and (1.3.14)
simplifies to
m
δ n = δki
∂xm ∂xk
.
∂xi ∂xn
Therefore, the Kronecker delta transforms as a mixed second order tensor.
Conjugate Metric Tensor
Let g denote the determinant of the matrix having the metric tensor gij , i, j = 1, . . . , N as its elements.
In our study of cofactor elements of a matrix we have shown that
cof (g1j )g1k + cof (g2j )g2k + . . . + cof (gN j )gN k = gδkj .
(1.3.15)
We can use this fact to find the elements in the inverse matrix associated with the matrix having the
components gij . The elements of this inverse matrix are
g ij =
1
cof (gij )
g
(1.3.16)
and are called the conjugate metric components. We examine the summation g ij gik and find:
g ij gik = g 1j g1k + g 2j g2k + . . . + g N j gN k
1
= [cof (g1j )g1k + cof (g2j )g2k + . . . + cof (gN j )gN k ]
g
1 h ji
gδk = δkj
=
g
The equation
g ij gik = δkj
(1.3.17)
is an example where we can use the quotient law to show g ij is a second order contravariant tensor. Because
of the symmetry of g ij and gij the equation (1.3.17) can be represented in other forms.
EXAMPLE 1.3-5. Let Ai and Ai denote respectively the covariant and contravariant components of a
~ Show these components are related by the equations
vector A.
Ai = gij Aj
k
jk
A = g Aj
where gij and g ij are the metric and conjugate metric components of the space.
(1.3.18)
(1.3.19)
78
Solution: We multiply the equation (1.3.18) by g im (inner product) and use equation (1.3.17) to simplify
the results. This produces the equation g im Ai = g im gij Aj = δjm Aj = Am . Changing indices produces the
result given in equation (1.3.19). Conversely, if we start with equation (1.3.19) and multiply by gkm (inner
j
Aj = Am which is another form of the equation (1.3.18) with
product) we obtain gkm Ak = gkm g jk Aj = δm
the indices changed.
Notice the consequences of what the equations (1.3.18) and (1.3.19) imply when we are in an orthogonal
Cartesian coordinate system where

1
gij =  0
0
In this special case, we have

0 0
1 0
0 1

and g ij
1
= 0
0

0 0
1 0.
0 1
A1 = g11 A1 + g12 A2 + g13 A3 = A1
A2 = g21 A1 + g22 A2 + g23 A3 = A2
A3 = g31 A1 + g32 A2 + g33 A3 = A3 .
These equations tell us that in a Cartesian coordinate system the contravariant and covariant components
are identically the same.
EXAMPLE 1.3-6. We have previously shown that if Ai is a covariant tensor of rank 1 its components in
a barred system of coordinates are
Ai = Aj
∂xj
.
∂xi
(1.3.20)
Solve for the Aj in terms of the Aj . (i.e. find the inverse transformation).
Solution: Multiply equation (1.3.20) by
∂xi
∂xm
Ai
(inner product) and obtain
∂xi
∂xj ∂xi
=
A
.
j
∂xm
∂xi ∂xm
(1.3.21)
∂xj
∂xj ∂xi
j
=
= δm
since xj and xm are assumed to be independent
∂xm
∂xi ∂xm
coordinates. This reduces equation (1.3.21) to the form
In the above product we have
Ai
∂xi
j
= Aj δm
= Am
∂xm
(1.3.22)
which is the desired inverse transformation.
This result can be obtained in another way. Examine the transformation equation (1.3.20) and ask the
question, “When we have two coordinate systems, say a barred and an unbarred system, does it matter which
system we call the barred system?” With some thought it should be obvious that it doesn’t matter which
system you label as the barred system. Therefore, we can interchange the barred and unbarred symbols in
∂xj
equation (1.3.20) and obtain the result Ai = Aj i which is the same form as equation (1.3.22), but with
∂x
a different set of indices.
79
Associated Tensors
Associated tensors can be constructed by taking the inner product of known tensors with either the
metric or conjugate metric tensor.
Definition: (Associated tensor) Any tensor constructed by multiplying (inner
product) a given tensor with the metric or conjugate metric tensor is called an
associated tensor.
Associated tensors are different ways of representing a tensor. The multiplication of a tensor by the
metric or conjugate metric tensor has the effect of lowering or raising indices. For example the covariant
and contravariant components of a vector are different representations of the same vector in different forms.
These forms are associated with one another by way of the metric and conjugate metric tensor and
g ij Ai = Aj
gij Aj = Ai .
EXAMPLE 1.3-7. The following are some examples of associated tensors.
Aj = g ij Ai
mi
Am
Aijk
.jk = g
mk nj
A.nm
g Aijk
i.. = g
Aj = gij Ai
ijk
Ai.k
m = gmj A
Amjk = gim Ai.jk
Sometimes ‘dots’are used as indices in order to represent the location of the index that was raised or lowered.
If a tensor is symmetric, the position of the index is immaterial and so a dot is not needed. For example, if
Amn is a symmetric tensor, then it is easy to show that An.m and A.n
m are equal and therefore can be written
as Anm without confusion.
Higher order tensors are similarly related. For example, if we find a fourth order covariant tensor Tijkm
we can then construct the fourth order contravariant tensor T pqrs from the relation
T pqrs = g pi g qj g rk g sm Tijkm .
This fourth order tensor can also be expressed as a mixed tensor. Some mixed tensors associated with
the given fourth order covariant tensor are:
p
= g pi Tijkm ,
T.jkm
pq
p
T..km
= g qj T.jkm
.
80
Riemann Space VN
A Riemannian space VN is said to exist if the element of arc length squared has the form
ds2 = gij dxi dxj
(1.3.23)
where the metrices gij = gij (x1 , x2 , . . . , xN ) are continuous functions of the coordinates and are different
from constants. In the special case gij = δij the Riemannian space VN reduces to a Euclidean space EN .
The element of arc length squared defined by equation (1.3.23) is called the Riemannian metric and any
geometry which results by using this metric is called a Riemannian geometry. A space VN is called flat if
it is possible to find a coordinate transformation where the element of arclength squared is ds2 = i (dxi )2
where each i is either +1 or −1. A space which is not flat is called curved.
Geometry in VN
~ i and B
~ j , then their dot product can be represented
~ = Ai E
~ = Bj E
Given two vectors A
~i · E
~ j = gij Ai B j = Aj B j = Ai Bi = g ij Aj Bi = |A||
~ B|
~ cos θ.
~·B
~ = Ai B j E
A
(1.3.24)
~ and B
~
Consequently, in an N dimensional Riemannian space VN the dot or inner product of two vectors A
is defined:
gij Ai B j = Aj B j = Ai Bi = g ij Aj Bi = AB cos θ.
(1.3.25)
In this definition A is the magnitude of the vector Ai , the quantity B is the magnitude of the vector Bi and
θ is the angle between the vectors when their origins are made to coincide. In the special case that θ = 90◦
we have gij Ai B j = 0 as the condition that must be satisfied in order that the given vectors Ai and B i are
orthogonal to one another. Consider also the special case of equation (1.3.25) when Ai = B i and θ = 0. In
this case the equations (1.3.25) inform us that
g in An Ai = Ai Ai = gin Ai An = (A)2 .
(1.3.26)
From this equation one can determine the magnitude of the vector Ai . The magnitudes A and B can be
1
written A = (gin Ai An ) 2
1
and B = (gpq B p B q ) 2 and so we can express equation (1.3.24) in the form
cos θ =
gij Ai B j
(gmn
1
Am An ) 2 (g
pq B
p B q ) 12
.
(1.3.27)
An import application of the above concepts arises in the dynamics of rigid body motion. Note that if a
vector Ai has constant magnitude and the magnitude of
i
dA
dt
dAi
dt
is different from zero, then the vectors Ai and
j
must be orthogonal to one another due to the fact that gij Ai dA
dt = 0. As an example, consider the unit
e2 and b
e3 on a rotating system of Cartesian axes. We have for constants ci , i = 1, 6 that
vectors b
e1 , b
db
e1
= c1 b
e 2 + c2 b
e3
dt
db
e2
= c3 b
e 3 + c4 b
e1
dt
db
e3
= c5 b
e 1 + c6 b
e2
dt
because the derivative of any b
ei (i fixed) constant vector must lie in a plane containing the vectors b
ej and
b
ek , (j 6= i , k 6= i and j 6= k), since any vector in this plane must be perpendicular to b
ei .
81
The above definition of a dot product in VN can be used to define unit vectors in VN .
Definition: (Unit vector)
Whenever the magnitude of a vec-
i
tor A is unity, the vector is called a unit vector. In this case we
have
gij Ai Aj = 1.
(1.3.28)
EXAMPLE 1.3-8. (Unit vectors)
In VN the element of arc length squared is expressed ds2 = gij dxi dxj which can be expressed in the
dxi
dxi dxj
. This equation states that the vector
, i = 1, . . . , N is a unit vector. One application
form 1 = gij
ds ds
ds
of this equation is to consider a particle moving along a curve in VN which is described by the parametric
equations xi = xi (t), for i = 1, . . . , N. The vector V i =
dxi
dt ,
i = 1, . . . , N represents a velocity vector of the
particle. By chain rule differentiation we have
Vi =
where V =
ds
dt
dxi ds
dxi
dxi
=
=V
,
dt
ds dt
ds
is the scalar speed of the particle and
dxi
ds
(1.3.29)
is a unit tangent vector to the curve. The equation
(1.3.29) shows that the velocity is directed along the tangent to the curve and has a magnitude V. That is
ds
dt
2
= (V )2 = gij V i V j .
EXAMPLE 1.3-9. (Curvilinear coordinates)
Find an expression for the cosine of the angles between the coordinate curves associated with the
transformation equations
x = x(u, v, w),
y = y(u, v, w),
z = z(u, v, w).
82
Figure 1.3-12. Angles between curvilinear coordinates.
Solution: Let y 1 = x, y 2 = y, y 3 = z and x1 = u, x2 = v, x3 = w denote the Cartesian and curvilinear
coordinates respectively. With reference to the figure 1.3-12 we can interpret the intersection of the surfaces
v = c2 and w = c3 as the curve ~r = ~r(u, c2 , c3 ) which is a function of the parameter u. By moving only along
∂~r
du and consequently
this curve we have d~r =
∂u
∂~r ∂~r
·
dudu = g11 (dx1 )2 ,
ds2 = d~r · d~r =
∂u ∂u
or
1=
This equation shows that the vector
be represented by tr(1) =
dx1
ds
r
√ 1 δ1
.
g11
=
d~r d~r
·
= g11
ds ds
√1
g11
dx1
ds
2
.
is a unit vector along this curve. This tangent vector can
The curve which is defined by the intersection of the surfaces u = c1 and w = c3 has the unit tangent
vector tr(2) =
r
√ 1 δ2
.
g22
Similarly, the curve which is defined as the intersection of the surfaces u = c1 and
v = c2 has the unit tangent vector tr(3) =
r
√ 1 δ3
.
g33
The cosine of the angle θ12 , which is the angle between the
unit vectors tr(1) and tr(2) , is obtained from the result of equation (1.3.25). We find
g12
1
1
cos θ12 = gpq tp(1) tq(2) = gpq √ δ1p √ δ2q = √ √ .
g11
g22
g11 g22
For θ13 the angle between the directions ti(1) and ti(3) we find
g13
cos θ13 = √ √ .
g11 g33
Finally, for θ23 the angle between the directions ti(2) and ti(3) we find
g23
cos θ23 = √ √ .
g22 g33
When θ13 = θ12 = θ23 = 90◦ , we have g12 = g13 = g23 = 0 and the coordinate curves which make up the
curvilinear coordinate system are orthogonal to one another.
In an orthogonal coordinate system we adopt the notation
g11 = (h1 )2 ,
g22 = (h2 )2 ,
g33 = (h3 )2
and
gij = 0, i 6= j.
83
Epsilon Permutation Symbol
Associated with the e−permutation symbols there are the epsilon permutation symbols defined by the
relations
ijk =
√
geijk
1
ijk = √ eijk
g
and
(1.3.30)
where g is the determinant of the metrices gij .
It can be demonstrated that the eijk permutation symbol is a relative tensor of weight −1 whereas the
ijk permutation symbol is an absolute tensor. Similarly, the eijk permutation symbol is a relative tensor of
weight +1 and the corresponding ijk permutation symbol is an absolute tensor.
EXAMPLE 1.3-10. ( permutation symbol)
Show that eijk is a relative tensor of weight −1 and the corresponding ijk permutation symbol is an
absolute tensor.
Solution: Examine the Jacobian
J
x
x
and make the substitution
aij =
=
∂x1
∂x1
∂x2
∂x1
∂x3
∂x1
∂x1
∂x2
∂x2
∂x2
∂x3
∂x2
∂x1
∂x3
∂x2
∂x3
∂x3
∂x3
∂xi
, i, j = 1, 2, 3.
∂xj
From the definition of a determinant we may write
x
eijk aim ajn akp = J( )emnp .
x
(1.3.31)
By definition, emnp = emnp in all coordinate systems and hence equation (1.3.31) can be expressed in the
form
h x i−1
∂xi ∂xj ∂xk
eijk m n p = emnp
J( )
x
∂x ∂x ∂x
(1.3.32)
which demonstrates that eijk transforms as a relative tensor of weight −1.
We have previously shown the metric tensor gij is a second order covariant tensor and transforms
∂xm ∂xn
. Taking the determinant of this result we find
according to the rule gij = gmn
∂xi ∂xj
g = |gij | = |gmn |
∂xm
∂xi
2
h x i2
= g J( )
x
(1.3.33)
where g is the determinant of (gij ) and g is the determinant of (g ij ). This result demonstrates that g is a
scalar invariant of weight +2. Taking the square root of this result we find that
p
g=
x
√
gJ( ).
x
(1.3.34)
√
Consequently, we call g a scalar invariant of weight +1. Now multiply both sides of equation (1.3.32) by
√
g and use (1.3.34) to verify the relation
p
∂xi ∂xj ∂xk
√
g eijk m n p = g emnp .
∂x ∂x ∂x
√
This equation demonstrates that the quantity ijk = g eijk transforms like an absolute tensor.
(1.3.35)
84
Figure 1.3-14. Translation followed by rotation of axes
In a similar manner one can show eijk is a relative tensor of weight +1 and ijk =
√1 eijk
g
is an absolute
tensor. This is left as an exercise.
Another exercise found at the end of this section is to show that a generalization of the e − δ identity
is the epsilon identity
g ij ipt jrs = gpr gts − gps gtr .
(1.3.36)
Cartesian Tensors
Consider the motion of a rigid rod in two dimensions. No matter how complicated the movement of
the rod is we can describe the motion as a translation followed by a rotation. Consider the rigid rod AB
illustrated in the figure 1.3-13.
Figure 1.3-13. Motion of rigid rod
In this figure there is a before and after picture of the rod’s position. By moving the point B to B 0 we
have a translation. This is then followed by a rotation holding B fixed.
85
Figure 1.3-15. Rotation of axes
A similar situation exists in three dimensions. Consider two sets of Cartesian axes, say a barred and
unbarred system as illustrated in the figure 1.3-14. Let us translate the origin 0 to 0 and then rotate the
(x, y, z) axes until they coincide with the (x, y, z) axes. We consider first the rotation of axes when the
origins 0 and 0 coincide as the translational distance can be represented by a vector bk , k = 1, 2, 3. When
the origin 0 is translated to 0 we have the situation illustrated in the figure 1.3-15, where the barred axes
can be thought of as a transformation due to rotation.
Let
e2 + z b
e3
~r = x b
e1 + y b
(1.3.37)
denote the position vector of a variable point P with coordinates (x, y, z) with respect to the origin 0 and the
e2 , b
e3 . This same point, when referenced with respect to the origin 0 and the unit vectors
unit vectors b
e1 , b
ê1 , ê2 , ê3 , has the representation
~r = x ê1 + y ê2 + z ê3 .
(1.3.38)
By considering the projections of ~r upon the barred and unbarred axes we can construct the transformation
equations relating the barred and unbarred axes. We calculate the projections of ~r onto the x, y and z axes
and find:
e1 ) + y( ê2 · b
e1 ) + z( ê3 · b
e1 )
~r · b
e1 = x = x( ê1 · b
e2 ) + y( ê2 · b
e2 ) + z( ê3 · b
e2 )
~r · b
e2 = y = x( ê1 · b
(1.3.39)
e3 ) + y( ê2 · b
e3 ) + z( ê3 · b
e3 ).
~r · b
e3 = z = x( ê1 · b
We also calculate the projection of ~r onto the x, y, z axes and find:
e1 · ê1 ) + y( b
e2 · ê1 ) + z( b
e3 · ê1 )
~r · ê1 = x = x( b
e1 · ê2 ) + y( b
e2 · ê2 ) + z( b
e3 · ê2 )
~r · ê2 = y = x( b
(1.3.40)
e1 · ê3 ) + y( b
e2 · ê3 ) + z( b
e3 · ê3 ).
~r · ê3 = z = x( b
By introducing the notation (y1 , y2 , y3 ) = (x, y, z)
(y 1 , y2 , y3 ) = (x, y, z) and defining θij as the angle
between the unit vectors b
ei and êj , we can represent the above transformation equations in a more concise
86
form. We observe that the direction cosines can be written as
e1 · ê1 = cos θ11
`11 = b
`12 = b
e1 · ê2 = cos θ12
`13 = b
e1 · ê3 = cos θ13
`21 = b
e2 · ê1 = cos θ21
`22 = b
e2 · ê2 = cos θ22
`23 = b
e2 · ê3 = cos θ23
`31 = b
e3 · ê1 = cos θ31
`32 = b
e3 · ê2 = cos θ32
`33 = b
e3 · ê3 = cos θ33
(1.3.41)
which enables us to write the equations (1.3.39) and (1.3.40) in the form
yi = `ij y j
and
y i = `ji yj .
(1.3.42)
b
ep = `pr êr
(1.3.43)
Using the index notation we represent the unit vectors as:
ep
êr = `pr b
or
where `pr are the direction cosines. In both the barred and unbarred system the unit vectors are orthogonal
and consequently we must have the dot products
êr · êp = δrp
and
b
em · b
en = δmn
(1.3.44)
where δij is the Kronecker delta. Substituting equation (1.3.43) into equation (1.3.44) we find the direction
cosines `ij must satisfy the relations:
ep · `ms b
em = `pr `ms b
ep · b
êr · ês = `pr b
em = `pr `ms δpm = `mr `ms = δrs
and
b
er · b
es = `rm êm · `sn ên = `rm `sn êm · ên = `rm `sn δmn = `rm `sm = δrs .
The relations
`mr `ms = δrs
and
`rm `sm = δrs ,
(1.3.45)
with summation index m, are important relations which are satisfied by the direction cosines associated with
a rotation of axes.
Combining the rotation and translation equations we find
yi = `ij y j +
| {z }
rotation
bi
|{z}
.
(1.3.46)
translation
We multiply this equation by `ik and make use of the relations (1.3.45) to find the inverse transformation
yk = `ik (yi − bi ).
(1.3.47)
These transformations are called linear or affine transformations.
Consider the xi axes as fixed, while the xi axes are rotating with respect to the xi axes where both sets
~ = Ai b
ei denote a vector fixed in and rotating with the xi axes. We
of axes have a common origin. Let A
~
~
dA
dA
~ with respect to the fixed (f) and rotating (r) axes. We can
and
the derivatives of A
denote by
dt f
dt r
87
~
db
ei
dA
dAi
db
ei
b
ei + Ai
. Note that
is the derivative of a
=
dt f
dt
dt
dt
vector with constant magnitude. Therefore there exists constants ωi , i = 1, . . . , 6 such that
write, with respect to the fixed axes, that
db
e1
= ω3 b
e2 − ω 2 b
e3
dt
db
e2
= ω1 b
e3 − ω 4 b
e1
dt
db
e3
= ω5 b
e1 − ω 6 b
e2
dt
be2 + d be1 · b
i.e. see page 80. From the dot product b
e1 · b
e2 = 0 we obtain by differentiation b
e1 · ddt
e2 = 0
dt
e1 · b
e3 and b
e2 · b
e3 we obtain by differentiation the
which implies ω4 = ω3 . Similarly, from the dot products b
~
additional relations ω5 = ω2 and ω6 = ω1 . The derivative of A with respect to the fixed axes can now be
represented
~
dA
dt
=
f
~
dAi
dA
b
ei + (ω2 A3 − ω3 A2 ) b
e1 + (ω3 A1 − ω1 A3 ) b
e2 + (ω1 A2 − ω2 A1 ) b
e3 =
dt
dt
~
+ ~ω × A
r
~ represents the
ei is called an angular velocity vector of the rotating system. The term ~ω × A
where ~ω = ωi b
i
~
dA
dA
b
ei represents the derivative with
=
velocity of the rotating system relative to the fixed system and
dt r
dt
respect to the rotating system.
Employing the special transformation equations (1.3.46) let us examine how tensor quantities transform
when subjected to a translation and rotation of axes. These are our special transformation laws for Cartesian
tensors. We examine only the transformation laws for first and second order Cartesian tensor as higher order
transformation laws are easily discerned. We have previously shown that in general the first and second order
tensor quantities satisfy the transformation laws:
∂yj
∂y i
∂y
i
A = Aj i
∂yj
∂y ∂y
mn
A
= Aij m n
∂yi ∂yj
∂yi ∂yj
Amn = Aij
∂y m ∂yn
∂y ∂yj
m
An = Aij m
∂yi ∂y n
Ai = Aj
(1.3.48)
(1.3.49)
(1.3.50)
(1.3.51)
(1.3.52)
For the special case of Cartesian tensors we assume that yi and y i , i = 1, 2, 3 are linearly independent. We
differentiate the equations (1.3.46) and (1.3.47) and find
∂yj
∂yi
= `ij
= `ij δjk = `ik ,
∂yk
∂yk
and
∂y k
∂yi
= `ik
= `ik δim = `mk .
∂ym
∂ym
Substituting these derivatives into the transformation equations (1.3.48) through (1.3.52) we produce the
transformation equations
Ai = Aj `ji
i
A = Aj `ji
mn
A
= Aij `im `jn
Amn = Aij `im `jn
m
An = Aij `im `jn .
88
Figure 1.3-16. Transformation to curvilinear coordinates
These are the transformation laws when moving from one orthogonal system to another. In this case the
direction cosines `im are constants and satisfy the relations given in equation (1.3.45). The transformation
laws for higher ordered tensors are similar in nature to those given above.
In the unbarred system (y1 , y2 , y3 ) the metric tensor and conjugate metric tensor are:
gij = δij
and
g ij = δij
where δij is the Kronecker delta. In the barred system of coordinates, which is also orthogonal, we have
g ij =
∂ym ∂ym
.
∂y i ∂yj
From the orthogonality relations (1.3.45) we find
g ij = `mi `mj = δij
and
gij = δij .
We examine the associated tensors
Ai = g ij Aj
Aij = g im g jn Amn
Ain = g im Amn
Ai = gij Aj
Amn = gmi gnj Aij
Ain = gnj Aij
and find that the contravariant and covariant components are identical to one another. This holds also in
the barred system of coordinates. Also note that these special circumstances allow the representation of
contractions using subscript quantities only. This type of a contraction is not allowed for general tensors. It
is left as an exercise to try a contraction on a general tensor using only subscripts to see what happens. Note
that such a contraction does not produce a tensor. These special situations are considered in the exercises.
Physical Components
~ can be represented in many forms depending upon
We have previously shown an arbitrary vector A
the coordinate system and basis vectors selected. For example, consider the figure 1.3-16 which illustrates a
Cartesian coordinate system and a curvilinear coordinate system.
89
Figure 1.3-17. Physical components
~ as
In the Cartesian coordinate system we can represent a vector A
~ = Ax b
e1 + Ay b
e2 + Az b
e3
A
where ( b
e1 , b
e2 , b
e3 ) are the basis vectors. Consider a coordinate transformation to a more general coordinate
1
~ can be represented with contravariant components as
system, say (x , x2 , x3 ). The vector A
~ 1 + A2 E
~ 2 + A3 E
~3
~ = A1 E
A
(1.3.53)
~ 2, E
~ 3 ). Alternatively, the same vector A
~ can be represented
~ 1, E
with respect to the tangential basis vectors (E
in the form
~ 1 + A2 E
~ 2 + A3 E
~3
~ = A1 E
A
(1.3.54)
~ 2, E
~ 3 ). These equations are
~ 1, E
having covariant components with respect to the gradient basis vectors (E
just different ways of representing the same vector. In the above representations the basis vectors need not
be orthogonal and they need not be unit vectors. In general, the physical dimensions of the components Ai
and Aj are not the same.
~ in a direction is defined as the projection of A
~ upon a unit
The physical components of the vector A
~ in the direction E
~ 1 is
vector in the desired direction. For example, the physical component of A
~
~ on E
~ 1.
~ · E1 = A1 = projection of A
A
~ 1|
~ 1|
|E
|E
(1.3.58)
~ in the direction E
~ 1 is
Similarly, the physical component of A
1
~1
~ on E
~ 1.
~ · E = A = projection of A
A
~ 1|
~ 1|
|E
|E
(1.3.59)
EXAMPLE 1.3-11. (Physical components) Let α, β, γ denote nonzero positive constants such that the
product relation αγ = 1 is satisfied. Consider the nonorthogonal basis vectors
~1 = α b
e1 ,
E
illustrated in the figure 1.3-17.
~2 = β b
E
e1 + γ b
e2 ,
~3 = b
E
e3
90
It is readily verified that the reciprocal basis is
~1 = γ b
e1 − β b
e2 ,
E
~2 = αb
E
e2 ,
~3 = b
E
e3 .
~ = Ax b
e1 + Ay b
e2 in the contravariant vector form
Consider the problem of representing the vector A
~ 1 + A2 E
~2
~ = A1 E
A
or tensor form Ai , i = 1, 2.
This vector has the contravariant components
~ ·E
~ 1 = γAx − βAy
A1 = A
and
~ ·E
~ 2 = αAy .
A2 = A
Alternatively, this same vector can be represented as the covariant vector
~ 1 + A2 E
~2
~ = A1 E
A
which has the tensor form
Ai , i = 1, 2.
The covariant components are found from the relations
~ ·E
~ 1 = αAx
A1 = A
~ ·E
~ 2 = βAx + γAy .
A2 = A
~ 2 are found to be:
~ in the directions E
~ 1 and E
The physical components of A
1
~1
x − βAy
~ · E = A = γA
p
= A(1)
A
1
1
~
~
|E |
|E |
γ2 + β2
2
~2
~ · E = A = αAy = Ay = A(2).
A
~ 2|
~ 2|
α
|E
|E
~
Note that these same results are obtained from the dot product relations using either form of the vector A.
For example, we can write
and
~1 ~1
~2 ~1
~1
~ · E = A1 (E · E ) + A2 (E · E ) = A(1)
A
~ 1|
~ 1|
|E
|E
~2
~1 ~2
~2 ~2
~ · E = A1 (E · E ) + A2 (E · E ) = A(2).
A
~ 2|
~ 2|
|E
|E
~ in a direction of a unit vector λi is the generalized
In general, the physical components of a vector A
dot product in VN . This dot product is an invariant and can be expressed
~ in direction of λi
gij Ai λj = Ai λi = Ai λi = projection of A
91
Physical Components For Orthogonal Coordinates
In orthogonal coordinates observe the element of arc length squared in V3 is
ds2 = gij dxi dxj = (h1 )2 (dx1 )2 + (h2 )2 (dx2 )2 + (h3 )2 (dx3 )2

where
(h1 )2

0
gij =
0
0
(h2 )2
0

0
0 .
(h3 )2
(1.3.60)
In this case the curvilinear coordinates are orthogonal and
h2(i) = g(i)(i)
i not summed and
gij = 0, i 6= j.
At an arbitrary point in this coordinate system we take λi , i = 1, 2, 3 as a unit vector in the direction
of the coordinate x1 . We then obtain
λ1 =
dx1
,
ds
λ2 = 0,
λ3 = 0.
This is a unit vector since
1 = gij λi λj = g11 λ1 λ1 = h21 (λ1 )2
or λ1 =
1
h1 .
Here the curvilinear coordinate system is orthogonal and in this case the physical component
of a vector Ai , in the direction xi , is the projection of Ai on λi in V3 . The projection in the x1 direction is
determined from
A(1) = gij Ai λj = g11 A1 λ1 = h21 A1
1
= h1 A1 .
h1
Similarly, we choose unit vectors µi and ν i , i = 1, 2, 3 in the x2 and x3 directions. These unit vectors
can be represented
1
dx2
=
,
ds
h2
ν 2 =0,
µ1 =0,
µ2 =
ν 1 =0,
µ3 =0
ν3 =
1
dx3
=
ds
h3
and the physical components of the vector Ai in these directions are calculated as
A(2) = h2 A2
and
A(3) = h3 A3 .
In summary, we can say that in an orthogonal coordinate system the physical components of a contravariant
tensor of order one can be determined from the equations
A(i) = h(i) A(i) =
√
g(i)(i) A(i) ,
i = 1, 2 or 3 no summation on i,
which is a short hand notation for the physical components (h1 A1 , h2 A2 , h3 A3 ). In an orthogonal coordinate
system the nonzero conjugate metric components are
g (i)(i) =
1
,
g(i)(i)
i = 1, 2, or 3 no summation on i.
92
These components are needed to calculate the physical components associated with a covariant tensor of
order one. For example, in the x1 −direction, we have the covariant components
λ1 = g11 λ1 = h21
1
= h1 ,
h1
λ2 = 0,
λ3 = 0
and consequently the projection in V3 can be represented
gij Ai λj = gij Ai g jm λm = Aj g jm λm = A1 λ1 g 11 = A1 h1
A1
1
=
= A(1).
h21
h1
In a similar manner we calculate the relations
A(2) =
A2
h2
and
A(3) =
A3
h3
for the other physical components in the directions x2 and x3 . These physical components can be represented
in the short hand notation
A(i) =
A(i)
A(i)
=√
,
h(i)
g(i)(i)
i = 1, 2 or 3
no summation on i.
In an orthogonal coordinate system the physical components associated with both the contravariant and
covariant components are the same. To show this we note that when Ai gij = Aj is summed on i we obtain
A1 g1j + A2 g2j + A3 g3j = Aj .
Since gij = 0 for i 6= j this equation reduces to
A(i) g(i)(i) = A(i) ,
i not summed.
Another form for this equation is
A(i)
√
A(i) = A(i) g(i)(i) = √
g(i)(i)
i not summed,
which demonstrates that the physical components associated with the contravariant and covariant components are identical.
NOTATION The physical components are sometimes expressed by symbols with subscripts which represent
the coordinate curve along which the projection is taken. For example, let H i denote the contravariant
components of a first order tensor. The following are some examples of the representation of the physical
components of H i in various coordinate systems:
orthogonal
coordinate
tensor
physical
coordinates
system
components
components
general
(x1 , x2 , x3 )
Hi
H(1), H(2), H(3)
rectangular
cylindrical
spherical
general
(x, y, z)
(r, θ, z)
(ρ, θ, φ)
(u, v, w)
H
i
Hx , Hy , Hz
H
i
Hr , Hθ , Hz
H
i
Hρ , Hθ , Hφ
H
i
Hu , Hv , Hw
93
Higher Order Tensors
The physical components associated with higher ordered tensors are defined by projections in VN just
like the case with first order tensors. For an nth ordered tensor Tij...k we can select n unit vectors λi , µi , . . . , ν i
and form the inner product (projection)
Tij...k λi µj . . . ν k .
When projecting the tensor components onto the coordinate curves, there are N choices for each of the unit
vectors. This produces N n physical components.
The above inner product represents the physical component of the tensor Tij...k along the directions of
the unit vectors λi , µi , . . . , ν i . The selected unit vectors may or may not be orthogonal. In the cases where
the selected unit vectors are all orthogonal to one another, the calculation of the physical components is
greatly simplified. By relabeling the unit vectors λi(m) , λi(n) , . . . , λi(p) where (m), (n), ..., (p) represent one of
the N directions, the physical components of a general nth order tensor is represented
T (m n . . . p) = Tij...k λi(m) λj(n) . . . λk(p)
EXAMPLE 1.3-12. (Physical components)
In an orthogonal curvilinear coordinate system V3 with metric gij , i, j = 1, 2, 3, find the physical components of
(i) the second order tensor Aij . (ii) the second order tensor Aij . (iii) the second order tensor Aij .
Solution: The physical components of Amn , m, n = 1, 2, 3 along the directions of two unit vectors λi and
µi is defined as the inner product in V3 . These physical components can be expressed
n
A(ij) = Amn λm
(i) µ(j)
i, j = 1, 2, 3,
where the subscripts (i) and (j) represent one of the coordinate directions. Dropping the subscripts (i) and
(j), we make the observation that in an orthogonal curvilinear coordinate system there are three choices for
the direction of the unit vector λi and also three choices for the direction of the unit vector µi . These three
choices represent the directions along the x1 , x2 or x3 coordinate curves which emanate from a point of the
curvilinear coordinate system. This produces a total of nine possible physical components associated with
the tensor Amn .
For example, we can obtain the components of the unit vector λi , i = 1, 2, 3 in the x1 direction directly
from an examination of the element of arc length squared
ds2 = (h1 )2 (dx1 )2 + (h2 )2 (dx2 )2 + (h3 )2 (dx3 )2 .
By setting dx2 = dx3 = 0, we find
1
dx1
=
= λ1 ,
ds
h1
λ2 = 0,
λ3 = 0.
This is the vector λi(1) , i = 1, 2, 3. Similarly, if we choose to select the unit vector λi , i = 1, 2, 3 in the x2
direction, we set dx1 = dx3 = 0 in the element of arc length squared and find the components
λ1 = 0,
λ2 =
1
dx2
=
,
ds
h2
λ3 = 0.
94
This is the vector λi(2) , i = 1, 2, 3. Finally, if we select λi , i = 1, 2, 3 in the x3 direction, we set dx1 = dx2 = 0
in the element of arc length squared and determine the unit vector
λ1 = 0,
λ2 = 0,
λ3 =
1
dx3
=
.
ds
h3
This is the vector λi(3) , i = 1, 2, 3. Similarly, the unit vector µi can be selected as one of the above three
directions. Examining all nine possible combinations for selecting the unit vectors, we calculate the physical
components in an orthogonal coordinate system as:
A12
h1 h2
A22
A(22) =
h2 h2
A32
A(32) =
h3 h2
A11
h1 h1
A21
A(21) =
h1 h2
A31
A(31) =
h3 h1
A(12) =
A(11) =
A13
h1 h3
A23
A(23) =
h2 h3
A33
A(33) =
h3 h3
A(13) =
These results can be written in the more compact form
A(i)(j)
h(i) h(j)
no summation on i or j .
(1.3.61)
Aij = g im Amj = g i1 A1j + g i2 A2j + g i3 A3j .
(1.3.62)
A(ij) =
For mixed tensors we have
From the fact g ij = 0 for i 6= j, together with the physical components from equation (1.3.61), the equation
(1.3.62) reduces to
(i)
A(j) = g (i)(i) A(i)(j) =
1
· h(i) h(j) A(ij) no summation on i and i, j = 1, 2 or 3.
h2(i)
This can also be written in the form
(i)
A(ij) = A(j)
h(i)
h(j)
no summation on i or j.
(1.3.63)
Hence, the physical components associated with the mixed tensor Aij in an orthogonal coordinate system
can be expressed as
A(11) = A11
h2
h1
h3
A(31) = A31
h1
A(21) = A21
A(12) = A12
A(22) = A22
A(32) = A32
h1
h2
h3
h2
h1
h3
h2
A(23) = A23
h3
A(33) = A33 .
A(13) = A13
For second order contravariant tensors we may write
Aij gjm = Aim = Ai1 g1m + Ai2 g2m + Ai3 g3m .
95
We use the fact gij = 0 for i 6= j together with the physical components from equation (1.3.63) to reduce the
(i)
above equation to the form A(m) = A(i)(m) g(m)(m)
no summation on m . In terms of physical components
we have
h(m)
A(im) = A(i)(m) h2(m)
h(i)
or A(im) = A(i)(m) h(i) h(m) . no summation i, m = 1, 2, 3
(1.3.64)
Examining the results from equation (1.3.64) we find that the physical components associated with the
contravariant tensor Aij , in an orthogonal coordinate system, can be written as:
A(11) = A11 h1 h1
A(12) = A12 h1 h2
A(13) = A13 h1 h3
A(21) = A21 h2 h1
A(22) = A22 h2 h2
A(23) = A23 h2 h3
A(31) = A31 h3 h1
A(32) = A32 h3 h2
A(33) = A33 h3 h3 .
Physical Components in General
In an orthogonal curvilinear coordinate system, the physical components associated with the nth order
tensor Tij...kl along the curvilinear coordinate directions can be represented:
T (ij . . . kl) =
T(i)(j)...(k)(l)
h(i) h(j) . . . h(k) h(l)
no summations.
These physical components can be related to the various tensors associated with Tij...kl . For example, in
ij...m
can be
an orthogonal coordinate system, the physical components associated with the mixed tensor Tn...kl
expressed as:
(i)(j)...(m) h(i) h(j)
T (ij . . . m n . . . kl) = T(n)...(k)(l)
. . . h(m)
h(n) . . . h(k) h(l)
no summations.
(1.3.65)
EXAMPLE 1.3-13. (Physical components) Let xi = xi (t), i = 1, 2, 3 denote the position vector of a
particle which moves as a function of time t. Assume there exists a coordinate transformation xi = xi (x), for
i = 1, 2, 3, of the form given by equations (1.2.33). The position of the particle when referenced with respect
to the barred system of coordinates can be found by substitution. The generalized velocity of the particle
in the unbarred system is a vector with components
vi =
dxi
, i = 1, 2, 3.
dt
The generalized velocity components of the same particle in the barred system is obtained from the chain
rule. We find this velocity is represented by
vi =
∂xi dxj
∂xi j
dxi
=
=
v .
j
dt
∂x dt
∂xj
This equation implies that the contravariant quantities
(v 1 , v 2 , v 3 ) = (
dx1 dx2 dx3
,
,
)
dt dt dt
96
are tensor quantities. These quantities are called the components of the generalized velocity. The coordinates
x1 , x2 , x3 are generalized coordinates. This means we can select any set of three independent variables for
the representation of the motion. The variables selected might not have the same dimensions. For example,
in cylindrical coordinates we let (x1 = r, x2 = θ, x3 = z). Here x1 and x3 have dimensions of distance but x2
has dimensions of angular displacement. The generalized velocities are
v1 =
dr
dx1
=
,
dt
dt
v2 =
dθ
dx2
=
,
dt
dt
v3 =
dz
dx3
=
.
dt
dt
Here v 1 and v 3 have units of length divided by time while v 2 has the units of angular velocity or angular
change divided by time. Clearly, these dimensions are not all the same. Let us examine the physical
components of the generalized velocities. We find in cylindrical coordinates h1 = 1, h2 = r, h3 = 1 and the
physical components of the velocity have the forms:
vr = v(1) = v 1 h1 =
dr
,
dt
vθ = v(2) = v 2 h2 = r
dθ
,
dt
vz = v(3) = v 3 h3 =
dz
.
dt
Now the physical components of the velocity all have the same units of length divided by time.
Additional examples of the use of physical components are considered later. For the time being, just
remember that when tensor equations are derived, the equations are valid in any generalized coordinate
system. In particular, we are interested in the representation of physical laws which are to be invariant and
independent of the coordinate system used to represent these laws. Once a tensor equation is derived, we
can chose any type of generalized coordinates and expand the tensor equations. Before using any expanded
tensor equations we must replace all the tensor components by their corresponding physical components in
order that the equations are dimensionally homogeneous. It is these expanded equations, expressed in terms
of the physical components, which are used to solve applied problems.
Tensors and Multilinear Forms
Tensors can be thought of as being created by multilinear forms defined on some vector space V. Let
us define on a vector space V a linear form, a bilinear form and a general multilinear form. We can then
illustrate how tensors are created from these forms.
Definition: (Linear form)
Let V denote a vector space which
contains vectors ~x, ~x1 , ~x2 , . . . . A linear form in ~x is a scalar function
ϕ(~x) having a single vector argument ~x which satisfies the linearity
properties:
(i)
(ii)
ϕ(~x1 + ~x2 ) = ϕ(~x1 ) + ϕ(~x2 )
(1.3.66)
ϕ(µ~x1 ) = µϕ(~x1 )
for all arbitrary vectors ~x1 , ~x2 in V and all real numbers µ.
97
An example of a linear form is the dot product relation
~ · ~x
ϕ(~x) = A
(1.3.67)
~ is a constant vector and ~x is an arbitrary vector belonging to the vector space V.
where A
Note that a linear form in ~x can be expressed in terms of the components of the vector ~x and the base
e2 , b
e3 ) used to represent ~x. To show this, we write the vector ~x in the component form
vectors ( b
e1 , b
ei = x1 b
e1 + x2 b
e2 + x3 b
e3 ,
~x = xi b
e1 , b
e2 , b
e3 ). By the linearity
where xi , i = 1, 2, 3 are the components of ~x with respect to the basis vectors ( b
property of ϕ we can write
ei ) = ϕ(x1 b
e1 + x2 b
e2 + x3 b
e3 )
ϕ(~x) = ϕ(xi b
e1 ) + ϕ(x2 b
e2 ) + ϕ(x3 b
e3 )
= ϕ(x1 b
e1 ) + x2 ϕ( b
e2 ) + x3 ϕ( b
e3 ) = xi ϕ( b
ei )
= x1 ϕ( b
ei ) and by defining the quantity ϕ( b
ei ) = ai as a tensor we obtain ϕ(~x) = xi ai .
Thus we can write ϕ(~x) = xi ϕ( b
~ 1, E
~ 2, E
~ 3 ) then the components of ~x also must change.
e2 , b
e3 ) to (E
Note that if we change basis from ( b
e1 , b
Letting xi denote the components of ~x with respect to the new basis, we would have
~i
~x = xi E
~ i ) = xi ϕ(E
~ i ).
and ϕ(~x) = ϕ(xi E
~ i ) so that ϕ(~x) = xi ai . Whenever there is a definite relation
The linear form ϕ defines a new tensor ai = ϕ(E
~ 1, E
~ 2, E
~ 3 ), say,
e2 , b
e3 ) and (E
between the basis vectors ( b
e1 , b
j
~ i = ∂x b
ej ,
E
∂xi
then there exists a definite relation between the tensors ai and ai . This relation is
~ i ) = ϕ(
ai = ϕ(E
∂xj
∂xj
∂xj
b
b
aj .
i ej ) =
i ϕ( ej ) =
∂x
∂x
∂xi
This is the transformation law for an absolute covariant tensor of rank or order one.
The above idea is now extended to higher order tensors.
Definition: ( Bilinear form) A bilinear form in ~x and ~y is a
scalar function ϕ(~x, ~y) with two vector arguments, which satisfies
the linearity properties:
(i) ϕ(~x1 + ~x2 , ~y1 ) = ϕ(~x1 , ~y1 ) + ϕ(~x2 , ~y1 )
(ii) ϕ(~x1 , ~y1 + ~y2 ) = ϕ(~x1 , ~y1 ) + ϕ(~x1 , ~y2 )
(1.3.68)
(iii) ϕ(µ~x1 , ~y1 ) = µϕ(~x1 , ~y1 )
(iv) ϕ(~x1 , µ~y1 ) = µϕ(~x1 , ~y1 )
for arbitrary vectors ~x1 , ~x2 , ~y1 , ~y2 in the vector space V and for all
real numbers µ.
98
Note in the definition of a bilinear form that the scalar function ϕ is linear in both the arguments ~x and
~y . An example of a bilinear form is the dot product relation
ϕ(~x, ~y) = ~x · ~y
(1.3.69)
where both ~x and ~y belong to the same vector space V.
The definition of a bilinear form suggests how multilinear forms can be defined.
Definition: (Multilinear forms) A multilinear form of degree M or a M degree
linear form in the vector arguments
~x1 , ~x2 , . . . , ~xM
is a scalar function
ϕ(~x1 , ~x2 , . . . , ~xM )
of M vector arguments which satisfies the property that it is a linear form in each of its
arguments. That is, ϕ must satisfy for each j = 1, 2, . . . , M the properties:
(i) ϕ(~x1 , . . . , ~xj1 + ~xj2 , . . . ~xM ) = ϕ(~x1 , . . . , ~xj1 , . . . , ~xM ) + ϕ(~x1 , . . . , ~xj2 , . . . , ~xM )
(ii)
ϕ(~x1 , . . . , µ~xj , . . . , ~xM ) = µϕ(~x1 , . . . , ~xj , . . . , ~xM )
(1.3.70)
for all arbitrary vectors ~x1 , . . . , ~xM in the vector space V and all real numbers µ.
An example of a third degree multilinear form or trilinear form is the triple scalar product
ϕ(~x, ~y , ~z) = ~x · (~y × ~z).
(1.3.71)
Note that multilinear forms are independent of the coordinate system selected and depend only upon the
e2 , b
e3 ) and represent
vector arguments. In a three dimensional vector space we select the basis vectors ( b
e1 , b
all vectors with respect to this basis set. For example, if ~x, ~y , ~z are three vectors we can represent these
vectors in the component forms
ei ,
~x = xi b
ej ,
~y = y j b
ek
~z = z k b
(1.3.72)
where we have employed the summation convention on the repeated indices i, j and k. Substituting equations
(1.3.72) into equation (1.3.71) we obtain
ei , y j b
ej , z k b
ek ) = xi y j z k ϕ( b
ei , b
ej , b
ek ),
ϕ(xi b
(1.3.73)
since ϕ is linear in all its arguments. By defining the tensor quantity
ej , b
ek ) = eijk
ϕ( b
ei , b
(1.3.74)
99
(See exercise 1.1, problem 15) the trilinear form, given by equation (1.3.71), with vectors from equations
(1.3.72), can be expressed as
ϕ(~x, ~y , ~z) = eijk xi y j z k ,
i, j, k = 1, 2, 3.
(1.3.75)
The coefficients eijk of the trilinear form is called a third order tensor. It is the familiar permutation symbol
considered earlier.
In a multilinear form of degree M , ϕ(~x, ~y , . . . , ~z), the M arguments can be represented in a component
e2 , b
e3 ). Let these vectors have components xi , y i , z i , i = 1, 2, 3
form with respect to a set of basis vectors ( b
e1 , b
with respect to the selected basis vectors. We then can write
ei ,
~x = xi b
ej ,
~y = y j b
ek .
~z = z k b
Substituting these vectors into the M degree multilinear form produces
ei , y j b
ej , . . . , z k b
ek ) = xi y j · · · z k ϕ( b
ei , b
ej , . . . , b
ek ).
ϕ(xi b
(1.3.76)
Consequently, the multilinear form defines a set of coefficients
ei , b
ej , . . . , b
ek )
aij...k = ϕ( b
(1.3.77)
which are referred to as the components of a tensor of order M. The tensor is thus created by the multilinear
form and has M indices if ϕ is of degree M.
~ 2, E
~ 3 ) the multilinear form defines
~ 1, E
Note that if we change to a different set of basis vectors, say, (E
a new tensor
~i, E
~j, . . . , E
~ k ).
aij...k = ϕ(E
(1.3.78)
This new tensor has a bar over it to distinguish it from the previous tensor. A definite relation exists between
the new and old basis vectors and consequently there exists a definite relation between the components of
the barred and unbarred tensors components. Recall that if we are given a set of transformation equations
y i = y i (x1 , x2 , x3 ), i = 1, 2, 3,
(1.3.79)
from rectangular to generalized curvilinear coordinates, we can express the basis vectors in the new system
by the equations
j
~ i = ∂y b
ej , i = 1, 2, 3.
(1.3.80)
E
∂xi
For example, see equations (1.3.11) with y 1 = x, y 2 = y, y 3 = z, x1 = u, x2 = v, x3 = w. Substituting
equations (1.3.80) into equations (1.3.78) we obtain
∂y α
∂y β
∂y γ
b
eα , j b
eβ , . . . , k b
eγ ).
i
∂x
∂x
∂x
By the linearity property of ϕ, this equation is expressible in the form
aij...k = ϕ(
∂y α ∂y β
∂y γ
. . . k ϕ( b
eα , b
eβ , . . . , b
eγ )
i
j
∂x ∂x
∂x
∂y α ∂y β
∂y γ
aij...k =
.
.
.
aαβ...γ
∂xi ∂xj
∂xk
This is the familiar transformation law for a covariant tensor of degree M. By selecting reciprocal basis
aij...k =
vectors the corresponding transformation laws for contravariant vectors can be determined.
The above examples illustrate that tensors can be considered as quantities derivable from multilinear
forms defined on some vector space.
100
Dual Tensors
The e-permutation symbol is often used to generate new tensors from given tensors. For Ti1 i2 ...im a
skew-symmetric tensor, we define the tensor
T̂ j1 j2 ...jn−m =
1 j1 j2 ...jn−m i1 i2 ...im
e
Ti1 i2 ...im
m!
m≤n
(1.3.81)
as the dual tensor associated with Ti1 i2 ...im . Note that the e-permutation symbol or alternating tensor has
a weight of +1 and consequently the dual tensor will have a higher weight than the original tensor.
The e-permutation symbol has the following properties
ei1 i2 ...iN ei1 i2 ...iN = N !
N
ei1 i2 ...iN ej1 j2 ...jN = δji11 ij22...i
...jN
m
ek1 k2 ...km i1 i2 ...iN −m ej1 j2 ...jm i1 i2 ...iN −m = (N − m)!δkj11jk22...j
...km
(1.3.82)
m
δkj11jk22...j
...km Tj1 j2 ...jm = m!Tk1 k2 ...km .
Using the above properties we can solve for the skew-symmetric tensor in terms of the dual tensor. We find
Ti1 i2 ...im =
1
ei i ...i j j ...j
T̂ j1 j2 ...jn−m .
(n − m)! 1 2 m 1 2 n−m
(1.3.83)
For example, if Aij i, j = 1, 2, 3 is a skew-symmetric tensor, we may associate with it the dual tensor
Vi =
1 ijk
e Ajk ,
2!
which is a first order tensor or vector. Note that Aij has the components


0
A12 A13
 −A12
0
A23 
−A13 −A23
0
(1.3.84)
~ are
and consequently, the components of the vector V
(V 1 , V 2 , V 3 ) = (A23 , A31 , A12 ).
(1.3.85)
Note that the vector components have a cyclic order to the indices which comes from the cyclic properties
of the e-permutation symbol.
As another example, consider the fourth order skew-symmetric tensor Aijkl , i, j, k, l = 1, . . . , n. We can
associate with this tensor any of the dual tensor quantities
1 ijkl
e Aijkl
4!
1
V i = eijklm Ajklm
4!
1
V ij = eijklmn Aklmn
4!
1
V ijk = eijklmnp Almnp
4!
1
V ijkl = eijklmnpr Amnpr
4!
V =
Applications of dual tensors can be found in section 2.2.
(1.3.86)
101
EXERCISE 1.3
I 1.
(a) From the transformation law for the second order tensor g ij = gab
solve for the gab in terms of gij .
∂xa ∂xb
∂xi ∂xj
(b) Show that if gij is symmetric in one coordinate system it is symmetric in all coordinate systems.
p
x
√
(c) Let g = det(gij ) and g = det(gij ) and show that g = gJ 2 ( xx ) and consequently g = gJ( ). This
x
√
shows that g is a scalar invariant of weight 2 and g is a scalar invariant of weight 1.
I 2.
For
gij =
I 3.
∂y m ∂y m
∂xi ∂xj
show that g ij =
∂xi ∂xj
∂y m ∂y m
Show that in a curvilinear coordinate system which is orthogonal we have:
(a)
g = det(gij ) = g11 g22 g33
(b)
gmn = g mn = 0
for m 6= n
1
for N = 1, 2, 3 (no summation on N)
gN N =
gN N
(c)
∂y i
∂xj
2
= J 2 , where J is the Jacobian.
I 4.
Show that g = det(gij ) =
I 5.
Define the quantities h1 = hu = |
∂~r
|,
∂u
h2 = h v = |
∂~r
|,
∂v
h3 = h w = |
∂~r
| and construct the unit
∂w
vectors
1 ∂~r
1 ∂~r
1 ∂~r
b
b
,
ev =
,
ew =
.
h1 ∂u
h2 ∂v
h3 ∂w
(a) Assume the coordinate system is orthogonal and show that
b
eu =
g11 = h21 =
g22
g33
2
+
∂x
∂v
2
∂y
∂u
2
+
∂z
∂u
2
,
2 2
∂y
∂z
=
=
+
+
,
∂v
∂v
2 2 2
∂x
∂y
∂z
= h23 =
+
+
.
∂w
∂w
∂w
h22
∂x
∂u
eu du + h2 b
ev dv + h3 b
ew dw.
(b) Show that d~r can be expressed in the form d~r = h1 b
(c) Show that the volume of the elemental parallelepiped having d~r as diagonal can be represented
dτ =
∂(x, y, z)
√
dudvdw.
g dudvdw = J dudvdw =
∂(u, v, w)
Hint:
A1
~ · (B
~ × C)|
~ = B1
|A
C1
A2
B2
C2
A3
B3
C3
102
Figure 1.3-18 Oblique cylindrical coordinates.
I 6.
For the change d~r given inqproblem 5, show the elemental parallelepiped with diagonal d~r has:
2 dvdw in the u =constant surface.
(a) the element of area dS1 = g22 g33 − g23
(b) The element of area dS2 =
(c) the element of area dS3 =
q
2 dudw in the v =constant surface.
g33 g11 − g13
q
2 dudv in the w =constant surface.
g11 g22 − g12
(d) What do the above elements
q of area reduce to in the special case the curvilinear coordinates are orthog~ × B)
~ · (A
~ × B)
~
~ × B|
~ = (A
|A
.
onal? Hint:
q
~ · A)(
~ B
~ · B)
~ − (A
~ · B)(
~ A
~ · B)
~
= (A
I 7.
In Cartesian coordinates you are given the affine transformation. xi = `ij xj where
x1 =
1
(5x1 − 14x2 + 2x3 ),
15
1
x2 = − (2x1 + x2 + 2x3 ),
3
x3 =
1
(10x1 + 2x2 − 11x3 )
15
(a) Show the transformation is orthogonal.
~ 1 , x2 , x3 ) in the unbarred system has the components
(b) A vector A(x
A1 = (x1 )2 ,
A2 = (x2 )2
A3 = (x3 )2 .
Find the components of this vector in the barred system of coordinates.
I 8.
Calculate the metric and conjugate metric tensors in cylindrical coordinates (r, θ, z).
I 9.
Calculate the metric and conjugate metric tensors in spherical coordinates (ρ, θ, φ).
I 10.
Calculate the metric and conjugate metric tensors in parabolic cylindrical coordinates (ξ, η, z).
I 11.
Calculate the metric and conjugate metric components in elliptic cylindrical coordinates (ξ, η, z).
I 12.
Calculate the metric and conjugate metric components for the oblique cylindrical coordinates (r, φ, η),
illustrated in figure 1.3-18, where x = r cos φ,
0<α≤
π
2.
Note: When α =
π
2
y = r sin φ + η cos α,
cylindrical coordinates result.
z = η sin α and α is a parameter
103
I 13.
Calculate the metric and conjugate metric tensor associated with the toroidal surface coordinates
(ξ, η) illustrated in the figure 1.3-19, where
x = (a + b cos ξ) cos η
a>b>0
y = (a + b cos ξ) sin η
0 < ξ < 2π
z = b sin ξ
0 < η < 2π
Figure 1.3-19. Toroidal surface coordinates
I 14.
Calculate the metric and conjugate metric tensor associated with the spherical surface coordinates
(θ, φ), illustrated in the figure 1.3-20, where
x = a sin θ cos φ
a>0
y = a sin θ sin φ
0 < φ < 2π
π
0<θ<
2
z = a cos θ
is constant
I 15.
Consider gij , i, j = 1, 2
g22
−g12
g11
,
g 12 = g 21 =
,
g 22 =
where ∆ = g11 g22 − g12 g21 .
(a) Show that g 11 =
∆
∆
∆
ik
k
(b) Use the results in part (a) and verify that gij g = δj , i, j, k = 1, 2.
I 16.
Let Ax , Ay , Az denote the constant components of a vector in Cartesian coordinates. Using the
transformation laws (1.2.42) and (1.2.47) to find the contravariant and covariant components of this vector
upon changing to (a) cylindrical coordinates (r, θ, z). (b) spherical coordinates (ρ, θ, φ) and (c) Parabolic
cylindrical coordinates.
I 17.
Find the relationship which exists between the given associated tensors.
(a) Apqk
r.
(b) Ap.mrs
and Apq
rs
and Apq
..rs
(c)
Ai.j.
.l.m
and A.s.p
r.t.
(d)
Amnk
and Aij
..k
104
Figure 1.3-20. Spherical surface coordinates
I 18.
Given the fourth order tensor Cikmp = λδik δmp + µ(δim δkp + δip δkm ) + ν(δim δkp − δip δkm ) where λ, µ
and ν are scalars and δij is the Kronecker delta. Show that under an orthogonal transformation of rotation of
axes with xi = `ij xj
where `rs `is = `mr `mi = δri the components of the above tensor are unaltered. Any
tensor whose components are unaltered under an orthogonal transformation is called an ‘isotropic’ tensor.
Another way of stating this problem is to say “Show Cikmp is an isotropic tensor.”
I 19.
Assume Aijl is a third order covariant tensor and B pqmn is a fourth order contravariant tensor. Prove
that Aikl B klmn is a mixed tensor of order three, with one covariant and two contravariant indices.
I 20.
Assume that Tmnrs is an absolute tensor. Show that if Tijkl + Tijlk = 0 in the coordinate system xr
then T ijkl + T ijlk = 0 in any other coordinate system xr .
I 21.
Show that
gir
ijk rst = gjr
gkr
gis
gjs
gks
git
gjt
gkt
Hint: See problem 38, Exercise 1.1
I 22.
Determine if the tensor equation mnp mij + mnj mpi = mni mpj is true or false. Justify your answer.
I 23.
Prove the epsilon identity g ij ipt jrs = gpr gts − gps gtr . Hint: See problem 38, Exercise 1.1
I 24.
rmn
1
Let Ars denote a skew-symmetric contravariant tensor and let cr = rmn Amn where
2
√
= germn . Show that cr are the components of a covariant tensor. Write out all the components.
1 rmn
1
Amn where rmn = √ ermn .
2
g
Show that cr are the components of a contravariant tensor. Write out all the components.
I 25.
Let Ars denote a skew-symmetric covariant tensor and let cr =
105
I 26.
s
s
Let Apq Brqs = Cpr
where Brqs is a relative tensor of weight ω1 and Cpr
is a relative tensor of weight
ω2 . Prove that Apq is a relative tensor of weight (ω2 − ω1 ).
I 27.
When Aij is an absolute tensor prove that
√ i
gAj is a relative tensor of weight +1.
I 28.
When Aij is an absolute tensor prove that
√1 Ai
g j
is a relative tensor of weight −1.
I 29.
(a) Show eijk is a relative tensor of weight +1.
(b) Show ijk =
I 30.
√1 eijk
g
is an absolute tensor. Hint: See example 1.1-25.
The equation of a surface can be represented by an equation of the form Φ(x1 , x2 , x3 ) = constant.
Show that a unit normal vector to the surface can be represented by the vector
ni =
∂Φ
g ij ∂x
j
1
∂Φ ∂Φ 2
(g mn ∂x
m ∂xn )
.
I 31.
Assume that gij = λgij with λ a nonzero constant. Find and calculate gij in terms of g ij .
I 32.
Determine if the following tensor equation is true. Justify your answer.
rjk Ari + irk Arj + ijr Ark = ijk Arr .
Hint: See problem 21, Exercise 1.1.
I 33.
Show that for Ci and C i associated tensors, and C i = ijk Aj Bk , then Ci = ijk Aj B k
I 34.
Prove that ijk and ijk are associated tensors.
I 35.
Show ijk Ai Bj Ck = ijk Ai B j C k .
I 36.
Let Tji , i, j = 1, 2, 3 denote a second order mixed tensor. Show that the given quantities are scalar
invariants.
Hint: Consider the determinant of gij .
(i) I1 = Tii
1 i 2
i
(Ti ) − Tm
(ii) I2 =
Tim
2
(iii) I3 = det|Tji |
I 37.
(a) Assume Aij and B ij , i, j = 1, 2, 3 are absolute contravariant tensors, and determine if the inner product
C ik = Aij B jk is an absolute tensor?
∂xj ∂xj
= δnm is satisfied, and determine whether the inner product in
(b) Assume that the condition
∂xn ∂xm
part (a) is a tensor?
(c) Consider only transformations which are a rotation and translation of axes y i = `ij yj + bi , where `ij are
∂y j ∂yj
= δnm
direction cosines for the rotation of axes. Show that
∂yn ∂ym
106
I 38.
For Aijk a Cartesian tensor, determine if a contraction on the indices i and j is allowed. That
is, determine if the quantity Ak = Aiik ,
(summation on i) is a tensor. Hint: See part(c) of the previous
problem.
I 39.
I 40.
j k
k
δn − δnj δm
.
Prove the e-δ identity eijk eimn = δm
Consider the vector Vk , k = 1, 2, 3 and define the matrix (aij ) having the elements aij = eijk Vk ,
where eijk is the e−permutation symbol.
(a) Solve for Vi in terms of amn by multiplying both sides of the given equation by eijl and note the e − δ
identity allows us to simplify the result.
(b) Sum the given expression on k and then assign values to the free indices (i,j=1,2,3) and compare your
results with part (a).
(c) Is aij symmetric, skew-symmetric, or neither?
I 41.
It can be shown that the continuity equation of fluid dynamics can be expressed in the tensor form
∂%
1 ∂ √
= 0,
( g%V r ) +
√
r
g ∂x
∂t
where % is the density of the fluid, t is time, V r , with r = 1, 2, 3 are the velocity components and g = |gij |
is the determinant of the metric tensor. Employing the summation convention and replacing the tensor
components of velocity by their physical components, express the continuity equation in
(a) Cartesian coordinates (x, y, z) with physical components Vx , Vy , Vz .
(b) Cylindrical coordinates (r, θ, z) with physical components Vr , Vθ , Vz .
(c) Spherical coordinates (ρ, θ, φ) with physical components Vρ , Vθ , Vφ .
Let x1 , x2 , x3 denote a set of skewed coordinates with respect to the Cartesian coordinates y 1 , y 2 , y 3 .
~ 2, E
~ 3 are unit vectors in the directions of the x1 , x2 and x3 axes respectively. If the unit
~ 1, E
Assume that E
I 42.
vectors satisfy the relations
~1 = 1
~1 · E
E
~1 · E
~ 2 = cos θ12
E
~2 = 1
~2 · E
E
~1 · E
~ 3 = cos θ13
E
~3 = 1
~3 · E
E
~2 · E
~ 3 = cos θ23 ,
E
then calculate the metrices gij and conjugate metrices g ij .
I 43.
Let Aij , i, j = 1, 2, 3, 4 denote the skew-symmetric second rank tensor

0
a
b
d
 −a 0
Aij = 
−b −d 0
−c −e −f

c
e
,
f
0
where a, b, c, d, e, f are complex constants. Calculate the components of the dual tensor
V ij =
1 ijkl
e Akl .
2
107
I 44.
In Cartesian coordinates the vorticity tensor at a point in a fluid medium is defined
1 ∂Vj
∂Vi
−
ωij =
2 ∂xi
∂xj
where Vi are the velocity components of the fluid at the point. The vorticity vector at a point in a fluid
1
medium in Cartesian coordinates is defined by ω i = eijk ωjk . Show that these tensors are dual tensors.
2
I 45.
Write out the relation between each of the components of the dual tensors
1
T̂ ij = eijkl Tkl i, j, k, l = 1, 2, 3, 4
2
and show that if ijkl is an even permutation of 1234, then T̂ ij = Tkl .
I 46.
Consider the general affine transformation x̄i = aij xj where (x1 , x2 , x3 ) = (x, y, z) with inverse
transformation xi = bij x̄j . Determine (a) the image of the plane Ax + By + Cz + D = 0 under this
transformation and (b) the image of a second degree conic section
Ax2 + 2Bxy + Cy 2 + Dx + Ey + F = 0.
I 47.
Using a multilinear form of degree M, derive the transformation law for a contravariant vector of
degree M.
I 48.
Let g denote the determinant of gij and show that
∂g
∂gij
= gg ij k .
∂xk
∂x
I 49.
We have shown that for a rotation of xyz axes with respect to a set of fixed x̄ȳ z̄ axes, the derivative
~ with respect to an observer on the barred axes is given by
of a vector A
~
~
dA
dA
~
=
+ω
~ × A.
dt f
dt r
Introduce the operators
~=
Df A
~
dA
dt
~
~ = dA
Dr A
dt
~ = (Dr + ~
~
ω ×)A.
(a) Show that Df A
= derivative in fixed system
f
= derivative in rotating system
r
~ is the position vector ~r. Show that Df ~r = (Dr + ~ω×)~r
(b) Consider the special case that the vector A
= V~
~
produces V
f
and V~
~
+~
ω × ~r where V
r
represents the velocity of a particle relative to the fixed system
f
represents the velocity of a particle with respect to the rotating system of coordinates.
r
+~
ω × (~
ω × ~r) where ~a
= ~a
(c) Show that ~a
f
r
represents the acceleration of a particle relative to the
f
represents the acceleration of a particle with respect to the rotating system.
fixed system and ~a
r
(d) Show in the special case ~
ω is a constant that
~ +ω
= 2~ω × V
~ × (~ω × ~r)
~a
f
~ is the velocity of the particle relative to the rotating system. The term 2~ω × V
~ is referred to
where V
as the Coriolis acceleration and the term ~ω × (~ω × ~r) is referred to as the centripetal acceleration.
108
§1.4 DERIVATIVE OF A TENSOR
In this section we develop some additional operations associated with tensors. Historically, one of the
basic problems of the tensor calculus was to try and find a tensor quantity which is a function of the metric
∂gij
∂ 2 gij
,
, . . . . A solution of this problem is the fourth order
tensor gij and some of its derivatives
∂xm
∂xm ∂xn
Riemann Christoffel tensor Rijkl to be developed shortly. In order to understand how this tensor was arrived
at, we must first develop some preliminary relationships involving Christoffel symbols.
Christoffel Symbols
Let us consider the metric tensor gij which we know satisfies the transformation law
g αβ = gab
∂xa ∂xb
.
∂xα ∂xβ
Define the quantity
(α, β, γ) =
∂gαβ
∂gab ∂xc ∂xa ∂xb
∂ 2 xa ∂xb
∂xa ∂ 2 xb
+
g
+
g
ab
ab
γ =
γ
α
α
γ
∂x
∂xc ∂x ∂x ∂xβ
∂x ∂x ∂xβ
∂xα ∂xβ ∂xγ
1
[(α, β, γ) + (β, γ, α) − (γ, α, β)] to obtain the result
2
∂g γα
∂gbc
∂gca ∂xa ∂xb ∂xc
∂xb ∂ 2 xa
1 ∂gab
−
+
−
+
g
.
=
ab
α
γ
2 ∂xc
∂xa
∂xb ∂x ∂xβ ∂x
∂xβ
∂xβ ∂xα ∂xγ
and form the combination of terms
∂gβγ
1 ∂gαβ
+
2 ∂xγ
∂xα
(1.4.1)
In this equation the combination of derivatives occurring inside the brackets is called a Christoffel symbol
of the first kind and is defined by the notation
[ac, b] = [ca, b] =
∂gbc
∂gac
1 ∂gab
+
−
.
2 ∂xc
∂xa
∂xb
(1.4.2)
The equation (1.4.1) defines the transformation for a Christoffel symbol of the first kind and can be expressed
as
[α γ, β] = [ac, b]
∂xa ∂xb ∂xc
∂ 2 xa ∂xb
+
g
.
ab
∂xα ∂xβ ∂xγ
∂xα ∂xγ ∂xβ
(1.4.3)
Observe that the Christoffel symbol of the first kind [ac, b] does not transform like a tensor. However, it is
symmetric in the indices a and c.
At this time it is convenient to use the equation (1.4.3) to develop an expression for the second derivative
term which occurs in that equation as this second derivative term arises in some of our future considerations.
∂xβ de
g and simplify the result to the
To solve for this second derivative we can multiply equation (1.4.3) by
∂xd
form
∂xa ∂xc
∂xβ de
∂ 2 xe
de
=
−g
[ac,
d]
+
[α
γ,
β]
g .
(1.4.4)
∂xα ∂xγ
∂xα ∂xγ
∂xd
The transformation g de = g λµ
∂xd ∂xe
allows us to express the equation (1.4.4) in the form
∂xλ ∂xµ
∂xa ∂xc
∂xe
∂ 2 xe
de
βµ
[α γ, β] µ .
α
γ = −g [ac, d]
α
γ +g
∂x ∂x
∂x ∂x
∂x
(1.4.5)
109
Define the Christoffel symbol of the second kind as
i
jk
=
i
kj
1
= g [jk, α] = g iα
2
iα
∂gkα
∂gjα
∂gjk
+
−
∂xj
∂xk
∂xα
.
(1.4.6)
This Christoffel symbol of the second kind is symmetric in the indices j and k and from equation (1.4.5) we
see that it satisfies the transformation law
µ
αγ
∂xe
=
∂xµ
e
ac
∂xa ∂xc
∂ 2 xe
.
α
γ +
∂x ∂x
∂xα ∂xγ
(1.4.7)
Observe that the Christoffel symbol of the second kind does not transform like a tensor quantity. We can use
the relation defined by equation (1.4.7) to express the second derivative of the transformation equations in
terms of the Christoffel symbols of the second kind. At times it will be convenient to represent the Christoffel
symbols with a subscript to indicate the metric from which they are calculated. Thus, an alternative notation
i
i
for j k is the notation j k .
g
EXAMPLE 1.4-1. (Christoffel symbols) Solve for the Christoffel symbol of the first kind in terms of
the Christoffel symbol of the second kind.
Solution: By the definition from equation (1.4.6) we have
i
jk
= g iα [jk, α].
We multiply this equation by gβi and find
gβi
and so
[jk, α] = gαi
i
jk
i
jk
= δβα [jk, α] = [jk, β]
= gα1
1
jk
+ · · · + gαN
N
jk
.
EXAMPLE 1.4-2. (Christoffel symbols of first kind)
Derive formulas to find the Christoffel symbols of the first kind in a generalized orthogonal coordinate
system with metric coefficients
gij = 0
for
i 6= j
and
g(i)(i) = h2(i) ,
i = 1, 2, 3
where i is not summed.
Solution: In an orthogonal coordinate system where gij = 0 for i 6= j we observe that
1
[ab, c] =
2
∂gac ∂gbc
∂gab
+
−
b
a
∂x
∂x
∂xc
.
Here there are 33 = 27 quantities to calculate. We consider the following cases:
(1.4.8)
110
CASE I Let a = b = c = i, then the equation (1.4.8) simplifies to
[ab, c] = [ii, i] =
1 ∂gii
2 ∂xi
(no summation on i).
(1.4.9)
From this equation we can calculate any of the Christoffel symbols
[11, 1],
[22, 2],
or [33, 3].
CASE II Let a = b = i 6= c, then the equation (1.4.8) simplifies to the form
[ab, c] = [ii, c] = −
1 ∂gii
2 ∂xc
(no summation on i and i 6= c).
(1.4.10)
since, gic = 0 for i 6= c. This equation shows how we may calculate any of the six Christoffel symbols
[11, 2],
[11, 3],
[22, 1],
[22, 3],
[33, 1],
[33, 2].
CASE III Let a = c = i 6= b, and noting that gib = 0 for i 6= b, it can be verified that the equation (1.4.8)
simplifies to the form
[ab, c] = [ib, i] = [bi, i] =
1 ∂gii
2 ∂xb
(no summation on i and i 6= b).
(1.4.11)
From this equation we can calculate any of the twelve Christoffel symbols
[12, 1] = [21, 1]
[31, 3] = [13, 3]
[32, 3] = [23, 3]
[21, 2] = [12, 2]
[13, 1] = [31, 1]
[23, 2] = [32, 2]
CASE IV Let a 6= b 6= c and show that the equation (1.4.8) reduces to
[ab, c] = 0,
(a 6= b 6= c.)
This represents the six Christoffel symbols
[12, 3] = [21, 3] = [23, 1] = [32, 1] = [31, 2] = [13, 2] = 0.
From the Cases I,II,III,IV all twenty seven Christoffel symbols of the first kind can be determined. In
practice, only the nonzero Christoffel symbols are listed.
EXAMPLE 1.4-3. (Christoffel symbols of the first kind)Find the nonzero Christoffel symbols of the
first kind in cylindrical coordinates.
Solution: From the results of example 1.4-2 we find that for x1 = r,
g11 = 1,
g22 = (x1 )2 = r2 ,
x2 = θ,
g33 = 1
the nonzero Christoffel symbols of the first kind in cylindrical coordinates are:
1 ∂g22
= −x1 = −r
2 ∂x1
1 ∂g22
= x1 = r.
[21, 2] = [12, 2] =
2 ∂x1
[22, 1] = −
x3 = z and
111
EXAMPLE 1.4-4. (Christoffel symbols of the second kind)
Find formulas for the calculation of the Christoffel symbols of the second kind in a generalized orthogonal
coordinate system with metric coefficients
gij = 0
for
i 6= j
and
g(i)(i) = h2(i) ,
i = 1, 2, 3
where i is not summed.
Solution: By definition we have
i
jk
= g im [jk, m] = g i1 [jk, 1] + g i2 [jk, 2] + g i3 [jk, 3]
(1.4.12)
By hypothesis the coordinate system is orthogonal and so
g ij = 0 for
i 6= j
and g ii =
1
gii
i not summed.
The only nonzero term in the equation (1.4.12) occurs when m = i and consequently
i
jk
= g ii [jk, i] =
[jk, i]
gii
no summation on i.
(1.4.13)
We can now consider the four cases considered in the example 1.4-2.
CASE I Let j = k = i and show
i
ii
=
1 ∂gii
1 ∂
[ii, i]
=
=
ln gii
gii
2gii ∂xi
2 ∂xi
no summation on i.
(1.4.14)
CASE II Let k = j 6= i and show
i
jj
=
−1 ∂gjj
[jj, i]
=
gii
2gii ∂xi
no summation on i or j.
(1.4.15)
CASE III Let i = j 6= k and verify that
j
jk
=
j
kj
=
1 ∂gjj
1 ∂
[jk, j]
=
=
ln gjj
gjj
2gjj ∂xk
2 ∂xk
no summation on i or j.
CASE IV For the case i 6= j 6= k we find
i
jk
=
[jk, i]
= 0,
gii
The above cases represent all 27 terms.
i 6= j 6= k
no summation on i.
(1.4.16)
112
EXAMPLE 1.4-5.
(Notation) In the case of cylindrical coordinates we can use the above relations and
find the nonzero Christoffel symbols of the second kind:
1
1 ∂g22
= −x1 = −r
=−
2g11 ∂x1
22
1
1
2
2
1 ∂g22
= 1 =
=
=
2g22 ∂x1
x
r
12
21
Note 1: The notation for the above Christoffel symbols are based upon the assumption that x1 = r, x2 = θ
and x3 = z. However, in tensor calculus the choice of the coordinates can be arbitrary. We could just as well
have defined x1 = z, x2 = r and x3 = θ. In this latter case, the numbering system of the Christoffel symbols
changes. To avoid confusion, an alternate method of writing the Christoffel symbols is to use coordinates in
place of the integers 1,2 and 3. For example, in cylindrical coordinates we can write
r
θ
θ
1
and
= −r.
=
=
θθ
r
rθ
θr
If we define x1 = r, x2 = θ, x3 = z, then the nonzero Christoffel symbols are written as
1
2
2
1
and
= −r.
=
=
22
r
12
21
In contrast, if we define x1 = z, x2 = r, x3 = θ, then the nonzero Christoffel symbols are written
2
3
3
1
and
= −r.
=
=
33
r
23
32
Note 2: Some textbooks use the notation Γa,bc for Christoffel symbols of the first kind and Γdbc = g da Γa,bc for
Christoffel symbols of the second kind. This notation is not used in these notes since the notation suggests
that the Christoffel symbols are third order tensors, which is not true. The Christoffel symbols of the first
and second kind are not tensors. This fact is clearly illustrated by the transformation equations (1.4.3) and
(1.4.7).
Covariant Differentiation
Let Ai denote a covariant tensor of rank 1 which obeys the transformation law
Aα = Ai
∂xi
.
∂xα
(1.4.17)
Differentiate this relation with respect to xβ and show
∂Ai ∂xj ∂xi
∂ 2 xi
∂Aα
= Ai α β +
.
β
∂xj ∂xβ ∂xα
∂x
∂x ∂x
(1.4.18)
Now use the relation from equation (1.4.7) to eliminate the second derivative term from (1.4.18) and express
it in the form
∂Aα
= Ai
∂xβ
"
σ
αβ
∂xi
−
∂xσ
i
jk
∂xj ∂xk
∂xα ∂xβ
#
+
∂Ai ∂xj ∂xi
.
∂xj ∂xβ ∂xα
(1.4.19)
113
Employing the equation (1.4.17), with α replaced by σ, the equation (1.4.19) is expressible in the form
σ
i
∂xj ∂xk
∂Aα
∂Aj ∂xj ∂xk
−
A
−
A
=
σ
i
α
β
β
k
∂x ∂x ∂x
αβ
j k ∂xα ∂xβ
∂x
or alternatively
"
# σ
∂Aj
i
∂xj ∂xk
∂Aα
−
A
−
A
.
=
σ
i
∂xk
∂xα ∂xβ
αβ
jk
∂xβ
Define the quantity
Aj,k
∂Aj
=
− Ai
∂xk
(1.4.20)
(1.4.21)
i
jk
(1.4.22)
as the covariant derivative of Aj with respect to xk . The equation (1.4.21) demonstrates that the covariant
derivative of a covariant tensor produces a second order tensor which satisfies the transformation law
Aα,β = Aj,k
∂xj ∂xk
.
∂xα ∂xβ
(1.4.23)
Other notations frequently used to denote the covariant derivative are:
Aj,k = Aj;k = Aj/k = ∇k Aj = Aj |k .
(1.4.24)
In the special case where gij are constants the Christoffel symbols of the second kind are zero, and conse∂Aj
. That is, under the special circumstances where the
quently the covariant derivative reduces to Aj,k =
∂xk
Christoffel symbols of the second kind are zero, the covariant derivative reduces to an ordinary derivative.
Covariant Derivative of Contravariant Tensor
i
A contravariant tensor Ai obeys the transformation law A = Aα
form
α
Ai = A
∂xi
which can be expressed in the
∂xα
∂xi
∂xα
(1.4.24)
by interchanging the barred and unbarred quantities. We write the transformation law in the form of equation
(1.4.24) in order to make use of the second derivative relation from the previously derived equation (1.4.7).
Differentiate equation (1.4.24) with respect to xj to obtain the relation
α
2 i
∂A ∂xβ ∂xi
∂xβ
∂Ai
α ∂ x
=
A
+
.
∂xj
∂xα ∂xβ ∂xj
∂xβ ∂xj ∂xα
(1.4.25)
Changing the indices in equation (1.4.25) and substituting for the second derivative term, using the relation
from equation (1.4.7), produces the equation
∂Ai
α
=A
j
∂x
"
σ
αβ
∂xi
−
∂xσ
i
mk
∂xm ∂xk
∂xα ∂xβ
#
α
∂A ∂xβ ∂xi
∂xβ
+
.
∂xj
∂xβ ∂xj ∂xα
Applying the relation found in equation (1.4.24), with i replaced by m, together with the relation
∂xβ ∂xk
= δjk ,
∂xj ∂xβ
(1.4.26)
114
we simplify equation (1.4.26) to the form
∂Ai
+
∂xj
i
mj
m
A
"
σ
∂A
+
=
∂xβ
Define the quantity
i
A
,j
∂Ai
=
+
∂xj
# β
i
σ
α ∂x ∂x
A
σ.
j
∂x ∂x
αβ
i
Am
mj
(1.4.27)
(1.4.28)
as the covariant derivative of the contravariant tensor Ai . The equation (1.4.27) demonstrates that a covariant
derivative of a contravariant tensor will transform like a mixed second order tensor and
β
σ ∂x
,β
j
Ai ,j = A
∂xi
.
∂x ∂xσ
(1.4.29)
∂Ai
and the
∂xj
covariant derivative of a contravariant tensor reduces to an ordinary derivative in this special case.
Again it should be observed that for the condition where gij are constants we have Ai ,j =
In a similar manner the covariant derivative of second rank tensors can be derived. We find these
derivatives have the forms:
Aij,k
Aij ,k
Aij ,k
σ
σ
∂Aij
=
− Aσj
− Aiσ
k
∂x
ik
jk
∂Aij
i
σ
=
+ Aσj
− Aiσ
∂xk
σk
jk
∂Aij
i
j
σj
iσ
=
+
A
+
A
.
∂xk
σk
σk
(1.4.30)
In general, the covariant derivative of a mixed tensor
Aij...k
lm...p
of rank n has the form
Aij...k
lm...p,q =
∂Aij...k
lm...p
∂xq
i
j
k
ij...σ
+ Aiσ...k
+
·
·
·
+
A
lm...p
lm...p
σq
σq
σq
σ
σ
σ
ij...k
ij...k
−
A
−
·
·
·
−
A
− Aij...k
σm...p
lσ...p
lm...σ
lq
mq
pq
+ Aσj...k
lm...p
(1.4.31)
and this derivative is a tensor of rank n + 1. Note the pattern of the + signs for the contravariant indices
and the − signs for the covariant indices.
Observe that the covariant derivative of an nth order tensor produces an n+ 1st order tensor, the indices
of these higher order tensors can also be raised and lowered by multiplication by the metric or conjugate
metric tensor. For example we can write
g im Ajk |m = Ajk |i
and g im Ajk |m = Ajk |i
115
Rules for Covariant Differentiation
The rules for covariant differentiation are the same as for ordinary differentiation. That is:
(i) The covariant derivative of a sum is the sum of the covariant derivatives.
(ii) The covariant derivative of a product of tensors is the first times the covariant derivative of the second
plus the second times the covariant derivative of the first.
(iii) Higher derivatives are defined as derivatives of derivatives. Be careful in calculating higher order derivatives as in general
Ai,jk 6= Ai,kj .
EXAMPLE 1.4-6. (Covariant differentiation)
Calculate the second covariant derivative Ai,jk .
Solution: The covariant derivative of Ai is
Ai,j
σ
∂Ai
=
− Aσ
.
∂xj
ij
By definition, the second covariant derivative is the covariant derivative of a covariant derivative and hence
Ai,jk = (Ai,j ) ,k =
∂
∂xk
∂Ai
σ
m
m
−
A
−
A
−
A
.
σ
m,j
i,m
∂xj
ij
ik
jk
Simplifying this expression one obtains
σ
∂ 2 Ai
∂Aσ σ
∂
−
−
A
σ
∂xj ∂xk
∂xk i j
∂xk i j
σ
m
∂Ai
σ
m
∂Am
−
A
−
A
−
.
−
σ
σ
∂xj
∂xm
mj
ik
im
jk
Ai,jk =
Rearranging terms, the second covariant derivative can be expressed in the form
∂ 2 Ai
∂Aσ σ
∂Am m
∂Ai m
−
−
−
∂xj ∂xk
∂xk i j
∂xj i k
∂xm j k
∂
σ
σ
m
m
σ
−
−
.
− Aσ
∂xk i j
im
jk
ik
mj
Ai,jk =
(1.4.32)
116
Riemann Christoffel Tensor
Utilizing the equation (1.4.32), it is left as an exercise to show that
σ
Ai,jk − Ai,kj = Aσ Rijk
where
σ
=
Rijk
∂
∂xj
σ
ik
−
∂
∂xk
σ
ij
+
m
ik
σ
mj
−
m
ij
σ
mk
(1.4.33)
is called the Riemann Christoffel tensor. The covariant form of this tensor is
i
.
Rhjkl = gih Rjkl
(1.4.34)
It is an easy exercise to show that this covariant form can be expressed in either of the forms
Rinjk
or
Rijkl
s
s
∂
∂
=
[nk, i] − k [nj, i] + [ik, s]
− [ij, s]
nj
nk
∂xj
∂x
2
∂ gil
1
∂ 2 gjl
∂ 2 gik
∂ 2 gjk
=
− i k − j l + i l + g αβ ([jk, β][il, α] − [jl, β][ik, α]) .
2 ∂xj ∂xk
∂x ∂x
∂x ∂x
∂x ∂x
From these forms we find that the Riemann Christoffel tensor is skew symmetric in the first two indices
and the last two indices as well as being symmetric in the interchange of the first pair and last pairs of
indices and consequently
Rjikl = −Rijkl
Rijlk = −Rijkl
Rklij = Rijkl .
In a two dimensional space there are only four components of the Riemann Christoffel tensor to consider.
These four components are either +R1212 or −R1212 since they are all related by
R1212 = −R2112 = R2121 = −R1221 .
In a Cartesian coordinate system Rhijk = 0. The Riemann Christoffel tensor is important because it occurs
in differential geometry and relativity which are two areas of interest to be considered later. Additional
properties of this tensor are found in the exercises of section 1.5.
117
Physical Interpretation of Covariant Differentiation
~ 1, E
~ 2, E
~ 3 ). These
In a system of generalized coordinates (x1 , x2 , x3 ) we can construct the basis vectors (E
basis vectors change with position. That is, each basis vector is a function of the coordinates at which they
are evaluated. We can emphasize this dependence by writing
~ i (x1 , x2 , x3 ) = ∂~r
~i = E
E
∂xi
i = 1, 2, 3.
Associated with these basis vectors we have the reciprocal basis vectors
~ i (x1 , x2 , x3 ),
~i = E
E
i = 1, 2, 3
~ can be represented in terms of contravariant components as
which are also functions of position. A vector A
~ 1 + A2 E
~ 2 + A3 E
~ 3 = Aj E
~j
~ = A1 E
A
(1.4.35)
or it can be represented in terms of covariant components as
~ 1 + A2 E
~ 2 + A3 E
~ 3 = Aj E
~ j.
~ = A1 E
A
(1.4.36)
~ is represented as
A change in the vector A
~=
dA
~
∂A
dxk
∂xk
where from equation (1.4.35) we find
~
~
∂Aj ~
∂A
j ∂ Ej
Ej
=
A
+
∂xk
∂xk
∂xk
(1.4.37)
or alternatively from equation (1.4.36) we may write
~j
~
∂Aj ~ j
∂E
∂A
E .
=
A
+
j
k
k
∂x
∂x
∂xk
(1.4.38)
We define the covariant derivative of the covariant components as
Ai,k =
~
~j
∂A
~ i = ∂Ai + Aj ∂ E · E
~ i.
·E
k
k
∂x
∂x
∂xk
(1.4.39)
The covariant derivative of the contravariant components are defined by the relation
Ai ,k =
i
~
~
∂A
~ i = ∂A + Aj ∂ Ej · E
~ i.
·E
k
k
∂x
∂x
∂xk
(1.4.40)
Introduce the notation
~j
∂E
=
∂xk
We then have
m ~
Em
jk
~
~ i · ∂ Ej =
E
∂xk
and
~j
j
∂E
~ m.
E
=−
mk
∂xk
m ~
i
~ i = m δi =
Em · E
jk
jk m
jk
(1.4.41)
(1.4.42)
118
and
~j
~m · E
~ i = − j δim = − j .
~ i · ∂E = − j E
E
mk
mk
ik
∂xk
(1.4.43)
Then equations (1.4.39) and (1.4.40) become
j
Aj
ik
i
∂Ai
=
+
Aj ,
k
jk
∂x
∂Ai
=
−
∂xk
Ai,k
Ai ,k
which is consistent with our earlier definitions from equations (1.4.22) and (1.4.28). Here the first term of
the covariant derivative represents the rate of change of the tensor field as we move along a coordinate curve.
The second term in the covariant derivative represents the change in the local basis vectors as we move
along the coordinate curves. This is the physical interpretation associated with the Christoffel symbols of
the second kind.
We make the observation that the derivatives of the basis vectors in equations (1.4.39) and (1.4.40) are
related since
~ j = δj
~i · E
E
i
and consequently
~j
~
∂ ~ ~j
~ i · ∂ E + ∂ Ei · E
~j = 0
(Ei · E ) = E
k
k
∂x
∂x
∂xk
~j
~
~ j · ∂ Ei
~ i · ∂ E = −E
or
E
k
∂x
∂xk
Hence we can express equation (1.4.39) in the form
Ai,k =
~
∂Ai
~ j · ∂ Ei .
− Aj E
k
∂x
∂xk
(1.4.44)
We write the first equation in (1.4.41) in the form
~j
∂E
=
∂xk
m
~ i = [jk, i]E
~i
gim E
jk
(1.4.45)
and consequently
~j
i ~ ~m
i
m
∂E
m
m
~
Ei · E =
·E =
δi =
k
jk
jk
jk
∂x
(1.4.46)
~
∂ Ej ~
i
i
~ ·E
~ m = [jk, i]δ = [jk, m].
· Em =[jk, i]E
and
m
∂xk
These results also reduce the equations (1.4.40) and (1.4.44) to our previous forms for the covariant derivatives.
The equations (1.4.41) are representations of the vectors
~i
∂E
∂xk
and
~j
∂E
∂xk
in terms of the basis vectors and
reciprocal basis vectors of the space. The covariant derivative relations then take into account how these
vectors change with position and affect changes in the tensor field.
The Christoffel symbols in equations (1.4.46) are symmetric in the indices j and k since
~j
∂
∂E
=
∂xk
∂xk
∂~r
∂xj
=
∂
∂xj
∂~r
∂xk
=
~k
∂E
.
∂xj
(1.4.47)
119
The equations (1.4.46) and (1.4.47) enable us to write
"
#
~j
~j
~k
∂
E
1
∂
E
∂
E
~m ·
~m ·
~m ·
E
=
+E
[jk, m] =E
∂xk
2
∂xk
∂xj
"
#
~m
~m
∂ ~
∂E
∂E
∂ ~
1
~
~
~
~
− Ek ·
Em · Ej + j Em · Ek − Ej ·
=
2 ∂xk
∂x
∂xk
∂xj
"
#
~k
~j
∂
E
∂
E
∂
∂ ~
1
~j +
~k − E
~k ·
~m · E
~j ·
−E
Em · E
E
=
2 ∂xk
∂xj
∂xm
∂xm
1 ∂ ~
~j + ∂ E
~k − ∂
~k
~m · E
~j · E
·
E
E
E
=
m
2 ∂xk
∂xj
∂xm
∂gmk
∂gjk
1 ∂gmj
+
− m = [kj, m]
=
k
j
2 ∂x
∂x
∂x
which again agrees with our previous result.
~j,
~ is represented in the form A
~ = Aj E
For future reference we make the observation that if the vector A
involving contravariant components, then we may write
!
~
∂Aj ~
j ∂ Ej
Ej + A
dxk
∂xk
∂xk
j
∂A ~
i ~
j
Ej + A
Ei dxk
=
∂xk
jk
j ∂A
j
m
~j.
~ j dxk = Aj dxk E
E
=
+
A
,k
mk
∂xk
~
~ = ∂ A dxk =
dA
∂xk
(1.4.48)
~ j involving covariant components it is left as
~ is represented in the form A
~ = Aj E
Similarly, if the vector A
an exercise to show that
~j
~ = Aj,k dxk E
dA
(1.4.49)
Ricci’s Theorem
Ricci’s theorem states that the covariant derivative of the metric tensor vanishes and gik,l = 0.
Proof: We have
gik,l
gik,l
gik,l
m
m
∂gik
=
−
gim −
gmk
kl
il
∂xl
∂gik
=
− [kl, i] − [il, k]
∂xl
∂gik
1 ∂gik
∂gil
∂gkl
∂gkl
∂gil
1 ∂gik
=
−
+ k −
+
− k = 0.
−
∂xl
2 ∂xl
∂x
∂xi
2 ∂xl
∂xi
∂x
Because of Ricci’s theorem the components of the metric tensor can be regarded as constants during covariant
differentiation.
i
= 0.
EXAMPLE 1.4-7. (Covariant differentiation) Show that δj,k
Solution
i
δj,k
∂δji
i
σ
i
i
σ
i
=
+ δj
− δσ
=
−
= 0.
∂xk
σk
jk
jk
jk
120
EXAMPLE 1.4-8. (Covariant differentiation) Show that g ij,k = 0.
Solution: Since gij g jk = δik we take the covariant derivative of this expression and find
k
=0
(gij g jk ),l = δi,l
gij g jk,l + gij,l g jk = 0.
But gij,l = 0 by Ricci’s theorem and hence gij g jk,l = 0. We multiply this expression by g im and obtain
g im gij g jk,l = δjm g jk,l = g mk
,l = 0
which demonstrates that the covariant derivative of the conjugate metric tensor is also zero.
EXAMPLE 1.4-9. (Covariant differentiation) Some additional examples of covariant differentiation
are:
(i) (gil Al ),k = gil Al ,k = Ai,k
(ii) (gim gjn Aij ) ,k = gim gjn Aij,k = Amn,k
Intrinsic or Absolute Differentiation
The intrinsic or absolute derivative of a covariant vector Ai taken along a curve xi = xi (t), i = 1, . . . , N
is defined as the inner product of the covariant derivative with the tangent vector to the curve. The intrinsic
derivative is represented
dxj
δAi
= Ai,j
δt
dt
j
∂Ai
α
dx
δAi
=
− Aα
δt
∂xj
dt
ij
j
dAi
α dx
δAi
=
− Aα
.
δt
dt
i j dt
(1.4.50)
Similarly, the absolute or intrinsic derivative of a contravariant tensor Ai is represented
dAi
dxj
δAi
= Ai ,j
=
+
δt
dt
dt
i
dxj
Ak
.
jk
dt
The intrinsic or absolute derivative is used to differentiate sums and products in the same manner as used
in ordinary differentiation. Also if the coordinate system is Cartesian the intrinsic derivative becomes an
ordinary derivative.
The intrinsic derivative of higher order tensors is similarly defined as an inner product of the covariant
derivative with the tangent vector to the given curve. For example,
δAij
dxp
klm
= Aij
klm,p
δt
dt
is the intrinsic derivative of the fifth order mixed tensor Aij
klm .
121
EXAMPLE 1.4-10. (Generalized velocity and acceleration) Let t denote time and let xi = xi (t)
for i = 1, . . . , N , denote the position vector of a particle in the generalized coordinates (x1 , . . . , xN ). From
the transformation equations (1.2.30), the position vector of the same particle in the barred system of
coordinates, (x1 , x2 , . . . , xN ), is
xi = xi (x1 (t), x2 (t), . . . , xN (t)) = xi (t),
The generalized velocity is v i =
dxi
dt ,
i = 1, . . . , N.
i = 1, . . . , N. The quantity v i transforms as a tensor since by definition
vi =
∂xi dxj
∂xi j
dxi
=
=
v .
dt
∂xj dt
∂xj
(1.4.51)
Let us now find an expression for the generalized acceleration. Write equation (1.4.51) in the form
vj = v i
∂xj
∂xi
(1.4.52)
and differentiate with respect to time to obtain
dv i ∂xj
∂ 2 xj dxk
dv j
= vi i k
+
dt
dt ∂xi
∂x ∂x dt
The equation (1.4.53) demonstrates that
dv i
dt
(1.4.53)
does not transform like a tensor. From the equation (1.4.7)
previously derived, we change indices and write equation (1.4.53) in the form
dxk
dv j
= vi
dt
dt
"
σ
ik
∂xj
−
∂xσ
#
j ∂xa ∂xc
∂xj dv i
.
+
i
k
a c ∂x ∂x
∂xi dt
Rearranging terms we find
∂v j dxk
+
∂xk dt
j
ac
c k
∂xa i
∂x dx
σ
∂xj dxk
∂xj ∂v i dxk
+
v
vi σ
=
i
i
k dt
k dt
∂x dt
ik
∂x
∂x ∂x
∂x
"
j k
# k
σ
j
j
∂v
∂v
σ
a dx
i dx ∂x
=
+
v
+
v
ak
∂xk
dt
dt ∂xσ
ik
∂xk
or
δv σ ∂xj
δv j
=
.
δt
δt ∂xσ
The above equation illustrates that the intrinsic derivative of the velocity is a tensor quantity. This derivative
is called the generalized acceleration and is denoted
dv i
δv i
dxj
= v i,j
=
+
f =
δt
dt
dt
i
m n
i
i
dx dx
d2 xi
m n
v v =
,
+
2
mn
m n dt dt
dt
i = 1, . . . , N
To summarize, we have shown that if
xi = xi (t),
i
i = 1, . . . , N
is the generalized position vector, then
dx
, i = 1, . . . , N is the generalized velocity, and
dt
i
δv
dxj
= v i,j
, i = 1, . . . , N is the generalized acceleration.
fi =
δt
dt
vi =
(1.4.54)
122
Parallel Vector Fields
Let y i = y i (t), i = 1, 2, 3 denote a space curve C in a Cartesian coordinate system and let Y i define a
constant vector in this system. Construct at each point of the curve C the vector Y i . This produces a field
of parallel vectors along the curve C. What happens to the curve and the field of parallel vectors when we
transform to an arbitrary coordinate system using the transformation equations
y i = y i (x1 , x2 , x3 ),
i = 1, 2, 3
xi = xi (y 1 , y 2 , y 3 ),
i = 1, 2, 3?
with inverse transformation
The space curve C in the new coordinates is obtained directly from the transformation equations and can
be written
xi = xi (y 1 (t), y 2 (t), y 3 (t)) = xi (t),
i = 1, 2, 3.
The field of parallel vectors Y i become X i in the new coordinates where
Y i = Xj
∂y i
.
∂xj
(1.4.55)
Since the components of Y i are constants, their derivatives will be zero and consequently we obtain by
differentiating the equation (1.4.55), with respect to the parameter t, that the field of parallel vectors X i
must satisfy the differential equation
dY i
∂ 2 y i dxm
dX j ∂y i
=
= 0.
+ Xj j m
j
dt ∂x
∂x ∂x dt
dt
(1.4.56)
Changing symbols in the equation (1.4.7) and setting the Christoffel symbol to zero in the Cartesian system
of coordinates, we represent equation (1.4.7) in the form
∂ 2yi
=
∂xj ∂xm
i
α
∂y
j m ∂xα
and consequently, the equation (1.4.56) can be reduced to the form
dX j
δX j
=
+
δt
dt
j
dxm
Xk
= 0.
km
dt
(1.4.57)
The equation (1.4.57) is the differential equation which must be satisfied by a parallel field of vectors X i
along an arbitrary curve xi (t).
123
EXERCISE 1.4
I 1.
1
Find the nonzero Christoffel symbols of the first and second kind in cylindrical coordinates
(x , x , x3 ) = (r, θ, z), where x = r cos θ,
I 2.
1
2
y = r sin θ,
z = z.
Find the nonzero Christoffel symbols of the first and second kind in spherical coordinates
2
(x , x , x3 ) = (ρ, θ, φ), where x = ρ sin θ cos φ,
y = ρ sin θ sin φ,
z = ρ cos θ.
I 3.
Find the nonzero Christoffel symbols of the first and second kind in parabolic cylindrical coordinates
1
(x , x , x3 ) = (ξ, η, z), where x = ξη, y = (ξ 2 − η 2 ), z = z.
2
1
2
I 4.
Find the nonzero Christoffel symbols of the first and second kind in parabolic coordinates
1
(x , x , x3 ) = (ξ, η, φ), where x = ξη cos φ, y = ξη sin φ, z = (ξ 2 − η 2 ).
2
1
I 5.
1
2
Find the nonzero Christoffel symbols of the first and second kind in elliptic cylindrical coordinates
(x , x , x3 ) = (ξ, η, z), where x = cosh ξ cos η,
I 6.
2
y = sinh ξ sin η,
z = z.
Find the nonzero Christoffel symbols of the first and second kind for the oblique cylindrical coordinates
1
(x , x2 , x3 ) = (r, φ, η), where x = r cos φ,
y = r sin φ+η cos α,
z = η sin α with 0 < α <
π
2
and α constant.
Hint: See figure 1.3-18 and exercise 1.3, problem 12.
I 7.
I 8.
Show [ij, k] + [kj, i] =
∂gik
.
∂xj
r
(a) Let
= g ri [st, i] and solve for the Christoffel symbol of the first kind in terms of the Christoffel
st
symbol of the secondkind.
n
and solve for the Christoffel symbol of the second kind in terms of the
(b) Assume [st, i] = gni
st
Christoffel symbol of the first kind.
I 9.
(a) Write down the transformation law satisfied by the fourth order tensor ijk,m .
(b) Show that ijk,m = 0 in all coordinate systems.
√
(c) Show that ( g),k = 0.
I 10.
Show ijk
,m = 0.
I 11.
Calculate the second covariant derivative Ai ,kj .
∂φ
.
∂xi
(a) Find the physical components associated with the covariant components φ ,i
Ai φ,i
dφ
=
(b) Show the directional derivative of φ in a direction Ai is
.
dA
(gmn Am An )1/2
I 12.
~i
The gradient of a scalar field φ(x1 , x2 , x3 ) is the vector grad φ = E
124
I 13.
(a) Show
√
g is a relative scalar of weight +1.
(b) Use the results from
9(c) and problem 44, Exercise 1.4, to show that
problem
√
∂ g
m √
√
−
g = 0.
( g),k =
k
∂x
km
m
1 ∂g
∂
√
ln( g) =
.
(c) Show that
=
k
∂x
2g ∂xk
km
I 14.
Use the result from problem 9(b) to show
Hint: Expand the covariant derivative rst,p
multiplication with
I 15.
rst
e√
g
1 ∂g
m
∂
√
ln( g) =
.
=
∂xk
2g ∂xk√
km
and then substitute rst = gerst . Simplify by inner
and note the Exercise 1.1, problem 26.
Calculate the covariant derivative Ai,m and then contract on m and i to show that
1 ∂ √ i
gA .
Ai,i = √
g ∂xi
I 16.
1 ∂ √ ij gg +
Show √
g ∂xj
I 17.
Prove that the covariant derivative of a sum equals the sum of the covariant derivatives.
i
pq
g pq = 0. Hint: See problem 14.
Hint: Assume Ci = Ai + Bi and write out the covariant derivative for Ci,j .
I 18.
Let Cji = Ai Bj and prove that the covariant derivative of a product equals the first term times the
covariant derivative of the second term plus the second term times the covariant derivative of the first term.
∂xα ∂xβ
and take an ordinary derivative of both sides
∂ x̄i ∂ x̄j
k
with respect to x̄ and hence derive the relation for Aij,k given in (1.4.30).
I 19.
Start with the transformation law Āij = Aαβ
∂xi ∂xj
and take an ordinary derivative of both sides
∂ x̄α ∂ x̄β
with respect to xk and hence derive the relation for Aij,k given in (1.4.30).
I 20.
Start with the transformation law Aij = Āα β
I 21.
Find the covariant derivatives of
(a) Aijk
I 22.
(b) Aijk
(c) Aijk
Find the intrinsic derivative along the curve xi = xi (t),
(a) Aijk
(b) Aijk
(c) Aijk
I 23.
~ i and show that dA
~ i.
~ = Ai dxk E
~ = Ai E
(a) Assume A
,k
i
k
~ and show that dA
~ i.
~ = Ai,k dx E
~ = Ai E
(b) Assume A
(d)
Aijk
i = 1, . . . , N for
(d)
Aijk
125
(parallel vector field) Imagine a vector field Ai = Ai (x1 , x2 , x3 ) which is a function of position.
I 24.
Assume that at all points along a curve xi = xi (t), i = 1, 2, 3 the vector field points in the same direction,
~ is a constant, then
we would then have a parallel vector field or homogeneous vector field. Assume A
~=
dA
~
∂A
∂xk
dxk = 0. Show that for a parallel vector field the condition Ai,k = 0 must be satisfied.
σ
ik
σ
.
ik
I 25.
∂
∂[ik, n]
= gnσ j
Show that
∂xj
∂x
I 26.
Show Ar,s − As,r =
I 27.
In cylindrical coordinates you are given the contravariant vector components
+ ([nj, σ] + [σj, n])
∂Ar
∂As
−
.
∂xs
∂xr
A1 = r
(a) Find the physical components Ar ,
A2 = cos θ
A3 = z sin θ
Aθ , and Az .
Arr
Arθ
Arz
(b) Denote the physical components of Ai,j , i, j = 1, 2, 3, by Aθr
Aθθ
Aθz
Azr
Azθ
Azz .
Find these physical components.
I 28.
Find the covariant form of the contravariant tensor C i = ijk Ak,j .
Express your answer in terms of Ak,j .
1
In Cartesian coordinates let x denote the magnitude of the position vector xi . Show that (a) x ,j = xj
x
1
1
1
2
−δij
3xi xj
(b) x ,ij = δij − 3 xi xj (c) x ,ii = . (d) LetU = , x 6= 0, and show that U ,ij =
+
and
3
x
x
x
x
x
x5
U ,ii = 0.
I 29.
I 30.
Consider a two dimensional space with element of arc length squared
g11
2
1 2
2 2
ds = g11 (du ) + g22 (du ) and metric gij =
0
0
g22
where u1 , u2 are surface coordinates.
(a) Find formulas to calculate the Christoffel symbols of the first kind.
(b) Find formulas to calculate the Christoffel symbols of the second kind.
I 31.
Find the metric tensor and Christoffel symbols of the first and second kind associated with the
two dimensional space describing points on a cylinder of radius a. Let u1 = θ and u2 = z denote surface
coordinates where
x = a cos θ = a cos u1
y = a sin θ = a sin u1
z = z = u2
126
I 32.
Find the metric tensor and Christoffel symbols of the first and second kind associated with the
two dimensional space describing points on a sphere of radius a. Let u1 = θ and u2 = φ denote surface
coordinates where
x = a sin θ cos φ = a sin u1 cos u2
y = a sin θ sin φ = a sin u1 sin u2
z = a cos θ = a cos u1
I 33.
Find the metric tensor and Christoffel symbols of the first and second kind associated with the
two dimensional space describing points on a torus having the parameters a and b and surface coordinates
u1 = ξ,
u2 = η. illustrated in the figure 1.3-19. The points on the surface of the torus are given in terms
of the surface coordinates by the equations
x = (a + b cos ξ) cos η
y = (a + b cos ξ) sin η
z = b sin ξ
I 34.
Prove that eijk am bj ck ui,m + eijk ai bm ck uj,m + eijk ai bj cm uk,m = ur,r eijk ai bj ck . Hint: See Exercise 1.3,
problem 32 and Exercise 1.1, problem 21.
I 35.
Calculate the second covariant derivative Ai,jk .
I 36.
1 ∂ √ ij gσ + σ mn
Show that σ ij,j = √
g ∂xj
I 37.
Find the contravariant, covariant and physical components of velocity and acceleration in (a) Cartesian
i
mn
coordinates and (b) cylindrical coordinates.
I 38.
Find the contravariant, covariant and physical components of velocity and acceleration in spherical
coordinates.
I 39.
In spherical coordinates (ρ, θ, φ) show that the acceleration components can be represented in terms
of the velocity components as
fρ = v̇ρ −
vθ2 + vφ2
,
ρ
fθ = v̇θ +
vφ2
vρ vθ
−
,
ρ
ρ tan θ
fφ = v̇φ +
vθ vφ
vρ vφ
+
ρ
ρ tan θ
Hint: Calculate v̇ρ , v̇θ , v̇φ .
I 40.
The divergence of a vector Ai is Ai,i . That is, perform a contraction on the covariant derivative
Ai,j to obtain Ai,i . Calculate the divergence in (a) Cartesian coordinates (b) cylindrical coordinates and (c)
spherical coordinates.
I 41.
If S is a scalar invariant of weight one and Aijk is a third order relative tensor of weight W , show
that S −W Aijk is an absolute tensor.
127
I 42.
Let Ȳ i ,i = 1, 2, 3 denote the components of a field of parallel vectors along the curve C defined by
the equations y i = ȳ i (t), i = 1, 2, 3 in a space with metric tensor ḡij , i, j = 1, 2, 3. Assume that Ȳ i and
dȳ i
dt
are unit vectors such that at each point of the curve C̄ we have
ḡij Ȳ i
dȳ j
= cos θ = Constant.
dt
(i.e. The field of parallel vectors makes a constant angle θ with the tangent to each point of the curve C̄.)
Show that if Ȳ i and ȳ i (t) undergo a transformation xi = xi (ȳ 1 , ȳ 2 , ȳ 3 ), i = 1, 2, 3 then the transformed
m
vector X m = Ȳ i ∂x
∂ ȳ j makes a constant angle with the tangent vector to the transformed curve C given by
xi = xi (ȳ 1 (t), ȳ 2 (t), ȳ 3 (t)).
I 43.
Let J denote the Jacobian determinant |
∂xi
|. Differentiate J with respect to xm and show that
∂xj
r
α ∂xp
∂J
=
J
−
J
.
∂xm
rm
α p ∂xm
Hint: See Exercise 1.1, problem 27 and (1.4.7).
I 44.
Assume that φ is a relative scalar of weight W so that φ = J W φ. Differentiate this relation with
respect to xk . Use the result from problem 43 to obtain the transformation law:
"
m
#
r
∂x
α
∂φ
∂φ
W
−W
φ =J
−W
φ
.
mr
∂xm
αk
∂xk
∂xk
The quantity inside the brackets is called the covariant derivative of a relative scalar of weight W. The
covariant derivative of a relative scalar of weight W is defined as
φ ,k
∂φ
=
−W
∂xk
r
φ
kr
and this definition has an extra term involving the weight.
It can be shown that similar results hold for relative tensors of weight W. For example, the covariant
derivative of first and second order relative tensors of weight W have the forms
i
T ,k
Tji ,k
i
r
∂T i
m
=
+
T −W
Ti
km
kr
∂xk
∂Tji
i
σ
r
σ
i
=
+
Tj −
Tσ − W
Ti
k
kσ
jk
kr j
∂x
When the weight term is zero these covariant derivatives reduce to the results given in our previous definitions.
I 45.
Let
dxi
dt
= v i denote a generalized velocity and define the scalar function of kinetic energy T of a
particle with mass m as
T =
1
1
m gij v i v j = m gij ẋi ẋj .
2
2
Show that the intrinsic derivative of T is the same as an ordinary derivative of T. (i.e. Show that
δT
δT
=
dT
dt
.)
128
I 46. Verify the relations
∂gij
= −gmj gni
∂xk
∂g in
= −g mn g ij
∂xk
∂g nm
∂xk
∂gjm
∂xk
1 ∂ √ ijk Assume that B ijk is an absolute tensor. Is the quantity T jk = √
gB
a tensor? Justify
g ∂xi
your answer. If your answer is “no”, explain your answer and determine if there any conditions you can
I 47.
impose upon B ijk such that the above quantity will be a tensor?
I 48.
The e-permutation symbol can be used to define various vector products. Let Ai , Bi , Ci , Di
i = 1, . . . , N denote vectors, then expand and verify the following products:
(a) In two dimensions
R =eij Ai Bj
Ri =eij Aj
a scalar determinant.
a vector (rotation).
(b) In three dimensions
S =eijk Ai Bj Ck
Si =eijk Bj Ck
Sij =eijk Ck
a scalar determinant.
a vector cross product.
a skew-symmetric matrix
(c) In four dimensions
T =eijkm Ai Bj Ck Dm
Ti =eijkm Bj Ck Dm
Tij =eijkm Ck Dm
Tijk =eikm Dm
Expand the curl operator for:
(a) Two dimensions B = eij Aj,i
(b) Three dimensions Bi = eijk Ak,j
(c) Four dimensions Bij = eijkm Am,k
4-dimensional cross product.
skew-symmetric matrix.
skew-symmetric tensor.
with similar products in higher dimensions.
I 49.
a scalar determinant.
129
§1.5 DIFFERENTIAL GEOMETRY AND RELATIVITY
In this section we will examine some fundamental properties of curves and surfaces. In particular, at
each point of a space curve we can construct a moving coordinate system consisting of a tangent vector, a
normal vector and a binormal vector which is perpendicular to both the tangent and normal vectors. How
these vectors change as we move along the space curve brings up the subjects of curvature and torsion
associated with a space curve. The curvature is a measure of how the tangent vector to the curve is changing
and the torsion is a measure of the twisting of the curve out of a plane. We will find that straight lines have
zero curvature and plane curves have zero torsion.
In a similar fashion, associated with every smooth surface there are two coordinate surface curves and
a normal surface vector through each point on the surface. The coordinate surface curves have tangent
vectors which together with the normal surface vectors create a set of basis vectors. These vectors can be
used to define such things as a two dimensional surface metric and a second order curvature tensor. The
coordinate curves have tangent vectors which together with the surface normal form a coordinate system at
each point of the surface. How these surface vectors change brings into consideration two different curvatures.
A normal curvature and a tangential curvature (geodesic curvature). How these curvatures are related to
the curvature tensor and to the Riemann Christoffel tensor, introduced in the last section, as well as other
interesting relationships between the various surface vectors and curvatures, is the subject area of differential
geometry.
Also presented in this section is a brief introduction to relativity where again the Riemann Christoffel
tensor will occur. Properties of this important tensor are developed in the exercises of this section.
Space Curves and Curvature
For xi = xi (s),i = 1, 2, 3, a 3-dimensional space curve in a Riemannian space Vn with metric tensor gij ,
and arc length parameter s, the vector T i =
dxi
ds
represents a tangent vector to the curve at a point P on
i
the curve. The vector T is a unit vector because
gij T i T j = gij
dxi dxj
= 1.
ds ds
(1.5.1)
Differentiate intrinsically, with respect to arc length, the relation (1.5.1) and verify that
gij T i
δT j
δT i j
+ gij
T = 0,
δs
δs
(1.5.2)
δT i
= 0.
δs
(1.5.3)
which implies that
gij T j
Hence, the vector
δT i
δs
is perpendicular to the tangent vector T i . Define the unit normal vector N i to the
space curve to be in the same direction as the vector
Ni =
δT i
δs
and write
1 δT i
κ δs
(1.5.4)
where κ is a scale factor, called the curvature, and is selected such that
gij N i N j = 1
which implies
gij
δT i δT j
= κ2 .
δs δs
(1.5.5)
130
The reciprocal of curvature is called the radius of curvature. The curvature measures the rate of change of
the tangent vector to the curve as the arc length varies. By differentiating intrinsically, with respect to arc
length s, the relation gij T i N j = 0 we find that
gij T i
δN j
δT i j
+ gij
N = 0.
δs
δs
(1.5.6)
Consequently, the curvature κ can be determined from the relation
gij T i
δN j
δT i j
= −gij
N = −gij κN i N j = −κ
δs
δs
(1.5.7)
which defines the sign of the curvature. In a similar fashion we differentiate the relation (1.5.5) and find that
gij N i
This later equation indicates that the vector
δN j
δs
δN j
= 0.
δs
(1.5.8)
is perpendicular to the unit normal N i . The equation
(1.5.3) indicates that T i is also perpendicular to N i and hence any linear combination of these vectors will
also be perpendicular to N i . The unit binormal vector is defined by selecting the linear combination
δN j
+ κT j
δs
(1.5.9)
and then scaling it into a unit vector by defining
1
B =
τ
j
δN j
+ κT j
δs
(1.5.10)
where τ is a scalar called the torsion. The sign of τ is selected such that the vectors T i , N i and B i form a
right handed system with ijk T i N j B k = 1 and the magnitude of τ is selected such that B i is a unit vector
satisfying
gij B i B j = 1.
(1.5.11)
The triad of vectors T i , N i , B i at a point on the curve form three planes. The plane containing T i and B i is
called the rectifying plane. The plane containing N i and B i is called the normal plane. The plane containing
T i and N i is called the osculating plane. The reciprocal of the torsion is called the radius of torsion. The
torsion measures the rate of change of the osculating plane. The vectors T i , N i and B i form a right-handed
orthogonal system at a point on the space curve and satisfy the relation
B i = ijk Tj Nk .
(1.5.12)
By using the equation (1.5.10) it can be shown that B i is perpendicular to both the vectors T i and N i since
gij B i T j = 0
and gij B i N j = 0.
It is left as an exercise to show that the binormal vector B i satisfies the relation
relations
δT i
= κN i
δs
δN i
= τ B i − κT i
δs
δB i
= −τ N i
δs
δB i
δs
= −τ N i . The three
(1.5.13)
131
are known as the Frenet-Serret formulas of differential geometry.
Surfaces and Curvature
Let us examine surfaces in a Cartesian frame of reference and then later we can generalize our results
to other coordinate systems. A surface in Euclidean 3-dimensional space can be defined in several different
ways. Explicitly, z = f (x, y), implicitly, F (x, y, z) = 0 or parametrically by defining a set of parametric
equations of the form
x = x(u, v),
y = y(u, v),
z = z(u, v)
which contain two independent parameters u, v called surface coordinates. For example, the equations
x = a sin θ cos φ,
y = a sin θ sin φ,
z = a cos θ
are the parametric equations which define a spherical surface of radius a with parameters u = θ and v = φ.
See for example figure 1.3-20 in section 1.3. By eliminating the parameters u, v one can derive the implicit
form of the surface and by solving for z one obtains the explicit form of the surface. Using the parametric
form of a surface we can define the position vector to a point on the surface which is then represented in
terms of the parameters u, v as
e2 + z(u, v) b
e3 .
~r = ~r(u, v) = x(u, v) b
e1 + y(u, v) b
(1.5.14)
The coordinates (u, v) are called the curvilinear coordinates of a point on the surface. The functions
x(u, v), y(u, v), z(u, v) are assumed to be real and differentiable such that
~r(u, c2 )
and
∂~
r
∂u
×
∂~
r
∂v
6= 0. The curves
~r(c1 , v)
(1.5.15)
with c1 , c2 constants, then define two surface curves called coordinate curves, which intersect at the surface
coordinates (c1 , c2 ). The family of curves defined by equations (1.5.15) with equally spaced constant values
ci , ci + ∆ci , ci + 2∆ci , . . . define a surface coordinate grid system. The vectors
∂~
r
∂u
∂~
r
and ∂v
evaluated at the
surface coordinates (c1 , c2 ) on the surface, are tangent vectors to the coordinate curves through the point
and are basis vectors for any vector lying in the surface. Letting (x, y, z) = (y 1 , y 2 , y 3 ) and (u, v) = (u1 , u2 )
and utilizing the summation convention, we can write the position vector in the form
ei .
~r = ~r(u1 , u2 ) = y i (u1 , u2 ) b
(1.5.16)
The tangent vectors to the coordinate curves at a point P can then be represented as the basis vectors
i
~ α = ∂~r = ∂y b
ei ,
E
α
α
∂u
∂u
α = 1, 2
(1.5.17)
where the partial derivatives are to be evaluated at the point P where the coordinate curves on the surface
intersect. From these basis vectors we construct a unit normal vector to the surface at the point P by
calculating the cross product of the tangent vector ~ru =
n
b=n
b(u, v) =
∂~
r
∂u
and ~rv =
∂~
r
∂v .
~1 × E
~2
~ru × ~rv
E
=
~
~
|~ru × ~rv |
|E1 × E2 |
A unit normal is then
(1.5.18)
132
~ 1, E
~ 2 and n
and is such that the vectors E
b form a right-handed system of coordinates.
If we transform from one set of curvilinear coordinates (u, v) to another set (ū, v̄), which are determined
by a set of transformation laws
u = u(ū, v̄),
v = v(ū, v̄),
the equation of the surface becomes
e2 + z(u(ū, v̄), v(ū, v̄)) b
e3
~r = ~r(ū, v̄) = x(u(ū, v̄), v(ū, v̄)) b
e1 + y(u(ū, v̄), v(ū, v̄)) b
and the tangent vectors to the new coordinate curves are
∂~r ∂u ∂~r ∂v
∂~r
=
+
∂ ū
∂u ∂ ū ∂v ∂ ū
and
∂~r
∂~r ∂u ∂~r ∂v
=
+
.
∂v̄
∂u ∂v̄
∂v ∂v̄
Using the indicial notation this result can be represented as
∂y i ∂uβ
∂y i
=
.
α
∂ ū
∂uβ ∂ ūα
This is the transformation law connecting the two systems of basis vectors on the surface.
A curve on the surface is defined by a relation f (u, v) = 0 between the curvilinear coordinates. Another
way to represent a curve on the surface is to represent it in a parametric form where u = u(t) and v = v(t),
where t is a parameter. The vector
∂~r du ∂~r dv
d~r
=
+
dt
∂u dt
∂v dt
is tangent to the curve on the surface.
An element of arc length with respect to the surface coordinates is represented by
ds2 = d~r · d~r =
where aαβ =
∂~
r
∂uα
·
∂~
r
∂uβ
∂~r
∂~r
·
duα duβ = aαβ duα duβ
α
∂u
∂uβ
(1.5.19)
with α, β = 1, 2 defines a surface metric. This element of arc length on the surface is
often written as the quadratic form
A = ds2 = E(du)2 + 2F du dv + G(dv)2 =
1
EG − F 2 2
(E du + F dv)2 +
dv
E
E
(1.5.20)
and called the first fundamental form of the surface. Observe that for ds2 to be positive definite the quantities
E and EG − F 2 must be positive.
The surface metric associated with the two dimensional surface is defined by
~α · E
~β =
aαβ = E
∂~r
∂~r
∂y i ∂y i
·
=
,
α
β
∂u
∂u
∂uα ∂uβ
α, β = 1, 2
(1.5.21)
with conjugate metric tensor aαβ defined such that aαβ aβγ = δγα . Here the surface is embedded in a three
dimensional space with metric gij and aαβ is the two dimensional surface metric. In the equation (1.5.20)
the quantities E, F, G are functions of the surface coordinates u, v and are determined from the relations
∂y i ∂y i
∂~r ∂~r
·
=
∂u ∂u
∂u1 ∂u1
∂y i ∂y i
∂~r ∂~r
·
=
=
∂u ∂v
∂u1 ∂u2
∂y i ∂y i
∂~r ∂~r
·
=
=
∂v ∂v
∂u2 ∂u2
E =a11 =
F =a12
G =a22
(1.5.22)
133
Here and throughout the remainder of this section, we adopt the convention that Greek letters have the
range 1,2, while Latin letters have the range 1,2,3.
Construct at a general point P on the surface the unit normal vector n
b at this point. Also construct a
plane which contains this unit surface normal vector n
b. Observe that there are an infinite number of planes
which contain this unit surface normal. For now, select one of these planes, then later on we will consider
all such planes. Let ~r = ~r(s) denote the position vector defining a curve C which is the intersection of the
selected plane with the surface, where s is the arc length along the curve, which is measured from some fixed
point on the curve. Let us find the curvature of this curve of intersection. The vector Tb = d~r , evaluated
ds
at the point P, is a unit tangent vector to the curve C and lies in the tangent plane to the surface at the
point P. Here we are using ordinary differentiation rather than intrinsic differentiation because we are in
a Cartesian system of coordinates. Differentiating the relation Tb · Tb = 1, with respect to arc length s we
find that Tb · dTb = 0 which implies that the vector dTb is perpendicular to the tangent vector Tb. Since the
ds
ds
coordinate system is Cartesian we can treat the curve of intersection C as a space curve, then the vector
~ = κ and radius of
~ = dTb , evaluated at point P, is defined as the curvature vector with curvature |K|
K
ds
b to the space curve is taken in the same direction as dTb so that the
curvature R = 1/κ. A unit normal N
ds
dTb
~
b
. Consider the geometry of figure 1.5-1
curvature will always be positive. We can then write K = κN =
ds
and define on the surface a unit vector u
b=n
b × Tb which is perpendicular to both the surface tangent vector
Tb and the surface normal vector n
b, such that the vectors T i ,ui and ni forms a right-handed system.
Figure 1.5-1 Surface curve with tangent plane and a normal plane.
134
~ 1 and E
~ 2 . Note that
The direction of u
b in relation to Tb is in the same sense as the surface tangents E
b
the vector ddsTb is perpendicular to the tangent vector Tb and lies in the plane which contains the vectors n
~
and u
b. We can therefore write the curvature vector K in the component form
b
~n +K
~g
~ = dT = κ(n) n
b + κ(g) u
b=K
K
ds
(1.5.23)
where κ(n) is called the normal curvature and κ(g) is called the geodesic curvature. The subscripts are not
indices. These curvatures can be calculated as follows. From the orthogonality condition n
b · Tb = 0 we obtain
b
db
n
dT
+ Tb ·
= 0. Consequently, the normal
by differentiation with respect to arc length s the result n
b·
ds
ds
curvature is determined from the dot product relation
d~r db
n
n
~ = κ(n) = −Tb · db
=− ·
.
n
b·K
ds
ds ds
(1.5.24)
By taking the dot product of u
b with equation (1.5.23) we find that the geodesic curvature is determined
from the triple scalar product relation
b·
κ(g) = u
dTb
dTb
= (b
n × Tb) ·
.
ds
ds
(1.5.25)
Normal Curvature
The equation (1.5.24) can be expressed in terms of a quadratic form by writing
n.
κ(n) ds2 = −d~r · db
(1.5.26)
The unit normal to the surface n
b and position vector ~r are functions of the surface coordinates u, v with
d~r =
∂~r
∂~r
du +
dv
∂u
∂v
and db
n=
∂b
n
∂b
n
du +
dv.
∂u
∂v
(1.5.27)
We define the quadratic form
B = −d~r · db
n=−
∂b
n
∂~r
∂b
n
∂~r
du +
dv ·
du +
dv
∂u
∂v
∂u
∂v
2
2
α
B = e(du) + 2f du dv + g(dv) = bαβ du du
where
n
∂~r ∂b
·
,
e=−
∂u ∂u
and bαβ
2f = −
∂~r ∂b
n ∂b
n ∂~r
·
+
·
∂u ∂v
∂u ∂v
,
(1.5.28)
β
g=−
n
∂~r ∂b
·
∂v ∂v
(1.5.29)
α, β = 1, 2 is called the curvature tensor and aαγ bαβ = bγβ is an associated curvature tensor.
The quadratic form of equation (1.5.28) is called the second fundamental form of the surface. Alternative
methods for calculating the coefficients of this quadratic form result from the following considerations. The
unit surface normal is perpendicular to the tangent vectors to the coordinate curves at the point P and
therefore we have the orthogonality relationships
∂~r
·n
b=0
∂u
and
∂~r
·n
b = 0.
∂v
(1.5.30)
135
Observe that by differentiating the relations in equation (1.5.30), with respect to both u and v, one can
derive the results
n
∂~r ∂b
∂ 2~r
·
= b11
·n
b=−
2
∂u
∂u ∂u
∂~r ∂b
n
∂b
n ∂~r
∂ 2~r
·n
b=−
·
=−
·
= b21 = b12
f=
∂u∂v
∂u ∂v
∂u ∂v
n
∂ 2~r
∂~r ∂b
·
= b22
g=
·n
b=−
2
∂v
∂v ∂v
and consequently the curvature tensor can be expressed as
e=
bαβ = −
(1.5.31)
∂~r
∂b
n
·
.
α
∂u
∂uβ
(1.5.32)
The quadratic forms from equations (1.5.20) and (1.5.28) enable us to represent the normal curvature
in the form of a ratio of quadratic forms. We find from equation (1.5.26) that the normal curvature in the
direction
du
dv
is
κ(n) =
e(du)2 + 2f du dv + g(dv)2
B
=
.
A
E(du)2 + 2F du dv + G(dv)2
(1.5.33)
r
∂~
r duα
If we write the unit tangent vector to the curve in the form Tb = d~
ds = ∂uα ds and express the derivative
n
∂b
n duβ
= ∂u
of the unit surface normal with respect to arc length as ddsb
β ds , then the normal curvature can be
expressed in the form
κ(n)
db
n
=−
= −Tb ·
ds
∂~r
∂b
n
·
∂uα ∂uβ
duα duβ
ds ds
(1.5.34)
bαβ duα duβ
bαβ duα duβ
=
.
=
ds2
aαβ duα duβ
Observe that the curvature tensor is a second order symmetric tensor.
In the previous discussions, the plane containing the unit normal vector was arbitrary. Let us now
consider all such planes that pass through this unit surface normal. As we vary the plane containing the unit
surface normal n
b at P we get different curves of intersection with the surface. Each curve has a curvature
associated with it. By examining all such planes we can find the maximum and minimum normal curvatures
associated with the surface. We write equation (1.5.33) in the form
κ(n) =
where λ =
dv
du .
e + 2f λ + gλ2
E + 2F λ + Gλ2
(1.5.35)
From the theory of proportions we can also write this equation in the form
κ(n) =
f + gλ
e + fλ
(e + f λ) + λ(f + gλ)
=
=
.
(E + F λ) + λ(F + Gλ)
F + Gλ
E + Fλ
(1.5.36)
Consequently, the curvature κ will satisfy the differential equations
(e − κE)du + (f − κF )dv = 0
and (f − κF )du + (g − κG)dv = 0.
The maximum and minimum curvatures occur in those directions λ where
dκ(n)
dλ
= 0. Calculating the deriva-
tive of κ(n) with respect to λ and setting the derivative to zero we obtain a quadratic equation in λ
(F g − Gf )λ2 + (Eg − Ge)λ + (Ef − F e) = 0,
(1.5.37)
(F g − Gf ) 6= 0.
136
This equation has two roots λ1 and λ2 which satisfy
λ1 + λ2 = −
Eg − Ge
F g − Gf
and
λ1 λ2 =
Ef − F e
,
F g − Gf
(1.5.38)
where F g − Gf 6= 0. The curvatures κ(1) ,κ(2) corresponding to the roots λ1 and λ2 are called the principal
curvatures at the point P. Several quantities of interest that are related to κ(1) and κ(2) are: (1) the principal
radii of curvature Ri = 1/κi ,i = 1, 2; (2) H =
1
2 (κ(1)
+ κ(2) ) called the mean curvature and K = κ(1) κ(2)
called the total curvature or Gaussian curvature of the surface. Observe that the roots λ1 and λ2 determine
two directions on the surface
∂~r
∂~r
d~r1
=
+
λ1
du
∂u ∂v
and
∂~r
∂~r
d~r2
=
+
λ2 .
du
∂u ∂v
If these directions are orthogonal we will have
∂~r
∂~r
∂~r
∂~r
d~r1 d~r2
·
=(
+
λ1 )(
+
λ2 ) = 0.
du du
∂u ∂v
∂u ∂v
This requires that
Gλ1 λ2 + F (λ1 + λ2 ) + E = 0.
(1.5.39)
It is left as an exercise to verify that this is indeed the case and so the directions determined by the principal
curvatures must be orthogonal. In the case where F g − Gf = 0 we have that F = 0 and f = 0 because the
coordinate curves are orthogonal and G must be positive. In this special case there are still two directions
determined by the differential equations (1.5.37) with dv = 0, du arbitrary, and du = 0, dv arbitrary. From
the differential equations (1.5.37) we find these directions correspond to
κ(1) =
We let λα =
duα
ds
e
E
and
κ(2) =
g
.
G
denote a unit vector on the surface satisfying aαβ λα λβ = 1. Then the equation (1.5.34)
can be written as κ(n) = bαβ λα λβ or we can write (bαβ − κ(n) aαβ )λα λβ = 0. The maximum and minimum
normal curvature occurs in those directions λα where
(bαβ − κ(n) aαβ )λα = 0
and so κ(n) must be a root of the determinant equation |bαβ − κ(n) aαβ | = 0 or
|aαγ bαβ − κ(n) δβγ | =
b11 − κ(n)
b21
b22
b
b12
= κ2(n) − bαβ aαβ κ(n) + = 0.
− κ(n)
a
(1.5.40)
This is a quadratic equation in κ(n) of the form κ2(n) − (κ(1) + κ(2) )κ(n) + κ(1) κ(2) = 0. In other words the
principal curvatures κ(1) and κ(2) are the eigenvalues of the matrix with elements bγβ = aαγ bαβ . Observe that
from the determinant equation in κ(n) we can directly find the total curvature or Gaussian curvature which
αγ
is an invariant given by K = κ(1) κ(2) = |bα
β | = |a bγβ | = b/a. The mean curvature is also an invariant
obtained from H =
1
2 (κ(1)
+ κ(2) ) =
1 αβ
bαβ ,
2a
where a = a11 a22 − a12 a21 and b = b11 b22 − b12 b21 are the
determinants formed from the surface metric tensor and curvature tensor components.
137
The equations of Gauss, Weingarten and Codazzi
~ and unit binormal
At each point on a space curve we can construct a unit tangent T~ , a unit normal N
~ The derivatives of these vectors, with respect to arc length, can also be represented as linear combinations
B.
~ , B.
~ See for example the Frenet-Serret formulas from equations (1.5.13). In a similar
of the base vectors T~ , N
b form a basis and the derivatives of these basis vectors with respect to
fashion the surface vectors ~ru , ~rv , n
b. For
the surface coordinates u, v can also be expressed as linear combinations of the basis vectors ~ru , ~rv , n
b. We can write
example, the derivatives ~ruu , ~ruv , ~rvv can be expressed as linear combinations of ~ru , ~rv , n
b
~ruu = c1~ru + c2~rv + c3 n
b
~ruv = c4~ru + c5~rv + c6 n
(1.5.41)
b
~rvv = c7~ru + c8~rv + c9 n
where c1 , . . . , c9 are constants to be determined. It is an easy exercise (see exercise 1.5, problem 8) to show
that these equations can be written in the indicial notation as
γ
∂~r
∂ 2~r
=
+ bαβ n
b.
α
β
α β ∂uγ
∂u ∂u
(1.5.42)
These equations are known as the Gauss equations.
In a similar fashion the derivatives of the normal vector can be represented as linear combinations of
the surface basis vectors. If we write
∂b
n
= c1~ru + c2~rv
∂u
∂b
n
= c3~ru + c4~rv
∂v
∂~r
∂b
n
∂b
n
= c∗1
+ c∗2
∂u
∂u
∂v
∂b
n
∂b
n
∂~r
= c∗3
+ c∗4
∂v
∂u
∂v
or
(1.5.43)
where c1 , . . . , c4 and c∗1 , . . . , c∗4 are constants. These equations are known as the Weingarten equations. It
is easily demonstrated (see exercise 1.5, problem 9) that the Weingarten equations can be written in the
indicial form
where bβα = aβγ bγα
∂~r
∂b
n
= −bβα β
∂uα
∂u
is the mixed second order form of the curvature tensor.
(1.5.44)
The equations of Gauss produce a system of partial differential equations defining the surface coordinates
i
x as a function of the curvilinear coordinates u and v. The equations are not independent as certain
compatibility conditions must be satisfied. In particular, it is required that the mixed partial derivatives
must satisfy
∂ 3~r
∂uα ∂uβ ∂uδ
=
∂ 3~r
∂uα ∂uδ ∂uβ
We calculate
∂ 3~r
=
∂uα ∂uβ ∂uδ
γ
αβ
∂ 2~r
+
∂uγ ∂uδ
∂
.
γ
αβ
∂uδ
∂bαβ
∂~r
∂b
n
+ bαβ δ +
n
b
∂uγ
∂u
∂uδ
and use the equations of Gauss and Weingarten to express this derivative in the form
 
ω
∂
αβ

 r
γ
γ
ω
∂bαβ
∂ 3~r
ω  ∂~

=
+
−
b
b
+
b
+
n
b.
αβ δ 
γδ
 ∂uδ
αβ
γδ
αβ
∂uα ∂uβ ∂uδ
∂uω
∂uδ
138
Forming the difference
∂ 3~r
∂uα ∂uβ ∂uδ
−
∂ 3~r
∂uα ∂uδ ∂uβ
we find that the coefficients of the independent vectors n
b and
∂~
r
∂uω
=0
must be zero. Setting the coefficient of n
b
equal to zero produces the Codazzi equations
γ
αβ
bγδ −
γ
αδ
bγβ +
∂bαβ
∂bαδ
−
= 0.
∂uδ
∂uβ
(1.5.45)
These equations are sometimes referred to as the Mainardi-Codazzi equations. Equating to zero the coefficient
of
∂~
r
∂uω
we find that Rδαγβ = bαβ bδγ − bαγ bδβ or changing indices we have the covariant form
aωδ Rδαβγ = Rωαβγ = bωβ bαγ − bωγ bαβ ,
where
Rδαγβ =
∂
∂uγ
δ
αβ
−
∂
∂uβ
δ
αγ
+
ω
αβ
δ
ωγ
(1.5.46)
−
ω
αγ
δ
ωβ
(1.5.47)
is the mixed Riemann curvature tensor.
EXAMPLE 1.5-1
Show that the Gaussian or total curvature K = κ(1) κ(2) depends only upon the metric aαβ and is
R1212
where a = det(aαβ ).
K=
a
Solution:
Utilizing the two-dimensional alternating tensor eαβ and the property of determinants we can write
eγδ K = eαβ bγα bδβ
where from page 137, K = |bγβ | = |aαγ bαβ |. Now multiply by eγζ and then contract on
ζ and δ to obtain
eγδ eγδ K = eγδ eαβ bγα bδβ = 2K
But eγδ aγµ aδν = aeµν so that
2K = eγδ eαβ (aγµ bαu ) aδν bβν
√
2K = eαβ a eµν bαµ bβν . Using aeµν = µν we have 2K = µν αβ bαµ bβν .
Interchanging indices we can write
2K = βγ ωα bωβ bαγ
and 2K = γβ ωα bωγ bαβ .
Adding these last two results we find that 4K = βγ ωγ (bωβ bαγ − bωγ bαβ ) = βγ ωγ Rωαβγ . Now multiply
βγ ωα
δλν Rωαβγ . From exercise 1.5, problem 16, the Riemann
both sides by στ λν to obtain 4Kστ λν = δστ
curvature tensor Rijkl is skew symmetric in the (i, j), (k, l) as well as being symmetric in the (ij), (kl) pair
βγ ωα
δλν Rωαβγ = 4Rλνστ and hence Rλνστ = Kστ λν and we have the special case
of indices. Consequently, δστ
√
√
b
R1212
. A much simpler way to obtain this result is to observe K =
where K ae12 ae12 = R1212 or K =
a
a
(bottom of page 137) and note from equation (1.5.46) that R1212 = b11 b22 − b12 b21 = b.
Note that on a surface ds2 = aαβ duα duβ where aαβ are the metrices for the surface. This metric is a
∂uα ∂uβ
and by taking determinants we find
tensor and satisfies āγδ = aαβ γ
∂ ū ∂ ūδ
ā = āγδ
∂uα ∂uβ
= aJ 2
∂ ūγ ∂ ūδ
139
where J is the Jacobian of the surface coordinate transformation. Here the curvature tensor for the surface
Rαβγδ has only one independent component since R1212 = R2121 = −R1221 = −R2112 (See exercises 20,21).
From the transformation law
R̄ηλµ = Rαβγδ
∂uα ∂uβ ∂uγ ∂uδ
∂ ū ∂ ūη ∂ ūλ ∂ ūµ
one can sum over the repeated indices and show that R̄1212 = R1212 J 2 and consequently
R1212
R̄1212
=
=K
ā
a
which shows that the Gaussian curvature is a scalar invariant in V2 .
Geodesic Curvature
~ associated with this curve, is
For C an arbitrary curve on a given surface the curvature vector K,
b and geodesic curvature κ(g) u
b and lies in a plane which
the vector sum of the normal curvature κ(n) n
is perpendicular to the tangent vector to the given curve on the surface. The geodesic curvature κ(g) is
obtained from the equation (1.5.25) and can be represented
κ(g)
dT~
dT~
~ =u
= (b
n × T~ ) ·
=
=u
b·K
b·
ds
ds
dT~
T~ ×
ds
!
·n
b.
Substituting into this expression the vectors
du
dv
d~r
= ~ru
+ ~rv
T~ =
ds
ds
ds
~
dT
~ = ~ruu (u 0 )2 + 2~ruv u 0 v 0 + ~rvv (v 0 )2 + ~ru u 00 + ~rv v 00 ,
=K
ds
where
0
=
d
ds ,
and by utilizing the results from problem 10 of the exercises following this section, we find
that the geodesic curvature can be represented as
κ(g)
2
2
1
0 3
=
(u ) + 2
−
(u 0 )2 v 0 +
11
12
11
p
1
2
1
(v 0 )3 + (u 0 v 00 − u 00 v 0 )
EG − F 2 .
−2
u 0 (v 0 )2 −
22
22
12
(1.5.48)
This equation indicates that the geodesic curvature is only a function of the surface metrices E, F, G and
the derivatives u 0 , v 0 , u 00 , v 00 . When the geodesic curvature is zero the curve is called a geodesic curve. Such
curves are often times, but not always, the lines of shortest distance between two points on a surface. For
example, the great circle on a sphere which passes through two given points on the sphere is a geodesic curve.
If you erase that part of the circle which represents the shortest distance between two points on the circle
you are left with a geodesic curve connecting the two points, however, the path is not the shortest distance
between the two points.
For plane curves we let u = x and v = y so that the geodesic curvature reduces to
kg = u 0 v 00 − u 00 v 0 =
dφ
ds
140
where φ is the angle between the tangent T~ to the curve and the unit vector b
e1 .
Geodesics are curves on the surface where the geodesic curvature is zero. Since kg = 0 along a geodesic
~ to the curve will be in the same
surface curve, then at every point on this surface curve the normal N
b = 0 and ~rv · n
b = 0 which reduces to
direction as the normal n
b to the surface. In this case, we have ~ru · n
dT~
· ~ru = 0
ds
since the vectors n
b and
~
dT
ds
and
dT~
· ~rv = 0,
ds
(1.5.49)
have the same direction. In particular, we may write
∂~r du ∂~r dv
d~r
=
+
= ~ru u0 + ~rv v 0
T~ =
ds
∂u ds
∂v ds
dT~
= ~ruu (u 0 )2 + 2~ruv u 0 v 0 + ~rvv (v 0 )2 + ~ru u 00 + ~rv v 00
ds
Consequently, the equations (1.5.49) become
dT~
· ~ru = (~ruu · ~ru ) (u 0 )2 + 2(~ruv · ~ru ) u 0 v 0 + (~rvv · ~ru ) (v 0 )2 + Eu 00 + F v 00 = 0
ds
.
dT~
0 2
0 0
0 2
00
00
· ~rv = (~ruu · ~rv ) (u ) + 2(~ruv · ~rv ) u v + (~rvv · ~rv ) (v ) + F u + Gv = 0.
ds
(1.5.50)
Utilizing the results from exercise 1.5,(See problems 4,5 and 6), we can eliminate v 00 from the equations
(1.5.50) to obtain
d2 u
+
ds2
1
11
du
ds
2
+2
1
12
du dv
+
ds ds
1
22
dv
ds
2
=0
and eliminating u 00 from the equations (1.5.50) produces the equation
d2 v
+
ds2
2
11
du
ds
2
+2
2
12
du dv
+
ds ds
2
22
dv
ds
2
= 0.
In tensor form, these last two equations are written
d2 uα
+
ds2
α
βγ
a
duβ duγ
= 0,
ds ds
α, β, γ = 1, 2
(1.5.51)
where u = u1 and v = u2 . The equations (1.5.51) are the differential equations defining a geodesic curve on
a surface. We will find that these same type of equations arise in considering the shortest distance between
two points in a generalized coordinate system. See for example problem 18 in exercise 2.2.
141
Tensor Derivatives
Let uα = uα (t) denote the parametric equations of a curve on the surface defined by the parametric
equations xi = xi (u1 , u2 ). We can then represent the surface curve in the spatial geometry since the surface
curve can be represented in the spatial coordinates through the representation xi = xi (u1 (t), u2 (t)) = xi (t).
Recall that for xi = xi (t) a given curve C , the intrinsic derivative of a vector field Ai along C is defined as
the inner product of the covariant derivative of the vector field with the tangent vector to the curve. This
intrinsic derivative is written
"
#
j
j
i
∂Ai
δAi
i dx
k dx
= A ,j
=
+
A
j
jk g
δt
dt
∂x
dt
or
dAi
δAi
=
+
δt
dt
i
jk
Ak
g
dxj
dt
where the subscript g indicates that the Christoffel symbol is formed from the spatial metric gij . If Aα is a
surface vector defined along the curve C, the intrinsic derivative is represented
α β
∂A
α
duβ
δAα
γ du
= Aα,β
=
+
A
βγ a
δt
dt
∂uβ
dt
or
dAα
δAα
=
+
δt
dt
α
βγ
Aγ
a
duβ
dt
where the subscript a denotes that the Christoffel is formed from the surface metric aαβ .
Similarly, the formulas for the intrinsic derivative of a covariant spatial vector Ai or covariant surface
vector Aα are given by
dAi
δAi
=
−
δt
dt
and
dAα
δAα
=
−
δt
dt
Consider a mixed tensor
Tαi
k
ij
γ
αβ
Ak
g
dxj
dt
Aα
a
duβ
.
dt
which is contravariant with respect to a transformation of space coordinates
i
x and covariant with respect to a transformation of surface coordinates uα . For Tαi defined over the surface
curve C, which can also be viewed as a space curve C, define the scalar invariant Ψ = Ψ(t) = Tαi Ai B α where
Ai is a parallel vector field along the curve C when it is viewed as a space curve and B α is also a parallel
vector field along the curve C when it is viewed as a surface curve. Recall that these parallel vector fields
must satisfy the differential equations
dAi
δAi
=
−
δt
dt
k
ij
Ak
g
dxj
= 0 and
dt
δB α
dB α
=
+
δt
dt
α
βγ
Bγ
a
duβ
= 0.
dt
(1.5.52)
The scalar invariant Ψ is a function of the parameter t of the space curve since both the tensor and the
parallel vector fields are to be evaluated along the curve C. By differentiating the function Ψ with respect
to the parameter t there results
dT i
dAi α
dB α
dΨ
= α Ai B α + Tαi
B + Tαi Ai
.
dt
dt
dt
dt
(1.5.53)
142
But the vectors Ai and B α are parallel vector fields and must satisfy the relations given by equations (1.5.52).
This implies that equation (1.5.53) can be written in the form
"
#
j
β
i
γ
dTαi
dΨ
k dx
i du
=
+
−
Ai B α .
T
T
k j g α dt
β α a γ dt
dt
dt
(1.5.54)
The quantity inside the brackets of equation (1.5.54) is defined as the intrinsic tensor derivative with respect
to the parameter t along the curve C. This intrinsic tensor derivative is written
j
dTαi
i
γ
duβ
δTαi
k dx
=
+
−
.
Tα
Tγi
dt
dt
kj g
dt
βα a
dt
(1.5.55)
The spatial representation of the curve C is related to the surface representation of the curve C through the
defining equations. Therefore, we can express the equation (1.5.55) in the form
"
#
j
β
i
γ
∂Tαi
δTαi
k ∂x
i du
=
+
T
−
T
k j g α ∂uβ
β α a γ dt
dt
∂uβ
(1.5.56)
The quantity inside the brackets is a mixed tensor which is defined as the tensor derivative of Tαi with
respect to the surface coordinates uβ . The tensor derivative of the mixed tensor Tαi with respect to the
surface coordinates uβ is written
i
Tα,β
∂Tαi
=
+
∂uβ
i
kj
Tαk
g
∂xj
−
∂uβ
γ
βα
Tγi .
a
i...j
which is contravariant with respect to transformations of the
In general, given a mixed tensor Tα...β
space coordinates and covariant with respect to transformations of the surface coordinates, then we can
define the scalar field along the surface curve C as
i...j
Ai · · · Aj B α · · · B β
Ψ(t) = Tα...β
(1.5.57)
where Ai , . . . , Aj and B α , . . . , B β are parallel vector fields along the curve C. The intrinsic tensor derivative
is then derived by differentiating the equation (1.5.57) with respect to the parameter t.
Tensor derivatives of the metric tensors gij , aαβ and the alternating tensors ijk , αβ and their associated
tensors are all zero. Hence, they can be treated as constants during the tensor differentiation process.
Generalizations
In a Riemannian space Vn with metric gij and curvilinear coordinates xi , i = 1, 2, 3, the equations of a
surface can be written in the parametric form xi = xi (u1 , u2 ) where uα , α = 1, 2 are called the curvilinear
coordinates of the surface. Since
dxi =
∂xi α
du
∂uα
(1.5.58)
then a small change duα on the surface results in change dxi in the space coordinates. Hence an element of
arc length on the surface can be represented in terms of the curvilinear coordinates of the surface. This same
element of arc length can also be represented in terms of the curvilinear coordinates of the space. Thus, an
element of arc length squared in terms of the surface coordinates is represented
ds2 = aαβ duα duβ
(1.5.59)
143
where aαβ is the metric of the surface. This same element when viewed as a spatial element is represented
ds2 = gij dxi dxj .
(1.5.60)
By equating the equations (1.5.59) and (1.5.60) we find that
gij dxi dxj = gij
∂xi ∂xj α β
du du = aαβ duα duβ .
∂uα ∂uβ
(1.5.61)
The equation (1.5.61) shows that the surface metric is related to the spatial metric and can be calculated
∂xi ∂xj
from the relation aαβ = gij α β . This equation reduces to the equation (1.5.21) in the special case of
∂u ∂u
Cartesian coordinates. In the surface coordinates we define the quadratic form A = aαβ duα duβ as the first
fundamental form of the surface. The tangent vector to the coordinate curves defining the surface are given
by
∂xi
∂uα
and can be viewed as either a covariant surface vector or a contravariant spatial vector. We define
this vector as
xiα =
∂xi
,
∂uα
i = 1, 2, 3,
α = 1, 2.
(1.5.62)
Any vector which is a linear combination of the tangent vectors to the coordinate curves is called a surface
vector. A surface vector Aα can also be viewed as a spatial vector Ai . The relation between the spatial
representation and surface representation is Ai = Aα xiα . The surface representation Aα , α = 1, 2 and the
spatial representation Ai , i = 1, 2, 3 define the same direction and magnitude since
gij Ai Aj = gij Aα xiα Aβ xjβ = gij xiα xjβ Aα Aβ = aαβ Aα Aβ .
Consider any two surface vectors Aα and B α and their spatial representations Ai and B i where
Ai = Aα xiα
and
B i = B α xiα .
(1.5.63)
These vectors are tangent to the surface and so a unit normal vector to the surface can be defined from the
cross product relation
ni AB sin θ = ijk Aj B k
(1.5.64)
where A, B are the magnitudes of Ai , B i and θ is the angle between the vectors when their origins are made
to coincide. Substituting equations (1.5.63) into the equation (1.5.64) we find
ni AB sin θ = ijk Aα xjα B β xkβ .
(1.5.65)
In terms of the surface metric we have AB sin θ = αβ Aα B β so that equation (1.5.65) can be written in the
form
(ni αβ − ijk xjα xkβ )Aα B β = 0
(1.5.66)
which for arbitrary surface vectors implies
ni αβ = ijk xjα xkβ
or ni =
1 αβ
ijk xjα xkβ .
2
(1.5.67)
The equation (1.5.67) defines a unit normal vector to the surface in terms of the tangent vectors to the
coordinate curves. This unit normal vector is related to the covariant derivative of the surface tangents as
144
is now demonstrated. By using the results from equation (1.5.50), the tensor derivative of equation (1.5.59),
with respect to the surface coordinates, produces
xiα,β
∂ 2 xi
=
+
∂uα ∂uβ
i
pq
xpα xqβ
g
−
σ
αβ
xiσ
(1.5.68)
a
where the subscripts on the Christoffel symbols refer to the metric from which they are calculated. Also the
tensor derivative of the equation (1.5.57) produces the result
gij xiα,γ xjβ + gij xiα xjβ,γ = aαβ,γ = 0.
(1.5.69)
Interchanging the indices α, β, γ cyclically in the equation (1.5.69) one can verify that
gij xiα,β xjγ = 0.
(1.5.70)
The equation (1.5.70) indicates that in terms of the space coordinates, the vector xiα,β is perpendicular to
the surface tangent vector xiγ and so must have the same direction as the unit surface normal ni . Therefore,
there must exist a second order tensor bαβ such that
bαβ ni = xiα,β .
(1.5.71)
By using the relation gij ni nj = 1 we can transform equation (1.5.71) to the form
bαβ = gij nj xiα,β =
1 γδ
ijk xiα,β xjγ xkδ .
2
(1.5.72)
The second order symmetric tensor bαβ is called the curvature tensor and the quadratic form
B = bαβ duα duβ
(1.5.73)
is called the second fundamental form of the surface.
Consider also the tensor derivative with respect to the surface coordinates of the unit normal vector to
the surface. This derivative is
ni, α =
∂ni
+
∂uα
i
jk
nj xkα .
(1.5.74)
g
Taking the tensor derivative of gij ni nj = 1 with respect to the surface coordinates produces the result
gij ni nj,α = 0 which shows that the vector nj,α is perpendicular to ni and must lie in the tangent plane to the
surface. It can therefore be expressed as a linear combination of the surface tangent vectors xiα and written
in the form
ni,α = ηαβ xiβ
(1.5.75)
where the coefficients ηαβ can be written in terms of the surface metric components aαβ and the curvature
components bαβ as follows. The unit vector ni is normal to the surface so that
gij ni xjα = 0.
(1.5.76)
145
The tensor derivative of this equation with respect to the surface coordinates gives
gij niβ xjα + gij ni xjα,β = 0.
(1.5.77)
Substitute into equation (1.5.77) the relations from equations (1.5.57), (1.5.71) and (1.5.75) and show that
bαβ = −aαγ ηβγ .
(1.5.78)
Solving the equation (1.5.78) for the coefficients ηβγ we find
ηβγ = −aαγ bαβ .
(1.5.79)
Now substituting equation (1.5.79) into the equation (1.5.75) produces the Weingarten formula
ni,α = −aγβ bγα xiβ .
(1.5.80)
This is a relation for the derivative of the unit normal in terms of the surface metric, curvature tensor and
surface tangents.
A third fundamental form of the surface is given by the quadratic form
C = cαβ duα duβ
(1.5.81)
where cαβ is defined as the symmetric surface tensor
cαβ = gij ni,α nj,β .
(1.5.82)
By using the Weingarten formula in the equation (1.5.81) one can verify that
cαβ = aγδ bαγ bβδ .
(1.5.83)
Geodesic Coordinates
In a Cartesian coordinate system the metric tensor gij is a constant and consequently the Christoffel
symbols are zero at all points of the space. This is because the Christoffel symbols are dependent upon
the derivatives of the metric tensor which is constant. If the space VN is not Cartesian then the Christoffel
symbols do not vanish at all points of the space. However, it is possible to find a coordinate system where
the Christoffel symbols will all vanish at a given point P of the space. Such coordinates are called geodesic
coordinates of the point P.
Consider a two dimensional surface with surface coordinates uα and surface metric aαβ . If we transform
to some other two dimensional coordinate system, say ūα with metric āαβ , where the two coordinates are
related by transformation equations of the form
uα = uα (ū 1 , ū 2 ),
α = 1, 2,
(1.5.84)
146
then from the transformation equation (1.4.7) we can write, after changing symbols,
α
∂ 2 uα
∂uα
∂uδ ∂u
δ
=
+
.
δ a ∂ ū β ∂ ū γ
∂ ū β ∂ ū γ
β γ ā ∂ ū δ
(1.5.85)
This is a relationship between the Christoffel symbols in the two coordinate systems. If
a point P , then for that particular point the equation (1.5.85) reduces to
α
∂uδ ∂u
∂ 2 uα
=
−
β
γ
δ a ∂ ū β ∂ ū γ
∂ ū ∂ ū
δ
βγ
vanishes at
ā
(1.5.86)
where all terms are evaluated
the point P. Conversely, if the equation (1.5.86) is satisfied at the point P,
at δ
must be zero at this point. Consider the special coordinate transformathen the Christoffel symbol β γ
ā
tion
α
uα = uα
0 + ū −
1 α
ū β ū α
2 βγ a
(1.5.87)
where uα
0 are the surface coordinates of the point P. The point P in the new coordinates is given by
ū α = 0. We now differentiate the relation (1.5.87) to see if it satisfies the equation (1.5.86). We calculate
the derivatives
1 α
1 α
∂uα
α
β
=
δ
−
ū
−
ūγ
τ
∂ ū τ
2 βτ a
2 τγ a
α
∂ 2 uα
=−
τσ a
∂ ū τ ∂ ū σ
and
uα =0
uα =0
(1.5.88)
(1.5.89)
where these derivative are evaluated at ū α = 0. We find the derivative equations (1.5.88) and (1.5.89) do
satisfy the equation (1.5.86) locally at the point P. Hence, the Christoffel symbols will all be zero at this
particular point. The new coordinates can then be called geodesic coordinates.
Riemann Christoffel Tensor
Consider the Riemann Christoffel tensor defined by the equation (1.4.33). Various properties of this
tensor are derived in the exercises at the end of this section. We will be particularly interested in the
Riemann Christoffel tensor in a two dimensional space with metric aαβ and coordinates uα . We find the
Riemann Christoffel tensor has the form
δ
δ
τ
δ
τ
δ
∂
∂
δ
−
+
−
R. αβγ =
∂uβ α γ
∂uγ α β
αγ
βτ
αβ
γτ
(1.5.90)
where the Christoffel symbols are evaluated with respect to the surface metric. The above tensor has the
associated tensor
Rσαβγ = aσδ R.δαβγ
(1.5.91)
which is skew-symmetric in the indices (σ, α) and (β, γ) such that
Rσαβγ = −Rασβγ
and
Rσαβγ = −Rσαγβ .
(1.5.92)
The two dimensional alternating tensor is used to define the constant
K=
1 αβ γδ
Rαβγδ
4
(1.5.93)
147
(see example 1.5-1) which is an invariant of the surface and called the Gaussian curvature or total curvature.
In the exercises following this section it is shown that the Riemann Christoffel tensor of the surface can be
expressed in terms of the total curvature and the alternating tensors as
Rαβγδ = Kαβ γδ .
(1.5.94)
Consider the second tensor derivative of xrα which is given by
xrα,βγ =
∂xrα,β
+
∂uγ
r
mn
xrα,β xnγ −
g
δ
αγ
xrδ,β −
a
δ
βγ
xrα,γ
(1.5.95)
a
which can be shown to satisfy the relation
xrα,βγ − xrα,γβ = Rδ.αβγ xrδ .
(1.5.96)
Using the relation (1.5.96) we can now derive some interesting properties relating to the tensors aαβ , bαβ ,
cαβ ,
Rαβγδ , the mean curvature H and the total curvature K.
Consider the tensor derivative of the equation (1.5.71) which can be written
xiα,βγ = bαβ,γ ni + bαβ ni,γ
where
bαβ,γ =
∂bαβ
−
∂uα
σ
αγ
bσβ −
a
σ
βγ
(1.5.97)
bασ .
(1.5.98)
a
By using the Weingarten formula, given in equation (1.5.80), the equation (1.5.97) can be expressed in the
form
xiα,βγ = bαβ,γ ni − bαβ aτ σ bτ γ xiσ
(1.5.99)
and by using the equations (1.5.98) and (1.5.99) it can be established that
xrα,βγ − xrα,γβ = (bαβ,γ − bαγ,β )nr − aτ δ (bαβ bτ γ − bαγ bτ β )xrδ .
(1.5.100)
Now by equating the results from the equations (1.5.96) and (1.5.100) we arrive at the relation
Rδ.αβγ xrδ = (bαβ,γ − bαγ,β )nr − aτ δ (bαβ bτ γ − bαγ bτ β )xrδ .
(1.5.101)
Multiplying the equation (1.5.101) by nr and using the results from the equation (1.5.76) there results the
Codazzi equations
bαβ,γ − bαγ,β = 0.
(1.5.102)
Multiplying the equation (1.5.101) by grm xm
σ and simplifying one can derive the Gauss equations of the
surface
Rσαβγ = bαγ bσβ − bαβ bσγ .
(1.5.103)
By using the Gauss equations (1.5.103) the equation (1.5.94) can be written as
Kσα βγ = bαγ bσβ − bαβ bσγ .
(1.5.104)
148
Another form of equation (1.5.104) is obtained by using the equation (1.5.83) together with the relation
aαβ = −aσγ σα βγ . It is left as an exercise to verify the resulting form
−Kaαβ = cαβ − aσγ bσγ bαβ .
(1.5.106)
Define the quantity
H=
1 σγ
a bσγ
2
(1.5.107)
as the mean curvature of the surface, then the equation (1.5.106) can be written in the form
cαβ − 2H bαβ + K aαβ = 0.
(1.5.108)
By multiplying the equation (1.5.108) by duα duβ and summing, we find
C − 2H B + K A = 0
(1.5.109)
is a relation connecting the first, second and third fundamental forms.
EXAMPLE 1.5-2
In a two dimensional space the Riemann Christoffel tensor has only one nonzero independent component
R1212 . ( See Exercise 1.5, problem number 21.) Consequently, the equation (1.5.104) can be written in the
√
√
form K ae12 ae12 = b22 b11 − b21 b12 and solving for the Gaussian curvature K we find
K=
R1212
b
b22 b11 − b12 b21
.
= =
a11 a22 − a12 a21
a
a
(1.5.110)
Surface Curvature
For a surface curve uα = uα (s),α = 1, 2 lying upon a surface xi = xi (u1 , u2 ),i = 1, 2, 3, we have a two
duα
is a unit tangent vector to
dimensional space embedded in a three dimensional space. Thus, if tα =
ds
α
β
du du
= aαβ tα tβ = 1. This same vector can be represented as the unit tangent
the surface curve then aαβ
ds ds
dxi
dxi dxj
. That is we will have gij
= gij T i T j = 1.
vector to the space curve xi = xi (u1 (s), u2 (s)) with T i =
ds
ds ds
The surface vector tα and the space vector T i are related by
∂xi duα
= xiα tα .
∂uα ds
Ti =
(1.5.111)
The surface vector tα is a unit vector so that aαβ tα tβ = 1. If we differentiate this equation intrinsically with
β
respect to the parameter s, we find that aαβ tα δtδs = 0. This shows that the surface vector
α
δtα
δs
is perpendicular
α
to the surface vector t . Let u denote a unit normal vector in the surface plane which is orthogonal to the
tangent vector tα . The direction of uα is selected such that αβ tα uβ = 1. Therefore, there exists a scalar κ(g)
such that
δtα
= κ(g) uα
δs
(1.5.112)
149
where κ(g) is called the geodesic curvature of the curve. In a similar manner it can be shown that
is a surface vector orthogonal to tα . Let
δuα
δs
δuα
δs
= αtα where α is a scalar constant to be determined. By
differentiating the relation aαβ tα uβ = 0 intrinsically and simplifying we find that α = −κ(g) and therefore
δuα
= −κ(g) tα .
δs
(1.5.113)
The equations (1.5.112) and (1.5.113) are sometimes referred to as the Frenet-Serret formula for a curve
relative to a surface.
Taking the intrinsic derivative of equation (1.5.111), with respect to the parameter s, we find that
δtα
duβ α
δT i
= xiα
+ xiα,β
t .
δs
δs
ds
(1.5.114)
Treating the curve as a space curve we use the Frenet formulas (1.5.13). If we treat the curve as a surface
curve, then we use the Frenet formulas (1.5.112) and (1.5.113). In this way the equation (1.5.114) can be
written in the form
κN i = xiα κ(g) uα + xiα,β tβ tα .
(1.5.115)
By using the results from equation (1.5.71) in equation (1.5.115) we obtain
κN i = κ(g) ui + bαβ ni tα tβ
(1.5.116)
where ui is the space vector counterpart of the surface vector uα . Let θ denote the angle between the surface
normal ni and the principal normal N i , then we have that cos θ = ni N i . Hence, by multiplying the equation
(1.5.116) by ni we obtain
κ cos θ = bαβ tα tβ .
(1.5.117)
Consequently, for all curves on the surface with the same tangent vector tα , the quantity κ cos θ will remain
constant. This result is known as Meusnier’s theorem. Note also that κ cos θ = κ(n) is the normal component
of the curvature and κ sin θ = κ(g) is the geodesic component of the curvature. Therefore, we write the
equation (1.5.117) as
κ(n) = bαβ tα tβ
(1.5.118)
which represents the normal curvature of the surface in the direction tα . The equation (1.5.118) can also be
written in the form
κ(n) = bαβ
B
duα duβ
=
ds ds
A
(1.5.119)
which is a ratio of quadratic forms.
The surface directions for which κ(n) has a maximum or minimum value is determined from the equation
(1.5.119) which is written as
(bαβ − κ(n) aαβ )λα λβ = 0.
(1.5.120)
The direction giving a maximum or minimum value to κ(n) must then satisfy
(bαβ − κ(n) aαβ )λβ = 0
(1.5.121)
150
so that κ(n) must be a root of the determinant equation
det(bαβ − κ(n) aαβ ) = 0.
(1.5.122)
The expanded form of equation (1.5.122) can be written as
κ2(n) − aαβ bαβ κ(n) +
b
=0
a
(1.5.123)
where a = a11 a22 − a12 a21 and b = b11 b22 − b12 b21 . Using the definition given in equation (1.5.107) and using
the result from equation (1.5.110), the equation (1.5.123) can be expressed in the form
κ2(n) − 2H κ(n) + K = 0.
(1.5.124)
The roots κ(1) and κ(2) of the equation (1.5.124) then satisfy the relations
H=
1
(κ(1) + κ(2) )
2
(1.5.125)
and
K = κ(1) κ(2) .
(1.5.126)
Here H is the mean value of the principal curvatures and K is the Gaussian or total curvature which is the
product of the principal curvatures. It is readily verified that
H=
Eg − 2f F + eG
2(EG − F 2 )
and K =
eg − f 2
EG − F 2
are invariants obtained from the surface metric and curvature tensor.
Relativity
Sir Isaac Newton and Albert Einstein viewed the world differently when it came to describing gravity and
the motion of the planets. In this brief introduction to relativity we will compare the Newtonian equations
with the relativistic equations in describing planetary motion. We begin with an examination of Newtonian
systems.
Newton’s viewpoint of planetary motion is a multiple bodied problem, but for simplicity we consider
only a two body problem, say the sun and some planet where the motion takes place in a plane. Newton’s
law of gravitation states that two masses m and M are attracted toward each other with a force of magnitude
GmM
ρ2 ,
where G is a constant, ρ is the distance between the masses, m is the mass of the planet and M is the
mass of the sun. One can construct an x, y plane containing the two masses with the origin located at the
e1 + sin φ b
e2 denote a unit vector at the origin of this coordinate
center of mass of the sun. Let b
eρ = cos φ b
system and pointing in the direction of the mass m. The vector force of attraction of mass M on mass m is
given by the relation
−GmM
b
eρ .
F~ =
ρ2
(1.5.127)
151
Figure 1.5-2. Parabolic and elliptic conic sections
The equation of motion of mass m with respect to mass M is obtained from Newton’s second law. Let
ρ
~ = ρb
eρ denote the position vector of mass m with respect to the origin. Newton’s second law can then be
written in any of the forms
~
−GmM
d2 ρ
dV
~
−GmM
b
=
e
=
m
=
m
ρ~
F~ =
ρ
ρ2
dt2
dt
ρ3
(1.5.128)
and from this equation we can show that the motion of the mass m can be described as a conic section.
Recall that a conic section is defined as a locus of points p(x, y) such that the distance of p from a fixed
point (or points), called a focus (foci), is proportional to the distance of the point p from a fixed line, called
a directrix, that does not contain the fixed point. The constant of proportionality is called the eccentricity
and is denoted by the symbol . For = 1 a parabola results; for 0 ≤ ≤ 1 an ellipse results; for > 1 a
hyperbola results; and if = 0 the conic section is a circle.
With reference to figure 1.5-2, a conic section is defined in terms of the ratio
FP
PD
= where F P = ρ and
P D = 2q − ρ cos φ. From the ratio we solve for ρ and obtain the polar representation for the conic section
ρ=
p
1 + cos φ
(1.5.129)
152
where p = 2q and the angle φ is known as the true anomaly associated with the orbit. The quantity p is
called the semi-parameter of the conic section. (Note that when φ =
π
2,
then ρ = p.) A more general form
of the above equation is
ρ=
p
1 + cos(φ − φ0 )
or u =
1
= A[1 + cos(φ − φ0 )],
ρ
(1.5.130)
where φ0 is an arbitrary starting anomaly. An additional symbol a, known as the semi-major axes of an
elliptical orbit can be introduced where q, p, , a are related by
p
= q = a(1 − )
1+
or p = a(1 − 2 ).
(1.5.131)
To show that the equation (1.5.128) produces a conic section for the motion of mass m with respect to
mass M we will show that one form of the solution of equation (1.5.128) is given by the equation (1.5.129).
To verify this we use the following vector identities:
ρ
~× b
eρ =0
d~
ρ
d2 ρ
~
d
ρ
~×
=~
ρ× 2
dt
dt
dt
db
eρ
b
=0
eρ ·
dt
db
eρ
db
eρ
b
eρ ×
.
eρ × b
=−
dt
dt
(1.5.132)
From the equation (1.5.128) we find that
d
dt
GM
~
d~
ρ
d2 ρ
~× b
eρ = ~0
ρ
~×
=ρ
~× 2 =− 2 ρ
dt
dt
ρ
(1.5.133)
so that an integration of equation (1.5.133) produces
ρ
~×
d~
ρ ~
= h = constant.
dt
(1.5.134)
ρ
~
~ = ρ
~ = ρ
The quantity H
~ × mV
~ × m d~
dt is the angular momentum of the mass m so that the quantity h
represents the angular momentum per unit mass. The equation (1.5.134) tells us that ~h is a constant for our
two body system. Note that because ~h is constant we have
~
GM
d~
ρ
d ~ ~ dV
~
V ×h =
×h=− 2 b
eρ × ρ
~×
dt
dt
ρ
dt
dρ
db
eρ
GM
b
+
eρ )]
eρ × [~
ρb
eρ × (ρ
=− 2 b
ρ
dt
dt
db
eρ 2
db
eρ
GM
)ρ = GM
eρ × ( b
eρ ×
=− 2 b
ρ
dt
dt
and consequently an integration produces
~
~ × ~h = GM b
V
eρ + C
153
~ is a vector constant of integration. The triple scalar product formula gives us
where C
d~
ρ
~
~ × ~h) = ~h · (~
~· b
eρ + ρ~ · C
ρ~ · (V
ρ × ) = h2 = GM ρ
dt
or
h2 = GM ρ + Cρ cos φ
(1.5.135)
~ and ρ~. From the equation (1.5.135) we find that
where φ is the angle between the vectors C
ρ=
p
1 + cos φ
(1.5.136)
where p = h2 /GM and = C/GM. This result is known as Kepler’s first law and implies that when < 1
the mass m describes an elliptical orbit with the sun at one focus.
We present now an alternate derivation of equation (1.5.130) for later use. From the equation (1.5.128)
we have
2
d
~
d~
ρ d2 ρ
· 2 =
dt dt
dt
d~
ρ d~
ρ
·
dt dt
= −2
GM d
d~
ρ
GM
=− 3
(~
ρ·ρ
~) .
ρ
~·
3
ρ
dt
ρ dt
(1.5.137)
Consider the equation (1.5.137) in spherical coordinates ρ, θ, φ. The tensor velocity components are V 1 =
V2 =
dθ
dt ,
V3 =
dφ
dt
and the physical components of velocity are given by Vρ =
dρ
dt ,
Vθ = ρ dθ
dt , Vφ =
dρ
dt ,
ρ sin θ dφ
dt
so that the velocity can be written
dρ
dθ
dφ
d~
ρ
b
b
=
eρ + ρ b
eθ + ρ sin θ
eφ .
V~ =
dt
dt
dt
dt
(1.5.138)
Substituting equation (1.5.138) into equation (1.5.137) gives the result
" 2 #
2
2
dρ
dφ
d 1
dθ
2GM dρ
GM d 2
d
2
2
2
(ρ ) = − 2
= 2GM
+ρ
+ ρ sin θ
=− 3
dt
dt
dt
dt
ρ dt
ρ dt
dt ρ
which can be integrated directly to give
dρ
dt
2
+ ρ2
dθ
dt
2
+ ρ2 sin2 θ
dφ
dt
2
=
2GM
−E
ρ
where −E is a constant of integration. In the special case of a planar orbit we set θ =
(1.5.139)
π
2
constant so that
the equation (1.5.139) reduces to
dρ
dt
dρ dφ
dφ dt
2
+ρ
2
+ρ
2
2
dφ
dt
dφ
dt
2
=
2
2GM
−E
ρ
(1.5.140)
2GM
− E.
=
ρ
Also for this special case of planar motion we have
|~
ρ×
By eliminating
dφ
dt
dφ
d~
ρ
| = ρ2
= h.
dt
dt
(1.5.141)
from the equation (1.5.140) we obtain the result
dρ
dφ
2
+ ρ2 =
2GM 3
E
ρ − 2 ρ4 .
h2
h
(1.5.142)
154
Figure 1.5-3. Relative motion of two inertial systems.
The substitution ρ =
1
u
can be used to represent the equation (1.5.142) in the form
du
dφ
2
+ u2 −
2GM
E
u+ 2 =0
2
h
h
(1.5.143)
which is a form we will return to later in this section. Note that we can separate the variables in equations
(1.5.142) or (1.5.143). The results can then be integrate to produce the equation (1.5.130).
Newton also considered the relative motion of two inertial systems, say S and S. Consider two such
systems as depicted in the figure 1.5-3 where the S system is moving in the x−direction with speed v relative
to the system S.
For a Newtonian system, if at time t = 0 we have clocks in both systems which coincide, than at time t
a point P (x, y, z) in the S system can be described by the transformation equations
x =x − vt
x =x + vt
y =y
y =y
or
(1.5.144)
z =z
z =z
t =t
t =t.
These are the transformation equation of Newton’s relativity sometimes referred to as a Galilean transformation.
Before Einstein the principle of relativity required that velocities be additive and obey Galileo’s velocity
addition rule
VP/R = VP/Q + VQ/R .
(1.5.145)
155
That is, the velocity of P with respect to R equals the velocity of P with respect to Q plus the velocity of Q
with respect to R. For example, a person (P ) running north at 3 km/hr on a train (Q) moving north at 60
km/hr with respect to the ground (R) has a velocity of 63 km/hr with respect to the ground. What happens
when (P ) is a light wave moving on a train (Q) which is moving with velocity V relative to the ground? Are
the velocities still additive? This type of question led to the famous Michelson-Morley experiment which
has been labeled as the starting point for relativity. Einstein’s answer to the above question was ”NO” and
required that VP/R = VP/Q = c =speed of light be a universal constant.
In contrast to the Newtonian equations, Einstein considered the motion of light from the origins 0 and
0 of the systems S and S. If the S system moves with velocity v relative to the S system and at time t = 0
a light signal is sent from the S system to the S system, then this light signal will move out in a spherical
wave front and lie on the sphere
x2 + y 2 + z 2 = c2 t2
(1.5.146)
where c is the speed of light. Conversely, if a light signal is sent out from the S system at time t = 0, it will
lie on the spherical wave front
2
x2 + y 2 + z 2 = c2 t .
(1.5.147)
Observe that the Newtonian equations (1.5.144) do not satisfy the equations (1.5.146) and (1.5.147) identically. If y = y and z = z then the space variables (x, x) and time variables (t, t) must somehow be related.
Einstein suggested the following transformation equations between these variables
x = γ(x − vt) and x = γ(x + vt)
(1.5.148)
where γ is a constant to be determined. The differentials of equations (1.5.148) produce
dx = γ(dx − vdt)
and dx = γ(dx + vdt)
(1.5.149)
from which we obtain the ratios
γ(dx − v dt)
dx
=
dx
γ(dx + v dt)
or
1
γ(1 +
v
dx )
= γ(1 −
dt
When
v
dx
dt
).
(1.5.150)
dx
dx
=
= c, the speed of light, the equation (1.5.150) requires that
dt
dt
γ 2 = (1 −
v 2 −1
)
c2
or γ = (1 −
v 2 −1/2
)
.
c2
(1.5.151)
From the equations (1.5.148) we eliminate x and find
t = γ(t −
v
x).
c2
(1.5.152)
We can now replace the Newtonian equations (1.5.144) by the relativistic transformation equations
x =γ(x − vt)
x =γ(x + vt)
y =y
y =y
or
z =z
t =γ(t +
v
x)
c2
(1.5.153)
z =z
t =γ(t −
v
x)
c2
156
where γ is given by equation (1.5.151). These equations are also known as the Lorentz transformation.
v
Note that for v << c, then 2 ≈ 0, γ ≈ 1 , then the equations (1.5.153) closely approximate the equations
c
(1.5.144). The equations (1.5.153) also satisfy the equations (1.5.146) and (1.5.147) identically as can be
readily verified by substitution. Further, by using chain rule differentiation we obtain from the relations
(1.5.148) that
dx
=
dt
dx
dt
1+
+v
dx
dt
.
(1.5.154)
v
c c
The equation (1.5.154) is the Einstein relative velocity addition rule which replaces the previous Newtonian
rule given by equation (1.5.145). We can rewrite equation (1.5.154) in the notation of equation (1.5.145) as
VP/R =
VP/Q + VQ/R
1+
VP/Q VQ/R
c
c
.
(1.5.155)
Observe that when VP/Q << c and VQ/R << c then equation (1.5.155) approximates closely the equation
(1.5.145). Also as VP/Q and VQ/R approach the speed of light we have
lim
VP/Q + VQ/R
VP/Q →C
VQ/R →C
VP/Q VQ/R
c
c
1+
=c
(1.5.156)
which agrees with Einstein’s hypothesis that the speed of light is an invariant.
Let us return now to the viewpoint of what gravitation is. Einstein thought of space and time as being
related and viewed the motion of the planets as being that of geodesic paths in a space-time continuum.
Recall the equations of geodesics are given by
d2 xi
+
ds2
i
jk
dxj dxk
= 0,
ds ds
(1.5.157)
where s is arc length. These equations are to be associated with a 4-dimensional space-time metric gij
where the indices i, j take on the values 1, 2, 3, 4 and the xi are generalized coordinates. Einstein asked
the question, ”Can one introduce a space-time metric gij such that the equations (1.5.157) can somehow
reproduce the law of gravitational attraction
d2 ρ
~
dt2
+
GM
~
ρ3 ρ
= 0?” Then the motion of the planets can be
viewed as optimized motion in a space-time continuum where the metrices of the space simulate the law of
gravitational attraction. Einstein thought that this motion should be related to the curvature of the space
which can be obtained from the Riemann-Christoffel tensor Rijkl . The metric we desire gij , i, j = 1, 2, 3, 4
has 16 components. The conjugate metric tensor g ij is defined such that g ij gjk = δki and an element of
arc length squared is given by ds2 = gij dxi dxj . Einstein thought that the metrices should come from the
Riemann-Christoffel curvature tensor which, for n = 4 has 256 components, but only 20 of these are linearly
independent. This seems like a large number of equations from which to obtain the law of gravitational
attraction and so Einstein considered the contracted tensor
n
n
m
n
m
n
∂
∂
−
+
−
.
Gij = Rtijt =
∂xj i n
∂xn i j
in
mj
ij
mn
Spherical coordinates (ρ, θ, φ) suggests a metric similar to
ds2 = −(dρ)2 − ρ2 (dθ)2 − ρ2 sin2 θ(dφ)2 + c2 (dt)2
(1.5.158)
157
where g11 = −1, g22 = −ρ2 , g33 = −ρ2 sin2 θ, g44 = c2 and gij = 0 for i 6= j. The negative signs are
2
= c2 − v 2 is positive when v < c and the velocity is not greater than c. However,
introduced so that ds
dt
this metric will not work since the curvature tensor vanishes. The spherical symmetry of the problem suggest
that g11 and g44 change while g22 and g33 remain fixed. Let (x1 , x2 , x3 , x4 ) = (ρ, θ, φ, t) and assume
g11 = −eu ,
g33 = −ρ2 sin2 θ,
g22 = −ρ2 ,
g44 = ev
(1.5.159)
where u and v are unknown functions of ρ to be determined. This gives the conjugate metric tensor
g 11 = −e−u ,
g 22 =
−1
,
ρ2
g 33 =
ρ2
−1
,
sin2 θ
g 44 = e−v
(1.5.160)
and g ij = 0 for i 6= j. This choice of a metric produces
ds2 = −eu (dρ)2 − ρ2 (dθ)2 − ρ2 sin2 θ(dφ)2 + ev (dt)2
(1.5.161)
together with the nonzero Christoffel symbols
1
11
1
22
1
33
1
44
1 du
2 dρ
= − ρe−u
= − ρe−u sin2 θ
=
dv
1
= ev−u
2
dr
2
12
2
21
2
33
1
ρ
1
=
ρ
= − sin θ cos θ
=
3
13
3
23
3
31
3
32
=
1
ρ
=
cos θ
sin θ
=
1
ρ
=
cos θ
sin θ
4
14
4
41
=
1 dv
2 dρ
=
1 dv
.
2 dρ
(1.5.162)
The equation (1.5.158) is used to calculate the nonzero Gij and we find that
2
1 du
1 d2 v
1 dv
1 du dv
−
+
−
2
2 dρ
4 dρ
4 dρ dρ ρ dρ
dv
1
du
1
− ρ
− eu
=e−u 1 + ρ
2 dρ 2 dρ
1 du
1 dv
− ρ
− eu sin2 θ
=e−u 1 + ρ
2 dρ 2 dρ
!
2
2
d
du
dv
1
dv
dv
1
1
v
1
+
= − ev−u
−
+
2 dρ2
4 dρ dρ 4 dρ
ρ dρ
G11 =
G22
G33
G44
(1.5.163)
and Gij = 0 for i 6= j. The assumption that Gij = 0 for all i, j leads to the differential equations
d2 v 1
+
dρ2
2
dv
dρ
2
2 du
1 du dv
−
=0
2 dρ dρ ρ dρ
1 du
1 dv
− ρ
− eu =0
1+ ρ
2 dρ 2 dρ
2
2 dv
1 dv
1 du dv
d2 v
+
=0.
+
−
2
dρ
2 dρ
2 dρ dρ ρ dρ
−
(1.5.164)
158
Subtracting the first equation from the third equation gives
du dv
+
=0
dρ dρ
or u + v = c1 = constant.
(1.5.165)
du
= 1 − eu
dρ
(1.5.166)
The second equation in (1.5.164) then becomes
ρ
Separate the variables in equation (1.5.166) and integrate to obtain the result
eu =
1
1 − cρ2
(1.5.167)
where c2 is a constant of integration and consequently
v
e =e
c1 −u
c2
=e
.
1−
ρ
c1
(1.5.168)
The constant c1 is selected such that g44 approaches c2 as ρ increases without bound. This produces the
metrices
g11 =
−1
,
1 − cρ2
g22 = −ρ2 ,
g33 = −ρ2 sin2 θ,
g44 = c2 (1 −
c2
)
ρ
(1.5.169)
where c2 is a constant still to be determined. The metrices given by equation (1.5.169) are now used to
expand the equations (1.5.157) representing the geodesics in this four dimensional space. The differential
equations representing the geodesics are found to be
1 du
d2 ρ
+
ds2
2 dρ
dρ
ds
2
− ρe−u
2 dθ dρ
d2 θ
+
− sin θ cos θ
ds2
ρ ds ds
dθ
ds
dφ
ds
2
2
− ρe−u sin2 θ
dφ
ds
2
+
1 v−u dv
e
2
dρ
dt
ds
2
=0
=0
(1.5.170)
(1.5.171)
d2 φ
2 dφ dρ
cos θ dφ dθ
+
+2
=0
ds2
ρ ds ds
sin θ ds ds
(1.5.172)
d2 t
dv dt dρ
+
= 0.
ds2
dρ ds ds
(1.5.173)
The equation (1.5.171) is identically satisfied if we examine planar orbits where θ =
π
2
is a constant. This
value of θ also simplifies the equations (1.5.170) and (1.5.172). The equation (1.5.172) becomes an exact
differential equation
d
ds
ρ2
dφ
ds
=0
dφ
= c4 ,
ds
(1.5.174)
dt v
e = c5 ,
ds
(1.5.175)
or ρ2
and the equation (1.5.173) also becomes an exact differential
d
ds
dt v
e
ds
=0
or
where c4 and c5 are constants of integration. This leaves the equation (1.5.170) which determines ρ. Substituting the results from equations (1.5.174) and (1.5.175), together with the relation (1.5.161), the equation
(1.5.170) reduces to
c2
c2 c24
c2 c24
d2 ρ
) = 0.
+
+
−
(1
−
ds2
2ρ2
2ρ4
ρ ρ3
(1.5.176)
159
By the chain rule we have
d2 ρ
d2 ρ
= 2
2
ds
dφ
dφ
ds
2
+
dρ d2 φ
d2 ρ c24
=
+
2
dφ ds
dφ2 ρ4
dρ
dφ
2 −2c24
ρ5
and so equation (1.5.176) can be written in the form
2
d2 ρ
−
2
dφ
ρ
The substitution ρ =
1
u
dρ
dφ
2
c2
c2 ρ 2
c2
− 1−
+
+
ρ = 0.
2 c24
2
ρ
(1.5.177)
reduces the equation (1.5.177) to the form
c2
3
d2 u
+ u − 2 = c2 u 2 .
dφ2
2c4
2
(1.5.178)
du
and integrate with respect to φ to obtain
Multiply the equation (1.5.178) by 2 dφ
du
dφ
2
+ u2 −
c2
u = c2 u 3 + c 6 .
c24
(1.5.179)
where c6 is a constant of integration. To determine the constant c6 we write the equation (1.5.161) in the
special case θ =
π
2
and use the substitutions from the equations (1.5.174) and (1.5.175) to obtain
e
or
The substitution ρ =
u
dρ
ds
1
u
2
=e
dρ
dφ
2
u
dρ dφ
dφ ds
2
=1−ρ
2
dφ
ds
2
+e
v
dt
ds
c2 ρ 4
c2
c2
− 52
+ 1−
= 0.
ρ2 + 1 −
ρ
ρ
c
c24
2
(1.5.180)
reduces the equation (1.5.180) to the form
du
dφ
2
+ u 2 − c2 u 3 +
1
c2
c2
− 2 u − 2 5 2 = 0.
2
c4
c4
c c4
(1.5.181)
Now comparing the equations (1.5.181) and (1.5.179) we select
2
c5
1
−
1
c6 =
c2
c24
so that the equation (1.5.179) takes on the form
du
dφ
2
c2
c25 1
+ u − 2u + 1 − 2
= c2 u 3
c4
c
c24
2
(1.5.182)
Now we can compare our relativistic equation (1.5.182) with our Newtonian equation (1.5.143). In order
that the two equations almost agree we select the constants c2 , c4 , c5 so that
2GM
c2
=
2
c4
h2
c2
and
1 − c52
E
= 2.
2
c4
h
(1.5.183)
The equations (1.5.183) are only two equations in three unknowns and so we use the additional equation
lim ρ2
ρ→∞
dφ
dφ ds
= lim ρ2
=h
ρ→∞
dt
ds dt
(1.5.184)
160
which is obtained from equation (1.5.141). Substituting equations (1.5.174) and (1.5.175) into equation
(1.5.184), rearranging terms and taking the limit we find that
c4 c2
= h.
c5
(1.5.185)
From equations (1.5.183) and (1.5.185) we obtain the results that
c25 =
c2
,
1 + cE2
c2 =
2GM
c2
1
1 + E/c2
h
c4 = p
c 1 + E/c2
,
(1.5.186)
These values substituted into equation (1.5.181) produce the differential equation
Let α =
c2
c24
=
2GM
h2
du
dφ
2
and β = c2 =
2GM
E
2GM
+u −
u+ 2 =
2
h
h
c2
2
2GM
1
c2 ( 1+E/c2 )
1
1 + E/c2
u3 .
(1.5.187)
then the differential equation (1.5.178) can be written as
3
α
d2 u
+ u − = βu2 .
dφ2
2
2
(1.5.188)
We know the solution to equation (1.5.143) is given by
u=
1
= A(1 + cos(φ − φ0 ))
ρ
(1.5.189)
and so we assume a solution to equation (1.5.188) of this same general form. We know that A is small and so
we make the assumption that the solution of equation (1.5.188) given by equation (1.5.189) is such that φ0 is
approximately constant and varies slowly as a function of Aφ. Observe that if φ0 = φ0 (Aφ), then
and
d 2 φ0
dφ2
dφ0
dφ
= φ00 A
= φ000 A2 , where primes denote differentiation with respect to the argument of the function. (i.e.
Aφ for this problem.) The derivatives of equation (1.5.189) produce
du
= − A sin(φ − φ0 )(1 − φ00 A)
dφ
d2 u
=A3 sin(φ − φ0 )φ000 − A cos(φ − φ0 )(1 − 2Aφ00 + A2 (φ00 )2 )
dφ2
= − A cos(φ − φ0 ) + 2A2 φ00 cos(φ − φ0 ) + O(A3 ).
Substituting these derivatives into the differential equation (1.5.188) produces the equations
2A2 φ00 cos(φ − φ0 ) + A −
3β
α
=
A2 + 2A2 cos(φ − φ0 ) + 2 A2 cos2 (φ − φ0 ) + O(A3 ).
2
2
Now A is small so that terms O(A3 ) can be neglected. Equating the constant terms and the coefficient of
the cos(φ − φ0 ) terms we obtain the equations
A−
3β 2
α
=
A
2
2
2A2 φ00 = 3βA2 +
3β 2 2
A cos(φ − φ0 ).
2
Treating φ0 as essentially constant, the above system has the approximate solutions
A≈
α
2
φ0 ≈
3β
3β
Aφ +
A sin(φ − φ0 )
2
4
(1.5.190)
161
The solutions given by equations (1.5.190) tells us that φ0 varies slowly with time. For less than 1, the
elliptical motion is affected by this change in φ0 . It causes the semi-major axis of the ellipse to slowly rotate
at a rate given by
dφ0
dt .
Using the following values for the planet Mercury
G =6.67(10−8) dyne cm2 /g2
M =1.99(1033 ) g
a =5.78(1012 ) cm
=0.206
c =3(1010 ) cm/sec
2GM
β ≈ 2 = 2.95(105) cm
pc
h ≈ GM a(1 − 2 ) = 2.71(1019) cm2 /sec
1/2
GM
dφ
≈
sec−1 Kepler’s third law
dt
a3
(1.5.191)
we calculate the slow rate of rotation of the semi-major axis to be approximately
dφ0 dφ
3
dφ
dφ0
=
≈ βA
≈3
dt
dφ dt
2
dt
GM
ch
2 GM
a3
1/2
=6.628(10−14) rad/sec
(1.5.192)
=43.01 seconds of arc per century.
This slow variation in Mercury’s semi-major axis has been observed and measured and is in agreement with
the above value. Newtonian mechanics could not account for the changes in Mercury’s semi-major axis, but
Einstein’s theory of relativity does give this prediction. The resulting solution of equation (1.5.188) can be
viewed as being caused by the curvature of the space-time continuum.
The contracted curvature tensor Gij set equal to zero is just one of many conditions that can be assumed
in order to arrive at a metric for the space-time continuum. Any assumption on the value of Gij relates to
imposing some kind of curvature on the space. Within the large expanse of our universe only our imaginations
limit us as to how space, time and matter interact. You can also imagine the existence of other tensor metrics
in higher dimensional spaces where the geodesics within the space-time continuum give rise to the motion
of other physical quantities.
This short introduction to relativity is concluded with a quote from the NASA News@hg.nasa.gov news
release, spring 1998, Release:98-51. “An international team of NASA and university researchers has found
the first direct evidence of a phenomenon predicted 80 years ago using Einstein’s theory of general relativity–
that the Earth is dragging space and time around itself as it rotates.”The news release explains that the
effect is known as frame dragging and goes on to say “Frame dragging is like what happens if a bowling
ball spins in a thick fluid such as molasses. As the ball spins, it pulls the molasses around itself. Anything
stuck in the molasses will also move around the ball. Similarly, as the Earth rotates it pulls space-time in
its vicinity around itself. This will shift the orbits of satellites near the Earth.”This research is reported in
the journal Science.
162
EXERCISE 1.5
~
δT
δs
~ and τ = δN~ · B.
~ Assume in turn that each of the intrinsic derivatives of T~ , N
~,B
~ are
·N
δs
~,B
~ and hence derive the Frenet-Serret formulas of differential geometry.
some linear combination of T~ , N
I 1.
Let κ =
I 2.
Determine the given surfaces. Describe and sketch the curvilinear coordinates upon each surface.
e2
(a) ~r(u, v) = u b
e1 + v b
I 3.
(b) ~r(u, v) = u cos v b
e1 + u sin v b
e2
(c) ~r(u, v) =
2uv 2
2u2 v
b
b
e
e2 .
+
1
u2 + v 2
u2 + v 2
Determine the given surfaces and describe the curvilinear coordinates upon the surface. Use some
graphics package to plot the surface and illustrate the coordinate curves on the surface. Find element of
area dS in terms of u and v.
e2 + c cos u b
e3
a, b, c constants 0 ≤ u, v ≤ 2π
(a) ~r(u, v) = a sin u cos v b
e1 + b sin u sin v b
u
u
u
e1 + (4 + v sin ) sin u b
e2 + v cos b
e3 − 1 ≤ v ≤ 1, 0 ≤ u ≤ 2π
(b) ~r(u, v) = (4 + v sin ) cos u b
2
2
2
e2 + cu b
e3
(c) ~r(u, v) = au cos v b
e1 + bu sin v b
(d) ~r(u, v) = u cos v b
e1 + u sin v b
e2 + αv b
e3
e2 + u b
e3
(e) ~r(u, v) = a cos v b
e1 + b sin v b
α constant
a, b constant
2
e3
e2 + u b
(f ) ~r(u, v) = u cos v b
e1 + u sin v b
E F
. Assume that the surface is
F G
described by equations of the form y i = y i (u, v) and that any point on the surface is given by the position
I 4.
Consider a two dimensional space with metric tensor (aαβ ) =
ei . Show that the metrices E, F, G are functions of the parameters u, v and are given
vector ~r = ~r(u, v) = y i b
by
E = ~ru · ~ru ,
I 5.
F = ~ru · ~rv ,
G = ~rv · ~rv
where ~ru =
∂~r
∂u
and ~rv =
∂~r
.
∂v
For the metric given in problem 4 show that the Christoffel symbols of the first kind are given by
[1 1, 1] = ~ru · ~ruu
[1 2, 1] = [2 1, 1] = ~ru · ~ruv
[2 2, 1] = ~ru · ~rvv
[1 1, 2] = ~rv · ~ruu
[1 2, 2] = [2 1, 2] = ~rv · ~ruv
[2 2, 2] = ~rv · ~rvv
∂ 2~r
∂~r
·
, α, β, γ = 1, 2.
α
β
∂u ∂u ∂uγ
Show that the results in problem 5 can also be written in the form
which can be represented [α β, γ] =
I 6.
1
Ev
2
1
1
[1 1, 2] = Fu − Ev
[1 2, 2] = [2 1, 2] = Gu
2
2
where the subscripts indicate partial differentiation.
[1 1, 1] =
1
Eu
2
[1 2, 1] = [2 1, 1] =
I 7.
1
[2 2, 1] = Fv − Gu
2
1
[2 2, 2] = Gv
2
For the metricgivenin problem 4, show that the Christoffel symbols of the second kind can be
γ
α, β, γ = 1, 2 and produce the results
expressed in the form
= aγδ [α β, δ],
αβ
1
1
2
GEv − F Gu
2EFu − EEv − F Eu
1
GEu − 2F Fu + F Ev
=
=
=
=
12
21
11
2(EG − F 2 )
2(EG − F 2 )
2(EG − F 2 )
11
1
2
2
2
2GFv − GGu − F Gv
EGu − F Ev
EGv − 2F Fv + F Gu
=
=
=
=
2
2
2(EG − F )
2(EG − F )
2(EG − F 2 )
22
12
21
22
where the subscripts indicate partial differentiation.
163
I 8.
Derive the Gauss equations by assuming that
b,
~ruu = c1~ru + c2~rv + c3 n
~ruv = c4~ru + c5~rv + c6 n
b,
~rvv = c7~ru + c8~rv + c9 n
b
with the vectors ~ru , ~rv ,
where c1 , . . . , c9 are constants
determined
by
taking dot products
of
the above
vectors
1
2
1
2
, c2 =
, c3 = e, c4 =
, c5 =
, c6 = f,
and n
b. Show that c1 =
11
12
12
1 1
1
2
γ
∂~r
∂ 2~r
, c8 =
, c9 = g Show the Gauss equations can be written
=
+ bαβ n
b.
c7 =
α
β
22
22
α β ∂uγ
∂u ∂u
I 9. Derive the Weingarten equations
n
b u = c1~ru + c2~rv
n
b v = c3~ru + c4~rv
and show
f F − eG
EG − F 2
eF − f E
c2 =
EG − F 2
c1 =
gF − f G
EG − F 2
f F − gE
c4 =
EG − F 2
c3 =
and
~ru = c∗1 n
bu + c∗2 n
bv
~rv = c∗3 n
bu + c∗4 n
bv
f F − gE
eg − f 2
f E − eF
c∗2 =
eg − f 2
c∗1 =
f G − gF
eg − f 2
f F − eG
c∗4 =
eg − f 2
c∗3 =
The constants in the above equations are determined in a manner similar to that suggested in problem 8.
Show that the Weingarten equations can be written in the form
∂~r
∂b
n
= −bβα β .
∂uα
∂u
~ru × ~rv
, the results from exercise 1.1, problem 9(a), and the results from problem 5,
Using n
b= √
EG − F 2
verify that
I 10.
p
b=
EG − F 2
(~ru × ~ruu ) · n
p
2
(~ru × ~ruv ) · n
b=
EG − F 2
12
p
1
(~rv × ~ruu ) · n
b=−
EG − F 2
11
p
2
(~ru × ~rvv ) · n
b=
EG − F 2
22
2
11
p
EG − F 2
p
1
(~rv × ~rvv ) · n
b=−
EG − F 2
22
p
(~ru × ~rv ) · n
b = EG − F 2
(~rv × ~ruv ) · n
b=−
1
21
and then derive the formula for the geodesic curvature given by equation (1.5.48).
α
dT~
dT~
= (T~ ×
)·n
b and aαδ ]β γ, δ] =
.
Hint:(b
n × T~ ) ·
βγ
ds
ds
164
I 11.
Verify the equation (1.5.39) which shows that the normal curvature directions are orthogonal. i.e.
verify that Gλ1 λ2 + F (λ1 + λ2 ) + E = 0.
I 12.
βγ ωα
δλν Rωαβγ = 4Rλνστ .
Verify that δστ
I 13.
Find the first fundamental form and unit normal to the surface defined by z = f (x, y).
I 14.
Verify
σ
Ai,jk − Ai,kj = Aσ R.ijk
where
σ
=
R.ijk
∂
∂xj
σ
ik
∂
∂xk
−
σ
ij
+
n
ik
σ
nj
−
n
ij
σ
.
nk
which is sometimes written
∂
∂xj
Rinjk =
∂
∂xk
I 15.
s
nk
[ij, s]
[ik, s]
+
[nj, k] [nk, i]
s
nj
σ
show
For Rijkl = giσ R.jkl
Rinjk =
∂
∂
s
s
[nk,
i]
−
[nj,
i]
+
[ik,
s]
−
[ij,
s]
nj
nk
∂xj
∂xk
which is sometimes written
∂
∂xj
σ
= R.ijk
I 16.
n
ik
σ
ik
σ
nk
+ n
ij
σ
nj
Show
Rijkl
I 17.
σ
ij
∂
∂xk
1
=
2
∂ 2 gil
∂ 2 gjl
∂ 2 gik
∂ 2 gjk
−
−
+
∂xj ∂xk
∂xi ∂xk
∂xj ∂xl
∂xi ∂xl
+ g αβ ([jk, β][il, α] − [jl, β][ik, α]) .
Use the results from problem 15 to show
(i) Rjikl = −Rijkl ,
(ii) Rijlk = −Rijkl ,
(iii) Rklij = Rijkl
Hence, the tensor Rijkl is skew-symmetric in the indices i, j and k, l. Also the tensor Rijkl is symmetric with
respect to the (ij) and (kl) pair of indices.
I 18.
Verify the following cyclic properties of the Riemann Christoffel symbol:
(i) Rnijk + Rnjki + Rnkij = 0
(ii) Rinjk + Rjnki + Rknij = 0
(iii) Rijnk + Rjkni + Rkinj = 0
(iv)
I 19.
Riijk ,
Rikjn + Rkjin + Rjikn = 0
first index fixed
second index fixed
third index fixed
fourth index fixed
By employing the results from the previous problems, show all components of the form:
Rinjj ,
Riijj ,
Riiii ,
(no summation on i or j) must be zero.
165
I 20.
Find the number of independent components associated with the Riemann Christoffel tensor
Rijkm ,
i, j, k, m = 1, 2, . . . , N. There are N 4 components to examine in an N −dimensional space. Many of
these components are zero and many of the nonzero components are related to one another by symmetries
or the cyclic properties. Verify the following cases:
CASE I We examine components of the form Rinin ,
i 6= n
with no summation of i or n. The first index
can be chosen in N ways and therefore with i 6= n the second index can be chosen in N − 1 ways. Observe
that Rinin = Rnini ,
leaves M1 =
1
2 N (N
(no summation on i or n) and so one half of the total combinations are repeated. This
− 1) components of the form Rinin . The quantity M1 can also be thought of as the
number of distinct pairs of indices (i, n).
CASE II We next examine components of the form Rinji ,
i 6= n 6= j where there is no summation on
the index i. We have previously shown that the first pair of indices can be chosen in M1 ways. Therefore,
the third index can be selected in N − 2 ways and consequently there are M2 = 12 N (N − 1)(N − 2) distinct
components of the form Rinji with i 6= n 6= j.
CASE III Next examine components of the form Rinjk where i 6= n 6= j 6= k. From CASE I the first pairs
of indices (i, n) can be chosen in M1 ways. Taking into account symmetries, it can be shown that the second
pair of indices can be chosen in 12 (N − 2)(N − 3) ways. This implies that there are 14 N (N − 1)(N − 2)(N − 3)
ways of choosing the indices i, n, j and k with i 6= n 6= j 6= k. By symmetry the pairs (i, n) and (j, k) can be
interchanged and therefore only one half of these combinations are distinct. This leaves
1
N (N − 1)(N − 2)(N − 3)
8
distinct pairs of indices. Also from the cyclic relations we find that only two thirds of the above components
are distinct. This produces
M3 =
N (N − 1)(N − 2)(N − 3)
12
distinct components of the form Rinjk with i 6= n 6= j 6= k.
Adding the above components from each case we find there are
M4 = M1 + M2 + M3 =
N 2 (N 2 − 1)
12
distinct and independent components.
Verify the entries in the following table:
Dimension of space N
1
2
3
4
5
Number of components N 4
1
16
81
256
625
M4 = Independent components of Rijkm
0
1
6
20
50
Note 1: A one dimensional space can not be curved and all one dimensional spaces are Euclidean. (i.e. if we have
an element of arc length squared given by ds2 = f (x)(dx)2 , we can make the coordinate transformation
p
f (x)dx = du and reduce the arc length squared to the form ds2 = du2 .)
Note 2: In a two dimensional space, the indices can only take on the values 1 and 2. In this special case there
are 16 possible components. It can be shown that the only nonvanishing components are:
R1212 = −R1221 = −R2112 = R2121 .
166
For these nonvanishing components only one independent component exists. By convention, the component R1212 is selected as the single independent component and all other nonzero components are
expressed in terms of this component.
Find the nonvanishing independent components Rijkl for i, j, k, l = 1, 2, 3, 4 and show that
R1212
R3434
R2142
R4124
R1313
R1231
R2342
R4314
R2323
R1421
R3213
R4234
R1414
R1341
R3243
R1324
R2424
R2132
R3143
R1432
can be selected as the twenty independent components.
I 21.
(a) For N = 2 show R1212 is the only nonzero independent component and
R1212 = R2121 = −R1221 = −R2112 .
(b) Show that on the surface of a sphere of radius r0 we have R1212 = r02 sin2 θ.
I 22.
Show for N = 2 that
R1212 = R1212 J 2 = R1212
I 23.
∂x
∂x
2
s
Define Rij = R.ijs
as the Ricci tensor and Gij = Rji − 12 δji R as the Einstein tensor, where Rji = g ik Rkj
and R = Rii . Show that
Rjk = g ab Rjabk
√
√
∂ 2 log g
b ∂ log g
a
b
a
∂
(b) Rij =
−
− a
+
i
j
b
ij
∂x ∂x
∂x
∂x i j
ia
jb
(a)
(c)
I 24.
i
=0
Rijk
By employing the results from the previous problem show that in the case N = 2 we have
R22
R12
R1212
R11
=
=
=−
g11
g22
g12
g
where g is the determinant of gij .
I 25.
Consider the case N = 2 where we have g12 = g21 = 0 and show that
(a)
R12 = R21 = 0
(c)
(b)
R11 g22 = R22 g11 = R1221
(d)
2R1221
g11 g22
1
Rij = Rgij ,
2
R=
where
R = g ij Rij
The scalar invariant R is known as the Einstein curvature of the surface and the tensor Gij = Rji − 12 δji R is
known as the Einstein tensor.
I 26.
For N = 3 show that R1212 , R1313 , R2323 , R1213 , R2123 , R3132 are independent components of the
Riemann Christoffel tensor.
167
I 27.
For N = 2 and aαβ =
K=
For N = 2 and aαβ =
1
K= √
2 a
0
a22
show that
1
∂
1 ∂a22
∂
1 ∂a11
R1212
√
√
=− √
+
.
a
2 a ∂u1
a ∂u1
∂u2
a ∂u2
I 28.
a11
0
a11
a21
a12
a22
show that
∂
a12 ∂a11
2 ∂a12
1 ∂a22
1 ∂a11
a12 ∂a11
∂
√
√
√
−√
−√
−
+
.
∂u1 a11 a ∂u2
a ∂u1
∂u2
a ∂u1
a ∂u2
a11 a ∂u1
Check your results by setting a12 = a21 = 0 and comparing this answer with that given in the problem 27.
I 29.
Write out the Frenet-Serret formulas (1.5.112)(1.5.113) for surface curves in terms of Christoffel
symbols of the second kind.
I 30.
(a) Use the fact that for n = 2 we have R1212 = R2121 = −R2112 = −R1221 together with eαβ , eαβ the two
dimensional alternating tensors to show that the equation (1.5.110) can be written as
Rαβγδ = Kαβ γδ
where
αβ =
√
aeαβ
1
and αβ = √ eαβ
a
are the corresponding epsilon tensors.
1
Rαβγδ αβ γδ = K.
4
Hint: See equations (1.3.82),(1.5.93) and (1.5.94).
(b) Show that from the result in part (a) we obtain
I 31.
Verify the result given by the equation (1.5.100).
I 32.
Show that aαβ cαβ = 4H 2 − 2K.
I 33.
Find equations for the principal curvatures associated with the surface
x = u,
y = v,
z = f (u, v).
I 34. Geodesics on a sphere Let (θ, φ) denote the surface coordinates of the sphere of radius ρ defined
by the parametric equations
x = ρ sin θ cos φ, y = ρ sin θ sin φ, z = ρ cos θ.
(1)
Consider also a plane which passes through the origin with normal having the direction numbers (n1 , n2 , n3 ).
This plane is represented by n1 x + n2 y + n3 z = 0 and intersects the sphere in a great circle which is described
by the relation
n1 sin θ cos φ + n2 sin θ sin φ + n3 cos θ = 0.
(2)
This is an implicit relation between the surface coordinates θ, φ which describes the great circle lying on the
sphere. We can write this later equation in the form
n1 cos φ + n2 sin φ =
−n3
tan θ
(3)
168
and in the special case where n1 = cos β, n2 = sin β,n3 = − tan α is expressible in the form
tan α
tan α
or φ − β = cos−1
.
cos(φ − β) =
tan θ
tan θ
(4)
The above equation defines an explicit relationship between the surface coordinates which defines a great
circle on the sphere. The arc length squared relation satisfied by the surface coordinates together with the
equation obtained by differentiating equation (4) with respect to arc length s gives the relations
tan α
dφ
dθ
=q
ds
tan2 α ds
1 − tan2 θ
(5)
ds2 = ρ2 dθ2 + ρ2 sin2 θ dφ2
(6)
sin2 θ
The above equations (1)-(6) are needed to consider the following problem.
(a) Show that the differential equations defining the geodesics on the surface of a sphere (equations (1.5.51))
are
2
dφ
d2 θ
− sin θ cos θ
=0
ds2
ds
dθ dφ
d2 φ
=0
+ 2 cot θ
ds2
ds ds
(7)
(8)
(b) Multiply equation (8) by sin2 θ and integrate to obtain
sin2 θ
dφ
= c1
ds
(9)
where c1 is a constant of integration.
(c) Multiply equation (7) by
dθ
ds
and use the result of equation (9) to show that an integration produces
dθ
ds
2
=
−c21
+ c22
sin2 θ
(10)
where c22 is a constant of integration.
(d) Use the equations (5)(6) to show that c2 = 1/ρ and c1 =
sin α
ρ .
(e) Show that equations (9) and (10) imply that
sec2 θ
tan α
dφ
q
=
dθ
tan2 θ 1 − tan2 α
tan2 θ
and making the substitution u =
tan α
tan θ
this equation can be integrated to obtain the equation (4). We
can now expand the equation (4) and express the results in terms of x, y, z to obtain the equation (3).
This produces a plane which intersects the sphere in a great circle. Consequently, the geodesics on a
sphere are great circles.
169
I 35.
I 36.
Find the differential equations defining the geodesics on the surface of a cylinder.
Find the differential equations defining the geodesics on the surface of a torus. (See problem 13,
Exercise 1.3)
I 37.
Find the differential equations defining the geodesics on the surface of revolution
x = r cos φ,
y = r sin φ,
z = f (r).
Note the curve z = f (x) gives a profile of the surface. The curves r = Constant are the parallels, while the
curves φ = Constant are the meridians of the surface and
ds2 = (1 + f 02 ) dr2 + r2 dφ2 .
I 38.
Find the unit normal and tangent plane to an arbitrary point on the right circular cone
x = u sin α cos φ,
y = u sin α sin φ,
z = u cos α.
This is a surface of revolution with r = u sin α and f (r) = r cot α with α constant.
I 39.
Let s denote arc length and assume the position vector ~r(s) is analytic about a point s0 . Show that
h2
h3
the Taylor series ~r(s) = ~r(s0 ) + h~r 0 (s0 ) + ~r 00 (s0 ) + ~r 000 (s0 ) + · · · about the point s0 , with h = s − s0 is
2!
3!
~ + 1 h3 (−κ2 T~ + κ0 N
~ + κτ B)
~ + · · · which is obtained by differentiating
given by ~r(s) = ~r(s0 ) + hT~ + 12 κh2 N
6
the Frenet formulas.
I 40.
(a) Show that the circular helix defined by x = a cos t,
y = a sin t,
z = bt with a, b constants, has the
property that any tangent to the curve makes a constant angle with the line defining the z-axis.
(i.e. T~ · b
e3 = cos α = constant.)
~ · b
e3 is parallel to the rectifying plane, which implies that
(b) Show also that N
e3 = 0 and consequently b
~ sin α.
b
e3 = T~ cos α + B
(c) Differentiate the result in part (b) and show that κ/τ = tan α is a constant.
I 41.
Consider a space curve xi = xi (s) in Cartesian coordinates.
p
dT~
= x0i x0i
(a) Show that κ =
ds
(b) Show that τ =
1
eijk x0i x00j x000
r 0 · ~r 00 × ~r 000
k . Hint: Consider ~
κ2
I 42.
(a) Find the direction cosines of a normal to a surface z = f (x, y).
(b) Find the direction cosines of a normal to a surface F (x, y, z) = 0.
(c) Find the direction cosines of a normal to a surface x = x(u, v), y = y(u, v), z = z(u, v).
I 43.
Show that for a smooth surface z = f (x, y) the Gaussian curvature at a point on the surface is given
by
K=
2
fxx fyy − fxy
.
(fx2 + fy2 + 1)2
170
I 44.
Show that for a smooth surface z = f (x, y) the mean curvature at a point on the surface is given by
H=
(1 + fy2 )fxx − 2fx fy fxy + (1 + fx2 )fyy
.
2(fx2 + fy2 + 1)3/2
I 45.
Express the Frenet-Serret formulas (1.5.13) in terms of Christoffel symbols of the second kind.
I 46.
Verify the relation (1.5.106).
I 47.
In Vn assume that Rij = ρgij and show that ρ =
R
n
where R = g ij Rij . This result is known as
Einstein’s gravitational equation at points where matter is present. It is analogous to the Poisson equation
∇2 V = ρ from the Newtonian theory of gravitation.
I 48.
In Vn assume that Rijkl = K(gik gjl − gil gjk ) and show that R = Kn(1 − n). (Hint: See problem 23.)
I 49.
Assume gij = 0 for i 6= j and verify the following.
(a) Rhijk = 0 for h 6= i 6= j 6= k
2√
√
√
√
√
∂ gii ∂ log ghh
∂ gii ∂ log gkk
∂ gii
√
−
−
(b) Rhiik = gii
for h, i, k unequal.
∂xh∂xk
∂xh
∂xk
∂xk
∂xh

√ √
√
√
n
X
∂ gii ∂ ghh 
1 ∂ gii
∂
1 ∂ ghh
√ √
 ∂
+ i √
+
(c) Rhiih = gii ghh  h √
 where h 6= i.
∂x
ghh ∂xh
∂x
gii ∂xi
∂xm ∂xm
m=1
m6=h m6=i
I 50.
Consider a surface of revolution where x = r cos θ, y = r sin θ and z = f (r) is a given function of r.
(a) Show in this V2 we have ds2 = (1 + (f 0 )2 )dr2 + r2 dθ2 where 0 =
d
ds .
(b) Show the geodesic equations in this V2 are
2
2
dr
dθ
f 0 f 00
r
d2 r
+
−
=0
2
0
2
0
2
ds
1 + (f )
ds
1 + (f )
ds
d2 θ 2 dθ dr
=0
+
ds2
r ds ds
a
dθ
= 2 . Substitute this result for ds in part (a) to show
(c) Solve the second equation in part (b) to obtain
ds
r
p
a 1 + (f 0 )2
dr which theoretically can be integrated.
dθ = ± √
r r 2 − a2
171
PART 2: INTRODUCTION TO CONTINUUM MECHANICS
In the following sections we develop some applications of tensor calculus in the areas of dynamics,
elasticity, fluids and electricity and magnetism. We begin by first developing generalized expressions for the
vector operations of gradient, divergence, and curl. Also generalized expressions for other vector operators
are considered in order that tensor equations can be converted to vector equations. We construct a table to
aid in the translating of generalized tensor equations to vector form and vice versa.
The basic equations of continuum mechanics are developed in the later sections. These equations are
developed in both Cartesian and generalized tensor form and then converted to vector form.
§2.1 TENSOR NOTATION FOR SCALAR AND VECTOR QUANTITIES
We consider the tensor representation of some vector expressions. Our goal is to develop the ability to
convert vector equations to tensor form as well as being able to represent tensor equations in vector form.
In this section the basic equations of continuum mechanics are represented using both a vector notation and
the indicial notation which focuses attention on the tensor components. In order to move back and forth
between these notations, the representation of vector quantities in tensor form is now considered.
Gradient
For Φ = Φ(x1 , x2 , . . . , xN ) a scalar function of the coordinates xi , i = 1, . . . , N , the gradient of Φ is
defined as the covariant vector
Φ,i =
∂Φ
,
∂xi
i = 1, . . . , N.
(2.1.1)
The contravariant form of the gradient is
g im Φ,m .
(2.1.2)
Note, if C i = g im Φ,m , i = 1, 2, 3 are the tensor components of the gradient then in an orthogonal coordinate
system we will have
C 1 = g 11 Φ,1 ,
C 2 = g 22 Φ,2 ,
C 3 = g 33 Φ,3 .
We note that in an orthogonal coordinate system that g ii = 1/h2i , (no sum on i), i = 1, 2, 3 and hence
replacing the tensor components by their equivalent physical components there results the equations
1 ∂Φ
C(1)
= 2 1,
h1
h1 ∂x
C(2)
1 ∂Φ
= 2 2,
h2
h2 ∂x
C(3)
1 ∂Φ
= 2 3.
h3
h3 ∂x
Simplifying, we find the physical components of the gradient are
C(1) =
1 ∂Φ
,
h1 ∂x1
C(2) =
1 ∂Φ
,
h2 ∂x2
C(3) =
1 ∂Φ
.
h3 ∂x3
These results are only valid when the coordinate system is orthogonal and gij = 0 for i 6= j and gii = h2i ,
with i = 1, 2, 3, and where i is not summed.
172
Divergence
The divergence of a contravariant tensor Ar is obtained by taking the covariant derivative with respect
to xk and then performing a contraction. This produces
div Ar = Ar,r .
(2.1.3)
Still another form for the divergence is obtained by simplifying the expression (2.1.3). The covariant derivative can be represented
Ar,k
∂Ar
=
+
∂xk
r
Am .
mk
Upon contracting the indices r and k and using the result from Exercise 1.4, problem 13, we obtain
√
∂Ar
1 ∂( g) m
+
A
√
∂xr
g ∂xm
√ 1 √ ∂Ar
r∂ g
= √
g r +A
g
∂x
∂xr
1 ∂ √ r
= √
( gA ) .
g ∂xr
Ar,r =
Ar,r
Ar,r
EXAMPLE 2.1-1. (Divergence)
coordinates (ρ, θ, φ). Solution:
x1 = ρ,
(2.1.4)
Find the representation of the divergence of a vector Ar in spherical
In spherical coordinates we have
x2 = θ,
g11 = h21 = 1,
x3 = φ with
gij = 0 for
g22 = h22 = ρ2 ,
The determinant of gij is g = |gij | = ρ4 sin2 θ and
√
i 6= j
and
g33 = h23 = ρ2 sin2 θ.
g = ρ2 sin θ. Employing the relation (2.1.4) we find
∂ √ 1
1
∂ √ 2
∂ √ 3
( gA ) + 2 ( gA ) + 3 ( gA ) .
div A = √
g ∂x1
∂x
∂x
r
In terms of the physical components this equation becomes
∂ √ A(1)
1
∂ √ A(2)
∂ √ A(3)
( g
( g
( g
)+
)+
) .
div A = √
g ∂ρ
h1
∂θ
h2
∂φ
h3
r
By using the notation
A(1) = Aρ ,
A(2) = Aθ ,
A(3) = Aφ
for the physical components, the divergence can be expressed in either of the forms:
∂ 2
1
∂ 2
Aθ
Aφ
∂ 2
(ρ
(ρ
)
+
(ρ
)
sin
θA
)
+
sin
θ
sin
θ
ρ
ρ2 sin θ ∂ρ
∂θ
ρ
∂φ
ρ sin θ
∂
∂A
1
1
1
∂
φ
(ρ2 Aρ ) +
(sin θAθ ) +
.
div Ar = 2
ρ ∂ρ
ρ sin θ ∂θ
ρ sin θ ∂φ
div Ar =
or
173
Curl
~ = curl A
~ are represented
The contravariant components of the vector C
C i = ijk Ak,j .
(2.1.5)
In expanded form this representation becomes:
1
C =√
g
1
1
C =√
g
2
1
C3 = √
g
EXAMPLE 2.1-2. (Curl)
∂A3
∂A2
−
∂x2
∂x3
∂A3
∂A1
−
3
∂x
∂x1
∂A2
∂A1
−
1
∂x
∂x2
(2.1.6)
.
~ in spherical coordinates
Find the representation for the components of curl A
(ρ, θ, φ).
Solution:
In spherical coordinates we have :x1 = ρ,
g11 = h21 = 1,
x2 = θ,
g22 = h22 = ρ2 ,
x3 = φ with gij = 0 for i 6= j and
g33 = h23 = ρ2 sin2 θ.
√
The determinant of gij is g = |gij | = ρ4 sin2 θ with g = ρ2 sin θ. The relations (2.1.6) are tensor equations
~ To find the components of curl A
~ in spherical components
representing the components of the vector curl A.
we write the equations (2.1.6) in terms of their physical components. These equations take on the form:
∂
1
∂
C(1)
(h3 A(3)) −
(h2 A(2))
=√
h1
g ∂θ
∂φ
∂
1
∂
C(2)
(h1 A(1)) −
(h3 A(3))
=√
h2
g ∂φ
∂ρ
∂
1
∂
C(3)
(h2 A(2)) −
(h1 A(1)) .
=√
h3
g ∂ρ
∂θ
(2.1.7)
We employ the notations
C(1) = Cρ ,
C(2) = Cθ ,
C(3) = Cφ ,
A(1) = Aρ ,
A(2) = Aθ ,
A(3) = Aφ
~ in spherical coordinates,
to denote the physical components, and find the components of the vector curl A,
are expressible in the form:
∂
1
∂
(ρ sin θAφ ) −
(ρAθ )
Cρ = 2
ρ sin θ ∂θ
∂φ
∂
1
∂
(Aρ ) −
(ρ sin θAφ )
Cθ =
ρ sin θ ∂φ
∂ρ
1 ∂
∂
(ρAθ ) −
(Aρ ) .
Cφ =
ρ ∂ρ
∂θ
(2.1.8)
174
Laplacian
The Laplacian ∇2 U has the contravariant form
2
ij
∇ U = g U,ij
ij ∂U
= (g U,i ),j = g
.
∂xi ,j
ij
(2.1.9)
Expanding this expression produces the equations:
j
∂
ij ∂U
im ∂U
g
+
g
∂xj
∂xi
∂xi m j
√
∂
1 ∂ g ij ∂U
ij ∂U
g
g
+
∇2 U =
√
∂xj
∂xi
g ∂xj
∂xi
√ 1 √ ∂
∂U
∂U ∂ g
∇2 U = √
g j g ij i + g ij i
g
∂x
∂x
∂x ∂xj
1 ∂
√ ij ∂U
gg
.
∇2 U = √
g ∂xj
∂xi
∇2 U =
(2.1.10)
In orthogonal coordinates we have g ij = 0 for i 6= j and
g11 = h21 ,
g22 = h22 ,
g33 = h23
and so (2.1.10) when expanded reduces to the form
∂
h2 h3 ∂U
h1 h3 ∂U
h1 h2 ∂U
1
∂
∂
+ 2
+ 3
.
∇ U=
h1 h2 h3 ∂x1
h1 ∂x1
∂x
h2 ∂x2
∂x
h3 ∂x3
2
(2.1.11)
This representation is only valid in an orthogonal system of coordinates.
EXAMPLE 2.1-3. (Laplacian)
Find the Laplacian in spherical coordinates.
Solution: Utilizing the results given in the previous example we find the Laplacian in spherical coordinates
has the form
∂
1 ∂U
1
∂U
∂
∂U
∂
2
ρ sin θ
+
sin θ
+
.
∇ U= 2
ρ sin θ ∂ρ
∂ρ
∂θ
∂θ
∂φ sin θ ∂φ
2
This simplifies to
∇2 U =
1 ∂ 2U
1
∂2U
∂2U
2 ∂U
cot θ ∂U
+ 2
+ 2 2
+
+ 2
.
2
2
∂ρ
ρ ∂ρ
ρ ∂θ
ρ ∂θ
ρ sin θ ∂φ2
The table 1 gives the vector and tensor representation for various quantities of interest.
(2.1.12)
(2.1.13)
175
VECTOR
GENERAL TENSOR
~
A
~·B
~
A
CARTESIAN TENSOR
Ai or Ai
Ai
Ai Bi = gij Ai B j = Ai B i
Ai Bi = g ij Ai Bj
Ai Bi
~ =A
~×B
~
C
1
C i = √ eijk Aj Bk
g
Ci = eijk Aj Bk
∇ Φ = grad Φ
g im Φ,m
Φ,i =
∂Φ
∂xi
~ = div A
~
∇·A
1 ∂ √ r
g mn Am,n = Ar,r = √
( gA )
g ∂xr
Ai,i =
∂Ai
∂xi
C i = ijk Ak,j
Ci = eijk
~=C
~ = curl A
~
∇×A
∇2 U
∂Ak
∂xj
∂
∂U
∂xi ∂xi
√ ij ∂U
gg
∂xi
1 ∂
g mn U ,mn = √
g ∂xj
~ = (A
~ · ∇)B
~
C
C i = Am B i,m
Ci = Am
~ = A(∇
~
~
C
· B)
C i = Ai B j,j
Ci = Ai
~ = ∇2 A
~
C
~·∇ φ
A
~
∇ ∇·A
~
∇× ∇×A
C i = g jm Ai ,mj
or Ci = g jm Ai,mj
g im Ai φ ,m
g im Ar,r
∂
∂xm
∂Bm
∂xm
∂Ai
∂xm
Ai φ,i
∂ 2 Ar
∂xi ∂xr
,m
ijk g jm kst At,s
Ci =
∂Bi
∂xm
,m
Table 1 Vector and tensor representations.
∂ 2 Ai
∂ 2 Aj
−
∂xj ∂xi
∂xj ∂xj
176
EXAMPLE 2.1-4. (Maxwell’s equations)
vectors and scalars:
In the study of electrodynamics there arises the following
~ =Electric force vector, [E]
~ = Newton/coulomb
E
~ =Magnetic force vector, [B]
~ = Weber/m2
B
~ =Displacement vector, [D]
~ = coulomb/m2
D
~ =Auxilary magnetic force vector, [H]
~ = ampere/m
H
~ = ampere/m2
J~ =Free current density, [J]
% =free charge density, [%] = coulomb/m3
The above quantities arise in the representation of the following laws:
Faraday’s Law This law states the line integral of the electromagnetic force around a loop is proportional
to the rate of flux of magnetic induction through the loop. This gives rise to the first electromagnetic field
equation:
~ =−
∇×E
Ampere’s Law
~
∂B
∂t
ijk Ek,j = −
or
∂B i
.
∂t
(2.1.15)
This law states the line integral of the magnetic force vector around a closed loop is
proportional to the sum of the current through the loop and the rate of flux of the displacement vector
through the loop. This produces the second electromagnetic field equation:
~ = J~ +
∇×H
~
∂D
∂t
ijk Hk,j = J i +
or
∂Di
.
∂t
(2.1.16)
Gauss’s Law for Electricity This law states that the flux of the electric force vector through a closed
surface is proportional to the total charge enclosed by the surface. This results in the third electromagnetic
field equation:
~ =%
∇·D
or
1 ∂ √ i
gD = %.
√
g ∂xi
(2.1.17)
Gauss’s Law for Magnetism This law states the magnetic flux through any closed volume is zero. This
produces the fourth electromagnetic field equation:
~ =0
∇·B
or
1 ∂ √ i
gB = 0.
√
g ∂xi
(2.1.18)
The four electromagnetic field equations are referred to as Maxwell’s equations. These equations arise
in the study of electrodynamics and can be represented in other forms. These other forms will depend upon
such things as the material assumptions and units of measurements used. Note that the tensor equations
(2.1.15) through (2.1.18) are representations of Maxwell’s equations in a form which is independent of the
coordinate system chosen.
In applications, the tensor quantities must be expressed in terms of their physical components. In a
general orthogonal curvilinear coordinate system we will have
g11 = h21 ,
This produces the result
g22 = h22 ,
g33 = h23 ,
and gij = 0
for i 6= j.
√
g = h1 h2 h3 . Further, if we represent the physical components of
Di , Bi , Ei , Hi
by D(i), B(i), E(i), and H(i)
177
the Maxwell equations can be represented by the equations in table 2. The tables 3, 4 and 5 are the
representation of Maxwell’s equations in rectangular, cylindrical, and spherical coordinates. These latter
tables are special cases associated with the more general table 2.
1
h1 h2 h3
1
h1 h2 h3
1
h1 h2 h3
1
h1 h2 h3
1
h1 h2 h3
1
h1 h2 h3
∂
∂
1 ∂B(1)
(h3 E(3)) − 3 (h2 E(2)) = −
2
∂x
∂x
h1 ∂t
∂
∂
1 ∂B(2)
(h1 E(1)) − 1 (h3 E(3)) = −
3
∂x
∂x
h2 ∂t
∂
∂
1 ∂B(3)
(h2 E(2)) − 2 (h1 E(1)) = −
1
∂x
∂x
h3 ∂t
∂
∂
J(1)
1 ∂D(1)
(h3 H(3)) − 3 (h2 H(2)) =
+
2
∂x
∂x
h1
h1 ∂t
∂
∂
J(2)
1 ∂D(2)
(h1 H(1)) − 1 (h3 H(3)) =
+
∂x3
∂x
h2
h2 ∂t
∂
∂
J(3)
1 ∂D(3)
(h2 H(2)) − 2 (h1 H(1)) =
+
∂x1
∂x
h3
h3 ∂t
∂
D(1)
D(2)
D(3)
1
∂
∂
h1 h2 h3
+ 2 h1 h2 h3
+ 3 h1 h2 h3
=%
h1 h2 h3 ∂x1
h1
∂x
h2
∂x
h3
∂
B(1)
B(2)
B(3)
∂
∂
1
h1 h2 h3
+ 2 h1 h2 h3
+ 3 h1 h2 h3
=0
h1 h2 h3 ∂x1
h1
∂x
h2
∂x
h3
Table 2 Maxwell’s equations in generalized orthogonal coordinates.
Note that all the tensor components have been replaced by their physical components.
178
∂Hz
∂Hy
−
= Jx +
∂y
∂z
∂Hz
∂Hx
−
= Jy +
∂z
∂x
∂Hx
∂Hy
−
= Jz +
∂x
∂y
∂Ey
∂Bx
∂Ez
−
=−
∂y
∂z
∂t
∂Ez
∂By
∂Ex
−
=−
∂z
∂x
∂t
∂Ex
∂Bz
∂Ey
−
=−
∂x
∂y
∂t
∂Dx
∂t
∂Dy
∂t
∂Dz
∂t
∂Dx
∂Dy
∂Dz
+
+
=%
∂x
∂y
∂z
∂By
∂Bz
∂Bx
+
+
=0
∂x
∂y
∂z
Here we have introduced the notations:
with x1 = x,
Dx = D(1)
Bx = B(1)
Hx = H(1)
Jx = J(1)
Ex = E(1)
Dy = D(2)
By = B(2)
Hy = H(2)
Jy = J(2)
Ey = E(2)
Dz = D(3)
Bz = B(3)
Hz = H(3)
Jz = J(3)
Ez = E(3)
x2 = y,
x3 = z,
h1 = h 2 = h 3 = 1
Table 3 Maxwell’s equations Cartesian coordinates
∂Eθ
∂Br
1 ∂Ez
−
=−
r ∂θ
∂z
∂t
∂Ez
∂Bθ
∂Er
−
=−
∂z
∂r
∂t
∂Bz
1 ∂Er
1 ∂
(rEθ ) −
=−
r ∂r
r ∂θ
∂t
∂Dz
1 ∂Dθ
1 ∂
(rDr ) +
+
=%
r ∂r
r ∂θ
∂z
1 ∂Hz
∂Hθ
∂Dr
−
= Jr +
r ∂θ
∂z
∂t
∂Hz
∂Dθ
∂Hr
−
= Jθ +
∂z
∂r
∂t
1 ∂Hr
∂Dz
1 ∂
(rHθ ) −
= Jz +
r ∂r
r ∂θ
∂t
1 ∂
∂Bz
1 ∂Bθ
(rBr ) +
+
=0
r ∂r
r ∂θ
∂z
Here we have introduced the notations:
with x1 = r,
Dr = D(1)
Br = B(1)
Hr = H(1)
Jr = J(1)
Er = E(1)
Dθ = D(2)
Bθ = B(2)
Hθ = H(2)
Jθ = J(2)
Eθ = E(2)
Dz = D(3)
Bz = B(3)
Hz = H(3)
Jz = J(3)
Ez = E(3)
x2 = θ,
x3 = z,
h1 = 1,
h2 = r,
h3 = 1.
Table 4 Maxwell’s equations in cylindrical coordinates.
179
∂
∂Eθ
1
∂Bρ
(sin θEφ ) −
=−
ρ sin θ ∂θ
∂φ
∂t
1 ∂
∂Bθ
1 ∂Eρ
−
(ρEφ ) = −
ρ sin θ ∂φ
ρ ∂ρ
∂t
∂Bφ
1 ∂Eρ
1 ∂
(ρEθ ) −
=−
ρ ∂ρ
ρ ∂θ
∂t
1
∂
∂Hθ
∂Dρ
(sin θHφ ) −
= Jρ +
ρ sin θ ∂θ
∂φ
∂t
1 ∂
∂Dθ
1 ∂Hρ
−
(ρHφ ) = Jθ +
ρ sin θ ∂φ
ρ ∂ρ
∂t
1 ∂Hρ
∂Dφ
1 ∂
(ρHθ ) −
= Jφ +
ρ ∂ρ
ρ ∂θ
∂t
∂
1
1 ∂Dφ
1 ∂ 2
(ρ Dρ ) +
(sin θDθ ) +
=%
2
ρ ∂ρ
ρ sin θ ∂θ
ρ sin θ ∂φ
∂
1
1 ∂Bφ
1 ∂ 2
(ρ Bρ ) +
(sin θBθ ) +
=0
ρ2 ∂ρ
ρ sin θ ∂θ
ρ sin θ ∂φ
Here we have introduced the notations:
with x1 = ρ,
Dρ = D(1)
Bρ = B(1)
Hρ = H(1)
Jρ = J(1)
Eρ = E(1)
Dθ = D(2)
Bθ = B(2)
Hθ = H(2)
Jθ = J(2)
Eθ = E(2)
Dφ = D(3)
Bφ = B(3)
Hφ = H(3)
Jφ = J(3)
Eφ = E(3)
x2 = θ,
x3 = φ,
h1 = 1,
h2 = ρ,
h3 = ρ sin θ
Table 5 Maxwell’s equations spherical coordinates.
Eigenvalues and Eigenvectors of Symmetric Tensors
Consider the equation
Tij Aj = λAi ,
i, j = 1, 2, 3,
(2.1.19)
where Tij = Tji is symmetric, Ai are the components of a vector and λ is a scalar. Any nonzero solution
Ai of equation (2.1.19) is called an eigenvector of the tensor Tij and the associated scalar λ is called an
eigenvalue. When expanded these equations have the form
(T11 − λ)A1 +
T12 A2 +
T13 A3 = 0
T21 A1 + (T22 − λ)A2 +
T23 A3 = 0
T31 A1 +
T32 A2 + (T33 − λ)A3 = 0.
The condition for equation (2.1.19) to have a nonzero solution Ai is that the characteristic equation
should be zero. This equation is found from the determinant equation
f (λ) =
T11 − λ
T21
T31
T12
T13
T22 − λ
T23
= 0,
T32
T33 − λ
(2.1.20)
180
which when expanded is a cubic equation of the form
f (λ) = −λ3 + I1 λ2 − I2 λ + I3 = 0,
(2.1.21)
where I1 , I2 and I3 are invariants defined by the relations
I1 = Tii
1
1
I2 = Tii Tjj − Tij Tij
2
2
I3 = eijk Ti1 Tj2 Tk3 .
(2.1.22)
When Tij is subjected to an orthogonal transformation, where T̄mn = Tij `im `jn , then
`im `jn (Tmn − λ δmn ) = T̄ij − λ δij
and
det (Tmn − λ δmn ) = det T̄ij − λ δij .
Hence, the eigenvalues of a second order tensor remain invariant under an orthogonal transformation.
If Tij is real and symmetric then
• the eigenvalues of Tij will be real, and
• the eigenvectors corresponding to distinct eigenvalues will be orthogonal.
Proof: To show a quantity is real we show that the conjugate of the quantity equals the given quantity. If
(2.1.19) is satisfied, we multiply by the conjugate Ai and obtain
Ai Tij Aj = λAi Ai .
(2.1.25)
The right hand side of this equation has the inner product Ai Ai which is real. It remains to show the left
hand side of equation (2.1.25) is also real. Consider the conjugate of this left hand side and write
Ai Tij Aj = Ai T ij Aj = Ai Tji Aj = Ai Tij Aj .
Consequently, the left hand side of equation (2.1.25) is real and the eigenvalue λ can be represented as the
ratio of two real quantities.
Assume that λ(1) and λ(2) are two distinct eigenvalues which produce the unit eigenvectors L̂1 and L̂2
with components `i1 and `i2 , i = 1, 2, 3 respectively. We then have
Tij `j1 = λ(1) `i1
and
Tij `j2 = λ(2) `i2 .
(2.1.26)
Consider the products
λ(1) `i1 `i2 = Tij `j1 `i2 ,
(2.1.27)
λ(2) `i1 `i2 = `i1 Tij `j2 = `j1 Tji `i2 .
and subtract these equations. We find that
[λ(1) − λ(2) ]`i1 `i2 = 0.
(2.1.28)
By hypothesis, λ(1) is different from λ(2) and consequently the inner product `i1 `i2 must be zero. Therefore,
the eigenvectors corresponding to distinct eigenvalues are orthogonal.
181
Therefore, associated with distinct eigenvalues λ(i) , i = 1, 2, 3 there are unit eigenvectors
L̂(i) = `i1 ê1 + `i2 ê2 + `i3 ê3
with components `im , m = 1, 2, 3 which are direction cosines and satisfy
`in `im = δmn
and
`ij `jm = δim .
(2.1.23)
The unit eigenvectors satisfy the relations
Tij `j1 = λ(1) `i1
Tij `j2 = λ(2) `i2
Tij `j3 = λ(3) `i3
and can be written as the single equation
Tij `jm = λ(m) `im ,
m = 1, 2, or 3
m not summed.
Consider the transformation
xi = `ij xj
or
xm = `mj xj
which represents a rotation of axes, where `ij are the direction cosines from the eigenvectors of Tij . This is a
linear transformation where the `ij satisfy equation (2.1.23). Such a transformation is called an orthogonal
transformation. In the new x coordinate system, called principal axes, we have
T mn = Tij
∂xi ∂xj
= Tij `im `jn = λ(n) `in `im = λ(n) δmn
∂xm ∂xn
(no sum on n).
(2.1.24)
This equation shows that in the barred coordinate system there are the components
T mn

λ(1)
= 0
0
0
λ(2)
0

0
0 .
λ(3)
That is, along the principal axes the tensor components Tij are transformed to the components T ij where
T ij = 0 for i 6= j. The elements T (i)(i) , i not summed, represent the eigenvalues of the transformation
(2.1.19).
182
EXERCISE 2.1
I 1. In cylindrical coordinates (r, θ, z) with f = f (r, θ, z) find the gradient of f.
~ = A(r,
~ θ, z) find div A.
~
I 2. In cylindrical coordinates (r, θ, z) with A
~ = A(r,
~ θ, z) find curl A.
~
I 3. In cylindrical coordinates (r, θ, z) for A
I 4. In cylindrical coordinates (r, θ, z) for f = f (r, θ, z) find ∇2 f.
I 5. In spherical coordinates (ρ, θ, φ) with f = f (ρ, θ, φ) find the gradient of f.
~ = A(ρ,
~ θ, φ) find div A.
~
I 6. In spherical coordinates (ρ, θ, φ) with A
~ = A(ρ,
~ θ, φ) find curl A.
~
I 7. In spherical coordinates (ρ, θ, φ) for A
I 8. In spherical coordinates (ρ, θ, φ) for f = f (ρ, θ, φ) find ∇2 f.
I 9. Let ~r = x ê1 + y ê2 + z ê3 denote the position vector of a variable point (x, y, z) in Cartesian coordinates.
Let r = |~r| denote the distance of this point from the origin. Find in terms of ~r and r:
(a) grad (r)
(b)
grad (rm )
(c)
1
grad ( )
r
(d) grad (ln r)
(e)
grad (φ)
where φ = φ(r) is an arbitrary function of r.
I 10. Let ~r = x ê1 +y ê2 +z ê3 denote the position vector of a variable point (x, y, z) in Cartesian coordinates.
Let r = |~r| denote the distance of this point from the origin. Find:
(a)
div (~r) (b) div (rm~r) (c)
div (r−3 ~r) (d)
div (φ ~r)
where φ = φ(r) is an arbitrary function or r.
I 11.
Let ~r = x ê1 + y ê2 + z ê3 denote the position vector of a variable point (x, y, z) in Cartesian
coordinates. Let r = |~r| denote the distance of this point from the origin. Find: (a)
curl ~r
where φ = φ(r) is an arbitrary function of r.
~
I 12. Expand and simplify the representation for curl (curl A).
I 13. Show that the curl of the gradient is zero in generalized coordinates.
I 14. Write out the physical components associated with the gradient of φ = φ(x1 , x2 , x3 ).
I 15. Show that
1 ∂ √ im
1 ∂ √ i gg Am = Ai,i = √
gA .
g im Ai,m = √
i
g ∂x
g ∂xi
(b) curl (φ ~r)
183
I 16. Let r = (~r · ~r)1/2 =
I 17.
p
x2 + y 2 + z 2 ) and calculate (a) ∇2 (r)
Given the tensor equations Dij =
1
2 (vi,j
+ vj,i ),
(b) ∇2 (1/r) (c) ∇2 (r2 )
(d) ∇2 (1/r2 )
i, j = 1, 2, 3. Let v(1), v(2), v(3) denote the
physical components of v1 , v2 , v3 and let D(ij) denote the physical components associated with Dij . Assume
the coordinate system (x1 , x2 , x3 ) is orthogonal with metric coefficients g(i)(i) = h2i , i = 1, 2, 3 and gij = 0
for i 6= j.
(a) Find expressions for the physical components D(11), D(22) and D(33) in terms of the physical compo1 ∂V (i) X V (j) ∂hi
+
no sum on i.
nents v(i), i = 1, 2, 3. Answer: D(ii) =
hi ∂xi
hi hj ∂xj
j6=i
(b) Find expressions for the physical components
D(12),
D(13)
and D(23)
in terms
of the physical compoV (i)
hj ∂
V (j)
1 hi ∂
+
nents v(i), i = 1, 2, 3. Answer: D(ij) =
2 hj ∂xj
hi
hi ∂xi
hj
I 18. Write out the tensor equations in problem 17 in Cartesian coordinates.
I 19. Write out the tensor equations in problem 17 in cylindrical coordinates.
I 20. Write out the tensor equations in problem 17 in spherical coordinates.
I 21. Express the vector equation (λ + 2µ)∇Φ − 2µ∇ × ~ω + F~ = ~0 in tensor form.
I 22. Write out the equations in problem 21 for a generalized orthogonal coordinate system in terms of
physical components.
I 23. Write out the equations in problem 22 for cylindrical coordinates.
I 24. Write out the equations in problem 22 for spherical coordinates.
I 25. Use equation (2.1.4) to represent the divergence in parabolic cylindrical coordinates (ξ, η, z).
I 26. Use equation (2.1.4) to represent the divergence in parabolic coordinates (ξ, η, φ).
I 27. Use equation (2.1.4) to represent the divergence in elliptic cylindrical coordinates (ξ, η, z).
Change the given equations from a vector notation to a tensor notation.
I 28.
I 29.
I 30.
I 31.
I 32.
~ = ~v ∇ · A
~ + (∇ · ~v ) A
~
B
~
~
~
d ~ ~
~ = dA · (B
~ × C)
~ +A
~ · ( dB × C)
~ +A
~ · (B
~ × dC )
[A · (B × C)]
dt
dt
dt
dt
∂~v
d~v
=
+ (~v · ∇)~v
dt
∂t
~
1 ∂H
~
= −curl E
c ∂t
~
dB
~ · ∇)~v + B(∇
~
− (B
· ~v ) = ~0
dt
184
Change the given equations from a tensor notation to a vector notation.
I 33.
ijk Bk,j + F i = 0
I 34.
gij jkl Bl,k + Fi = 0
∂%
+ (%vi ), i = 0
∂t
∂P
∂ 2 vi
∂vi
∂vi
+ vm m ) = − i + µ m m + Fi
%(
∂t
∂x
∂x
∂x ∂x
I 35.
I 36.
Z Z
I 37. The moment of inertia of an area or second moment of area is defined by Iij =
A
(ym ym δij −yi yj ) dA
where dA is an element of area. Calculate
of inertia
Iij , i, j = 1, 2 for the triangle illustrated in
1 the3 moment
1 2 2
bh
−
b
h
12
24
the figure 2.1-1 and show that Iij =
.
1 2 2
1 3
− 24
b h
12 b h
Figure 2.1-1 Moments of inertia for a triangle
I 38. Use the results from problem 37 and rotate the axes in figure 2.1-1 through an angle θ to a barred
system of coordinates.
(a) Show that in the barred system of coordinates
I11 + I22
I11 − I22
+
cos 2θ + I12 sin 2θ
2
2
I11 − I22
=−
sin 2θ + I12 cos 2θ
2
I11 + I22
I11 − I22
=
−
cos 2θ − I12 sin 2θ
2
2
I 11 =
I 12 = I 21
I 22
(b) For what value of θ will I 11 have a maximum value?
(c) Show that when I 11 is a maximum, we will have I 22 a minimum and I 12 = I 21 = 0.
185
Figure 2.1-2 Mohr’s circle
I 39. Otto Mohr1 gave the following physical interpretation to the results obtained in problem 38:
• Plot the points A(I11 , I12 ) and B(I22 , −I12 ) as illustrated in the figure 2.1-2
• Draw the line AB and calculate the point C where this line intersects the I axes. Show the point C
has the coordinates
I11 + I22
, 0)
2
• Calculate the radius of the circle with center at the point C and with diagonal AB and show this
(
s
radius is
r=
I11 − I22
2
2
2
+ I12
• Show the maximum and minimum values of I occur where the constructed circle intersects the I axes.
I11 + I22
I11 + I22
+r
Imin = I 22 =
− r.
Show that Imax = I 11 =
2
2
I11 I12
I 40. Show directly that the eigenvalues of the symmetric matrix Iij =
are λ1 = Imax and
I21 I22
λ2 = Imin where Imax and Imin are given in problem 39.
I 41. Find the principal axes and moments of inertia for the triangle given in problem 37 and summarize
your results from problems 37,38,39, and 40.
I 42. Verify for orthogonal coordinates the relations
h
3
i
X
e(i)jk
∂(h(k) A(k))
~ · ê(i) =
h(i)
∇×A
h1 h2 h3
∂xj
k=1
or
~=
∇×A
1
h1 h2 h3
h1 ê1
∂
∂x1
h1 A(1)
h2 ê2
I 43. Verify for orthogonal coordinates the relation
h
1
3
i
X
h(i)
∂
~ · ê(i) =
∇ × (∇ × A)
e(i)jr ersm
h1 h2 h3 ∂xj
m=1
Christian Otto Mohr (1835-1918) German civil engineer.
h3 ê3
∂
∂x2
h2 A(2)
∂
∂x3
h3 A(3)
"
.
h2(r) ∂(h(m) A(m))
h1 h2 h3
∂xs
#
186
I 44. Verify for orthogonal coordinates the relation
h i
1
∂
1
∂(h2 h3 A(1)) ∂(h1 h3 A(2)) ∂(h1 h2 A(3))
~
+
+
∇ ∇·A
· ê(i) =
h(i) ∂x(i) h1 h2 h3
∂x1
∂x2
∂x3
I 45. Verify the relation
3
h
i
X
∂h(i)
∂hk
A(k) ∂B(i) X B(k)
~ · ∇)B
~ · ê(i) =
+
− A(k)
A(i)
(A
h(k) ∂xk
hk h(i)
∂xk
∂x(i)
k=1
k6=i
I 46. The Gauss divergence theorem is written
ZZ
ZZZ 1
∂F 2
∂F 3
∂F
+
+
dτ =
n1 F 1 + n2 F 2 + n3 F 3 dσ
∂x
∂y
∂z
V
S
where V is the volume within a simple closed surface S. Here it is assumed that F i = F i (x, y, z) are
continuous functions with continuous first order derivatives throughout V and ni are the direction cosines
of the outward normal to S, dτ is an element of volume and dσ is an element of surface area.
(a) Show that in a Cartesian coordinate system
∂F 2
∂F 3
∂F 1
+
+
∂x
∂y
∂z
ZZ
ZZZ
F,ii dτ =
F i ni dσ.
and that the tensor form of this theorem is
F,ii =
V
(b) Write the vector form of this theorem.
S
(c) Show that if we define
∂u
∂v
, vr =
∂xr
∂xr
= g im (uvi,m + um vi )
ur =
then F,ii = g im Fi,m
and Fr = grm F m = uvr
(d) Show that another form of the Gauss divergence theorem is
ZZ
ZZZ
ZZZ
im
m
g um vi dτ =
uvm n dσ −
ug im vi,m dτ
V
S
Write out the above equation in Cartesian coordinates.
I 47.
Show
I 48.
Show
I 49.
Show
V

1
Find the eigenvalues and eigenvectors associated with the matrix A =  1
2
that the eigenvectors are orthogonal.

1
Find the eigenvalues and eigenvectors associated with the matrix A =  2
1
that the eigenvectors are orthogonal.

1
Find the eigenvalues and eigenvectors associated with the matrix A =  1
0
that the eigenvectors are orthogonal.

1 2
2 1.
1 1

2 1
1 0.
0 1

1 0
1 1.
1 1
I 50. The harmonic and biharmonic functions or potential functions occur in the mathematical modeling
of many physical problems. Any solution of Laplace’s equation ∇2 Φ = 0 is called a harmonic function and
any solution of the biharmonic equation ∇4 Φ = 0 is called a biharmonic function.
(a) Expand the Laplace equation in Cartesian, cylindrical and spherical coordinates.
(b) Expand the biharmonic equation in two dimensional Cartesian and polar coordinates.
Hint: Consider ∇4 Φ = ∇2 (∇2 Φ). In Cartesian coordinates ∇2 Φ = Φ,ii and ∇4 Φ = Φ,iijj .
187
§2.2 DYNAMICS
Dynamics is concerned with studying the motion of particles and rigid bodies. By studying the motion
of a single hypothetical particle, one can discern the motion of a system of particles. This in turn leads to
the study of the motion of individual points in a continuous deformable medium.
Particle Movement
The trajectory of a particle in a generalized coordinate system is described by the parametric equations
xi = xi (t),
i = 1, . . . , N
(2.2.1)
where t is a time parameter. If the coordinates are changed to a barred system by introducing a coordinate
transformation
xi = xi (x1 , x2 , . . . , xN ),
i = 1, . . . , N
then the trajectory of the particle in the barred system of coordinates is
xi = xi (x1 (t), x2 (t), . . . , xN (t)),
i = 1, . . . , N.
(2.2.2)
The generalized velocity of the particle in the unbarred system is defined by
vi =
dxi
,
dt
i = 1, . . . , N.
(2.2.3)
By the chain rule differentiation of the transformation equations (2.2.2) one can verify that the velocity in
the barred system is
vr =
∂xr dxj
∂xr j
dxr
=
=
v ,
j
dt
∂x dt
∂xj
r = 1, . . . , N.
(2.2.4)
Consequently, the generalized velocity v i is a first order contravariant tensor. The speed of the particle is
obtained from the magnitude of the velocity and is
v 2 = gij v i v j .
The generalized acceleration f i of the particle is defined as the intrinsic derivative of the generalized velocity.
The generalized acceleration has the form
n
dv i
δv i
i dx
= v,n
=
+
f =
δt
dt
dt
i
m n
i
d2 xi
i
dx dx
m n
v v =
+
mn
m n dt dt
dt2
and the magnitude of the acceleration is
f 2 = gij f i f j .
(2.2.5)
188
Figure 2.2-1 Tangent, normal and binormal to point P on curve.
Frenet-Serret Formulas
The parametric equations (2.2.1) describe a curve in our generalized space. With reference to the figure
2.2-1 we wish to define at each point P of the curve the following orthogonal unit vectors:
T i = unit tangent vector at each point P.
N i = unit normal vector at each point P.
B i = unit binormal vector at each point P.
These vectors define the osculating, normal and rectifying planes illustrated in the figure 2.2-1.
In the generalized coordinates the arc length squared is
ds2 = gij dxi dxj .
Define T i =
dxi
ds
as the tangent vector to the parametric curve defined by equation (2.2.1). This vector is a
unit tangent vector because if we write the element of arc length squared in the form
1 = gij
dxi dxj
= gij T i T j ,
ds ds
(2.2.6)
we obtain the generalized dot product for T i . This generalized dot product implies that the tangent vector
is a unit vector. Differentiating the equation (2.2.6) intrinsically with respect to arc length s along the curve
produces
gmn
δT m n
δT n
T + gmn T m
= 0,
δs
δs
which simplifies to
gmn T n
δT m
= 0.
δs
(2.2.7)
189
The equation (2.2.7) is a statement that the vector
vector is defined as
Ni =
1 δT i
κ δs
δT m
δs
is orthogonal to the vector T m . The unit normal
or
Ni =
1 δTi
,
κ δs
(2.2.8)
where κ is a scalar called the curvature and is chosen such that the magnitude of N i is unity. The reciprocal
of the curvature is R =
1
κ,
which is called the radius of curvature. The curvature of a straight line is zero
while the curvature of a circle is a constant. The curvature measures the rate of change of the tangent vector
as the arc length varies.
The equation (2.2.7) can be expressed in the form
gij T i N j = 0.
(2.2.9)
Taking the intrinsic derivative of equation (2.2.9) with respect to the arc length s produces
gij T i
or
gij T i
δN j
δT i j
+ gij
N =0
δs
δs
δN j
δT i j
= −gij
N = −κgij N i N j = −κ.
δs
δs
(2.2.10)
The generalized dot product can be written
gij T i T j = 1,
and consequently we can express equation (2.2.10) in the form
gij T
i δN
δs
j
= −κgij T T
i
j
or
gij T
i
δN j
+ κT j
δs
= 0.
(2.2.11)
Consequently, the vector
δN j
+ κT j
δs
(2.2.12)
is orthogonal to T i . In a similar manner, we can use the relation gij N i N j = 1 and differentiate intrinsically
with respect to the arc length s to show that
gij N i
δN j
= 0.
δs
This in turn can be expressed in the form
gij N
i
δN j
+ κT j
δs
= 0.
This form of the equation implies that the vector represented in equation (2.2.12) is also orthogonal to the
unit normal N i . We define the unit binormal vector as
1 δN i
i
i
+ κT
B =
or
τ
δs
1
Bi =
τ
δNi
+ κTi
δs
(2.2.13)
where τ is a scalar called the torsion. The torsion is chosen such that the binormal vector is a unit vector.
The torsion measures the rate of change of the osculating plane and consequently, the torsion τ is a measure
190
of the twisting of the curve out of a plane. The value τ = 0 corresponds to a plane curve. The vectors
T i , N i , B i , i = 1, 2, 3 satisfy the cross product relation
B i = ijk Tj Nk .
If we differentiate this relation intrinsically with respect to arc length s we find
δTj
δNk
δB i
= ijk Tj
+
Nk
δs
δs
δs
= ijk [Tj (τ Bk − κTk ) + κNj Nk ]
(2.2.14)
= τ ijk Tj Bk = −τ ikj Bk Tj = −τ N i .
The relations (2.2.8),(2.2.13) and (2.2.14) are now summarized and written
δT i
= κN i
δs
δN i
= τ B i − κT i
δs
δB i
= −τ N i .
δs
(2.2.15)
These equations are known as the Frenet-Serret formulas of differential geometry.
Velocity and Acceleration
Chain rule differentiation of the generalized velocity is expressible in the form
vi =
where v =
ds
dt
dxi ds
dxi
=
= T i v,
dt
ds dt
(2.2.16)
is the speed of the particle and is the magnitude of v i . The vector T i is the unit tangent vector
to the trajectory curve at the time t. The equation (2.2.16) is a statement of the fact that the velocity of a
particle is always in the direction of the tangent vector to the curve and has the speed v.
By chain rule differentiation, the generalized acceleration is expressible in the form
fr =
dv r
δv r
δT r
=
T +v
δt
dt
δt
δT r ds
dv r
T +v
=
dt
δs dt
dv r
T + κv 2 N r .
=
dt
(2.2.17)
The equation (2.2.17) states that the acceleration lies in the osculating plane. Further, the equation (2.2.17)
indicates that the tangential component of the acceleration is
2
eration is κv .
dv
dt ,
while the normal component of the accel-
191
Work and Potential Energy
Define M as the constant mass of the particle as it moves along the curve defined by equation (2.2.1).
Also let Qr denote the components of a force vector (in appropriate units of measurements) which acts upon
the particle. Newton’s second law of motion can then be expressed in the form
Qr = M f r
or
Qr = M fr .
(2.2.18)
The work done W in moving a particle from a point P0 to a point P1 along a curve xr = xr (t), r = 1, 2, 3,
with parameter t, is represented by a summation of the tangential components of the forces acting along the
path and is defined as the line integral
Z
P1
W =
P0
dxr
ds =
Qr
ds
Z
P1
Z
r
t1
Qr dx =
P0
t0
dxr
dt =
Qr
dt
Z
t1
Qr v r dt
(2.2.19)
t0
where Qr = grs Qs is the covariant form of the force vector, t is the time parameter and s is arc length along
the curve.
Conservative Systems
If the force vector is conservative it means that the force is derivable from a scalar potential function
V = V (x1 , x2 , . . . , xN )
such that
Qr = −V ,r = −
∂V
,
∂xr
r = 1, . . . , N.
(2.2.20)
In this case the equation (2.2.19) can be integrated and we find that to within an additive constant we will
have V = −W. The potential function V is called the potential energy of the particle and the work done
becomes the change in potential energy between the starting and end points and is independent of the path
connecting the points.
Lagrange’s Equations of Motion
The kinetic energy T of the particle is defined as one half the mass times the velocity squared and can
be expressed in any of the forms
1
T = M
2
ds
dt
2
=
1
1
1
M v 2 = M gmn v m v n = M gmn ẋm ẋn ,
2
2
2
(2.2.21)
where the dot notation denotes differentiation with respect to time. It is an easy exercise to calculate the
derivatives
∂T
= M grmẋm
r
∂
ẋ
∂grm n m
d ∂T
m
ẍ
+
ẋ
ẋ
=
M
g
rm
dt ∂ ẋr
∂xn
1 ∂gmn m n
∂T
= M
ẋ ẋ ,
r
∂x
2
∂xr
(2.2.22)
and thereby verify the relation
d
dt
∂T
∂ ẋr
−
∂T
= M fr = Qr ,
∂xr
r = 1, . . . , N.
(2.2.23)
192
This equation is called the Lagrange’s form of the equations of motion.
EXAMPLE 2.2-1. (Equations of motion in spherical coordinates)
Find the Lagrange’s form of
the equations of motion in spherical coordinates.
Solution: Let x1 = ρ, x2 = θ, x3 = φ then the element of arc length squared in spherical coordinates has
the form
ds2 = (dρ)2 + ρ2 (dθ)2 + ρ2 sin2 θ(dφ)2 .
The element of arc length squared can be used to construct the kinetic energy. For example,
2
i
ds
1 h
1
= M (ρ̇)2 + ρ2 (θ̇)2 + ρ2 sin2 θ(φ̇)2 .
T = M
2
dt
2
The Lagrange form of the equations of motion of a particle are found from the relations (2.2.23) and are
calculated to be:
d ∂T
−
dt ∂ ρ̇
d ∂T
−
M f2 = Q2 =
dt ∂ θ̇
d ∂T
−
M f3 = Q3 =
dt ∂ φ̇
M f1 = Q1 =
h
i
∂T
= M ρ̈ − ρ(θ̇)2 − ρ sin2 θ(φ̇)2
∂ρ
d
∂T
=M
ρ2 θ̇ − ρ2 sin θ cos θ(φ̇)2
∂θ
dt
d
∂T
=M
ρ2 sin2 θφ̇ .
∂φ
dt
In terms of physical components we have
h
i
Qρ = M ρ̈ − ρ(θ̇)2 − ρ sin2 θ(φ̇)2
M d 2 ρ θ̇ − ρ2 sin θ cos θ(φ̇)2
Qθ =
ρ dt
d
M
ρ2 sin2 θφ̇ .
Qφ =
ρ sin θ dt
Euler-Lagrange Equations of Motion
Starting with the Lagrange’s form of the equations of motion from equation (2.2.23), we assume that
the external force Qr is derivable from a potential function V as specified by the equation (2.2.20). That is,
we assume the system is conservative and express the equations of motion in the form
∂V
∂T
d ∂T
− r = − r = Qr , r = 1, . . . , N
dt ∂ ẋr
∂x
∂x
(2.2.24)
The Lagrangian is defined by the equation
L = T − V = T (x1 , . . . , xN , ẋ1 , . . . , ẋN ) − V (x1 , . . . , xN ) = L(xi , ẋi ).
(2.2.25)
Employing the defining equation (2.2.25), it is readily verified that the equations of motion are expressible
in the form
d
dt
∂L
∂ ẋr
−
∂L
= 0,
∂xr
r = 1, . . . , N,
which are called the Euler-Lagrange form for the equations of motion.
(2.2.26)
193
Figure 2.2-2 Simply pulley system
EXAMPLE 2.2-2. (Simple pulley system) Find the equation of motion for the simply pulley system
illustrated in the figure 2.2-2.
Solution: The given system has only one degree of freedom, say y1 . It is assumed that
y1 + y2 = ` = a constant.
The kinetic energy of the system is
T =
1
(m1 + m2 )ẏ12 .
2
Let y1 increase by an amount dy1 and show the work done by gravity can be expressed as
dW = m1 g dy1 + m2 g dy2
dW = m1 g dy1 − m2 g dy1
dW = (m1 − m2 )g dy1 = Q1 dy1 .
Here Q1 = (m1 − m2 )g is the external force acting on the system where g is the acceleration of gravity. The
Lagrange equation of motion is
d
dt
∂T
∂ ẏ1
−
∂T
= Q1
∂y1
or
(m1 + m2 )ÿ1 = (m1 − m2 )g.
Initial conditions must be applied to y1 and ẏ1 before this equation can be solved.
194
EXAMPLE 2.2-3. (Simple pendulum) Find the equation of motion for the pendulum system illustrated in the figure 2.2-3.
Solution: Choose the angle θ illustrated in the figure 2.2-3 as the generalized coordinate. If the pendulum
is moved from a vertical position through an angle θ, we observe that the mass m moves up a distance
h = ` − ` cos θ. The work done in moving this mass a vertical distance h is
W = −mgh = −mg`(1 − cos θ),
since the force is −mg in this coordinate system. In moving the pendulum through an angle θ, the arc length
s swept out by the mass m is s = `θ. This implies that the kinetic energy can be expressed
2
ds
1 2
1
1
= m `θ̇ = m`2 (θ̇)2 .
T = m
2
dt
2
2
Figure 2.2-3 Simple pendulum system
The Lagrangian of the system is
L=T −V =
1 2 2
m` (θ̇) − mg`(1 − cos θ)
2
and from this we find the equation of motion
∂L
d ∂L
=0
or
−
dt ∂ θ̇
∂θ
d 2 m` θ̇ − mg`(− sin θ) = 0.
dt
This in turn simplifies to the equation
θ̈ +
g
sin θ = 0.
`
This equation together with a set of initial conditions for θ and θ̇ represents the nonlinear differential equation
which describes the motion of a pendulum without damping.
195
EXAMPLE 2.2-4. (Compound pendulum)
Find the equations of motion for the compound pendulum
illustrated in the figure 2.2-4.
Solution: Choose for the generalized coordinates the angles x1 = θ1 and x2 = θ2 illustrated in the figure
2.2-4. To find the potential function V for this system we consider the work done as the masses m1 and
m2 are moved. Consider independent motions of the angles θ1 and θ2 . Imagine the compound pendulum
initially in the vertical position as illustrated in the figure 2.2-4(a). Now let m1 be displaced due to a change
in θ1 and obtain the figure 2.2-4(b). The work done to achieve this position is
W1 = −(m1 + m2 )gh1 = −(m1 + m2 )gL1 (1 − cos θ1 ).
Starting from the position in figure 2.2-4(b) we now let θ2 undergo a displacement and achieve the configuration in the figure 2.2-4(c).
Figure 2.2-4 Compound pendulum
The work done due to the displacement θ2 can be represented
W2 = −m2 gh2 = −m2 gL2 (1 − cos θ2 ).
Since the potential energy V satisfies V = −W to within an additive constant, we can write
V = −W = −W1 − W2 = −(m1 + m2 )gL1 cos θ1 − m2 gL2 cos θ2 + constant,
where the constant term in the potential energy has been neglected since it does not contribute anything to
the equations of motion. (i.e. the derivative of a constant is zero.)
The kinetic energy term for this system can be represented
2
2
1
ds1
ds2
1
m1
+ m2
2
dt
2
dt
1
1
T = m1 (ẋ21 + ẏ12 ) + m2 (ẋ22 + ẏ22 ),
2
2
T =
(2.2.27)
196
where
(x1 , y1 ) = (L1 sin θ1 , −L1 cos θ1 )
(x2 , y2 ) = (L1 sin θ1 + L2 sin θ2 , −L1 cos θ1 − L2 cos θ2 )
(2.2.28)
are the coordinates of the masses m1 and m2 respectively. Substituting the equations (2.2.28) into equation
(2.2.27) and simplifying produces the kinetic energy expression
T =
1
1
(m1 + m2 )L21 θ̇12 + m2 L1 L2 θ̇1 θ̇2 cos(θ1 − θ2 ) + m2 L22 θ̇22 .
2
2
(2.2.29)
Writing the Lagrangian as L = T − V , the equations describing the motion of the compound pendulum
are obtained from the Lagrangian equations
d
dt
∂L
∂ θ̇1
∂L
=0
−
∂θ1
d
dt
and
∂L
∂ θ̇2
−
∂L
= 0.
∂θ2
Calculating the necessary derivatives, substituting them into the Lagrangian equations of motion and then
simplifying we derive the equations of motion
L1 θ̈1 +
m2
m2
L2 θ̈2 cos(θ1 − θ2 ) +
L2 (θ̇2 )2 sin(θ1 − θ2 ) + g sin θ1 = 0
m1 + m2
m1 + m2
L1 θ̈1 cos(θ1 − θ2 ) + L2 θ̈2 − L1 (θ̇1 )2 sin(θ1 − θ2 ) + g sin θ2 = 0.
These equations are a set of coupled, second order nonlinear ordinary differential equations. These equations
are subject to initial conditions being imposed upon the angular displacements (θ1 , θ2 ) and the angular
velocities (θ̇1 , θ̇2 ).
Alternative Derivation of Lagrange’s Equations of Motion
Let c denote a given curve represented in the parametric form
xi = xi (t),
i = 1, . . . , N,
t0 ≤ t ≤ t1
and let P0 , P1 denote two points on this curve corresponding to the parameter values t0 and t1 respectively.
Let c denote another curve which also passes through the two points P0 and P1 as illustrated in the figure
2.2-5.
The curve c is represented in the parametric form
xi = xi (t) = xi (t) + η i (t),
i = 1, . . . , N,
t0 ≤ t ≤ t1
in terms of a parameter . In this representation the function η i (t) must satisfy the end conditions
η i (t0 ) = 0 and η i (t1 ) = 0
i = 1, . . . , N
since the curve c is assumed to pass through the end points P0 and P1 .
Consider the line integral
Z
t1
I() =
t0
L(t, xi + η i , ẋi + η̇ i ) dt,
(2.2.30)
197
Figure 2.2-5. Motion along curves c and c
where
i
L = T − V = L(t, xi , ẋ )
is the Lagrangian evaluated along the curve c. We ask the question, “What conditions must be satisfied by
the curve c in order that the integral I() have an extremum value when is zero?”If the integral I() has
a minimum value when is zero it follows that its derivative with respect to will be zero at this value and
we will have
dI()
d
Employing the definition
dI
d
= 0.
=0
I() − I(0)
= I 0 (0) = 0
→0
= lim
=0
we expand the Lagrangian in equation (2.2.30) in a series about the point = 0. Substituting the expansion
∂L i
∂L i
η + i η̇ + 2 [ ] + · · ·
L(t, x + η , ẋ + η̇ ) = L(t, x , ẋ ) + ∂xi
∂ ẋ
i
i
i
i
i
i
into equation (2.2.30) we calculate the derivative
I() − I(0)
= lim
I (0) = lim
→0
→0
0
Z
t1
t0
∂L i
∂L i
η (t) + i η̇ (t) dt + [ ] + · · · = 0,
∂xi
∂ ẋ
where we have neglected higher order powers of since is approaching zero. Analysis of this equation
informs us that the integral I has a minimum value at = 0 provided that the integral
Z
t1
δI =
t0
∂L i
∂L i
η
(t)
+
η̇
(t)
dt = 0
∂xi
∂ ẋi
(2.2.31)
198
is satisfied. Integrating the second term of this integral by parts we find
Z
t1
δI =
t0
t1 Z t1
∂L i
∂L i
d ∂L
η dt +
η (t)
−
η i (t) dt = 0.
∂xi
∂ ẋi
∂ ẋi
t0 dt
t0
(2.2.32)
The end condition on η i (t) makes the middle term in equation (2.2.32) vanish and we are left with the
integral
Z
t1
δI =
η i (t)
t0
∂L
d
−
∂xi
dt
∂L
∂ ẋi
dt = 0,
(2.2.33)
which must equal zero for all η i (t). Since η i (t) is arbitrary, the only way the integral in equation (2.2.33) can
be zero for all η i (t) is for the term inside the brackets to vanish. This produces the result that the integral
of the Lagrangian is an extremum when the Euler-Lagrange equations
d
dt
∂L
∂ ẋi
∂L
= 0,
∂xi
−
i = 1, . . . , N
(2.2.34)
are satisfied. This is a necessary condition for the integral I() to have a minimum value.
In general, any line integral of the form
Z
t1
I=
φ(t, xi , ẋi ) dt
(2.2.35)
t0
has an extremum value if the curve c defined by xi = xi (t), i = 1, . . . , N satisfies the Euler-Lagrange
equations
d
dt
∂φ
∂ ẋi
−
∂φ
= 0,
∂xi
i = 1, . . . , N.
(2.2.36)
The above derivation is a special case of (2.2.36) when φ = L. Note that the equations of motion equations
(2.2.34) are just another form of the equations (2.2.24). Note also that
δ
δT
=
δt
δt
1
mgij v i v j
2
= mgij v i f j = mfi v i = mfi ẋ i
∂V
and if we assume that the force Qi is derivable from a potential function V , then mfi = Qi = − i , so
∂x
δ
∂V i
δV
δT
i
i
= mfi ẋ = Qi ẋ = − i ẋ = −
or (T + V ) = 0 or T + V = h = constant called the energy
that
δt
∂x
δt
δt
constant of the system.
Action Integral
The equations of motion (2.2.34) or (2.2.24) are interpreted as describing geodesics in a space whose
line-element is
ds2 = 2m(h − V )gjk dxj dxk
where V is the potential function for the force system and T + V = h is the energy constant of the motion.
The integral of ds along a curve C between two points P1 and P2 is called an action integral and is
Z
√
A = 2m
P2
P1
1/2
dxj dxk
dτ
(h − V )gjk
dτ dτ
199
where τ is a parameter used to describe the curve C. The principle of stationary action states that of all
curves through the points P1 and P2 the one which makes the action an extremum is the curve specified by
Newton’s second law. The extremum is usually a minimum. To show this let
φ=
√
1/2
dxj dxk
2m (h − V )gjk
dτ dτ
in equation (2.2.36). Using the notation ẋ k =
dxk
dτ
we find that
2m
∂φ
(h − V )gik ẋ k
=
i
∂ ẋ
φ
∂gjk j k 2m ∂V
∂φ 2m
(h − V )
=
ẋ ẋ −
gjk ẋ j ẋ k .
∂xi 2φ
∂xi
2φ ∂xi
The equation (2.2.36) which describe the extremum trajectories are found to be
∂gjk j k 2m ∂V
2m
d 2m
(h − V )gik ẋ k −
(h − V )
ẋ ẋ +
gjk ẋ j ẋ k = 0.
dt φ
2φ
∂xi
φ ∂xi
By changing variables from τ to t where
satisfy the equation
d
m
dt
dt
dτ
=
√
√
mφ
2(h−V )
we find that the trajectory for an extremum must
∂V
dxk
m ∂gjk dxj dxk
+ i =0
gik
−
i
dt
2 ∂x dt dt
∂x
which are the same equations as (2.2.24). (i.e. See also the equations (2.2.22).)
Dynamics of Rigid Body Motion
Let us derive the equations of motion of a rigid body which is rotating due to external forces acting
upon it. We neglect any translational motion of the body since this type of motion can be discerned using
our knowledge of particle dynamics. The derivation of the equations of motion is restricted to Cartesian
tensors and rotational motion.
Consider a system of N particles rotating with angular velocity ωi , i = 1, 2, 3, about a line L through
the center of mass of the system. Let V~ (α) denote the velocity of the αth particle which has mass m(α) and
(α)
position xi , i = 1, 2, 3 with respect to an origin on the line L. Without loss of generality we can assume
that the origin of the coordinate system is also at the center of mass of the system of particles, as this choice
of an origin simplifies the derivation. The velocity components for each particle is obtained by taking cross
products and we can write
~ (α) = ~
ω × ~r (α)
V
or
(α)
Vi
(α)
= eijk ωj xk .
(2.2.37)
The kinetic energy of the system of particles is written as the sum of the kinetic energies of each
individual particle and is
N
N
1X
1X
(α) (α)
(α)
m(α) Vi Vi =
m(α) eijk ωj xk eimn ωm x(α)
T =
n .
2 α=1
2 α=1
(2.2.38)
200
Employing the e − δ identity the equation (2.2.38) can be simplified to the form
T =
N
1X
(α) (α)
(α)
m(α) ωm ωm xk xk − ωn ωk xk x(α)
.
n
2 α=1
Define the second moments and products of inertia by the equation
Iij =
N
X
(α) (α)
(α) (α)
m(α) xk xk δij − xi xj
(2.2.39)
α=1
and write the kinetic energy in the form
T =
1
Iij ωi ωj .
2
(2.2.40)
Similarly, the angular momentum of the system of particles can also be represented in terms of the
second moments and products of inertia. The angular momentum of a system of particles is defined as a
summation of the moments of the linear momentum of each individual particle and is
Hi =
N
X
(α) (α)
m(α) eijk xj vk
=
α=1
N
X
(α)
m(α) eijk xj ekmn ωm x(α)
n .
(2.2.41)
α=1
The e − δ identity simplifies the equation (2.2.41) to the form
Hi = ω j
N
X
(α) (α)
(α)
m(α) x(α)
= ωj Iji .
n xn δij − xj xi
(2.2.42)
α=1
The equations of motion of a rigid body is obtained by applying Newton’s second law of motion to the
system of N particles. The equation of motion of the αth particle is written
(α)
m(α) ẍi
(α)
= Fi
.
(2.2.43)
Summing equation (2.2.43) over all particles gives the result
N
X
(α)
m(α) ẍi
=
α=1
N
X
(α)
Fi
.
(2.2.44)
α=1
This represents the translational equations of motion of the rigid body. The equation (2.2.44) represents the
rate of change of linear momentum being equal to the total external force acting upon the system. Taking
(α)
the cross product of equation (2.2.43) with the position vector xj
(α)
produces
(α)
= erst x(α)
m(α) ẍt erst x(α)
s
s Ft
and summing over all particles we find the equation
N
X
α=1
(α)
m(α) erst x(α)
s ẍt
=
N
X
α=1
(α)
erst x(α)
s Ft
.
(2.2.45)
201
The equations (2.2.44) and (2.2.45) represent the conservation of linear and angular momentum and can be
written in the forms
d
dt
and
d
dt
!
m(α) ẋ(α)
r
=
α=1
N
X
Fr(α)
(2.2.46)
α=1
!
(α)
m(α) erst x(α)
s ẋt
=
α=1
N
X
(α)
erst x(α)
s Ft
.
(2.2.47)
α=1
P (α)
representing the linear momentum, Fr =
Fr the total force
P
(α) (α)
acting on the system of particles, Hr = m(α) erst xs ẋt is the angular momentum of the system relative
P
(α) (α)
erst xs Ft is the total moment of the system relative to the origin. We can
to the origin, and Mr =
By definition we have Gr =
P
N
X
N
X
(α)
m(α) ẋr
therefore express the equations (2.2.46) and (2.2.47) in the form
dGr
= Fr
dt
(2.2.48)
dHr
= Mr .
dt
(2.2.49)
and
The equation (2.2.49) expresses the fact that the rate of change of angular momentum is equal to the
moment of the external forces about the origin. These equations show that the motion of a system of
particles can be studied by considering the motion of the center of mass of the system (translational motion)
and simultaneously considering the motion of points about the center of mass (rotational motion).
We now develop some relations in order to express the equations (2.2.49) in an alternate form. Toward
this purpose we consider first the concepts of relative motion and angular velocity.
Relative Motion and Angular Velocity
Consider two different reference frames denoted by S and S. Both reference frames are Cartesian
coordinates with axes xi and xi , i = 1, 2, 3, respectively. The reference frame S is fixed in space and is
called an inertial reference frame or space-fixed reference system of axes. The reference frame S is fixed
to and rotates with the rigid body and is called a body-fixed system of axes. Again, for convenience, it
is assumed that the origins of both reference systems are fixed at the center of mass of the rigid body.
ei , i = 1, 2, 3, while the reference system S has the basis
Further, we let the system S have the basis vectors b
vectors êi , i = 1, 2, 3. The transformation equations between the two sets of reference axes are the affine
transformations
xi = `ji xj
and
xi = `ij xj
(2.2.50)
where `ij = `ij (t) are direction cosines which are functions of time t (i.e. the `ij are the cosines of the
angles between the barred and unbarred axes where the barred axes are rotating relative to the space-fixed
unbarred axes.) The direction cosines satisfy the relations
`ij `ik = δjk
and
`ij `kj = δik .
(2.2.51)
202
EXAMPLE 2.2-5. (Euler angles φ, θ, ψ)
Consider the following sequence of transformations which
are used in celestial mechanics. First a rotation about the x3 axis taking the xi axes to the yi axes
 
  
cos φ sin φ 0
x1
y1
 y2  =  − sin φ cos φ 0   x2 
y3
x3
0
0
1
where the rotation angle φ is called the longitude of the ascending node. Second, a rotation about the y1
axis taking the yi axes to the yi0 axes
 
 0 
1
0
0
y1
y1
 y20  =  0 cos θ sin θ   y2 
y30
y3
0 − sin θ cos θ
where the rotation angle θ is called the angle of inclination of the orbital plane. Finally, a rotation about
the y30 axis taking the yi0 axes to the x̄i axes
 0 
  
cos ψ
sin ψ 0
y1
x̄1
 x̄2  =  − sin ψ cos ψ 0   y20 
x̄3
y30
0
0
1
where the rotation angle ψ is called the argument of perigee. The Euler angle θ is the angle x̄3 0x3 , the angle
φ is the angle x1 0y1 and ψ is the angle y1 0x̄1 . These angles are illustrated in the figure 2.2-6. Note also that
the rotation vectors associated with these transformations are vectors of magnitude φ̇, θ̇, ψ̇ in the directions
indicated in the figure 2.2-6.
Figure 2.2-6. Euler angles.
By combining the above transformations there results the transformation equations (2.2.50)
 
  
cos ψ cos φ − cos θ sin φ sin ψ
cos ψ sin φ + cos θ cos φ sin ψ sin ψ sin θ
x1
x̄1
 x̄2  =  − sin ψ cos φ − cos θ sin φ cos ψ − sin ψ sin φ + cos θ cos φ cos ψ cos ψ sin θ   x2  .
x̄3
x3
sin θ sin φ
− sin θ cos φ
cos θ
It is left as an exercise to verify that the transformation matrix is orthogonal and the components `ji
satisfy the relations (2.2.51).
203
Consider the velocity of a point which is rotating with the rigid body. Denote by vi = vi (S), for
i = 1, 2, 3, the velocity components relative to the S reference frame and by v i = v i (S), i = 1, 2, 3 the
velocity components of the same point relative to the body-fixed axes. In terms of the basis vectors we can
write
~ = v1 (S) ê1 + v2 (S) ê2 + v3 (S) ê3 = dxi êi
V
dt
(2.2.52)
as the velocity in the S reference frame. Similarly, we write
dxi b
~ = v (S)b
V
e1 + v 2 (S)b
e2 + v 3 (S)b
e3 =
ei
1
dt
(2.2.53)
as the velocity components relative to the body-fixed reference frame. There are occasions when it is desirable
~ in the S frame of reference and V
~ in the S frame of reference. In these instances we can write
to represent V
~ = v1 (S)b
e1 + v2 (S)b
e2 + v3 (S)b
e3
V
(2.2.54)
~ = v (S) ê + v (S) ê + v (S) ê .
V
1
1
2
2
3
3
(2.2.55)
and
Here we have adopted the notation that vi (S) are the velocity components relative to the S reference frame
and vi (S) are the same velocity components relative to the S reference frame. Similarly, v i (S) denotes the
velocity components relative to the S reference frame, while v i (S) denotes the same velocity components
relative to the S reference frame.
~ are vectors and so their components are first order tensors and satisfy the transfor~ and V
Here both V
mation laws
v i (S) = `ji vj (S) = `ji ẋj
and
vi (S) = `ij v j (S) = `ij ẋj .
(2.2.56)
The equations (2.2.56) define the relative velocity components as functions of time t. By differentiating the
equations (2.2.50) we obtain
dxi
= v i (S) = `ji ẋj + `˙ji xj
dt
(2.2.57)
dxi
= vi (S) = `ij ẋj + `˙ij xj .
dt
(2.2.58)
and
Multiply the equation (2.2.57) by `mi and multiply the equation (2.2.58) by `im and derive the relations
vm (S) = vm (S) + `mi `˙ji xj
(2.2.59)
v m (S) = v m (S) + `im `˙ij xj .
(2.2.60)
and
The equations (2.2.59) and (2.2.60) describe the transformation laws of the velocity components upon changing from the S to the S reference frame. These equations can be expressed in terms of the angular velocity
by making certain substitutions which are now defined.
The first order angular velocity vector ωi is related to the second order skew-symmetric angular velocity
tensor ωij by the defining equation
ωmn = eimn ωi .
(2.2.61)
204
The equation (2.2.61) implies that ωi and ωij are dual tensors and
ωi =
1
eijk ωjk .
2
Also the velocity of a point which is rotating about the origin relative to the S frame of reference is vi (S) =
eijk ωj xk which can also be written in the form vm (S) = −ωmk xk . Since the barred axes rotate with the rigid
body, then a particle in the barred reference frame will have vm (S) = 0, since the coordinates of a point
in the rigid body will be constants with respect to this reference frame. Consequently, we write equation
(2.2.59) in the form 0 = vm (S) + `mi `˙ji xj which implies that
vm (S) = −`mi`˙ji xj = −ωmk xk
or ωmj = ωmj (S, S) = `mi `˙ji .
This equation is interpreted as describing the angular velocity tensor of S relative to S. Since ωij is a tensor,
it can be represented in the barred system by
ω mn (S, S) = `im `jn ωij (S, S)
= `im `jn `is `˙js
= δms `jn `˙js
(2.2.62)
= `jn `˙jm
By differentiating the equations (2.2.51) it is an easy exercise to show that ωij is skew-symmetric. The
second order angular velocity tensor can be used to write the equations (2.2.59) and (2.2.60) in the forms
vm (S) = vm (S) + ωmj (S, S)xj
(2.2.63)
v m (S) = v m (S) + ωjm (S, S)xj
The above relations are now employed to derive the celebrated Euler’s equations of motion of a rigid body.
Euler’s Equations of Motion
We desire to find the equations of motion of a rigid body which is subjected to external forces. These
equations are the formulas (2.2.49), and we now proceed to write these equations in a slightly different form.
Similar to the introduction of the angular velocity tensor, given in equation (2.2.61), we now introduce the
following tensors
1. The fourth order moment of inertia tensor Imnst which is related to the second order moment of
inertia tensor Iij by the equations
Imnst =
1
ejmn eist Iij
2
or Iij =
1
Ipqrs eipq ejrs
2
(2.2.64)
2. The second order angular momentum tensor Hjk which is related to the angular momentum vector
Hi by the equation
Hi =
1
eijk Hjk
2
or Hjk = eijk Hi
(2.2.65)
3. The second order moment tensor Mjk which is related to the moment Mi by the relation
Mi =
1
eijk Mjk
2
or Mjk = eijk Mi .
(2.2.66)
205
Now if we multiply equation (2.2.49) by erjk , then it can be written in the form
dHij
= Mij .
(2.2.67)
dt
Similarly, if we multiply the equation (2.2.42) by eimn , then it can be expressed in the alternate form
Hmn = eimn ωj Iji = Imnst ωst
and because of this relation the equation (2.2.67) can be expressed as
d
(Iijst ωst ) = Mij .
(2.2.68)
dt
We write this equation in the barred system of coordinates where I pqrs will be a constant and consequently
its derivative will be zero. We employ the transformation equations
Iijst = `ip `jq `sr `tk I pqrk
ω ij = `si `tj ωst
M pq = `ip `jq Mij
and then multiply the equation (2.2.68) by `ip `jq and simplify to obtain
d
`iα `jβ I αβrk ωrk = M pq .
dt
Expand all terms in this equation and take note that the derivative of the I αβrk is zero. The expanded
`ip `jq
equation then simplifies to
dω rk
+ (δαu δpv δβq + δpα δβu δqv ) I αβrk ω rk ωuv = M pq .
(2.2.69)
dt
Substitute into equation (2.2.69) the relations from equations (2.2.61),(2.2.64) and (2.2.66), and then multiply
I pqrk
by empq and simplify to obtain the Euler’s equations of motion
dω i
− etmj I ij ω i ωt = M m .
(2.2.70)
dt
Dropping the bar notation and performing the indicated summations over the range 1,2,3 we find the
I im
Euler equations have the form
dω1
dω2
dω3
+ I21
+ I31
dt
dt
dt
dω1
dω2
dω3
+ I22
+ I32
I12
dt
dt
dt
dω1
dω2
dω3
+ I23
+ I33
I13
dt
dt
dt
In the special case where
I11
+ (I13 ω1 + I23 ω2 + I33 ω3 ) ω2 − (I12 ω1 + I22 ω2 + I32 ω3 ) ω3 = M1
+ (I11 ω1 + I21 ω2 + I31 ω3 ) ω3 − (I13 ω1 + I23 ω2 + I33 ω3 ) ω1 = M2
(2.2.71)
+ (I12 ω1 + I22 ω2 + I32 ω3 ) ω1 − (I11 ω1 + I21 ω2 + I31 ω3 ) ω2 = M3 .
the barred axes are principal axes, then Iij = 0 for i 6= j and the Euler’s
equations reduces to the system of nonlinear differential equations
dω1
+ (I33 − I22 )ω2 ω3 = M1
dt
dω2
(2.2.72)
+ (I11 − I33 )ω3 ω1 = M2
I22
dt
dω3
+ (I22 − I11 )ω1 ω2 = M3 .
I33
dt
In the case of constant coefficients and constant moments the solutions of the above differential equations
I11
can be expressed in terms of Jacobi elliptic functions.
206
EXERCISE 2.2
I 1. Find a set of parametric equations for the straight line which passes through the points P1 (1, 1, 1) and
P2 (2, 3, 4). Find the unit tangent vector to any point on this line.
I 2. Consider the space curve x =
1
2
sin2 t, y = 12 t − 14 sin 2t, z = sin t where t is a parameter. Find the unit
vectors T i , B i , N i , i = 1, 2, 3 at the point where t = π.
I 3. A claim has been made that the space curve x = t, y = t2 , z = t3 intersects the plane 11x-6y+z=6 in
three distinct points. Determine if this claim is true or false. Justify your answer and find the three points
of intersection if they exist.
I 4. Find a set of parametric equations xi = xi (s1 , s2 ), i = 1, 2, 3 for the plane which passes through the
points P1 (3, 0, 0), P2 (0, 4, 0) and P3 (0, 0, 5). Find a unit normal to this plane.
2
t find the equation of the tangent plane to the curve at the
π
point where t = π/4. Find the equation of the tangent line to the curve at the point where t = π/4.
I 5. For the helix x = sin t
y = cos t
z=
∂T
= M grm ẋm .
∂ ẋr
∂grm n m
d ∂T
m
ẋ ẋ .
= M grm ẍ +
I 7. Verify the derivative
dt ∂ ẋr
∂xn
I 6. Verify the derivative
I 8. Verify the derivative
1 ∂gmn m n
∂T
= M
ẋ ẋ .
∂xr
2
∂xr
I 9. Use the results from problems 6,7 and 8 to derive the Lagrange’s form for the equations of motion
defined by equation (2.2.23).
I 10. Express the generalized velocity and acceleration in cylindrical coordinates (x1 , x2 , x3 ) = (r, θ, z) and
show
2
dθ
d2 r
δV 1
f =
= 2 −r
δt
dt
dt
2
2
d
dr
dθ
δV
2
θ
= 2 +
f2 =
δt
dt
r dt dt
d2 z
δV 3
3
= 2
f =
δt
dt
Find the physical components of velocity and acceleration in cylindrical coordinates and show
dr
dx1
=
V =
dt
dt
dθ
dx2
2
=
V =
dt
dt
3
dz
dx
=
V3 =
dt
dt
1
dr
Vr =
dt
dθ
Vθ =r
dt
dz
Vz =
dt
1
2
dθ
d2 r
fr = 2 − r
dt
dt
2
d θ
dr dθ
fθ =r 2 + 2
dt
dt dt
2
d z
fz = 2
dt
207
I 11. Express the generalized velocity and acceleration in spherical coordinates (x2 , x2 , x3 ) = (ρ, θ, φ) and
show
dρ
dx1
=
V =
dt
dt
2
dθ
dx
=
V2 =
dt
dt
3
dφ
dx
=
V3 =
dt
dt
1
f1 =
d2 ρ
δV 1
= 2 −ρ
δt
dt
dθ
dt
2
− ρ sin2 θ
dφ
dt
2
2
dφ
d2 θ
δV 2
2 dρ dθ
= 2 − sin θ cos θ
f =
+
δt
dt
dt
ρ dt dt
3
2
d
dρ
dφ
dθ
dφ
δV
2
φ
= 2 +
+ 2 cot θ
f3 =
δt
dt
ρ dt dt
dt dt
2
Find the physical components of velocity and acceleration in spherical coordinates and show
d2 ρ
fρ = 2 − ρ
dt
dρ
Vρ =
dt
dθ
Vθ =ρ
dt
Vφ =ρ sin θ
dθ
dt
2
2
− ρ sin θ
dφ
dt
2
2
d2 θ
dφ
dρ dθ
−
ρ
sin
θ
cos
θ
+2
2
dt
dt
dt dt
2
dθ dφ
d φ
dρ dφ
+ 2ρ cos θ
fφ =ρ sin θ 2 + 2 sin θ
dt
dt dt
dt dt
fθ =ρ
dφ
dt
I 12. Expand equation (2.2.39) and write out all the components of the moment of inertia tensor Iij .
I 13. For ρ the density of a continuous material and dτ an element of volume inside a region R where the
material is situated, we write ρdτ as an element of mass inside R. Find an equation which describes the
center of mass of the region R.
I 14. Use the equation (2.2.68) to derive the equation (2.2.69).
I 15. Drop the bar notation and expand the equation (2.2.70) and derive the equations (2.2.71).
I 16. Verify the Euler transformation, given in example 2.2-5, is orthogonal.
I 17. For the pulley and mass system illustrated in the figure 2.2-7 let
a = the radius of each pulley.
`1 = the length of the upper chord.
`2 = the length of the lower chord.
Neglect the weight of the pulley and find the equations of motion for the pulley mass system.
208
Figure 2.2-7. Pulley and mass system
I 18. Let φ =
ds
dt ,
where s is the arc length between two points on a curve in generalized coordinates.
p
(a) Write the arc length in general coordinates as ds = gmn ẋm ẋn dt and show the integral I, defined by
equation (2.2.35), represents the distance between two points on a curve.
(b) Using the Euler-Lagrange equations (2.2.36) show that the shortest distance between two points in a
d2 s
i
2
ẋj ẋk = ẋi dt
generalized space is the curve defined by the equations: ẍi +
ds
jk
dt
i dxj dxk
d2 xi
= 0, for
(c) Show in the special case t = s the equations in part (b) reduce to
+
j k ds ds
ds2
i = 1, . . . , N. An examination of equation (1.5.51) shows that the above curves are geodesic curves.
(d) Show that the shortest distance between two points in a plane is a straight line.
(e) Consider two points on the surface of a cylinder of radius a. Let u1 = θ and u2 = z denote surface
coordinates in the two dimensional space defined by the surface of the cylinder. Show that the shortest
p
distance between the points where θ = 0, z = 0 and θ = π, z = H is L = a2 π 2 + H 2 .
I 19. For T = 12 mgij v i v j the kinetic energy of a particle and V the potential energy of the particle show
that T + V = constant.
Hint:
∂V
mfi = Qi = − ∂x
i,
i = 1, 2, 3 and
dxi
dt
= ẋi = v i , i = 1, 2, 3.
I 20. Define H = T + V as the sum of the kinetic energy and potential energy of a particle. The quantity
H = H(xr , pr ) is called the Hamiltonian of the particle and it is expressed in terms of:
• the particle position xi and
• the particle momentum pi = mvi = mgij ẋj . Here xr and pr are treated as independent variables.
(a) Show that the particle momentum is a covariant tensor of rank 1.
(b) Express the kinetic energy T in terms of the particle momentum.
∂T
.
(c) Show that pi =
∂ ẋi
209
Figure 2.2-8. Compound pendulum
∂H
∂H
dpi
dxi
=
= − i . These are a set of differential equations describing the
and
dt
∂pi
dt
∂x
position change and momentum change of the particle and are known as Hamilton’s equations of motion
(d) Show that
for a particle.
I 21.
Let
δT i
δs
= κN i and
B i = ijk Tj Nk and find
δB i
δs
δN i
δs
= τ B i − κT i and calculate the intrinsic derivative of the cross product
in terms of the unit normal vector.
I 22. For T the kinetic energy of a particle and V the potential energy of a particle, define the Lagrangian
1
L = L(xi , ẋi ) = T − V = M gij ẋi ẋj − V as a function of the independent variables xi , ẋi . Define the
2
1 ij
g pi pj + V, as a function of the independent variables xi , pi ,
Hamiltonian H = H(xi , pi ) = T + V =
2M
where pi is the momentum vector of the particle and M is the mass of the particle.
∂T
.
(a) Show that pi =
∂ ẋi
∂L
∂H
=− i
(b) Show that
i
∂x
∂x
I 23.
When the Euler angles, figure 2.2-6, are applied to the motion of rotating objects, θ is the angle
of nutation, φ is the angle of precession and ψ is the angle of spin. Take projections and show that the
time derivative of the Euler angles are related to the angular velocity vector components ωx , ωy , ωz by the
relations
ωx = θ̇ cos ψ + φ̇ sin θ sin ψ
ωy = −θ̇ sin ψ + φ̇ sin θ cos ψ
ωz = ψ̇ + φ̇ cos θ
where ωx , ωy , ωz are the angular velocity components along the x1 , x2 , x3 axes.
I 24. Find the equations of motion for the compound pendulum illustrated in the figure 2.2-8.
210
GM m
I 25. Let F~ = − 3 ~r denote the inverse square law force of attraction between the earth and sun, with
r
G a universal constant, M the mass of the sun, m the mass of the earth and ~rr a unit vector from origin
at the center of the sun pointing toward the earth. (a) Write down Newton’s second law, in both vector
d
(~r × ~v ) = ~0 and
and tensor form, which describes the motion of the earth about the sun. (b) Show that
dt
r
~
consequently ~r × ~v = ~r × d~
dt = h = a constant.
I 26. Construct a set of axes fixed and attached to an airplane. Let the x axis be a longitudinal axis running
from the rear to the front of the plane along its center line. Let the y axis run between the wing tips and
let the z axis form a right-handed system of coordinates. The y axis is called a lateral axis and the z axis is
called a normal axis. Define pitch as any angular motion about the lateral axis. Define roll as any angular
motion about the longitudinal axis. Define yaw as any angular motion about the normal axis. Consider two
sets of axes. One set is the x, y, z axes attached to and moving with the aircraft. The other set of axes is
denoted X, Y, Z and is fixed in space ( an inertial set of axes). Describe the pitch, roll and yaw of an aircraft
with respect to the inertial set of axes. Show the transformation is orthogonal. Hint: Consider pitch with
respect to the fixed axes, then consider roll with respect to the pitch axes and finally consider yaw with
respect to the roll axes. This produces three separate transformation matrices which can then be combined
to describe the motions of pitch, roll and yaw of an aircraft.
xi (t) denote a curve
I 27. In Cartesian coordinates let Fi = Fi (x1 , x2 , x3 ) denote a force field and let xi =
i 2 !
dx
dxi
d 1
m
C. (a) Show Newton’s second law implies that along the curve C
= Fi (x1 , x2 , x3 )
dt 2
dt
dt
(no summation on i) and hence
"
1 2 2 2 3 2 !#
dx
dx
dx
dx1
dx2
dx3
d 1
d 1
m
mv 2 = F1
+ F2
+ F3
+
+
=
dt 2
dt
dt
dt
dt 2
dt
dt
dt
(b) Consider two points on the curve C, say point A, xi (tA ) and point B, xi (tB ) and show that the work
done in moving from A to B in the force field Fi is
tB Z B
1
2
mv
=
Fi dx1 + F2 dx2 + F3 dx3
2
A
tA
where the right hand side is a line integral along the path C from A to B. (c) Show that if the force field is
derivable from a potential function U (x1 , x2 , x3 ) by taking the gradient, then the work done is independent
of the path C and depends only upon the end points A and B.
I 28. Find the Lagrangian equations of motion of a spherical pendulum which consists of a bob of mass m
suspended at the end of a wire of length `, which is free to swing in any direction subject to the constraint
that the wire length is constant. Neglect the weight of the wire and show that for the wire attached to the
origin of a right handed x, y, z coordinate system, with the z axis downward, φ the angle between the wire
and the z axis and θ the angle of rotation of the bob from the y axis, that there results the equations of
2
dθ
dθ
g
d2 φ
d
2
−
sin φ cos φ + sin φ = 0
sin φ
=0
and
motion
dt
dt
dt2
dt
`
211
I 29. In Cartesian coordinates show the Frenet formulas can be written
dT~
= ~δ × T~ ,
ds
~
dN
~,
= ~δ × N
ds
~
dB
~
= ~δ × B
ds
~
where ~δ is the Darboux vector and is defined ~δ = τ T~ + κB.
I 30. Consider the following two cases for rigid body rotation.
Case 1: Rigid body rotation about a fixed line which is called the fixed axis of rotation. Select a point 0
on this fixed axis and denote by b
e a unit vector from 0 in the direction of the fixed line and denote by êR
a unit vector which is perpendicular to the fixed axis of rotation. The position vector of a general point
in the rigid body can then be represented by a position vector from the point 0 given by ~r = h b
e + r0 êR
(a)
(b)
(a)
(b)
(c)
(d)
(e)
where h, r0 and b
e are all constants and the vector êR is fixed in and rotating with the rigid body.
dθ
the scalar angular change with respect to time of the vector êR as it rotates about
Denote by ω =
dt
dθ
d
b
e) =
e where θ b
e is defined as the
the fixed line and define the vector angular velocity as ~ω = (θ b
dt
dt
vector angle of rotation.
d êR
= b
e × êR .
Show that
dθ
~ = d~r = r0 d êR = r0 d êR dθ = ~ω × (r0 êR ) = ~ω × (h b
e + r0 êR ) = ~ω × ~r.
Show that V
dt
dt
dθ dt
Case 2: Rigid body rotation about a fixed point 0. Construct at point 0 the unit vector ê1 which is
d ê1
must be perpendicular
fixed in and rotating with the rigid body. From pages 80,87 we know that
dt
d ê1
such that
to ê1 and so we can define the vector ê2 as a unit vector which is in the direction of
dt
d ê1
= α ê2 for some constant α. We can then define the unit vector ê3 from ê3 = ê1 × ê2 .
dt
d ê3
, which must be perpendicular to ê3 , is also perpendicular to ê1 .
Show that
dt
d ê3
d ê3
can be written as
= β ê2 for some constant β.
Show that
dt
dt
d ê2
= (α ê3 − β ê1 ) × ê2
From ê2 = ê3 × ê1 show that
dt
d ê1
d ê2
d ê3
= ~ω × ê1 ,
=ω
~ × ê2 ,
= ~ω × ê3
Define ~ω = α ê3 − β ê1 and show that
dt
dt
dt
Let ~r = x ê1 + y ê2 + z ê3 denote an arbitrary point within the rigid body with respect to the point 0.
d~r
=~
ω × ~r.
Show that
dt
Note that in Case 2 the direction of ω
~ is not fixed as the unit vectors ê3 and ê1 are constantly changing.
In this case the direction ~
ω is called an instantaneous axis of rotation and ~ω , which also can change in
magnitude and direction, is called the instantaneous angular velocity.
211
§2.3 BASIC EQUATIONS OF CONTINUUM MECHANICS
Continuum mechanics is the study of how materials behave when subjected to external influences.
External influences which affect the properties of a substance are such things as forces, temperature, chemical
reactions, and electric phenomena. Examples of forces are gravitational forces, electromagnetic forces, and
mechanical forces. Solids deform under external forces and so deformations are studied. Fluids move under
external forces and so the velocity of the fluid is studied.
A material is considered to be a continuous media which is a collection of material points interconnected
by internal forces (forces between the atoms making up the material). We concentrate upon the macroscopic
properties rather than the microscopic properties of the material. We treat the material as a body which is
homogeneous and continuous in its makeup.
In this introduction we will only consider solid media and liquid media. In general, most of the ideas
and concepts developed in this section can be applied to any type of material which is assumed to be a
collection of material points held together by some kind of internal forces.
An elastic material is one which deforms under applied forces in such a way that it will return to its
original unloaded state when the applied forces are removed. When a linear relation exists between the
applied forces and material displacements, then the material is called a linear elastic material. In contrast, a
plastic material is one which deforms under applied forces in such a way that it does not return to its original
state after removal of the applied forces. Plastic materials will always exhibit some permanent deformation
after removal of the applied forces. An elastic material is called homogeneous if it has the same properties
throughout. An isotropic material has the same properties, at a point, in all directions about the point.
In this introduction we develop the basic mathematical equations which describe how a continuum
behaves when subjected to external forces. We shall discover that there exists a set of basic equations
associated with all continuous material media. These basic equations are developed for linear elastic materials
and applied to solids and fluids in later sections.
Introduction to Elasticity
Take a rubber band, which has a rectangular cross section, and mark on it a parallelepiped having a
length `, a width w and a height h, as illustrated in the figure 2.3-1.
Now apply a force F to both ends of the parallelepiped cross section on the rubber band and examine
what happens to the parallelepiped. You will see that:
1.
` increases by an amount ∆`.
2.
w decreases by an amount ∆w.
3.
h decreases by an amount ∆h.
There are many materials which behave in a manner very similar to the rubber band. Most materials,
when subjected to tension forces will break if the change ∆` is only one or two percent of the original length.
The above example introduces us to several concepts which arise in the study of materials when they are
subjected to external forces. The first concept is that of strain which is defined as
strain =
change in length
,
original length
(dimensionless).
212
Figure 2.3-1. Section of a rubber band
When the force F is applied to our rubber band example there arises the strains
∆`
,
`
∆w
,
w
∆h
.
h
The second concept introduced by our simple example is stress. Stress is defined as a force per unit area. In
particular,
force
Force
,
with dimension of
.
Area over which force acts
unit area
We will be interested in studying stress and strain in homogeneous, isotropic materials which are in equilibstress =
rium with respect to the force system acting on the material.
Hooke’s Law
For linear elastic materials, where the forces are all one dimensional, the stress and strains are related
by Hooke’s law which has two parts. The Hooke’s law, part one, states that stress is proportional to strain
in the stretch direction, where the Young’s modulus E is the proportionality constant. This is written
∆`
F
=E
.
(2.3.1)
Hooke’s law part 1
A
`
A graph of stress vs strain is a straight line with slope E in the linear elastic range of the material.
The Hooke’s law, part two, involves the fact that there is a strain contraction perpendicular to the
stretch direction. The strain contraction is the same for both the width and height and is proportional to
the strain in the stretch direction. The proportionality constant being the Poisson’s ratio ν.
Hooke’s law part 2
∆h
∆`
∆w
=
= −ν
,
w
h
`
0<ν<
1
.
2
(2.3.2)
The proportionality constants E and ν depend upon the material being considered. The constant ν is called
the Poisson’s ratio and it is always a positive number which is less than one half. Some representative values
for E and ν are as follows.
Various types of steel
Various types of aluminium
28 (10)6 psi ≤ E ≤ 30 (10)6 psi
9.0 (10)6 psi ≤ E ≤ 11.0 (10)6 psi
0.26 ≤ ν ≤ 0.31
0.3 ≤ ν ≤ 0.35
213
Figure 2.3-2. Typical Stress-strain curve.
Consider a typical stress-strain curve, such as the one illustrated in the figure 2.3-2, which is obtained
by placing a material in the shape of a rod or wire in a machine capable of performing tensile straining at a
low rate. The engineering stress is the tensile force F divided by the original cross sectional area A0 . Note
that during a tensile straining the cross sectional area A of the sample is continually changing and getting
smaller so that the actual stress will be larger than the engineering stress. Observe in the figure 2.3-2 that
the stress-strain relation remains linear up to a point labeled the proportional limit. For stress-strain points
in this linear region the Hooke’s law holds and the material will return to its original shape when the loading
is removed. For points beyond the proportional limit, but less than the yield point, the material no longer
obeys Hooke’s law. In this nonlinear region the material still returns to its original shape when the loading
is removed. The region beyond the yield point is called the plastic region. At the yield point and beyond,
there is a great deal of material deformation while the loading undergoes only small changes. For points
in this plastic region, the material undergoes a permanent deformation and does not return to its original
shape when the loading is removed. In the plastic region there usually occurs deformation due to slipping of
atomic planes within the material. In this introductory section we will restrict our discussions of material
stress-strain properties to the linear region.
EXAMPLE 2.3-1. (One dimensional elasticity) Consider a circular rod with cross sectional area A
which is subjected to an external force F applied to both ends. The figure 2.3-3 illustrates what happens to
the rod after the tension force F is applied. Consider two neighboring points P and Q on the rod, where P
is at the point x and Q is at the point x + ∆x. When the force F is applied to the rod it is stretched and
P moves to P 0 and Q moves to Q0 . We assume that when F is applied to the rod there is a displacement
function u = u(x, t) which describes how each point in the rod moves as a function of time t. If we know the
displacement function u = u(x, t) we would then be able to calculate the following distances in terms of the
displacement function
P P 0 = u(x, t),
0P 0 = x + u(x, t),
QQ0 = u(x + ∆x, t)
0Q0 = x + ∆x + u(x + ∆x, t).
214
Figure 2.3-3. One dimensional rod subjected to tension force
The strain associated with the distance ` = ∆x = P Q is
(0Q0 − 0P 0 ) − (0Q − 0P )
P 0 Q0 − P Q
∆`
=
=
`
PQ
PQ
[x + ∆x + u(x + ∆x, t) − (x + u(x, t))] − [(x + ∆x) − x]
e=
∆x
u(x + ∆x, t) − u(x, t)
.
e=
∆x
e=
Use the Hooke’s law part(i) and write
u(x + ∆x, t) − u(x, t)
F
=E
.
A
∆x
Taking the limit as ∆x approaches zero we find that
∂u(x, t)
F
=E
.
A
∂x
Hence, the stress is proportional to the spatial derivative of the displacement function.
Normal and Shearing Stresses
Let us consider a more general situation in which we have some material which can be described as
having a surface area S which encloses a volume V. Assume that the density of the material is % and the
material is homogeneous and isotropic. Further assume that the material is subjected to the forces ~b and ~t (n)
where ~b is a body force per unit mass [f orce/mass], and ~t (n) is a surface traction per unit area [f orce/area].
The superscript (n) on the vector is to remind you that we will only be interested in the normal component
of the surface forces. We will neglect body couples, surface couples, and concentrated forces or couples that
act at a single point. If the forces described above are everywhere continuous we can calculate the resultant
~ acting on the material by constructing various surface and volume integrals
force F~ and resultant moment M
which sum the forces acting upon the material. In particular, the resultant force F~ acting on our material
can be described by the surface and volume integrals:
ZZZ
ZZ
~t (n) dS +
%~b dτ
F~ =
S
V
(2.3.3)
215
Figure 2.3-4. Stress vectors acting upon an element of volume
which is a summation of all the body forces and surface tractions acting upon our material. Here % is the
density of the material, dS is an element of surface area, and dτ is an element of volume.
~ about the origin is similarly expressed as
The resultant moment M
ZZZ
ZZ
~ =
M
%(~r × ~b) dτ.
~r × ~t (n) dS +
S
(2.3.4)
V
The global motion of the material is governed by the Euler equations of motion.
• The time rate of change of linear momentum equals the resultant force or
d
dt
ZZ
ZZZ
~t (n) dS +
%~v dτ = F~ =
%~b dτ.
Z Z Z
V
S
(2.3.5)
V
This is a statement concerning the conservation of linear momentum.
• The time rate of change of angular momentum equals the resultant moment or
d
dt
Z Z Z
ZZ
~ =
%~r × ~v dτ = M
V
ZZZ
%(~r × ~b) dτ.
~r × ~t (n) dS +
S
(2.3.6)
V
This is a statement concerning conservation of angular momentum.
The Stress Tensor
Define the stress vectors
~t 1 = σ 11 ê1 + σ 12 ê2 + σ 13 ê3
~t 2 = σ 21 ê1 + σ 22 ê2 + σ 23 ê3
(2.3.7)
~t 3 = σ 31 ê1 + σ 32 ê2 + σ 33 ê3 ,
where σ ij , i, j = 1, 2, 3 is the stress tensor acting at each point of the material. The index i indicates the
coordinate surface xi = a constant, upon which ~t i acts. The second index j denotes the direction associated
with the components of ~t i .
216
Figure 2.3-5. Stress distribution at a point
For i = 1, 2, 3 we adopt the convention of sketching the components of ~t i in the positive directions if
the exterior normal to the surface xi = constant also points in the positive direction. This gives rise to the
figure 2.3-4 which illustrates the stress vectors acting upon an element of volume in rectangular Cartesian
coordinates. The components σ 11 , σ 22 , σ 33 are called normal stresses while the components σ ij , i 6= j are
called shearing stresses. The equations (2.3.7) can be written in the more compact form using the indicial
notation as
~t i = σ ij êj ,
i, j = 1, 2, 3.
(2.3.8)
If we know the stress distribution at three orthogonal interfaces at a point P in a solid body, we can then
determine the stress at the point P with respect to any plane passing through the point P. With reference to
the figure 2.3-5, consider an arbitrary plane passing through the point P which lies within the material body
being considered. Construct the elemental tetrahedron with orthogonal axes parallel to the x1 = x, x2 = y
and x3 = z axes. In this figure we have the following surface tractions:
−~t 1
on the surface 0BC
−~t 2
on the surface 0AC
−~t 3
on the surface 0AB
~ (n)
on the surface ABC
t
The superscript parenthesis n is to remind you that this surface traction depends upon the orientation of
the plane ABC which is determined by a unit normal vector having the direction cosines n1 , n2 and n3 .
217
Let
∆S1 = the surface area 0BC
∆S2 = the surface area 0AC
∆S3 = the surface area 0AB
∆S = the surface area ABC .
These surface areas are related by the relations
∆S1 = n1 ∆S,
∆S2 = n2 ∆S,
∆S3 = n3 ∆S
(2.3.9)
which can be thought of as projections of ∆S upon the planes xi =constant for i = 1, 2, 3.
Cauchy Stress Law
Let tj (n) denote the components of the surface traction on the surface ABC. That is, we let
~t (n) = t1 (n) ê1 + t2 (n) ê2 + t3 (n) ê3 = tj (n) êj .
(2.3.10)
It will be demonstrated that the components tj (n) of the surface traction forces ~t (n) associated with a plane
through P and having the unit normal with direction cosines n1 , n2 and n3 , must satisfy the relations
tj (n) = ni σ ij ,
i, j = 1, 2, 3.
(2.3.11)
This relation is known as the Cauchy stress law.
Proof: Sum the forces acting on the elemental tetrahedron in the figure 2.3-5. If the body is in equilibrium,
then the sum of these forces must equal zero or
1
(−~t ∆S1 ) + (−~t 2 ∆S2 ) + (−~t 3 ∆S3 ) + ~t (n) ∆S = 0.
(2.3.12)
The relations in the equations (2.3.9) are used to simplify the sum of forces in the equation (2.3.12). It is
readily verified that the sum of forces simplifies to
~t (n) = n1~t 1 + n2~t 2 + n3~t 3 = ni~t i .
(2.3.13)
Substituting in the relations from equation (2.3.8) we find
~t (n) = tj (n) êj = ni σ ij êj ,
i, j = 1, 2, 3
(2.3.14)
or in component form
tj (n) = ni σ ij
which is the Cauchy stress law.
(2.3.15)
218
Conservation of Linear Momentum
Let R denote a region in space where there exists a material volume with density % having surface
tractions and body forces acting upon it. Let v i denote the velocity of the material volume and use Newton’s
second law to set the time rate of change of linear momentum equal to the forces acting upon the volume as
in (2.3.5). We find
δ
δt
Z Z Z
ZZ
j
ZZZ
σ ij ni dS +
%v dτ =
R
S
%bj dτ.
R
Here dτ is an element of volume, dS is an element of surface area, bj are body forces per unit mass, and σ ij
are the stresses. Employing the Gauss divergence theorem, the surface integral term is replaced by a volume
integral and Newton’s second law is expressed in the form
ZZZ
j
%f − %bj − σ ij ,i dτ = 0,
(2.3.16)
R
where f j is the acceleration from equation (1.4.54). Since R is an arbitrary region, the equation (2.3.16)
implies that
σ ij ,i + %bj = %f j .
(2.3.17)
This equation arises from a balance of linear momentum and represents the equations of motion for material
in a continuum. If there is no velocity term, then equation (2.3.17) reduces to an equilibrium equation which
can be written
σ ij ,i + %bj = 0.
(2.3.18)
This equation can also be written in the covariant form
g si σms,i + %bm = 0,
which reduces to σij,j + %bi = 0 in Cartesian coordinates. The equation (2.3.18) is an equilibrium equation
and is one of our fundamental equations describing a continuum.
Conservation of Angular Momentum
The conservation of angular momentum equation (2.3.6) has the Cartesian tensors representation
d
dt
ZZ
ZZZ
%eijk xj vk dτ =
eijk xj σpk np dS +
%eijk xj bk dτ.
Z Z Z
R
S
(2.3.19)
R
Employing the Gauss divergence theorem, the surface integral term is replaced by a volume integral to obtain
ZZZ R
d
∂
dτ = 0.
eijk % (xj vk ) − eijk %xj bk + p (xj σpk )
dt
∂x
Since equation (2.3.20) must hold for all arbitrary volumes R we conclude that
d
∂σpk
+ σjk
eijk % (xj vk ) = eijk %xj bk + xj
dt
∂xp
(2.3.20)
219
Figure 2.3-6. Shearing parallel to the y axis
which can be rewritten in the form
∂σpk
dvk
) − %vj vk = 0.
eijk σjk + xj ( p + %bk − %
∂x
dt
(2.3.21)
In the equation (2.3.21) the middle term is zero because of the equation (2.3.17). Also the last term in
(2.3.21) is zero because eijk vj vk represents the cross product of a vector with itself. The equation (2.3.21)
therefore reduces to
eijk σjk = 0,
(2.3.22)
which implies (see exercise 1.1, problem 22) that σij = σji for all i and j. Thus, the conservation of angular
momentum requires that the stress tensor be symmetric. Consequently, there are only 6 independent stress
components to be determined. This is another fundamental law for a continuum.
Strain in Two Dimensions
Consider the matrix equation
1
x
=
y
β
0
1
x
y
(2.3.23)
which can be used to transform points (x, y) to points (x, y). When this transformation is applied to the
unit square illustrated in the figure 2.3-6(a) we obtain the geometry illustrated in the figure 2.3-6(b) which
represents a shearing parallel to the y axis. If β is very small, we can use the approximation tan β ≈ β and
then this transformation can be thought of as a rotation of the element P1 P2 through an angle β to the
position P10 P20 when the barred axes are placed atop the unbarred axes.
Similarly, the matrix equation
x
1
=
y
0
α
1
x
y
(2.3.24)
can be used to represent a shearing of the unit square parallel to the x axis as illustrated in the figure
2.3-7(b).
220
Figure 2.3-7. Shearing parallel to the x axis
Figure 2.3-8. Shearing parallel to x and y axes
Again, if α is very small, we may use the approximation tan α ≈ α and interpret α as an angular rotation
of the element P1 P4 to the position P10 P40 . Now let us multiply the matrices given in equations (2.3.23) and
(2.3.24). Note that the order of multiplication is important as can be seen by an examination of the products
1 0
1 α
x
1
α
x
x
=
=
y
β 1
0 1
y
β 1 + αβ
y
1 α
1 0
x
1 + αβ α
x
x
=
=
.
y
0 1
β 1
y
β
1
y
(2.3.25)
In equation (2.3.25) we will assume that the product αβ is very, very small and can be neglected. Then the
order of matrix multiplication will be immaterial and the transformation equation (2.3.25) will reduce to
1
x
=
y
β
α
1
x
.
y
(2.3.26)
Applying this transformation to our unit square we obtain the simultaneous shearing parallel to both the x
and y axes as illustrated in the figure 2.3-8.
This transformation can then be interpreted as the superposition of the two shearing elements depicted
in the figure 2.3-9.
For comparison, we consider also the transformation equation
1 0
x
x
=
y
−α 1
y
(2.3.27)
221
Figure 2.3-9. Superposition of shearing elements
Figure 2.3-10. Rotation of element P1 P2
where α is very small. Applying this transformation to the unit square previously considered we obtain the
results illustrated in the figure 2.3-10.
Note the difference in the direction of shearing associated with the transformation equations (2.3.27)
and (2.3.23) illustrated in the figures 2.3-6 and 2.3-10. If the matrices appearing in the equations (2.3.24)
and (2.3.27) are multiplied and we neglect product terms because α is assumed to be very small, we obtain
the matrix equation
1 α
x
1
x
=
=
y
−α 1
y
0
|
0 α
x
0
x
+
.
−α 0
y
1
y
{z
} |
{z
}
rotation
identity
(2.3.28)
This can be interpreted as a superposition of the transformation equations (2.3.24) and (2.3.27) which
represents a rotation of the unit square as illustrated in the figure 2.3-11.
The matrix on the right-hand side of equation (2.3.28) is referred to as a rotation matrix. The ideas
illustrated by the above simple transformations will appear again when we consider the transformation of an
arbitrary small element in a continuum when it under goes a strain. In particular, we will be interested in
extracting the rigid body rotation from a deformed element and treating this rotation separately from the
strain displacement.
222
Figure 2.3-11. Rotation of unit square
Transformation of an Arbitrary Element
In two dimensions, we consider a rectangular element ABCD as illustrated in the figure 2.3-12.
Let the points ABCD have the coordinates
A(x, y),
B(x + ∆x, y),
C(x, y + ∆y),
D(x + ∆x, y + ∆y)
(2.3.29)
and denote by
u = u(x, y),
v = v(x, y)
the displacement field associated with each of the points in the material continuum when it undergoes a
deformation. Assume that the deformation of the element ABCD in figure 2.3-12 can be represented by the
matrix equation
x
b11
=
y
b21
b12
b22
x
y
(2.3.30)
where the coefficients bij , i, j = 1, 2, 3 are to be determined. Let us define u = u(x, y) as the horizontal
displacement of the point (x, y) and v = v(x, y) as the vertical displacement of the same point. We can now
express the displacement of each of the points A, B, C and D in terms of the displacement field u = u(x, y)
and v = v(x, y). Consider first the displacement of the point A to A0 . Here the coordinates (x, y) deform to
the new coordinates
x = x + u,
y = y + v.
That is, the coefficients bij must be chosen such that the equation
x+u
y+v
=
b11
b21
b12
b22
x
y
(2.3.31)
is satisfied. We next examine the displacement of the point B to B 0 . This displacement is described by the
coordinates (x + ∆x, y) transforming to (x, y), where
x = x + ∆x + u(x + ∆x, y),
y = y + v(x + ∆x, y).
(2.3.32)
223
Figure 2.3-12. Displacement of element ABCD to A0 B 0 C 0 D0
Expanding u and v in (2.3.32) in Taylor series about the point (x, y) we find
x = x + ∆x + u +
y =y+v+
∂u
∆x + h.o.t.
∂x
(2.3.33)
∂v
∆x + h.o.t.,
∂x
where h.o.t. denotes higher order terms which have been neglected. The equations (2.3.33) require that the
coefficients bij satisfy the matrix equation
x + u + ∆x + ∂u
∂x ∆x
∂v
y + v + ∂x
∆x
=
b11
b21
b12
b22
x + ∆x
y
.
(2.3.34)
224
The displacement of the point C to C 0 is described by the coordinates (x, y + ∆y) transforming to (x, y)
where
x = x + u(x, y + ∆y),
y = y + ∆y + v(x, y + ∆y).
(2.3.35)
Again we expand the displacement field components u and v in a Taylor series about the point (x, y) and
find
∂u
∆y + h.o.t.
∂y
∂v
∆y + h.o.t.
y = y + ∆y + v +
∂y
x = x+u+
(2.3.36)
This equation implies that the coefficients bij must be chosen such that
x + u + ∂u
∂y ∆y
y + v + ∆y + ∂v
∂y ∆y
=
b11
b21
b12
b22
x
y + ∆y
.
(2.3.37)
Finally, it can be verified that the point D with coordinates (x + ∆x, y + ∆y) moves to the point D0 with
coordinates
x = x + ∆x + u(x + ∆x, y + ∆y),
y = y + ∆y + v(x + ∆x, y + ∆y).
(2.3.38)
Expanding u and v in a Taylor series about the point (x, y) we find the coefficients bij must be chosen to
satisfy the matrix equation
x + ∆x + u +
y + ∆y + v +
∂u
∂x ∆x +
∂v
∂x ∆x +
∂u
∂y ∆y
∂v
∂y ∆y
=
b11
b21
b12
b22
x + ∆x
y + ∆y
.
(2.3.39)
The equations (2.3.31),(2.3.34),(2.3.37) and (2.3.39) give rise to the simultaneous equations
b11 x + b12 y = x + u
b21 x + b22 y = y + v
b11 (x + ∆x) + b12 y = x + u + ∆x +
∂u
∆x
∂x
∂v
∆x
∂x
∂u
∆y
b11 x + b12 (y + ∆y) = x + u +
∂y
∂v
∆y
b21 x + b22 (y + ∆y) = y + v + ∆y +
∂y
∂u
∂u
∆x +
∆y
b11 (x + ∆x) + b12 (y + ∆y) = x + ∆x + u +
∂x
∂y
∂v
∂v
∆x +
∆y.
b21 (x + ∆x) + b22 (y + ∆y) = y + ∆y + v +
∂x
∂y
b21 (x + ∆x) + b22 y = y + v +
(2.3.40)
It is readily verified that the system of equations (2.3.40) has the solution
b11 = 1 +
b21
∂v
=
∂x
∂u
∂x
b12 =
b22
∂u
∂y
∂v
.
=1+
∂y
(2.3.41)
225
Figure 2.3-13. Change in 45◦ line
Hence the transformation equation (2.3.30) can be written as
∂u
1 + ∂u
x
x
∂x
∂y
=
.
∂v
∂v
y
y
1
+
∂x
∂y
(2.3.42)
A physical interpretation associated with this transformation is obtained by writing it in the form:
1 0
x
e11 e12
ω11 ω12
x
x
x
=
+
+
,
(2.3.43)
y
0 1
y
e21 e22
ω21 ω22
y
y
|
{z
} |
{z
} |
{z
}
strain matrix
identity
rotation matrix
1 ∂u ∂v
e21 =
+
e11
2 ∂y
∂x
∂v
e12
e22 =
∂y
are the elements of a symmetric matrix called the strain matrix and
1 ∂u ∂v
ω11 = 0
−
ω
=
12
2 ∂y
∂x
∂u
1 ∂v
−
ω21 =
ω22 = 0
2 ∂x ∂y
where
∂u
=
∂x
∂u
1 ∂v
+
=
2 ∂x ∂y
(2.3.44)
(2.3.45)
are the elements of a skew symmetric matrix called the rotation matrix.
The strain per unit length in the x-direction associated with the point A in the figure 2.3-12 is
e11 =
∆x +
∂u
∂x ∆x
− ∆x
=
∆x
and the strain per unit length of the point A in the y direction is
∆y +
∂v
∂y ∆y
− ∆y
∂u
∂x
(2.3.46)
∂v
.
(2.3.47)
∆y
∂y
These are the terms along the main diagonal in the strain matrix. The geometry of the figure 2.3-12 implies
e22 =
=
that
tan β =
∂v
∂x ∆x
,
∆x + ∂u
∂x ∆x
and
tan α =
∂u
∂y ∆y
.
∂v
∆y + ∂y
∆y
(2.3.48)
For small derivatives associated with the displacements u and v it is assumed that the angles α and β are
small and the equations (2.3.48) therefore reduce to the approximate equations
∂u
∂v
tan α ≈ α =
.
(2.3.49)
∂x
∂y
For a physical interpretation of these terms we consider the deformation of a small rectangular element which
tan β ≈ β =
undergoes a shearing as illustrated in the figure 2.3-13.
226
Figure 2.3-14. Displacement field due to state of strain
The quantity
α+β =
∂u ∂v
+
∂y
∂x
= 2e12 = 2e21
(2.3.50)
is the change from a ninety degree angle due to the deformation and hence we can write 12 (α + β) = e12 = e21
as representing a change from a 45◦ angle due to the deformation. The quantities e21 , e12 are called the
shear strains and the quantity
γ12 = 2e12
(2.3.51)
is called the shear angle.
In the equation (2.3.45), the quantities ω21 = −ω12 are the elements of the rigid body rotation matrix
and are interpreted as angles associated with a rotation. The situation is analogous to the transformations
and figures for the deformation of the unit square which was considered earlier.
Strain in Three Dimensions
The development of strain in three dimensions is approached from two different viewpoints. The first
approach considers the derivation using Cartesian tensors and the second approach considers the derivation
of strain using generalized tensors.
Cartesian Tensor Derivation of Strain.
Consider a material which is subjected to external forces such that all the points in the material undergo
a deformation. Let (y1 , y2 , y3 ) denote a set of orthogonal Cartesian coordinates, fixed in space, which is
used to describe the deformations within the material. Further, let ui = ui (y1 , y2 , y3 ), i = 1, 2, 3 denote a
displacement field which describes the displacement of each point within the material. With reference to the
figure 2.3-14 let P and Q denote two neighboring points within the material while it is in an unstrained state.
These points move to the points P 0 and Q0 when the material is in a state of strain. We let yi , i = 1, 2, 3
represent the position vector to the general point P in the material, which is in an unstrained state, and
denote by yi + ui , i = 1, 2, 3 the position vector of the point P 0 when the material is in a state of strain.
227
For Q a neighboring point of P which moves to Q0 when the material is in a state of strain, we have
from the figure 2.3-14 the following vectors:
position of P : yi ,
i = 1, 2, 3
position of P 0 : yi + ui (y1 , y2 , y3 ),
position of Q : yi + ∆yi ,
i = 1, 2, 3
(2.3.52)
i = 1, 2, 3
position of Q0 : yi + ∆yi + ui (y1 + ∆y1 , y2 + ∆y2 , y3 + ∆y3 ),
i = 1, 2, 3
Employing our earlier one dimensional definition of strain, we define the strain associated with the point P
L − L0
, where L0 = P Q and L = P 0 Q0 . To calculate the strain we need to first
in the direction P Q as e =
L0
calculate the distances L0 and L. The quantities L20 and L2 are easily calculated by considering dot products
of vectors. For example, we have L20 = ∆yi ∆yi , and the distance L = P 0 Q0 is the magnitude of the vector
yi + ∆yi + ui (y1 + ∆y1 , y2 + ∆y2 , y3 + ∆y3 ) − (yi + ui (y1 , y2 , y3 )),
i = 1, 2, 3.
Expanding the quantity ui (y1 + ∆y1 , y2 + ∆y2 , y3 + ∆y3 ) in a Taylor series about the point P and neglecting
higher order terms of the expansion we find that
L2 = (∆yi +
∂ui
∂ui
∆ym )(∆yi +
∆yn ).
∂ym
∂yn
Expanding the terms in this expression produces the equation
L2 = ∆yi ∆yi +
∂ui
∂ui
∂ui ∂ui
∆yi ∆yn +
∆ym ∆yi +
∆ym ∆yn .
∂yn
∂ym
∂ym ∂yn
Note that L and L0 are very small and so we express the difference L2 − L20 in terms of the strain e. We can
write
L2 − L20 = (L + L0 )(L − L0 ) = (L − L0 + 2L0 )(L − L0 ) = (e + 2)eL20 .
Now for e very small, and e2 negligible, the above equation produces the approximation
1 ∂um
L2 − L20
∂un
∂ur ∂ur
=
+
+
eL20 ≈
∆ym ∆yn .
2
2 ∂yn
∂ym
∂ym ∂yn
The quantities
emn =
1 ∂um
∂un
∂ur ∂ur
+
+
2 ∂yn
∂ym
∂ym ∂yn
(2.3.53)
is called the Green strain tensor or Lagrangian strain tensor. To show that eij is indeed a tensor, we consider
the transformation yi = `ij yj +bi , where `ji `ki = δjk = `ij `ik . Note that from the derivative relation
∂yi
∂y j
= `ij
and the transformation equations ui = `ij uj , i = 1, 2, 3 we can express the strain in the barred system of
coordinates. Performing the necessary calculations produces
1 ∂ui
∂uj
∂ur ∂ur
eij =
+
+
2 ∂yj
∂yi
∂y i ∂yj
∂yn
∂
∂ym
∂
∂yk ∂
∂yt
1 ∂
(`ik uk )
+
(`jk uk )
+
(`rs us )
(`rm um )
=
2 ∂yn
∂yj
∂ym
∂yi
∂yk
∂y i ∂yt
∂y j
∂um
∂uk
∂us ∂up
1
+ `jk `mi
+ `rs `rp `ki `tj
`im `nj
=
2
∂yn
∂ym
∂yk ∂yt
∂un
∂us ∂us
1 ∂um
+
+
`im `nj
=
2 ∂yn
∂ym
∂ym ∂yn
or
eij = emn `im `nj .Consequently, the strain eij transforms like a second order Cartesian tensor.
228
Lagrangian and Eulerian Systems
Let xi denote the initial position of a material particle in a continuum. Assume that at a later time the
particle has moved to another point whose coordinates are xi . Both sets of coordinates are referred to the
same coordinate system. When the final position can be expressed as a function of the initial position and
time we can write xi = xi (x1 , x2 , x3 , t). Whenever the changes of any physical quantity is represented in terms
of its initial position and time, the representation is referred to as a Lagrangian or material representation of
the quantity. This can be thought of as a transformation of the coordinates. When the Jacobian J( xx ) of this
transformation is different from zero, the above set of equations have a unique inverse xi = xi (x1 , x2 , x3 , t),
where the position of the particle is now expressed in terms of its instantaneous position and time. Such a
representation is referred to as an Eulerian or spatial description of the motion.
Let (x1 , x2 , x3 ) denote the initial position of a particle whose motion is described by xi = xi (x1 , x2 , x3 , t),
then ui = xi − xi denotes the displacement vector which can by represented in a Lagrangian or Eulerian
form. For example, if
x1 = 2(x1 − x2 )(et − 1) + (x2 − x1 )(e−t − 1) + x1
x2 = (x1 − x2 )(et − 1) + (x2 − x1 )(e−t − 1) + x2
x3 = x3
then the displacement vector can be represented in the Lagrangian form
u1 = 2(x1 − x2 )(et − 1) + (x2 − x1 )(e−t − 1)
u2 = (x1 − x2 )(et − 1) + (x2 − x1 )(e−t − 1)
u3 = 0
or the Eulerian form
u1 = x1 − (2x2 − x1 )(1 − e−t ) − (x1 − x2 )(e−2t − e−t ) − x1 e−t
u2 = x2 − (2x2 − x1 )(1 − e−t ) − (x2 − x1 )(e−2t − e−t ) − x2 e−t
u3 = 0.
Note that in the Lagrangian system the displacements are expressed in terms of the initial position and
time, while in the Eulerian system the independent variables are the position coordinates and time. Euler
equations describe, as a function of time, how such things as density, pressure, and fluid velocity change at
a fixed point in the medium. In contrast, the Lagrangian viewpoint follows the time history of a moving
individual fluid particle as it moves through the medium.
229
General Tensor Derivation of Strain.
With reference to the figure 2.3-15 consider the deformation of a point P within a continuum. Let
1
(y , y 2 , y 3 ) denote a Cartesian coordinate system which is fixed in space. We can introduce a coordinate
transformation y i = y i (x1 , x2 , x3 ),
i = 1, 2, 3 and represent all points within the continuum with respect
to a set of generalized coordinates (x1 , x2 , x3 ). Let P denote a general point in the continuum while it is
in an unstrained state and assume that this point gets transformed to a point P 0 when the continuum
experiences external forces. If P moves to P 0 , then all points Q which are near P will move to points Q0
near P 0 . We can imagine that in the unstrained state all the points of the continuum are referenced with
respect to the set of generalized coordinates (x1 , x2 , x3 ). After the strain occurs, we can imagine that it will
be convenient to represent all points of the continuum with respect to a new barred system of coordinates
(x1 , x2 , x3 ). We call the original set of coordinates the Lagrangian system of coordinates and the new set
of barred coordinates the Eulerian coordinates. The Eulerian coordinates are assumed to be described by
a set of coordinate transformation equations xi = xi (x1 , x2 , x3 ),
i
i
1
2
3
x = x (x , x , x ),
i = 1, 2, 3 with inverse transformations
i = 1, 2, 3, which are assumed to exist. The barred and unbarred coordinates can
be related to a fixed set of Cartesian coordinates y i , i = 1, 2, 3, and we may assume that there exists
transformation equations
y i = y i (x1 , x2 , x3 ),
i = 1, 2, 3 and y i = y i (x1 , x2 , x3 ),
i = 1, 2, 3
which relate the barred and unbarred coordinates to the Cartesian axes. In the discussion that follows
be sure to note whether there is a bar over a symbol, as we will be jumping back and forth between the
Lagrangian and Eulerian reference frames.
Figure 2.3-15. Strain in generalized coordinates
~ i = ∂~r which produce
In the Lagrangian system of unbarred coordinates we have the basis vectors E
∂xi
~
~
the metrices gij = Ei · Ej . Similarly, in the Eulerian system of barred coordinates we have the basis vectors
~ ·E
~ . These basis vectors are illustrated in the figure 2.3-15.
~ = ∂~r which produces the metrices G = E
E
i
ij
i
j
∂xi
230
We assume that an element of arc length squared ds2 in the unstrained state is deformed to the element
of arc length squared ds2 in the strained state. An element of arc length squared can be expressed in terms
of the barred or unbarred coordinates. For example, in the Lagrangian system, let d~r = P Q so that
L20 = d~r · d~r = ds2 = gij dxi dxj ,
(2.3.54)
where gij are the metrices in the Lagrangian coordinate system. This same element of arc length squared
can be expressed in the barred system by
L20 = ds2 = gij dxi dxj ,
where g ij = gmn
∂xm ∂xn
.
∂xi ∂xj
(2.3.55)
Similarly, in the Eulerian system of coordinates the deformed arc length squared is
L2 = d~r · d~r = ds2 = Gij dxi dxj ,
(2.3.56)
where Gij are the metrices in the Eulerian system of coordinates. This same element of arc length squared
can be expressed in the Lagrangian system by the relation
L2 = ds2 = Gij dxi dxj ,
where Gij = Gmn
∂xm ∂xn
.
∂xi ∂xj
(2.3.57)
In the Lagrangian system we have
ds2 − ds2 = (Gij − gij )dxi dxj = 2eij dxi dxj
where
eij =
1
(Gij − gij )
2
(2.3.58)
is called the Green strain tensor or Lagrangian strain tensor. Alternatively, in the Eulerian system of
coordinates we may write
ds2 − ds2 = Gij − g ij dxi dxj = 2eij dxi dxj
where
eij =
1
Gij − gij
2
is called the Almansi strain tensor or Eulerian strain tensor.
(2.3.59)
231
Note also in the figure 2.3-15 there is the displacement vector ~u. This vector can be represented in any
of the following forms:
~i
~u = ui E
contravariant, Lagrangian basis
~i
~u = ui E
~
~u = ui E
covariant, Lagrangian reciprocal basis
i
~i
~u = ui E
contravariant, Eulerian basis
covariant, Eulerian reciprocal basis.
By vector addition we have ~r + ~u = ~r and consequently d~r + d~u = d~r . In the Lagrangian frame of reference
~ i and write d~r in the form d~r = dxi E
~ i . By
at the point P we represent ~u in the contravariant form ~u = ui E
~ i . These substitutions produce the
use of the equation (1.4.48) we can express d~u in the form d~u = ui,k dxk E
~ i in the Lagrangian coordinate system. We can then express ds2 in the
representation d~r = (dxi + ui,k dxk )E
Lagrangian system. We find
~ i · (dxj + uj dxm )E
~j
d~r · d~r = ds2 = (dxi + ui,k dxk )E
,m
= (dxi dxj + uj,m dxm dxi + ui,k dxk dxj + ui,k uj,m dxk dxm )gij
and consequently from the relation (2.3.58) we derive the representation
eij =
1
ui,j + uj,i + um,i um
,j .
2
(2.3.60)
This is the representation of the Lagrangian strain tensor in any system of coordinates. The strain tensor
eij is symmetric. We will restrict our study to small deformations and neglect the product terms in equation
(2.3.60). Under these conditions the equation (2.3.60) reduces to eij = 12 (ui,j + uj,i ).
If instead, we chose to represent the displacement ~u with respect to the Eulerian basis, then we can
write
~
~u = ui E
i
with
~ .
d~u = ui,k dxk E
i
These relations imply that
~ .
d~r = d~r − d~u = (dxi − ui,k dxk )E
i
This representation of d~r in the Eulerian frame of reference can be used to calculate the strain eij from the
relation ds2 − ds2 . It is left as an exercise to show that there results
eij =
1
ui,j + uj,i − um,i um
,j .
2
(2.3.61)
The equation (2.3.61) is the representation of the Eulerian strain tensor in any system of coordinates. Under
conditions of small deformations both the equations (2.3.60) and (2.3.61) reduce to the linearized Lagrangian
and Eulerian strain tensor eij = 12 (ui,j + uj,i ). In the case of large deformations the equations (2.3.60) and
(2.3.61) describe the strains. In the case of linear elasticity, where the deformations are very small, the
product terms in equations (2.3.60) and (2.3.61) are neglected and the Lagrangian and Eulerian strains
reduce to their linearized forms
eij =
1
[ui,j + uj,i ]
2
eij =
1
[ui,j + uj,i ] .
2
(2.3.62)
232
Figure 2.3-16. Displacement due to strain
Compressible and Incompressible Material With reference to figure 2.3-16, let xi , i = 1, 2, 3 denote
the position vector of an arbitrary point P in a continuum before there is a state of strain. Let Q be
a neighboring point of P with position vector xi + dxi , i = 1, 2, 3. Also in the figure 2.3-16 there is the
displacement vector ~u. Here it is assumed that ~u = ~u(x1 , x2 , x3 ) denotes the displacement field when the
continuum is in a state of strain. The figure 2.3-16 illustrates that in a state of strain P moves to P 0 and Q
moves to Q0 . Let us find a relationship between the distance P Q before the strain and the distance P 0 Q0 when
~ 2, E
~ 3 basis functions constructed at P we have previously
~ 1, E
the continuum is in a state of strain. For E
shown that if
~i
~u(x1 , x2 , x3 ) = ui E
then
~ i.
d~u = ui,j dxj E
Now for ~u + d~u the displacement of the point Q we may use vector addition and write
P Q + ~u + d~u = ~u + P 0 Q0 .
(2.3.63)
~ i = ai E
~ i denote an arbitrary small change in the continuum. This arbitrary displacement
Let P Q = dxi E
~ i due to the state of strain in the continuum. Employing the equation (2.3.63)
gets deformed to P 0 Q0 = Ai E
we write
dxi + ui,j dxj = ai + ui,j aj = Ai
which can be written in the form
δai = Ai − ai = ui,j aj
where
dxi = ai , i = 1, 2, 3
(2.3.64)
denotes an arbitrary small change. The tensor ui,j and the associated tensor ui,j = git ut,j are in general
not symmetric tensors. However, we know we can express ui,j as the sum of a symmetric (eij ) and skewsymmetric(ωij ) tensor. We therefore write
ui,j = eij + ωij
or ui,j = eij + ω ij ,
where
eij =
1
1
(ui,j + uj,i ) = (gim um ,j + gjm um ,i )
2
2
and
ωij =
1
1
(ui,j − uj,i ) = (gim um ,j − gjm um ,i ) .
2
2
The deformation of a small quantity ai can therefore be represented by a pure strain Ai − ai = eis as followed
by a rotation Ai − ai = ωsi as .
233
Consider now a small element of volume inside a material medium. With reference to the figure 2.317(a) we let ~a, ~b, ~c denote three small arbitrary independent vectors constructed at a general point P within
the material before any external forces are applied. We imagine ~a, ~b, ~c as representing the sides of a small
parallelepiped before any deformation has occurred. When the material is placed in a state of strain the
~ B,
~ C
~ as illustrated in
point P will move to P 0 and the vectors ~a, ~b, ~c will become deformed to the vectors A,
~ B,
~ C
~ represent the sides of the parallelepiped after the deformation.
the figure 2.3-17(b). The vectors A,
Figure 2.3-17. Deformation of a parallelepiped
Let ∆V denote the volume of the parallelepiped with sides ~a, ~b, ~c at P before the strain and let ∆V 0
~ B,
~ C
~ at the
denote the volume of the deformed parallelepiped after the strain, when it then has sides A,
point P 0 . We define the ratio of the change in volume due to the strain divided by the original volume as
the dilatation at the point P. The dilatation is thus expressed as
Θ=
∆V 0 − ∆V
= dilatation.
∆V
(2.3.65)
Since ui , i = 1, 2, 3 represents the displacement field due to the strain, we use the result from equation
~ B,
~ C
~ in the form
(2.3.64) and represent the displaced vectors A,
Ai = ai + ui,j aj
B i = bi + ui,j bj
i
i
C =c +
(2.3.66)
ui,j cj
where ~a, ~b, ~c are arbitrary small vectors emanating from the point P in the unstrained state. The element of
volume ∆V, before the strain, is calculated from the triple scalar product relation
∆V = ~a · (~b × ~c) = eijk ai bj ck .
The element of volume ∆V 0 , which occurs due to the strain, is calculated from the triple scalar product
~ · (B
~ × C)
~ = eijk Ai B j C k .
∆V 0 = A
234
Substituting the relations from the equations (2.3.66) into the triple scalar product gives
∆V 0 = eijk (ai + ui,m am )(bj + uj,n bn )(ck + uk,p cp ).
Expanding the triple scalar product and employing the result from Exercise 1.4, problem 34, we find the
simplified result gives us the dilatation
Θ=
∆V 0 − ∆V
= ur,r = div (~u).
∆V
(2.3.67)
That is, the dilatation is the divergence of the displacement field. If the divergence of the displacement field
is zero, there is no volume change and the material is said to be incompressible. If the divergence of the
displacement field is different from zero, the material is said to be compressible.
Note that the strain eij is expressible in terms of the displacement field by the relation
eij =
1
(ui,j + uj,i ),
2
and consequently g mn emn = ur,r .
(2.3.68)
Hence, for an orthogonal system of coordinates the dilatation can be expressed in terms of the strain elements
along the main diagonal.
Conservation of Mass
Consider the material in an arbitrary region R of a continuum. Let % = %(x, y, z, t) denote the density
of the material within the region. Assume that the dimension of the density % is gm/cm3 in the cgs system
of units. We shall assume that the region R is bounded by a closed surface S with exterior unit normal ~n
defined everywhere on the surface. Further, we let ~v = ~v (x, y, z, t) denote a velocity field associated with all
points within the continuum. The velocity field has units of cm/sec in the cgs system of units. Neglecting
sources and sinks, the law of conservation of mass examines
Z Z Z all the material entering and leaving a region R.
% dτ with dimensions of gm in the cgs system of
Enclosed within R is the material mass m where m =
R
units. Here dτ denotes an element of volume inside the region R. The change of mass with time is obtained
by differentiating the above relation. Differentiating the mass produces the equation
ZZZ
∂%
∂m
=
dτ
∂t
R ∂t
(2.3.69)
and has the dimensions of gm/sec.
ZZ
Consider also the surface integral
%~v · n̂ dσ
I=
(2.3.70)
S
where dσ is an element of surface area on the surface S which encloses R and n̂ is the exterior unit normal
vector to the surface S. The dimensions of the integral I is determined by examining the dimensions of each
term in the integrand of I. We find that
[I] =
gm
gm cm
· cm2 =
·
3
cm
sec
sec
and so the dimension of I is the same as the dimensions for the change of mass within the region R. The
surface integral I is the flux rate of material crossing the surface of R and represents the change of mass
235
entering the region if ~v · n̂ is negative and the change of mass leaving the region if ~v · n̂ is positive, as n̂ is
always an exterior unit normal vector. Equating the relations from equations (2.3.69) and (2.3.70) we obtain
a mathematical statement for mass conservation
ZZZ
ZZ
∂%
∂m
=
dτ = −
%~v · ~n dσ.
∂t
R ∂t
S
(2.3.71)
The equation (2.3.71) implies that the rate at which the mass contained in R increases must equal the rate
at which the mass flows into R through the surface S. The negative sign changes the direction of the exterior
normal so that we consider flow of material into the region. Employing the Gauss divergence theorem, the
surface integral in equation (2.3.71) can be replaced by a volume integral and the law of conservation of
mass is then expressible in the form
ZZZ R
∂%
+ div (%~v ) dτ = 0.
∂t
(2.3.72)
Since the region R is an arbitrary volume we conclude that the term inside the brackets must equal zero.
This gives us the continuity equation
∂%
+ div (%~v ) = 0
∂t
(2.3.73)
which represents the mass conservation law in terms of velocity components. This is the Eulerian representation of continuity of mass flow.
Equivalent forms of the continuity equation are:
∂%
+ ~v · grad % + % div ~v = 0
∂t
∂vi
∂%
∂%
+ vi i + % i = 0
∂t
∂x
∂x
∂vi
D%
+% i =0
Dt
∂x
∂%
∂% dxi
∂%
∂%
D%
=
+
=
+
vi is called the material derivative of the density %. Note that the
Dt
∂t
∂xi dt
∂t
∂xi
∂%
material derivative contains the expression ∂x
i vi which is known as the convective or advection term. If the
where
density % = %(x, y, z, t) is a constant we have
∂% ∂% dx ∂% dy
∂% dz
∂%
∂% dxi
D%
=
+
+
+
=
+ i
=0
Dt
∂t
∂x dt
∂y dt
∂z dt
∂t
∂x dt
(2.3.74)
and hence the continuity equation reduces to div (~v ) = 0. Thus, if div (~v ) is zero, then the material is
incompressible.
EXAMPLE 2.3-2. (Continuity Equation) Find the Lagrangian representation of mass conservation.
Solution: Let (X, Y, Z) denote the initial position of a fluid particle and denote the density of the fluid by
%(X, Y, Z, t) so that %(X, Y, Z, 0) denotes the density at the time t = 0. Consider a simple closed region in
our continuum and denote this region by R(0) at time t = 0 and by R(t) at some later time t. That is, all
the points
Z Z in
Z R(0) move in a one-to-one fashion to points in R(t). Initially the mass of material in R(0) is
%(X, Y, Z, 0) dτ (0) where dτ (0) = dXdY dZ is an element of volume in R(0). We have after a
m(0) =
R(0)
236
ZZZ
time t has elapsed the mass of material in the region R(t) given by m(t) =
%(X, Y, Z, t) dτ (t) where
x,y,z
dτ (0) where J is
dτ (t) = dxdydz is a deformed element of volume related to the dτ (0) by dτ (t) = J X,Y,Z
R(t)
the Jacobian of the Eulerian (x, y, z) variables with respect to the Lagrangian (X, Y, Z) representation. For
mass conservation we require that m(t) = m(0) for all t. This implies that
%(X, Y, Z, t)J = %(X, Y, Z, 0)
(2.3.75)
for all time, since the initial region R(0) is arbitrary. The right hand side of equation (2.3.75) is independent
of time and so
d
(%(X, Y, Z, t)J) = 0.
dt
(2.3.76)
This is the Lagrangian form of the continuity equation which expresses mass conservation. Using the result
dJ
= Jdiv V~ , (see problem 28, Exercise 2.3), the equation (2.3.76) can be expanded and written in the
that
dt
form
D%
~ =0
+ % div V
(2.3.77)
Dt
where
D%
Dt
is from equation (2.3.74). The form of the continuity equation (2.3.77) is one of the Eulerian forms
previously developed.
In the Eulerian coordinates the continuity equation is written
system the continuity equation is written
d(% J)
dt
∂%
∂t
+ div (%~v ) = 0, while in the Lagrangian
= 0. Note that the velocity carries the Lagrangian axes and
the density change grad %. This is reflective of the advection term ~v · grad %. Thus, in order for mass to
be conserved it need not remain stationary. The mass can flow and the density can change. The material
derivative is a transport rule depicting the relation between the Eulerian and Lagrangian viewpoints.
In general, from a Lagrangian viewpoint, any quantity Q(x, y, z, t) which is a function of both position
and time is seen as being transported by the fluid velocity (v1 , v2 , v3 ) to Q(x + v1 dt, y + v2 dt, z + v3 dt, t + dt).
Then the time derivative of Q contains both
∂Q
∂t
and the advection term ~v · ∇Q. In terms of mass flow, the
Eulerian viewpoint sees flow into and out of a fixed volume in space, as depicted by the equation (2.3.71),
In contrast, the Lagrangian viewpoint sees the same volume moving with the fluid and consequently
Z Z Z
D
ρ dτ = 0,
Dt
R(t)
where R(t) represents the volume moving with the fluid. Both viewpoints produce the same continuity
equation reflecting the conservation of mass.
Summary of Basic Equations
Let us summarize the basic equations which are valid for all types of a continuum. We have derived:
• Conservation of mass (continuity equation)
∂%
+ (%v i ),i = 0
∂t
237
• Conservation of linear momentum sometimes called the Cauchy equation of motion.
σ ij ,i + %bj = %f j ,
j = 1, 2, 3.
• Conservation of angular momentum
σij = σji
• Strain tensor for linear elasticity
eij =
1
(ui,j + uj,i ).
2
If we assume that the continuum is in equilibrium, and there is no motion, then the velocity and
acceleration terms above will be zero. The continuity equation then implies that the density is a constant.
The conservation of angular momentum equation requires that the stress tensor be symmetric and we need
find only six stresses. The remaining equations reduce to a set of nine equations in the fifteen unknowns:
3 displacements u1 , u2 , u3
6 strains
e11 , e12 , e13 , e22 , e23 , e33
6 stresses σ11 , σ12 , σ13 , σ22 , σ23 , σ33
Consequently, we still need additional information if we desire to determine these unknowns.
Note that the above equations do not involve any equations describing the material properties of the
continuum. We would expect solid materials to act differently from liquid material when subjected to external
forces. An equation or equations which describe the material properties are called constitutive equations.
In the following sections we will investigate constitutive equations for solids and liquids. We will restrict
our study to linear elastic materials over a range where there is a linear relationship between the stress and
strain. We will not consider plastic or viscoelastic materials. Viscoelastic materials have the property that
the stress is not only a function of strain but also a function of the rates of change of the stresses and strains
and consequently properties of these materials are time dependent.
238
EXERCISE 2.3
I 1.
Assume an orthogonal coordinate system with metric tensor gij = 0 for i 6= j and g(i)(i) = h2i (no
summation on i). Use the definition of strain
ers =
1
1
(ur,s + us,r ) =
grt ut,s + gst ut,r
2
2
and show that in terms of the physical components
e(ij) =
eij
hi hj
u(i) = hi ui
no summation on i or j
no summation on i
there results the equations:
t
∂ut
m
+
u
no summation on i
eii = git
mi
∂xi
∂ut
∂ut
i 6= j
2eij = git j + gjt i ,
∂x
∂x
3
u(i)
1 X u(m) ∂
∂
h2
no summation on i
+
e(ii) =
∂xi
hi
2h2i m=1 hm ∂xm i
u(i)
u(j)
hj ∂
hi ∂
+
, no summation on i or j, i 6= j.
2e(ij) =
hj ∂xj
hi
hi ∂xi
hj
I 2. Use the results from problem 1 to write out all components of the strain tensor in Cartesian coordinates.
Use the notation u(1) = u,u(2) = v,u(3) = w and
e(11) = exx ,
e(22) = eyy ,
e(33) = ezz ,
e(12) = exy ,
to verify the relations:
exx
eyy
ezz
∂u
=
∂x
∂v
=
∂y
∂w
=
∂z
e(13) = exz ,
e(23) = eyz
∂u
1 ∂v
+
=
2 ∂x ∂y
1 ∂u ∂w
+
=
2 ∂z
∂x
1 ∂w ∂v
+
=
2 ∂y
∂z
exy
exz
ezy
I 3. Use the results from problem 1 to write out all components of the strain tensor in cylindrical coordinates.
Use the notation u(1) = ur , u(2) = uθ , u(3) = uz and
e(11) = err ,
e(22) = eθθ ,
e(33) = ezz ,
e(12) = erθ ,
e(13) = erz ,
to verify the relations:
err
eθθ
ezz
∂ur
=
∂r
ur
1 ∂uθ
+
=
r ∂θ
r
∂uz
=
∂z
∂uθ
uθ
1 1 ∂ur
+
−
2 r ∂θ
∂r
r
∂ur
1 ∂uz
+
=
2 ∂r
∂z
1 ∂uz
1 ∂uθ
+
=
2 ∂z
r ∂θ
erθ =
erz
eθz
e(23) = eθz
239
I 4. Use the results from problem 1 to write out all components of the strain tensor in spherical coordinates.
Use the notation u(1) = uρ ,u(2) = uθ ,u(3) = uφ and
e(11) = eρρ ,
e(22) = eθθ ,
e(33) = eφφ ,
e(12) = eρθ ,
e(13) = eρφ ,
e(23) = eθφ
to verify the relations
eρρ
eθθ
eφφ
∂uρ
=
∂ρ
uρ
1 ∂uθ
+
=
ρ ∂θ
ρ
uθ
1 ∂uφ uρ
+
+
cot θ
=
ρ sin θ ∂φ
ρ
ρ
eρθ
eρφ
eθφ
uθ
∂uθ
1 1 ∂uρ
−
+
=
2 ρ ∂θ
ρ
∂ρ
1 ∂uρ
uφ ∂uφ
1
−
+
=
2 ρ sin θ ∂φ
ρ
∂ρ
uφ
1 ∂uθ
1 1 ∂uφ
−
cot θ +
=
2 ρ ∂θ
ρ
ρ sin θ ∂φ
I 5. Expand equation (2.3.67) and find the dilatation in terms of the physical components of an orthogonal
system and verify that
Θ=
∂(h2 h3 u(1)) ∂(h1 h3 u(2)) ∂(h1 h2 u(3))
1
+
+
h1 h2 h3
∂x1
∂x2
∂x3
I 6. Verify that the dilatation in Cartesian coordinates is
Θ = exx + eyy + ezz =
∂u ∂v ∂w
+
+
.
∂x ∂y
∂z
I 7. Verify that the dilatation in cylindrical coordinates is
Θ = err + eθθ + ezz =
1 ∂uθ
1
∂ur
∂uz
+
+ ur +
.
∂r
r ∂θ
r
∂z
I 8. Verify that the dilatation in spherical coordinates is
Θ = eρρ + eθθ + eφφ =
1 ∂uθ
2
uθ cot θ
∂uρ
1 ∂uφ
+
+ uρ +
+
.
∂ρ
ρ ∂θ
ρ
ρ sin θ ∂φ
ρ
I 9. Show that in an orthogonal set of coordinates the rotation tensor ωij can be written in terms of physical
components in the form
1
ω(ij) =
2hi hj
∂(hi u(i)) ∂(hj u(j))
−
,
∂xj
∂xi
no summations
Hint: See problem 1.
I 10. Use the result from problem 9 to verify that in Cartesian coordinates
∂u
1 ∂v
−
ωyx =
2 ∂x ∂y
1 ∂u ∂w
−
ωxz =
2 ∂z
∂x
1 ∂w ∂v
−
ωzy =
2 ∂y
∂z
240
I 11. Use the results from problem 9 to verify that in cylindrical coordinates
1 ∂(ruθ ) ∂ur
−
ωθr =
2r
∂r
∂θ
∂uz
1 ∂ur
−
ωrz =
2 ∂z
∂r
∂uθ
1 1 ∂uz
−
ωzθ =
2 r ∂θ
∂z
I 12. Use the results from problem 9 to verify that in spherical coordinates
1 ∂(ρuθ ) ∂uρ
−
ωθρ =
2ρ
∂ρ
∂θ
1 ∂uρ
∂(ρuφ )
1
−
ωρφ =
2ρ sin θ ∂φ
∂ρ
∂(uφ sin θ) ∂uθ
1
−
ωφθ =
2ρ sin θ
∂θ
∂φ
I 13. The conditions for static equilibrium in a linear elastic material are determined from the conservation
law
σij ,j + %bi = 0,
i, j = 1, 2, 3,
where σji are the stress tensor components, bi are the external body forces per unit mass and % is the density
of the material. Assume an orthogonal coordinate system and verify the following results.
(a) Show that
1 ∂ √ j
( gσi ) − [ij, m]σ mj
σij ,j = √
g ∂xj
(b) Use the substitutions
σ(ij) = σij
hj
hi
no summation on i or j
bi
no summation on i
hi
σ(ij) = σ ij hi hj no summation on i or j
b(i) =
and express the equilibrium equations in terms of physical components and verify the relations
√
3
3
X
ghi σ(ij)
1 ∂
1 X σ(jj) ∂(h2j )
+ hi %b(i) = 0,
−
√
g ∂xj
hj
2 j=1 h2j ∂xi
j=1
where there is no summation on i.
I 14. Use the results from problem 13 and verify that the equilibrium equations in Cartesian coordinates
can be expressed
∂σxy
∂σxz
∂σxx
+
+
+ %bx = 0
∂x
∂y
∂z
∂σyy
∂σyz
∂σyx
+
+
+ %by = 0
∂x
∂y
∂z
∂σzy
∂σzz
∂σzx
+
+
+ %bz = 0
∂x
∂y
∂z
241
I 15. Use the results from problem 13 and verify that the equilibrium equations in cylindrical coordinates
can be expressed
1 ∂σrθ
∂σrz
1
∂σrr
+
+
+ (σrr − σθθ ) + %br = 0
∂r
r ∂θ
∂z
r
1 ∂σθθ
∂σθz
2
∂σθr
+
+
+ σθr + %bθ = 0
∂r
r ∂θ
∂z
r
1 ∂σzθ
∂σzz
1
∂σzr
+
+
+ σzr + %bz = 0
∂r
r ∂θ
∂z
r
I 16. Use the results from problem 13 and verify that the equilibrium equations in spherical coordinates
can be expressed
1 ∂σρθ
1 ∂σρφ
1
∂σρρ
+
+
+ (2σρρ − σθθ − σφφ + σρθ cot θ) + %bρ = 0
∂ρ
ρ ∂θ
ρ sin θ ∂φ
ρ
1 ∂σθθ
1 ∂σθφ
1
∂σθρ
+
+
+ (3σρθ + [σθθ − σφφ ] cot θ) + %bθ = 0
∂ρ
ρ ∂θ
ρ sin θ ∂φ
ρ
1 ∂σφθ
1 ∂σφφ
1
∂σφρ
+
+
+ (3σρφ + 2σθφ cot θ) + %bφ = 0
∂ρ
ρ ∂θ
ρ sin θ ∂φ
ρ
I 17. Derive the result for the Lagrangian strain defined by the equation (2.3.60).
I 18. Derive the result for the Eulerian strain defined by equation (2.3.61).
I 19.
The equation δai = ui,j aj , describes the deformation in an elastic solid subjected to forces. The
quantity δai denotes the difference vector Ai − ai between the undeformed and deformed states.
(a) Let |a| denote the magnitude of the vector ai and show that the strain e in the direction ai can be
represented
δ|a|
= eij
e=
|a|
ai
|a|
aj
|a|
= eij λi λj ,
where λi is a unit vector in the direction ai .
(b) Show that for λ1 = 1, λ2 = 0, λ3 = 0 there results e = e11 , with similar results applying to vectors λi in
the y and z directions.
Hint: Consider the magnitude squared |a|2 = gij ai aj .
I 20. At the point (1, 2, 3) of an elastic solid construct the small vector ~a = ( 23 ê1 +
2
3
ê2 +
1
3
ê3 ), where
> 0 is a small positive quantity. The solid is subjected to forces such that the following displacement field
results.
~u = (xy ê1 + yz ê2 + xz ê3 ) × 10−2
~ after the displacement field has been imposed.
Calculate the deformed vector A
I 21. For the displacement field
~u = (x2 + yz) ê1 + (xy + z 2 ) ê2 + xyz ê3
(a) Calculate the strain matrix at the point (1, 2, 3).
(b) Calculate the rotation matrix at the point (1, 2, 3).
242
I 22. Show that for an orthogonal coordinate system the ith component of the convective operator can be
written
~ · ∇) A]
~i=
[(V
3
3
X
∂hm
V (m) ∂A(i) X A(m)
∂hi
+
V (i) m − V (m) i
hm ∂xm
hm hi
∂x
∂x
m=1
m=1
m6=i
I 23.
Consider a parallelepiped with dimensions `, w, h which has a uniform pressure P applied to each
face. Show that the volume strain can be expressed as
∆` ∆w
∆h
−3P (1 − 2ν)
∆V
=
+
+
=
.
V
`
w
h
E
The quantity k = E/3(1 − 2ν) is called the bulk modulus of elasticity.
I 24. Show in Cartesian coordinates the continuity equation is
∂% ∂(%u) ∂(%v) ∂(%w)
+
+
+
= 0,
∂t
∂x
∂y
∂z
where (u, v, w) are the velocity components.
I 25. Show in cylindrical coordinates the continuity equation is
∂% 1 ∂(r%Vr ) 1 ∂(%Vθ ) ∂(%Vz )
+
+
+
=0
∂t
r ∂r
r ∂θ
∂z
where Vr , Vθ , Vz are the velocity components.
I 26. Show in spherical coordinates the continuity equation is
1 ∂(ρ2 %Vρ )
1 ∂(%Vθ sin θ)
1 ∂(%Vφ )
∂%
+ 2
+
+
=0
∂t
ρ
∂ρ
ρ sin θ
∂θ
ρ sin θ ∂φ
where Vρ , Vθ , Vφ are the velocity components.
I 27. (a) Apply a stress σyy to both ends of a square element in a x, y continuum. Illustrate and label
all changes that occur due to this stress. (b) Apply a stress σxx to both ends of a square element in a
x, y continuum. Illustrate and label all changes that occur due to this stress. (c) Use superposition of your
results in parts (a) and (b) and explain each term in the relations
exx =
σyy
σxx
−ν
E
E
and
eyy =
I 28. Show that the time derivative of the Jacobian J = J
div V~ =
∂V2
∂V3
∂V1
+
+
∂x
∂y
∂z
and V1 =
σxx
σyy
−ν
.
E
E
x, y, z
X, Y, Z
dx
,
dt
V2 =
satisfies
dy
,
dt
dJ
~ where
= J div V
dt
V3 =
dz
.
dt
Hint: Let (x, y, z) = (x1 , x2 , x3 ) and (X, Y, Z) = (X1 , X2 , X3 ), then note that
eijk
∂V1 ∂x2 ∂x3
∂V1 ∂xm ∂x2 ∂x3
∂x1 ∂x2 ∂x3 ∂V1
= eijk
= eijk
,
∂Xi ∂Xj ∂Xk
∂xm ∂Xi ∂Xj ∂Xk
∂Xi ∂Xj ∂Xk ∂x1
etc.
243
§2.4 CONTINUUM MECHANICS (SOLIDS)
In this introduction to continuum mechanics we consider the basic equations describing the physical
effects created by external forces acting upon solids and fluids. In addition to the basic equations that
are applicable to all continua, there are equations which are constructed to take into account material
characteristics. These equations are called constitutive equations. For example, in the study of solids the
constitutive equations for a linear elastic material is a set of relations between stress and strain. In the study
of fluids, the constitutive equations consists of a set of relations between stress and rate of strain. Constitutive
equations are usually constructed from some basic axioms. The resulting equations have unknown material
parameters which can be determined from experimental investigations.
One of the basic axioms, used in the study of elastic solids, is that of material invariance. This axiom requires that certain symmetry conditions of solids are to remain invariant under a set of orthogonal
transformations and translations. This axiom is employed in the next section to simplify the constitutive
equations for elasticity. We begin our study of continuum mechanics by investigating the development of
constitutive equations for linear elastic solids.
Generalized Hooke’s Law
If the continuum material is a linear elastic material, we introduce the generalized Hooke’s law in
Cartesian coordinates
σij = cijkl ekl ,
i, j, k, l = 1, 2, 3.
(2.4.1)
The Hooke’s law is a statement that the stress is proportional to the gradient of the deformation occurring
in the material. These equations assume a linear relationship exists between the components of the stress
tensor and strain tensor and we say stress is a linear function of strain. Such relations are referred to as a
set of constitutive equations. Constitutive equations serve to describe the material properties of the medium
when it is subjected to external forces.
Constitutive Equations
The equations (2.4.1) are constitutive equations which are applicable for materials exhibiting small
deformations when subjected to external forces. The 81 constants cijkl are called the elastic stiffness of the
material. The above relations can also be expressed in the form
eij = sijkl σkl ,
i, j, k, l = 1, 2, 3
(2.4.2)
where sijkl are constants called the elastic compliance of the material. Since the stress σij and strain eij
have been shown to be tensors we can conclude that both the elastic stiffness cijkl and elastic compliance
sijkl are fourth order tensors. Due to the symmetry of the stress and strain tensors we find that the elastic
stiffness and elastic compliance tensor must satisfy the relations
cijkl = cjikl = cijlk = cjilk
(2.4.3)
sijkl = sjikl = sijlk = sjilk
and consequently only 36 of the 81 constants are actually independent. If all 36 of the material (crystal)
constants are independent the material is called triclinic and there are no material symmetries.
244
Restrictions on Elastic Constants due to Symmetry
The equations (2.4.1) and (2.4.2) can be replaced by an equivalent set of equations which are easier to
analyze. This is accomplished by defining the quantities
where
e1 ,
e2 ,
e3 ,
σ1 ,
σ2 ,
σ3 ,

e4
e2
e6
e1
 e4
e5

and
σ4
σ2
σ6
σ1
 σ4
σ5
e4 ,
σ4 ,
e5 ,
e6
σ5 ,
σ6
 
e5
e11
e6  =  e21
e3
e31
e12
e22
e32

e13
e23 
e33
 
σ5
σ11
σ6  =  σ21
σ3
σ31
σ12
σ22
σ32

σ13
σ23  .
σ33
Then the generalized Hooke’s law from the equations (2.4.1) and (2.4.2) can be represented in either of
the forms
σi = cij ej
or ei = sij σj
where i, j = 1, . . . , 6
(2.4.4)
where cij are constants related to the elastic stiffness and sij are constants related to the elastic compliance.
These constants satisfy the relation
smi cij = δmj
Here
eij =
and similarly
σij =
where
i, m, j = 1, . . . , 6
ei ,
e1+i+j ,
i = j = 1, 2, 3
i 6= j, and i = 1, or, 2
σi ,
σ1+i+j ,
i = j = 1, 2, 3
i 6= j, and i = 1, or, 2.
(2.4.5)
These relations show that the constants cij are related to the elastic stiffness coefficients cpqrs by the
relations
cm1 = cij11
cm4 = 2cij12
cm2 = cij22
cm5 = 2cij13
cm3 = cij33
cm6 = 2cij23
where
m=
i,
1 + i + j,
if i = j = 1, 2, or 3
if i 6= j and i = 1 or 2.
A similar type relation holds for the constants sij and spqrs . The above relations can be verified by expanding
the equations (2.4.1) and (2.4.2) and comparing like terms with the expanded form of the equation (2.4.4).
245
The generalized Hooke’s law can now be expressed in a form where the 36 independent constants can
be examined in more detail under special material symmetries. We will examine the form
 
  
σ1
s11 s12 s13 s14 s15 s16
e1
 e2   s21 s22 s23 s24 s25 s26   σ2 
 
  
 e3   s31 s32 s33 s34 s35 s36   σ3 
 .
 =
 e4   s41 s42 s43 s44 s45 s46   σ4 
 
  
e5
s51 s52 s53 s54 s55 s56
σ5
e6
s61 s62 s63 s64 s65 s66
σ6
Alternatively, in the arguments that follow, one can examine
  
c11 c12 c13 c14 c15
σ1
 σ2   c21 c22 c23 c24 c25
  
 σ3   c31 c32 c33 c34 c35
 =
 σ4   c41 c42 c43 c44 c45
  
σ5
c51 c52 c53 c54 c55
σ6
c61 c62 c63 c64 c65
(2.4.6)
the equivalent form
 
e1
c16
c26   e2 
 
c36   e3 
 .
c46   e4 
 
c56
e5
c66
e6
Material Symmetries
A material (crystal) with one plane of symmetry is called an aelotropic material. If we let the x1 x2 plane be a plane of symmetry then the equations (2.4.6) must remain invariant under the coordinate
transformation
 
x1
1 0
 x2  =  0 1
x3
0 0

 
0
x1
0   x2 
x3
−1
(2.4.7)
which represents an inversion of the x3 axis. That is, if the x1 -x2 plane is a plane of symmetry we should be
able to replace x3 by −x3 and the equations (2.4.6) should remain unchanged. This is equivalent to saying
that a transformation of the type from equation (2.4.7) changes the Hooke’s law to the form ei = sij σ j where
the sij remain unaltered because it is the same material. Employing the transformation equations
x1 = x1 ,
x2 = x2 ,
x3 = −x3
(2.4.8)
we examine the stress and strain transformation equations
σ ij = σpq
∂xp ∂xq
∂xi ∂xj
and
eij = epq
∂xp ∂xq
.
∂xi ∂xj
(2.4.9)
If we expand both of the equations (2.4.9) and substitute in the nonzero derivatives
∂x1
= 1,
∂x1
∂x2
= 1,
∂x2
∂x3
= −1,
∂x3
(2.4.10)
we obtain the relations
σ 11 = σ11
e11 = e11
σ 22 = σ22
e22 = e22
σ 33 = σ33
e33 = e33
σ 21 = σ21
e21 = e21
σ 31 = −σ31
e31 = −e31
σ 23 = −σ23
e23 = −e23 .
(2.4.11)
246
We conclude that if the material undergoes a strain, with the x1 -x2 plane as a plane of symmetry then
e5 and e6 change sign upon reversal of the x3 axis and e1 , e2 , e3 , e4 remain unchanged. Similarly, we find σ5
and σ6 change sign while σ1 , σ2 , σ3 , σ4 remain unchanged. The equation (2.4.6) then becomes
 
e1
s11
 e2   s21
 

 e3   s31
=

 e4   s41
 

−e5
s51
−e6
s61

s12
s22
s32
s42
s52
s62
s13
s23
s33
s43
s53
s63
s14
s24
s34
s44
s54
s64
s15
s25
s35
s45
s55
s65


s16
σ1
s26   σ2 


s36   σ3 

.
s46   σ4 


s56
−σ5
s66
−σ6
(2.4.12)
If the stress strain relation for the new orientation of the x3 axis is to have the same form as the
old orientation, then the equations (2.4.6) and (2.4.12) must give the same results. Comparison of these
equations we find that
s15 = s16 = 0
s25 = s26 = 0
s35 = s36 = 0
(2.4.13)
s45 = s46 = 0
s51 = s52 = s53 = s54 = 0
s61 = s62 = s63 = s64 = 0.
In summary, from an examination of the equations (2.4.6) and (2.4.12) we find that for an aelotropic
material (crystal), with one plane of symmetry, the 36 constants sij reduce to 20 constants and the generalized
Hooke’s law (constitutive equation) has the form
 
s11
e1
 e2   s21
  
 e3   s31
 =
 e4   s41
  
e5
0
e6
0

s12
s22
s32
s42
0
0
s13
s23
s33
s43
0
0
s14
s24
s34
s44
0
0
0
0
0
0
s55
s65
 
σ1
0
0   σ2 
 
0   σ3 
 .
0   σ4 
 
s56
σ5
s66
σ6
Alternatively, the Hooke’s law can be represented in the form
 
c11
σ1
 σ2   c21
  
 σ3   c31
 =
 σ4   c41
  
σ5
0
σ6
0

c12
c22
c32
c42
0
0
c13
c23
c33
c43
0
0
c14
c24
c34
c44
0
0
0
0
0
0
c55
c65
 
0
e1
0   e2 
 
0   e3 
 .
0   e4 
 
c56
e5
c66
e6
(2.4.14)
247
Additional Symmetries
If the material (crystal) is such that there is an additional plane of symmetry, say the x2 -x3 plane, then
reversal of the x1 axis should leave the equations (2.4.14) unaltered. If there are two planes of symmetry
then there will automatically be a third plane of symmetry. Such a material (crystal) is called orthotropic.
Introducing the additional transformation
x1 = −x1 ,
x2 = x2 ,
x3 = x3
which represents the reversal of the x1 axes, the expanded form of equations (2.4.9) are used to calculate the
effect of such a transformation upon the stress and strain tensor. We find σ1 , σ2 , σ3 , σ6 , e1 , e2 , e3 , e6 remain
unchanged while σ4 , σ5 , e4 , e5 change sign. The equation (2.4.14) then becomes
 
s11
e1
e
s
 2   21
 

 e3   s31
=

 −e4   s41
 

−e5
0
e6
0

s12
s22
s32
s42
0
0
s13
s23
s33
s43
0
0
s14
s24
s34
s44
0
0
0
0
0
0
s55
s65


σ1
0
0   σ2 


0   σ3 

.
0   −σ4 


s56
−σ5
s66
σ6
(2.4.15)
Note that if the constitutive equations (2.4.14) and (2.4.15) are to produce the same results upon reversal
of the x1 axes, then we require that the following coefficients be equated to zero:
s14 = s24 = s34 = 0
s41 = s42 = s43 = 0
s56 = s65 = 0.
This then produces the constitutive equation
 
s11
e1
 e2   s21
  
 e3   s31
 =
 e4   0
  
e5
0
e6
0
s12
s22
s32
0
0
0
s13
s23
s33
0
0
0
0
0
0
s44
0
0
0
0
0
0
s55
0
 
σ1
0
0   σ2 
 
0   σ3 
 
0   σ4 
 
0
σ5
σ6
s66
 
c11
σ1
 σ2   c21
  
 σ3   c31
 =
 σ4   0
  
σ5
0
σ6
0
c12
c22
c32
0
0
0
c13
c23
c33
0
0
0
0
0
0
c44
0
0
0
0
0
0
c55
0
 
e1
0
0   e2 
 
0   e3 
 
0   e4 
 
0
e5
e6
c66

or its equivalent form

(2.4.16)
and the original 36 constants have been reduced to 12 constants. This is the constitutive equation for
orthotropic material (crystals).
248
Axis of Symmetry
If in addition to three planes of symmetry there is an axis of symmetry then the material (crystal) is
termed hexagonal. Assume that the x1 axis is an axis of symmetry and consider the effect of the transformation
x1 = x1 ,
x2 = x3
x3 = −x2
upon the constitutive equations. It is left as an exercise to verify that the constitutive equations reduce to
the form where there are 7 independent constants having either of the forms
 
s11
e1
 e2   s21
  
 e3   s21
 =
 e4   0
  
e5
0
e6
0
s12
s22
s23
0
0
0
s12
s23
s22
0
0
0
0
0
0
s44
0
0
0
0
0
0
s44
0
 
σ1
0
0   σ2 
 
0   σ3 
 
0   σ4 
 
0
σ5
σ6
s66
 
c11
σ1
 σ2   c21
  
 σ3   c21
 =
 σ4   0
  
σ5
0
σ6
0
c12
c22
c23
0
0
0
c12
c23
c22
0
0
0
0
0
0
c44
0
0
0
0
0
0
c44
0
 
0
e1
0   e2 
 
0   e3 
 .
0   e4 
 
0
e5
e6
c66

or

Finally, if the material is completely symmetric, the x2 axis is also an axis of symmetry and we can
consider the effect of the transformation
x1 = −x3 ,
x2 = x2 ,
x3 = x1
upon the constitutive equations. It can be verified that these transformations reduce the Hooke’s law
constitutive equation to the form
 
s11
e1
e
s
 2   12
  
 e3   s12
 =
 e4   0
  
e5
0
e6
0

s12
s11
s12
0
0
0
s12
s12
s11
0
0
0
0
0
0
s44
0
0
0
0
0
0
s44
0
 
σ1
0
0   σ2 
 
0   σ3 
 .
0   σ4 
 
0
σ5
σ6
s44
(2.4.17)
Materials (crystals) with atomic arrangements that exhibit the above symmetries are called isotropic
materials. An equivalent form of (2.4.17) is the relation
 
c11
σ1
 σ2   c12
  
 σ3   c12
 =
 σ4   0
  
σ5
0
σ6
0

c12
c11
c12
0
0
0
c12
c12
c11
0
0
0
0
0
0
c44
0
0
0
0
0
0
c44
0
 
0
e1
0   e2 
 
0   e3 
 .
0   e4 
 
0
e5
e6
c44
The figure 2.4-1 lists values for the elastic stiffness associated with some metals which are isotropic1
1
Additional constants are given in “International Tables of Selected Constants”, Metals: Thermal and
Mechanical Data, Vol. 16, Edited by S. Allard, Pergamon Press, 1969.
249
Metal
Na
Pb
Cu
Ni
Cr
Mo
W
c11
0.074
0.495
1.684
2.508
3.500
4.630
5.233
c12
0.062
0.423
1.214
1.500
0.678
1.610
2.045
c44
0.042
0.149
0.754
1.235
1.008
1.090
1.607
Figure 2.4-1. Elastic stiffness coefficients for some metals which are cubic.
Constants are given in units of 1012 dynes/cm2
Under these conditions the stress strain constitutive relations can be written as
σ1 = σ11 = (c11 − c12 )e11 + c12 (e11 + e22 + e33 )
σ2 = σ22 = (c11 − c12 )e22 + c12 (e11 + e22 + e33 )
σ3 = σ33 = (c11 − c12 )e33 + c12 (e11 + e22 + e33 )
(2.4.18)
σ4 = σ12 = c44 e12
σ5 = σ13 = c44 e13
σ6 = σ23 = c44 e23 .
Isotropic Material
Materials (crystals) which are elastically the same in all directions are called isotropic. We have shown
that for a cubic material which exhibits symmetry with respect to all axes and planes, the constitutive
stress-strain relation reduces to the form found in equation (2.4.17). Define the quantities
s11 =
1
,
E
s12 = −
ν
,
E
s44 =
1
2µ
where E is the Young’s Modulus of elasticity, ν is the Poisson’s ratio, and µ is the shear or rigidity modulus.
For isotropic materials the three constants E, ν, µ are not independent as the following example demonstrates.
EXAMPLE 2.4-1. (Elastic constants)
For an isotropic material, consider a cross section of material in
the x1 -x2 plane which is subjected to pure shearing so that σ4 = σ12 is the only nonzero stress as illustrated
in the figure 2.4-2.
For the above conditions, the equation (2.4.17) reduces to the single equation
e4 = e12 = s44 σ4 = s44 σ12
or
µ=
σ12
γ12
and so the shear modulus is the ratio of the shear stress to the shear angle. Now rotate the axes through a
45 degree angle to a barred system of coordinates where
x1 = x1 cos α − x2 sin α
x2 = x1 sin α + x2 cos α
250
Figure 2.4-2. Element subjected to pure shearing
where α =
π
4.
Expanding the transformation equations (2.4.9) we find that
σ 1 = σ 11 = cos α sin α σ12 + sin α cos α σ21 = σ12 = σ4
σ 2 = σ 22 = − sin α cos α σ12 − sin α cos α σ21 = −σ12 = −σ4 ,
and similarly
e2 = e22 = −e4 .
e1 = e11 = e4 ,
In the barred system, the Hooke’s law becomes
e1 = s11 σ 1 + s12 σ 2
or
e4 = s11 σ4 − s12 σ4 = s44 σ4 .
Hence, the constants s11 , s12 , s44 are related by the relation
s11 − s12 = s44
or
ν
1
1
+
=
.
E
E
2µ
(2.4.19)
This is an important relation connecting the elastic constants associated with isotropic materials. The
above transformation can also be applied to triclinic, aelotropic, orthotropic, and hexagonal materials to
find relationships between the elastic constants.
Observe also that some texts postulate the existence of a strain energy function U ∗ which has the
property that σij =
∗
∂U ∗
∂eij .
In this case the strain energy function, in the single index notation, is written
U = cij ei ej where cij and consequently sij are symmetric. In this case the previous discussed symmetries
give the following results for the nonzero elastic compliances sij : 13 nonzero constants instead of 20 for
aelotropic material, 9 nonzero constants instead of 12 for orthotropic material, and 6 nonzero constants
instead of 7 for hexagonal material. This is because of the additional property that sij = sji be symmetric.
251
The previous discussion has shown that for an isotropic material the generalized Hooke’s law (constitutive equations) have the form
e11 =
e22 =
e33 =
e21 = e12 =
e32 = e23 =
e31 = e13 =
1
[σ11 − ν(σ22 + σ33 )]
E
1
[σ22 − ν(σ33 + σ11 )]
E
1
[σ33 − ν(σ11 + σ22 )]
E
,
1+ν
σ12
E
1+ν
σ23
E
1+ν
σ13
E
(2.4.20)
where equation (2.4.19) holds. These equations can be expressed in the indicial notation and have the form
eij =
1+ν
ν
σij − σkk δij ,
E
E
(2.4.21)
where σkk = σ11 + σ22 + σ33 is a stress invariant and δij is the Kronecker delta. We can solve for the stress
in terms of the strain by performing a contraction on i and j in equation (2.4.21). This gives the dilatation
eii =
1+ν
3ν
1 − 2ν
σii −
σkk =
σkk .
E
E
E
Note that from the result in equation (2.4.21) we are now able to solve for the stress in terms of the strain.
We find
1+ν
ν
σij −
ekk δij
E
1 − 2ν
νE
E
eij = σij −
ekk δij
1+ν
(1 + ν)(1 − 2ν)
E
νE
eij +
ekk δij .
or
σij =
1+ν
(1 + ν)(1 − 2ν)
eij =
The tensor equation (2.4.22) represents the six scalar equations
E
[(1 − ν)e11 + ν(e22 + e33 )]
(1 + ν)(1 − 2ν)
E
[(1 − ν)e22 + ν(e33 + e11 )]
=
(1 + ν)(1 − 2ν)
E
[(1 − ν)e33 + ν(e22 + e11 )]
=
(1 + ν)(1 − 2ν)
E
e12
1+ν
E
e13
=
1+ν
E
e23 .
=
1+ν
σ11 =
σ12 =
σ22
σ13
σ33
σ23
(2.4.22)
252
Alternative Approach to Constitutive Equations
The constitutive equation defined by Hooke’s generalized law for isotropic materials can be approached
from another point of view. Consider the generalized Hooke’s law
σij = cijkl ekl ,
i, j, k, l = 1, 2, 3.
If we transform to a barred system of coordinates, we will have the new Hooke’s law
σ ij = cijkl ekl ,
i, j, k, l = 1, 2, 3.
For an isotropic material we require that
cijkl = cijkl .
Tensors whose components are the same in all coordinate systems are called isotropic tensors. We have
previously shown in Exercise 1.3, problem 18, that
cpqrs = λδpq δrs + µ(δpr δqs + δps δqr ) + κ(δpr δqs − δps δqr )
is an isotropic tensor when we consider affine type transformations. If we further require the symmetry
conditions found in equations (2.4.3) be satisfied, we find that κ = 0 and consequently the generalized
Hooke’s law must have the form
σpq = cpqrs ers = [λδpq δrs + µ(δpr δqs + δps δqr )] ers
(2.4.23)
σpq = λδpq err + µ(epq + eqp )
σpq = 2µepq + λerr δpq ,
or
where err = e11 + e22 + e33 = Θ is the dilatation. The constants λ and µ are called Lame’s constants.
Comparing the equation (2.4.22) with equation (2.4.23) we find that the constants λ and µ satisfy the
relations
µ=
E
2(1 + ν)
λ=
νE
.
(1 + ν)(1 − 2ν)
(2.4.24)
In addition to the constants E, ν, µ, λ, it is sometimes convenient to introduce the constant k, called the bulk
modulus of elasticity, (Exercise 2.3, problem 23), defined by
k=
E
.
3(1 − 2ν)
(2.4.25)
The stress-strain constitutive equation (2.4.23) was derived using Cartesian tensors. To generalize the
equation (2.4.23) we consider a transformation from a Cartesian coordinate system y i , i = 1, 2, 3 to a general
coordinate system xi , i = 1, 2, 3. We employ the relations
g ij =
and
σ mn = σij
∂y i ∂y j
,
∂xm ∂xn
∂y m ∂y m
,
∂xi ∂xj
g ij =
emn = eij
∂y i ∂y j
,
∂xm ∂xn
∂xi ∂xj
∂y m ∂y m
or
erq = eij
∂xi ∂xj
∂y r ∂y q
253
and convert equation (2.4.23) to a more generalized form. Multiply equation (2.4.23) by
the result
σ mn = λ
∂y p ∂y q
and verify
∂xm ∂xn
∂y q ∂y q
err + µ (emn + enm ) ,
∂xm ∂xn
which can be simplified to the form
σ mn = λg mn eij g ij + µ (emn + enm ) .
Dropping the bar notation, we have
σmn = λgmn g ij eij + µ (emn + enm ) .
The contravariant form of this equation is
σ sr = λg sr g ij eij + µ (g ms g nr + g ns g mr ) emn .
Employing the equations (2.4.24) the above result can also be expressed in the form
σ
rs
E
=
2(1 + ν)
g ms g nr + g ns g mr +
2ν
sr mn
g g
emn .
1 − 2ν
(2.4.26)
This is a more general form for the stress-strain constitutive equations which is valid in all coordinate systems.
Multiplying by gsk and employing the use of associative tensors, one can verify
σji
or
E
=
1+ν
eij +
ν
em δ i
1 − 2ν m j
i
σji = 2µeij + λem
m δj ,
are alternate forms for the equation (2.4.26). As an exercise, solve for the strains in terms of the stresses
and show that
m i
δj .
Eeij = (1 + ν)σji − νσm
EXAMPLE 2.4-2.
(Hooke’s law)
Let us construct a simple example to test the results we have
developed so far. Consider the tension in a cylindrical bar illustrated in the figure 2.4-3.
Figure 2.4-3. Stress in a cylindrical bar
254
F
Assume that
A
σij =  0
0
0
0
0

0
0
0
where F is the constant applied force and A is the cross sectional area of the cylinder. Consequently, the
generalized Hooke’s law (2.4.21) produces the nonzero strains
1+ν
ν
σ11
σ11 − (σ11 + σ22 + σ33 ) =
E
E
E
−ν
σ11
=
E
−ν
σ11
=
E
e11 =
e22
e33
From these equations we obtain:
The first part of Hooke’s law
σ11 = Ee11 or
F
= Ee11 .
A
The second part of Hooke’s law
−e22
−e33
lateral contraction
=
=
= ν = Poisson’s ratio.
longitudinal extension
e11
e11
This example demonstrates that the generalized Hooke’s law for homogeneous and isotropic materials
reduces to our previous one dimensional result given in (2.3.1) and (2.3.2).
Basic Equations of Elasticity
Assuming the density % is constant, the basic equations of elasticity reduce to the equations representing
conservation of linear momentum and angular momentum together with the strain-displacement relations
and constitutive equations. In these equations the body forces are assumed known. These basic equations
produce 15 equations in 15 unknowns and are a formidable set of equations to solve. Methods for solving
these simultaneous equations are: 1) Express the linear momentum equations in terms of the displacements
ui and obtain a system of partial differential equations. Solve the system of partial differential equations
for the displacements ui and then calculate the corresponding strains. The strains can be used to calculate
the stresses from the constitutive equations. 2) Solve for the stresses and from the stresses calculate the
strains and from the strains calculate the displacements. This converse problem requires some additional
considerations which will be addressed shortly.
255
Basic Equations of Linear Elasticity
• Conservation of linear momentum.
σ ij,i + %bj = %f j
j = 1, 2, 3.
(2.4.27(a))
where σ ij is the stress tensor, bj is the body force per unit mass and f j is
the acceleration. If there is no motion, then f j = 0 and these equations
reduce to the equilibrium equations
σ ij,i + %bj = 0
• Conservation of angular momentum.
• Strain tensor.
eij =
j = 1, 2, 3.
(2.4.27(b))
σij = σji
1
(ui,j + uj,i )
2
(2.4.28)
where ui denotes the displacement field.
• Constitutive equation. For a linear elastic isotropic material we have
σji =
E i
E
e +
ek δ i
1 + ν j (1 + ν)(1 − 2ν) k j
i, j = 1, 2, 3
(2.4.29(a))
or its equivalent form
σji = 2µeij + λerr δji
i, j = 1, 2, 3,
(2.4.29(b))
where err is the dilatation. This produces 15 equations for the 15 unknowns
u1 , u2 , u3 , σ11 , σ12 , σ13 , σ22 , σ23 , σ33 , e11 , e12 , e13 , e22 , e23 , e33 ,
which represents 3 displacements, 6 strains and 6 stresses. In the above
equations it is assumed that the body forces are known.
Navier’s Equations
The equations (2.4.27) through (2.4.29) can be combined and written as one set of equations. The
resulting equations are known as Navier’s equations for the displacements ui over the range i = 1, 2, 3. To
derive the Navier’s equations in Cartesian coordinates, we write the equations (2.4.27),(2.4.28) and (2.4.29)
in Cartesian coordinates. We then calculate σij,j in terms of the displacements ui and substitute the results
into the momentum equation (2.4.27(a)). Differentiation of the constitutive equations (2.4.29(b)) produces
σij,j = 2µeij,j + λekk,j δij .
(2.4.30)
256
A contraction of the strain produces the dilatation
err =
1
(ur,r + ur,r ) = ur,r
2
(2.4.31)
From the dilatation we calculate the covariant derivative
ekk,j = uk,kj .
(2.4.32)
Employing the strain relation from equation (2.4.28), we calculate the covariant derivative
eij,j =
1
(ui,jj + uj,ij ).
2
(2.4.33)
These results allow us to express the covariant derivative of the stress in terms of the displacement field. We
find
σij,j = µ [ui,jj + uj,ij ] + λδij uk,kj
or
(2.4.34)
σij,j = (λ + µ)uk,ki + µui,jj .
Substituting equation (2.4.34) into the linear momentum equation produces the Navier equations:
(λ + µ)uk,ki + µui,jj + %bi = %fi ,
i = 1, 2, 3.
(2.4.35)
In vector form these equations can be expressed
(λ + µ)∇ (∇ · ~u) + µ∇2 ~u + %~b = %f~,
(2.4.36)
where ~u is the displacement vector, ~b is the body force per unit mass and f~ is the acceleration. In Cartesian
coordinates these equations have the form:
(λ + µ)
∂ 2 u2
∂ 2 u3
∂ 2 u1
+
+
∂x1 ∂xi
∂x2 ∂xi
∂x3 ∂xi
for i = 1, 2, 3, where
∇2 ui =
+ µ∇2 ui + %bi = %
∂ 2 ui
,
∂t2
∂ 2 ui
∂ 2 ui
∂ 2 ui
+
+
.
∂x1 2
∂x2 2
∂x3 2
The Navier equations must be satisfied by a set of functions ui = ui (x1 , x2 , x3 ) which represent the
displacement at each point inside some prescribed region R. Knowing the displacement field we can calculate
the strain field directly using the equation (2.4.28). Knowledge of the strain field enables us to construct the
corresponding stress field from the constitutive equations.
In the absence of body forces, such as gravity, the solution to equation (2.4.36) can be represented
in the form ~u = ~u (1) + ~u (2) , where ~u (1) satisfies div ~u (1) = ∇ · ~u (1) = 0 and the vector ~u (2) satisfies
curl ~u (2) = ∇ × ~u (2) = 0. The vector field ~u (1) is called a solenoidal field, while the vector field ~u (2) is
called an irrotational field. Substituting ~u into the equation (2.4.36) and setting ~b = 0, we find in Cartesian
coordinates that
%
∂ 2 ~u (2)
∂ 2 ~u (1)
+
2
∂t
∂t2
= (λ + µ)∇ ∇ · ~u (2) + µ∇2 ~u (1) + µ∇2 ~u (2) .
(2.4.37)
257
The vector field ~u (1) can be eliminated from equation (2.4.37) by taking the divergence of both sides of the
equation. This produces
%
∂ 2 ∇ · ~u (2)
= (λ + µ)∇2 (∇ · ~u (2) ) + µ∇ · ∇2 ~u (2) .
∂t2
The displacement field is assumed to be continuous and so we can interchange the order of the operators ∇2
and ∇ and write
∂ 2~u (2)
− (λ + 2µ)∇2 ~u (2)
∇· %
∂t2
This last equation implies that
%
∂ 2~u (2)
= (λ + 2µ)∇2 ~u(2)
∂t2
and consequently, ~u (2) is a vector wave which moves with the speed
= 0.
p
(λ + 2µ)/%. Similarly, when the vector
field ~u (2) is eliminated from the equation (2.4.37), by taking the curl of both sides, we find the vector ~u (1)
also satisfies a wave equation having the form
%
This later wave moves with the speed
p
∂ 2~u (1)
= µ∇2 ~u (1) .
∂t2
µ/%. The vector ~u (2) is a compressive wave, while the wave u (1) is
a shearing wave.
The exercises 30 through 38 enable us to write the Navier’s equations in Cartesian, cylindrical or
spherical coordinates. In particular, we have for cartesian coordinates
∂2w
∂2u ∂2u ∂2u
∂2v
∂2u
∂2u
+
)
+
µ(
+
+
+
)
+
%b
=%
x
∂x2
∂x∂y ∂x∂z
∂x2
∂y 2
∂z 2
∂t2
2
2
2
2
2
2
∂ v
∂ v
∂ u
∂ w
∂ v ∂ v
∂2v
+ 2+
) + µ( 2 + 2 + 2 ) + %by =% 2
(λ + µ)(
∂x∂y ∂y
∂y∂z
∂x
∂y
∂z
∂t
∂2v
∂2w
∂2u
∂2w ∂2w ∂2w
∂2w
+
+
(λ + µ)(
) + µ( 2 +
+
) + %bz =% 2
∂x∂z ∂y∂z
∂z 2
∂x
∂y 2
∂z 2
∂t
(λ + µ)(
and in cylindrical coordinates
∂uz
1 ∂uθ
∂ 1 ∂
(rur ) +
+
+
∂r r ∂r
r ∂θ
∂z
1 ∂ 2 ur
1 ∂ur
∂ 2 ur
ur
2 ∂uθ
∂ 2 ur
∂ 2 ur
+ 2
) + %br =% 2
+
− 2 − 2
µ( 2 +
2
2
∂r
r ∂r
r ∂θ
r
r ∂θ
∂t
∂z
∂uz
1 ∂ 1 ∂
1 ∂uθ
(rur ) +
+
(λ + µ)
+
r ∂θ r ∂r
r ∂θ
∂z
1 ∂ 2 uθ
uθ
1 ∂uθ
∂ 2 uθ
2 ∂ur
∂ 2 uθ
∂ 2 uθ
+ 2
−
+
+
)
+
%b
=%
µ( 2 +
θ
∂r
r ∂r
r ∂θ2 ∂z 2
r2 ∂θ
r2
∂t2
∂uz
∂ 1 ∂
1 ∂uθ
(rur ) +
+
(λ + µ)
+
∂z r ∂r
r ∂θ
∂z
1 ∂ 2 uz
1 ∂uz
∂ 2 uz
∂ 2 uz
∂ 2 uz
+ 2
+
)
+
%b
=%
µ( 2 +
z
∂r
r ∂r
r ∂θ2
∂z 2
∂t2
(λ + µ)
258
and in spherical coordinates
1 ∂ 2
∂
1
1 ∂uφ
(ρ
(u
u
)
+
sin
θ)
+
+
ρ
θ
ρ2 ∂ρ
ρ sin θ ∂θ
ρ sin θ ∂φ
2uθ cot θ
2
2 ∂uθ
2 ∂uφ
∂ 2 uρ
−
) + %bρ =% 2
− 2
µ(∇2 uρ − 2 uρ − 2
2
ρ
ρ ∂θ
ρ
ρ sin θ ∂φ
∂t
1 ∂ 2
∂
1 ∂
1
1 ∂uφ
(ρ uρ ) +
(uθ sin θ) +
(λ + µ)
+
ρ ∂θ ρ2 ∂ρ
ρ sin θ ∂θ
ρ sin θ ∂φ
uθ
2 ∂uρ
∂ 2 uθ
2 cos θ ∂uφ
− 2 2 − 2
) + %bθ =% 2
µ(∇2 uθ + 2
2
ρ ∂θ
∂t
ρ sin θ ρ sin θ ∂φ
∂
1 ∂ 2
∂
1
1
1 ∂uφ
(ρ uρ ) +
(uθ sin θ) +
(λ + µ)
+
ρ sin θ ∂φ ρ2 ∂ρ
ρ sin θ ∂θ
ρ sin θ ∂φ
2 cos θ ∂uθ
1
2 ∂uρ
∂ 2 uφ
+ 2 2
) + %bφ =% 2
µ(∇2 uφ − 2 2 uφ + 2
ρ sin θ ∂φ
∂t
ρ sin θ
ρ sin θ ∂φ
(λ + µ)
∂
∂ρ
where ∇2 is determined from either equation (2.1.12) or (2.1.13).
Boundary Conditions
In elasticity the body forces per unit mass (bi , i = 1, 2, 3) are assumed known. In addition one of the
following type of boundary conditions is usually prescribed:
• The displacements ui ,
i = 1, 2, 3 are prescribed on the boundary of the region R over which a solution
is desired.
• The stresses (surface tractions) are prescribed on the boundary of the region R over which a solution is
desired.
• The displacements ui , i = 1, 2, 3 are given over one portion of the boundary and stresses (surface
tractions) are specified over the remaining portion of the boundary. This type of boundary condition is
known as a mixed boundary condition.
General Solution of Navier’s Equations
There has been derived a general solution to the Navier’s equations. It is known as the Papkovich-Neuber
solution. In the case of a solid in equilibrium one must solve the equilibrium equations
(λ + µ)∇ (∇ · ~u) + µ∇2 ~u + %~b = 0 or
%
1
1
∇(∇ · ~u) + ~b = 0 (ν =
6
)
∇2 ~u +
1 − 2ν
µ
2
(2.4.38)
259
THEOREM
~
A general elastostatic solution of the equation (2.4.38) in terms of harmonic potentials φ,ψ
is
~ − 4(1 − ν)ψ
~
~u = grad (φ + ~r · ψ)
(2.4.39)
~ are continuous solutions of the equations
where φ and ψ
∇2 φ =
−%~r · ~b
4µ(1 − ν)
and
~=
∇2 ψ
%~b
4µ(1 − ν)
(2.4.40)
with ~r = x ê1 + y ê2 + z ê3 a position vector to a general point (x, y, z) within the continuum.
Proof: First we write equation (2.4.38) in the tensor form
ui,kk +
1
%
(uj,j ) ,i + bi = 0
1 − 2ν
µ
(2.4.41)
Now our problem is to show that equation (2.4.39), in tensor form,
ui = φ,i + (xj ψj ),i − 4(1 − ν)ψi
(2.4.42)
is a solution of equation (2.4.41). Toward this purpose, we differentiate equation (2.4.42)
ui,k = φ,ik + (xj ψj ),ik − 4(1 − ν)ψi,k
(2.4.43)
and then contract on i and k giving
ui,i = φ,ii + (xj ψj ),ii − 4(1 − ν)ψi,i .
(2.4.44)
Employing the identity (xj ψj ),ii = 2ψi,i + xi ψi,kk the equation (2.4.44) becomes
ui,i = φ,ii + 2ψi,i + xi ψi,kk − 4(1 − ν)ψi,i .
(2.4.45)
By differentiating equation (2.4.43) we establish that
ui,kk = φ,ikk + (xj ψj ),ikk − 4(1 − ν)ψi,kk
= (φ,kk ),i + ((xj ψj ),kk ),i − 4(1 − ν)ψi,kk
(2.4.46)
= [φ,kk + 2ψj,j + xj ψj,kk ],i − 4(1 − ν)ψi,kk .
We use the hypothesis
φ,kk =
−%xj Fj
4µ(1 − ν)
and
ψj,kk =
%Fj
,
4µ(1 − ν)
and simplify the equation (2.4.46) to the form
ui,kk = 2ψj,ji − 4(1 − ν)ψi,kk .
(2.4.47)
Also by differentiating (2.4.45) one can establish that
uj,ji = (φ,jj ),i + 2ψj,ji + (xj ψj,kk ),i − 4(1 − ν)ψj,ji
−%xj Fj
%xj Fj
=
+ 2ψj,ji +
− 4(1 − ν)ψj,ji
4µ(1 − ν) ,i
4µ(1 − ν) ,i
= −2(1 − 2ν)ψj,ji .
(2.4.48)
260
Finally, from the equations (2.4.47) and (2.4.48) we obtain the desired result that
ui,kk +
1
%Fi
uj,ji +
= 0.
1 − 2ν
µ
Consequently, the equation (2.4.39) is a solution of equation (2.4.38).
As a special case of the above theorem, note that when the body forces are zero, the equations (2.4.40)
become
∇2 φ = 0
and
~ = ~0.
∇2 ψ
In this case, we find that equation (2.4.39) is a solution of equation (2.4.38) provided φ and each component of
~ are harmonic functions. The Papkovich-Neuber potentials are used together with complex variable theory
ψ
to solve various two-dimensional elastostatic problems of elasticity. Note also that the Papkovich-Neuber
~ can produce the same value for ~u.
potentials are not unique as different combinations of φ and ψ
Compatibility Equations
If we know or can derive the displacement field ui , i = 1, 2, 3 we can then calculate the components of
the strain tensor
eij =
1
(ui,j + uj,i ).
2
(2.4.49)
Knowing the strain components, the stress is found using the constitutive relations.
Consider the converse problem where the strain tensor is given or implied due to the assigned stress
field and we are asked to determine the displacement field ui , i = 1, 2, 3. Is this a realistic request? Is it even
possible to solve for three displacements given six strain components? It turns out that certain mathematical
restrictions must be placed upon the strain components in order that the inverse problem have a solution.
These mathematical restrictions are known as compatibility equations. That is, we cannot arbitrarily assign
six strain components eij and expect to find a displacement field ui , i = 1, 2, 3 with three components which
satisfies the strain relation as given in equation (2.4.49).
EXAMPLE 2.4-3. Suppose we are given the two partial differential equations,
∂u
=x+y
∂x
and
∂u
= x3 .
∂y
Can we solve for u = u(x, y)? The answer to this question is “no”, because the given equations are inconsistent. The inconsistency is illustrated if we calculate the mixed second derivatives from each equation. We
∂2u
∂2u
= 1 and from the second equation we calculate
= 3x2 . These
find from the first equation that
∂x∂y
∂y∂x
√
mixed second partial derivatives are unequal for all x different from 3/3. In general, if we have two first
∂u
∂u
= f (x, y) and
= g(x, y), then for consistency (integrability of
order partial differential equations
∂x
∂y
the equations) we require that the mixed partial derivatives
∂f
∂2u
∂g
∂2u
=
=
=
∂x∂y
∂y
∂y∂x
∂x
be equal to one another for all x and y values over the domain for which the solution is desired. This is an
example of a compatibility equation.
261
A similar situation occurs in two dimensions for a material in a state of strain where ezz = ezx = ezy = 0,
called plane strain. In this case, are we allowed to arbitrarily assign values to the strains exx , eyy and exy and
from these strains determine the displacement field u = u(x, y) and v = v(x, y) in the x− and y−directions?
Let us try to answer this question. Assume a state of plane strain where ezz = ezx = ezy = 0. Further, let
us assign 3 arbitrary functional values f, g, h such that
exx =
∂u
= f (x, y),
∂x
exy =
1
2
∂u ∂v
+
∂y
∂x
= g(x, y),
eyy =
∂v
= h(x, y).
∂y
We must now decide whether these equations are consistent. That is, will we be able to solve for the
displacement field u = u(x, y) and v = v(x, y)? To answer this question, let us derive a compatibility equation
(integrability condition). From the given equations we can calculate the following partial derivatives
∂3u
∂2f
∂ 2 exx
=
=
2
2
∂y
∂x∂y
∂y 2
2
3
∂ v
∂2h
∂ eyy
=
=
∂x2
∂y∂x2
∂x2
2
3
∂ u
∂ exy
∂3v
∂2g
=
.
2
+
=
2
∂x∂y
∂x∂y 2
∂y∂x2
∂x∂y
This last equation gives us the compatibility equation
2
∂ 2 exx
∂ 2 eyy
∂ 2 exy
=
+
∂x∂y
∂y 2
∂x2
or the functions g, f, h must satisfy the relation
2
∂2f
∂2h
∂2g
=
+
.
∂x∂y
∂y 2
∂x2
Cartesian Derivation of Compatibility Equations
If the displacement field ui , i = 1, 2, 3 is known we can derive the strain and rotation tensors
eij =
1
(ui,j + uj,i )
2
and
ωij =
1
(ui,j − uj,i ).
2
(2.4.50)
Now work backwards. Assume the strain and rotation tensors are given and ask the question, “Is it possible
to solve for the displacement field ui , i = 1, 2, 3?” If we view the equation (2.4.50) as a system of equations
with unknowns eij , ωij and ui and if by some means we can eliminate the unknowns ωij and ui then we
will be left with equations which must be satisfied by the strains eij . These equations are known as the
compatibility equations and they represent conditions which the strain components must satisfy in order
that a displacement function exist and the equations (2.4.37) are satisfied. Let us see if we can operate upon
the equations (2.4.50) to eliminate the quantities ui and ωij and hence derive the compatibility equations.
Addition of the equations (2.4.50) produces
ui,j =
∂ui
= eij + ωij .
∂xj
(2.4.51)
262
Differentiate this expression with respect to xk and verify the result
∂eij
∂ωij
∂ 2 ui
=
+
.
∂xj ∂xk
∂xk
∂xk
(2.4.52)
We further assume that the displacement field is continuous so that the mixed partial derivatives are equal
and
∂ 2 ui
∂ 2 ui
=
.
∂xj ∂xk
∂xk ∂xj
(2.4.53)
Interchanging j and k in equation (2.4.52) gives us
∂eik
∂ωik
∂ 2 ui
=
+
.
∂xk ∂xj
∂xj
∂xj
(2.4.54)
Equating the second derivatives from equations (2.4.54) and (2.4.52) and rearranging terms produces the
result
Making the observation that ωij
∂eik
∂ωik
∂ωij
∂eij
−
=
−
(2.4.55)
∂xk
∂xj
∂xj
∂xk
∂ωik ∂ωij
∂ωjk
satisfies
−
=
, the equation (2.4.55) simplifies to the
∂xj
∂xk
∂xi
form
The term involving ωjk
∂eik
∂ωjk
∂eij
−
=
.
∂xk
∂xj
∂xi
can be eliminated by using the mixed partial derivative relation
(2.4.56)
∂ 2 ωjk
∂ 2 ωjk
=
.
(2.4.57)
∂xi ∂xm
∂xm ∂xi
To derive the compatibility equations we differentiate equation (2.4.56) with respect to xm and then
interchanging the indices i and m and substitute the results into equation (2.4.57). This will produce the
compatibility equations
∂ 2 emk
∂ 2 eik
∂ 2 emj
∂ 2 eij
+
−
−
= 0.
(2.4.58)
∂xm ∂xk
∂xi ∂xj
∂xm ∂xj
∂xi ∂xk
This is a set of 81 partial differential equations which must be satisfied by the strain components. Fortunately,
due to symmetry considerations only 6 of these 81 equations are distinct. These 6 distinct equations are
known as the St. Venant’s compatibility equations and can be written as
∂ 2 e12
∂ 2 e23
∂ 2 e31
∂ 2 e11
=
−
+
2
∂x2 ∂x3
∂x1 ∂x3
∂x1
∂x1 ∂x2
2
2
2
∂ e23
∂ e31
∂ 2 e12
∂ e22
=
−
+
2
∂x1 ∂x3
∂x2 ∂x1
∂x2
∂x2 ∂x3
∂ 2 e31
∂ 2 e12
∂ 2 e23
∂ 2 e33
=
−
+
∂x1 ∂x2
∂x3 ∂x2
∂x3 2
∂x3 ∂x1
2
2
2
∂ e12
∂ e11
∂ e22
2
=
+
∂x1 ∂x2
∂x2 2
∂x1 2
2
2
∂ e23
∂ e22
∂ 2 e33
2
=
+
∂x2 ∂x3
∂x3 2
∂x2 2
2
2
∂ e31
∂ e33
∂ 2 e11
2
=
+
.
2
∂x3 ∂x1
∂x1
∂x3 2
Observe that the fourth compatibility equation is the same as that derived in the example 2.4-3.
(2.4.59)
These compatibility equations can also be expressed in the indicial form
eij,km + emk,ji − eik,jm − emj,ki = 0.
(2.4.60)
263
Compatibility Equations in Terms of Stress
In the generalized Hooke’s law, equation (2.4.29), we can solve for the strain in terms of stress. This
in turn will give rise to a representation of the compatibility equations in terms of stress. The resulting
equations are known as the Beltrami-Michell equations. Utilizing the strain-stress relation
eij =
1+ν
ν
σij − σkk δij
E
E
we substitute for the strain in the equations (2.4.60) and rearrange terms to produce the result
σij,km + σmk,ji − σik,jm − σmj,ki =
ν
[δij σnn,km + δmk σnn,ji − δik σnn,jm − δmj σnn,ki ] .
1+ν
(2.4.61)
Now only 6 of these 81 equations are linearly independent. It can be shown that the 6 linearly independent
equations are equivalent to the equations obtained by setting k = m and summing over the repeated indices.
We then obtain the equations
σij,mm + σmm,ij − (σim,m ),j − (σmj,m ),i =
ν
[δij σnn,mm + σnn,ij ] .
1+ν
Employing the equilibrium equation σij,i + %bj = 0 the above result can be written in the form
σij,mm +
1
ν
σkk,ij −
δij σnn,mm = −(%bi ),j − (%bj ),i
1+ν
1+ν
∇2 σij +
1
ν
σkk,ij −
δij σnn,mm = −(%bi ),j − (%bj ),i .
1+ν
1+ν
or
This result can be further simplified by observing that a contraction on the indices k and i in equation
(2.4.61) followed by a contraction on the indices m and j produces the result
σij,ij =
1−ν
σnn,jj .
1+ν
Consequently, the Beltrami-Michell equations can be written in the form
∇2 σij +
1
ν
σpp,ij = −
δij (%bk ) ,k − (%bi ) ,j − (%bj ) ,i .
1+ν
1−ν
(2.4.62)
Their derivation is left as an exercise. The Beltrami-Michell equations together with the linear momentum
(equilibrium) equations σij,i + %bj = 0 represent 9 equations in six unknown stresses. This combinations
of equations is difficult to handle. An easier combination of equations in terms of stress functions will be
developed shortly.
The Navier equations with boundary conditions are difficult to solve in general. Let us take the momentum equations (2.4.27(a)), the strain relations (2.4.28) and constitutive equations (Hooke’s law) (2.4.29)
and make simplifying assumptions so that a more tractable systems results.
264
Plane Strain
The plane strain assumption usually is applied in situations where there is a cylindrical shaped body
whose axis is parallel to the z axis and loads are applied along the z−direction. In any x-y plane we assume
that the surface tractions and body forces are independent of z. We set all strains with a subscript z equal
to zero. Further, all solutions for the stresses, strains and displacements are assumed to be only functions
of x and y and independent of z. Note that in plane strain the stress σzz is different from zero.
In Cartesian coordinates the strain tensor is expressible in terms of its physical components which can
be represented in the matrix form

e11
 e21
e31
e12
e22
e32
 
e13
exx
e23  =  eyx
e33
ezx
exy
eyy
ezy

exz
eyz  .
ezz
If we assume that all strains which contain a subscript z are zero and the remaining strain components are
functions of only x and y, we obtain a state of plane strain. For a state of plane strain, the stress components
are obtained from the constitutive equations. The condition of plane strain reduces the constitutive equations
to the form:
exx =
eyy =
0=
exy = eyx =
ezy = eyz =
ezx = exz =
where σxx ,
σyy ,
1
[σxx − ν(σyy + σzz )]
E
1
[σyy − ν(σzz + σxx )]
E
1
[σzz − ν(σxx + σyy )]
E
1+ν
σxy
E
1+ν
σyz = 0
E
1+ν
σxz = 0
E
σzz ,
σxy ,
σxz ,
E
[(1 − ν)exx + νeyy ]
(1 + ν)(1 − 2ν)
E
[(1 − ν)eyy + νexx ]
=
(1 + ν)(1 − 2ν)
E
[ν(eyy + exx )]
=
(1 + ν)(1 − 2ν)
E
exy
=
1+ν
=0
σxx =
σyy
σzz
σxy
σxz
(2.4.63)
σyz = 0
σyz are the physical components of the stress. The above constitutive
equations imply that for a state of plane strain we will have
σzz = ν(σxx + σyy )
1+ν
[(1 − ν)σxx − νσyy ]
exx =
E
1+ν
[(1 − ν)σyy − νσxx ]
eyy =
E
1+ν
σxy .
exy =
E
Also under these conditions the compatibility equations reduce to
∂ 2 eyy
∂ 2 exy
∂ 2 exx
.
+
=2
2
2
∂y
∂x
∂x∂y
265
Plane Stress
An assumption of plane stress is usually applied to thin flat plates. The plate thinness is assumed to be
in the z−direction and loads are applied perpendicular to z. Under these conditions all stress components
with a subscript z are assumed to be zero. The remaining stress components are then treated as functions
of x and y.
In Cartesian coordinates the stress tensor is expressible in terms of its physical components and can be
represented by the matrix

σ11
 σ21
σ31
σ12
σ22
σ32
 
σ13
σxx
σ23  =  σyx
σ33
σzx
σxy
σyy
σzy

σxz
σyz  .
σzz
If we assume that all the stresses with a subscript z are zero and the remaining stresses are only functions of
x and y we obtain a state of plane stress. The constitutive equations simplify if we assume a state of plane
stress. These simplified equations are
1
ν
σxx − σyy
E
E
1
ν
= σyy − σxx
E
E
ν
= − (σxx + σyy )
E
1+ν
σxy
=
E
=0
E
[exx + νeyy ]
1 − ν2
E
=
[eyy + νexx ]
1 − ν2
= 0 = (1 − ν)ezz + ν(exx + eyy )
E
exy
=
1+ν
=0
exx =
σxx =
eyy
σyy
ezz
exy
exz
σzz
σxy
σyz
(2.4.64)
σxz = 0
eyz = 0.
For a state of plane stress the compatibility equations reduce to
∂ 2 exy
∂ 2 exx ∂ 2 eyy
+
=2
2
2
∂y
∂x
∂x∂y
(2.4.65)
and the three additional equations
∂ 2 ezz
= 0,
∂x2
∂ 2 ezz
= 0,
∂y 2
∂ 2 ezz
= 0.
∂x∂y
These three additional equations complicate the plane stress problem.
Airy Stress Function
In Cartesian coordinates we examine the equilibrium equations (2.4.25(b)) under the conditions of plane
strain. In terms of physical components we find that these equations reduce to
∂σxy
∂σxx
+
+ %bx = 0,
∂x
∂y
∂σyy
∂σyx
+
+ %by = 0,
∂x
∂y
∂σzz
= 0.
∂z
The last equation is satisfied since σzz is a function of x and y. If we further assume that the body forces
are conservative and derivable from a potential function V by the operation %~b = −grad V or %bi = −V ,i
we can express the above equilibrium equations in the form:
∂σxy
∂V
∂σxx
+
−
=0
∂x
∂y
∂x
∂σyy
∂V
∂σyx
+
−
=0
∂x
∂y
∂y
(2.4.66)
266
We will consider these equations together with the compatibility equations (2.4.65). The equations
(2.4.66) will be automatically satisfied if we introduce a scalar function φ = φ(x, y) and assume that the
stresses are derivable from this function and the potential function V according to the rules:
σxx =
∂2φ
+V
∂y 2
σxy = −
∂2φ
∂x∂y
σyy =
∂2φ
+ V.
∂x2
(2.4.67)
The function φ = φ(x, y) is called the Airy stress function after the English astronomer and mathematician
Sir George Airy (1801–1892). Since the equations (2.4.67) satisfy the equilibrium equations we need only
consider the compatibility equation(s).
For a state of plane strain we substitute the relations (2.4.63) into the compatibility equation (2.4.65)
and write the compatibility equation in terms of stresses. We then substitute the relations (2.4.67) and
express the compatibility equation in terms of the Airy stress function φ. These substitutions are left as
exercises. After all these substitutions the compatibility equation, for a state of plane strain, reduces to the
form
∂4φ
∂ 4 φ 1 − 2ν
∂4φ
+2 2 2 + 4 +
4
∂x
∂x ∂y
∂y
1−ν
∂2V
∂2V
+
2
∂x
∂y 2
= 0.
(2.4.68)
In the special case where there are no body forces we have V = 0 and equation (2.4.68) is further simplified
to the biharmonic equation.
∂4φ
∂4φ
∂4φ
+
2
+
= 0.
∂x4
∂x2 ∂y 2
∂y 4
In polar coordinates the biharmonic equation is written
2
2
∂
1 ∂2
1 ∂2φ
∂ φ 1 ∂φ
1 ∂
+
+
+
+
= 0.
∇4 φ = ∇2 (∇2 φ) =
∂r2
r ∂r r2 ∂θ2
∂r2
r ∂r
r2 ∂θ2
∇4 φ =
(2.4.69)
For conditions of plane stress, we can again introduce an Airy stress function using the equations (2.4.67).
However, an exact solution of the plane stress problem which satisfies all the compatibility equations is
difficult to obtain. By removing the assumptions that σxx , σyy , σxy are independent of z, and neglecting
body forces, it can be shown that for symmetrically distributed external loads the stress function φ can be
represented in the form
φ=ψ−
νz 2
∇2 ψ
2(1 + ν)
(2.4.70)
where ψ is a solution of the biharmonic equation ∇4 ψ = 0. Observe that if z is very small, (the condition
of a thin plate), then equation (2.4.70) gives the approximation φ ≈ ψ. Under these conditions, we obtain
the approximate solution by using only the compatibility equation (2.4.65) together with the stress function
defined by equations (2.4.67) with V = 0. Note that the solution we obtain from equation (2.4.69) does not
satisfy all the compatibility equations, however, it does give an excellent first approximation to the solution
in the case where the plate is very thin.
In general, for plane strain or plane stress problems, the equation (2.4.68) or (2.4.69) must be solved for
the Airy stress function φ which is defined over some region R. In addition to specifying a region of the x, y
plane, there are certain boundary conditions which must be satisfied. The boundary conditions specified for
the stress will translate through the equations (2.4.67) to boundary conditions being specified for φ. In the
special case where there are no body forces, both the problems for plane stress and plane strain are governed
by the biharmonic differential equation with appropriate boundary conditions.
267
EXAMPLE 2.4-4 Assume there exist a state of plane strain with zero body forces. For F11 , F12 , F22
constants, consider the function defined by
φ = φ(x, y) =
1
F22 x2 − 2F12 xy + F11 y 2 .
2
This function is an Airy stress function because it satisfies the biharmonic equation ∇4 φ = 0. The resulting
stress field is
σxx =
∂2φ
= F11
∂y 2
σyy =
∂ 2φ
= F22
∂x2
σxy = −
∂2φ
= F12 .
∂x∂y
This example, corresponds to stresses on an infinite flat plate and illustrates a situation where all the stress
components are constants for all values of x and y. In this case, we have σzz = ν(F11 +F22 ). The corresponding
strain field is obtained from the constitutive equations. We find these strains are
exx =
1+ν
[(1 − ν)F11 − νF22 ]
E
eyy =
1+ν
[(1 − ν)F22 − νF11 ]
E
exy =
1+ν
F12 .
E
The displacement field is found to be
1+ν
1+ν
[(1 − ν)F11 − νF22 ] x +
F12 y + c1 y + c2
u = u(x, y) =
E
E
1+ν
1+ν
[(1 − ν)F22 − νF11 ] y +
v = v(x, y) =
F12 x − c1 x + c3 ,
E
E
with c1 , c2 , c3 constants, and is obtained by integrating the strain displacement equations given in Exercise
2.3, problem 2.
EXAMPLE 2.4-5. A special case from the previous example is obtained by setting F22 = F12 = 0.
This is the situation of an infinite plate with only tension in the x−direction. In this special case we have
φ = 12 F11 y 2 . Changing to polar coordinates we write
φ = φ(r, θ) =
F11 2
F11 2 2
r sin θ =
r (1 − cos 2θ).
2
4
The Exercise 2.4, problem 20, suggests we utilize the Airy equations in polar coordinates and calculate the
stresses
1 ∂2φ
1 ∂φ
F11
+ 2 2 = F11 cos2 θ =
(1 + cos 2θ)
r ∂r
r ∂θ
2
2
∂ φ
F11
(1 − cos 2θ)
=
= F11 sin2 θ =
2
∂r
2
F11
1 ∂φ 1 ∂ 2 φ
−
=−
sin 2θ.
= 2
r ∂θ
r ∂r∂θ
2
σrr =
σθθ
σrθ
268
EXAMPLE 2.4-6. We now consider an infinite plate with a circular hole x2 + y 2 = a2 which is traction
free. Assume the plate has boundary conditions at infinity defined by σxx = F11 ,
σyy = 0,
σxy = 0. Find
the stress field.
Solution:
The traction boundary condition at r = a is ti = σmi nm or
t1 = σ11 n1 + σ12 n2
and
t2 = σ12 n1 + σ22 n2 .
For polar coordinates we have n1 = nr = 1, n2 = nθ = 0 and so the traction free boundary conditions at
the surface of the hole are written σrr |r=a = 0 and σrθ |r=a = 0. The results from the previous example
are used as the boundary conditions at infinity.
Our problem is now to solve for the Airy stress function φ = φ(r, θ) which is a solution of the biharmonic
equation. The previous example 2.4-5 and the form of the boundary conditions at infinity suggests that we
assume a solution to the biharmonic equation which has the form φ = φ(r, θ) = f1 (r) + f2 (r) cos 2θ, where
f1 , f2 are unknown functions to be determined. Substituting the assumed solution into the biharmonic
equation produces the equation
1 d
d2
+
dr2
r dr
2
4
d
1
1 d
1 0
f2
00
−
f
+
+
−
4
f100 + f10 +
f
cos 2θ = 0.
2
r
dr2
r dr r2
r 2
r2
We therefore require that f1 , f2 be chosen to satisfy the equations
or
1 d
d2
+
2
dr
r dr
(iv)
r 4 f1
1 0
00
f1 + f1 = 0
r
+ 2r3 f1000 − r2 f100 + rf10 = 0
d2
4
1 d
− 2
+
2
dr
r dr r
(iv)
r 4 f2
1 0
f2
00
f2 + f2 − 4 2 = 0
r
r
+ 2r3 f2000 − 9r2 f200 + 9rf20 = 0
These equations are Cauchy type equations. Their solutions are obtained by assuming a solution of the form
f1 = rλ and f2 = rm and then solving for the constants λ and m. We find the general solutions of the above
equations are
f1 = c1 r2 ln r + c2 r2 + c3 ln r + c4
and f2 = c5 r2 + c6 r4 +
c7
+ c8 .
r2
The constants ci , i = 1, . . . , 8 are now determined from the boundary conditions. The constant c4 can be
arbitrary since the derivative of a constant is zero. The remaining constants are determined from the stress
conditions. Using the results from Exercise 2.4, problem 20, we calculate the stresses
c3 c7
c8 −
2c
+
6
+
4
cos 2θ
5
r2 r4
r2 c3
c7
= c1 (3 + 2 ln r) + 2c2 − 2 + 2c5 + 12c6 r2 + 6 4 cos 2θ
r
r
c7
c8 = 2c5 + 6c6 r2 − 6 4 − 2 2 sin 2θ.
r
r
σrr = c1 (1 + 2 ln r) + 2c2 +
σθθ
σrθ
269
The stresses are to remain bounded for all values of r and consequently we require c1 and c6 to be zero
to avoid infinite stresses for large values of r. The stress σrr |r=a = 0 requires that
2c2 +
c3
=0
a2
and 2c5 + 6
The stress σrθ |r=a = 0 requires that
2c5 − 6
c7
c8
+ 4 2 = 0.
4
a
a
c7
c8
− 2 2 = 0.
4
a
a
In the limit as r → ∞ we require that the stresses must satisfy the boundary conditions from the previous
F11
F11
and 2c5 = −
. Solving the above system of equations
example 2.4-5. This leads to the equations 2c2 =
2
2
produces the Airy stress function
φ = φ(r, θ) =
F11 2 a2
F11
+
r − F11 ln r + c4 +
4
4
2
F11 a2
F11 2 F11 a4
−
r −
2
4
4r2
cos 2θ
and the corresponding stress field is
σrr
σrθ
σθθ
F11
a2
a2
a4
F11
=
1− 2 +
1 + 3 4 − 4 2 cos 2θ
2
r
2
r
r
F11
a2
a4
=−
1 − 3 4 + 2 2 sin 2θ
2
r
r
F11
a2
a4
F11
=
1+ 2 −
1 + 3 4 cos 2θ.
2
r
2
r
There is a maximum stress σθθ = 3F11 at θ = π/2, 3π/2 and a minimum stress σθθ = −F11 at θ = 0, π.
The effect of the circular hole has been to magnify the applied stress. The factor of 3 is known as a stress
concentration factor. In general, sharp corners and unusually shaped boundaries produce much higher stress
concentration factors than rounded boundaries.
EXAMPLE 2.4-7. Consider an infinite cylindrical tube, with inner radius R1 and the outer radius R0 ,
which is subjected to an internal pressure P1 and an external pressure P0 as illustrated in the figure 2.4-7.
Find the stress and displacement fields.
Solution: Let ur , uθ , uz denote the displacement field. We assume that uθ = 0 and uz = 0 since the
cylindrical surface r equal to a constant does not move in the θ or z directions. The displacement ur = ur (r)
is assumed to depend only upon the radial distance r. Under these conditions the Navier equations become
d
(λ + 2µ)
dr
This equation has the solution ur = c1
err =
dur
,
dr
1 d
(rur ) = 0.
r dr
c2
r
+
and the strain components are found from the relations
2
r
eθθ =
ur
,
r
ezz = erθ = erz = ezθ = 0.
The stresses are determined from Hooke’s law (the constitutive equations) and we write
σij = λδij Θ + 2µeij ,
270
where
Θ=
ur
1 ∂
∂ur
+
=
(rur )
∂r
r
r ∂r
is the dilatation. These stresses are found to be
σrr = (λ + µ)c1 −
2µ
c2
r2
σθθ = (λ + µ)c1 +
We now apply the boundary conditions
2µ
σrr |r=R1 nr = − (λ + µ)c1 − 2 c2 = +P1
R1
2µ
c2
r2
σzz = λc1
σrθ = σrz = σzθ = 0.
2µ
and σrr |r=R0 nr = (λ + µ)c1 − 2 c2 = −P0 .
R0
Solving for the constants c1 and c2 we find
c1 =
R12 P1 − R02 P0
,
(λ + µ)(R02 − R12 )
c2 =
R12 R02 (P1 − P0 )
.
2µ(R02 − R12 )
This produces the displacement field
r
R02
r
R12
R12 P1
R02 P0
+
+
−
,
ur =
2(R02 − R12 ) λ + µ
µr
2(R02 − R12 ) λ + µ
µr
and stress fields
uθ = 0,
uz = 0,
R12 P1
R02
R02 P0
R12
1
−
−
1
−
R02 − R12
r2
R02 − R12
r2
R2 P1
R2
R2 P0
R2
= 2 1 2 1 + 20 − 2 0 2 1 + 21
R − R1
r
R0 − R1
r
0
2
2
λ
R1 P1 − R0 P0
=
λ+µ
R02 − R12
σrr =
σθθ
σzz
σrz = σzθ = σrθ = 0
EXAMPLE 2.4-8. By making simplifying assumptions the Navier equations can be reduced to a more
tractable form. For example, we can reduce the Navier equations to a one dimensional problem by making
the following assumptions
1. Cartesian coordinates x1 = x,
2. u1 = u1 (x, t),
x2 = y,
x3 = z
u2 = u3 = 0.
3. There are no body forces.
∂u1 (x, 0)
=0
∂t
5. Boundary conditions of the displacement type u1 (0, t) = f (t),
4. Initial conditions of
u1 (x, 0) = 0 and
where f (t) is a specified function. These assumptions reduce the Navier equations to the single one dimensional wave equation
∂ 2 u1
∂ 2 u1
= α2
,
2
∂t
∂x2
α2 =
λ + 2µ
.
ρ
The solution of this equation is
u1 (x, t) =
f (t − x/α),
0,
x ≤ αt
.
x > αt
271
The solution represents a longitudinal elastic wave propagating in the x−direction with speed α. The stress
wave associated with this displacement is determined from the constitutive equations. We find
σxx = (λ + µ)exx = (λ + µ)
∂u1
.
∂x
This produces the stress wave
σxx =
0
− (λ+µ)
α f (t − x/α),
x ≤ αt
0,
x > αt
.
Here there is a discontinuity in the stress wave front at x = αt.
Summary of Basic Equations of Elasticity
The equilibrium equations for a continuum have been shown to have the form σij,j + %bi = 0, where
bi are the body forces per unit mass and σ ij is the stress tensor. In addition to the above equations we
have the constitutive equations σij = λekk δij + 2µeij which is a generalized Hooke’s law relating stress to
strain for a linear elastic isotropic material. The strain tensor is related to the displacement field ui by
1
the strain equations eij = (ui,j + uj,i ) . These equations can be combined to obtain the Navier equations
2
µui,jj + (λ + µ)uj,ji + %bi = 0.
The above equations must be satisfied at all interior points of the material body. A boundary value
problem results when conditions on the displacement of the boundary are specified. That is, the Navier
equations must be solved subject to the prescribed displacement boundary conditions. If conditions on
the stress at the boundary are specified, then these prescribed stresses are called surface tractions and
must satisfy the relations ti (n) = σ ij nj , where ni is a unit outward normal vector to the boundary. For
surface tractions, we need to use the compatibility equations combined with the constitutive equations and
equilibrium equations. This gives rise to the Beltrami-Michell equations of compatibility
σij,kk +
1
ν
σkk,ij + %(bi,j + bj,i ) +
%bk,k = 0.
1+ν
1−ν
Here we must solve for the stress components throughout the continuum where the above equations hold
subject to the surface traction boundary conditions. Note that if an elasticity problem is formed in terms of
the displacement functions, then the compatibility equations can be ignored.
For mixed boundary value problems we must solve a system of equations consisting of the equilibrium
equations, constitutive equations, and strain displacement equations. We must solve these equations subject
to conditions where the displacements ui are prescribed on some portion(s) of the boundary and stresses are
prescribed on the remaining portion(s) of the boundary. Mixed boundary value problems are more difficult
to solve.
For elastodynamic problems, the equilibrium equations are replaced by equations of motion. In this
case we need a set of initial conditions as well as boundary conditions before attempting to solve our basic
system of equations.
272
EXERCISE 2.4
I 1. Verify the generalized Hooke’s law constitutive equations for hexagonal materials.
In the following problems the Young’s modulus E, Poisson’s ratio ν, the shear modulus or modulus
of rigidity µ (sometimes denoted by G in Engineering texts), Lame’s constant λ and the bulk modulus of
elasticity k are assumed to satisfy the equations (2.4.19), (2.4.24) and (2.4.25). Show that these relations
imply the additional relations given in the problems 2 through 6.
I 2.
µ(3λ + 2µ)
µ+λ
λ(1 + ν)(1 − 2ν)
E=
ν
E=
I 3.
I 4.
E=
p
(E + λ)2 + 8λ2 + (E + 3λ)
6
2µ + 3λ
k=
3
I 6.
3k(1 − 2ν)
µ=
2(1 + ν)
3Ek
µ=
9k − E
3kν
1+ν
µ(2µ − E)
λ=
E − 3µ
λ=
9kµ
µ + 3k
E = 3(1 − 2ν)k
p
(E + λ)2 + 8λ2 − (E + λ)
ν=
4λ
3k − 2µ
ν=
2(µ + 3k)
3k − E
ν=
6k
λ
ν=
2(µ + λ)
3(k − λ)
µ=
2
λ(1 − 2ν)
µ=
2ν
E=
E = 2µ(1 + ν)
k=
I 5.
9k(k − λ)
3k − λ
E
3(1 − 2ν)
µE
k=
3(3µ − E)
k=
E − 2µ
2µ
λ
ν=
3k − λ
ν=
2µ(1 + ν)
3(1 − 2ν)
λ(1 + ν)
k=
3ν
k=
p
(E + λ)2 + 8λ2 + (E − 3λ)
µ=
4
E
µ=
2(1 + ν)
3k − 2µ
3
3k(3k − E)
λ=
9k − E
λ=
νE
(1 + ν)(1 − 2ν)
2µν
λ=
1 − 2ν
λ=
I 7. The previous exercises 2 through 6 imply that the generalized Hooke’s law
σij = 2µeij + λδij ekk
is expressible in a variety of forms. From the set of constants (µ,λ,ν,E,k) we can select any two constants
and then express Hooke’s law in terms of these constants.
(a) Express the above Hooke’s law in terms of the constants E and ν.
(b) Express the above Hooke’s law in terms of the constants k and E.
(c) Express the above Hooke’s law in terms of physical components. Hint: The quantity ekk is an invariant
hence all you need to know is how second order tensors are represented in terms of physical components.
See also problems 10,11,12.
273
I 8. Verify the equations defining the stress for plane strain in Cartesian coordinates are
E
[(1 − ν)exx + νeyy ]
(1 + ν)(1 − 2ν)
E
[(1 − ν)eyy + νexx ]
=
(1 + ν)(1 − 2ν)
Eν
[exx + eyy ]
=
(1 + ν)(1 − 2ν)
E
exy
=
1+ν
=0
σxx =
σyy
σzz
σxy
σyz = σxz
I 9. Verify the equations defining the stress for plane strain in polar coordinates are
E
[(1 − ν)err + νeθθ ]
(1 + ν)(1 − 2ν)
E
[(1 − ν)eθθ + νerr ]
=
(1 + ν)(1 − 2ν)
νE
[err + eθθ ]
=
(1 + ν)(1 − 2ν)
E
erθ
=
1+ν
=0
σrr =
σθθ
σzz
σrθ
σrz = σθz
I 10. Write out the independent components of Hooke’s generalized law for strain in terms of stress, and
stress in terms of strain, in Cartesian coordinates. Express your results using the parameters ν and E.
(Assume a linear elastic, homogeneous, isotropic material.)
I 11. Write out the independent components of Hooke’s generalized law for strain in terms of stress, and
stress in terms of strain, in cylindrical coordinates. Express your results using the parameters ν and E.
(Assume a linear elastic, homogeneous, isotropic material.)
I 12. Write out the independent components of Hooke’s generalized law for strain in terms of stress, and
stress in terms of strain in spherical coordinates. Express your results using the parameters ν and E. (Assume
a linear elastic, homogeneous, isotropic material.)
I 13. For a linear elastic, homogeneous, isotropic material assume there exists a state of plane strain in
Cartesian coordinates. Verify the equilibrium equations are
∂σxy
∂σxx
+
+ %bx = 0
∂x
∂y
∂σyy
∂σyx
+
+ %by = 0
∂x
∂y
∂σzz
+ %bz = 0
∂z
Hint: See problem 14, Exercise 2.3.
274
I 14 . For a linear elastic, homogeneous, isotropic material assume there exists a state of plane strain in
polar coordinates. Verify the equilibrium equations are
1 ∂σrθ
1
∂σrr
+
+ (σrr − σθθ ) + %br = 0
∂r
r ∂θ
r
1 ∂σθθ
2
∂σrθ
+
+ σrθ + %bθ = 0
∂r
r ∂θ
r
∂σzz
+ %bz = 0
∂z
Hint: See problem 15, Exercise 2.3.
I 15. For a linear elastic, homogeneous, isotropic material assume there exists a state of plane stress in
Cartesian coordinates. Verify the equilibrium equations are
∂σxy
∂σxx
+
+ %bx = 0
∂x
∂y
∂σyy
∂σyx
+
+ %by = 0
∂x
∂y
I 16. Determine the compatibility equations in terms of the Airy stress function φ when there exists a state
of plane stress. Assume the body forces are derivable from a potential function V.
I 17. For a linear elastic, homogeneous, isotropic material assume there exists a state of plane stress in
polar coordinates. Verify the equilibrium equations are
1 ∂σrθ
1
∂σrr
+
+ (σrr − σθθ ) + %br = 0
∂r
r ∂θ
r
1 ∂σθθ
2
∂σrθ
+
+ σrθ + %bθ = 0
∂r
r ∂θ
r
275
I 18. Figure 2.4-4 illustrates the state of equilibrium on an element in polar coordinates assumed to be of
unit length in the z-direction. Verify the stresses given in the figure and then sum the forces in the r and θ
directions to derive the same equilibrium laws developed in the previous exercise.
Figure 2.4-4. Polar element in equilibrium.
Hint: Resolve the stresses into components in the r and θ directions. Use the results that sin dθ
2 ≈
cos
dθ
2
dθ
2
and
≈ 1 for small values of dθ. Sum forces and then divide by rdr dθ and take the limit as dr → 0 and
dθ → 0.
I 19.
Express each of the physical components of plane stress in polar coordinates, σrr , σθθ , and σrθ
in terms of the physical components of stress in Cartesian coordinates σxx , σyy , σxy . Hint: Consider the
∂xa ∂xb
.
transformation law σ ij = σab i
∂x ∂xj
I 20. Use the results from problem 19 and assume the stresses are derivable from the relations
σxx = V +
∂2φ
,
∂y 2
σxy = −
∂2φ
,
∂x∂y
σyy = V +
∂2φ
∂x2
where V is a potential function and φ is the Airy stress function. Show that upon changing to polar
coordinates the Airy equations for stress become
σrr = V +
1 ∂2φ
1 ∂φ
+ 2 2,
r ∂r
r ∂θ
σrθ =
1 ∂φ 1 ∂ 2 φ
−
,
r2 ∂θ
r ∂r∂θ
σθθ = V +
∂2φ
.
∂r2
I 21. Verify that the Airy stress equations in polar coordinates, given in problem 20, satisfy the equilibrium
equations in polar coordinates derived in problem 17.
276
I 22.
In Cartesian coordinates show that the traction boundary conditions, equations (2.3.11), can be
written in terms of the constants λ and µ as
∂u1
∂u1
∂u2
∂u3
∂u1
+
+
+
n
T1 = λn1 ekk + µ 2n1 1 + n2
3
∂x
∂x2
∂x1
∂x3
∂x1
∂u2
∂u2
∂u1
∂u3
∂u2
+
+
n
+
+
2n
T2 = λn2 ekk + µ n1
2
3
∂x1
∂x2
∂x2
∂x3
∂x2
∂u3
∂u3
∂u1
∂u2
∂u3
+
+
+
n
+
2n
T3 = λn3 ekk + µ n1
2
3
∂x1
∂x3
∂x2
∂x3
∂x3
where (n1 , n2 , n3 ) are the direction cosines of the unit normal to the surface, u1 , u2 , u3 are the components
of the displacements and T1 , T2 , T3 are the surface tractions.
I 23. Consider an infinite plane subject to tension in the x−direction only. Assume a state of plane strain
and let σxx = T with σxy = σyy = 0. Find the strain components exx , eyy and exy . Also find the displacement
field u = u(x, y) and v = v(x, y).
I 24. Consider an infinite plane subject to tension in the y-direction only. Assume a state of plane strain
and let σyy = T with σxx = σxy = 0. Find the strain components exx , eyy and exy . Also find the displacement
field u = u(x, y) and v = v(x, y).
I 25. Consider an infinite plane subject to tension in both the x and y directions. Assume a state of plane
strain and let σxx = T , σyy = T and σxy = 0. Find the strain components exx , eyy and exy . Also find the
displacement field u = u(x, y) and v = v(x, y).
I 26. An infinite cylindrical rod of radius R0 has an external pressure P0 as illustrated in figure 2.5-5. Find
the stress and displacement fields.
Figure 2.4-5. External pressure on a rod.
277
Figure 2.4-6. Internal pressure on circular hole.
Figure 2.4-7. Tube with internal and external pressure.
I 27. An infinite plane has a circular hole of radius R1 with an internal pressure P1 as illustrated in the
figure 2.4-6. Find the stress and displacement fields.
I 28. A tube of inner radius R1 and outer radius R0 has an internal pressure of P1 and an external pressure
of P0 as illustrated in the figure 2.4-7. Verify the stress and displacement fields derived in example 2.4-7.
I 29. Use Cartesian tensors and combine the equations of equilibrium σij,j + %bi = 0, Hooke’s law σij =
1
λekk δij + 2µeij and the strain tensor eij = (ui,j + uj,i ) and derive the Navier equations of equilibrium
2
σij,j + %bi = (λ + µ)
∂Θ
∂ 2 ui
+
µ
+ %bi = 0,
∂xi
∂xk ∂xk
where Θ = e11 + e22 + e33 is the dilatation.
I 30. Show the Navier equations in problem 29 can be written in the tensor form
µui,jj + (λ + µ)uj,ji + %bi = 0
or the vector form
µ∇2 ~u + (λ + µ)∇ (∇ · ~u) + %~b = ~0.
278
I 31. Show that in an orthogonal coordinate system the components of ∇(∇ · ~u) can be expressed in terms
of physical components by the relation
1
∂(h2 h3 u(1)) ∂(h1 h3 u(2)) ∂(h1 h2 u(3))
1 ∂
+
+
[∇ (∇ · ~u)]i =
hi ∂xi h1 h2 h3
∂x1
∂x2
∂x3
I 32. Show that in orthogonal coordinates the components of ∇2 ~u can be written
2 ∇ ~u i = g jk ui,jk = Ai
and in terms of physical components one can write

3
3 3 X
X
m ∂(hm u(m)) X m ∂(hi u(i))
1  ∂ 2 (hi u(i))
−
2
−
hi A(i) =
h2
∂xj ∂xj
∂xj
∂xm
ij
jj
m=1
m=1
j=1 j

X
X
!
3
3 3 X
∂
m
m
p
m
p

−
hm u(m)
−
−
∂xj i j
ip
jj
jp
ij
m=1
p=1
p=1
I 33. Use the results in problem 32 to show in Cartesian coordinates the physical components of [∇2 ~u]i = Ai
can be represented
∂2u ∂2u ∂2u
+ 2 + 2
∇2 ~u · ê1 = A(1) =
∂x2
∂y
∂z
2
2
2 ∂ v
∂ v
∂2v
+ 2+ 2
∇ ~u · ê2 = A(2) =
2
∂x
∂y
∂z
2
2
2 ∂ w ∂ w ∂2w
+
+
∇ ~u · ê3 = A(3) =
∂x2
∂y 2
∂z 2
where (u, v, w) are the components of the displacement vector ~u.
I 34. Use the results in problem 32 to show in cylindrical coordinates the physical components of [∇2 ~u]i = Ai
can be represented
1
2 ∂uθ
∇2 ~u · êr = A(1) = ∇2 ur − 2 ur − 2
r
r ∂θ
2 1
2 ∂ur
2
− 2 uθ
∇ ~u · êθ = A(2) = ∇ uθ + 2
r
∂θ
r
2 ∇ ~u · êz = A(3) = ∇2 uz
1 ∂2α ∂2α
∂ 2 α 1 ∂α
+
+
+
∂r2
r ∂r
r2 ∂θ2
∂z 2
I 35. Use the results in problem 32 to show in spherical coordinates the physical components of [∇2 ~u]i = Ai
where ur , uθ , uz are the physical components of ~u and ∇2 α =
can be represented
2 cot θ
2
2 ∂uθ
2 ∂uφ
−
uθ − 2
∇2 ~u · êρ = A(1) = ∇2 uρ − 2 uρ − 2
2
ρ
ρ ∂θ
ρ
ρ sin θ ∂φ
2 ∂uθ
1
2
2
cos
θ
∂u
ρ
− 2
uθ − 2 2
∇ ~u · êθ = A(2) = ∇2 uθ + 2
ρ ∂θ
ρ sin θ
ρ sin θ ∂φ
2 2 cos θ ∂uθ
1
2 ∂uρ
+ 2 2
∇ ~u · êφ = A(3) = ∇2 uφ − 2 2 uφ + 2
ρ sin θ ∂φ
ρ sin θ
ρ sin θ ∂φ
where uρ , uθ , uφ are the physical components of ~u and where
∇2 α =
∂2α
1 ∂ 2 α cot θ ∂α
1
∂ 2 α 2 ∂α
+
+
+
+
∂ρ2
ρ ∂ρ
ρ2 ∂θ2
ρ2 ∂θ
ρ2 sin2 θ ∂φ2
279
I 36. Combine the results from problems 30,31,32 and 33 and write the Navier equations of equilibrium
in Cartesian coordinates. Alternatively, write the stress-strain relations (2.4.29(b)) in terms of physical
components and then use these results, together with the results from Exercise 2.3, problems 2 and 14, to
derive the Navier equations.
I 37. Combine the results from problems 30,31,32 and 34 and write the Navier equations of equilibrium
in cylindrical coordinates. Alternatively, write the stress-strain relations (2.4.29(b)) in terms of physical
components and then use these results, together with the results from Exercise 2.3, problems 3 and 15, to
derive the Navier equations.
I 38. Combine the results from problems 30,31,32 and 35 and write the Navier equations of equilibrium
in spherical coordinates. Alternatively, write the stress-strain relations (2.4.29(b)) in terms of physical
components and then use these results, together with the results from Exercise 2.3, problems 4 and 16, to
derive the Navier equations.
I 39. Assume %~b = −grad V and let φ denote the Airy stress function defined by
∂2φ
σxx =V + 2
∂y
∂2φ
σyy =V +
∂x2
∂2φ
σxy = −
∂x∂y
(a) Show that for conditions of plane strain the equilibrium equations in two dimensions are satisfied by the
above definitions. (b) Express the compatibility equation
∂ 2 exy
∂ 2 exx ∂ 2 eyy
+
=
2
∂y 2
∂x2
∂x∂y
in terms of φ and V and show that
1 − 2ν 2
∇ V = 0.
∇4 φ +
1−ν
I 40. Consider the case where the body forces are conservative and derivable from a scalar potential function
such that %bi = −V,i . Show that under conditions of plane strain in rectangular Cartesian coordinates the
1
∇2 V
, i = 1, 2
compatibility equation e11,22 + e22,11 = 2e12,12 can be reduced to the form ∇2 σii =
1−ν
involving the stresses and the potential. Hint: Differentiate the equilibrium equations.
i
I 41. Use the relation σji = 2µeij + λem
m δj and solve for the strain in terms of the stress.
I 42. Derive the equation (2.4.26) from the equation (2.4.23).
I 43.
In two dimensions assume that the body forces are derivable from a potential function V and
%b = −g ij V ,j . Also assume that the stress is derivable from the Airy stress function and the potential
i
function by employing the relations σ ij = im jn um,n + g ij V
pq
i, j, m, n = 1, 2 where um = φ ,m and
is the two dimensional epsilon permutation symbol and all indices have the range 1,2.
(a) Show that im jn (φm ) ,nj = 0.
(b) Show that σ ij,j = −%bi .
(c) Verify the stress laws for cylindrical and Cartesian coordinates given in problem 20 by using the above
expression for σ ij . Hint: Expand the contravariant derivative and convert all terms to physical components. Also recall that ij =
√1 eij .
g
280
I 44. Consider a material with body forces per unit volume ρF i , i = 1, 2, 3 and surface tractions denoted by
σ r = σ rj nj , where nj is a unit surface normal. Further, let δui denote a small displacement vector associated
with a small variation in the strain δeij .
Z
ρF i δui dτ
(a) Show the work done during a small variation in strain is δW = δWB + δWS where δWB =
V
Z
σ r δur dS is a surface
is a volume integral representing the work done by the body forces and δWS =
S
integral representing the work done by the surface forces.
(b) Using the Gauss divergence theorem show that the work done can be represented as
Z
Z
1
1
cijmn δ[emn eij ] dτ or W =
σ ij eij dτ.
δW =
2 V
2 V
The scalar quantity 12 σ ij eij is called the strain energy density or strain energy per unit volume.
R
Hint: Interchange subscripts, add terms and calculate 2W = V σ ij [δui,j + δuj,i ] dτ.
I 45. Consider a spherical shell subjected to an internal pressure pi and external pressure po . Let a denote
the inner radius and b the outer radius of the spherical shell. Find the displacement and stress fields in
spherical coordinates (ρ, θ, φ).
Hint: Assume symmetry in the θ and φ directions and let the physical components of displacements satisfy
the relations uρ = uρ (ρ),
I 46.
uθ = uφ = 0.
(a) Verify the average normal stress is proportional to the dilatation, where the proportionality
constant is the bulk modulus of elasticity. i.e. Show that 13 σii =
E
1 i
1−2ν 3 ei
= keii where k is the bulk modulus
of elasticity.
(b) Define the quantities of strain deviation and stress deviation in terms of the average normal stress
s = 13 σii and average cubic dilatation e = 13 eii as follows
strain deviator
εij = eij − eδji
stress deviator
sij = σji − sδji
Show that zero results when a contraction is performed on the stress and strain deviators. (The above
definitions are used to split the strain tensor into two parts. One part represents pure dilatation and
the other part represents pure distortion.)
(c) Show that (1 − 2ν)s = Ee
or s = (3λ + 2µ)e
(d) Express Hooke’s law in terms of the strain and stress deviator and show
E(εij + eδji ) = (1 + ν)sij + (1 − 2ν)sδji
which simplifies to sij = 2µεij .
I 47. Show the strain energy density (problem 44) can be written in terms of the stress and strain deviators
(problem 46) and
1
W =
2
Z
V
and from Hooke’s law
W =
1
σ eij dτ =
2
Z
ij
3
2
Z
V
(3se + sij εij ) dτ
V
((3λ + 2µ)e2 +
2µ ij
ε εij ) dτ.
3
281
I 48. Find the stress σrr ,σrθ and σθθ in an infinite plate with a small circular hole, which is traction free,
when the plate is subjected to a pure shearing force F12 . Determine the maximum stress.
I 49. Show that in terms of E and ν
C1111 =
E(1 − ν)
(1 + ν)(1 − 2ν)
C1122 =
Eν
(1 + ν)(1 − 2ν)
C1212 =
E
2(1 + ν)
I 50. Show that in Cartesian coordinates the quantity
S = σxx σyy + σyy σzz + σzz σxx − (σxy )2 − (σyz )2 − (σxz )2
1
(σii σjj − σij σij ).
2
I 51. Show that in Cartesian coordinates for a state of plane strain where the displacements are given by
is a stress invariant. Hint: First verify that in tensor form S =
u = u(x, y),v = v(x, y) and w = 0, the stress components must satisfy the equations
∂σxy
∂σxx
+
+ %bx =0
∂x
∂y
∂σyy
∂σyx
+
+ %by =0
∂x
∂y
−%
∇ (σxx + σyy ) =
1−ν
2
∂bx ∂by
+
∂x
∂y
I 52. Show that in Cartesian coordinates for a state of plane stress where σxx = σxx (x, y), σyy = σyy (x, y),
σxy = σxy (x, y) and σxz = σyz = σzz = 0 the stress components must satisfy
∂σxy
∂σxx
+
+ %bx =0
∂x
∂y
∂σyy
∂σyx
+
+ %by =0
∂x
∂y
2
∇ (σxx + σyy ) = − %(ν + 1)
∂bx
∂by
+
∂x
∂y
282
§2.5 CONTINUUM MECHANICS (FLUIDS)
Let us consider a fluid medium and use Cartesian tensors to derive the mathematical equations that
describe how a fluid behaves. A fluid continuum, like a solid continuum, is characterized by equations
describing:
1. Conservation of linear momentum
σij,j + %bi = %v̇i
(2.5.1)
2. Conservation of angular momentum σij = σji .
3. Conservation of mass (continuity equation)
∂%
∂vi
∂%
+
vi + %
=0
∂t
∂xi
∂xi
or
D%
~ = 0.
+ %∇ · V
Dt
(2.5.2)
In the above equations vi , i = 1, 2, 3 is a velocity field, % is the density of the fluid, σij is the stress tensor
and bj is an external force per unit mass. In the cgs system of units of measurement, the above quantities
have dimensions
[v̇j ] = cm/sec2 ,
[σij ] = dyne/cm2 ,
[bj ] = dynes/g,
[%] = g/cm3 .
(2.5.3)
The displacement field ui , i = 1, 2, 3 can be represented in terms of the velocity field vi , i = 1, 2, 3, by
Z
the relation
ui =
0
t
vi dt.
(2.5.4)
The strain tensor components of the medium can then be represented in terms of the velocity field as
Z t
Z t
1
1
(vi,j + vj,i ) dt =
Dij dt,
(2.5.5)
eij = (ui,j + uj,i ) =
2
0 2
0
where
1
(vi,j + vj,i )
2
is called the rate of deformation tensor , velocity strain tensor, or rate of strain tensor.
Dij =
(2.5.6)
Note the difference in the equations describing a solid continuum compared with those for a fluid
continuum. In describing a solid continuum we were primarily interested in calculating the displacement
field ui , i = 1, 2, 3 when the continuum was subjected to external forces. In describing a fluid medium, we
calculate the velocity field vi , i = 1, 2, 3 when the continuum is subjected to external forces. We therefore
replace the strain tensor relations by the velocity strain tensor relations in all future considerations concerning
the study of fluid motion.
Constitutive Equations for Fluids
In addition to the above basic equations, we will need a set of constitutive equations which describe the
material properties of the fluid. Toward this purpose consider an arbitrary point within the fluid medium
and pass an imaginary plane through the point. The orientation of the plane is determined by a unit normal
(n)
ni , i = 1, 2, 3 to the planar surface. For a fluid at rest we wish to determine the stress vector ti
on the plane element passing through the selected point P. We desire to express
(n)
ti
acting
in terms of the stress
tensor σij . The superscript (n) on the stress vector is to remind you that the stress acting on the planar
element depends upon the orientation of the plane through the point.
283
(n)
We make the assumption that ti
is colinear with the normal vector to the surface passing through
the selected point. It is also assumed that for fluid elements at rest, there are no shear forces acting on the
planar element through an arbitrary point and therefore the stress tensor σij should be independent of the
orientation of the plane. That is, we desire for the stress vector σij to be an isotropic tensor. This requires
σij to have a specific form. To find this specific form we let σij denote the stress components in a general
coordinate system xi , i = 1, 2, 3 and let σ ij denote the components of stress in a barred coordinate system
xi , i = 1, 2, 3. Since σij is a tensor, it must satisfy the transformation law
σ mn = σij
∂xi ∂xj
,
∂xm ∂xn
i, j, m, n = 1, 2, 3.
(2.5.7)
We desire for the stress tensor σij to be an invariant under an arbitrary rotation of axes. Consider
therefore the special coordinate transformations illustrated in the figures 2.5-1(a) and (b).
Figure 2.5-1. Coordinate transformations due to rotations
For the transformation equations given in figure 2.5-1(a), the stress tensor in the barred system of
coordinates is
σ 11 = σ22
σ 21 = σ32
σ 31 = σ12
σ 12 = σ23
σ 22 = σ33
σ 32 = σ13
σ 13 = σ21
σ 23 = σ31
σ 33 = σ11 .
(2.5.8)
If σij is to be isotropic, we desire that σ 11 = σ11 , σ 22 = σ22 and σ 33 = σ33 . If the equations (2.5.8) are
to produce these results, we require that σ11 , σ22 and σ33 must be equal. We denote these common values
by (−p). In particular, the equations (2.5.8) show that if σ 11 = σ11 , σ 22 = σ22 and σ 33 = σ33 , then we must
require that σ11 = σ22 = σ33 = −p. If σ 12 = σ12 and σ 23 = σ23 , then we also require that σ12 = σ23 = σ31 .
We note that if σ 13 = σ13 and σ 32 = σ32 , then we require that σ21 = σ32 = σ13 . If the equations (2.5.7) are
expanded using the transformation given in figure 2.5-1(b), we obtain the additional requirements that
σ 11 = σ22
σ 12 = −σ21
σ 13 = σ23
σ 21 = −σ12
σ 22 = σ11
σ 23 = −σ13
σ 31 = σ32
σ 32 = −σ31
σ 33 = σ33 .
(2.5.9)
284
Analysis of these equations implies that if σij is to be isotropic, then σ 21 = σ21 = −σ12 = −σ21
or σ21 = 0 which implies
σ12 = σ23 = σ31 = σ21 = σ32 = σ13 = 0.
(2.5.10)
The above analysis demonstrates that if the stress tensor σij is to be isotropic, it must have the form
σij = −pδij .
(2.5.11)
Use the traction condition (2.3.11), and express the stress vector as
(n)
tj
= σij ni = −pnj .
(2.5.12)
This equation is interpreted as representing the stress vector at a point on a surface with outward unit
normal ni , where p is the pressure (hydrostatic pressure) stress magnitude assumed to be positive. The
negative sign in equation (2.5.12) denotes a compressive stress.
Imagine a submerged object in a fluid medium. We further imagine the object to be covered with unit
normal vectors emanating from each point on its surface. The equation (2.5.12) shows that the hydrostatic
pressure always acts on the object in a compressive manner. A force results from the stress vector acting on
the object. The direction of the force is opposite to the direction of the unit outward normal vectors. It is
a compressive force at each point on the surface of the object.
The above considerations were for a fluid at rest (hydrostatics). For a fluid in motion (hydrodynamics)
a different set of assumptions must be made. Hydrodynamical experiments show that the shear stress
components are not zero and so we assume a stress tensor having the form
σij = −pδij + τij ,
i, j = 1, 2, 3,
(2.5.13)
where τij is called the viscous stress tensor. Note that all real fluids are both viscous and compressible.
Definition: (Viscous/inviscid fluid) If the viscous stress tensor τij is zero for all i, j, then the fluid is called an inviscid, nonviscous, ideal or perfect fluid. The fluid is called viscous when τij
is different from zero.
In these notes it is assumed that the equation (2.5.13) represents the basic form for constitutive equations
describing fluid motion.
285
Figure 2.5-2. Viscosity experiment.
Viscosity
Most fluids are characterized by the fact that they cannot resist shearing stresses. That is, if you put a
shearing stress on the fluid, the fluid gives way and flows. Consider the experiment illustrated in the figure
2.5-2 which illustrates a fluid moving between two parallel plane surfaces. Let S denote the distance between
the two planes. Now keep the lower surface fixed or stationary and move the upper surface parallel to the
lower surface with a constant velocity V~0 . If you measure the force F required to maintain the constant
velocity of the upper surface, you discover that the force F varies directly as the area A of the surface and
the ratio V0 /S. This is expressed in the form
V0
F
= µ∗ .
A
S
(2.5.14)
The constant µ∗ is a proportionality constant called the coefficient of viscosity. The viscosity usually depends
upon temperature, but throughout our discussions we will assume the temperature is constant. A dimensional
analysis of the equation (2.5.14) implies that the basic dimension of the viscosity is [µ∗ ] = M L−1 T −1 . For
example, [µ∗ ] = gm/(cm sec) in the cgs system of units. The viscosity is usually measured in units of
centipoise where one centipoise represents one-hundredth of a poise, where the unit of 1 poise= 1 gram
per centimeter per second. The result of the above experiment shows that the stress is proportional to the
change in velocity with change in distance or gradient of the velocity.
Linear Viscous Fluids
The above experiment with viscosity suggest that the viscous stress tensor τij is dependent upon both
the gradient of the fluid velocity and the density of the fluid.
In Cartesian coordinates, the simplest model suggested by the above experiment is that the viscous
stress tensor τij is proportional to the velocity gradient vi,j and so we write
τik = cikmp vm,p ,
(2.5.15)
where cikmp is a proportionality constant which is dependent upon the fluid density.
The viscous stress tensor must be independent of any reference frame, and hence we assume that the
proportionality constants cikmp can be represented by an isotropic tensor. Recall that an isotropic tensor
has the basic form
cikmp = λ∗ δik δmp + µ∗ (δim δkp + δip δkm ) + ν ∗ (δim δkp − δip δkm )
(2.5.16)
286
where λ∗ , µ∗ and ν ∗ are constants. Examining the results from equations (2.5.11) and (2.5.13) we find that if
the viscous stress is symmetric, then τij = τji . This requires ν ∗ be chosen as zero. Consequently, the viscous
stress tensor reduces to the form
τik = λ∗ δik vp,p + µ∗ (vk,i + vi,k ).
(2.5.17)
The coefficient µ∗ is called the first coefficient of viscosity and the coefficient λ∗ is called the second coefficient
of viscosity. Sometimes it is convenient to define
2
ζ = λ∗ + µ∗
3
(2.5.18)
as “another second coefficient of viscosity,” or “bulk coefficient of viscosity.” The condition of zero bulk
viscosity is known as Stokes hypothesis. Many fluids problems assume the Stoke’s hypothesis. This requires
that the bulk coefficient be zero or very small. Under these circumstances the second coefficient of viscosity
is related to the first coefficient of viscosity by the relation λ∗ = − 32 µ∗ . In the study of shock waves and
acoustic waves the Stoke’s hypothesis is not applicable.
There are many tables and empirical formulas where the viscosity of different types of fluids or gases
can be obtained. For example, in the study of the kinetic theory of gases the viscosity can be calculated
C1 gT 3/2
where C1 , C2 are constants for a specific gas. These constants
from the Sutherland formula µ∗ =
T + C2
can be found in certain tables. The quantity g is the gravitational constant and T is the temperature in
degrees Rankine (o R = 460 + o F ). Many other empirical formulas like the above exist. Also many graphs
and tabular values of viscosity can be found. The table 5.1 lists the approximate values of the viscosity of
some selected fluids and gases.
Table 5.1
Substance
Water
Alcohol
Ethyl Alcohol
Glycol
Mercury
Air
Helium
Nitrogen
Viscosity of selected fluids and gases
gram
in units of cm−sec = Poise
at Atmospheric Pressure.
0◦ C
0.01798
0.01773
0.017
1.708(10−4)
1.86(10−4)
1.658(10−4)
20◦ C
0.01002
60◦ C
0.00469
0.012
0.199
0.0157
0.00592
0.0495
0.013
1.94(10−4 )
1.74(10−4 )
1.92(10−4)
100◦ C
0.00284
0.0199
0.0100
2.175(10−4)
2.28(10−4)
2.09(10−4)
The viscous stress tensor given in equation (2.5.17) may also be expressed in terms of the rate of
deformation tensor defined by equation (2.5.6). This representation is
τij = λ∗ δij Dkk + 2µ∗ Dij ,
(2.5.19)
where 2Dij = vi,j + vj,i and Dkk = D11 + D22 + D33 = v1,1 + v2,2 + v3,3 = vi,i = Θ is the rate of change
of the dilatation considered earlier. In Cartesian form, with velocity components u, v, w, the viscous stress
287
tensor components are
τyy
τzz
∂u
∂v
∂w
+ λ∗
+
∂x
∂y
∂z
∂u
∂w
∗
∗ ∂v
∗
=(λ + 2µ )
+λ
+
∂y
∂x
∂z
∂u ∂v ∂w
+ λ∗
+
=(λ∗ + 2µ∗ )
∂z
∂x
∂y
τxx =(λ∗ + 2µ∗ )
τzx = τxz
τzy = τyz
∂u
∂v
∂y
∂x
∂w
∂u
=µ∗
+
∂x
∂z
∂v
∂w
=µ∗
+
∂z
∂y
τyx = τxy =µ∗
+
In cylindrical form, with velocity components vr , vθ , vz , the viscous stess tensor components are
∂vr
~
+ λ∗ ∇ · V
∂r
1 ∂vθ
vr
~
τθθ =2µ∗
+ λ∗ ∇ · V
+
r ∂θ
r
∂vz
~
+ λ∗ ∇ · V
τzz =2µ∗
∂z
~ = 1 ∂ (rvr ) + 1 ∂vθ + ∂vz
∇·V
r ∂r
r ∂θ
∂z
τrr =2µ∗
where
τrz = τzr
τzθ = τθz
1 ∂vr
∂vθ
vθ
+
−
r ∂θ
∂r
r
∂vr
∂vz
∗
=µ
+
∂z
∂r
1 ∂vz
∂vθ
∗
=µ
+
r ∂θ
∂z
τθr = τrθ =µ∗
In spherical coordinates, with velocity components vρ , vθ , vφ , the viscous stress tensor components have the
form
∂vρ
~
+ λ∗ ∇ · V
∂ρ
1 ∂vθ
vρ
~
τθθ =2µ∗
+ λ∗ ∇ · V
+
ρ ∂θ
ρ
1 ∂vφ
vρ
vθ cot θ
~
τφφ =2µ∗
+ λ∗ ∇ · V
+
+
ρ sin θ ∂φ
ρ
ρ
∂
1 ∂vφ
~ = 1 ∂ ρ2 vρ + 1
(sin θvθ ) +
∇·V
2
ρ ∂ρ
ρ sin θ ∂θ
ρ sin θ ∂φ
τρρ =2µ∗
where
τφρ = τρφ
τθφ = τφθ
∂
1 ∂vρ
vθ
+
∂ρ ρ
ρ ∂θ
1 ∂vr
vθ
∂
=µ∗
+ρ
ρ sin θ ∂φ
∂ρ ρ
v 1 ∂vθ
sin θ ∂
φ
=µ∗
+
ρ ∂θ sin θ
ρ sin θ ∂φ
τρθ = τθρ =µ∗ ρ
Note that the viscous stress tensor is a linear function of the rate of deformation tensor Dij . Such a
fluid is called a Newtonian fluid. In cases where the viscous stress tensor is a nonlinear function of Dij the
fluid is called non-Newtonian.
Definition: (Newtonian Fluid)
If the viscous stress tensor τij
is expressible as a linear function of the rate of deformation tensor
Dij , the fluid is called a Newtonian fluid. Otherwise, the fluid is
called a non-Newtonian fluid.
Important note: Do not assume an arbitrary form for the constitutive equations unless there is experimental evidence to support your assumption. A constitutive equation is a very important step in the
modeling processes as it describes the material you are working with. One cannot arbitrarily assign a form
to the viscous stress and expect the mathematical equations to describe the correct fluid behavior. The form
of the viscous stress is an important part of the modeling process and by assigning different forms to the
viscous stress tensor then various types of materials can be modeled. We restrict our study in these notes
to Newtonian fluids.
In Cartesian coordinates the rate of deformation-stress constitutive equations for a Newtonian fluid can
be written as
σij = −pδij + λ∗ δij Dkk + 2µ∗ Dij
(2.5.20)
288
which can also be written in the alternative form
σij = −pδij + λ∗ δij vk,k + µ∗ (vi,j + vj,i )
(2.5.21)
involving the gradient of the velocity.
Upon transforming from a Cartesian coordinate system y i , i = 1, 2, 3 to a more general system of
coordinates xi , i = 1, 2, 3, we write
σ mn = σij
∂y i ∂y j
.
∂xm ∂xn
(2.5.22)
Now using the divergence from equation (2.1.3) and substituting equation (2.5.21) into equation (2.5.22) we
obtain a more general expression for the constitutive equation. Performing the indicated substitutions there
results
∂y i ∂y j
σ mn = −pδij + λ∗ δij v k,k + µ∗ (vi,j + vj,i )
∂xm ∂xn
∗
k
∗
σ mn = −pgmn + λ gmn v ,k + µ (v m,n + v n,m ).
Dropping the bar notation, the stress-velocity strain relationships in the general coordinates xi , i = 1, 2, 3, is
σmn = −pgmn + λ∗ gmn g ik vi,k + µ∗ (vm,n + vn,m ).
(2.5.23)
Summary
The basic equations which describe the motion of a Newtonian fluid are :
Continuity equation (Conservation of mass)
∂%
+ %v i ,i = 0,
∂t
Conservation of linear momentum
or
D%
~ =0
+ %∇ · V
Dt
σ ij,j + %bi = %v̇ i ,
1 equation.
(2.5.24)
3 equations
DV~
= %~b + ∇ · σ = %~b − ∇p + ∇ · τ
(2.5.25)
Dt
P3 P3
P3 P3
= i=1 j=1 (−pδij + τij ) êi êj and τ = i=1 j=1 τij êi êj are second order tensors. Conseror in vector form %
where
σ
vation of angular momentum σ ij = σ ji ,
(Reduces the set of equations (2.5.23) to 6 equations.) Rate of
deformation tensor (Velocity strain tensor)
Dij =
1
(vi,j + vj,i ) ,
2
6 equations.
(2.5.26)
Constitutive equations
σmn = −pgmn + λ∗ gmn g ik vi,k + µ∗ (vm,n + vn,m ),
6 equations.
(2.5.27)
289
In the cgs system of units the above quantities have the following units of measurements in Cartesian
coordinates
is the velocity field , i = 1, 2, 3,
vi
σij
is the stress tensor, i, j = 1, 2, 3,
%
bi
is the fluid density
is the external body forces per unit mass
Dij
is the rate of deformation tensor
p is the pressure
λ∗ , µ∗
are coefficients of viscosity
[vi ] = cm/sec
[σij ] = dyne/cm2
[%] = gm/cm3
[bi ] = dyne/gm
[Dij ] = sec−1
[p] = dyne/cm2
[λ∗ ] = [µ∗ ] = Poise
where 1 Poise = 1gm/cm sec
If we assume the external body forces per unit mass are known, then the equations (2.5.24), (2.5.25),
(2.5.26), and (2.5.27) represent 16 equations in the 16 unknowns
%, v1 , v2 , v3 , σ11 , σ12 , σ13 , σ22 , σ23 , σ33 , D11 , D12 , D13 , D22 , D23 , D33 .
Navier-Stokes-Duhem Equations of Fluid Motion
Substituting the stress tensor from equation (2.5.27) into the linear momentum equation (2.5.25), and
assuming that the viscosity coefficients are constants, we obtain the Navier-Stokes-Duhem equations for fluid
motion. In Cartesian coordinates these equations can be represented in any of the equivalent forms
%v̇i = %bi − p,j δij + (λ∗ + µ∗ )vk,ki + µ∗ vi,jj
%
∂vi
+ %vj vi,j = %bi + (−pδij + τij ) ,j
∂t
∂%vi
+ (%vi vj + pδij − τij ) ,j = %bi
∂t
D~v
= %~b − ∇ p + (λ∗ + µ∗ )∇ (∇ · ~v ) + µ∗ ∇2 ~v
%
Dt
(2.5.28)
∂~v
D~v
=
+ (~v · ∇) ~v is the material derivative, substantial derivative or convective derivative. This
Dt
∂t
derivative is represented as
where
v̇i =
∂vi dxj
∂vi
∂vi
∂vi
∂vi
+ j
=
+ j vj =
+ vi,j v j .
∂t
∂x dt
∂t
∂x
∂t
(2.5.29)
In the vector form of equations (2.5.28), the terms on the right-hand side of the equation represent force
terms. The term %~b represents external body forces per unit volume. If these forces are derivable from a
potential function φ, then the external forces are conservative and can be represented in the form −%∇ φ.
The term −∇ p is the gradient of the pressure and represents a force per unit volume due to hydrostatic
pressure. The above statement is verified in the exercises that follow this section. The remaining terms can
be written
f~viscous = (λ∗ + µ∗ )∇ (∇ · ~v ) + µ∗ ∇2~v
(2.5.30)
290
and are given the physical interpretation of an internal force per unit volume. These internal forces arise
from the shearing stresses in the moving fluid. If f~viscous is zero the vector equation in (2.5.28) is called
Euler’s equation.
If the viscosity coefficients are nonconstant, then the Navier-Stokes equations can be written in the
Cartesian form
∂vi
∂
∂vj
∂vi
∂vk
∂vi
∗
∗
+ vj
] =%bi +
+µ
+
−pδij + λ δij
%[
∂t
∂xj
∂xj
∂xk
∂xj
∂xi
∂v
∂p
∂
∂vj
∂v
∂
k
i
∗
∗
+
+
λ
+ j µ
=%bi −
∂xi
∂xi
∂xk
∂x
∂xj
∂xi
which can also be written in terms of the bulk coefficient of viscosity ζ = λ∗ + 23 µ∗ as
%[
∂vi
∂p
∂
∂vk
∂vj
∂vi
2
∂
∂vi
+ vj
] =%bi −
+
+
(ζ − µ∗ )
+ j µ∗
∂t
∂xj
∂xi
∂xi
3
∂xk
∂x
∂xj
∂xi
∂vi
∂p
∂
∂vj
2 ∂vk
∂vk
∂
+
+
− δij
ζ
+ j µ∗
=%bi −
∂xi
∂xi
∂xk
∂x
∂xj
∂xi
3 ∂xk
These equations form the basics of viscous flow theory.
In the case of orthogonal coordinates, where g(i)(i) = h2i (no summation) and gij = 0 for i 6= j, general
expressions for the Navier-Stokes equations in terms of the physical components v(1), v(2), v(3) are:
Navier-Stokes-Duhem equations for compressible fluid in terms of physical components: (i 6= j 6= k)
h
%
v(2) ∂v(i)
v(3) ∂v(i)
∂v(i)
v(1) ∂v(i)
+
+
+
∂t
h1 ∂x1
h2 ∂x2
h3 ∂x3
−
%
v(j)
hi hj
v(j)
∂hj
∂hi
− v(i)
∂xi
∂xj
+
∗
b(i)
1 ∂p
1 ∂
~ + µ
−
+
λ∗ ∇ · V
hi
hi ∂xi
hi ∂xi
hi hj
+
µ∗
hi hk
−
2µ∗
hi hk
+
∂
∂xj
h
hi ∂
hk ∂xk
v(i)
hi
+
hk ∂
hi ∂xi
v(k)
hk
i
1 ∂v(k)
v(i) ∂hk
v(k) ∂hk
+
+
hk ∂xk
hi hk ∂xi
hk hj ∂xi
µ∗ hi hk
hj ∂
hi ∂xi
v(j)
hj
+
v(k)
hi hk
hi ∂
hj ∂xj
v(i)
hj ∂
hi ∂xi
∂hi
∂hk
− v(k)
∂xk
∂xi
v(j)
hj
2µ∗
∂hi
−
∂xk
hi hj
+
v(i)
hi
+
hi ∂
hj ∂xj
=
v(i)
hi
∂hi
∂hj
v(k) ∂hj
v(i) ∂hj
1 ∂v(j)
+
+
hj ∂xj
hj hk ∂xk
hi hj ∂xi
∂hk
1
+
∂xi
hi hj hk
∂
∂xk
∂
∂xi
n
2µ∗ hj hk
µ∗ hi hj
1 ∂v(i)
v(j) ∂hi
v(k) ∂hi
+
+
hi ∂xi
hi hj ∂hj
hi hk ∂xk
hi ∂
hk ∂xk
v(i)
hi
+
hk ∂
hi ∂xi
v(k)
hk
o
(2.5.31)
where ∇ · ~v is found in equation (2.1.4).
In the above equation, cyclic values are assigned to i, j and k. That is, for the x1 components assign
the values i = 1, j = 2, k = 3; for the x2 components assign the values i = 2, j = 3, k = 1; and for the x3
components assign the values i = 3, j = 1, k = 2.
The tables 5.2, 5.3 and 5.4 show the expanded form of the Navier-Stokes equations in Cartesian, cylindrical and spherical coordinates respectively.
291
h
%
DVx
∂p
∂Vx
∂
~
2µ∗
=%bx −
+
+ λ∗ ∇ · V
Dt
∂x
∂x
∂x
%
DVy
∂p
∂
µ∗
=%by −
+
Dt
∂y
∂x
%
DVz
∂p
∂
µ∗
=%bz −
+
Dt
∂z
∂x
h
h
i
∂Vy
∂Vx
+
∂x
∂y
∂Vz
∂Vx
+
∂x
∂z
i
i
+
h
∂
µ∗
∂y
∂Vx
∂Vy
+
∂y
∂x
h
i
+
∂
∂Vy
~
2µ∗
+ λ∗ ∇ · V
∂y
∂y
+
∂
µ∗
∂y
h
∂Vz
∂Vy
+
∂y
∂z
i
i
h
h
+
∂
µ∗
∂z
+
∂
µ∗
∂z
+
∂Vx
∂Vz
+
∂z
∂x
∂Vy
∂Vz
+
∂z
∂y
h
i
i
∂
∂Vz
~
2µ∗
+ λ∗ ∇ · V
∂z
∂z
i
D
∂( )
∂( )
∂( )
∂( )
() =
+ Vx
+ Vy
+ Vz
Dt
∂t
∂x
∂y
∂z
where
~ = ∂Vx + ∂Vy + ∂Vz
∇·V
∂x
∂y
∂z
and
(2.5.31a)
Table 5.2 Navier-Stokes equations for compressible fluids in Cartesian coordinates.
%
V2
DVr
− θ
Dt
r
=%br −
+
h
%
DVθ
Vr Vθ
+
Dt
r
i
h
∂p
∂Vr
∂
~
2µ∗
+
+ λ∗ ∇ · V
∂r
∂r
∂r
h
∂
µ∗
∂z
∂V
r
∂z
+
∂Vz
∂r
h
where
and
+
2µ∗
r
h
1 ∂
µ∗
r ∂θ
+
∂V
r
∂r
−
1 ∂Vr
∂Vθ
Vθ
+
−
r ∂θ
∂r
r
1 ∂Vθ
Vr
−
r ∂θ
r
i
h
DVz
∂p
1 ∂
µ∗ r
=%bz −
+
Dt
∂z
r ∂r
∂Vr
∂Vz
+
∂z
∂r
i
+
h
1 ∂
µ∗
r ∂θ
i
1 ∂p
1 ∂
1 ∂Vr
∂
∂Vθ
Vθ
µ∗
+
2µ∗
+
+
−
r ∂θ
∂r
r ∂θ
∂r
r
r ∂θ
h 1 ∂V
i 2µ∗ h 1 ∂V
i
∂Vθ
∂Vθ
Vθ
∂
z
r
+
+
−
+
µ∗
+
∂z
r ∂θ
∂z
r
r ∂θ
∂r
r
=%bθ −
h
%
i
i
1 ∂Vθ
Vr
+
r ∂θ
r
1 ∂Vz
∂Vθ
+
r ∂θ
∂z
i
+
~
+ λ∗ ∇ · V
i
h
∂
∂Vz
~
2µ∗
+ λ∗ ∇ · V
∂z
∂z
i
D
∂( )
∂( )
∂( )
Vθ ∂( )
() =
+ Vr
+
+ Vz
Dt
∂t
∂r
r ∂θ
∂z
~ =
∇·V
1 ∂(rVr )
1 ∂Vθ
∂Vz
+
+
r ∂r
r ∂θ
∂z
Table 5.3 Navier-Stokes equations for compressible fluids in cylindrical coordinates.
(2.5.31b)
292
Observe that for incompressible flow
D%
Dt
= 0 which implies ∇ · V~ = 0. Therefore, the assumptions
of constant viscosity and incompressibility of the flow will simplify the above equations. If on the other
hand the viscosity is temperature dependent and the flow is compressible, then one should add to the above
equations the continuity equation, an energy equation and an equation of state. The energy equation comes
from the first law of thermodynamics applied to a control volume within the fluid and will be considered
in the sections ahead. The equation of state is a relation between thermodynamic variables which is added
so that the number of equations equals the number of unknowns. Such a system of equations is known as
a closed system. An example of an equation of state is the ideal gas law where pressure p is related to gas
density % and temperature T by the relation p = %RT where R is the universal gas constant.
h
Vθ2 + Vφ2
DVρ
%
−
Dt
ρ
= %bρ −
h
h
∂p
∂Vρ
∂
~
2µ∗
+
+ λ∗ ∇ · V
∂ρ
∂ρ
∂ρ
i
+
i
h
1 ∂
∂
µ∗ ρ
ρ ∂θ
∂ρ
Vφ
µ∗ ∂Vρ
∂
1
∂
+ µ∗ ρ
ρ sin θ ∂φ ρ sin θ ∂φ
∂ρ
ρ
h ∂V
∂Vφ
2
∂V
4V
2
2Vθ cot θ
∂
µ∗
ρ
ρ
θ
−
−
−
−
+ ρ cot θ
+
4
ρ
∂ρ
ρ ∂θ
ρ
ρ sin θ ∂φ
ρ
∂ρ
+
h
%
Vφ2 cot θ
DVθ
Vρ Vθ
+
−
Dt
ρ
ρ
h
= %bθ −
h
1 ∂p
∂
∂
µ∗ ρ
+
ρ ∂θ
∂ρ
∂ρ
i
Vθ
ρ
+
µ∗ ∂Vρ
ρ ∂θ
1 ∂ 2µ∗ ∂Vθ
~
+ Vρ + λ∗ ∇ · V
ρ ∂θ
ρ
∂θ
h
V i
∂ µ∗ sin θ ∂
1
µ∗ ∂Vθ
φ
+
+
ρ sin θ ∂φ
ρ
∂θ sin θ
ρ sin θ ∂φ
h
1 ∂V
∂
1 ∂Vφ
Vθ cot θ
µ∗
θ
−
−
+
2 cot θ
+3 ρ
ρ
ρ ∂θ
ρ sin θ ∂φ
ρ
∂ρ
+
%
where
and
h DV
φ
Dt
+
i
h
ρ
Vθ
ρ
V θ
ρ
+
1 ∂Vρ
ρ ∂θ
+
Vθ Vφ cot θ
= %bφ −
i
+
µ∗ ∂Vρ
ρ ∂θ
+
cot θ ∂Vρ
ρ ∂θ
i
i
V i
1 ∂p
∂
µ∗ ∂Vρ
∂
+
+ µ∗ ρ
ρ
ρ
ρ sin θ ∂φ
∂ρ ρ sin θ ∂φ
∂ρ
h
V i
1 ∂ µ∗ sin θ ∂
µ∗ ∂Vθ
φ
+
+
ρ ∂θ
ρ
∂θ sin θ
ρ sin θ ∂φ
h
i
1
1 ∂Vφ
∂ 2µ∗
~
+
+ Vρ + Vθ cot θ + λ∗ ∇ · V
ρ sin θ ∂φ
ρ
sin θ ∂φ
h
V V µ∗
1
3 ∂Vρ
sin θ ∂
∂
φ
φ
+
+ 2 cot θ
+
+ 3ρ
ρ ρ sin θ ∂φ
∂ρ
ρ
ρ ∂θ sin θ
ρ sin θ
Vθ Vφ
i
V θ
φ
ρ
∂Vθ
∂φ
i
Vφ ∂( )
D
∂( )
∂( )
Vθ ∂( )
() =
+ Vρ
+
+
Dt
∂t
∂ρ
ρ ∂θ
ρ sin θ ∂φ
~ =
∇·V
1 ∂(ρ2 Vρ )
1 ∂Vθ sin θ
1 ∂Vφ
+
+
ρ2
∂ρ
ρ sin θ
∂θ
ρ sin θ ∂φ
(2.5.31c)
Table 5.4 Navier-Stokes equations for compressible fluids in spherical coordinates.
293
We now consider various special cases of the Navier-Stokes-Duhem equations.
Special Case 1: Assume that ~b is a conservative force such that ~b = −∇ φ. Also assume that the viscous
v
force terms are zero. Consider steady flow ( ∂~
∂t = 0) and show that equation (2.5.28) reduces to the equation
(~v · ∇) ~v =
−1
∇ p − ∇ φ % is constant.
%
(2.5.32)
Employing the vector identity
1
(~v · ∇) ~v = (∇ × ~v ) × ~v + ∇(~v · ~v ),
2
(2.5.33)
we take the dot product of equation (2.5.32) with the vector ~v . Noting that ~v · [(∇ × ~v ) × ~v ] = ~0 we obtain
1 2
p
+ φ + v = 0.
(2.5.34)
~v · ∇
%
2
This equation shows that for steady flow we will have
1
p
+ φ + v 2 = constant
%
2
(2.5.35)
along a streamline. This result is known as Bernoulli’s theorem. In the special case where φ = gh is a
v2
p
+ gh = constant. This equation is known as
force due to gravity, the equation (2.5.35) reduces to +
%
2
Bernoulli’s equation. It is a conservation of energy statement which has many applications in fluids.
Special Case 2: Assume that ~b = −∇ φ is conservative and define the quantity Ω by
~ = ∇ × ~v = curl ~v
Ω
ω=
1
Ω
2
(2.5.36)
as the vorticity vector associated with the fluid flow and observe that its magnitude is equivalent to twice
the angular velocity of a fluid particle. Then using the identity from equation (2.5.33) we can write the
Navier-Stokes-Duhem equations in terms of the vorticity vector. We obtain the hydrodynamic equations
1
1
1
∂~v ~
+ Ω × ~v + ∇ v 2 = − ∇ p − ∇ φ + f~viscous ,
∂t
2
%
%
(2.5.37)
where f~viscous is defined by equation (2.5.30). In the special case of nonviscous flow this further reduces to
the Euler equation
1
1
∂~v ~
+ Ω × ~v + ∇ v 2 = − ∇ p − ∇ φ.
∂t
2
%
If the density % is a function of the pressure only it is customary to introduce the function
Z p
dP
1
dp
so that ∇P =
∇p = ∇p
P =
%
dp
%
c
then the Euler equation becomes
1
∂~v ~
+ Ω × ~v = −∇(P + φ + v 2 ).
∂t
2
Some examples of vorticies are smoke rings, hurricanes, tornadoes, and some sun spots. You can create
a vortex by letting water stand in a sink and then remove the plug. Watch the water and you will see that
a rotation or vortex begins to occur. Vortices are associated with circulating motion.
294
Pick
I an arbitrary simple closed curve C and place it in the fluid flow and define the line integral
~v · êt ds, where ds is an element of arc length along the curve C, ~v is the vector field defining the
K =
C
velocity, and êt is a unit tangent vector to the curve C. The integral K is called the circulation of the fluid
around the closed curve C. The circulation is the summation of the tangential components of the velocity
field along the curve C. The local vorticity at a point is defined as the limit
lim
Area→0
Circulation around C
= circulation per unit area.
Area inside C
By Stokes theorem, if curl ~v = ~0, then the fluid is called irrotational and the circulation is zero. Otherwise
the fluid is rotational and possesses vorticity.
If we are only interested in the velocity field we can eliminate the pressure by taking the curl of both
sides of the equation (2.5.37). If we further assume that the fluid is incompressible we obtain the special
equations
∇ · ~v = 0
Incompressible fluid, % is constant.
~ = curl ~v
Ω
∗
~
∂Ω
~ × ~v ) = µ ∇2 Ω
~
+ ∇ × (Ω
∂t
%
Definition of vorticity vector.
(2.5.38)
Results because curl of gradient is zero.
Note that when Ω is identically zero, we have irrotational motion and the above equations reduce to the
~ × ~v ) is neglected, then the last equation in
Cauchy-Riemann equations. Note also that if the term ∇ × (Ω
equation (2.5.38) reduces to a diffusion equation. This suggests that the vorticity diffuses through the fluid
once it is created.
Vorticity can be caused by a rigid rotation or by shear flow. For example, in cylindrical coordinates let
~ = ∇×V
~ = 2ω êz , which shows the
~
V = rω êθ , with r, ω constants, denote a rotational motion, then curl V
vorticity is twice the rotation vector. Shear can also produce vorticity. For example, consider the velocity
~ | increases as y increases.
~ = y ê1 with y ≥ 0. Observe that this type of flow produces shear because |V
field V
~ = − ê3 . The right-hand rule tells us that if an imaginary paddle
For this flow field we have curl V~ = ∇ × V
wheel is placed in the flow it would rotate clockwise because of the shear effects.
Scaled Variables
In the Navier-Stokes-Duhem equations for fluid flow we make the assumption that the external body
forces are derivable from a potential function φ and write ~b = −∇ φ [dyne/gm] We also want to write the
Navier-Stokes equations in terms of scaled variables
~v = ~v
v0
p
p=
p0
%
%0
t
t=
τ
%=
φ
,
gL
x
x=
L
φ=
y
L
z
z=
L
y=
which can be referred to as the barred system of dimensionless variables. Dimensionless variables are introduced by scaling each variable associated with a set of equations by an appropriate constant term called a
characteristic constant associated with that variable. Usually the characteristic constants are chosen from
various parameters used in the formulation of the set of equations. The characteristic constants assigned to
each variable are not unique and so problems can be scaled in a variety of ways. The characteristic constants
295
assigned to each variable are scales, of the appropriate dimension, which act as reference quantities which
reflect the order of magnitude changes expected of that variable over a certain range or area of interest
associated with the problem. An inappropriate magnitude selected for a characteristic constant can result
in a scaling where significant information concerning the problem can be lost. This is analogous to selecting
an inappropriate mesh size in a numerical method. The numerical method might give you an answer but
details of the answer might be lost.
In the above scaling of the variables occurring in the Navier-Stokes equations we let v0 denote some
characteristic speed, p0 a characteristic pressure, %0 a characteristic density, L a characteristic length, g the
acceleration of gravity and τ a characteristic time (for example τ = L/v0 ), then the barred variables v, p,
%,φ, t, x, y and z are dimensionless. Define the barred gradient operator by
∇=
∂
∂
∂
ê1 +
ê2 +
ê3
∂x
∂y
∂z
where all derivatives are with respect to the barred variables. The above change of variables reduces the
Navier-Stokes-Duhem equations
%
to the form
∂~v
+ %(~v · ∇) ~v = −%∇φ − ∇ p + (λ∗ + µ∗ )∇ (∇ · ~v ) + µ∗ ∇2 ~v ,
∂t
p % v ∂~v % v 2 0 0
0 0
0
+
%
% ~v · ∇ ~v = −%0 g%∇ φ −
∇p
τ
L
L
∂t
∗ µ v0
(λ∗ + µ∗ )
2
~
v0 ∇ ∇ · v +
∇ ~v.
+
2
2
L
L
(2.5.39)
(2.5.40)
Now if each term in the equation (2.5.40) is divided by the coefficient %0 v02 /L, we obtain the equation
S%
∂~v
−1
+ % ~v · ∇ ~v =
%∇ φ − E∇p +
F
∂t
λ∗
+1
µ∗
1 2
1
∇ ∇ · ~v + ∇ ~v
R
R
(2.5.41)
which has the dimensionless coefficients
p0
= Euler number
%0 v02
v2
F = 0 = Froude number, g is acceleration of gravity
gL
E=
%0 V0 L
= Reynolds number
µ∗
L
= Strouhal number.
S=
τ v0
R=
Dropping the bars over the symbols, we write the dimensionless equation using the above coefficients.
The scaled equation is found to have the form
S%
1
∂~v
+ %(~v · ∇)~v = − %∇φ − E∇p +
∂t
F
λ∗
+1
µ∗
1
1
∇ (∇ · ~v ) + ∇2~v
R
R
(2.5.42)
296
Boundary Conditions
Fluids problems can be classified as internal flows or external flows. An example of an internal flow
problem is that of fluid moving through a converging-diverging nozzle. An example of an external flow
problem is fluid flow around the boundary of an aircraft. For both types of problems there is some sort of
boundary which influences how the fluid behaves. In these types of problems the fluid is assumed to adhere
to a boundary. Let ~rb denote the position vector to a point on a boundary associated with a moving fluid,
and let ~r denote the position vector to a general point in the fluid. Define ~v (~r ) as the velocity of the fluid at
the point ~r and define ~v (~rb ) as the known velocity of the boundary. The boundary might be moving within
the fluid or it could be fixed in which case the velocity at all points on the boundary is zero. We define the
boundary condition associated with a moving fluid as an adherence boundary condition.
Definition: (Adherence Boundary Condition)
An adherence boundary condition associated with a fluid in motion
is defined as the limit lim ~v (~r) = ~v (~rb ) where ~rb is the position
~
r →~
rb
vector to a point on the boundary.
Sometimes, when no finite boundaries are present, it is necessary to impose conditions on the components
of the velocity far from the origin. Such conditions are referred to as boundary conditions at infinity.
Summary and Additional Considerations
Throughout the development of the basic equations of continuum mechanics we have neglected thermodynamical and electromagnetic effects. The inclusion of thermodynamics and electromagnetic fields adds
additional terms to the basic equations of a continua. These basic equations describing a continuum are:
Conservation of mass
The conservation of mass is a statement that the total mass of a body is unchanged during its motion.
This is represented by the continuity equation
∂%
+ (%v k ),k = 0 or
∂t
D%
~ =0
+ %∇ · V
Dt
where % is the mass density and v k is the velocity.
Conservation of linear momentum
The conservation of linear momentum requires that the time rate of change of linear momentum equal
the resultant of all forces acting on the body. In symbols, we write
D
Dt
where
Dv i
Dt
=
∂v i
∂t
+
∂v i
∂xk
Z
Z
Z
%v i dτ =
V
S
i
F(s)
ni dS +
V
i
%F(b)
dτ +
n
X
i
F(α)
(2.5.43)
α=1
i
i
v k is the material derivative, F(s)
are the surface forces per unit area, F(b)
are the
i
represents isolated external forces. Here S represents the surface and
body forces per unit mass and F(α)
V represents the volume of the control volume. The right-hand side of this conservation law represents the
resultant force coming from the consideration of all surface forces and body forces acting on a control volume.
297
Surface forces acting upon the control volume involve such things as pressures and viscous forces, while body
forces are due to such things as gravitational, magnetic and electric fields.
Conservation of angular momentum
The conservation of angular momentum states that the time rate of change of angular momentum
(moment of linear momentum) must equal the total moment of all forces and couples acting upon the body.
In symbols,
D
Dt
Z
Z
j k
V
%eijk x v dτ =
Z
j
S
eijk x
k
F(s)
j
dS +
V
%eijk x
k
F(b)
dτ +
n
X
k
i
(eijk xj(α) F(α)
+ M(α)
)
(2.5.44)
α=1
i
k
represents concentrated couples and F(α)
represents isolated forces.
where M(α)
Conservation of energy
The conservation of energy law requires that the time rate of change of kinetic energy plus internal
energies is equal to the sum of the rate of work from all forces and couples plus a summation of all external
energies that enter or leave a control volume per unit of time. The energy equation results from the first law
of thermodynamics and can be written
D
(E + K) = Ẇ + Q̇h
Dt
(2.5.45)
where E is the internal energy, K is the kinetic energy, Ẇ is the rate of work associated with surface and
Z
body forces, and Q̇h is the input heat rate from surface and internal effects.
%e dτ represents
Let e denote the internal specific energy density within a control volume, then E =
V
the total
Z internal energy of the control volume. The kinetic energy of the control volume is expressed as
1
%gij v i v j dτ where v i is the velocity, % is the density and dτ is a volume element. The energy (rate
K=
2 V
of work) associated with the body and surface forces is represented
Z
Z
Ẇ =
S
i
gij F(s)
v j dS +
i
%gij F(b)
v j dτ +
V
n
X
i
i
(gij F(α)
v j + gij M(α)
ωj )
α=1
i
i
are isolated forces, and M(α)
are isolated couples.
where ω j is the angular velocity of the point xi(α) , F(α)
Two external energy sources due to thermal sources are heat flow q i and rate of internal heat production
∂Q
∂t
per unit volume. The conservation of energy can thus be represented
D
Dt
Z
1
%(e + gij v i v j ) dτ =
2
V
Z
S
+
Z
i
(gij F(s)
v j − qi ni ) dS +
n
X
V
i
(%gij F(b)
vj +
∂Q
) dτ
∂t
(2.5.46)
i
i
(gij F(α)
v j + gij M(α)
ω j + U(α) )
α=1
where U(α) represents all other energies resulting from thermal, mechanical, electric, magnetic or chemical
sources which influx the control volume and D/Dt is the material derivative.
In equation (2.5.46) the left hand side is the material derivative of an integral of the total energy
et = %(e + 12 gij v i v j ) over the control volume. Material derivatives are not like ordinary derivatives and so
298
we cannot interchange the order of differentiation and integration in this term. Here we must use the result
Z Z
that
D
Dt
V
et dτ =
V
∂et
~
+ ∇ · (et V ) dτ.
∂t
To prove this result we consider a more general problem. Let A denote the amount of some quantity per
unit mass. The quantity A can be a scalar, vector or tensor. The total amount of this quantity inside the
R
control volume is A = V %A dτ and therefore the rate of change of this quantity is
∂A
=
∂t
Z
D
∂(%A)
dτ =
∂t
Dt
V
Z
V
Z
%A dτ −
S
~ · n̂ dS,
%AV
which represents the rate of change of material within the control volume plus the influx into the control
volume. The minus sign is because n̂ is always a unit outward normal. By converting the surface integral to
a volume integral, by the Gauss divergence theorem, and rearranging terms we find that
Z
Z ∂(%A)
D
~
+ ∇ · (%AV ) dτ.
%A dτ =
Dt V
∂t
V
i
i
= σ ij nj , F(b)
= bi where
In equation (2.5.46) we neglect all isolated external forces and substitute F(s)
σij = −pδij + τij . We then replace all surface integrals by volume integrals and find that the conservation of
energy can be represented in the form
∂et
~ ) = ∇(σ · V~ ) − ∇ · ~q + %~b · V
~ + ∂Q
+ ∇ · (et V
(2.5.47)
∂t
∂t
P3 P3
where et = %e + %(v12 + v22 + v32 )/2 is the total energy and σ = i=1 j=1 σij êi êj is the second order stress
tensor. Here
σ · V~
~ +
= −pV
3
X
j=1
∗
τ1j vj ê1 +
3
X
τ2j vj ê2 +
j=1
3
X
~ +τ ·V
~
τ3j vj ê3 = −pV
j=1
∗
and τij = µ (vi,j + vj,i ) + λ δij vk,k is the viscous stress tensor. Using the identities
%
∂et
D(et /%)
=
+ ∇ · (et V~ )
Dt
∂t
and %
De
D(V 2 /2)
D(et /%)
=%
+%
Dt
Dt
Dt
~ as
together with the momentum equation (2.5.25) dotted with V
%
DV~ ~
~ − ∇p · V~ + (∇ · τ ) · V
~
· V = %~b · V
Dt
the energy equation (2.5.47) can then be represented in the form
%
De
~ ) = −∇ · ~q + ∂Q + Φ
+ p(∇ · V
Dt
∂t
(2.5.48)
where Φ is the dissipation function and can be represented
~.
Φ = (τij vi ) ,j − vi τij,j = ∇ · (τ · V~ ) − (∇ · τ ) · V
As an exercise it can be shown that the dissipation function can also be represented as Φ = 2µ∗ Dij Dij +λ∗ Θ2
where Θ is the dilatation. The heat flow vector is determined from the Fourier law of heat conduction in
299
terms of the temperature T as ~
q = −κ∇ T , where κ is the thermal conductivity. Consequently, the energy
equation can be written as
%
De
~ ) = ∂Q + Φ + ∇(k∇T ).
+ p(∇ · V
Dt
∂t
(2.5.49)
In Cartesian coordinates (x, y, z) we use
∂
∂
∂
∂
D
= + Vx
+ Vy
+ Vz
Dt ∂t
∂x
∂y
∂z
∂V
∂V
∂V
x
y
z
~ =
+
+
∇·V
∂x
∂y
∂z
∂T
∂
∂T
∂
∂T
∂
κ
+
κ
+
κ
∇ · (κ∇T ) =
∂x
∂x
∂y
∂y
∂z
∂z
In cylindrical coordinates (r, θ, z)
∂
Vθ ∂
∂
∂
D
= + Vr
+
+ Vz
Dt ∂t
∂r
r ∂θ
∂z
∂
∂V
1
∂V
1
θ
z
~ =
(rVr ) + 2
+
∇·V
r ∂r r
∂θ
∂z
1 ∂
∂T
1 ∂
∂T
∂
∂T
∇ · (κ∇T ) =
rκ
+ 2
κ
+
κ
r ∂r
∂r
r ∂θ
∂θ
∂z
∂z
and in spherical coordinates (ρ, θ, φ)
∂
Vθ ∂ Vφ ∂
∂
D
= + Vρ
+
Dt ∂t
∂ρ
ρ ∂θ ρ sin θ ∂φ
∂
1
1 ∂Vφ
∂
1
(ρVρ ) +
(Vθ sin θ) +
∇ · V~ = 2
ρ ∂ρ
ρ sin θ ∂θ
ρ sin θ ∂φ
∂
∂
∂T
1
∂T
1
∂T
1 ∂
2
ρ κ
+ 2
κ sin θ
+ 2 2
κ
∇ · (κ∇T ) = 2
ρ ∂ρ
∂ρ
ρ sin θ ∂θ
∂θ
∂φ
ρ sin θ ∂φ
The combination of terms h = e + p/% is known as enthalpy and at times is used to express the energy
equation in the form
%
D p ∂Q
Dh
=
+
− ∇ · ~q + Φ.
Dt
Dt
∂t
The derivation of this equation is left as an exercise.
Conservative Systems
Let Q denote some physical quantity per unit volume. Here Q can be either a scalar, vector or tensor
field. Place within this field an imaginary simple closed surface S which encloses a volume V. The total
RRR
amount of Q within the surface is given by
V Q dτ and the rate of change of this amount with respect
RRR
∂
Q dτ. The total amount of Q within S changes due to sources (or sinks) within the volume
to time is ∂t
~ called current, which represents a
and by transport processes. Transport processes introduce a quantity J,
RR
flow per unit area across the surface S. The inward flux of material into the volume is denoted S −J~ · n̂ dσ
(n̂ is a unit outward normal.) The sources (or sinks) SQ denotes a generation (or loss) of material per unit
RRR
S dτ denotes addition (or loss) of material to the volume. For a fixed volume we then
volume so that
V Q
have the material balance
ZZZ
V
∂Q
dτ = −
∂t
ZZ
ZZZ
J~ · n̂ dσ +
S
SQ dτ.
V
300
Using the divergence theorem of Gauss one can derive the general conservation law
∂Q
+ ∇ · J~ = SQ
∂t
(2.5.50)
The continuity equation and energy equations are examples of a scalar conservation law in the special case
where SQ = 0. In Cartesian coordinates, we can represent the continuity equation by letting
Q=%
~ = %(Vx ê1 + Vy ê2 + Vz ê3 )
and J~ = %V
(2.5.51)
The energy equation conservation law is represented by selecting Q = et and neglecting the rate of internal
heat energy we let
"
J~ = (et + p)v1 −
3
X
#
vi τxi + qx ê1 +
i=1
"
(et + p)v2 −
3
X
#
"
(et + p)v3 −
3
X
(2.5.52)
vi τyi + qy ê2 +
i=1
#
vi τzi + qz
ê3 .
i=1
In a general orthogonal system of coordinates (x1 , x2 , x3 ) the equation (2.5.50) is written
∂
∂
∂
∂
((h1 h2 h3 Q)) +
((h2 h3 J1 )) +
((h1 h3 J2 )) +
((h1 h2 J3 )) = 0,
∂t
∂x1
∂x2
∂x3
where h1 , h2 , h3 are scale factors obtained from the transformation equations to the general orthogonal
coordinates.
The momentum equations are examples of a vector conservation law having the form
∂~a
+ ∇ · (T ) = %~b
∂t
where ~a is a vector and T is a second order symmetric tensor T =
(2.5.53)
3
3 X
X
Tjk êj êk . In Cartesian coordinates
k=1 j=1
we let ~a = %(Vx ê1 + Vy ê2 + Vz ê3 ) and Tij = %vi vj + pδij − τij . In general coordinates (x1 , x2 , x3 ) the
~ and Tij = %vi vj + pδij − τij . In a general orthogonal system
momentum equations result by selecting ~a = %V
the conservation law (2.5.53) has the general form
∂ ∂ ∂ ∂
((h1 h2 h3~a)) +
(h2 h3 T · ê1 ) +
(h1 h3 T · ê2 ) +
(h1 h2 T · ê3 ) = %~b.
∂t
∂x1
∂x2
∂x3
(2.5.54)
Neglecting body forces and internal heat production, the continuity, momentum and energy equations
can be expressed in the strong conservative form
where
∂E
∂F
∂G
∂U
+
+
+
=0
∂t
∂x
∂y
∂z
(2.5.55)

ρ
 ρVx 


U =  ρVy 


ρVz
et
(2.5.56)

301


ρVx
+ p − τxx




E=
ρVx Vy − τxy



ρVx Vz − τxz
(et + p)Vx − Vx τxx − Vy τxy − Vz τxz + qx


ρVy
ρVx Vy − τxy




ρVy2 + p − τyy
F =



ρVy Vz − τyz
(et + p)Vy − Vx τyx − Vy τyy − Vz τyz + qy


ρVz
ρVx Vz − τxz




ρVy Vz − τyz
G=



2
ρVz + p − τzz
(et + p)Vz − Vx τzx − Vy τzy − Vz τzz + qz
ρVx2
(2.5.57)
(2.5.58)
(2.5.59)
where the shear stresses are τij = µ∗ (Vi,j + Vj,i ) + δij λ∗ Vk,k for i, j, k = 1, 2, 3.
Computational Coordinates
To transform the conservative system (2.5.55) from a physical (x, y, z) domain to a computational (ξ, η, ζ)
domain requires that a general change of variables take place. Consider the following general transformation
of the independent variables
ξ = ξ(x, y, z)
η = η(x, y, z)
ζ = ζ(x, y, z)
(2.5.60)
with Jacobian different from zero. The chain rule for changing variables in equation (2.5.55) requires the
operators
∂( )
∂( )
∂( ) ∂( )
=
ξx +
ηx +
ζx
∂x
∂ξ
∂η
∂ζ
∂( )
∂( )
∂( ) ∂( )
=
ξy +
ηy +
ζy
∂y
∂ξ
∂η
∂ζ
∂( )
∂( )
∂( ) ∂( )
=
ξz +
ηz +
ζz
∂z
∂ξ
∂η
∂ζ
(2.5.61)
The partial derivatives in these equations occur in the differential expressions

dξ =ξx dx + ξy dy + ξz dz
dη =ηx dx + ηy dy + ηz dz
or
dζ =ζx dx + ζy dy + ζz dz
 
dξ
ξx
 dη  =  ηx
dζ
ζx
ξy
ηy
ζy
 
ξz
dx
ηz   dy 
ζz
dz
(2.5.62)
In a similar mannaer from the inverse transformation equations
x = x(ξ, η, ζ)
y = y(ξ, η, ζ)
z = z(ξ, η, ζ)
(2.5.63)
we can write the differentials

dx =xξ dξ + xη dη + xζ dζ
dy =yξ dξ + yη dη + yζ dζ
dz =zξ dξ + zζ dζ + zζ dζ
or
 
dx
xξ
 dy  =  yξ
dz
zξ
xη
yη
zη
 
xζ
dξ
yζ   dη 
zζ
dζ
(2.5.64)
302
The transformations (2.5.62) and (2.5.64) are inverses of each other and so we can write

ξx
 ηx
ζx
ξy
ηy
ζy
 
ξz
xξ
ηz  =  yξ
ζz
zξ

xη
yη
zη
−1
xζ
yζ 
zζ

−(xη zζ − xζ zη )
xη yζ − xζ yη
yη zζ − yζ zη
=J  −(yξ zζ − yζ zξ )
xξ zζ − xζ zξ
−(xξ yζ − xζ yξ ) 
−(xξ zη − xη zξ )
xξ yη − xη yξ
yξ zη − yη zξ
(2.5.65)
By comparing like elements in equation (2.5.65) we obtain the relations
ξx =J(yη zζ − yζ zη )
ηx = − J(yξ zζ − yζ zξ )
ζx =J(yξ zη − yη zξ )
ξy = − J(xη zζ − xζ zη )
ηy =J(xξ zζ − zζ zξ )
ζy = − J(xξ zη − xη zξ )
ξz =J(xη yζ − xζ yη )
ηz = − J(xξ yζ − xζ yξ )
ζz =J(xξ yη − xη yξ )
(2.5.66)
The equations (2.5.55) can now be written in terms of the new variables (ξ, η, ζ) as
∂E
∂E
∂E
∂F
∂F
∂F
∂G
∂G
∂G
∂U
+
ξx +
ηx +
ζx +
ξy +
ηy +
ζy +
ξz +
ηz +
ζz = 0
∂t
∂ξ
∂η
∂ζ
∂ξ
∂η
∂ζ
∂ξ
∂η
∂ζ
(2.5.67)
Now divide each term by the Jacobian J and write the equation (2.5.67) in the form
∂
∂t
U
J
∂ Eξx + F ξy + Gξz
∂ξ
J
∂ Eηx + F ηy + Gηz
+
∂η
J
∂ Eζx + F ζy + Gζz
+
∂ζ
J
∂ ζx
∂ ξx
∂ ηx +
+
−E
∂ξ J
∂η J
∂ζ J
∂ ζy
∂ ξy
∂ ηy
+
+
−F
∂ξ J
∂η J
∂ζ J
∂ ζz
∂ ξz
∂ ηz +
+
=0
−G
∂ξ J
∂η J
∂ζ J
+
(2.5.68)
Using the relations given in equation (2.5.66) one can show that the curly bracketed terms above are all zero
and so the transformed equations (2.5.55) can also be written in the conservative form
where
b ∂ Fb ∂ G
b
b
∂E
∂U
+
+
+
=0
∂t
∂ξ
∂η
∂ζ
(2.5.69)
b =U
U
J
Eξx + F ξy + Gξz
b
E=
J
Eη
+
F
ηy + Gηz
x
Fb =
J
+
F
ζy + Gζz
Eζ
x
b=
G
J
(2.5.70)
303
Fourier law of heat conduction
The Fourier law of heat conduction can be written qi = −κT,i for isotropic material and qi = −κij T,j
for anisotropic material. The Prandtl number is a nondimensional constant defined as P r =
cp µ∗
κ
so that
the heat flow terms can be represented in Cartesian coordinates as
qx = −
cp µ∗ ∂T
P r ∂x
qy = −
cp µ∗ ∂T
P r ∂y
qz = −
cp µ∗ ∂T
P r ∂z
Now one can employ the equation of state relations P = %e(γ − 1), cp =
above equations in the alternate forms
∂ γP
∂ γP
µ∗
µ∗
qy = −
qx = −
P r(γ − 1) ∂x
%
P r(γ − 1) ∂y
%
s
The speed of sound is given by a =
γR
γ−1
, cp T =
γRT
γ−1
∂
µ∗
qz = −
P r(γ − 1) ∂z
and write the
γP
%
p
γP
γP
= γRT and so one can substitute a2 in place of the ratio
%
%
in the above equations.
Equilibrium and Nonequilibrium Thermodynamics
High temperature gas flows require special considerations. In particular, the specific heat for monotonic
and diatomic gases are different and are in general a function of temperature. The energy of a gas can be
written as e = et + er + ev + ee + en where et represents translational energy, er is rotational energy, ev is
vibrational energy, ee is electronic energy, and en is nuclear energy. The gases follow a Boltzmann distribution
for each degree of freedom and consequently at very high temperatures the rotational, translational and
vibrational degrees of freedom can each have their own temperature. Under these conditions the gas is said
to be in a state of nonequilibrium. In such a situation one needs additional energy equations. The energy
equation developed in these notes is for equilibrium thermodynamics where the rotational, translational and
vibrational temperatures are the same.
Equation of state
It is assumed that an equation of state such as the universal gas law or perfect gas law pV = nRT
holds which relates pressure p [N/m2 ], volume V [m3 ], amount of gas n [mol],and temperature T [K] where
R [J/mol − K] is the universal molar gas constant. If the ideal gas law is represented in the form p = %RT
where % [Kg/m3 ] is the gas density, then the universal gas constant must be expressed in units of [J/Kg − K]
(See Appendix A). Many gases deviate from this ideal behavior. In order to account for the intermolecular
forces associated with high density gases, an empirical equation of state of the form
p = ρRT +
M1
X
2
βn ρn+r1 + e−γ1 ρ−γ2 ρ
n=1
M2
X
cn ρn+r2
n=1
involving constants M1 , M2 , βn , cn , r1 , r2 , γ1 , γ2 is often used. For a perfect gas the relations
e = cv T
γ=
cp
cv
cv =
R
γ−1
cp =
γR
γ−1
h = cp T
hold, where R is the universal gas constant, cv is the specific heat at constant volume, cp is the specific
heat at constant pressure, γ is the ratio of specific heats and h is the enthalpy. For cv and cp constants the
relations p = (γ − 1)%e and RT = (γ − 1)e can be verified.
304
EXAMPLE 2.5-1. (One-dimensional fluid flow)
Construct an x-axis running along the center line of a long cylinder with cross sectional area A. Consider
the motion of a gas driven by a piston and moving with velocity v1 = u in the x-direction. From an Eulerian
point of view we imagine a control volume fixed within the cylinder and assume zero body forces. We require
the following equations be satisfied.
∂%
~ ) = 0 which in one-dimension reduces to ∂% + ∂ (%u) = 0.
+ div(%V
Conservation of mass
∂t
∂t
∂x
∂p
∂
∂
(%u) +
%u2 +
= 0.
Conservation of momentum, equation (2.5.28) reduces to
∂t
∂x
∂x
Conservation of energy, equation
(2.5.48) in the absence of heat flow and internal heat production,
∂e
∂u
∂e
+u
+p
= 0. Using the conservation of mass relation this
becomes in one dimension %
∂t
∂x
∂x
∂
∂u
∂
(%e) +
(%eu) + p
= 0.
equation can be written in the form
∂t
∂x
∂x
In contrast, from a Lagrangian point of view we let the control volume move with the flow and consider
advection terms. This gives the following three equations which can then be compared with the above
Eulerian equations of motion.
D%
∂u
d
+%
= 0.
Conservation of mass (%J) = 0 which in one-dimension is equivalent to
dt
Dt
∂x
Du ∂p
+
= 0.
Conservation of momentum, equation (2.5.25) in one-dimension %
Dt
∂x
∂u
De
+p
= 0. In the above equations
Conservation of energy, equation (2.5.48) in one-dimension %
Dt
∂x
D( )
∂
∂
Dt = ∂t ( ) + u ∂x ( ). The Lagrangian viewpoint gives three equations in the three unknowns ρ, u, e.
In both the Eulerian and Lagrangian equations the pressure p represents the total pressure p = pg + pv
where pg is the gas pressure and pv is the viscous pressure which causes loss of kinetic energy. The gas pressure
c
is a function of %, e and is determined from the ideal gas law pg = %RT = %(cp − cv )T = %( cpv − 1)cv T or
pg = %(γ − 1)e. Some kind of assumption is usually made to represent the viscous pressure pv as a function
of e, u. The above equations are then subjected to boundary and initial conditions and are usually solved
numerically.
Entropy inequality
Energy transfer is not always reversible. Many energy transfer processes are irreversible. The second
law of thermodynamics allows energy transfer to be reversible only in special circumstances. In general,
the second law of thermodynamics can be written as an entropy inequality, known as the Clausius-Duhem
inequality. This inequality states that the time rate of change of the total entropy is greater than or equal to
the total entropy change occurring across the surface and within the body of a control volume. The ClausiusDuhem inequality places restrictions on the constitutive equations. This inequality can be expressed in the
form
Z
Z
Z
n
X
D
i
%s dτ
s ni dS +
ρb dτ +
B(α)
≥
Dt V
S
V
α=1
{z
}
|
|
{z
}
Rate of entropy increase Entropy input rate into control volume
where s is the specific entropy density, si is an entropy flux, b is an entropy source and B(α) are isolated
entropy sources. Irreversible processes are characterized by the use of the inequality sign while for reversible
305
Figure 2.5-3. Interaction of various fields.
processes the equality sign holds. The Clausius-Duhem inequality is assumed to hold for all independent
thermodynamical processes.
If in addition there are electric and magnetic fields to consider, then these fields place additional forces
upon the material continuum and we must add all forces and moments due to these effects. In particular we
must add the following equations
Gauss’s law for magnetism
Gauss’s law for electricity
1 ∂ √ i
( gB ) = 0.
√
g ∂xi
~ = %e
∇·D
1 ∂ √ i
( gD ) = %e .
√
g ∂xi
~ =−
∇×E
Faraday’s law
Ampere’s law
~ =0
∇·B
~
∂B
∂t
~
~ = J~ + ∂ D
∇×H
∂t
ijk Ek,j = −
∂B i
.
∂t
ijk Hk,j = J i +
∂Di
.
∂t
where %e is the charge density, J i is the current density, Di = ji Ej + Pi is the electric displacement vector,
Hi is the magnetic field, Bi = µji Hj + Mi is the magnetic induction, Ei is the electric field, Mi is the
magnetization vector and Pi is the polarization vector. Taking the divergence of Ampere’s law produces the
law of conservation of charge which requires that
∂%e
+ ∇ · J~ = 0
∂t
∂%e
1 ∂ √ i
+√
( gJ ) = 0.
∂t
g ∂xi
The figure 2.5-3 is constructed to suggest some of the interactions that can occur between various
variables which define the continuum. Pyroelectric effects occur when a change in temperature causes
changes in the electrical properties of a material. Temperature changes can also change the mechanical
properties of materials. Similarly, piezoelectric effects occur when a change in either stress or strain causes
changes in the electrical properties of materials. Photoelectric effects are said to occur if changes in electric
or mechanical properties effect the refractive index of a material. Such changes can be studied by modifying
the constitutive equations to include the effects being considered.
From figure 2.5-3 we see that there can exist a relationship between the displacement field Di and
electric field Ei . When this relationship is linear we can write Di = ji Ej and Ej = βjn Dn , where ji are
306
dielectric constants and βjn are dielectric impermabilities. Similarly, when linear piezoelectric effects exist
we can write linear relations between stress and electric fields such as σij = −gkij Ek and Ei = −eijk σjk ,
where gkij and eijk are called piezoelectric constants. If there is a linear relation between strain and an
electric fields, this is another type of piezoelectric effect whereby eij = dijk Ek and Ek = −hijk ejk , where
dijk and hijk are another set of piezoelectric constants. Similarly, entropy changes can cause pyroelectric
effects. Piezooptical effects (photoelasticity) occurs when mechanical stresses change the optical properties of
the material. Electrical and heat effects can also change the optical properties of materials. Piezoresistivity
occurs when mechanical stresses change the electric resistivity of materials. Electric field changes can cause
variations in temperature, another pyroelectric effect. When temperature effects the entropy of a material
this is known as a heat capacity effect. When stresses effect the entropy in a material this is called a
piezocaloric effect. Some examples of the representation of these additional effects are as follows. The
piezoelectric effects are represented by equations of the form
σij = −hmij Dm
Di = dijk σjk
eij = gkij Dk
Di = eijk ejk
where hmij , dijk , gkij and eijk are piezoelectric constants.
Knowledge of the material or electric interaction can be used to help modify the constitutive equations.
For example, the constitutive equations can be modified to included temperature effects by expressing the
constitutive equations in the form
σij = cijkl ekl − βij ∆T
and
eij = sijkl σkl + αij ∆T
where for isotropic materials the coefficients αij and βij are constants. As another example, if the strain is
modified by both temperature and an electric field, then the constitutive equations would take on the form
eij = sijkl σkl + αij ∆T + dmij Em .
Note that these additional effects are additive under conditions of small changes. That is, we may use the
principal of superposition to calculate these additive effects.
If the electric field and electric displacement are replaced by a magnetic field and magnetic flux, then
piezomagnetic relations can be found to exist between the variables involved. One should consult a handbook
to determine the order of magnitude of the various piezoelectric and piezomagnetic effects. For a large
majority of materials these effects are small and can be neglected when the field strengths are weak.
The Boltzmann Transport Equation
The modeling of the transport of particle beams through matter, such as the motion of energetic protons
or neutrons through bulk material, can be approached using ideas from the classical kinetic theory of gases.
Kinetic theory is widely used to explain phenomena in such areas as: statistical mechanics, fluids, plasma
physics, biological response to high-energy radiation, high-energy ion transport and various types of radiation
shielding. The problem is basically one of describing the behavior of a system of interacting particles and their
distribution in space, time and energy. The average particle behavior can be described by the Boltzmann
equation which is essentially a continuity equation in a six-dimensional phase space (x, y, z, Vx , Vy , Vz ). We
307
will be interested in examining how the particles in a volume element of phase space change with time. We
introduce the following notation:
(i) ~r the position vector of a typical particle of phase space and dτ = dxdydz the corresponding spatial
volume element at this position.
~ the velocity vector associated with a typical particle of phase space and dτv = dVx dVy dVz the
(ii) V
corresponding velocity volume element.
~ a unit vector in the direction of the velocity V
~ = v Ω.
~
(iii) Ω
(iv) E = 12 mv 2 kinetic energy of particle.
~ is a solid angle about the direction Ω
~ and dτ dE dΩ
~ is a volume element of phase space involving the
(v) dΩ
solid angle about the direction Ω.
~ t) the number of particles in phase space per unit volume at position ~r per unit velocity
(vi) n = n(~r, E, Ω,
~ at time t and N = N (~r, E, Ω,
~ t) = vn(~r, E, Ω,
~ t)
at position V~ per unit energy in the solid angle dΩ
~ at time t. The quantity
the number of particles per unit volume per unit energy in the solid angle dΩ
~ t)dτ dE dΩ
~ represents the number of particles in a volume element around the position ~r with
N (~r, E, Ω,
~ in the solid angle dΩ
~ at time t.
energy between E and E + dE having direction Ω
~ t) = vN (~r, E, Ω,
~ t) is the particle flux (number of particles/cm2 − Mev − sec).
(vii) φ(~r , E, Ω,
~ 0 → Ω)
~ a scattering cross-section which represents the fraction of particles with energy E 0
(viii) Σ(E 0 → E, Ω
~ 0 that scatter into the energy range between E and E + dE having direction Ω
~ in the
and direction Ω
~ per particle flux.
solid angle dΩ
(ix) Σs (E, ~r) fractional number of particles scattered out of volume element of phase space per unit volume
per flux.
(x) Σa (E, ~r) fractional number of particles absorbed in a unit volume of phase space per unit volume per
flux.
Consider a particle at time t having a position ~r in phase space as illustrated in the figure 2.5-4. This
~ in a direction Ω
~ and has an energy E. In terms of dτ = dx dy dz, Ω
~ and E an
particle has a velocity V
~ where dΩ
~ = dΩ(θ,
~
element of volume of phase space can be denoted dτ dEdΩ,
ψ) = sin θdθdψ is a solid angle
~
about the direction Ω.
The Boltzmann transport equation represents the rate of change of particle density in a volume element
~ of phase space and is written
dτ dE dΩ
d
~ t) dτ dE dΩ
~ = DC N (~r, E, Ω,
~ t)
N (~r, E, Ω,
dt
(2.5.71)
where DC is a collision operator representing gains and losses of particles to the volume element of phase
space due to scattering and absorption processes. The gains to the volume element are due to any sources
~ t) per unit volume of phase space, with units of number of particles/sec per volume of phase space,
S(~r, E, Ω,
together with any scattering of particles into the volume element of phase space. That is particles entering
the volume element of phase space with energy E, which experience a collision, leave with some energy
E − ∆E and thus will be lost from our volume element. Particles entering with energies E 0 > E may,
308
Figure 2.5-4. Volume element and solid angle about position ~r.
depending upon the cross-sections, exit with energy E 0 − ∆E = E and thus will contribute a gain to the
volume element. In terms of the flux φ the gains due to scattering into the volume element are denoted by
Z
Z
~ 0 → Ω)φ(~
~
~ t) dτ dE dΩ
~
~ 0 dE 0 Σ(E 0 → E, Ω
r , E 0 , Ω,
dΩ
and represents the particles at position ~r experiencing a scattering collision with a particle of energy E 0 and
~ in dΩ.
~
~ 0 which causes the particle to end up with energy between E and E + dE and direction Ω
direction Ω
The summations are over all possible initial energies.
In terms of φ the losses are due to those particles leaving the volume element because of scattering and
are
~ t)dτ dE dΩ.
~
Σs (E, ~r)φ(~r , E, Ω,
The particles which are lost due to absorption processes are
~ t) dτ dE dΩ.
~
Σa (E, ~r)φ(~r , E, Ω,
The total change to the number of particles in an element of phase space per unit of time is obtained by
summing all gains and losses. This total change is
Z
Z
dN
~ 0 dE 0 Σ(E 0 → E, Ω
~ 0 → Ω)φ(~
~
~ t) dτ dE dΩ
~
dτ dE dΩ = dΩ
r , E 0 , Ω,
dt
~ t)dτ dE dΩ
− Σs (E, ~r)φ(~r , E, Ω,
~ t) dτ dE dΩ
~
− Σa (E, ~r)φ(~r , E, Ω,
~ t)dτ dE dΩ.
~
+ S(~r, E, Ω,
The rate of change
dN
dt
on the left-hand side of equation (2.5.72) expands to
∂N ∂N dx ∂N dy ∂N dz
dN
=
+
+
+
dt
∂t ∂x dt
∂y dt
∂z dt
∂N dVy
∂N dVz
∂N dVx
+
+
+
∂Vx dt
∂Vy dt
∂Vz dt
(2.5.72)
309
which can be written as
where
~
dV
dt
=
~
F
m
~
∂N
dN
~ · ∇~r N + F · ∇ ~ N
=
+V
V
dt
∂t
m
(2.5.73)
represents any forces acting upon the particles. The Boltzmann equation can then be
expressed as
~
∂N
~ · ∇~r N + F · ∇ ~ N = Gains − Losses.
+V
V
∂t
m
(2.5.74)
If the right-hand side of the equation (2.5.74) is zero, the equation is known as the Liouville equation. In
the special case where the velocities are constant and do not change with time the above equation (2.5.74)
can be written in terms of the flux φ and has the form
1 ∂
~
~ t) = DC φ
+ Ω · ∇~r + Σs (E, ~r) + Σa (E, ~r) φ(~r , E, Ω,
v ∂t
Z
where
DC φ =
~0
dΩ
Z
(2.5.75)
~ 0 → Ω)φ(~
~
~ 0 , t) + S(~r, E, Ω,
~ t).
r, E 0, Ω
dE 0 Σ(E 0 → E, Ω
The above equation represents the Boltzmann transport equation in the case where all the particles are
the same. In the case of atomic collisions of particles one must take into consideration the generation of
secondary particles resulting from the collisions.
Let there be a number of particles of type j in a volume element of phase space. For example j = p
(protons) and j = n (neutrons). We consider steady state conditions and define the quantities
~ as the flux of the particles of type j.
(i) φj (~r , E, Ω)
~ Ω
~ 0 , E, E 0 ) the collision cross-section representing processes where particles of type k moving in
(ii) σjk (Ω,
~ with energy E.
~ 0 with energy E 0 produce a type j particle moving in the direction Ω
direction Ω
(iii) σj (E) = Σs (E, ~r) + Σa (E, ~r) the cross-section for type j particles.
The steady state form of the equation (2.5.64) can then be written as
~
~
~ · ∇φj (~r, E, Ω)+σ
r , E, Ω)
Ω
j (E)φj (~
Z
X
~ Ω
~ 0 , E, E 0 )φk (~r, E 0 , Ω
~ 0 )dΩ
~ 0 dE 0
σjk (Ω,
=
(2.5.76)
k
where the summation is over all particles k 6= j.
The Boltzmann transport equation can be represented in many different forms. These various forms
are dependent upon the assumptions made during the derivation, the type of particles, and collision crosssections. In general the collision cross-sections are dependent upon three components.
(1) Elastic collisions. Here the nucleus is not excited by the collision but energy is transferred by projectile
recoil.
(2) Inelastic collisions. Here some particles are raised to a higher energy state but the excitation energy is
not sufficient to produce any particle emissions due to the collision.
(3) Non-elastic collisions. Here the nucleus is left in an excited state due to the collision processes and
some of its nucleons (protons or neutrons) are ejected. The remaining nucleons interact to form a stable
structure and usually produce a distribution of low energy particles which is isotropic in character.
310
Various assumptions can be made concerning the particle flux. The resulting form of Boltzmann’s
equation must be modified to reflect these additional assumptions. As an example, we consider modifications
to Boltzmann’s equation in order to describe the motion of a massive ion moving into a region filled with a
homogeneous material. Here it is assumed that the mean-free path for nuclear collisions is large in comparison
with the mean-free path for ion interaction with electrons. In addition, the following assumptions are made
(i) All collision interactions are non-elastic.
(ii) The secondary particles produced have the same direction as the original particle. This is called the
straight-ahead approximation.
(iii) Secondary particles never have kinetic energies greater than the original projectile that produced them.
(iv) A charged particle will eventually transfer all of its kinetic energy and stop in the media. This stopping
distance is called the range of the projectile. The stopping power Sj (E) =
dE
dx
represents the energy
loss per unit length traveled in the media and determines the range by the relation dEj = Sj 1(E) or
RE
0
1
Rj (E) = 0 SjdE
(E 0 ) . Using the above assumptions Wilson, et.al. show that the steady state linearized
dR
Boltzmann equation for homogeneous materials takes on the form
~ − 1 ∂ (Sj (E)φj (~r, E, Ω))
~ + σj (E)φj (~r, E, Ω)
~
~ · ∇φj (~r, E, Ω)
Ω
Aj ∂E
XZ
~ 0 σjk (Ω,
~ Ω
~ 0 , E, E 0 )φk (~r, E 0 , Ω
~ 0)
dE 0 dΩ
=
(2.5.77)
k6=j
~ is the flux of ions of type j moving in
where Aj is the atomic mass of the ion of type j and φj (~r, E, Ω)
~ with energy E.
the direction Ω
Observe that in most cases the left-hand side of the Boltzmann equation represents the time rate of
change of a distribution type function in a phase space while the right-hand side of the Boltzmann equation
represents the time rate of change of this distribution function within a volume element of phase space due
to scattering and absorption collision processes.
Boltzmann Equation for gases
~ , t) which can be
Consider the Boltzmann equation in terms of a particle distribution function f (~r, V
written as
~
∂
~ · ∇~r + F · ∇ ~
+V
V
∂t
m
!
~ , t)
~ , t) = DC f (~r, V
f (~r, V
(2.5.78)
for a single species of gas particles where there is only scattering and no absorption of the particles. An
element of volume in phase space (x, y, z, Vx , Vy , Vz ) can be thought of as a volume element dτ = dxdydz
for the spatial elements together with a volume element dτv = dVx dVy dVz for the velocity elements. These
elements are centered at position ~r and velocity V~ at time t. In phase space a constant velocity V1 can be
thought of as a sphere since V12 = Vx2 + Vy2 + Vz2 . The phase space volume element dτ dτv changes with time
~ change with time. The position vector ~r changes because of velocity
since the position ~r and velocity V
1
John W. Wilson, Lawrence W. Townsend, Walter Schimmerling, Govind S. Khandelwal, Ferdous Kahn,
John E. Nealy, Francis A. Cucinotta, Lisa C. Simonsen, Judy L. Shinn, and John W. Norbury, Transport
Methods and Interactions for Space Radiations, NASA Reference Publication 1257, December 1991.
311
and the velocity vector changes because of the acceleration
~
F
m.
~ , t)dτ dτv represents the expected
Here f (~r, V
number of particles in the phase space element dτ dτv at time t.
Assume there are no collisions, then each of the gas particles in a volume element of phase space centered
~1 move during a time interval dt to a phase space element centered at position
at position ~r and velocity V
~
F
~1 + dt. If there were no loss or gains of particles, then the number of particles must be
~r + V~1 dt and V
m
conserved and so these gas particles must move smoothly from one element of phase space to another without
any gains or losses of particles. Because of scattering collisions in dτ many of the gas particles move into or
~1 + dV
~1 . These collision scattering processes are denoted by the collision
~1 to V
out of the velocity range V
~ , t) in the Boltzmann equation.
operator DC f (~r, V
Consider two identical gas particles which experience a binary collision. Imagine that particle 1 with
~2 . Denote by σ(V
~1 → V~10 , V~2 → V
~20 ) dτV1 dτV2 the
velocity V~1 collides with particle 2 having velocity V
~ 0 and V~ 0 + dV
~ 0 and the
~1 to between V
conditional probability that particle 1 is scattered from velocity V
1
1
1
~ 0 + dV
~ 0 . We will be interested in collisions
~2 to between V~ 0 and V
struck particle 2 is scattered from velocity V
2
2
2
~ 0 ) → (V
~1 , V
~2 ) for a fixed value of V
~1 as this would represent the number of particles scattered
~ 0, V
of the type (V
1
2
~1 , V
~2 ) → (V
~10 , V
~20 ) for a fixed value V
~1 as this represents
into dτV1 . Also of interest are collisions of the type (V
particles scattered out of dτV1 . Imagine a gas particle in dτ with velocity V~10 subjected to a beam of particles
~10 − V
~20 |f (~r, V
~20 , t)dτV 0 and hence
~20 . The incident flux on the element dτ dτV 0 is |V
with velocities V
1
2
~20 ) dτV1 dτV2 dt |V
~10 − V
~20 |f (~r, V
~20 , t) dτV 0
~1 → V~10 , V~2 → V
σ(V
2
(2.5.79)
~1 and V
~1 + dV
~1
represents the number of collisions, in the time interval dt, which scatter from V~10 to between V
~ 0 to between V~2 and V
~2 + dV
~2 . Multiply equation (2.5.79) by the density of particles
as well as scattering V
2
~10 ,V
~20 and final velocities V~2 not equal
in the element dτ dτV10 and integrate over all possible initial velocities V
to V~1 . This gives the number of particles in dτ which are scattered into dτV1 dt as
Z
Z
~0 →V
~1 , V
~0 →V
~2 )|V
~ 0 − V~ 0 |f (~r, V~ 0 , t)f (~r, V
~ 0 , t).
0
(2.5.80)
N sin = dτ dτV1 dt dτV2 dτV2 dτV10 σ(V
1
2
1
2
1
2
In a similar manner the number of particles in dτ which are scattered out of dτV1 dt is
Z
Z
Z
~1 , t) dτV2 dτV 0 dτV 0 σ(V
~0 →V
~1 , V
~0 →V
~2 )|V
~2 − V
~1 |f (~r, V~2 , t).
N sout = dτ dτV1 dtf (~r, V
1
2
2
1
(2.5.81)
Let
~1 , V
~0 →V
~2 ) = |V
~1 − V
~2 | σ(V
~0 →V
~1 , V
~ 0 → V~2 )
~0 →V
W (V
1
2
1
2
(2.5.82)
~ , t) = N sin − N sout to represent the
define a symmetric scattering kernel and use the relation DC f (~r, V
Boltzmann equation for gas particles in the form
Z
~
∂
~ · ∇~r + F · ∇ ~
+V
V
∂t
m
Z
dτV 0
1
~1 , t) =
f (~r , V
Z
dτV 0
2
(2.5.83)
~1 → V
~ 0, V
~2 → V
~ 0 ) f (~r , V
~ 0 , t)f (~r , V
~ 0 , t) − f (~r , V
~1 , t)f (~r , V
~2 , t) .
dτV2 W (V
1
2
1
2
~1 ). That
Take the moment of the Boltzmann equation (2.5.83) with respect to an arbitrary function φ(V
~1 ) and then integrate over all elements of velocity space dτV1 . Define
is, multiply equation (2.5.83) by φ(V
the following averages and terminology:
312
• The particle density per unit volume
Z+∞
ZZ
Z
n = n(~r, t) =
~ , t) =
dτV f (~r, V
~ , t)dVx dVy dVz
f (~r, V
(2.5.84)
−∞
where ρ = nm is the mass density.
• The mean velocity
~1 = V
~ = 1
V
n
Z+∞
ZZ
~1 f (~r, V~1 , t)dV1x dV1y dV1z
V
−∞
~1 ) define the barred quantity
For any quantity Q = Q(V
1
Q = Q(~r, t) =
n(~r, t)
Further, assume that
~
F
m
Z
~ )f (~r, V
~ , t) dτV = 1
Q(V
n
Z+∞
ZZ
~ )f (~r, V
~ , t)dVx dVy dVz .
Q(V
(2.5.85)
−∞
is independent of V~ , then the moment of equation (2.5.83) produces the result
3
3
X
X
∂
Fi ∂φ
∂
nφ +
φ
−
n
=0
nV
1i
i
∂t
∂x
m ∂V1i
i=1
i=1
(2.5.86)
known as the Maxwell transfer equation. The first term in equation (2.5.86) follows from the integrals
Z
Z
~1 , t)
∂f (~r, V
~1 )dτV1 = ∂
~1 ) dτV1 = ∂ (nφ)
~1 , t)φ(V
φ(V
f (~r, V
∂t
∂t
∂t
(2.5.87)
where differentiation and integration have been interchanged. The second term in equation (2.5.86) follows
from the integral
Z
Z X
3
∂f
φ dτV1
∂xi
i=1
Z
3
X
∂
=
φf
dτ
V
1i
V1
∂xi
i=1
~1 )dτV1 =
V~1 ∇~r f φ(V
=
V1i
(2.5.88)
3
X
∂
nV1i φ .
i
∂x
i=1
The third term in equation (2.5.86) is obtained from the following integral where integration by parts is
employed
Z X
Z ~
3 Fi ∂f
F
∇V~1 f φ dτV1 =
φ dτV1
m
m ∂V1i
i=1
Z+∞
ZZ X
3
Fi ∂f
=
φ
dV1x dV1y dV1y
m ∂V1i
−∞ i=1
Z
Fi
∂
φ f dτV1
=−
∂V1i m
Fi ∂φ
Fi
∂
φ =−
= −n
∂V1i m
m ∂V1i
(2.5.89)
313
~1 and f (~r, V
~ , t) equals zero for Vi equal to ±∞. The right-hand side of
since Fi does not depend upon V
equation (2.5.86) represents the integral of (DC f )φ over velocity space. This integral is zero because of
the symmetries associated with the right-hand side of equation (2.5.83). Physically, the integral of (Dc f )φ
over velocity space must be zero since collisions with only scattering terms cannot increase or decrease the
number of particles per cubic centimeter in any element of phase space.
In equation (2.5.86) we write the velocities V1i in terms of the mean velocities (u, v, w) and random
velocities (Ur , Vr , Wr ) with
V11 = Ur + u,
V12 = Vr + v,
V13 = Wr + w
(2.5.90)
~1 = V
~r + V
~r = 0 (i.e. the average random velocity is zero.) For
~r + V
~ with V
~ = V
~ since V
or V~1 = V
future reference we write equation (2.5.86) in terms of these random velocities and the material derivative.
Substitution of the velocities from equation (2.5.90) in equation (2.5.86) gives
3
X
∂ ∂ ∂ Fi ∂φ
∂(nφ)
+
n(Vr + v)φ +
n(Wr + w)φ − n
n(Ur + u)φ +
=0
∂t
∂x
∂y
∂z
m ∂V1i
i=1
or
(2.5.91)
∂
∂
∂(nφ) ∂
+
nuφ +
nvφ +
nwφ
∂t
∂x
∂y
∂z
3
X
∂
∂
Fi ∂φ
∂
nUr φ +
nVr φ +
nWr φ − n
= 0.
+
∂x
∂y
∂z
m ∂V1i
i=1
Observe that
(2.5.92)
Z+∞
ZZ
~ , t)dVx dVy dVz = nuφ
uφf (~r, V
nuφ =
(2.5.93)
−∞
and similarly nvφ = nvφ, nwφ = nwφ. This enables the equation (2.5.92) to be written in the form
n
∂φ
∂φ
∂φ
∂φ
+ nu
+ nv
+ nw
∂t
∂x
∂y
∂z
∂n
∂
∂
∂
+
(nu) +
(nv) +
(nw)
+φ
∂t
∂x
∂y
∂z
+
(2.5.94)
3
X
∂
∂
Fi ∂φ
∂
nUr φ +
nVr φ +
nWr φ − n
= 0.
∂x
∂y
∂z
m ∂V1i
i=1
The middle bracketed sum in equation (2.5.94) is recognized as the continuity equation when multiplied by
m and hence is zero. The moment equation (2.5.86) now has the form
n
3
X
∂
∂
∂
Fi ∂φ
Dφ
+
nUr φ +
nVr φ +
nWr φ − n
= 0.
Dt
∂x
∂y
∂z
m ∂V1i
i=1
(2.5.95)
Note that from the equations (2.5.86) or (2.5.95) one can derive the basic equations of fluid flow from
continuum mechanics developed earlier. We consider the following special cases of the Maxwell transfer
equation.
314
(i) In the special case φ = m the equation (2.5.86) reduces to the continuity equation for fluids. That is,
equation (2.5.86) becomes
∂
~1 ) = 0
(nm) + ∇ · (nmV
∂t
(2.5.96)
∂ρ
~)=0
+ ∇ · (ρV
∂t
(2.5.97)
which is the continuity equation
~ is the mean velocity defined earlier.
where ρ is the mass density and V
~1 is momentum, the equation (2.5.86) reduces to the momentum equation
(ii) In the special case φ = mV
~1 V~1 in the form
for fluids. To show this, we write equation (2.5.86) in terms of the dyadic V
or
Let
σ = −ρV~r V~r
∂ ~1 + ∇ · (nmV
~1 V~1 ) − nF~ = 0
nmV
∂t
(2.5.98)
∂ ~
~ ) + ∇ · (ρ(V
~r + V
~ )(V
~r + V~ )) − nF~ = 0.
ρ(Vr + V
∂t
(2.5.99)
denote a stress tensor which is due to the random motions of the gas particles and
write equation (2.5.99) in the form
~
∂ρ
∂V
~ (∇ · V~ ) + V
~ (∇ · (ρV
~ )) − ∇ · σ − nF~ = 0.
+ V~
+ ρV
(2.5.100)
∂t
∂t
~ ) = 0 because of the continuity equation and so equation (2.5.100) reduces
+ ∇ · (ρV
ρ
~
The term V
∂ρ
∂t
to the momentum equation
ρ
~
∂V
~∇·V
~
+V
∂t
!
= nF~ + ∇ · σ .
(2.5.101)
~ + qV
~ ×B
~ + m~b, where q is charge, E
~ and B
~ are electric and magnetic fields, and ~b is a
For F~ = q E
body force per unit mass, together with
σ=
3
3 X
X
(−pδij + τij )b
ei ebj
(2.5.102)
i=1 j=1
the equation (2.5.101) becomes the momentum equation
ρ
DV~
~ +V
~ × B).
~
= ρ~b − ∇p + ∇ · τ + nq(E
Dt
(2.5.103)
~ and B
~ vanish, the equation (2.5.103) reduces to the previous momentum
In the special case were E
equation (2.5.25) .
(iii) In the special case φ =
m~ ~
2 V1 · V1
=
m
2
2
2
2 (V11 + V12 + V13 )
is the particle kinetic energy, the equation (2.5.86)
simplifies to the energy equation of fluid mechanics. To show this we substitute φ into equation (2.5.95)
and simplify. Note that
i
mh
(Ur + u)2 + (Vr + v)2 + (Wr + w)2
2
i
mh 2
φ=
Ur + Vr2 + Wr2 + u2 + v 2 + w2
2
φ=
(2.5.104)
315
since uUr = vVr = wWr = 0. Let V 2 = u2 + v 2 + w2 and Cr2 = Ur2 + Vr2 + Wr2 and write equation
(2.5.104) in the form
φ=
m 2
Cr + V 2 .
2
(2.5.105)
Also note that
i
nm h
Ur (Ur + u)2 + Ur (Vr + v)2 + Ur (Wr + w)2
2 "
#
nm Ur Cr2
+ uUr2 + vUr Vr + wUr Wr
=
2
2
nUr φ =
(2.5.106)
and that
i
nm h
Vr Cr2 + uVr Ur + vVr2 + wVr Wr
2
i
nm h
Wr Cr2 + uWr Ur + vWr Vr + wWr2
nWr φ =
2
nVr φ =
(2.5.107)
(2.5.108)
are similar results.
We use
∂
∂V1i
(φ) = mV1i together with the previous results substituted into the equation (2.5.95), and
find that the Maxwell transport equation can be expressed in the form
!
V2
∂ D Cr2
+
=−
ρ[uUr2 + vUr Vr + wUr Wr ]
ρ
Dt
2
2
∂x
∂ ρ[uVr Ur + vVr2 + wVr Wr ]
−
∂y
∂ ρ[uWr Ur + vWr Vr + wWr2 ]
−
∂z
!
!
!
∂
∂
Ur Cr2
Vr Cr2
Wr Cr2
∂
~.
ρ
−
ρ
−
ρ
+ nF~ · V
−
∂x
2
∂y
2
∂z
2
Compare the equation (2.5.109) with the energy equation (2.5.48)
D V2
De
~ ) − ∇ · ~q + ρ~b · V
~
+ρ
= ∇(σ · V
ρ
Dt
Dt
2
(2.5.109)
(2.5.110)
C2
where the internal heat energy has been set equal to zero. Let e = 2r denote the internal energy due to
random motion of the gas particles, F~ = m~b, and let
!
!
!
∂
∂
Ur Cr2
Vr Cr2
Wr Cr2
∂
ρ
−
ρ
−
ρ
∇·~
q =−
∂x
2
∂y
2
∂z
2
(2.5.111)
∂T
∂
∂T
∂
∂T
∂
k
−
k
−
k
=−
∂x
∂x
∂y
∂y
∂z
∂z
represent the heat conduction terms due to the transport of particle energy
mCr2
2
by way of the random
particle motion. The remaining terms are related to the rate of change of work and surface stresses giving
∂
∂ ρ[uUr2 + vUr Vr + wUr Wr ] =
(uσxx + vσxy + wσxz )
−
∂x
∂x
∂
∂ ρ[uVr Ur + vVr2 + wVr Wr ] =
(uσyx + vσyy + wσyz )
−
(2.5.112)
∂y
∂y
∂
∂ ρ[uWr Ur + vWr Vr + wWr2 ] = (uσzx + vσzy + wσzz ) .
−
∂z
∂z
316
This gives the stress relations due to random particle motion
σxx = − ρUr2
σyx = − ρVr Ur
σzx = − ρWr Ur
σxy = − ρUr Vr
σyy = − ρVr2
σzy = − ρWr Vr
σxz = − ρUr Wr
σyz = − ρVr Wr
σzz = − ρWr2 .
(2.5.113)
The Boltzmann equation is a basic macroscopic model used for the study of individual particle motion
where one takes into account the distribution of particles in both space, time and energy. The Boltzmann
equation for gases assumes only binary collisions as three-body or multi-body collisions are assumed to
rarely occur. Another assumption used in the development of the Boltzmann equation is that the actual
time of collision is thought to be small in comparison with the time between collisions. The basic problem
associated with the Boltzmann equation is to find a velocity distribution, subject to either boundary and/or
initial conditions, which describes a given gas flow.
The continuum equations involve trying to obtain the macroscopic variables of density, mean velocity,
stress, temperature and pressure which occur in the basic equations of continuum mechanics considered
earlier. Note that the moments of the Boltzmann equation, derived for gases, also produced these same
continuum equations and so they are valid for gases as well as liquids.
In certain situations one can assume that the gases approximate a Maxwellian distribution
f (~r, V~ , t) ≈ n(~r, t)
m
m 3/2
~ ·V
~
V
exp −
2πkT
2kT
(2.5.114)
thereby enabling the calculation of the pressure tensor and temperature from statistical considerations.
In general, one can say that the Boltzmann integral-differential equation and the Maxwell transfer
equation are two important formulations in the kinetic theory of gases. The Maxwell transfer equation
depends upon some gas-particle property φ which is assumed to be a function of the gas-particle velocity.
The Boltzmann equation depends upon a gas-particle velocity distribution function f which depends upon
~ and time t. These formulations represent two distinct and important viewpoints
position ~r, velocity V
considered in the kinetic theory of gases.
317
EXERCISE 2.5
I 1.
Let p = p(x, y, z), [dyne/cm2 ] denote the pressure at a point (x, y, z) in a fluid medium at rest
(hydrostatics), and let ∆V denote an element of fluid volume situated at this point as illustrated in the
figure 2.5-5.
Figure 2.5-5. Pressure acting on a volume element.
(a) Show that the force acting on the face ABCD is p(x, y, z)∆y∆z ê1 .
(b) Show that the force acting on the face EF GH is
∂ 2 p (∆x)2
∂p
∆x + 2
+ · · · ∆y∆z ê1 .
−p(x + ∆x, y, z)∆y∆z ê1 = − p(x, y, z) +
∂x
∂x
2!
(c) In part (b) neglect terms with powers of ∆x greater than or equal to 2 and show that the resultant force
∂p
in the x-direction is − ∆x∆y∆z ê1 .
∂x
(d) What has been done in the x-direction can also be done in the y and z-directions. Show that the
∂p
∂p
and
− ∆x∆y∆z ê3 . (e) Show that −∇p =
resultant forces in these directions are − ∆x∆y∆z ê2
∂y
∂z
∂p
∂p
∂p
ê1 +
ê2 +
ê3 is the force per unit volume acting at the point (x, y, z) of the fluid medium.
−
∂x
∂y
∂z
I 2. Follow the example of exercise 1 above but use cylindrical coordinates and find the force per unit volume
at a point (r, θ, z). Hint: An element of volume in cylindrical coordinates is given by ∆V = r∆r∆θ∆z.
I 3. Follow the example of exercise 1 above but use spherical coordinates and find the force per unit volume
at a point (ρ, θ, φ). Hint: An element of volume in spherical coordinates is ∆V = ρ2 sin θ∆ρ∆θ∆φ.
I 4. Show that if the density % = %(x, y, z, t) is a constant, then v r,r = 0.
I 5. Assume that λ∗ and µ∗ are zero. Such a fluid is called a nonviscous or perfect fluid. (a) Show the
Cartesian equations describing conservation of linear momentum are
∂u
∂u
∂u
1 ∂p
∂u
+u
+v
+w
= bx −
∂t
∂x
∂y
∂z
% ∂x
∂v
∂v
∂v
1 ∂p
∂v
+u
+v
+w
= by −
∂t
∂x
∂y
∂z
% ∂y
∂w
∂w
∂w
1 ∂p
∂w
+u
+v
+w
= bz −
∂t
∂x
∂y
∂z
% ∂z
where (u, v, w) are the physical components of the fluid velocity. (b) Show that the continuity equation can
be written
∂
∂
∂
∂%
+
(%u) +
(%v) +
(%w) = 0
∂t
∂x
∂y
∂z
318
I 6. Assume λ∗ = µ∗ = 0 so that the fluid is ideal or nonviscous. Use the results given in problem 5 and
make the following additional assumptions:
• The density is constant and so the fluid is incompressible.
• The body forces are zero.
• Steady state flow exists.
• Only two dimensional flow in the x-yplane is considered such that u = u(x, y), v = v(x, y) and
w = 0. (a) Employ the above assumptions and simplify the equations in problem 5 and verify the
results
∂u 1 ∂p
∂u
+v
+
=0
∂x
∂y
% ∂x
∂v
1 ∂p
∂v
+v
+
=0
u
∂x
∂y % ∂y
∂u ∂v
+
=0
∂x ∂y
u
(b) Make the additional assumption that the flow is irrotational and show that this assumption
produces the results
∂u
∂v
−
=0
∂x ∂y
and
1
1 2
u + v 2 + p = constant.
2
%
(c) Point out the Cauchy-Riemann equations and Bernoulli’s equation in the above set of equations.
I 7. Assume the body forces are derivable from a potential function φ such that bi = −φ,i . Show that for an
ideal fluid with constant density the equations of fluid motion can be written in either of the forms
1
∂v r
+ v r,s v s = − g rm p,m − g rm φ,m
∂t
%
or
∂vr
1
+ vr,s v s = − p,r − φ,r
∂t
%
1
∇ (~v · ~v ) − ~v × (∇ × ~v ) are
2
used to express the Navier-Stokes-Duhem equations in alternate forms involving the vorticity Ω = ∇ × ~v .
I 8. The vector identities ∇2~v = ∇ (∇ · ~v ) − ∇ × (∇ × ~v )
and
(~v · ∇) ~v =
(a) Use Cartesian tensor notation and derive the above identities. (b) Show the second identity can be written
∂v 2
mj k
v vk,j − mnp ijk gpi vn vk,j . Hint: Show that
= 2v k vk,j .
in generalized coordinates as v j v m
,j = g
∂xj
I 9. Use problem 8 and show that the results in problem 7 can be written
or
v2
p
∂
∂v r
− rnp Ωp vn = −g rm m
+φ+
∂t
∂x
%
2
2
v
p
∂
∂vi
− ijk v j Ωk = − i
+φ+
∂t
∂x %
2
I 10. In terms of physical components, show that in generalized
coordinates,
orthogonal
for i 6= j, the rate
v(i)
v(j)
1 hi ∂
hj ∂
+
, no summations
of deformation tensor Dij can be written D(ij) =
2 hj ∂xj
hi
hi ∂xi
hj
3
∂hi
1 ∂v(i) v(i) ∂hi X 1
−
+
v(k) k ,
no summations. (Hint: See
and for i = j there results D(ii) =
2
i
i
hi ∂x
hi ∂x
hi hk
∂x
k=1
Problem 17 Exercise 2.1.)
319
Figure 2.5-6. Plane Couette flow
I 11. Find the physical components of the rate of deformation tensor Dij in Cartesian coordinates. (Hint:
See problem 10.)
I 12. Find the physical components of the rate of deformation tensor in cylindrical coordinates. (Hint: See
problem 10.)
I 13. (Plane Couette flow) Assume a viscous fluid with constant density is between two plates as illustrated
in the figure 2.5-6.
(a) Define ν =
µ∗
%
as the kinematic viscosity and show the equations of fluid motion can be written
1
∂v i
+ v i,s v s = − g im p,m + νg jm v i,mj + g ij bj ,
∂t
%
i = 1, 2, 3
(b) Let ~v = (u, v, w) denote the physical components of the fluid flow and make the following assumptions
• u = u(y), v = w = 0
• Steady state flow exists
• The top plate, with area A, is a distance ` above the bottom plate. The bottom plate is fixed and
a constant force F is applied to the top plate to keep it moving with a velocity u0 = u(`).
• p and % are constants
• The body force components are zero.
Find the velocity u = u(y)
u0
F
= σ21 = σxy = σyx = µ∗ . This
A
`
example illustrates that the stress is proportional to u0 and inversely proportional to `.
(c) Show the tangential stress exerted by the moving fluid is
I 14. In the continuity equation make the change of variables
t=
t
,
τ
%=
%
,
%0
~v = ~v ,
v0
x=
x
,
L
y=
y
,
L
z=
z
L
and write the continuity equation in terms of the barred variables and the Strouhal parameter.
I 15. (Plane Poiseuille flow)
Consider two flat plates parallel to one another as illustrated in the figure
2.5-7. One plate is at y = 0 and the other plate is at y = 2`. Let ~v = (u, v, w) denote the physical components
of the fluid velocity and make the following assumptions concerning the flow The body forces are zero. The
∂p
∂p
∂p
= −p0 is a constant and
=
= 0. The velocity in the x-direction is a function of y only
derivative
∂x
∂y
∂z
320
Figure 2.5-7. Plane Poiseuille flow
with u = u(y) and v = w = 0 with boundary values u(0) = u(2`) = 0. The density is constant and ν = µ∗ /%
is the kinematic viscosity.
d2 u p0
= 0,
u(0) = u(2`) = 0
+
dy 2
%
(b) Find the velocity u = u(y) and find the maximum velocity in the x-direction. (c) Let M denote the
(a) Show the equation of fluid motion is ν
mass flow rate across the plane x = x0 = constant, , where 0 ≤ y ≤ 2`, and 0 ≤ z ≤ 1.
2
Show that M = ∗ %p0 `3 . Note that as µ∗ increases, M decreases.
3µ
∂(δcu)
, where c is the
∂t
3
specific heat [cal/gm C], δ is the volume density [gm/cm ], H is the rate of heat generation [cal/sec cm3 ], u
I 16. The heat equation (or diffusion equation) can be expressed div ( k grad u)+ H =
is the temperature [C], k is the thermal conductivity [cal/sec cm C]. Assume constant thermal conductivity,
volume density and specific heat and express the boundary value problem
∂u
∂2u
= δc , 0 < x < L
∂x2
∂t
u(L, t) = u1 ,
u(x, 0) = f (x)
k
u(0, t) = 0,
in a form where all the variables are dimensionless. Assume u1 is constant.
I 17. Simplify the Navier-Stokes-Duhem equations using the assumption that there is incompressible flow.
I 18. (Rayleigh impulsive flow)
The figure 2.5-8 illustrates fluid motion in the plane where y > 0 above a
plate located along the axis where y = 0. The plate along y = 0 has zero velocity for all negative time and
at time t = 0 the plate is given an instantaneous velocity u0 in the positive x-direction. Assume the physical
components of the velocity are ~v = (u, v, w) which satisfy u = u(y, t), v = w = 0. Assume that the density
of the fluid is constant, the gradient of the pressure is zero, and the body forces are zero. (a) Show that the
velocity in the x-direction is governed by the differential equation
∂ 2u
∂u
= ν 2,
∂t
∂y
with
ν=
µ∗
.
%
Assume u satisfies the initial condition u(0, t) = u0 H(t) where H is the Heaviside step function. Also assume
there exist a condition at infinity limy→∞ u(y, t). This latter condition requires a bounded velocity at infinity.
(b) Use any method to show the velocity is
u(y, t) = u0 − u0 erf
y
√
2 νt
= u0 erfc
y
√
2 νt
321
Figure 2.5-8. Rayleigh impulsive flow
where erf and erfc are the error function and complimentary error function respectively. Pick a point on the
√
line y = y0 = 2 ν and plot the velocity as a function of time. How does the viscosity effect the velocity of
the fluid along the line y = y0 ?
I 19. Simplify the Navier-Stokes-Duhem equations using the assumption that there is incompressible and
irrotational flow.
I 20. Let ζ = λ∗ + 23 µ∗ and show the constitutive equations (2.5.21) for fluid motion can be written in the
2
σij = −pδij + µ∗ vi,j + vj,i − δij vk,k + ζδij vk,k .
3
form
I 21. (a) Write out the Navier-Stokes-Duhem equation for two dimensional flow in the x-y direction under
the assumptions that
• λ∗ + 23 µ∗ = 0
(This condition is referred to as Stoke’s flow.)
• The fluid is incompressible
• There is a gravitational force ~b = −g∇ h Hint: Express your answer as two scalar equations
involving the variables v1 , v2 , h, g, %, p, t, µ∗ plus the continuity equation. (b) In part (a) eliminate
the pressure and body force terms by cross differentiation and subtraction. (i.e. take the derivative
of one equation with respect to x and take the derivative of the other equationwith respect
to y
∂v1
1 ∂v2
−
and
and then eliminate any common terms.) (c) Assume that ~ω = ω ê3 where ω =
2 ∂x
∂y
derive the vorticity-transport equation
dω
= ν∇2 ω
dt
where
∂ω
dω
∂ω
∂ω
=
+ v1
+ v2
.
dt
∂t
∂x
∂y
Hint: The continuity equation makes certain terms zero. (d) Define a stream function ψ = ψ(x, y)
∂ψ
∂ψ
and
v2 = −
and show the continuity equation is identically satisfied.
satisfying v1 =
∂y
∂x
Show also that ω = − 21 ∇2 ψ and that
∇4 ψ =
1 ∂∇2 ψ ∂ψ ∂∇2 ψ ∂ψ ∂∇2 ψ
+
−
.
ν
∂t
∂y ∂x
∂x ∂y
If ν is very large, show that ∇4 ψ ≈ 0.
322
I 22. In generalized orthogonal coordinates, show that the physical components of the rate of deformation
stress can be written, for i 6= j
σ(ij) = µ∗
hi ∂
hj ∂xj
v(i)
hi
+
hj ∂
hi ∂xi
v(j)
hj
,
no summation,
and for i 6= j 6= k
1 ∂v(i)
1
∂hi
1
∂hi
+
v(j)
+
v(k)
σ(ii) = −p + 2µ∗
hi ∂xi
hi hj
∂xj
hi hk
∂xk
∗
∂
∂
∂
λ
{h
h
v(1)}
+
{h
h
v(2)}
+
{h
h
v(3)}
,
+
2
3
1
3
1
2
h1 h2 h3 ∂x1
∂x2
∂x3
no summation
I 23. Find the physical components for the rate of deformation stress in Cartesian coordinates. Hint: See
problem 22.
I 24. Find the physical components for the rate of deformations stress in cylindrical coordinates. Hint: See
problem 22.
1
I 25. Verify the Navier-Stokes equations for an incompressible fluid can be written v̇i = − p,i + νvi,mm + bi
%
∗
where ν = µ% is called the kinematic viscosity.
I 26. Verify the Navier-Stokes equations for a compressible fluid with zero bulk viscosity can be written
∗
1
ν
v̇i = − p,i + vm,mi + νvi,mm + bi with ν = µ% the kinematic viscosity.
%
3
I 27. The constitutive equation for a certain non-Newtonian Stokesian fluid is σij = −pδij +βDij +γDik Dkj .
Assume that β and γ are constants (a) Verify that σij,j = −p,i + βDij,j + γ(Dik Dkj,j + Dik,j Dkj )
(b) Write out the Cauchy equations of motion in Cartesian coordinates. (See page 236).
I 28. Let the constitutive equations relating stress and strain for a solid material take into account thermal
1+ν
ν
σij − σkk δij +α T δij
stresses due to a temperature T . The constitutive equations have the form eij =
E
E
where α is a coefficient of linear expansion for the material and T is the absolute temperature. Solve for the
stress in terms of strains.
I 29. Derive equation (2.5.53) and then show that when the bulk coefficient of viscosity is zero, the NavierStokes equations, in Cartesian coordinates, can be written in the conservation form
∂(%u) ∂(%u2 + p − τxx ) ∂(%uv − τxy ) ∂(%uw − τxz )
+
+
+
= %bx
∂t
∂x
∂y
∂z
∂(%v) ∂(%uv − τxy ) ∂(%v 2 + p − τyy ) ∂(%vw − τyz )
+
+
+
= %by
∂t
∂x
∂y
∂z
∂(%w) ∂(%uw − τxz ) ∂(%vw − τyz ) ∂(%w2 + p − τzz )
+
+
+
= %bz
∂t
∂x
∂y
∂z
2
where v1 = u,v2 = v,v3 = w and τij = µ∗ (vi,j + vj,i − δij vk,k ). Hint: Alternatively, consider 2.5.29 and use
3
the continuity equation.
323
I 30. Show that for a perfect gas, where λ∗ = − 32 µ∗ and η = µ∗ is a function of position, the vector form
of equation (2.5.25) is
%
4
D~v
= %~b − ∇p + ∇(η∇ · ~v ) + ∇(~v · ∇η) − ~v ∇2 η + (∇η) × (∇ × ~v ) − (∇ · ~v )∇η − ∇ × (∇ × (η~v ))
Dt
3
D p ∂Q
Dh
=
+
− ∇ · ~q + Φ. Hint: Use the continuity equation.
Dt
Dt
∂t
I 32. Show that in Cartesian coordinates the Navier-Stokes equations of motion for a compressible fluid
I 31. Derive the energy equation %
can be written
∂
∂v
∂
∂u
∂p
∂
Du
∗ ∂u
∗
∗ ∂u
∗ ∂w
~
=ρbx −
+
2µ
+λ ∇·V +
µ (
+
) +
µ (
+
)
ρ
Dt
∂x ∂x
∂x
∂y
∂y
∂x
∂z
∂x
∂z
∂
∂w
∂
∂w
∂p
∂
Dv
∗ ∂v
∗
∗ ∂v
∗ ∂w
~
=ρby −
+
+λ ∇·V +
+
) +
+
)
2µ
µ (
µ (
ρ
Dt
∂y ∂y
∂y
∂z
∂z
∂y
∂x
∂y
∂x
∂
∂u
∂
∂w
∂p
∂
Dv
∗ ∂w
∗
∗ ∂w
∗ ∂v
~
=ρbz −
+
+λ ∇·V +
+
) +
+
)
ρ
2µ
µ (
µ (
Dt
∂z ∂z
∂z
∂x
∂x
∂z
∂y
∂z
∂y
where (Vx , Vy , Vz ) = (u, v, w).
I 33. Show that in cylindrical coordinates the Navier-Stokes equations of motion for a compressible fluid
can be written
V2
∂
∂p
∂Vr
DVr
~ + 1 ∂ µ∗ ( 1 ∂Vr + ∂Vθ − Vθ )
− θ =%br −
+
+ λ∗ ∇ · V
2µ∗
%
Dt
r
∂r
∂r
∂r
r ∂θ
r ∂θ
∂r
r
∗
∂V
2µ
∂V
1
∂V
V
∂V
∂
r
z
r
θ
r
+
) +
(
−
− )
+
µ∗ (
∂z
∂z
∂r
r ∂r
r ∂θ
r
Vr Vθ
Vr
DVθ
1 ∂p 1 ∂
1 ∂Vθ
~ + ∂ µ∗ ( 1 ∂Vz + ∂Vθ )
+
+
+ ) + λ∗ ∇ · V
=%bθ −
2µ∗ (
%
Dt
r
r ∂θ
r ∂θ
r ∂θ
r
∂z
r ∂θ
∂z
∗
∂V
∂V
V
2µ
1
∂V
∂V
V
1
∂
r
θ
θ
r
θ
θ
+
− ) +
(
+
− )
+
µ∗ (
∂r
r ∂θ
∂r
r
r r ∂θ
∂r
r
∂
∂Vz
∂p
1 ∂
DVz
∗ ∂Vz
∗
∗ ∂Vr
~
=%bz −
+
+λ ∇·V +
+
)
2µ
µ r(
%
Dt
∂z ∂z
∂z
r ∂r
∂z
∂r
∂Vθ
1 ∂Vz
1 ∂
+
)
+
µ∗ (
r ∂θ
r ∂θ
∂z
I 34. Show that the dissipation function Φ can be written as Φ = 2µ∗ Dij Dij + λ∗ Θ2 .
I 35. Verify the identities:
(a)
%
∂et
D
~)
(et /%) =
+ ∇ · (et V
Dt
∂t
(b) %
D
D
De
(et /%) = %
+%
V 2 /2 .
Dt
Dt
Dt
I 36. Show that the conservation law for heat flow is given by
∂T
+ ∇ · (T ~v − κ∇T ) = SQ
∂t
where κ is the thermal conductivity of the material, T is the temperature, J~advection = T ~v,
J~conduction = −κ∇T and SQ is a source term. Note that in a solid material there is no flow and so ~v = 0 and
324
the above equation reduces to the heat equation. Assign units of measurements to each term in the above
equation and make sure the equation is dimensionally homogeneous.
I 37. Show that in spherical coordinates the Navier-Stokes equations of motion for a compressible fluid can
be written
2
2
∂
∂p
∂Vρ
DVρ Vθ + Vφ
~ + 1 ∂ µ∗ (ρ ∂ (Vθ /ρ) + 1 ∂Vρ )
−
) = %bρ −
+
+ λ∗ ∇ · V
2µ∗
Dt
ρ
∂ρ ∂ρ
∂ρ
ρ ∂θ
∂ρ
ρ ∂θ
∂
∂
1 ∂Vρ
1
+ρ
(Vφ /ρ))
+
µ∗ (
ρ sin θ ∂φ
ρ sin θ ∂φ
∂ρ
2 ∂Vθ
4Vρ
2 ∂Vφ
2Vθ cot θ
∂
cot θ ∂Vρ
µ∗ ∂Vρ
(4
−
−
−
−
+ ρ cot θ
(Vθ /ρ) +
)
+
ρ
∂ρ
ρ ∂θ
ρ
ρ sin θ ∂φ
ρ
∂ρ
ρ ∂θ
Vρ Vθ Vφ2 cot θ
1 ∂p 1 ∂ 2µ∗ ∂Vθ
DVθ
~
+
−
) = %bθ −
+
(
+ Vρ ) + λ∗ ∇ · V
%(
Dt
ρ
ρ
ρ ∂θ ρ ∂θ
ρ ∂θ
∂
sin
θ
1 ∂Vθ
∂
1 ∂Vρ
∂
∂
1
(Vφ / sin θ) +
) +
(Vθ /ρ) +
)
µ∗ (
µ∗ (ρ
+
ρ sin θ ∂φ
ρ ∂θ
ρ sin θ ∂φ
∂ρ
∂ρ
ρ ∂θ
1 ∂Vθ
1 ∂Vφ
Vθ cot θ
1 ∂Vρ
µ∗
∂
−
−
(Vθ /ρ) +
+
2
cot θ + 3 ρ
ρ
ρ ∂θ
ρ sin θ ∂φ
ρ
∂ρ
ρ ∂θ
Vφ Vρ Vθ Vφ cot θ
∂
∂V
∂
1 ∂p
1
DVφ
ρ
+
+
= %bφ −
+
µ∗
+ρ
(Vφ /ρ)
%
Dt
ρ
ρ
ρ sin θ ∂φ ∂ρ
ρ sin θ ∂φ
∂ρ
∂ 2µ∗
1 ∂Vφ
1
~
+ Vρ + Vθ cot θ + λ∗ ∇ · V
+
ρ sin θ ∂φ
ρ
sin θ ∂φ
sin θ ∂
1 ∂Vθ
1 ∂
(Vφ / sin θ) +
µ∗
+
ρ ∂θ
ρ ∂θ
ρ sin θ ∂φ
sin θ ∂
1 ∂Vρ
∂
1 ∂Vθ
µ∗
+ρ
(Vφ /ρ) + 2 cot θ
(Vφ / sin θ) +
3
+
ρ
ρ sin θ ∂φ
∂ρ
ρ ∂θ
ρ sin θ ∂φ
%(
I 38. Verify all the equations (2.5.28).
I 39. Use the conservation of energy equation (2.5.47) together with the momentum equation (2.5.25) to
derive the equation (2.5.48).
I 40. Verify the equation (2.5.55).
I 41. Consider nonviscous flow and write the 3 linear momentum equations and the continuity equation
and make the following assumptions: (i) The density % is constant. (ii) Body forces are zero. (iii) Steady
state flow only. (iv) Consider only two dimensional flow with non-zero velocity components u = u(x, y) and
v = v(x, y). Show that there results the system of equations
u
∂u 1 ∂P
∂u
+v
+
= 0,
∂x
∂y
% ∂x
u
∂v
∂v
1 ∂P
+v
+
= 0,
∂x
∂y % ∂y
∂u ∂v
+
= 0.
∂x ∂y
Recognize that the last equation in the above set as one of the Cauchy-Riemann equations that f (z) = u − iv
be an analytic function of a complex variable. Further assume that the fluid flow is irrotational so that
P
1 2
∂v ∂u
−
= 0. Show that this implies that
u + v 2 + = Constant. If in addition u and v are derivable
∂x ∂y
2
%
and
v = ∂φ
from a potential function φ(x, y), such that u = ∂φ
∂x
∂y , then show that φ is a harmonic function.
By constructing the conjugate harmonic function ψ(x, y) the complex potential F (z) = φ(x, y) + iψ(x, y) is
such that F 0 (z) = u(x, y) − iv(x, y) and F 0 (z) gives the velocity. The family of curves φ(x, y) =constant are
called equipotential curves and the family of curves ψ(x, y) = constant are called streamlines. Show that
these families are an orthogonal family of curves.
325
§2.6 ELECTRIC AND MAGNETIC FIELDS
Introduction
In electromagnetic theory the mks system of units and the Gaussian system of units are the ones most
often encountered. In this section the equations will be given in the mks system of units. If you want the
equations in the Gaussian system of units make the replacements given in the column 3 of Table 1.
Table 1. MKS AND GAUSSIAN UNITS
volt/m
Replacement
symbol
~
E
statvolt/cm
~ (Magnetic field)
B
weber/m2
~
B
c
gauss
~ (Displacement field)
D
coulomb/m2
~
D
4π
statcoulomb/cm2
~ (Auxiliary Magnetic field)
H
ampere/m
~
cH
4π
oersted
J~ (Current density)
ampere/m2
J~
statampere/cm2
~ (Vector potential)
A
weber/m
~
A
c
gauss-cm
V (Electric potential)
volt
V
statvolt
MKS
symbol
MKS
units
~ (Electric field)
E
(Dielectric constant)
4π
µ (Magnetic permeability)
4πµ
c2
GAUSSIAN
units
Electrostatics
A basic problem in electrostatic theory is to determine the force F~ on a charge Q placed a distance r
from another charge q. The solution to this problem is Coulomb’s law
1 qQ
b
er
F~ =
4π0 r2
(2.6.1)
where q, Q are measured in coulombs, 0 = 8.85 × 10−12 coulomb2 /N · m2 is called the permittivity in a
vacuum, r is in meters, [F~ ] has units of Newtons and b
er is a unit vector pointing from q to Q if q, Q have
~ = F~ /Q is called the
the same sign or pointing from Q to q if q, Q are of opposite sign. The quantity E
~ = F~ and so Q = 1 is called
electric field produced by the charges. In the special case Q = 1, we have E
a test charge. This tells us that the electric field at a point P can be viewed as the force per unit charge
exerted on a test charge Q placed at the point P. The test charge Q is always positive and so is repulsed if
q is positive and attracted if q is negative.
The electric field associated with many charges is obtained by the principal of superposition. For
example, let q1 , q2 , . . . , qn denote n-charges having respectively the distances r1 , r2 , . . . , rn from a test charge
Q placed at a point P. The force exerted on Q is
F~ =F~1 + F~2 + · · · + F~n
1
q2 Q
qn Q
q1 Q
b
b
b
e
e
e
F~ =
+
+
·
·
·
+
r
r
r
1
2
n
4π0
r12
r22
rn2
n
X qi
~
~ = E(P
~ ) =F = 1
b
er
or E
Q
4π0 i=1 ri2 i
(2.6.2)
326
~ = E(P
~ ) is the electric field associated with the system of charges. The equation (2.6.2) can be genwhere E
eralized to other situations by defining other types of charge distributions. We introduce a line charge density
λ∗ , (coulomb/m), a surface charge density µ∗ , (coulomb/m2 ), a volume charge density ρ∗ , (coulomb/m3 ),
then we can calculate the electric field associated with these other types of charge distributions. For example,
if there is a charge distribution λ∗ = λ∗ (s) along a curve C, where s is an arc length parameter, then we
would have
~ )=
E(P
1
4π0
Z
b
er ∗
λ ds
r2
C
(2.6.3)
as the electric field at a point P due to this charge distribution. The integral in equation (2.6.3) being a
line integral along the curve C and where ds is an element of arc length. Here equation (2.6.3) represents a
continuous summation of the charges along the curve C. For a continuous charge distribution over a surface
S, the electric field at a point P is
~ )=
E(P
Z Z
1
4π0
b
er ∗
µ dσ
r2
S
(2.6.4)
where dσ represents an element of surface area on S. Similarly, if ρ∗ represents a continuous charge distribution throughout a volume V , then the electric field is represented
~ )=
E(P
1
4π0
Z Z Z
b
er ∗
ρ dτ
r2
V
(2.6.5)
where dτ is an element of volume. In the equations (2.6.3), (2.6.4), (2.6.5) we let (x, y, z) denote the position
of the test charge and let (x0 , y 0 , z 0 ) denote a point on the line, on the surface or within the volume, then
e1 + (y − y 0 ) b
e2 + (z − z 0 ) b
e3
~r = (x − x0 ) b
(2.6.6)
~r
er = .
represents the distance from the point P to an element of charge λ∗ ds, µ∗ dσ or ρ∗ dτ with r = |~r| and b
r
~ = 0, and so it is derivable from a potential function V
If the electric field is conservative, then ∇ × E
by taking the negative of the gradient of V and
~ = −∇V.
E
(2.6.7)
~ · d~r is an exact differential so that the potential function can
For these conditions note that ∇V · d~r = −E
be represented by the line integral
Z
P
~ · d~r
E
V = V(P ) = −
(2.6.8)
α
where α is some reference point (usually infinity, where V(∞) = 0). For a conservative electric field the line
integral will be independent of the path connecting any two points a and b so that
Z
b
V(b) − V(a) = −
α
Z
~
E · d~r − −
a
~ · d~r
E
α
Z
Z
b
~ · d~r =
E
=−
a
b
∇V · d~r.
(2.6.9)
a
Let α = ∞ in equation (2.6.8), then the potential function associated with a point charge moving in
the radial direction b
er is
Z
V(r) = −
~ · d~r = −q
E
4π0
∞
r
Z
r
∞
q 1 r
q
1
| =
.
dr =
r2
4π0 r ∞ 4π0 r
327
By superposition,
ofZcharges is given by
Z Z Zthe ∗potential at a point P for a continuous volume distribution Z
1
ρ
µ∗
1
dτ and for a surface distribution of charges V(P ) =
dσ and for a line
V(P ) =
4π0
4π0
V r
S r
Z
∗
1
λ
ds; and for a discrete distribution of point charges
distribution of charges V(P ) =
4π0 C r
N
1 X qi
. When the potential functions are defined from a common reference point, then the
V(P ) =
4π0 i=1 ri
principal of superposition applies.
The potential function V is related to the work done W in moving a charge within the electric field.
The work done in moving a test charge Q from point a to point b is an integral of the force times distance
~ and so the force F~ = −QE
~ is in opposition to this
moved. The electric force on a test charge Q is F~ = QE
force as you move the test charge. The work done is
Z
Z
b
F~ · d~r =
W =
a
Z
b
~ · d~r = Q
−QE
a
b
∇V · d~r = Q[V(b) − V(a)].
(2.6.10)
a
The work done is independent of the path joining the two points and depends only on the end points and
the change in the potential. If one moves Q from infinity to point b, then the above becomes W = QV (b).
~ = E(P
~ ) is a vector field which can be represented graphically by constructing vectors
An electric field E
at various selected points in the space. Such a plot is called a vector field plot. A field line associated with
a vector field is a curve such that the tangent vector to a point on the curve has the same direction as the
vector field at that point. Field lines are used as an aid for visualization of an electric field and vector fields
~ at that point.
in general. The tangent to a field line at a point has the same direction as the vector field E
e2 denote the position vector to a point on a field line. The
For example, in two dimensions let ~r = x b
e1 + y b
~ = E(x,
~
e2 . If E
y) = −N (x, y) b
e1 + M (x, y) b
e2
tangent vector to this point has the direction d~r = dx b
e1 + dy b
~ and d~r must be colinear. Thus, for each point (x, y)
is the vector field constructed at the same point, then E
~ for some constant K. Equating like components we find that the
on a field line we require that d~r = K E
field lines must satisfy the differential relation.
dy
dx
=
=K
−N (x, y)
M (x, y)
or
(2.6.11)
M (x, y) dx + N (x, y) dy =0.
In two dimensions, the family of equipotential curves V(x, y) = C1 =constant, are orthogonal to the family
of field lines and are described by solutions of the differential equation
N (x, y) dx − M (x, y) dy = 0
obtained from equation (2.6.11) by taking the negative reciprocal of the slope. The field lines are perpendicular to the equipotential curves because at each point on the curve V = C1 we have ∇V being perpendicular
~ at this same point. Field lines associated with electric
to the curve V = C1 and so it is colinear with E
fields are called electric lines of force. The density of the field lines drawn per unit cross sectional area are
proportional to the magnitude of the vector field through that area.
328
Figure 2.6-1. Electric forces due to a positive charge at (−a, 0) and negative charge at (a, 0).
EXAMPLE 2.6-1.
Find the field lines and equipotential curves associated with a positive charge q located at the point
(−a, 0) and a negative charge −q located at the point (a, 0).
~ on a test charge Q = 1 place
Solution: With reference to the figure 2.6-1, the total electric force E
at a general point (x, y) is, by superposition, the sum of the forces from each of the isolated charges and is
~ 2 . The electric force vectors due to each individual charge are
~ =E
~1 + E
E
e1 + kqy b
e2
~ 1 = kq(x + a) b
with r12 = (x + a)2 + y 2
E
3
r1
e1 − kqy b
e2
~ 2 = −kq(x − a) b
E
with r22 = (x − a)2 + y 2
r23
where k =
(2.6.12)
1
is a constant. This gives
4π0
kq(x + a) kq(x − a)
kqy kqy
~
~
~
b
e1 +
e2 .
−
− 3 b
E = E1 + E2 =
r13
r23
r13
r2
This determines the differential equation of the field lines
dx
kq(x+a)
r13
−
kq(x−a)
r23
=
kqy
r13
dy
.
− kqy
r3
(2.6.13)
2
To solve this differential equation we make the substitutions
cos θ1 =
x+a
r1
and
cos θ2 =
x−a
r2
(2.6.14)
329
Figure 2.6-2. Lines of electric force between two opposite sign charges.
as suggested by the geometry from figure 2.6-1. From the equations (2.6.12) and (2.6.14) we obtain the
relations
− sin θ1 dθ1 =
r1 dx − (x + a) dr1
r12
2r1 dr1 =2(x + a) dx + 2ydy
− sin θ2 dθ2 =
r2 dx − (x − a)dr2
r22
2r2 dr2 =2(x − a) dx + 2y dy
which implies that
(x + a)y dy y 2 dx
+ 3
r13
r1
2
(x − a)y dy y dx
− sin θ2 dθ2 = −
+ 3
r23
r2
− sin θ1 dθ1 = −
(2.6.15)
Now compare the results from equation (2.6.15) with the differential equation (2.6.13) and determine that
y is an integrating factor of equation (2.6.13) . This shows that the differential equation (2.6.13) can be
written in the much simpler form of the exact differential equation
− sin θ1 dθ1 + sin θ2 dθ2 = 0
(2.6.16)
in terms of the variables θ1 and θ2 . The equation (2.6.16) is easily integrated to obtain
cos θ1 − cos θ2 = C
(2.6.17)
where C is a constant of integration. In terms of x, y the solution can be written
x−a
x+a
p
−p
= C.
2
2
(x + a) + y
(x − a)2 + y 2
These field lines are illustrated in the figure 2.6-2.
(2.6.18)
330
The differential equation for the equipotential curves is obtained by taking the negative reciprocal of
the slope of the field lines. This gives
dy
=
dx
kq(x−a)
r23
kqy
r13
−
−
kq(x+a)
r13
kqy
r23
.
This result can be written in the form
(x − a)dx + ydy
(x + a)dx + ydy
+
=0
−
r13
r23
which simplifies to the easily integrable form
−
dr2
dr1
+ 2 =0
2
r1
r2
in terms of the new variables r1 and r2 . An integration produces the equipotential curves
or
1
1
−
=C2
r1
r2
1
1
p
−p
=C2 .
(x + a)2 + y 2
(x − a)2 + y 2
The potential function for this problem can be interpreted as a superposition of the potential functions
kq
kq
and V2 =
associated with the isolated point charges at the points (−a, 0) and (a, 0).
V1 = −
r1
r2
Observe that the electric lines of force move from positive charges to negative charges and they do not
cross one another. Where field lines are close together the field is strong and where the lines are far apart
the field is weak. If the field lines are almost parallel and equidistant from one another the field is said to be
~ If one moves along a field
uniform. The arrows on the field lines show the direction of the electric field E.
line in the direction of the arrows the electric potential is decreasing and they cross the equipotential curves
~ = 0.
at right angles. Also, when the electric field is conservative we will have ∇ × E
In three dimensions the situation is analogous to what has been done in two dimensions. If the electric
~ = E(x,
~
e2 + R(x, y, z) b
e3 and ~r = x b
e1 + y b
e2 + z b
e3 is the position
field is E
y, z) = P (x, y, z) b
e1 + Q(x, y, z) b
~ must be colinear so that
vector to a variable point (x, y, z) on a field line, then at this point d~r and E
~ for some constant K. Equating like coefficients gives the system of equations
d~r = K E
dy
dz
dx
=
=
= K.
P (x, y, z)
Q(x, y, z)
R(x, y, z)
(2.6.19)
From this system of equations one must try to obtain two independent integrals, call them u1 (x, y, z) = c1
and u2 (x, y, z) = c2 . These integrals represent one-parameter families of surfaces. When any two of these
~ These
surfaces intersect, the result is a curve which represents a field line associated with the vector field E.
type of field lines in three dimensions are more difficult to illustrate.
~ over a surface S is defined as the summation of the normal
The electric flux φE of an electric field E
~ over the surface and is represented
component of E
ZZ
~ · n̂ dσ
E
φE =
S
with units of
N m2
C
(2.6.20)
331
where n̂ is a unit normal to the surface. The flux φE can be thought of as being proportional to the number
of electric field lines passing through an element of surface area. If the surface is a closed surface we have
by the divergence theorem of Gauss
ZZZ
ZZ
~ · n̂ dσ
E
~ dτ =
∇·E
φE =
V
S
where V is the volume enclosed by S.
Gauss Law
Let dσ denote an element of surface area on a surface S. A cone is formed if all points on the boundary
of dσ are connected by straight lines to the origin. The cone need not be a right circular cone. The situation
is illustrated in the figure 2.6-3.
Figure 2.6-3. Solid angle subtended by element of area.
We let ~r denote a position vector from the origin to a point on the boundary of dσ and let n̂ denote a
unit outward normal to the surface at this point. We then have n̂ · ~r = r cos θ where r = |~r| and θ is the
angle between the vectors n̂ and ~r. Construct a sphere, centered at the origin, having radius r. This sphere
dΩ
intersects the cone in an element of area dΩ. The solid angle subtended by dσ is defined as dω = 2 . Note
r
that this is equivalent to constructing a unit sphere at the origin which intersect the cone in an element of
area dω. Solid angles are measured in steradians. The total solid angle about a point equals the area of the
sphere divided by its radius squared or 4π steradians. The element of area dΩ is the projection of dσ on the
n̂ · ~r
dΩ
n̂ · ~r
dσ so that dω = 3 dσ = 2 . Observe that sometimes the
constructed sphere and dΩ = dσ cos θ =
r
r
r
dot product n̂ · ~r is negative, the sign depending upon which of the normals to the surface is constructed.
(i.e. the inner or outer normal.)
The Gauss law for electrostatics in a vacuum states that the flux through any surface enclosing many
charges is the total charge enclosed by the surface divided by 0 . The Gauss law is written
Qe
ZZ
for charges inside S
~ · n̂ dσ = 0
E
0
for charges outside S
S
(2.6.21)
332
where Qe represents the total charge enclosed by the surface S with n̂ the unit outward normal to the surface.
The proof of Gauss’s theorem follows. Consider a single charge q within the closed surface S. The electric
~ = 1 q b
er and so the flux
field at a point on the surface S due to the charge q within S is represented E
4π0 r2
integral is
ZZ
ZZ
ZZ
er · n̂
q
q
q b
dΩ
~ · n̂ dσ =
dσ
=
=
(2.6.22)
E
φE =
2
r2
4π0
0
S
S 4π0
S r
ZZ
b
cos θ dσ
dΩ
er · n̂
=
=
=
dω
and
dω = 4π. By superposition of the charges, we obtain a similar
since
r2
r2
r2
S
n
X
qi . For a continuous
result for each of the charges within the surface. Adding these results gives Qe =
i=1
ZZZ
ρ∗ dτ , where ρ∗ is the charge distribution
distribution of charge inside the volume we can write Qe =
V
per unit volume. Note that charges outside of the closed surface do not contribute to the total flux across
the surface.
This is because the field lines go in one side of the surface and go out the other side. In this
ZZ
~
E · n̂ dσ = 0 for charges outside the surface. Also the position of the charge or charges within the
case
S
volume does not effect the Gauss law.
The equation (2.6.21) is the Gauss law in integral form. We can put this law in differential form as
follows. Using the Gauss divergence theorem we can write for an arbitrary volume that
ZZZ
ZZZ ∗
ZZZ
ZZ
Qe
1
ρ
~
~
E · n̂ dσ =
∇ · E dτ =
dτ =
=
ρ∗ dτ
0
0
S
V
V 0
V
which for an arbitrary volume implies
∗
~ = ρ .
∇·E
0
(2.6.23)
The equations (2.6.23) and (2.6.7) can be combined so that the Gauss law can also be written in the form
ρ∗
which is called Poisson’s equation.
∇2 V = −
0
EXAMPLE 2.6-2
Find the electric field associated with an infinite plane sheet of positive charge.
Solution: Assume there exists a uniform surface charge µ∗ and draw a circle at some point on the plane
surface. Now move the circle perpendicular to the surface to form a small cylinder which extends equal
distances above and below the plane surface. We calculate the electric flux over this small cylinder in the
limit as the height of the cylinder goes to zero. The charge inside the cylinder is µ∗ A where A is the area of
the circle. We find that the Gauss law requires that
ZZ
∗
~ · n̂ dσ = Qe = µ A
E
0
0
S
(2.6.24)
where n̂ is the outward normal to the cylinder as we move over the surface S. By the symmetry of the
situation the electric force vector is uniform and must point away from both sides to the plane surface in the
en and
direction of the normals to both sides of the surface. Denote the plane surface normals by b
en and − b
~ = −β b
~ = βb
en on the other side of the surface for some
assume that E
en on one side of the surface and E
constant β. Substituting this result into the equation (2.6.24) produces
ZZ
~ · n̂ dσ = 2βA
E
S
(2.6.25)
333
since only the ends of the cylinder contribute to the above surface integral. On the sides of the cylinder we
will have n̂ · ± b
en = 0 and so the surface integral over the sides of the cylinder is zero. By equating the
µ∗
and consequently we can write
results from equations (2.6.24) and (2.6.25) we obtain the result that β =
20
∗
~ = µ b
en where b
E
en represents one of the normals to the surface.
20
Note an electric field will always undergo a jump discontinuity when crossing a surface charge µ∗ . As in
∗
∗
~ down = − µ b
~ up = µ b
en so that the difference is
en and E
the above example we have E
20
2
∗
~ down = µ b
~ up − E
en
E
0
(1)
E i ni
or
(2)
+ E i ni
+
µ∗
= 0.
0
(2.6.26)
It is this difference which causes the jump discontinuity.
EXAMPLE 2.6-3.
Calculate the electric field associated with a uniformly charged sphere of radius a.
Solution: We proceed as in the previous example. Let µ∗ denote the uniform charge distribution over the
surface
er denote the unit normal to the sphere. The total charge then is written as
Z Z of the sphere and let b
∗
2 ∗
µ dσ = 4πa µ . If we construct a sphere of radius r > a around the charged sphere, then we have
q=
Sa
ZZZ
by the Gauss theorem
Qe
q
~·b
E
er dσ =
= .
0
0
Sr
(2.6.27)
~ and assume that it points radially outward in the direction of the
Again, we can assume symmetry for E
~ = βb
~ into the
er for some constant β. Substituting this value for E
surface normal b
er and has the form E
equation (2.6.27) we find that
ZZ
ZZ
~·b
E
er dσ = β
Sr
dσ = 4πβr2 =
Sr
q
.
0
(2.6.28)
1 q
b
er where b
er is the outward normal to the sphere. This shows that the electric field
4π0 r2
outside the sphere is the same as if all the charge were situated at the origin.
~ =
This gives E
For S a piecewise closed surface enclosing a volume V and F i = F i (x1 , x2 , x3 ) i = 1, 2, 3, a continuous
vector field with continuous derivatives the Gauss divergence theorem enables us to replace a flux integral
of F i over S by a volume integral of the divergence of F i over the volume V such that
ZZZ
ZZZ
ZZ
ZZ
i
i
~
F · n̂ dσ =
F ni dσ =
F ,i dτ or
div F~ dτ.
S
V
S
(2.6.29)
V
If V contains a simple closed surface Σ where F i is discontinuous we must modify the above Gauss divergence
theorem.
EXAMPLE 2.6-4.
We examine the modification of the Gauss divergence theorem for spheres in order to illustrate the
concepts. Let V have surface area S which encloses a surface Σ. Consider the figure 2.6-4 where the volume
V enclosed by S and containing Σ has been cut in half.
334
Figure 2.6-4. Sphere S containing sphere Σ.
Applying the Gauss divergence theorem to the top half of figure 2.6-4 gives
ZZ
ZZ
ZZZ
ZZ
i
T
F i nTi dσ +
F i nbi T dσ +
F i nΣ
dσ
=
F ,i
dτ
i
ST
ΣT
Sb1
(2.6.30)
VT
where the ni are the unit outward normals to the respective surfaces ST , Sb1 and ΣT . Applying the Gauss
divergence theorem to the bottom half of the sphere in figure 2.6-4 gives
ZZ
ZZ
ZZZ
ZZ
i B
i bB
i ΣB
F ni dσ +
F ni dσ +
F ni dσ =
SB
ΣB
Sb2
F i,i dτ
(2.6.31)
VB
Observe that the unit normals to the surfaces Sb1 and Sb2 are equal and opposite in sign so that adding the
equations (2.6.30) and (2.6.31) we obtain
ZZ
ZZ
i
F ni dσ +
S
Σ
(1)
F i ni
ZZZ
dσ =
VT +VB
i
F ,i
dτ
(2.6.32)
335
where S = ST + SB is the total surface area of the outside sphere and Σ = ΣT + ΣB is the total surface area
(1)
of the inside sphere, and ni
is the inward normal to the sphere Σ when the top and bottom volumes are
combined. Applying the Gauss divergence theorem to just the isolated small sphere Σ we find
ZZZ
ZZ
(2)
F i ni dσ =
F i,i dτ
Σ
(2)
where ni
(2.6.33)
VΣ
is the outward normal to Σ. By adding the equations (2.6.33) and (2.6.32) we find that
ZZ ZZ
i
F ni dσ +
S
Σ
(1)
F i ni
+
(2)
F i ni
ZZZ
F i,i dτ
(2.6.34)
ZZ (1)
(2)
F i ni + F i ni
dσ.
(2.6.35)
dσ =
V
where V = VT + VB + VΣ . The equation (2.6.34) can also be written as
ZZZ
ZZ
i
F ,i
dτ −
F i ni dσ =
S
Σ
V
In the case that V contains a surface Σ the total electric charge inside S is
ZZZ
ZZ
ρ∗ dτ +
µ∗ dσ
Qe =
(2.6.36)
Σ
V
where µ∗ is the surface charge density on Σ and ρ∗ is the volume charge density throughout V. The Gauss
theorem requires that
ZZ
Qe
1
E ni dσ =
=
0
0
S
ZZZ
1
ρ dτ +
0
∗
i
V
ZZ
µ∗ dσ.
(2.6.37)
Σ
In the case of a jump discontinuity across the surface Σ we use the results of equation (2.6.34) and write
ZZZ
ZZ
E i,i dτ −
E i ni dσ =
S
ZZ Σ
V
(1)
E i ni
(2)
+ E i ni
dσ.
(2.6.38)
Subtracting the equation (2.6.37) from the equation (2.6.38) gives
ZZ ZZZ ρ∗
µ∗
i
i (1)
i (2)
E ,i −
dτ −
E ni + E ni +
dσ = 0.
0
0
V
Σ
(2.6.39)
For arbitrary surfaces S and Σ, this equation implies the differential form of the Gauss law
E i,i =
ρ∗
.
0
(2.6.40)
Further, on the surface Σ, where there is a surface charge distribution we have
(1)
E i ni
(2)
+ E i ni
+
µ∗
=0
0
which shows the electric field undergoes a discontinuity when you cross a surface charge µ∗ .
(2.6.41)
336
Electrostatic Fields in Materials
When charges are introduced into materials it spreads itself throughout the material. Materials in
which the spreading occurs quickly are called conductors, while materials in which the spreading takes a
long time are called nonconductors or dielectrics. Another electrical property of materials is the ability to
hold local charges which do not come into contact with other charges. This property is called induction.
For example, consider a single atom within the material. It has a positively charged nucleus and negatively
~ the negative cloud
charged electron cloud surrounding it. When this atom experiences an electric field E
~ while the positively charged nucleus moves in the direction of E.
~ If E
~ is large enough it
moves opposite to E
can ionize the atom by pulling the electrons away from the nucleus. For moderately sized electric fields the
atom achieves an equilibrium position where the positive and negative charges are offset. In this situation
the atom is said to be polarized and have a dipole moment p~.
Definition: When a pair of charges +q and −q are separated by a distance 2d~ the electric dipole
~ where p~ has dimensions of [C m].
moment is defined by p~ = 2dq,
~ and the material is symmetric we say that p~
In the special case where d~ has the same direction as E
~ and write p~ = αE,
~ where α is called the atomic polarizability. If in a material subject
is proportional to E
to an electric field their results many such dipoles throughout the material then the dielectric is said to be
polarized. The vector quantity P~ is introduced to represent this effect. The vector P~ is called the polarization
vector having units of [C/m2 ], and represents an average dipole moment per unit volume of material. The
vectors Pi and Ei are related through the displacement vector Di such that
Pi = Di − 0 Ei .
(2.6.42)
For an anisotropic material (crystal)
Di = ji Ej
and
Pi = αji Ej
(2.6.43)
where ji is called the dielectric tensor and αji is called the electric susceptibility tensor. Consequently,
Pi = αji Ej = ji Ej − 0 Ei = (ji − 0 δij )Ej
so that
αji = ji − 0 δij .
(2.6.44)
A dielectric material is called homogeneous if the electric force and displacement vector are the same for any
two points within the medium. This requires that the electric force and displacement vectors be constant
parallel vector fields. It is left as an exercise to show that the condition for homogeneity is that ji,k = 0.
A dielectric material is called isotropic if the electric force vector and displacement vector have the same
direction. This requires that ji = δji where δji is the Kronecker delta. The term = 0 Ke is called the
dielectric constant of the medium. The constant 0 = 8.85(10)−12 coul2 /N · m2 is the permittivity of free
space and the quantity ke =
0
is called the relative dielectric constant (relative to 0 ). For free space ke = 1.
Similarly for an isotropic material we have αji = 0 αe δij where αe is called the electric susceptibility. For a
~ and E
~ are related by
linear medium the vectors P~ , D
Di = 0 Ei + Pi = 0 Ei + 0 αe Ei = 0 (1 + αe )Ei = 0 Ke Ei = Ei
(2.6.45)
337
where Ke = 1 + αe is the relative dielectric constant. The equation (2.6.45) are constitutive equations for
dielectric materials.
The effect of polarization is to produce regions of bound charges ρb within the material and bound
surface charges µb together with free charges ρf which are not a result of the polarization. Within dielectrics
en = µb for bound surface charges, where b
en is a
we have ∇ · P~ = ρb for bound volume charges and P~ · b
unit normal to the bounding surface of the volume. In these circumstances the expression for the potential
function is written
V=
1
4π0
ZZZ
V
1
ρb
dτ +
r
4π0
ZZ
S
µb
dσ
r
(2.6.46)
and the Gauss law becomes
~ = ρ∗ = ρb + ρf = −∇ · P~ + ρf
0 ∇ · E
or
~ + P~ ) = ρf .
∇(0 E
(2.6.47)
~ + P~ the Gauss law can also be written in the form
~ = 0 E
Since D
~ = ρf
∇·D
or
D,ii = ρf .
(2.6.48)
When no confusion arises we replace ρf by ρ. In integral form the Gauss law for dielectrics is written
ZZ
~ · n̂ dσ = Qf e
D
(2.6.49)
S
where Qf e is the total free charge density within the enclosing surface.
Magnetostatics
~ while a moving charge generates a magnetic field B.
~
A stationary charge generates an electric field E
Magnetic field lines associated with a steady current moving in a wire form closed loops as illustrated in the
figure 2.6-5.
Figure 2.6-5. Magnetic field lines.
The direction of the magnetic force is determined by the right hand rule where the thumb of the right
hand points in the direction of the current flow and the fingers of the right hand curl around in the direction
~ The force on a test charge Q moving with velocity V
~ in a magnetic field is
of the magnetic field B.
~ × B).
~
F~m = Q(V
(2.6.50)
The total electromagnetic force acting on Q is the electric force plus the magnetic force and is
h
i
~ + (V
~ × B)
~
F~ = Q E
(2.6.51)
338
which is known as the Lorentz force law. The magnetic force due to a line charge density λ∗ moving along
Z
a curve C is the line integral
~ × B)
~ =
λ∗ ds(V
F~mag =
C
Z
~
I~ × Bds.
(2.6.52)
C
Similarly, for a moving surface charge density moving on a surface
ZZ
ZZ
~ ×B
~ dσ
~ × B)
~ =
K
µ∗ dσ(V
F~mag =
S
(2.6.53)
S
and for a moving volume charge density
ZZZ
ZZ
~ dτ
~ × B)
~ =
J~ × B
ρ∗ dτ (V
F~mag =
V
(2.6.54)
V
~, K
~ = µ∗ V
~ and J~ = ρ∗ V
~ are respectively the current, the current per unit
where the quantities I~ = λ∗ V
length, and current per unit area.
A conductor is any material where the charge is free to move. The flow of charge is governed by Ohm’s
law. Ohm’s law states that the current density vector Ji is a linear function of the electric intensity or
Ji = σim Em , where σim is the conductivity tensor of the material. For homogeneous, isotropic conductors
σim = σδim so that Ji = σEi where σ is the conductivity and 1/σ is called the resistivity.
Surround a charge density ρ∗ with an arbitrary simple closed surface S having volume V and calculate
the flux of the current density across the surface. We find by the divergence theorem
ZZZ
ZZ
J~ · n̂ dσ =
∇ · J~ dτ.
S
(2.6.55)
V
If charge is to be conserved, the current flow out of the volume through the surface must equal the loss due
to the time rate of change of charge within the surface which implies
ZZZ
ZZZ
ZZZ
ZZ
d
∂ρ∗
dτ
J~ · n̂ dσ =
∇ · J~ dτ = −
ρ∗ dτ = −
dt
S
V
V
V ∂t
ZZZ or
V
∂ρ∗
∇ · J~ +
∂t
(2.6.56)
dτ = 0.
(2.6.57)
This implies that for an arbitrary volume we must have
∇ · J~ = −
∂ρ∗
.
∂t
(2.6.58)
Note that equation (2.6.58) has the same form as the continuity equation (2.3.73) for mass conservation and
so it is also called a continuity equation for charge conservation. For magnetostatics there exists steady line
∗
currents or stationary current so ∂ρ = 0. This requires that ∇ · J~ = 0.
∂t
339
Figure 2.6-6. Magnetic field around wire.
Biot-Savart Law
The Biot-Savart law for magnetostatics describes the magnetic field at a point P due to a steady line
current moving along a curve C and is
~ ) = µ0
B(P
4π
Z ~
I×b
er
ds
2
r
C
(2.6.59)
with units [N/amp · m] and where the integration is in the direction of the current flow. In the Biot-Savart
et is
law we have the constant µ0 = 4π × 10−7 N/amp2 which is called the permeability of free space, I~ = I b
er is a unit vector directed
the current flowing in the direction of the unit tangent vector b
et to the curve C, b
from a point on the curve C toward the point P and r is the distance from a point on the curve to the
general point P. Note that for a steady current to exist along the curve the magnitude of I~ must be the
same everywhere along the curve. Hence, this term can be brought out in front of the integral. For surface
~ and volume currents J~ the Biot-Savart law is written
currents K
ZZ
~ ×b
K
er
dσ
2
r
S
ZZZ ~
J×b
er
µ0
~
dτ.
B(P ) =
2
4π
r
V
~ ) = µ0
B(P
4π
and
EXAMPLE 2.6-5.
~ a distance h perpendicular to a wire carrying a constant current I.
~
Calculate the magnetic field B
Solution: The magnetic field circles around the wire. For the geometry of the figure 2.6-6, the magnetic
field points out of the page. We can write
et × b
er = Iê sin α
I~ × b
er = I b
where ê is a unit vector tangent to the circle of radius h which encircles the wire and cuts the wire perpendicularly.
340
For this problem the Biot-Savart law is
Z
~ ) = µ0 I
B(P
4π
ê
ds.
r2
In terms of θ we find from the geometry of figure 2.6-6
tan θ =
s
h
ds = h sec2 θ dθ
with
Therefore,
~ ) = µ0
B(P
π
Z
θ2
θ1
and
cos θ =
h
.
r
Iê sin α h sec2 θ
dθ.
h2 / cos2 θ
But, α = π/2 + θ so that sin α = cos θ and consequently
~ ) = µ0 Iê
B(P
4πh
Z
θ2
cos θ dθ =
θ1
µ0 Iê
(sin θ2 − sin θ1 ).
4πh
~ ) = µ0 Iê .
For a long straight wire θ1 → −π/2 and θ2 → π/2 to give the magnetic field B(P
2πh
For volume currents the Biot-Savart law is
~ ) = µ0
B(P
4π
ZZZ
V
J~ × b
er
dτ
r2
(2.6.60)
and consequently (see exercises)
~ = 0.
∇·B
(2.6.61)
∗
~ = ρ is known as the Gauss’s law for electric fields and so
Recall the divergence of an electric field is ∇ · E
0
~ = 0 is sometimes referred to as Gauss’s law for magnetic fields. If ∇ · B
~ = 0,
in analogy the divergence ∇ · B
~ such that B
~ = ∇ × A.
~ The vector field A
~ is called the vector potential of
then there exists a vector field A
~ Note that ∇ · B
~ = ∇ · (∇ × A)
~ = 0. Also the vector potential A
~ is not unique since B
~ is also derivable
B.
~ + ∇φ where φ is an arbitrary continuous and differentiable scalar.
from the vector potential A
Ampere’s Law
Ampere’s law is associated with the work done in moving around a simple closed path. For example,
~ around a circular path of radius h
consider the previous example 2.6-5. In this example the integral of B
which is centered at some point on the wire can be associated with the work done in moving around this
path. The summation of force times distance is
Z
Z
Z
~ · d~r =
~ · ê ds = µ0 I ds = µ0 I
B
B
2πh C
C
C
(2.6.62)
Z
ds = 2πh is the distance
where now d~r = ê ds is a tangent vector to the circle encircling the wire and
C
around this circle. The equation (2.6.62) holds not only for circles, but for any simple closed curve around
the wire. Using the Stoke’s theorem we have
ZZ
ZZ
Z
~ · d~r =
~ ·b
en dσ
B
(∇ × B)
en dσ = µ0 I =
µ0 J~ · b
C
S
S
(2.6.63)
341
ZZ
J~ · b
en dσ is the total flux (current) passing through the surface which is created by encircling
where
S
some curve about the wire. Equating like terms in equation (2.6.63) gives the differential form of Ampere’s
law
~
~ = µ0 J.
∇×B
(2.6.64)
Magnetostatics in Materials
Similar to what happens when charges are introduced into materials we have magnetic fields whenever
there are moving charges within materials. For example, when electrons move around an atom tiny current
loops are formed. These current loops create what are called magnetic dipole moments m
~ throughout the
~
material. When a magnetic field B is applied to a material medium there is a net alignment of the magnetic
~ , called the magnetization vector is introduced. Here M
~ is associated with a
dipoles. The quantity M
dielectric medium and has the units [amp/m] and represents an average magnetic dipole moment per unit
~
volume and is analogous to the polarization vector P~ used in electrostatics. The magnetization vector M
acts a lot like the previous polarization vector in that it produces bound volume currents J~b and surface
~ = J~b is a volume current density throughout some volume and M
~ ×b
~ b is a
~ b where ∇ × M
en = K
currents K
surface current on the boundary of this volume.
~
From electrostatics note that the time derivative of 0 ∂∂tE has the same units as current density. The
~
total current in a magnetized material is then J~t = J~b + J~f + 0 ∂ E where J~b is the bound current, J~f is the
∂t
~
free current and 0 ∂∂tE is the induced current. Ampere’s law, equation (2.6.64), in magnetized materials then
becomes
~
~
~ = µ0 J~t = µ0 (J~b + J~f + 0 ∂ E ) = µ0 J~ + µ0 0 ∂ E
∇×B
∂t
∂t
(2.6.65)
~
where J~ = J~b + J~f . The term 0 ∂∂tE is referred to as a displacement current or as a Maxwell correction to
the field equation. This term implies that a changing electric field induces a magnetic field.
~ defined by
An auxiliary magnet field H
Hi =
1
Bi − Mi
µ0
(2.6.66)
~ and magnetization vector M
~ . This is another conis introduced which relates the magnetic force vector B
stitutive equation which describes material properties. For an anisotropic material (crystal)
Bi = µji Hj
Mi = χji Hj
and
(2.6.67)
where µji is called the magnetic permeability tensor and χji is called the magnetic permeability tensor. Both
of these quantities are dimensionless. For an isotropic material
µji = µδij
where
µ = µ0 km .
Here µ0 = 4π × 10−7 N/amp2 is the permeability of free space and km =
coefficient. Similarly, for an isotropic material we have
χji
=
χm δij
(2.6.68)
µ
µ0
is the relative permeability
where χm is called the magnetic sus-
ceptibility coefficient and is dimensionless. The magnetic susceptibility coefficient has positive values for
342
materials called paramagnets and negative values for materials called diamagnets. For a linear medium the
~ M
~ and H
~ are related by
quantities B,
Bi = µ0 (Hi + Mi ) = µ0 Hi + µ0 χm Hi = µ0 (1 + χm )Hi = µ0 km Hi = µHi
(2.6.69)
where µ = µ0 km = µ0 (1 + χm ) is called the permeability of the material.
~ for magnetostatics in materials plays a role similar to the
Note: The auxiliary magnetic vector H
~ for electrostatics in materials. Be careful in using electromagnetic equations from
displacement vector D
~ and H.
~ Some authors call H
~ the magnetic field.
different texts as many authors interchange the roles of B
~ should be the fundamental quantity.1
However, the quantity B
Electrodynamics
In the nonstatic case of electrodynamics there is an additional quantity J~p =
current which satisfies
∂
∂ρb
∂ P~
= ∇ · P~ = −
∇ · J~p = ∇ ·
∂t
∂t
∂t
~
∂P
∂t
called the polarization
(2.6.70)
and the current density has three parts
~
~ + J~f + ∂ P
J~ = J~b + J~f + J~p = ∇ × M
∂t
(2.6.71)
consisting of bound, free and polarization currents.
Faraday’s law states that a changing magnetic field creates an electric field. In particular, the electromagnetic force induced in a closed loop circuit C is proportional to the rate of change of flux of the magnetic
field associated with any surface S connected with C. Faraday’s law states
ZZ
Z
∂
~·b
~
B
en dσ.
E · d~r = −
∂t S
C
Using the Stoke’s theorem, we find
ZZ
ZZ
~ ·b
(∇ × E)
en dσ = −
S
S
~
∂B
·b
en dσ.
∂t
The above equation must hold for an arbitrary surface and loop. Equating like terms we obtain the differential
form of Faraday’s law
~ =−
∇×E
~
∂B
.
∂t
(2.6.72)
This is the first electromagnetic field equation of Maxwell.
Ampere’s law, equation (2.6.65), written in terms of the total current from equation (2.6.71) , becomes
~
~
~ + J~f + ∂ P ) + µ0 0 ∂ E
~ = µ0 (∇ × M
∇×B
∂t
∂t
which can also be written as
∇×(
1
1 ~
~ ) = J~f + ∂ (P~ + 0 E)
~
B−M
µ0
∂t
D.J. Griffiths, Introduction to Electrodynamics, Prentice Hall, 1981. P.232.
(2.6.73)
343
or
~ = J~f +
∇×H
~
∂D
.
∂t
(2.6.74)
This is Maxwell’s second electromagnetic field equation.
To the equations (2.6.74) and (2.6.73) we add the Gauss’s law for magnetization, equation (2.6.61) and
Gauss’s law for electrostatics, equation (2.6.48). These four equations produce the Maxwell’s equations of
electrodynamics and are now summarized. The general form of Maxwell’s equations involve the quantities
Ei , Electric force vector, [Ei ] = Newton/coulomb
Bi , Magnetic force vector, [Bi ] = Weber/m2
Hi , Auxilary magnetic force vector, [Hi ] = ampere/m
Di , Displacement vector, [Di ] = coulomb/m2
Ji , Free current density, [Ji ] = ampere/m2
Pi , Polarization vector, [Pi ] = coulomb/m2
Mi , Magnetization vector, [Mi ] = ampere/m
for i = 1, 2, 3. There are also the quantities
%, representing the free charge density, with units [%] = coulomb/m3
0 , Permittivity of free space, [0 ] = farads/m or coulomb2 /Newton · m2 .
µ0 , Permeability of free space, [µ0 ] = henrys/m or kg · m/coulomb2
In addition, there arises the material parameters:
µij , magnetic permeability tensor, which is dimensionless
ij , dielectric tensor, which is dimensionless
αij , electric susceptibility tensor, which is dimensionless
χij , magnetic susceptibility tensor, which is dimensionless
These parameters are used to express variations in the electric field Ei and magnetic field Bi when
acting in a material medium. In particular, Pi , Di , Mi and Hi are defined from the equations
Di =ji Ej = 0 Ei + Pi
ij = 0 δji + αji
Bi =µji Hj = µ0 Hi + µ0 Mi ,
Pi =αji Ej ,
and
µij = µ0 (δji + χij )
Mi = χji Hj
for i = 1, 2, 3.
The above quantities obey the following laws:
Faraday’s Law
This law states the line integral of the electromagnetic force around a loop is proportional
to the rate of flux of magnetic induction through the loop. This gives rise to the first electromagnetic field
equation:
~
~ = − ∂B
∇×E
∂t
or
ijk Ek,j = −
∂B i
.
∂t
(2.6.75)
344
Ampere’s Law
This law states the line integral of the magnetic force vector around a closed loop is
proportional to the sum of the current through the loop and the rate of flux of the displacement vector
through the loop. This produces the second electromagnetic field equation:
~
~ = J~f + ∂ D
∇×H
∂t
ijk Hk,j = Jfi +
or
∂Di
.
∂t
(2.6.76)
Gauss’s Law for Electricity This law states that the flux of the electric force vector through a closed
surface is proportional to the total charge enclosed by the surface. This results in the third electromagnetic
field equation:
~ = ρf
∇·D
D,ii = ρf
or
or
1 ∂ √ i
gD = ρf .
√
g ∂xi
(2.6.77)
Gauss’s Law for Magnetism This law states the magnetic flux through any closed volume is zero. This
produces the fourth electromagnetic field equation:
~ =0
∇·B
B,ii = 0
or
or
1 ∂ √ i
gB = 0.
√
g ∂xi
(2.6.78)
When no confusion arises it is convenient to drop the subscript f from the above Maxwell equations.
Special expanded forms of the above Maxwell equations are given on the pages 176 to 179.
Electromagnetic Stress and Energy
Let V denote the volume of some simple closed surface S. Let us calculate the rate at which electromagnetic energy is lost from this volume. This represents the energy flow per unit volume. Begin with the
first two Maxwell’s equations in Cartesian form
∂Bi
∂t
∂Di
.
=Ji +
∂t
ijk Ek,j = −
(2.6.79)
ijk Hk,j
(2.6.80)
Now multiply equation (2.6.79) by Hi and equation (2.6.80) by Ei . This gives two terms with dimensions of
energy per unit volume per unit of time which we write
∂Bi
Hi
∂t
∂Di
Ei .
ijk Hk,j Ei =Ji Ei +
∂t
ijk Ek,j Hi = −
(2.6.81)
(2.6.82)
Subtracting equation (2.6.82) from equation (2.6.81) we find
∂Di
Ei −
∂t
∂Di
Ei −
+ Hi,j Ek ] = − Ji Ei −
∂t
ijk (Ek,j Hi − Hk,j Ei ) = − Ji Ei −
ijk [(Ek Hi ),j − Ek Hi,j
∂Bi
Hi
∂t
∂Bi
Hi
∂t
Observe that jki (Ek Hi ),j is the same as ijk (Ej Hk ),i so that the above simplifies to
ijk (Ej Hk ),i + Ji Ei = −
∂Di
∂Bi
Ei −
Hi .
∂t
∂t
(2.6.83)
345
Now integrate equation (2.6.83) over a volume and apply Gauss’s divergence theorem to obtain
ZZZ
ZZ Z
ZZ
∂Di
∂Bi
Ei +
Hi ) dτ.
ijk Ej Hk ni dσ +
Ji Ei dτ = −
(
∂t
∂t
S
V
V
(2.6.84)
The first term in equation (2.6.84) represents the outward flow of energy across the surface enclosing the
volume. The second term in equation (2.6.84) represents the loss by Joule heating and the right-hand side
is the rate of decrease of stored electric and magnetic energy. The equation (2.6.84) is known as Poynting’s
theorem and can be written in the vector form
ZZZ
ZZ
~
~
~ · ∂B − E
~ · J)
~ dτ.
~ × H)
~ · n̂ dσ =
~ · ∂D − H
(E
(−E
∂t
∂t
S
V
(2.6.85)
For later use we define the quantity
Si = ijk Ej Hk
~=E
~ ×H
~
or S
[Watts/m2 ]
(2.6.86)
as Poynting’s energy flux vector and note that Si is perpendicular to both Ei and Hi and represents units
of energy density per unit time which crosses a unit surface area within the electromagnetic field.
Electromagnetic Stress Tensor
Instead of calculating energy flow per unit volume, let us calculate force per unit volume. Consider a
region containing charges and currents but is free from dielectrics and magnetic materials. To obtain terms
with units of force per unit volume we take the cross product of equation (2.6.79) with Di and the cross
product of equation (2.6.80) with Bi and subtract to obtain
−irs ijk (Ek,j Ds + Hk,j Bs ) = ris Ji Bs + ris
∂Di
∂Bs
Bs +
Di
∂t
∂t
which simplifies using the e − δ identity to
−(δrj δsk − δrk δsj )(Ek,j Ds + Hk,j Bs ) = ris Ji Bs + ris
∂
(Di Bs )
∂t
which further simplifies to
−Es,r Ds + Er,s Ds − Hs,r Bs + Hr,s Bs = ris Ji Bs +
∂
(ris Di Bs ).
∂t
Observe that the first two terms in the equation (2.6.87) can be written
Er,s Ds − Es,r Ds =Er,s Ds − 0 Es,r Es
1
=(Er Ds ),s − Er Ds,s − 0 ( Es Es ),r
2
1
=(Er Ds ),s − ρEr − (Ej Dj δsr ),s
2
1
=(Er Ds − Ej Dj δrs ),s − ρEr
2
which can be expressed in the form
E
− ρEr
Er,s Ds − Es,r Ds = Trs,s
(2.6.87)
346
where
1
E
= Er Ds − Ej Dj δrs
Trs
2
(2.6.88)
is called the electric stress tensor. In matrix form the stress tensor is written


E1 D1 − 12 Ej Dj
E1 D2
E1 D3
E
.
=
E2 D1
E2 D2 − 12 Ej Dj
E2 D3
Trs
E3 D1
E3 D2
E3 D3 − 12 Ej Dj
(2.6.89)
By performing similar calculations we can transform the third and fourth terms in the equation (2.6.87) and
obtain
M
Hr,s Bs − Hs,r Bs = Trs,s
(2.6.90)
1
M
= Hr BS − Hj Bj δrs
Trs
2
(2.6.91)
where
is the magnetic stress tensor. In matrix form the magnetic stress tensor is written

M
Trs
B1 H1 − 12 Bj Hj
=
B2 H1
B3 H1
B1 H2
B2 H2 − 12 Bj Hj
B3 H2

B1 H3
.
B2 H3
B3 H3 − 12 Bj Hj
(2.6.92)
The total electromagnetic stress tensor is
E
M
+ Trs
.
Trs = Trs
(2.6.93)
Then the equation (2.6.87) can be written in the form
Trs,s − ρEr = ris Ji Bs +
∂
(ris Di Bs )
∂t
ρEr + ris Ji BS = Trs,s −
∂
(ris Di Bs ).
∂t
or
(2.6.94)
For free space Di = 0 Ei and Bi = µ0 Hi so that the last term of equation (2.6.94) can be written in terms
of the Poynting vector as
µ0 0
∂
∂Sr
= (ris Di Bs ).
∂t
∂t
(2.6.95)
Now integrate the equation (2.6.94) over the volume to obtain the total electromagnetic force
ZZZ
ZZZ
ρEr dτ +
V
ZZZ
V
ZZZ
Trs,s dτ − µ0 0
ris Ji Bs dτ =
V
V
∂Sr
dτ.
∂t
Applying the divergence theorem of Gauss gives
ZZZ
ZZ
ZZZ
ZZZ
∂Sr
dτ.
ρEr dτ +
ris Ji Bs dτ =
Trs ns dσ − µ0 0
V
V
S
V ∂t
(2.6.96)
The left side of the equation (2.6.96) represents the forces acting on charges and currents contained within
the volume element. If the electric and magnetic fields do not vary with time, then the last term on the
right is zero. In this case the forces can be expressed as an integral of the electromagnetic stress tensor.
347
EXERCISE 2.6
I 1.
Find the field lines and equipotential curves associated with a positive charge q located at (−a, 0) and
a positive charge q located at (a, 0). The field lines are illustrated in the figure 2.6-7.
Figure 2.6-7. Lines of electric force between two charges of the same sign.
I 2. Calculate the lines of force and equipotential curves associated with the electric field
~ = E(x,
~
e2 . Sketch the lines of force and equipotential curves. Put arrows on the lines of
E
y) = 2y b
e1 + 2x b
force to show direction of the field lines.
I 3.
A right circular cone is defined by
x = u sin θ0 cos φ,
y = u sin θ0 sin φ,
z = u cos θ0
with 0 ≤ φ ≤ 2π and u ≥ 0. Show the solid angle subtended by this cone is Ω =
A
r2
= 2π(1 − cos θ0 ).
I 4.
A charge +q is located at the point (0, a) and a charge −q is located at the point (0, −a). Show that
−2aq
~ at the position (x, 0), where x > a is E
~ = 1
b
e2 .
the electric force E
4π0 (a2 + x2 )3/2
Let the circle x2 + y 2 = a2 carry a line charge λ∗ . Show the electric field at the point (0, 0, z) is
∗
e3
~ = 1 λ az(2π) b
.
E
2
2
3/2
4π0 (a + z )
I 5.
I 6.
Use superposition to find the electric field associated with two infinite parallel plane sheets each
carrying an equal but opposite sign surface charge density µ∗ . Find the field between the planes and outside
∗
µ
and perpendicular to plates.
of each plane. Hint: Fields are of magnitude ± 2
0
ZZZ ~
J×b
er
µ0
~ = 0.
~
~
dτ. Show that ∇ · B
I 7. For a volume current J the Biot-Savart law gives B =
2
4π
r
V
~r
~r
Hint: Let b
er = and consider ∇ · (J~ × 3 ). Then use numbers 13 and 10 of the appendix C. Also note that
r
r
∇ × J~ = 0 because J~ does not depend upon position.
348
I 8.
A homogeneous dielectric is defined by Di and Ei having parallel vector fields. Show that for a
homogeneous dielectric ji,k = 0.
I 9.
I 10.
Show that for a homogeneous, isotropic dielectric medium that is a constant.
Show that for a homogeneous, isotropic linear dielectric in Cartesian coordinates
Pi,i =
I 11.
αe
ρf .
1 + αe
Verify the Maxwell’s equations in Gaussian units for a charge free isotropic homogeneous dielectric.
~ =0
~ =1∇ · D
∇·E
~ =µ∇H
~ =0
∇·B
I 12.
~
~
~ = − 1 ∂B = − µ ∂H
∇×E
c ∂t
c ∂t
~
~
∂
D
4π
∂E
4π ~
1
~ =
+
J~ =
+
σE
∇×H
c ∂t
c
c ∂t
c
Verify the Maxwell’s equations in Gaussian units for an isotropic homogeneous dielectric with a
charge.
~ =4πρ
∇·D
~ =0
∇·B
I 13.
~
1 ∂B
c ∂t
~
4π
~ = J~ + 1 ∂ D
∇×H
c
c ∂t
~ =−
∇×E
For a volume charge ρ in an element of volume dτ located at a point (ξ, η, ζ) Coulombs law is
ZZZ
ρ
1
~
b
e dτ
E(x,
y, z) =
2 r
4π0
r
V
(a) Show that r2 = (x − ξ)2 + (y − η)2 + (z − ζ)2 .
1
e1 + (y − η) b
e2 + (z − ζ) b
e3 ) .
(b) Show that b
er = ((x − ξ) b
r
(c) Show that
ZZZ
ZZZ
b
er
1
(x − ξ) b
e1 + (y − η) b
e2 + (z − ζ) b
e3
1
~
ρ dξdηdζ =
∇
E(x, y, z) =
ρ dξdηdζ
2
2
2
3/2
4π0
4π0
r2
V [(x − ξ) + (y − η) + (z − ζ)
V
Z Z ]Z
ρ(ξ, η, ζ)
~ is V = 1
dξdηdζ
(d) Show that the potential function for E
2
2
2 1/2
4π0
V [(x − ξ) + (y − η) + (z − ζ) ]
~ = −∇V.
(e) Show that E
ρ
(f) Show that ∇2 V = − Hint: Note that the integrand is zero everywhere except at the point where
(ξ, η, ζ) = (x, y, z). Consider the integral split into two regions. One region being a small sphere
about the
point
(x, y, z) in the
limit
as the radius of this sphere approaches zero. Observe the identity
b
b
er
er
= −∇(ξ, η, ζ)
enables one to employ the Gauss divergence theorem to obtain a
∇(x,y,z)
r2
r2
ZZ
b
er
ρ
ρ
· n̂dS =
4π since n̂ = − b
er .
surface integral. Use a mean value theorem to show −
2
4π0
r
4π
0
S
I 14.
Show that for a point charge in space ρ∗ = qδ(x − x0 )δ(y − y0 )δ(z − z0 ), where δ is the Dirac delta
function, the equation (2.6.5) can be reduced to the equation (2.6.1).
I 15.
1
r2
b
er is irrotational. Here b
er = ~rr is a unit vector in the direction of r.
~ = −∇V which satisfies V(r0 ) = 0 for r0 > 0.
(b) Find the potential function V such that E
~ =
(a) Show the electric field E
349
I 16.
~ is a conservative electric field such that E
~ = −∇V, then show that E
~ is irrotational and satisfies
(a) If E
~ = curl E
~ = 0.
∇×E
~ = curl E
~ = 0, show that E
~ is conservative. (i.e. Show E
~ = −∇V.)
(b) If ∇ × E
Hint: The work done on a test charge Q = 1 along the straight line segments from (x0 , y0 , z0 ) to
(x, y0 , z0 ) and then from (x, y0 , z0 ) to (x, y, z0 ) and finally from (x, y, z0 ) to (x, y, z) can be written
Z
Z
x
Z
y
E1 (x, y0 , z0 ) dx −
V = V(x, y, z) = −
x0
y0
Now note that
∂V
= −E2 (x, y, z0 ) −
∂y
z
E2 (x, y, z0 ) dy −
Z
E3 (x, y, z) dz.
z0
z
z0
∂E3 (x, y, z)
dz
∂y
~ = 0 we find ∂E3 = ∂E2 , which implies ∂V = −E2 (x, y, z). Similar results are obtained
and from ∇ × E
∂y
∂z
∂y
∂V
∂V
~
and
. Hence show −∇V = E.
for
∂x
∂z
I 17.
~ = 0, then there exists some vector field A
~ such that B
~ = ∇ × A.
~
(a) Show that if ∇ · B
~
~
The vector field A is called the vector potential of B.
Z 1
~
~
sB(sx,
sy, sz) × ~r ds where ~r = x b
e1 + y b
e2 + z b
e3
Hint: Let A(x, y, z) =
0
Z 1
dBi 2
s ds by parts.
and integrate
ds
0
~ = 0.
(b) Show that ∇ · (∇ × A)
I 18.
Use Faraday’s law and Ampere’s law to show
g im (E j,j ),m − g jm E i,mj = −µ0
I 19.
∂
∂E i
J i + 0
∂t
∂t
~ where σ is the conductivity. Show that for ρ = 0 Maxwell’s equations produce
Assume that J~ = σ E
~
~
∂E
∂2E
~
+ µ0 0 2 =∇2 E
∂t
∂t
~
~
∂B
∂2B
~
+ µ0 0 2 =∇2 B.
µ0 σ
∂t
∂t
µ0 σ
and
~ and B
~ satisfy the same equation which is known as the telegrapher’s equation.
Here both E
I 20.
Show that Maxwell’s equations (2.6.75) through (2.6.78) for the electric field under electrostatic
conditions reduce to
~ =0
∇×E
~ =ρf
∇·D
~ is irrotational so that E
~ = −∇V. Show that ∇2 V = − ρf .
Now E
350
I 21.
Show that Maxwell’s equations (2.6.75) through (2.6.78) for the magnetic field under magnetostatic
~ = J~ and ∇ · B
~ = 0. The divergence of B
~ being zero implies B
~ can be derived
conditions reduce to ∇ × H
~ such that B
~ = ∇ × A.
~ Here A
~ is not unique, see problem 24. If we select
from a vector potential function A
~ such that ∇ · A
~ = 0 then show for a homogeneous, isotropic material, free of any permanent magnets, that
A
~ = −µJ.
~
∇2 A
I 22.
Show that under nonsteady state conditions of electrodynamics the Faraday law from Maxwell’s
~ = −∇V. Why is this? Observe that
equations (2.6.75) through (2.6.78) does not allow one to set E
~ = 0 so we can write B
~ = ∇×A
~ for some vector potential A.
~ Using this vector potential show that
∇·B
!
~
~ + ∂ A = 0. This shows that the quantity inside the parenthesis is
Faraday’s law can be written ∇ × E
∂t
~
~ + ∂ A = −∇V for some scalar potential V. The representation
conservative and so we can write E
∂t
~
~ = −∇V − ∂ A
E
∂t
is a more general representation of the electric potential. Observe that for steady state conditions
~
∂A
∂t
=0
so that this potential representation reduces to the previous one for electrostatics.
~
~ = −∇V − ∂ A derived in problem 22, show that in a vacuum
Using the potential formulation E
∂t
~
ρ
∂∇
·
A
=−
(a) Gauss law can be written ∇2 V +
∂t
0
(b) Ampere’s law can be written
~
∂V
∂2A
~
~
− µ0 0 2
∇ × ∇ × A = µ0 J − µ0 0 ∇
∂t
∂t
I 23.
(c) Show the result in part (b) can also be expressed in the form
!
~
∂
A
∂V
2~
~
− ∇ ∇ · A + µ0 0
= −µ0 J~
∇ A − µ0 0
∂t
∂t
I 24.
The Maxwell equations in a vacuum have the form
~
~ = ∂ D + ρ V~
∇×H
∂t
~
~ = − ∂B
∇×E
∂t
~ =ρ
∇·D
~ =0
∇·B
~ = µ0 H
~ with 0 and µ0 constants satisfying 0 µ0 = 1/c2 where c is the speed of light.
B
~
~ and scalar potential V defined by B
~ = ∇×A
~ and E
~ = − ∂ A − ∇ V.
Introduce the vector potential A
∂t
Note that the vector potential is not unique. For example, given ψ as a scalar potential we can write
~ = ∇×A
~ = ∇ × (A
~ + ∇ ψ), since the curl of a gradient is zero. Therefore, it is customary to impose some
B
~
~ = 0 E,
where D
~ and B
~ are
kind of additional requirement on the potentials. These additional conditions are such that E
∂V
1
~ and V satisfy ∇ · A
~+
= 0. This relation is known as the
not changed. One such condition is that A
c2 ∂t
~ and V and show
Lorentz relation or Lorentz gauge. Find the Maxwell’s equations in a vacuum in terms of A
that
1 ∂2
∇ − 2 2
c ∂t
2
ρ
V=−
0
and
1 ∂2
∇ − 2 2
c ∂t
2
~ = −µ0 ρV
~.
A
351
I 25.
~ and B
~ satisfy
In a vacuum show that E
~ =
∇2 E
~
1 ∂2E
c2 ∂t2
~ =
∇2 B
~
1 ∂2B
c2 ∂t2
~ =0
∇·E
~ =0
∇B
I 26.
(a) Show that the wave equations in problem 25 have solutions in the form of waves traveling in the
x- direction given by
~ = E(x,
~
~ 0 ei(kx±ωt)
E
t) = E
and
~ = B(x,
~
~ 0 ei(kx±ωt)
B
t) = B
~ 0 and B
~ 0 are constants. Note that wave functions of the form u = Aei(kx±ωt) are called plane
where E
harmonic waves. Sometimes they are called monochromatic waves. Here i2 = −1 is an imaginary unit.
Euler’s identity shows that the real and imaginary parts of these type wave functions have the form
A cos(kx ± ωt)
and
A sin(kx ± ωt).
These represent plane waves. The constant A is the amplitude of the wave , ω is the angular frequency,
and k/2π is called the wave number. The motion is a simple harmonic motion both in time and space.
That is, at a fixed point x the motion is simple harmonic in time and at a fixed time t, the motion is
harmonic in space. By examining each term in the sine and cosine terms we find that x has dimensions of
length, k has dimension of reciprocal length, t has dimensions of time and ω has dimensions of reciprocal
time or angular velocity. The quantity c = ω/k is the wave velocity. The value λ = 2π/k has dimension
of length and is called the wavelength and 1/λ is called the wave number. The wave number represents
the number of waves per unit of distance along the x-axis. The period of the wave is T = λ/c = 2π/ω
and the frequency is f = 1/T. The frequency represents the number of waves which pass a fixed point
in a unit of time.
(b) Show that ω = 2πf
(c) Show that c = f λ
(d) Is the wave motion u = sin(kx − ωt) + sin(kx + ωt) a traveling wave? Explain.
1 ∂2φ
(e) Show that in general the wave equation ∇2 φ = 2 2 have solutions in the form of waves traveling in
c ∂t
either the +x or −x direction given by
φ = φ(x, t) = f (x + ct) + g(x − ct)
where f and g are arbitrary twice differentiable functions.
(f) Assume a plane electromagnetic wave is moving in the +x direction. Show that the electric field is in
the xy−plane and the magnetic field is in the xz−plane.
Hint: Assume solutions Ex = g1 (x − ct),
Ey = g2 (x − ct), Ez = g3 (x − ct), Bx = g4 (x − ct),
By = g5 (x − ct), Bz = g6 (x − ct) where gi ,i = 1, ..., 6 are arbitrary functions. Then show that Ex
~ = 0 which implies g1 must be independent of x and so not a wave function. Do
does not satisfy ∇ · E
~ Since both ∇ · E
~ = ∇·B
~ = 0 then Ex = Bx = 0. Such waves
the same for the components of B.
are called transverse waves because the electric and magnetic fields are perpendicular to the direction
~ and B
~ waves must be in phase and be mutually
of propagation. Faraday’s law implies that the E
perpendicular to each other.
352
BIBLIOGRAPHY
• Abramowitz, M. and Stegun, I.A., Handbook of Mathematical Functions, 10th ed,
New York:Dover, 1972.
• Akivis, M.A., Goldberg, V.V., An Introduction to Linear Algebra and Tensors, New York:Dover, 1972.
• Aris, Rutherford, Vectors, Tensors, and the Basic Equations of Fluid Mechanics,
Englewood Cliffs, N.J.:Prentice-Hall, 1962.
• Atkin, R.J., Fox, N., An Introduction to the Theory of Elasticity,
London:Longman Group Limited, 1980.
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• Borisenko, A.I., Tarapov, I.E., Vector and Tensor Analysis with Applications, New York:Dover, 1968.
• Chorlton, F., Vector and Tensor Methods, Chichester,England:Ellis Horwood Ltd, 1976.
• Dodson, C.T.J., Poston, T., Tensor Geometry, London:Pittman Publishing Co., 1979.
• Eisenhart, L.P., Riemannian Geometry, Princeton, N.J.:Univ. Princeton Press, 1960.
• Eringen, A.C., Mechanics of Continua, Huntington, N.Y.:Robert E. Krieger, 1980.
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• Flügge, W., Tensor Analysis and Continuum Mechanics, New York:Springer-Verlag, 1972.
• Fung, Y.C., A First Course in Continuum Mechanics, Englewood Cliffs,N.J.:Prentice-Hall, 1969.
• Goodbody, A.M., Cartesian Tensors, Chichester, England:Ellis Horwood Ltd, 1982.
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• Levi-Civita, T., The Absolute Differential Calculus, London:Blackie and Son Limited, 1954.
• Lovelock, D., Rund, H. ,Tensors, Differential Forms, and Variational Principles, New York:Dover, 1989.
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Englewood Cliffs, N.J.:Prentice-Hall, 1969.
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Bibliography
353
APPENDIX A
UNITS OF MEASUREMENT
The following units, abbreviations and prefixes are from the
Système International d’Unitès
(designated SI in all Languages.)
Prefixes.
Abreviations
Multiplication factor
1012
109
106
103
102
10
10−1
10−2
10−3
10−6
10−9
10−12
Symbol
T
G
M
K
h
da
d
c
m
µ
n
p
Basic units of measurement
Name
Length
meter
Mass
kilogram
Time
second
Electric current
ampere
Temperature
degree Kelvin
Luminous intensity
candela
Symbol
m
kg
s
A
◦
K
cd
Prefix
tera
giga
mega
kilo
hecto
deka
deci
centi
milli
micro
nano
pico
Basic Units.
Unit
Unit
Plane angle
Solid angle
Supplementary units
Name
radian
steradian
Symbol
rad
sr
354
Name
Area
Volume
Frequency
Density
Velocity
Angular velocity
Acceleration
Angular acceleration
Force
Pressure
Kinematic viscosity
Dynamic viscosity
Work, energy, quantity of heat
Power
Electric charge
Voltage, Potential difference
Electromotive force
Electric force field
Electric resistance
Electric capacitance
Magnetic flux
Inductance
Magnetic flux density
Magnetic field strength
Magnetomotive force
DERIVED UNITS
Units
square meter
cubic meter
hertz
kilogram per cubic meter
meter per second
radian per second
meter per second squared
radian per second squared
newton
newton per square meter
square meter per second
newton second per square meter
joule
watt
coulomb
volt
volt
volt per meter
ohm
farad
weber
henry
tesla
ampere per meter
ampere
Symbol
m2
m3
−1
Hz (s )
kg/m3
m/s
rad/s
m/s2
rad/s2
N (kg · m/s2 )
N/m2
m2 /s
N · s/m2
J (N · m)
W (J/s)
C (A · s)
V (W/A)
V (W/A)
V/m
Ω (V/A)
F (A · s/V)
Wb (V · s)
H (V · s/A)
T (Wb/m2 )
A/m
A
Physical constants.
4 arctan 1 = π = 3.14159 26535 89793 23846 2643 . . .
n
1
= e = 2.71828 18284 59045 23536 0287 . . .
lim 1 +
n→∞
n
Euler’s constant γ = 0.57721 56649 01532 86060 6512 . . .
1
1 1
γ = lim 1 + + + · · · + − log n
n→∞
2 3
n
speed of light in vacuum = 2.997925(10)8 m s−1
electron charge = 1.60210(10)−19 C
Avogadro’s constant = 6.02252(10)23 mol−1
Plank’s constant = 6.6256(10)−34 J s
Universal gas constant = 8.3143 J K −1 mol−1 = 8314.3 J Kg −1 K −1
Boltzmann constant = 1.38054(10)−23 J K −1
Stefan–Boltzmann constant = 5.6697(10)−8 W m−2 K −4
Gravitational constant = 6.67(10)−11 N m2 kg −2
355
APPENDIX B
CHRISTOFFEL SYMBOLS OF SECOND KIND
1. Cylindrical coordinates (r, θ, z) = (x1 , x2 , x3 )
x = r cos θ
r≥0
h1 = 1
y = r sin θ
0 ≤ θ ≤ 2π
h2 = r
z=z
−∞<z <∞
h3 = 1
The coordinate curves are formed by the intersection of the coordinate surfaces
x2 + y 2 = r2 ,
Cylinders
y/x = tan θ
Planes
z = Constant
1
22
= −r
2
12
Planes.
=
2
21
=
1
r
2. Spherical coordinates (ρ, θ, φ) = (x1 , x2 , x3 )
x = ρ sin θ cos φ
ρ≥0
h1 = 1
y = ρ sin θ sin φ
0≤θ≤π
h2 = ρ
z = ρ cos θ
0 ≤ φ ≤ 2π
h3 = ρ sin θ
The coordinate curves are formed by the intersection of the coordinate surfaces
x2 + y 2 + z 2 = ρ2
Spheres
x2 + y 2 = tan2 θ z
y = x tan φ
1
= −ρ
22
1
= −ρ sin2 θ
33
2
= − sin θ cos θ
33
Cones
Planes.
2
2
1
=
=
12
21
ρ
3
3
1
=
=
ρ
13
31
3
3
=
= cot θ
32
23
356
3. Parabolic cylindrical coordinates (ξ, η, z) = (x1 , x2 , x3 )
x = ξη
1
y = (ξ 2 − η 2 )
2
z=z
−∞<z <∞
p
ξ 2 + η2
p
h2 = ξ 2 + η 2
η≥0
h3 = 1
−∞<ξ <∞
h1 =
The coordinate curves are formed by the intersection of the coordinate surfaces
x2 = −2ξ 2 (y −
ξ2
)
2
Parabolic cylinders
η2
)
2
z = Constant
x2 = 2η 2 (y +
Parabolic cylinders
Planes.
1
−ξ
= 2
ξ + η2
22
1
1
η
=
= 2
ξ + η2
12
21
2
2
ξ
=
= 2
ξ + η2
21
12
1
ξ
= 2
ξ + η2
11
2
η
= 2
ξ + η2
22
2
−η
= 2
ξ + η2
11
4. Parabolic coordinates (ξ, η, φ) = (x1 , x2 , x3 )
x = ξη cos φ
ξ≥0
y = ξη sin φ
1
z = (ξ 2 − η 2 )
2
η≥0
p
ξ 2 + η2
p
h2 = ξ 2 + η 2
0 < φ < 2π
h3 = ξη
h1 =
The coordinate curves are formed by the intersection of the coordinate surfaces
x2 + y 2 = −2ξ 2 (z −
x2 + y 2 = 2η 2 (z +
y = x tan φ
1
=
11
2
=
22
1
=
22
2
=
11
2
=
33
ξ
2
ξ + η2
η
2
ξ + η2
−ξ
2
ξ + η2
−η
2
ξ + η2
−ηξ 2
ξ 2 + η2
ξ2
)
2
η2
)
2
Paraboloids
Paraboloids
Planes.
1
=
33
1
1
=
=
21
21
2
2
=
=
21
12
3
3
=
=
32
23
3
3
=
=
13
31
−ξη 2
ξ 2 + η2
η
2
ξ + η2
ξ
2
ξ + η2
1
η
1
ξ
357
5. Elliptic cylindrical coordinates (ξ, η, z) = (x1 , x2 , x3 )
x = cosh ξ cos η
ξ≥0
y = sinh ξ sin η
0 ≤ η ≤ 2π
q
sinh2 ξ + sin2 η
q
h2 = sinh2 ξ + sin2 η
z=z
−∞<z <∞
h3 = 1
h1 =
The coordinate curves are formed by the intersection of the coordinate surfaces
y2
x2
+
=1
cosh2 ξ
sinh2 ξ
y2
x2
−
=1
cos2 η sin2 η
Elliptic cylinders
Hyperbolic cylinders
z = Constant
1
=
11
1
=
22
1
1
=
=
12
21
Planes.
2
sin η cos η
=
22
sinh2 ξ + sin2 η
2
− sin η cos η
=
11
sinh2 ξ + sin2 η
2
2
sinh ξ cosh ξ
=
=
12
21
sinh2 ξ + sin2 η
sinh ξ cosh ξ
sinh2 ξ + sin2 η
− sinh ξ cosh ξ
sinh2 ξ + sin2 η
sin η cos η
sinh2 ξ + sin2 η
6. Elliptic coordinates (ξ, η, φ) = (x1 , x2 , x3 )
s
p
(1 − η 2 )(ξ 2 − 1) cos φ
p
y = (1 − η 2 )(ξ 2 − 1) sin φ
1≤ξ<∞
z = ξη
0 ≤ φ < 2π
x=
h1 =
s
−1≤η ≤1
h2 =
h3 =
ξ 2 − η2
ξ2 − 1
ξ 2 − η2
1 − η2
p
(1 − η 2 )(ξ 2 − 1)
The coordinate curves are formed by the intersection of the coordinate surfaces
y2
z2
x2
+
+
=1
ξ2 − 1 ξ2 − 1 ξ2
x2
y2
z2
−
−
=1
η2
1 − η2
1 − η2
y = x tan φ
ξ
1
ξ
+ 2
=−
2
−1 + ξ
ξ − η2
11
2
η
η
− 2
=
2
1−η
ξ − η2
22
ξ −1 + ξ 2
1
=−
(1 − η 2 ) (ξ 2 − η 2 )
22
ξ −1 + ξ 2 1 − η 2
1
=−
ξ 2 − η2
33
η 1 − η2
2
=
(−1 + ξ 2 ) (ξ 2 − η 2 )
11
Prolate ellipsoid
Two-sheeted hyperboloid
Planes
−1 + ξ 2 η 1 − η 2
2
=
33
ξ 2 − η2
1
η
=− 2
ξ − η2
12
2
ξ
= 2
ξ − η2
21
3
ξ
=
−1 + ξ 2
31
3
η
=−
1 − η2
32
358
7. Bipolar coordinates (u, v, z) = (x1 , x2 , x3 )
a sinh v
,
0 ≤ u < 2π
cosh v − cos u
a sin u
,
−∞ < v < ∞
y=
cosh v − cos u
z=z
−∞<z <∞
h21 = h22
x=
h22 =
a2
(cosh v − cos u)2
h23 = 1
The coordinate curves are formed by the intersection of the coordinate surfaces
a2
sinh2 v
a2
x2 + (y − a cot u)2 =
sin2 u
z = Constant
(x − a coth v)2 + y 2 =
Cylinders
Cylinders
Planes.
2
sinh v
=
11
− cos u + cosh v
1
sinh v
=
cos u − cosh v
12
2
sin u
=
cos u − cosh v
21
1
sin u
=
cos u − cosh v
11
2
sinh v
=
cos u − cosh v
22
1
sin u
=
− cos u + cosh v
22
8. Conical coordinates (u, v, w) = (x1 , x2 , x3 )
uvw
,
b 2 > v 2 > a2 > w 2 ,
ab
r
u (v 2 − a2 )(w2 − a2 )
y=
a
a2 − b 2
r
v (v 2 − b2 )(w2 − b2 )
z=
b
b 2 − a2
x=
u≥0
h21 = 1
u2 (v 2 − w2 )
− a2 )(b2 − v 2 )
u2 (v 2 − w2 )
h23 =
2
(w − a2 )(w2 − b2 )
h22 =
(v 2
The coordinate curves are formed by the intersection of the coordinate surfaces
x2 + y 2 + z 2 = u2
2
2
Spheres
2
y
z
x
+ 2
+ 2
= 0,
v2
v − a2
v − b2
y2
z2
x2
+ 2
+ 2
= 0,
2
2
w
w −a
w − b2
v
v
2
v
−
+ 2
= 2
b − v2
−a2 + v 2
v − w2
22
3
w
w
w
−
−
=− 2
v − w2
−a2 + w2
−b2 + w2
33
u v 2 − w2
1
=− 2
(b − v 2 ) (−a2 + v 2 )
22
u v 2 − w2
1
=−
(−a2 + w2 ) (−b2 + w2 )
33
v b2 − v 2 −a2 + v 2
2
=− 2
(v − w2 ) (−a2 + w2 ) (−b2 + w2 )
33
Cones
Cones.
w −a2 + w2 −b2 + w2
3
= 2
22
(b − v 2 ) (−a2 + v 2 ) (v 2 − w2 )
2
1
=
u
21
2
w
=− 2
v − w2
23
3
1
=
u
31
3
v
= 2
v − w2
32
359
9. Prolate spheroidal coordinates (u, v, φ) = (x1 , x2 , x3 )
x = a sinh u sin v cos φ,
u≥0
h21 = h22
y = a sinh u sin v sin φ,
0≤v≤π
h22 = a2 (sinh2 u + sin2 v)
z = a cosh u cos v,
h23 = a2 sinh2 u sin2 v
0 ≤ φ < 2π
The coordinate curves are formed by the intersection of the coordinate surfaces
y2
z2
x2
+
+
= 1,
2
2
(a sinh u)
a sinh u)
a cosh u)2
y2
z2
x2
−
−
= 1,
2
2
(a cos v)
(a sin v)
(a cos v)2
Prolate ellipsoids
Two-sheeted hyperpoloid
y = x tan φ,
1
cosh u sinh u
=
11
sin2 v + sinh2 u
2
cos v sin v
=
22
sin2 v + sinh2 u
1
cosh u sinh u
=− 2
22
sin v + sinh2 u
1
sin2 v cosh u sinh u
=−
33
sin2 v + sinh2 u
2
cos v sin v
=− 2
11
sin v + sinh2 u
Planes.
2
cos v sin vsinh2 u
=−
33
sin2 v + sinh2 u
1
cos v sin v
=
12
sin2 v + sinh2 u
2
cosh u sinh u
=
21
sin2 v + sinh2 u
3
cosh u
=
sinh u
31
3
cos v
=
sin v
32
10. Oblate spheroidal coordinates (ξ, η, φ) = (x1 , x2 , x3 )
x = a cosh ξ cos η cos φ,
y = a cosh ξ cos η sin φ,
z = a sinh ξ sin η,
ξ≥0
π
π
− ≤η≤
2
2
0 ≤ φ ≤ 2π
h21 = h22
h22 = a2 (sinh2 ξ + sin2 η)
h23 = a2 cosh2 ξ cos2 η
The coordinate curves are formed by the intersection of the coordinate surfaces
y2
z2
x2
+
+
= 1,
(a cosh ξ)2
(a cosh ξ)2
(a sinh ξ)2
y2
z2
x2
+
−
= 1,
(a cos η)2
(a cos η)2
(a sin η)2
y = x tan φ,
1
cosh ξ sinh ξ
=
11
sin2 η + sinh2 ξ
2
cos η sin η
=
22
sin2 η + sinh2 ξ
1
cosh ξ sinh ξ
=− 2
22
sin η + sinh2 ξ
1
cos2 η cosh ξ sinh ξ
=−
33
sin2 η + sinh2 ξ
2
cos η sin η
=− 2
11
sin η + sinh2 ξ
Oblate ellipsoids
One-sheet hyperboloids
Planes.
2
cos η sin ηcosh2 ξ
=
33
sin2 η + sinh2 ξ
1
cos η sin η
=
12
sin2 η + sinh2 ξ
2
cosh ξ sinh ξ
=
21
sin2 η + sinh2 ξ
3
sinh ξ
=
cosh ξ
31
3
sin η
=−
cos η
32
360
11. Toroidal coordinates (u, v, φ) = (x1 , x2 , x3 )
a sinh v cos φ
,
cosh v − cos u
a sinh v sin φ
,
y=
cosh v − cos u
a sin u
,
z=
cosh v − cos u
x=
0 ≤ u < 2π
−∞ < v < ∞
0 ≤ φ < 2π
h21 = h22
h22 =
a2
(cosh v − cos u)2
h23 =
a2 sinh2 v
(cosh v − cos u)2
The coordinate curves are formed by the intersection of the coordinate surfaces
a cos u 2
a2
,
=
x2 + y 2 + z −
sin u
sin2 u
2
p
cosh v
a2
,
x2 + y 2 − a
+ z2 =
sinh v
sinh2 v
y = x tan φ,
1
=
11
2
=
22
1
=
22
1
=
33
2
=
11
sin u
cos u − cosh v
sinh v
cos u − cosh v
sin u
− cos u + cosh v
sin usinh v 2
− cos u + cosh v
sinh v
− cos u + cosh v
Spheres
Tores
planes
2
sinh v (cos u cosh v − 1)
=−
33
cos u − cosh v
1
sinh v
=
cos u − cosh v
12
2
sin u
=
cos u − cosh v
21
3
sin u
=
cos u − cosh v
31
3
cos u cosh v − 1
=
cos u sinh v − cosh v sinh v
32
361
12. Confocal ellipsoidal coordinates (u, v, w) = (x1 , x2 , x3 )
(a2 − u)(a2 − v)(a2 − w)
,
(a2 − b2 )(a2 − c2 )
(b2 − u)(b2 − v)(b2 − w)
,
y2 =
(b2 − a2 )(b2 − c2 )
(c2 − u)(c2 − v)(c2 − w)
,
z2 =
(c2 − a2 )(c2 − b2 )
x2 =
u < c2 < b 2 < a 2
c2 < v < b 2 < a 2
c2 < b 2 < v < a 2
(u − v)(u − w)
4(a2 − u)(b2 − u)(c2 − u)
(v − u)(v − w)
h22 =
4(a2 − v)(b2 − v)(c2 − v)
(w − u)(w − v)
h23 =
4(a2 − w)(b2 − w)(c2 − w)
h21 =
1
1
1
1
1
1
+
+
+
+
=
2
2
2
2 (a − u) 2 (b − u) 2 (c − u) 2 (u − v) 2 (u − w)
11
1
1
1
1
2
1
+
+
+
+
=
2
2
2
2 (a − v) 2 (b − v) 2 (c − v) 2 (−u + v) 2 (v − w)
22
1
1
1
1
3
1
+
+
+
+
=
2
2
2
2 (a − w) 2 (b − w) 2 (c − w) 2 (−u + w) 2 (−v + w)
33
a2 − u b2 − u c2 − u (v − w)
1
1
−1
=
=
2 (a2 − v) (b2 − v) (c2 − v) (u − v) (u − w)
22
12
2 (u − v)
a2 − u b2 − u c2 − u (−v + w)
1
1
−1
=
=
2 (u − v) (a2 − w) (b2 − w) (c2 − w) (u − w)
33
2 (u − w)
13
a2 − v b2 − v c2 − v (u − w)
2
2
−1
=
=
2
2
2
2 (a − u) (b − u) (c − u) (−u + v) (v − w)
2 (−u + v)
11
21
2
2
2
2
−1
−
v
b
−
v
c
−
v
(−u
+
w)
a
2
=
=
2
2
2
2 (v − w)
23
2 (−u + v) (a − w) (b − w) (c − w) (v − w)
33
2
2
2
−1
3
(u − v) a − w b − w c − w
3
=
=
2 (−u + w)
31
2 (a2 − u) (b2 − u) (c2 − u) (−u + w) (−v + w)
11
2
2
2
3
−1
(−u + v) a − w b − w c − w
3
=
=
2 (−v + w)
32
2 (a2 − v) (b2 − v) (c2 − v) (−u + w) (−v + w)
22
362
APPENDIX C
VECTOR IDENTITIES
~ B,
~ C,
~ D
~ are differentiable vector functions of position while
The following identities assume that A,
f, f1 , f2 are differentiable scalar functions of position.
1.
~ · (B
~ × C)
~ =B
~ · (C
~ × A)
~ =C
~ · (A
~ × B)
~
A
2.
~ × (B
~ × C)
~ = B(
~ A
~ · C)
~ − C(
~ A
~ · B)
~
A
3.
~ × B)
~ · (C
~ × D)
~ = (A
~ · C)(
~ B
~ · D)
~ − (A
~ · D)(
~ B
~ · C)
~
(A
4.
~ × (B
~ × C)
~ +B
~ × (C
~ × A)
~ +C
~ × (A
~ × B)
~ = ~0
A
5.
~ × B)
~ × (C
~ × D)
~ = B(
~ A
~·C
~ × D)
~ − A(
~ B
~ ·C
~ × D)
~
(A
~ A
~·B
~ × C)
~ − D(
~ A
~·B
~ × C)
~
= C(
6.
~ × B)
~ · (B
~ × C)
~ × (C
~ × A)
~ = (A
~·B
~ × C)
~ 2
(A
7.
∇(f1 + f2 ) = ∇f1 + ∇f2
8.
~ + B)
~ = ∇·A
~ +∇·B
~
∇ · (A
9.
~ + B)
~ =∇×A
~+∇×B
~
∇ × (A
10.
~ = (∇f ) · A
~ + f∇ · A
~
∇(f A)
11.
∇(f1 f2 ) = f1 ∇f2 + f2 ∇f1
12.
~ =)∇f ) × A
~ + f (∇ × A)
~
∇ × (f A)
13.
14.
~ × B)
~ =B
~ · (∇ × A)
~ −A
~ · (∇ × B)
~
∇ · (A
!
~2
~ × (∇ × A)
~
~ · ∇)A
~ = ∇ |A|
−A
(A
2
15.
~ · B)
~ = (B
~ · ∇)A
~ + (A
~ · ∇)B
~ +B
~ × (∇ × A)
~ +A
~ × (∇ × B)
~
∇(A
16.
~ × B)
~ = (B
~ · ∇)A
~ − B(∇
~
~ − (A
~ · ∇)B
~ + A(∇
~ · B)
~
∇ × (A
· A)
17.
∇ · (∇f ) = ∇2 f
18.
∇ × (∇f ) = ~0
19.
~ =0
∇ · (∇ × A)
20.
~
~ = ∇(∇ · A)
~ − ∇2 A
∇ × (∇ × A)
363
INDEX
A
Absolute differentiation 120
Absolute scalar field 43
Absolute tensor 45,46,47,48
Acceleration 121, 190, 192
Action integral 198
Addition of systems 6, 51
Addition of tensors 6, 51
Adherence boundary condition 294
Aelotropic material 245
Affine transformation 86, 107
Airy stress function 264
Almansi strain tensor 229
Alternating tensor 6,7
Ampere’s law 176,301,337,341
Angle between vectors 80, 82
Angular momentum 218, 287
Angular velocity 86,87,201,203
Arc length 60, 67, 133
Associated tensors 79
Auxiliary Magnetic field 338
Axis of symmetry 247
Cauchy stress law 216
Cauchy-Riemann equations 293,321
Charge density 323
Christoffel symbols 108,110,111
Circulation 293
Codazzi equations 139
Coefficient of viscosity 285
Cofactors 25, 26, 32
Compatibility equations 259, 260, 262
Completely skew symmetric system 31
Compound pendulum 195,209
Compressible material 231
Conic sections 151
Conical coordinates 74
Conjugate dyad 49
Conjugate metric tensor 36, 77
Conservation of angular momentum 218, 295
Conservation of energy 295
Conservation of linear momentum 217, 295
Conservation of mass 233, 295
Conservative system 191, 298
Conservative electric field 323
B
Basic equations elasticity 236, 253, 270
Basic equations for a continuum 236
Basic equations of fluids 281, 287
Basis vectors 1,2,37,48
Beltrami 262
Bernoulli’s Theorem 292
Biharmonic equation 186, 265
Bilinear form 97
Binormal vector 130
Biot-Savart law 336
Bipolar coordinates 73
Boltzmann equation 302,306
Boundary conditions 257, 294
Bulk modulus 251
Bulk coefficient of viscosity 285
C
Cartesian coordinates 19,20,42, 67, 83
Cartesian tensors 84, 87, 226
Constitutive equations 242, 251,281, 287
Continuity equation 106,234, 287, 335
Contraction 6, 52
Contravariant components 36, 44
Contravariant tensor 45
Coordinate curves 37, 67
Coordinate surfaces 37, 67
Coordinate transformations 37
Coulomb law 322
Covariant components 36, 47
Covariant differentiation 113,114,117
Covariant tensor 46
Cross product 11
Curl 21, 173
Curvature 130, 131, 134, 149
Curvature tensor 134, 145
Curvilinear coordinates 66, 81
Cylindrical coordinates 18, 42, 69
364
INDEX
D
Deformation 222
Derivative of tensor 108
Derivatives and indicial notation 18, 31
Determinant 10, 25, 32, 33
Dielectric tensor 333
Differential geometry 129
Diffusion equation 303
Dilatation 232
Direction cosines 85
Displacement vector 333
Dissipation function 297
Equilibrium equations 273,300
Elastic constants 243,248
Equipotential curves 325
Euler number 294
Euler-Lagrange equations 192
Eulerian angles 201, 209
Eulerian form 287
Eulerian system 227
Eulers equations of motion 204
F
Distribution function 302
Faraday’s law 176,301, 340
Divergence 21, 172
Field lines 324, 327
Divergence theorem 24
Field electric 322
Dot product 5
First fundamental form 133,143
Double dot product 50, 62
Fourier law 297, 299
Dual tensor 100
Free indices 3
Dummy index 4, 5
Frenet-Serret formulas 131, 188
Dyads 48,62,63
Froude number 294
Dynamics 187
Fluids 281
E
G
e Permutation symbol 6, 7, 12
Gas law 300
e-δ identity 12
Gauss divergence theorem 24, 330
Eigenvalues 179,189
Gauss equations 138
Eigenvectors 179,186
Gauss’s law for electricity 176,301,328
Einstein tensor 156
Gauss’s law for magnetism 176,301,341
Elastic constants 248
Gaussian curvature 137,139, 149
Elastic stiffness 242
Geodesics 140, 146
Elasticity 211,213
Geodesic curvature 135, 140
Electrostatic field 322,333
General tensor 48
Electric flux 327
Generalized e − δ identity 84, 104
Electric units 322
Generalized Hooke’s law 242
Electrodynamics 339
Generalized Kronecker delta 13, 31
Electromagnetic energy 341
Generalized stress strain 242
Electromagnetic stress 341,342
Geometry in Riemannian Space 80
Elliptic coordinates 72
Gradient 20, 171
Elliptical cylindrical coordinates 71
Gradient basis 37
Enthalpy 298
Green’s theorem 24
Entropy 300
Group properties 41, 54
Epsilon permutation symbol 83
Generalized velocity 121
Equation of state 300
Generalized acceleration 121
365
INDEX
H
Hamiltonian 208
Heat equation 316
Hexagonal material 247
Higher order tensors 47, 93
Hooke’s law 212, 242, 252
Hydrodynamic equations 283
I
M
Magnitude of vector 80
Magnetostatics 334,338
Magnetic field 334
Magnetization vector 337
Magnetic permeability 337
Material derivative 234, 288
Material symmetry 244, 246
Ideal fluid 283
Maxwell equations 176, 339
Idemfactor 50
Maxwell transfer equation 308
Incompressible material 231
Maximum, minimum curvature 130, 140
Index notation 1, 2, 14
Mean curvature 137, 148
Indicial notation 1, 2, 14,24
Metric tensor 36, 65
Inner product 52
Meusnier’s Theorem 150
Inertia 30
Mixed tensor 49
Integral theorems 24
Mohr’s circle 185
Intrinsic derivative 120
Moment of inertia 30, 184, 200
Invariant 43
Momentum 217, 218
Inviscid fluid 283
Multilinear forms 96, 98
Isotropic material 248
Multiplication of tensors 6, 51
Isotropic tensor 104
N
J
Navier’s equations 254, 257
Jacobian 17, 30, 40, 101, 127
Jump discontinuity 330
K
Navier-Stokes equations 288, 290
Newtonian fluids 286
Nonviscous fluid 283
Normal curvature 135, 136
Kronecker delta 3, 8, 13, 31, 76
Normal plane 188
Kinetic energy 201
Normal stress 214
Kinematic viscosity 302
Normal vector 130, 132
L
Lagrange’s equation of motion 191, 196
Notation for physical components 92
O
Lagrangian 209
Laplacian 174
Linear form 96
Linear momentum 209, 287
Linear transformation 86
Linear viscous fluids 284
Lorentz transformation 57
Lame’s constants 251
Oblate Spheroidal coordinates 75
Oblique coordinates 60
Oblique cylindrical coordinates 102
Order 2
Orthogonal coordinates 78, 86
Orthotropic material 246
Outer product 6, 51
Osculating plane 188
366
INDEX
P
Parallel vector field 122
Rayleigh implusive flow 317
Pappovich-Neuber solution 263
Reciprocal basis 35, 38
Parabolic coordinates 70
Relative scalar 127
Parabolic cylindrical coordinates 69
Relative tensor 50, 121
Particle motion 190
Relative motion 202
Pendulum system 197, 210
Relativity 151
Perfect gas 283, 299
Relative motion 155
Permutations 6
Reynolds number 294
Phase space 302
Ricci’s theorem 119
Physical components 88, 91,93
Riemann Christoffel tensor 116, 129,139, 147
Piezoelectric 300
Riemann space 80
Pitch,roll, Yaw 209
Rectifying plane 188
Plane Couette flow 315
Rigid body rotation 199
Plane Poiseuille flow 316
Rotation of axes 85, 87, 107
Plane strain 263
Rules for indices 2
Plane stress 264
Poisson’s equation 329
S
Poisson’s ratio 212
Polar element 273
Scalar 40, 43
Polarization vector 333
Scalar invariant 43, 62, 105
Polyads 48
Scalar potential 191
Potential energy 191
Scaled variables 293
Potential function 323
Second fundamental form 135, 145
Poynting’s vector 341
Second order tensor 47
Pressure 283
Shearing stresses 214
Principal axes 183
Simple pulley system 193
Projection 35
Simple pendulum 194
Prolated Spheroidal coordinates 74
Skew symmetric system 3, 31
Pully system 194, 207
Skewed coordinates 60, 102
Solid angle 328
Q
Space curves 130
Special tensors 65
Quotient law 53
R
Spherical coordinates 18, 43, 56, 69, 103,194
Stokes flow 318
Stokes hypothesis 285
Radius of curvature 130, 136
Stokes theorem 24
Range convention 2, 3
Straight line 60
Rate of deformation 281, 286
Strain 218, 225, 228
Rate of strain 281
Strain deviator 279
367
INDEX
Stress 214
U
Stress deviator 279
Strong conservative form 298
Unit binormal 131, 192
Strouhal number 294
Unit normal 131, 191
St Venant 258
Unit tangent 131, 191
Subscripts 2
Unit vector 81, 105
Subtraction of tensors 51, 62
Summation convention 4, 9
V
Superscripts 2
Vector identities 15, 20, 315
Surface 62, 131
Vector transformation 45, 47
Surface area 59
Vector operators 20, 175
Surface curvature 149
Vector potential 188
Surface metric 125, 133
Velocity 95, 121, 190, 193
Susceptibility tensor 333
Velocity strain tensor 281
Sutherland formula 285
Viscosity 285
Symmetric system 3, 31, 51, 101
Viscosity table 285
Symmetry 243
Viscous fluid 283
System 2, 31
Viscous forces 288
T
Tangential basis 37
Tangent vector 130
Viscous stress tensor 285
Vorticity 107, 292
W
Tensor and vector forms 40, 150
Wave equation 255, 269
Tensor derivative 141
Weighted tensor 48, 127
Tensor general 48
Weingarten’s equation 138, 153
Tensor notation 92, 160
Work 191, 279
Tensor operations 6, 51, 175
Work done 324
Test charge 322
Thermodynamics 299
Y
Third fundamental form 146
Third order systems 31
Toroidal coordinates 75, 103
Torus 124
Transformation equations 17, 37, 86
Transitive property 45,46
Translation of coordinates 84
Transport equation 302
Transposition 6
Triad 50
Trilinear form 98
Triple scalar product 15
Young’s modulus 212