International Series in Pure and Applied Mathematics
William Ted Martin, CONSULTING EDITOR
VECTOR AND TENSOR ANALYSIS
a
International Series in Pure and Applied 'Mathematics
WILI.LA11 TED MARTIN, Consulting Editor
Complex Analysis
BELLMAN Stability Theory of Differential Equations
AIILFORS
Bucx Advanced Calculus
CODDnNUTON AND LEVINSON Theory of Ordinary Differential Equations
G01.011 13 AND SHANKS Elements of Ordinary Differential Equations
of Real Variables
GRAVES The Theory of
GRIFFIN Elementary Theory of Numbers
IIILDEBRAND rodurl ion to Numerical Analysis
Principles of Nunurriral Analysis
LAS b;leuu'nts of Pure aunt Applied Mathematics
LASS Vector and Tensor Analysis
LEIGIITON An Introduction to the Theory of Differential Equations
NEHAIU Conformal Mapping
NEWELL Vector Analysis
ROSSER Logic for 'Mathematicians
RUDIN Principles of Mathematical Analysis
SNEDDON Elciuents of Partial I)iiTerential Equations
Pourier Transforms
SNEDDON
STOLL Linear Algebra and 'Matrix Theory
WEINSTOCK Calculus of Variations
VECTOR AND TENSOR
ANALYSIS
BY
HARRY LASS
JET PROPULSION LABORATORY
Calif. Institute of Tech.
4800 Oak Grove Drive
Pasadena, California
New York Toronto London
McGRAW-HILL BOOK COMPANY, INC.
1950
VECTOR AND TENSOR. ANALYSIS
Copyright, 1950, by the '.11e(_;ratt-Bill Book Company, Inc. Printed in the
1 nited States of America. All rights reserved. 't'his hook, or parts thereof,
may not be reproduced in any form without, perrnis.sion of the publishers.
To MY
MOTHER AND FATHER
PREFACE
This text can be used in a variety of ways. The student
totally unfamiliar with vector analysis can peruse Chapters 1, 2,
and 4 to gain familiarity with the algebra and calculus of vectors.
These chapters cover the ordinary one-semester course in vector
analysis. Numerous examples in the fields of differential
geometry, electricity, mechanics, hydrodynamics, and elasticity
can be found in Chapters 3, 5, 6, and 7, respectively. Those
already acquainted with vector analysis who feel that they would
like to become better acquainted with the applications of vectors
can read the above-mentioned chapters with little difficulty:
only a most rudimentary knowledge of these fields is necessary
in order that the reader be capable of following their contents,
which are fairly complete from an elementary viewpoint. A
knowledge of these chapters should enable the reader to further
digest the more comprehensive treatises dealing with these subjects, some of which are listed in the reference section. It is
hoped that these chapters will give the mathematician a brief
introduction to elementary theoretical physics. Finally, the
author feels that Chapters 8 and 9 deal sufficiently with tensor
analysis and Riemannian geometry to enable the reader to study
the theory of relativity with a minimum of effort as far as the
mathematics involved is concerned.
In order to cover such a wide range of topics the treatment has
necessarily been brief. It is hoped, however, that nothing has
been sacrificed in the way of clearness of ideas. The author has
attempted to be as rigorous as is possible in a work of this nature.
Numerous examples have been worked out fully in the text.
The teacher who plans on using this book as a text can surely
arrange the topics to suit his needs for a one-, two-, or even threesemester course.
If the book is successful, it is due in no small measure to the
composite efforts of those men who have invented and who have
vii
viii
PREFACE
applied the vector and tensor analysis. The excellent works
listed in the reference section have been of great aid. Finally, I
wish to thank Professor Charles de Prima of the California
Institute of Technology for his kind interest in the development
of this text.
HARRY LASS
URBANA, ILL.
February, 1950
CONTENTS
PREFACE .
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vii
CHAPTER 1
THE ALGEBRA OF VECTORS .
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1
1. Definition of a vector 2. Equality of vectors 3. Multipli-
cation by a scalar 4. Addition of vectors 5. Subtraction of
6. Linear functions 7. Coordinate systems 8. Scalar,
or dot, product 9. Applications of the scalar product to space
geometry 10. Vector, or cross, product 11. The distributive
law for the vector product 12. Examples of the vector product
13. The triple scalar product 14. The triple vector product 15.
vectors
Applications to spherical trigonometry
CHAPTER 2
DIFFERENTIAL VECTOR CALCULUS . . . . . . . . . . . . . 29
16. Differentiation of vectors 17. Differentiation rules 18. The
gradient 19. The vector operator del, V 20. The divergence of a
vector 21. The curl of a vector 22. Recapitulation 23. Curvilinear coordinates
CHAPTER 3
DIFFERENTIAL GEOMETRY . . . . . . . . . . . . . . . . . 58
24. Frenet-Serret formulas 25. Fundamental planes 26. Intrinsic equations of a curve 27. Involutes 28. Evolutes 29.
Spherical indicatrices 30. Envelopes 31. Surfaces and curvilinear coordinates 32. Length of arc on a surface 33. Surface
curves 34. Normal to a surface 35. The second fundamental
form 36. Geometrical significance of the second fundamental
form 37. Principal directions 38. Conjugate directions 39.
Asymptotic lines 40. Geodesics
CHAPTER 4
INTEGRATION . . . . . . . . . . . . . . . . . . . . . . . . 89
41. Point-set theory 42. Uniform continuity 43. Some properties of continuous functions 44. Cauchy criterion for sequences
45. Regular area in the plane 46. Jordan curves 47. Functions
of bounded variation 48. Arc length 49. The Riemann integral
ix
CONTENTS
x
50. Connected and simply connected regions 51. The line inte53. Stokes's theorem 54.
gral 52. Line integral (continued)
Examples of Stokes's theorem 55. The divergence theorem
(Gauss)
56. Conjugate functions
CHAPTER 5
STATIC AND DYNAMIC ELECTRICITY . . . . . . . . . . . 127
57. Electrostatic forces 58. Gauss's law 59. Poisson's formula
60. Dielectrics 61. Energy of the electrostatic field 62. Discontinuities of D and E 63. Green's reciprocity theorem 64.
Method of images 65. Conjugate harmonic functions 66. Integration of Laplace's equation 67. Solution of Laplace's equation
in spherical coordinates 68. Applications 69. Integration of
Poisson's equation 70. Decomposition of a vector into a sum of
solenoidal and irrotational vectors 71. Dipoles 72. Electric
polarization 73. Magnetostatics 74. Solid angle 75. Moving
charges, or currents 76. Magnetic effect of currents (Oersted)
77. Mutual induction and action of two circuits 78. Law of induction (Faraday) 79. Maxwell's equations 80. Solution of
Maxwell's equations for electrically free space 81. Poynting's
theorem 82. Lorentz's electron theory 83. Retarded potentials
CHAPTER 6
MECHANICS . . . . . . . . . . . . . . . . . . . . . . . . . 184
84. Kinematics of a particle 85. Motion about a fixed axis 86.
Relative motion 87. Dynamics of a particle 88. Equations of
motion for a particle 89. System of particles 90. Momentum
and angular momentum 91. Torque, or force, moment 92. A
theorem relating angular momentum with torque 93. Moment of
momentum (continued) 94. Moment of relative momentum 95.
Kinetic energy 96. Work 97. Rigid bodies 98. Kinematics of
a rigid body 99. Relative time rate of change of vectors 100.
Velocity 101. Acceleration 102. Motion of a rigid body with
one point fixed 103. Applications 104. Euler's angular coordinates 105. Motion of a free top about a fixed point 106. The
top (continued) 107. Inertia tensor
CHAPTER 7
HYDRODYNAMICS AND ELASTICITY . . . . . . . . . . . . 230
108. Pressure 109. The equation of continuity 110. Equations
of motion for a perfect fluid 111. Equations of motion for an
incompressible fluid under the action of a conservative field 112.
The general motion of a fluid 113. Vortex motion 114. Applications 115. Small displacements. Strain tensor 116. The
stress tensor 117. Relationship between the strain and stress
tensors 118. Navier-Stokes equation
CONTENTS
xi
CHAPTER 8
TENSOR ANALYSIS AND RIEMANNIAN GEOMETRY. . . . . 259
119. Summation notation 120. The Kronecker deltas 121.
Determinants 122. Arithmetic, or vector, n-space 123. Contravariant vectors 124. Covariant vectors 125. Scalar product of
two vectors 126. Tensors 127. The line element 128. Geodesics in a Riemannian space 129. Law of transformation for the
Christoffel symbols 130. Covariant differentiation 131. Geodesic coordinates 132. The curvature tensor 133. RiemannChristoffel tensor 134. Euclidean space
CHAPTER 9
FURTHER APPLICATIONS OF TENSOR ANALYSIS . . . . . . 311
135. Frenet-Serret formulas 136. Parallel displacement of vectors
137. Parallelism in a subspace 138. Generalized covariant differentiation 139. Riemannian curvature. Schur's theorem 140.
Lagrange's equations 141. Einstein's law of gravitation 142.
Two-point tensors
REFERENCES .
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INDEX .
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339
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341
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CHAPTER 1
THE ALGEBRA OF VECTORS
1. Definition of a Vector. Our starting point for the definition
of a vector will be the intuitive one encountered in elementary
physics. Any directed line segment will be called a vector. The
length of the vector will be denoted by the word magnitude. Any
physical element that has magnitude and direction, and hence
can be represented by a vector, will also be designated as a vector.
In Chap. 8 we will give a more mathematically rigorous definition
of a vector.
Elementary examples of vectors are displacements, velocities,
forces, accelerations, etc. Physical concepts, such as speed,
temperature, distance, and specific gravity, and arithmetic numbers, such as 2, i, etc., are called scalars to distinguish them from
vectors. We note that no direction is associated with a scalar.
We shall represent vectors by
arrows and use boldface type to Ca-,
a]
a,
a,
a,
indicate that we are speaking of
a vector. In order to distina
guish between scalars and vec-
tors, the student will have to
Fia. 1.
adopt some notation for describing a vector in writing. The student may choose his mode of
representing a vector from Fig. 1 or may adopt his own notation.
To every vector will be associated a real nonnegative number
equal to the length of the vector. This number will depend, of
course, on the unit chosen to represent a given class of vectors.
A vector of length one will be called a unit vector. If a represents the length of the vector a, we shall write a = jal. If
jal = 0, we define a as the zero vector.
2. Equality of Vectors. Two vectors will be defined to be
equal if, and only if, they are parallel, have the same sense of
direction, and the same magnitude. The starting points of the
vectors are immaterial. It is the direction and magnitude which
1
VECTOR AND TENSOR ANALYSIS
2
[SEC. 3
are important. Equal vectors, however, may produce different
physical effects, as will be seen later. We write a = b if the
vectors are equal (see Fig. 2).
Fia. 2.
3. Multiplication by a Scalar.
If we multiply a vector a by a
real number x, we define the product xa to be a new vector
parallel to a whose magnitude has been multiplied by the factor x.
Thus 2a will be a vector which is twice as long as the vector a
Z-a
Fm. 4.
FIG. 3.
and which has the same direction as a (see Fig. 3). We define
-a as the vector obtained from a by reversing its direction (see
Fig. 4).
We note that
x(ya) = (xy)a = xya
(x + y)a = xa + ya
Oa = 0
(zero vector)
It is immediately seen that two vectors are parallel if, and only
if, one of them can be written as a scalar multiple of the other.
4. Addition of Vectors. Let us suppose we have two vectors
given, say a and b. We form a third vector by constructing a
triangle with a and b forming two sides of the triangle, b adjoined
to a (see Fig. 5). The vector starting from the origin of a and
ending at the arrow of b is defined as the vector sum a + b.
We see that a + 0 = a, and if a = b, c = d, then
a+c=b+d
THE ALGEBRA OF VECTORS
SEc. 6]
3
From Euclidean geometry we note that
a+b=b+a
(1)
(2)
(3)
(a+b)+c=a+(b+c)
x(a+b) =xa+xb
(1) is called the commutative law of vector addition; (2) is called
the associative law of vector addition; (3) is the distributive law
for multiplication by a scalar. The reader should have no
trouble proving these three results geometrically.
a+b+c
Fio. 5.
5. Subtraction of Vectors. Given the two vectors a and b,
we can ask ourselves the following question: What vector c must
be added to b to give a? The vector c is defined to be the vector
a - b. We can obtain the desired result by two methods.
First, construct -b and then add this vector to a, or second, let b
and a have a common origin and construct the third side of the
triangle. The two possible directions will give a - b and b - a
(see Fig. 6). Thus a - b = a + (-b).
a-b
a-b
Flo. 6.
6. Linear Functions. Let us consider all vectors in the twodimensional Euclidean plane. We choose a basis for this system
of vectors by considering any two nonparallel, nonzero vectors.
Call them a and b. Any third vector c can be written as a linear
VECTOR AND TENSOR ANALYSIS
4
[SEC. 6
combination or function of a and b,
c = xa + yb
(4)
The proof of (4) is by construction (see Fig. 7).
Let us now consider the following problem: Let a and b have
a common origin, 0, and let c be any vector starting from 0 whose
end point lies on the line joining the ends of a and b (see Fig. 8).
B
a
FIG. 8.
Let C divide BA in the ratio x: y where x + y = 1. In particular, if C is the mid-point of BA, then x = y = fr. Now
c=OB+BC
= b + x(a - b)
=xa+(1 - x)b
so that
c = xa + yb
Now conversely, assume c = xa + yb, x + y = 1.
(5)
Then
c=xa+(1 -x)b =x(a-b)+b
We now note that c is a vector that is obtained by adding to b the
vector x(a - b), this latter vector being parallel to the vector
a - b. This immediately implies that the end point of c lies
on the line joining A to B. We can rewrite (5) as
c-xa-yb=0
1-x-y=0
(6)
SEC. 61
THE ALGEBRA OF VECTORS
5
We have proved our first important theorem. A necessary
and sufficient condition that the end points of any three vectors
with common origin be on a straight line is that real constants
it m, n exist such that
la+mb+nc=0
l+m+n=O
(7)
with l2+m2+n2p,, 0.
We shall, however, find (5) more useful for solving problems.
Example 1. Let us prove that the medians of a triangle meet
at a point P which divides each median in the ratio 1:2.
C
0
FIG. 9.
Let ABC be the given triangle and let A', B', C' be the midChoose 0 anywhere in space and construct the vectors
from 0 to A, B, C, A', B', C', calling them a, b, c, a', b', c' (see
Fig. 9). From (5) we have
points.
a'=4b+ic
b' = Ja + 4c
(8)
Now P (the intersection of two of the medians) lies on the line
joining A and A' and on the line joining B and B'. We shall thus
find it expedient to find a relationship between the four vectors
a, b, a', b' associated with A, B, A', B'. From (8) we eliminate
the vector c and obtain
2a' + a = 2b' + b
or
*a' + '}a = *b' + b
(9)
6
VECTOR AND TENSOR ANALYSIS
[SFC. 6
But from (5), -sa' + -ia represents a vector whose origin is at 0
and whose end point lies on the line joining A to A. Similarly,
,'jb' + *b represents a vector whose origin is at 0 and whose end
point lies on the line joining B to B'. There can only be one
y
vector having both these properties, and this is the vector p = OP.
Hence p = tea' + is = -b' -fib. Note that P divides AA' and
BB' in the ratios 2: 1. Had we considered the median CC' in
connection with AA', we would have obtained that p
+ J e,
and this completes the proof of the theorem.
Example 2. To prove that the diagonals of a parallelogram
bisect each other. Let ABCD be the parallelogram and 0 any
C
FIG. 10.
point in space (see Fig.
10).
The equation d - a = c - b
implies that ABCD is a parallelogram.
Hence
Ja+4c=lb+Id =p
so that P bisects AC and BD.
Problems
1. Interpret I a
a
2. Give a geometric proof of (3).
3. a, b, c are consecutive vectors forming a triangle. What
is the vector sum a + b + c? Generalize this result for any
closed polygon.
4. Vectors are drawn from the center of a regular polygon to
its vertices. From symmetry considerations show that the vector sum is zero.
SEc. 61
THE ALGEBRA OF VECTORS
7
5. a and bare consecutive vectors of a parallelogram. Express
the diagonal vectors in terms of a and b.
6. a, b, c, d are consecutive vector sides of a quadrilateral.
Show that a necessary and sufficient condition that the figure be
a parallelogram is that a + c = 0 and show that this implies
b + d = 0.
7. Show graphically that lal + lbl >_ la + bl. From this
show that la - bl >_ lal - lbi.
8. a, b, c, d are vectors from 0 to A, B, C, D.
If
b-a=2(d-c)
show that the intersection point of the two lines joining A and D
and B and C trisects these lines.
9. a, b, c, d are four vectors with a common origin. Find a
necessary and sufficient condition that their end points lie in a
plane.
10. What is the vector condition that the end points of the
vectors of Prob. 9 form the vertices of a parallelogram?
11. Show that the mid-points of the lines which join the midpoints of the opposite sides of a quadrilateral coincide. The
four sides of the quadrilateral are not necessarily coplanar.
12. Show that the line which joins one vertex of a parallelogram
to the mid-point of an opposite side trisects the diagonal.
13. A line from a vertex of a triangle trisects the opposite side.
It intersects a similar line issuing from another vertex. In what
ratio do these lines intersect one another?
14. A line from a vertex of a triangle bisects the opposite side.
It is trisected by a similar line issuing from another vertex. How
does this latter line intersect the opposite side?
15. Show that the bisectors of a triangle meet in a point.
16. Show that if two triangles in space are so situated that the
three points of intersection of corresponding sides lie on a line,
then the lines joining the corresponding vertices pass through a
common point, and conversely. This is Desargues's theorem.
17. b = (sin t)a is a variable vector which always remains
parallel to the fixed vector a. What is
rically the meaning of
A.
fib?
Explain geomet-
8
VECTOR AND TENSOR ANALYSIS
[SEC. 7
18. Let v, be the velocity of A relative to B and let v2 be the
velocity of B relative to C. What is the velocity of A relative
to C? Of C relative to A? Are these results obvious?
19. Let a, b be constant vectors and let c be defined by the
equation
c = (cos t)a + (sin t)b
When is c parallel to a? Parallel to b? Can c ever be parallel
do d2c
If a and b
to a + b? Perpendicular to a + b? Find
9
dt dt2
are unit orthogonal vectors with common origin, describe the
positions of c and show that c is perpendicular to
dc
20. If a and b are not parallel, show that ma + nb = ka + 3b
implies m = k, n = j.
21. Theorem of Ceva. A necessary and sufficient condition that
the lines which join three points, one on each side of a triangle,
to the opposite vertices be concurrent is that the product of the
algebraic ratios in which the three points divide the sides be -1.
22. Theorem of Menelaus. Three points, one on each side of a
triangle ABC, are collinear if and only if the product of the
algebraic ratios in which they divide the sides BC, CA, AB is
unity.
7. Coordinate Systems. For a considerable portion of the
text we shall deal with the Euclidean space of three dimensions.
This is the ordinary space encountered by students of analytic
geometry and the calculus. We choose a right-handed coordinate system. If we rotate the x axis into the y axis, a right-hand
screw will advance along the positive z axis.
We let i, j, k be the three unit vectors along the positive x, y,
and z axes, respectively. The vectors i, j, k form a very simple
and elegant basis for our three-dimensional Euclidean space.
From Fig. 11 we observe that
r=xi+yj+zk
(10)
The numbers x, y, z are called the components of the vector r.
Note that they represent the projections of the vector r on the
x, y, and z axes. r is called the position vector of the point P
SEC. 71
THE ALGEBRA OF VECTORS
9
and will be used quite frequently in what follows. The most general space-time vector that we shall encounter will be of the form
u =u(x,y,z,t) =a(x,y,z,t)1+#(x,y,z,t)j
+ y(x, y, z, t)k
(11)
It is of the utmost importance that the student understand the
meaning of (11). To be more specific, let us consider a fluid in
motion. At any time t the particle which happens to be at the
z
Fm. 11.
point P(x, y, z) will have a velocity which depends on the coordinates x, y, z and on the time t. As time goes on, various particles
arrive at P(x, y, z) and have the velocities u(x, y, z, t) with components along the x, y, z axes given by a(x, y, z, t), Xx, y) z, t),
y(x) y, z, t).
Whenever we have a vector of the type (11), we say that we
An elementary example would be the vector
u = yi - xj. This vector field is time-independent and so is
have a vector field.
called a steady field.
At the point P(1, -2, 3) it has the value
- 2i - j. Another example would be u = 3xzeq - xyztj + 5xk.
We shall have more to say about this type of vector in later
VECTOR AND TENSOR ANALYSIS
10
[SEC. 8
chapters and will, for the present, be interested only in constant
vectors (uniform fields).
A moment's reflection shows that if a = ali + a2j + ask,
b = b1i + ba + b3k, then
a + b = (ai + b1)i + (a2 + b2)j + (as + b3)k
(12)
xa + yb = (xai + yb1)i + (xaz + ybz)j + (xaa + yb3)k
8. Scalar, or Dot, Product. We define the scalar or dot product of two vectors by the identity
a b = JaJJbJ cos 0
(13)
where 0 is the angle between the two vectors when drawn from a
common origin. It makes no difference whether we choose 0 or
- 0 since cos 0 = cos (- 0). This definition of the scalar product
arose in physics and will play a dominant role in the development of the text.
Fm. 12.
From (13) we at once verify that
a
a
ab
then
the
b
b=
a
a
a
b
b
ab
= (proj a)b (bI = (proj b)a (al
(17)
SEC. 8J
THE ALGEBRA OF VECTORS
11
With this in mind we proceed to prove the distributive law, which
states that
(18)
From Fig. 13 it is apparent that
[proj (b+c)]aIaF
f
I
= (proj b)a Iai + (proj C)a Ial
Fm. 13.
since the projection of the sum is the sum of the projections.
Let the reader now prove that
Example 3. To prove that the median to the base of an isosceles triangle is perpendicular to the base (see Fig. 14). From
(5) we see that
m=1
1
so that
0
FIG. 14.
which proves that OM is perpendicular to AB.
Example 4. To prove that an angle inscribed in a semicircle
is a right angle (see Fig. 15).
_j.k:g=o
1'1
_
C2
co$B
2ab
-b2+a2_.
6
Example
that
$o
.]a
PIG.
cb
_a)
(b
a -a
Ab
b2
_
C
C.
.
T'`iq°n°metr
of
Lau,
Cone
&
xample
E
18.
I°
j'i
c,a
c
z
a
0
angle
aright
4$CAis
that
so
0
_
AC
BC
-,
$
+c
s
[SEc.
BC AND
c a
'
yAC
YECr?,oR
12
THE ALGEBRA OF VECTORS
SEC. 8]
13
Hence if a = a1i + a2j + ask, b = b1i + b2j + b3k, then
a b = albs + a2b2 + a,b3
Formula (19) is of the utmost importance.
Example 7. Cauchy's Inequality
(19)
Notice that 0 a = 0.
(a - b)(a - b) _ 1a121b12 cost 0 5 1a12Ib12
so that from (19)
(albs + aab2 + aab3)2 S (a12 + a22 + as2)(b12 + b22 + b 32)
In general
n
n
n
I aaba s (Z aa2)1(I ba2)1
a-1
a-1
a-1
(20)
Let i' be a unit vector making angles a, fl, 7 with
the x, y, z axes. The projections of i' on the x, y, z axes are
cos a, cos ft, cos -y, so that
Example 8.
i' = cosai+cos0 j+cos7k
=pli+qlj+rlk
(21)
pi, q1, ri are called the direction cosines of the vector Y. Similarly, let j' and k' be unit
vectors with direction cosines p2, qa, r2 and pa, qa, r,. Thus
Notice that p12 + q12 + r12 = 1.
j' = p2i + q2j + r2k
k' = pai+qsj+rak
()
We also impose the condition that i', j', k' be mutually orthogonal,
so that the x', y', z' axes form a coordinate system similar to the
x-y-z coordinate system with common origin 0 (see Fig. 17).
We have r = r' so that xi + yj + zk = x'i' + y'j' + z'k',
where x, y, z are the coordinates of a point P as measured in the
x-y-z coordinate system and x', y', z' are the coordinates of the
same point P as measured in the x'-y'-z' coordinate system.
Making use of (21) and (22) and equating components, we find
that
x = p1x' + ply' + pgz'
y = q1x' + qty' + qaz'
z = r1x' + r2y' + raz'
(23)
14
VECTOR AND TENSOR ANALYSIS
1Src. 8
We now find it more convenient to rename the x-y-z coordinate
system. Let x = x1, y = x2, z = x3, where the superscripts do
not designate powers but are just labels which enable us to differentiate between the various axes. Similarly, let x' = x1, y' = x2'
y
FIG. 17.
Z' = 28.
Now let a,a represent the cosine of the angle between
the xa and V axes. We can write (23) as
3
xa = I a0a.V,
a = 1, 2, 3
(24)
a-1
By making use of the fact that i' j' = j' k' = k' i' = 0, we
can prove that
3
xa = I Asax#,
0-1
a = 1, 2, 3
(25)
where Asa = a). We leave this as an exercise for the reader.
Let us notice that differentiating (24) yields
axa
- = a,a,
of, or = 1, 2, 3
(26)
SEC. 91
THE ALGEBRA OF VECTORS
15
Example 9. The vector a = a'i + a2j + a'k may be represented by the number triple (a', a2, a'). Hence, without appealing to geometry we could develop an algebraic theory of vectors.
If b = (b', b2, b3), then a + b is defined by the number triple
(a' + b', a2 + b2, a3 + b3), and xa = x(al, a2, a') is defined by the
number triple (xa', xa2, xa3). From this the reader can prove
that
(a', a2, a3) = a'(1, 0, 0) + a2(0, 1, 0) + a3(0, 0, 1)
The triples (1, 0, 0), (0, 1, 0), (0, 0, 1) form a basis for our linear
vector space, that is, the space of number triples. We note that
the determinant formed from these triples, namely,
1
0
0
1
00
0
0 =1
1
does not vanish. Any three triples whose determinant does not
vanish can be used to form a basis. Let the reader prove this
result. We can define the scalar product (inner product) of
two triples by the law (a - b) = a'b' + a2b2 + a3b3.
9. Applications of the Scalar Product to Space Geometry
(a) We define a plane as the locus of lines passing through a
fixed point perpendicular to a fixed direction. Let the fixed
point be Po(xo, yo, zo) and let the fixed direction be given by the
vector N = Ai + Bj + Ck. Let r be the position vector to any
-4
point P(x, y, z) on the plane (Fig. 18). Now POP = r - ro is perpendicular to N so that
or
[(x - xo)i + (y - yo)j + (z - zo)kl - (Ai + Bj + Ck) = 0
and
A(x - xo) + B(y - yo) + C(z - zo) = 0
(27)
This is the equation of the plane. The point Po(xo, yo, zo) obviously lies in the plane since its coordinates satisfy (27). Equation (27) is linear in x, y, Z.
(b) Consider the surface Ax + By + Cz + D = 0. Let P(xo,
yo, zo) be any point on the surface. Of necessity,
Axo+Byo+Czo+D=0.
16
VECTOR AND TENSOR ANALYSIS
[SEC.9
Subtracting we have
A(x - xo) + B(y - yo) + C(z - zo) = 0
(28)
Now consider the two vectors Ai + Bj + Ck and
(x - xo)i + (y - yo)j + (z - zo)k
Fia. 18.
Equation (28) shows that these two vectors are perpendicular.
Hence the constant vector Ai + Bj + Ck is normal to the surface at every point so that the surface is a plane.
Fia. 19.
(c) Distance from a point to a plane. Let the equation of the
plane be Ax + By + Cz + D = 0, and let P(E, % r) be any point
in space. We wish to determine the shortest distance from P
THE ALGEBRA OF VECTORS
SEC. 91
17
to the plane. Choose any point Po lying in the plane. It is
apparent that the shortest distance will be the projection of
PoP on N, where N is a unit
vector normal to the plane (see
Fig. 19).
Now
d = JPoP NJ
=
IAZ+B, +Cr+DI
(29)
(A2 + B2 + C2)f
where use has been made of the
fact that
Axo+Byo+Czo+D=O.
FIG. 20.
(d) Equation of a straight line through the point Po(xo, yo, zo)
parallel to the vector T = li + mj + A. From Fig. 20 it is
z
Fia. 21.
VECTOR AND TENSOR ANALYSIS
18
[SEC. 9
apparent that r - ro is parallel to T so that r - ro = AT,
-- <A<+O0
Hence (x - xo)i + (y - yo)j + (z - zo)k = A(li + mj + nk), so
that equating components yields
x
xo
y - yo -z - zo=A
m
n
(30)
By allowing A to vary from - oo to + oo we generate every point
on the line.
(e) Equation of a sphere with center at Po(xo, yo, zo) and radius a.
In Fig. 21 obviously (r - ro) (r - ro) = a2, or
(x - xo) 2 + (y - yo) 2 + (z - zo) 2 = a2
Problems
1. Add and subtract the vectors a = 2i - 3j + 5k,
b = -2i+2j+2k
Show that the vectors are perpendicular.
2. Find the cosine of the angle between the two vectors
a = 2i - 3j + k and b = 3i - j - 2k.
3. If c is normal to a and b, show that c is normal to a + b,
a - b.
4. Let a and b be unit vectors in the x-y plane making angles
a and 6 with the x axis. Show that a = cos a i + sin a j,
b = cos 9 i + sin 3 j, and prove that
cos (a - 6) = cos a cos S + sin a sin 8
5. Find the equation of the cone whose generators make an
angle of 30° with the unit vector which makes equal angles with
the x, y, and z axes.
6. The position vectors of the foci of an ellipse are c and - c,
and the length of the major axis is 2a. Show that the equation
of the ellipse is a4 - a2(r2 + c2) + (c r)2 = O.
7. Prove that the altitudes of a triangle are concurrent.
8. Find the shortest distance from the point A(1, 0, 1) to the
line through the points B(2, 3, 4) and C(-1, 1, -2).
THE ALGEBRA OF VECTORS
SEC. 91
19
9. Let a = 2i - j + k, b =i-3j - 5k. Find a vector c
so that a, b, c form the sides of a right triangle.
10. Let r be the position vector of a point P(x, y, z), and let a
be a constant vector. Interpret the equation (r - a) a = 0.
11. Given a = 2i - 3j + k, b = 3j - 4k, find the projection
of a along b.
12. Show that the line joining the end points of the vectors
a =2i -j - k, b = -i + 3j - k with common origin at O is
parallel to the x-y plane, and find its length.
13. Prove that the sum of the squares of the diagonals of a
parallelogram is equal to the sum of the squares of its sides.
14. Let a = a, + a2 where a, b = 0 and a2 is parallel to b.
Show that a2 = (iblb
b, al = a -
(a . b) b.
JbJ2
15. Derive (25).
16. Verify (26).
17. Find a vector perpendicular to the vectors a = i - j + k,
b=2i+3j-k.
18. Let a = f(t)i + g(t)j + h(t)k, and define
h'(t)k
da = f'(t)i
+ g'(t)j +
dt
Show that
da
dt
a=
lal
dal
dt
19. Find the angle between the plane Ax + By + Cz + D = 0
and the plane ax + by + cz + d = 0.
20, a, b, c are coplanar. If a is not parallel to b, show that
aa ca
ab cb
Icb bbi
c=
aa ab
ab bb
ca ab
b
21. For the an", Asa defined by (24) and (25), show that
a, -Ap r = Sp-
y-1
where &0°= Iifa=l4,Se=Oif aPd f.
22. If B1, B2, B3 are the components of a vector B, that is,
B = B'i + B2j + B3k (see Example 8 in regard to the super-
VECTOR AND TENSOR ANALYSIS
20
[SEC. 10
scripts), show that for a rotation of axes the components of the
3
vector B become (BI, B2, B3) where B° _ I ayaB#, a = 1, 2, 3,
Ps1
and B = BIi' + B2j' + Mk'. Read Example 8 carefully.
23. Show that for a rotation of axes,
BICI + B2C2 + B3C3 = BICI + B2C2 +,&3103
This shows the invariance of the scalar product for rotations of
axes. The invariance here refers to both the numerical invariance of the scalar product and the formal invariance,
3
BaC' = Z BaC'a
aa1
amt
24. Prove the statements made in Example 9.
25. Generalize the statements of Example 9 for n-tuples
(a', a2
a")
10. Vector, or Cross, Product. Given any two nonparallel
vectors a and b, we may construct a third vector c as follows:
When translated so that they
have a common origin, the
two vectors a, b form two
sides of a parallelogram. We
define c to be perpendicular
to the plane of this parallelogram with magnitude equal
to the area of the parallelogram. We choose that normal obtained by the motion
of a right-hand screw when a
is rotated into b (angle of rotation less than 180°) (see Fig. 22). A cross is placed between the
vectors a and b to denote the vector c = a x b. The vector c
is called the cross, or vector, product of a and b and is given by
F ia. 22.
c=axb=lallbl sin0E
where JEl = 1.
The area of the parallelogram is
A = !alibi sin 0
(31)
TIIE ALGEBRA OF VECTORS
SEc. 121
21
The cross product will occur frequently in mechanics and electricity, but for the present Ave discuss its algebraic behavior. It
is obvious that a x b = -b x a, so that vector multiplication
is not commutative. If a and b are parallel, a x b = 0. In
particular, a x a = 0.
11. The Distributive Law for the Vector Product. We desire
to prove that a x (b + c) = a x b + a x c. Let
u= ax(b+c)-axb-axc
and form the scalar product of this vector with an arbitrary
vector v. We obtain
x(b+c)] -
x b)
In Sec. 13 we shall show that a (b x c) = (a x b) c. Hence
This implies either that u = 0 or that v is perpendicular to u.
Since v is arbitrary, we can choose it not perpendicular to u.
Hence u = 0 and
ax(b+c) =a xb+a xc
(32)
This proof is by Professor Morgan Ward of the California Institute of Technology.
12. Examples of the Vector Product
Example 10
ixi=jxj=kxk=0
jxk=i, kxl=j
ixj=k,
For the vectors a = all + a2j + ask, b = b1i + b2j + b3k we
obtain a x b = (a2b3 - a3b2)i + (a3b1 - alb3)j + (aib2 - a2b1)k
by making use of the distributive law of Sec. 11. Symbolically
we have
axb = a1
a2
k
as
bl
b2
b3
i
j
(33)
where (33) is to be expanded by the ordinary method of determinants.
VECTOR AND TENSOR ANALYSIS
22
Example 11.
a = 2i - 3j + 5k, b = -i + 2j - 3k, so that
k
i
j
2
-3
5
-1
2
-3
a xb =
Example 12.
[SEC. 12
Sine law of trigonometry
c=b - a
c xc=cx(b-a)
0=cxb-cxa
or
c xa = c xb
Fia. 23.
However, if two vectors are equal, their magnitudes are equal so
that
lellal sin I = lcilbl sin a
and
a
b
c
sin a
sin fl
sin ti
Example 13. Rotation of a Particle. Assume that a particle is
rotating about a fixed line L with angular speed w. We assume
that its distance from L remains constant. Let us define the
angular velocity of the particle as the vector w, whose direction
is along L and whose length is w. We choose the direction of w
in the usual sense of a right-hand screw advance (see Fig. 24).
THE ALGEBRA OF VECTORS
SEC. 13]
23
It is our aim to prove that the velocity vector v can be represented
by w x r, where r is the position vector of P from any origin
taken on the line L. Let the reader show that v and w x r are
parallel. Now lw x rl = wa = speed of
L
P, so that
a
13. The Triple Scalar Product. Let
us consider the scalar a (b x c). This
OP
%
v
scalar represents the volume of the parallelepiped formed by the coterminous
sides a, b, c, since
a- (b xc) =fallbllclsin0cosa
= hA = volume
(35)
A being the area of the parallelogram
with sides b and c, the altitude of the parallelepiped being
denoted by h (see Fig. 25).
Fia. 25.
Now
a (b x c) = (a,i + a2j + ask) .
= ai(b2ca - bac2) + a
i
j
k
b,
b2
bs
c1
C2
C3
(baci - bic3) + as(bic2 - b2c1)
so that
a1
a2
as
a (b x c) = b,
b2
bs
C1
C2
C3
(36)
24
VECTOR AND TENSOR ANALYSIS
[SEC. 14
Notice that a (b x c) = (a x b) c since both terms represent
the volume of the parallelepiped. It is also very easy to show
that the determinant of (36) represents (a x b) c. We usually
write a (b x c) = (abc) since there can be no confusion as to
where the dot and cross belong. We note that
(abc) = (cab) = (bca)
and that (abc) _ - (bac) = - (cba) = - (acb). These results
follow from elementary theorems on determinants. We are thus
allowed to interchange the dot and the cross when working with
the triple scalar product. This result was used to prove (32).
If the three vectors a, b, c are coplanar, no volume exists, and we
at once have (abc) = 0. In particular, if two of the three vectors
are equal, the triple scalar product vanishes.
14. The Triple Vector Product. The triple vector product
a x (b x c) plays an important role in the development of vector
analysis and in its applications. The result is a vector since
it is the vector product of a and (b x c). This vector is therefore perpendicular to b x c so that it lies in the plane of b and c.
If b is not parallel to c, a x (b x c) = xb + yc, from Sec. 6.
Now dot both sides with a and obtain x(a b) + y(a c) = 0,
since a [a x (b x c)] = 0. Hence
(a X
where
= (a b) = X,
X is a scalar, so that
a x (b x c) _ X[(a c)b - (a b)c]
(37)
In the special case when b = a, we can quickly prove that X = 1.
We dot (37) with c and obtain
c
[a x (a x c)] = X[(a c)2 - a2c2J
or
-(a x c)2 = X[(a c)2 - a2c2J
by an interchange of dot and cross.
Hence
-a2c' sine O= A(a'c2 cost 0- a2c2) = -Xa'c2 sin2 0
so that A = 1. Hence a x (a x c) = (a c)a - (a a)c. From
this it immediately follows that (a x b) x b = (a b)b - (b b)a.
Now we prove that A = 1 for the general case. We dot (37)
with b and obtain
THE ALGEBRA OF VECTORS
SEC. 15]
25
x(b xc)]
-[(a xb)
x
b
(a
x b) x
implying X = 1.
(a b)(b c)
Thus
a x (b x c) = (a c)b - (a b)c
(38)
We leave it to the reader to show that
(a x b) x c = (a c)b - (b c)a
Notice that a x (b x c) 5-1 (a x b) x c. If b is parallel to c,
(38) reduces to the identity 0 = 0, so that (38) holds for any three
vectors. The expansion (38) of a x (b x c) is often referred to
as the rule of the middle factor.
More complicated products are simplified by use of the triple
products. For example, we can expand (a x b) x (c x d) by
considering (a x b) as a single vector and applying (38).
(a x b) x (c x d) = (a x b d)c - (a x b c)d
= (abd)c - (abc)d
(39)
Also
(c x d)
d)
(b c)
d)
(40)
15. Applications to the Spherical Trigonometry. Consider the
spherical triangle ABC (sides are arcs of great circles) (see Fig.
26). Let the sphere be of radius 1. Now from (40) we see that
(a x b) (a x c) = (b
c) - (a
b) (a c)
The angle between a x b and a x c is the same as the dihedral
angle A between the planes OAC and OAB, since a x b is perpendicular to the plane of OAB and since a x C is perpendicular
to the plane of OAC. Hence
sin y sin # cos A = cos a - cos y cos,6.
VECTOR AND TENSOR ANALYSIS
26
[SEC. 15
FIG. 26.
Problems
1. Show by two methods that the vectors a = 2i - 3j - k,
b = -6i + 9j + 3k are parallel.
2. Find a unit vector perpendicular to the vectors
a=i-j+k
b=i+i - k
3. A particle has an angular speed of 2 radians per second, and
its axis of rotation passes through the points P(0, 1, 2),
Q(1, 3, -2)
Find the velocity of the particle when it is located at the point
R(3, 6, 4).
4. Find the equation of the plane passing through the end
points of the vectors a = a1i + a j + ask, b = b1i + bd + bsk,
c = c1i + cd + cak, all three vectors with origin at P(0, 0, 0).
5. Show that (a x b) x (c x d) = (acd)b - (bcd)a.
6. Prove that d x (a x b) (a x c) = (abc) (a d).
7. If a + b+ c= 0, prove that a x b= b x c= c x a,
and interpret this result trigonometrically.
THE ALGEBRA OF VECTORS
SEC. 15]
21
8. Four vectors have directions which are outward perpendiculars to the four faces of a tetrahedron, and their lengths are
equal to the areas of the faces they represent. Show that the
sum of these four vectors is the zero vector.
9. Prove that (a x b) (b x c) x (c x a) = (abc)2.
10. If a, b, c are not coplanar, show that
d = (dbc) a +
(abc)
(adc)
(abc)
b+
(abd)
c
(abc)
for any vector d.
11. If a, b, c are not coplanar, show that
d= (c(abc)d) a x b + - (a(abc)d) b x c -1-
(b d)
(abc)
cxa
for any vector d.
12. Prove that
a x(b xc) +b x(c x a)+ c x (a xb) =0
13. The four vectors a, b, c, d are coplanar. Show that
(a x b) x (c x d) = 0.
14. By considering the expansion for (a x b) x (a x c), derive
a spherical trigonometric identity (see Sec. 15).
15. Show that
(a x b) (c x d) x (e x f) = (abd) (cef) - (abc) (def)
= (abe) (fcd) - (abf) (ecd)
= (cda) (bef) - (cdb) (aef)
16. Find an expression for the shortest distance from the end
point of the vector r, to the plane passing through the end points
of the vectors r2, r8, r4. All four vectors have their origin at
P(0, 0, 0).
17. Consider the system of equations
a1x + b,y + c1z = d,
a2x + b2y + c2z = d2
(41)
asx + bsy + c3z = da
Let a = aui + a2j + aak, etc. Write (41) as a single vector equation, and assuming (abc) F6 0, solve for x, y, z.
18. Find the shortest distance between two straight lines in
space.
28
VECTOR AND TENSOR ANALYSIS
[Ssc. 15
19. Show directly that a x (b x c) _ (a c)b - (a b)c, where
a, b, c take on the values i, j, k in all possible ways. Now show
that, (38) is linear in a, b, c, that is,
a x [b x (ac + ad)] = as x (b x c) + #a x (b x d)
etc. Since any vector is a linear combination of i, j, k, explain
why (38) holds for all vectors.
20. If a and b lie in a plane normal to a plane containing c and
d, show that (a x b) (c x d) = 0.
21. If (a', a2, a3), (b', b2, b3) are the components of the vectors
a, b, show that three of the nine numbers Ca9 obtained by considering cab = aab# - a8ba, a, a = 1, 2, 3, represent the components
of a x b, that three others represent b x a, while the remaining
three vanish. Represent c,--P as a matrix and show that
cap = - G a
By considering a rotation of axes (see Example 8), show that the
EaP in the new coordinate system are related to the caP in the old
a
a
coordinate system by the equations &A = I I a aa,i9c°r,
r-1 a=1
a, 6 = 1, 2, 3. Show that caP _ -Cfta.
The numbers c' = a2b3 - a3b2, c2 = a3b' - a1b3,
c3 = a'b2 - a2b'
a
are the components of a x b. Show that ca = 1 apa#S under a
B=1
rotation of axes (see Prob. 22, Sec. 9).
22. We can construct a x b by three geometrical constructions.
We first construct a vector normal to b lying in the plane of a
and b. We project a onto this vector, and finally we rotate
this new vector through an angle of 90° about the axis parallel
to b, magnifying this newly constructed vector by the factor Ibl
The final result yields a x b. Use this to prove that
a x (b + c) = a x b + a x c
CHAPTER 2
DIFFERENTIAL VECTOR CALCULUS
16. Differentiation of Vectors. Let us consider the vector
field
u = a(x,y,z,t)i+13(x,y,z,t)j+ '(x,y,z,t)k
(42)
At any point P(x, y, z) and at any time t, (42) defines a vector.
If we keep P fixed, the vector u can still change because of the
time dependence of its components a, S, and y. If we keep the
time fixed, we note that the vector at the point P(x, y, z) will, in
general, be different from that at the point
Q(x + dx, y + dy, z + dz)
Now, in the calculus, the student has learned how to find the
change in a single function of x, y, z, t. What difficulties do we
encounter in the case of a vector? Actually none, since we easily
note that u will change if and only if its components change.
Thus a change in a(x, y, z, t) produces a change in u in the
x direction, and similarly changes in f3 and y produce changes in u
in the y and z directions, respectively. We are thus led to the
following definition:
dai.+d,6i+dyk
du =
du
as
(-aa
ax dx
+
ay
(
dy +
as
as
az
dz + at
d) i
-dz
+(axdx+aydy+a
+
(43)
ax dx +
atdt)j
dy -}-
az dz }- at dt J k
For example, let r = xi + yj + zk be the position vector of a
moving particle P(x, y, z) in three-space.
Then
dr = dx i + dy j + dz k
29
VECTOR AND TENSOR ANALYSIS
30
[SEC. 16
and
v
dt
a=
dt 1
d 2r
+
2y
z
dt2 = dt2 I
+
dt
+
dt
,
k
(44)
2
j + dt2 k
dt2
(45)
Equations (44) and (45) are, by definition, the velocity and
acceleration of the particle. We have assumed that the vectors
i, j, k remain fixed in space.
If the vector u depends on a single variable t, we can define
du
dt
lim
u(t + At) - u(t)
(46)
At
At-,o
It is easy to verify that (46) is equivalent to (43).
Example 14. Consider a
particle P moving on a circle
(see Fig. 27).
C -A-.
:Q/
-44-1k
-4- -
angular speed w = - (Fig
28).
Fio. 27.
We note that
r=rcos0i+rsin0j
so that
v = dt = (-r sin 6i+rcos Bj) de
and
a = dt
(dA2
(-r cos 0 i - r sin B j) dt/
dt2 =
C
Therefore the acceleration is
a = -w2r
(47)
The point P has an acceleration toward the origin of constant
magnitude w2r.
This acceleration is due to the fact that the
SEC. 16]
DIFFERENTIAL VECTOR CALCULUS
31
velocity vector is changing direction at a constant rate; it is
called the centripetal acceleration.
z
Y
Fra. 28.
Example 15.
FIG. 29.
Let P be any point on the space curve (Fig. 29)
x=x(s)
y = y(s)
z = z(s)
where s is are length measured from some fixed point Q.
r = x(s)i + y(s)j + z(s)k
Now
(48)
so that
dr _ dx
,
dy
,
dz
(49)
ds
and
dr dr
dx 2
dy z
(dz 2
ds ds = ds + \dsI + \ds)
=dx2+dy2+dz2=1
ds2
from the calculus. Hence ds is a unit vector.
As As -> 0, the
position of Ar approaches the tangent line at P.
Hence (49)
represents the unit tangent vector to the space curve (48).
VECTOR AND TENSOR ANALYSIS
32
[SEC. 17
17. Differentiation Rules. Consider
'p(t) = u(t) v(t)
'P(t + At) - ,P(t) = u(t + At) v(t + At) - u(t) o(t)
Now
u(t + At) = u(t) + Au
v(t + At) = v(t) + AV
(see Fig. 27), so that
c(t+
At
At
At
At
and passing to the limit, we obtain
d (u v)
dt
_
- udtdv + dudt'v
(50)
Similarly
d(u x v)
-
dt
d(fu)
dv
u
v
(51)
x dt + dt
-
du
f
dt
du
dt
d_f
+
u
(52)
dt
Notice how these formulas conform to the rules of the calculus.
Example 16. Let u(t) be a vector of constant magnitude.
Therefore
u u = u2 = constant
By differentiating we obtain
du
u'dt+
du
u'dt=0
DIFFERENTIAL VECTOR CALCULUS
SEC. 17]
3m
Hence either dt = 0 or dl is perpendicular to u. This is an
important result and should be fully understood by the student.
The reader should give a geometric proof of this theorem.
Example 17. In all cases
y
u u = u2 where u is the length
of u. Differentiation yields
2u
' dt =
2u d and
du
u'du
(53)
auat
-
This result is not trivial, for
FIG. 30.
du.
Example 18.
Idul
Now r = rR, where R is a
Motion in a Plane.
unit vector (see Fig. 30). Hence
V-dt dtR
Now
-I-r dR
R is perpendicular to R (see Example 16).
Also
IdRI
= de
We can easily verify this by differ-
since R is a unit vector.
entiating R = cos B i + sin 9 j.
Hence v =
Wt-
R + r d8 P, where
P is a unit vector perpendicular to R. Differentiating again we
obtain
a
_
_
dv
dt
d2r
dr dR
dt'R+dl
dt
dr d6
d20
+dtdt P+rdt2
dB dP
P+rdt dt
or
d2r
a_dt
since
R+2dtd6P+rd- P-r(d8)2R
P=
-d6R
(54)
[Sic. 17
VECTOR AND TENSOR ANALYSIS
34
Thus
I d2r
- (dO)2]Rid(odO\p
dt
r dt
dt
(55)
Problems
1. Prove (51) and (52).
2. Prove (54).
/
3. Differentiate l r - dr1 with respect to t.
4. Expand dt [p x (q x r)].
5. Show that
d
r x dt) = r x
d2r
6. Find the first and second derivatives of
dr d2r
(r dt dt2
7. r = a cos wt + b sin wt; a, b, w are constants. Prove that
r
xdt = wa xb andd
+w2r = 0.
8. If r x dt = 0, show that r has a constant direction.
9. R is a unit vector in the direction r. Show that
RxdR= r x dr
7.2
10. If dt = w x a,
d = w x b, show that
dt(axb)=0x(axb)
11. If r = aew' + be-,", show that dt; -- w2r = 0. a, bare constart vectors.
12. Find a vector u which satisfies dtu = at + b.
stant vectors.
Is u unique?
13. Show that
d
dt
()
r dt
r dt r.
a, b are con-
DIFFERENTIAL VECTOR CALCULUS
SEC. 17]
35
14. Let r° be the position vector to a fixed point P in space, and
let r be the position vector to a variable point Q lying on a space
curve r = r($). Show that if the distance Pte(- is a minimum, then
r - r° is perpendicular to the tangent at Q. Show also that
z
l2
r ds2 + ds,
d1r
r0
ds2
15. If u = a(x, y, z, t)i + 13(x, y, z, t)j + y(x, y, z, t)k, show
_
thatdu
_ au au dx au dy au dz
dt
at
+
+
ax dt
ay dt
+ az dt
z
Fia. 31.
16. The transformation between rectangular coordinates and
spherical coordinates is given by
x=rsin0cos0
y=rsin0sinrp
z = r cos 0
where 0 is the colatitude, (p is the longitudinal or azimuthal angle,
and r is the magnitude of the position vector r from the origin
to the particle in question. Find the components of the velocity
and acceleration of the particle along the unit orthogonal vectors
er, ee, e,, (see Fig. 31).
36
VECTOR AND TENSOR ANALYSIS
1SEc. 1b
17. Consider the differential equation
(i)
du+2Adu+Bu=0
where A, B are constants. Assume a solution of the form
u(t) = e'°'C, where C is a constant vector, and show that
u(t) = Clew,' + C2e'°2' is a solution of (i), w1, w2 being roots of
w2 + 2Aw + B = 0. Consider the cases for which A2 - B < 0,
A 2 - B = 0, A2 - B > 0.
18. Find the vector u which satisfies
a
d3u
d2u
such that u = i,
du
2
te
dt3
d
du
0
= j, d = k for t = 0.
19. If u, is a solution of
d8u+Ad2u+Bdu+Cu=0
and if u2 is a solution of
d
dtu+Adtu+Bdt+Cu=F(t)
show that u1 + U2 is a solution of (ii) provided A, B, C are
independent of u. Why is this necessary?
20. A particle moving in the plane of (r, 8) has no transverse
acceleration, that is,
dt
t r2
df) = 0. Show that the radius
vector from the originrto the particle sweeps out equal areas in
equal intervals of time.
18. The Gradient. Let p(x, y, z) be any continuous differentiable space function. From the calculus
dV
=
a dx + a dy + a dz
ax
ay
az
(56)
DIFFERENTIAL VECTOR CALCULUS
SEC. 181
37
Now let r be the position vector to the point P(x, y, z).
r = xi + yj + zk
If we move to the point Q(x + dx, y + dy, z + dz) (Fig. 32),
dr = dxi + dyj + dzk
Now notice that (56) contains the terms dx, dy, dz and the terms
aP, aSP,
aP
ax 8y az
We define a new vector formed from gp by taking its
three partial derivatives. Let del rp =
gradient p be defined by
Vsp=axi+ yJ+aZk
(57)
We immediately see that
d(p = dr VV
(58)
We shall now give a geometrical interpretation of IV-p.
At the point P(xo, yo, zo), so has
the value gp(xo, yo, zo) so that
go(x, y, z) = gp(xo, yo, zo)
represents a surface which
obviously contains the point
P(xo, yo, zo)
FIG. 32.
As long as we move along this surface, (p has the constant value
,p(xo, yo, zo) and d-r = 0. Consequently, from (58),
(59)
Now Vgp is a vector which is at once completely determined after
,p has been differentiated, and Eq. (59) states that Vgp is perpendicular to dr as long as dr represents a change from P to Q, where
Q remains on the surface v = constant. Thus V(p is normal to
all the possible tangents to the surface at P so that V(p must
necessarily be normal to the surface (p(x, y, z) = constant (see
38
VECTOR AND TENSOR ANALYSIS
Fig. 33). Let us now return to dip = dr Vv.
[Sec. 18
The vector Vp is
fixed at any point P(x, y, z), so that dip (the change in p) will
depend to a great extent on dr. Certainly dcp will be a maximum
when dr is parallel to Vp, since dr VV = Idrl IV pl cos 0, and cos 8
is a maximum for 0 = 0°.
VO
Thus Vp is in the direction of
maximum increase of p(x, y, z).
Let ldrl = ds so that
d(P
ds
=u
V(p
(60)
where u is a unit vector in
the direction dr. Hence the
change of (p in any direction
the projection of Vp on
the unit vector having this
is
Fia. 33.
Example 19.
direction.
To find a unit vector normal to the surface
x2 + y2 - z = 1 at the point P(1, 1, 1).
Here
,P(x,y,z) =x2+y2-z
V(p=2xi+2yj-k
=2i-2j-katP(1,1,1)
Thus
N=2i-2j-k
3
Example 20. We find Vr if r = (x2 + y2 + Z2)1. The surface
r = constant is a sphere. Hence Vr is normal to the sphere and
so is parallel to the position vector r. Thus Vr = kr. Now
dr = dr Vr = k dr r = kr dr
from (53)
Therefore
k=1
r
and
Vr = r = R
r
Example 21
Vf(u) = f'(u) Vu,
u = u(x, y, z)
(61)
39
DIFFERENTIAL VECTOR CALCULUS
SEC. 18]
Proof:
Vf(u) =
i+ayj+azk
a i + f'(u)
= r(u)
ax
au
au j
ay
+
au k
f'(u) az
= f'(u) Vu
Example 22
Vf(ul, u2,
. . .
, un)
=a i+ayj+azk
_
of aua
1
n
=
,
au, ax
of aua
,
aua ay
a
If Vua
aua
of au-aua az
)
(62)
Consider the ellipse given by rl + r2 = constant
(see Fig. 34). Now V(rl + r2) is normal to the ellipse. Let
Example 23.
Y
Fio. 34.
T be a unit tangent to the ellipse. Thus V(rl + r2) T = 0, and
W2 - T
(63)
But from Example 20, Vrj is a unit vector parallel to the vector
AP, and Vr2 is a unit vector parallel to the vector BP.
Equation
40
VECTOR AND TENSOR ANALYSIS
[SEC. 19
(63) shows that AP and BP make equal angles with the tangent
to the ellipse.
19. The Vector Operator V. We define
v=ia +iaay +kaz
(64)
Notice that V is an operator, just as dx is an operator in the differ-
ential calculus. Thus
v - (1-+jay +kaz)'p
a vector operator because of its components
a
a
a
ax ay az
It will help us in the future to keep in mind that V
acts both as a differential operator and as a vector.
Example 24
v (uv) =
i a (uv)
ax
q!!!) + k a (uv)
+ J ay
az
C'ax+j
y
a+kc1v )u+Clax +jau+k
az
V(uv) = U Vv + v Vu
y
)v
-az
(65)
This result is easily remembered if we keep in mind that V is a
differential operator, so that we can apply the ordinary rules of
calculus.
Problems
1. Find the equation of the tangent plane to the surface
xy - z = 1 at the point (2, 1, 1).
Sac. 191
DIFFERENTIAL VECTOR CALCULUS
41
2. Show that V(a r) = a, where a is a constant vector and r
is the position vector.
3. If r = (x2 + y2 + Z2)'1, find Vr" by explicit use of (57).
4. If So = (r x a) (r x b), show that
V,p = b x (r x a) + a x (r x b)
when a and b are constant vectors.
5. Let gp = x2 + y2. Find IVpI and show that it is the maximum change of V.
6. Find the cosine of the angle between the surfaces
x2y + z = 3
and x log z - y2 = -4 at the point of intersection P(- 1, 2, 1).
7. What is the value of Vv(x, y, z) at a point that makes So a
maximum?
8. The surfaces g(x, y, z) = constant and #(x, y, z) = constant are normal along a curve of intersection. What is the
value of V(p V4' along this curve?
9. What is the direction for the maximum change of the space
function gp(z, y, z) = x sin z - y cos z at the origin?
10. Expand V(u/v) where u = u(x, y, z), v = v(x, y, z).
11. Let r and x be the distances from the focus and directrix
to any point on a parabola. We know that r = x. Show that
(R - i) T = 0, where T is a unit tangent vector to the parabola,
and interpret this equation.
12. Show that the ellipse r, + rz = c, and the hyperbola
r, - ra = c2 intersect at right angles when they have the same
foci.
13. If Vv is always parallel to the position vector r, show that
rp=(p(r),r== x2+ysd-z2.
14. Find the change of g = xyz in the direction normal to the
surface yx2 + xy2 + z'y = 3 at the point P(1, 1, 1).
15. If f = f(xl, x', xa) (see Example 8), and if
x°t = x01(yl, y2, ya),
a = 1, 2, 3, show that
of
of axe
aye
aya
a = 1, 2, 3
VECTOR AND TENSOR ANALYSIS
42
Using the fact that
s
of
of ay a
axa
9-1 ay8 axa
axa
_
3
ax"
aye
1 ay° ax8
ax8
`
[SEC. 20
1 if« _ R
, show that
0 if a 0 ,e
'
a = 1, 2, 3.
16. Apply the results of Prob. 15 above to the transformation
x = r cos 0
y = r sin 0
z=z
1 of , of
are the components of Vf along the
and show that af,
Or r ae az
three mutually orthogonal unit vectors e, es, e. which occur in
cylindrical coordinates.
dr
17. If (_ v(x, y, z, t), show that d,P= ate +
V(p.
18. Ifr =xi+yj+zk,find
xyevzi:'
+ log
-X
z
(x + z + y)
19. Ifu =u(x,y,z,t)showthat dt = au, +Cad V>u.
20. If u(tx, ty, tz) = t"u(x, y, z), show that (r V)u = nu.
20. The Divergence of a Vector. Let us consider the motion
of a fluid of density p(x, y, z). We assume that the velocity
field is given by f = u(x, y, z)i + v(x, y, z)j + w(x, y, z)k. This
type of motion is called steady motion because of the explicit
independence of p and f on the time. We now concentrate on
the flow through a small parallelepiped ABCDEFGH (Fig. 35),
of dimensions dx, dy, dz.
Let us first calculate the amount of fluid passing through the
face ABCD per unit time. The x and z components of the
velocity f contribute nothing to the flow through ABCD. The
mass of fluid entering ABCD per unit time is given by pv dx dz.
The mass of fluid leaving the face EFGH per unit, time is
r
pv +
a(pv)
1
dy dx dz
ay
SEC. 20)
DIFFERENTIAL VECTOR CALCULUS
43
The loss of mass per unit time is thus seen to be equal to
a(pv)
ay
dx dy dz
If we also take into consideration the other two faces, we find
that the total loss of mass per unit time is
[a(Pu)+o(Pv)+a(Pw)]dddz
z
A
dy
E
ry
Fia. 35.
so tha t
a(Pu)
ax
+
a(pv)
ay
+
(66)
a(Pw)
az
represents the loss of mass per unit time per unit volume. This
quantity is called the divergence of the vector pf. We see at
once that
VV
. (pf )
= di v (pf) =
a(Pu)
ax
a(Pv)
+
ay
+
a(Pw)
az
= 1 dM
V dt
(67 )
VECTOR AND TENSOR ANALYSIS
44
[SEc. 20
since i, j, k are constant vectors. M and V are the mass and
volume of the fluid.
The divergence of any vector f is defined as V f. We now
calculate the divergence of p(x, y, z)f.
V
a(q.u)
+ a(te) +
ax
ay
_
(ax
au
av
x
+ay+
aw
az
a(cyw)
az
arp
a'p
sip
+ uax +v ay +u'aa
(68)
We remember this result easily enough if we consider V as a
vector differential operator. Thus, when operating on spf, we
first keep p fixed and let V operate on f, and then we keep f fixed
and let V operate on V(V V is nonsense), and since f and Vp are
vectors we complete their multiplication by taking their dot
product.
Example 25.
Compute V f if f = r/r' (inverse-square force).
V (r-'r) = r-3V r+r Vr-'
V (r- 3r) = 0
(69)
This is an important result. The divergence of an inverse-square
force is zero. We note that
y
Example 26.
What is the divergence of a gradient?
k/
axe+ayz+az,
DIFFERENTIAL VECTOR CALCULUS
SEc.21]
45
This important quantity is called the Laplacian of 0.
= v2 = axe
v
(70)
aye
az2
21. The Curl of a Vector. We postpone the physical meaning
of the curl and define
curl f=vxf=
_ -J
Caw
vxf = i
ay
+
j
a
a
ax
ay
az
u
v
w
_ awl
1
az
i
a
+k
ax
az
_ au
1
ax
(71)
ay,
Example 27
=0
Vxr=
Example 28
i
j
k
a
a
a
ax
ay
az
cpu
vv cv
1
v x (wf) =
[a(te) _ a()1
ay
az
JJ
[a(sou)
az
a(caw)
ax
+ k [a(;)
a(,Pu)]
ay
ax
v x (cOf) = cO
aw
[i (ay
av
az
+j
au
Ow
az
ax
U
Vx('vf)=(Pvxf+Via xf
v
w
(72)
46
VECTOR AND TENSOR ANALYSIS
[SEC. 21
This result is easily obtained by considering V as a vector differential operator.
Example 29. To show that the curl of a gradient is zero.
i
a
V X (VP) = ax
j
k
a
ay
a
az
i
av av av
ax
ay
- az ay
ay az
a2
a2IP
az
02,p
a2,p
+ i
\ az ax
ax
az/ + k \ax ay
a2,
C1 2,p
ay ax/
Hence
Vxvio =0
(73)
provided (p has continuous second derivatives.
Example 30. To show that the divergence of a curl is zero.
a au_aw'
a aw_av
V (v x f) =
ax ay
az
a2u
a2ul
ay az
az ay
+ ay az
a2V
ax /
a(av_au)
+ az \ax
a2w
a2v
+ az ax
ax az
+
ay/
a2w
ax ay - ay ax
Thus
(74)
Example 31.
V.
What does (u V)v mean? We first dot u with
This yields the scalar differential operator
a
a
a
ux ax + ua ay + U. az
Then we operate on v obtaining
uxax+Uyay+u$az
Thus
df=axdx+af dy+azdz
y
=dxa+dya+dzaz
y
DIFFERENTIAL VECTOR CALCULUS
SFC. 211
47
and
df = (dr V)f
(75)
since dr = dx i + dy j + dz k.
If f = f(x, y, z, t),
df = (dr V)f + a dt
(76)
Example 32
+vsa
= vJ + vyj + vZk
(v - V)r = v
(77)
where r is the position vector xi + yj + zk.
Example 33. Let us expand V(u v). Now
u x (V x v) = Vti(u
v) - (u V)v
Here we have applied the rule of the middle factor, noting also
that V operates only on v. Vn(u v) means that we keep the
components of u fixed and differentiate only the components of v.
v) - (v V)u. Adding, we
Similarly, v x (V x u) =
obtain
Vu(u v) + Vn(u v) = u x (V x v) + v
x u)
+ (u V)v + (v V)u
and
u x(V xv) + v x(V xu) +
(78)
Example 34
V x (u x v) = (v V)u - v(V u) + u(V v) - (u
V)v
(79)
48
VECTOR AND TENSOR ANALYSIS
[SEc. 22
Example 35
V. (u xv) = Vu (u xv) +Vv (u xv)
(V
(80)
(V
Example 36
V X (V x v) = V(V v) - V2v
Let A = V x ((pi) where V2,p = 0.
pute A V x A. Since A = V(p x i, we obtain
Example 37.
V x A = V x V xi =
V
iV2
(81)
We now com-
a
cix
from (7), Sec. 22. Thus
AVxA=
i
j
k
app
app
app
a2
ax
ay
az
ax2+j ay ax+kazax
1
0
0
app a2
= az ay ax
a2y
()2
-P
app a2g
ft azax
If also sp = X(x)Y(y)Z(z), we can immediately conclude that
22. Recapitulation. We relist the above results:
1. V (UV) = u Vv + v Vu
2.
V
V
xv+V(pxv
4. V x (V(p) = 0
5. V. (Vxv)=0
6. V. (u x v) _ (V x u) - v - (V x v) - u
7. V x (u x v) _ (v V)u - v(V u) + u(V v) - (u V)v
8.
V
Vxr
=0
13. df = (dr V)f + a dt
Sxc. 22]
DIFFERENTIAL VECTOR CALCULUS
14. dco = dr VV +
49
LP dt
15. V- (r 3r) =0
Problems
1. Show that V2(1/r) = 0 where r = (x2 + y2 + z2)}.
2. Compute V2 r, V2r2, V2(1/r2) where r = (x2 + y2 + z2)3.
3. Expand V(uvw).
4. Find the divergence and curl of (xi - yj)/(x + y); of
xcoszi+ylogxj -z2k.
5. If a = axi + fiyj + yzk, show that V(a r) = 2a.
6. Show that V x V(r)r] = 0 when r = (x2 + y2 + Z2) i and
r = xi + yj + zk.
7. Let u = u(x, y, z), v = v(x, y, z). Suppose u and v satisfy
an equation of the form f(u, v) = 0. Show that Vu x Vv = 0.
8. Assume Vu x Vv = 0 and assume that we move on the surface u (x, y, z) = constant. Show that v remains constant and
hence v = f(u) or F(u, v) = 0.
9. Prove that a necessary and sufficient condition that u, v, w
satisfy an equation f(u, v, w) = 0 is that Vu Vv x Vw = 0, or
au
au
au
ax
ay
az
av
ft
av
ax
ay
az
0
aw aw aw
ax
ay
az
This determinant is called the Jacobian of (u, v, w) with respect
to (x, y, z), written J[(u, v, w)/(x, y, z)].
10. If w is a constant vector, prove that V x (w x r) = 2w,
where r=xi+yJ+zk.
11. If pf = Vp, prove that f V x f = 0.
12. Prove that (v V)v = 4 Vv2 - v x (V x v).
13. If A is a constant unit vector, show that
V x (v xA)] =
14. If fl, f2, f3 are the components of the vector f in one set of
rectangular axes and 11, /, / are the components of f after a
50
VECTOR AND TENSOR ANALYSIS
[SEC. 23
rotation of axes (see Example 8), show that
a fa
a=1
xa
aa_
=
a=1 x
so that V f is a scalar invariant under a rotation of axes.
Also
see Prob. 21, Sec. 9.
15. Prove (79), (80), (81).
16. Let f = f1i + f2j + f3k and consider nine quantities
afi
axi
9is
of,
axis
i, j = 1, 2, 3
Show that gi; = -gi and that three of the nine quantities yield
the three components of V x f. Use this result to show that
V x((pf) =('V xf+Vcpxf.
23. Curvilinear Coordinates. Often the mathematician, physicist, or engineer finds it convenient to use a coordinate system
other than the familiar rectangular cartesian coordinate system.
If he is dealing with spheres, he will probably find it expedient to
describe the position of a point in space by the spherical coordinates r, 8, sp (see Fig. 31). Let us note the following: The sphere
x2 + y2 + z2 = r2, the cone z/(x2 + y2 + z2)'1 = cos 8, and the
plane y/x = tan sp pass through the point P(r, 0, Sp). We may
consider the transformations
r = (x2+y2+Z2)}
a = cos- '
Sp =
z
(x2 + y2 + z2)
tan-1 y
X
as a change of coordinates from the x-y-z coordinate system to
the r-B-#p coordinate system. The surfaces
r = (x2 + y2 + z2)} = cl
8 = cos-' [z/(x2 + y2 + z2)}1 = c2, O = tan-' y/x = c3 are respect
tively, a sphere, cone, and plane. Through any point P in space,
except the origin, there will pass exactly one surface of each type,
the coordinates of the point P determining the constants c1, c2, c3.
The intersection of the sphere and the cone is a circle, the circle
of latitude, having e, as its unit tangent vector at P. This circle
is called the (p-curve since r and 0 remain constant on this curve
SEC. 23]
DIFFERENTIAL VECTOR CALCULUS
51
so that only the coordinate p changes as we move along this
curve. The intersection of the sphere and the plane yields the
0-curve, the circle of longitude, while the intersection of the cone
and plane yields the straight line from the origin through. P, the
r curve. ee and er are the unit tangent vectors to the 0- and
r curves, respectively. The three unit vectors at P, er, ee, e, are
mutually perpendicular to each other and can be considered as
forming a basis for a coordinate system in the neighborhood of P.
Unlike i, j, k, they are not fixed, for as we move from point to
point their directions change. Thus we may expect to find more
complicated formulas for the gradient, divergence, curl, and
Laplacian when dealing with spherical coordinates.
Since Vv is perpendicular to the plane cp = constant, we must
have V(p parallel to e, Hence e9, = h3 Vsp, where hs is the scalar
factor of proportionality between e,, and Vp. If drs is a vector
tangent to the p-curve, of length dss = jdr3j, we have from (58)
dip = drs VP = dr3
e,p
hs
=
ds3
hs
so that ds3 = hs dip.
Hence h3 is
that quantity which must be multiplied into the differential change
of coordinate gyp, namely, d-p, to yield arc length along the p-curve.
Thus e,, = r sin 0 V(p, while similarly er = Vr and ee = r VB.
We note that er = ee x e,, = r2 sin 0 VO X VV,
eB=e,xe,.=rsin 0V(pxVr
e,=erxee=rVrxVB
Any vector at P may be represented as f = flex + flee + f3ec.
The scalars f1, f2, fs can be functions of r, 0, p. We may also
represent f as f = f1 Vr + f2r VO + far sin 0 VV and also by
f = f1r2 sin O VO x V(p + f2r sin BV(p x Vr + far Vr x V8.
We also
note that the triple scalar product Vr V9 x Vtp is equal to
(r2 sin 0)-1 and that dV = dsl ds2 ds3 = r2 sin 0 dr d9 dp.
Spherical coordinates are special cases of orthogonal curvilinear
coordinate systems so that we will proceed to discuss these more
general coordinate systems in order to obtain expressions for the
gradient, divergence, curl, and Laplacian.
Let us make a change of coordinates from the x-y-z system to a
u1-u2-u3 system as given by the equations
Ul = u1(x, y, z)
U2 = u2(x, y, z)
ug = u3(x, y, z)
(82)
52
VECTOR AND TENSOR ANALYSIS
[SEC. 23
We assume that the Jacobian J[(ul, u2, u3)/(x, y, z)] ;P,-, 0 so that
the transformation (82) is one to one in the neighborhood of a
point. A point in space is determined when x, y, z are known
and hence when ui, u2, u3 are known. By considering
ui(x, y, z) = ci
u2(x, y, z) = C2, u3(x, y, z) = es, we obtain a family of surfaces.
Through a point P(xo, yo, zo) will pass the three surfaces
ui(x, y, z) = ui(xo, yo, zo)
z
y
FIG. 36.
u2(x, y) z) = u2(xo, yo, Zo), and u3(x, y, z) = u8(xo, yo, zo). Let us
assume that the three surfaces intersect one another orthogonally.
The surfaces will intersect in pairs, yielding three curves which
intersect orthogonally at the point P(xo, yo, zo). The curve of
intersection of the surfaces uI = ci and u2 = C2 we shall call the
us curve, since along this curve only the variable us is allowed to
change. Let u1, u2, us be three unit vectors issuing from P
tangent to the ui, us, us curves, respectively (see Fig. 36).
DIFFERENTIAL VECTOR CALCULUS
SEc. 231
53
Now Vu3 is perpendicular to the surface
u3(x, y, z) = u3(xo, yo, zo)
so that Vu3 is parallel to the unit vector u3. Hence u3 = h3 Vu3
where h3 is the scalar factor of proportionality between us and
Vu3.
Now let dr3 be a tangent vector along the us curve,
dr3' = dss.
Obviously dr3 us = dss, and
so that from (58)
dss = h3 du3
(83)
We see that hs is that quantity which must be multiplied into
the differential coordinate dus so that are length will result. For
example, in polar coordinates ds = r do if we move on the 0-curve,
so that r = h2.
Similarly, ul = h1 Vu1j u2 = h2 Du2, so that
u1 = U2 X u3 = h2h3 Vu2 X Vu3
u2 = us x u1 = h3h1 Vu3 X Vu1
113 = u1 x u2 = h1h2 Vu1 x out
(84)
and
Du1 Du2 X Vu3 =
u1 U2
h1 h g
x
Us
s
= (h1h2ha)-1
(85)
Note that the differential of volume is
dV = ds1 ds2 ds3 = h1h2h3 du1 due du3
and making use of (85) as well as Prob. 9, Sec. 22,
dV = j (_x, y, z ) du1 due du3
u1, U2, u3
Example 38.
In cylindrical coordinates
ds2 = dr2 + r2 d02 + dz2
so that h1 = 1, h2 = r, h3 = 1.
(86)
VECTOR AND TENSOR ANALYSIS
54
[SEC. 23
If f = f (U1, u2, u3), then from Example 21,
Example 39.
of
yf =
vu3
Vul + au2
` f vu2 + of
au3
au1
of
of
of
Vf = h1 au1 u1+----u2+-''U3
h3 au3
h2 49u2
1
1
1
(87)
In cylindrical coordinates
of
ar
R+p+k
r a9
az
Our next attempt is to obtain an expression for the divergence
of a vector when its components are known in an orthogonal
curvilinear coordinate system. Now
f = flul + f2u2 + f3u3
= f lh2ha Vu2 x vu3 + f 2h3h1 Vu3 x Vul + fahlh2 Vu1 x vu2
from (84).
Consequently
V f = V (f 1h2h3) vu2 x Vu3 + f 1h2h3 V (Vu2 X Vu3)
+ V(f 2h3h1) Dua x Vu1 + f 2h3hly (Vu3 X Vu1)
+ V(f3h1h2) Vu1 X Vu2 + f3h1h2V (Vul x vu2)
Now V(f1h2h3) VU2 X Vu3 =
(88)
a(flh2ha )
Vu1 Vu2 x Vu3, and
out
V (vu2 x vua) = 0
ao that (88) reduces to
[o(h2h3fi)
1
h1h2h3
49U,
+
a(h3h1f2)
au2
h,h2f3)
+
(89
49u3
)
If we apply (89) to the vector VV as given by (87), we obtain
VV2
1
a
h1h2h3 au1
h2h3 a
{
h1 49U,
a (hAi a
+au2
h2 au2
a (h,h2 a
+ aus
h3
au]
(90)
55
DIFFERENTIAL VECTOR CALCULUS
SEC. 23]
This is the Laplacian in any orthogonal curvilinear coordinate
system.
Example 40.
V2V
In cylindrical coordinates
l
r
a
ar
r
a
ar
-I-
a
1a_
aB
r aB
+az- r aza
a
(91)
Example 41. Solve V2V = 0 assuming V = V(r),
r = (x2 + y2)i
From (91)
1 d (LV
r J =0
dr
r dr
r
or
ddV = c1
and
V =cllogr+C2
Finally we obtain the curl of f.
f = f1u1 + f2u2 + faun
= f 1h1 Vu1 + f2h2 Due + fshs Vua
and
V x f = V(f1hi) x Vul + V(f2h2) X Vu2 + V(faha) X Vus
since V x (Vul) = V X (Vu2) = V x (Vua) = 0. Now
V(f1h1) X VU, =
a(f1h1)
49U,
Vul X Vul +
a(f1h1)
Vu2 X Vu1
49U2
+ a(flhl
aus
Vus X Vul
us
Replacing Vu2 x Vul by -- hlh2, etc., we obtain
Vxf=
ul f a(hsf3)
h2h3 L
au2
u2 ra(hlfl) ^ a(hsfa)
)1
- a(h2f2
aul
aua J + hsh1 L aua
ua
` a(h1f1) (92)
La(h2f2)
+
1h2
49U1
49U2
56
VECTOR AND TENSOR ANALYSIS
[Sec. 23
Problems
1. For spherical coordinates, ds2 = dr2 + r2 d92 + r2 sine 9 d(p2
where B is the colatitude and (p the azimuthal angle. Show that
V Z V=
a I r2 sin B a
r2 sin 9 ar \\\
ar
1
I
a
}-
(sin
a9 `
6
aV
aB/J
+ a (sin 9
2. Solve V2V = 0 in spherical coordinates if V = V(r).
3. Express V - f and V x f in cylindrical coordinates.
4. Express V f and V x f in spherical coordinates by letting
a, b, c be unit vectors in the r, 9, (p directions, respectively.
5. Write Eq. (92) in terms of a determinant.
6. Show that V x [(r V9)/sin 0) = V(p where r, 0, (p are spherical coordinates.
7. If a, b, c are the vectors of Prob. 4, show that
as
as
ar=0'
a9= b,
ab
=0
ab
as =sin9c
= -a
ar
a9
ac
ac
ar=0'
a9=0'
ab
= cos 9 c
a(P
_
ac
= - sin0a - cos0b
8. If x = r sin 9 cos (p, y = r sin 9 sin (p, z = r cos 9, then
the form ds2 = dx2 + dy2 + dz2 becomes
ds2 = dr2 + r2 d92 + r2 sin2 9 d(p2
3
Prove this. If, in general, ds2 = I (dxa) 2, and if
a-1
xa = xa(y', y2, y3)
a = 1, 2, 3, show that
axa axa
ds2 =
a,
..ydyOdy
_ I go, dys
X-f
Y
dyo dyr
dy''
SEC. 231
DIFFERENTIAL VECTOR CALCULUS
57
where
3
90Y = aI
ax" axa
a" ay"
Check this result for the transformation to cylindrical coordinates:
x = r cos 0
y = r sin 0
z=z
and obtain ds2 = dr2 + r2 d02 + dz2.
9. By making use of V2V = V(V V) - V x (V x V), find
V2V for V = v(r)e,, V being purely radial (spherical coordinates).
Find V2V for V = f(r)e, + (p(z)e, in cylindrical coordinates.
10. Find V IV if V = w(r)k x r.
11. Consider the equations
a2S
(X -f 'U)V(V
X, µ, p constants.
s) + u V2s = p
ate
Assume s = eiP'sl, p constant, and show that
(X + p)V(V s1) + (A + pp2)sl = 0
Next show that [V2 + (µ + pp2)/(X + µ)](V s1) = 0,
X+µ
0.
12. If A = V x (¢r), V2¢ = 0, show that
1
a a2Y'
sin 0 appaOar
a¢ a2Y'
a0acpar
so that A V x A = 0 if, moreover, ¢ =
13. Show that Cpi = Ae9 + Bey + Ce° satisfies V2sp1 = ci, and
show that if 402 satisfies V°02 = 0, then rp = Cpl + 4p2 also satisfies
V2ip = gyp. Find a solution of V2Sp = -(p.
CHAPTER 3
DIFFERENTIAL GEOMETRY
24. Frenet-Serret Formulas. A three-dimensional curve in a
Euclidean space can be represented by the locus of the end point
of the position vector given by
r(t) = x(t)i + y(t)j + z(t)k
(93)
where t is a parameter ranging over a set of values to < t < ti.
We assume that x(t), y(t), z(t) have continuous derivatives of all
orders and that they can be expanded in a Taylor series in the
neighborhood of any point of the curve.
We have seen in Chap. 2, Sec. 16, that ds is the unit tangent
vector to the curve. Let t =
ds-
Now t is a unit vector so that
its derivative is perpendicular to t. Moreover, this derivative,
dse
tells us how fast the unit tangent vector is changing direction
as we move along the curve. The principal normal to the curve
is consequently defined by the equation
dt
= Kn
(94)
ds
where K is the magnitude of ds and is called the curvature.
The
reciprocal of the curvature, p =I 1K, is called the radius of curvature. It is important to note that (94) defines both K and n,
K being the length of ds while n is the unit vector parallel to
dt
At any point P of our curve we now have two vectors t, n at
ds
right angles to each other (see Fig. 37). This enables us to set up
58
DIFFERENTIAL GEOMETRY
SEC. 24]
59
a local coordinate system at P by defining a third vector at right
angles to t and n. We define as the binormal the vector
b = t xn
All vectors associated with the curve at the point P can be
written as a linear combination of the three fundamental vectors
t, n, b, which form a trihedral at P.
z
0
Y
x
Fia. 37.
Let us now evaluate - -- and
dn
Since b is a unit vector, its
derivative is perpendicular to b and so lies in the plane of t and n.
Moreover, b t = 0 so that on differentiating we obtain
t=
0.
Hence `
must be parallel to n.
is also perpendicular to t so that
dd_b
Consequently, ds = rn, where r by defini-
tion is the magnitude of -. r is called the torsion of the curve.
VECTOR AND TENSOR ANALYSIS
60
[SEC. 24
Finally, to obtain dn, we note that n = b x t so that
b
ds
b x ds +
x t = b x Kn + rn x t =- Kt- rb
The famous Frenet-Serret formulas are
dt
Kn
ds
do
- (Kt + Tb)
ds
95)
db
rn
ds
Successive derivatives are functions of t, n, b and the derivatives
of K and r.
Example 42.
The circular helix is given by
r = a cos ti+asintj+btk
t =ds = (-a sin ti+acostj + bk) st
and
(dl (a2sin2t+a2cos2t+b2)
2
1=
(.)2
(a2 + b2)
Hence
t = (-a sin ti+acostj+bk)(a2+b2) 4
Now
Kn = d = (-a cos t i - a sin t j)(a2 + b2)-t
so that
K = a(a2 +
b2)-i
Also
i
j
b = t x n = -a sin t a cost
-cos t
-sin t
k
b
(a2 + b2)-1
0
= (b sin t i - b cos t j + ak) (a2 +
b2)-+
DIFFERENTIAL GEOMETRY
SEC. 241
61
and
db
ds
= rn = (b cos t i + b sin t j)(a2 + b2)-1
so that
T = b(a2 + b2)-1
Problems
1. Show that the radius of curvature of the twisted curve
x = log cos 0, y = log sin 0, z = V2 0 is p =
csc 20.
2. Show that r = 0 is a necessary and sufficient condition that
a curve be a plane curve.
(r'r"r"').
3. Prove that T =
K
4. For the curve xz = a(3t - te), y = 3at2, z = a(3t + t3),
show that K = T = 1/3a(1 + t2)2.
5. Prove that
AA _=
ds . ds
Kr,
do db
d8 - ds
=
0,
dt dn
_
d8 . ds
= 0.
6. Prove that r"' = -K2t + K'n - rKb, where the primes
mean differentiation with respect to are length.
7. Prove that the shortest distance between the principal
normals at consecutive points at a distance ds apart (s measured
along the arc) is ds p(p2 +
8. Find the curvature and torsion of the curve
r`2)_;
y = all - cos u),
x = a(u - sin u),
z = bu
9. For a plane curve given by r = x(t)i + y(t)j, show that
x,y - y,x
[(x')2 + (y1)2]1
10. Prove that (t't"t"') = K5
ds (K)
11. Show that the line element ds2 = dx2 + dy2 + dz2 - c2 dt2
remains invariant in form under the Lorentz transformation
x=
- yt
[1
- (V2/c2)1I
y=I
z=2
t=
- (V/c2)x
[1 - (V2/c2)J*
62
VECTOR AND TENSOR ANALYSIS
[SEC. 25
V, c are constants. The transformation ct = iT, i
leads to the four-dimensional Euclidean line element
ds2=dx2+dy2+dz2+dr2
12. If xa = xa(s), a = 1, 2, . . . , n, represents a curve in an
n-dimensional Euclidean space for which
d82
= (d_-1)2 + (dx2) 2 +
.
. + (dxn) 2
define the unit tangent vector to this curve, this definition being a
generalization of the definition of the tangent vector for the case
n = 3. Show that the vector
d2xa
ds2
) a = 1, 2,
.
.
.
, n, is normal
to the tangent vector, and define the unit principal normal n,
and curvature K, by the equations
d2xa
dta
d82 - ds =
Klnla,
a = 1, 2,
..,n
a
Show that
to
a-1
ds
a = 1, 2,
dnla =
- K1.
!Is
. . .
Define the second curvature K2 and unit
normal n2 by th e equati ons d
..
, n, is normal to nl and that
d
la
= -Klt a + K2n2 a, a =
1, 2,
i
. , n, and show that n2a is normal to to and nla if K2 ;P'- 0.
Continue in this manner and obtain the generalization of the
Frenet-Serret formulas.
25. Fundamental Planes. The plane containing the tangent
and principal normal is called the osculating plane. Let s be a
variable vector to any point in this plane and let r be the vector
to the point P on the curve. s - r lies in the plane and is conse-
quently perpendicular to the binormal. The equation of the
osculating plane is
(s - r) b = 0
(96)
The normal plane to the curve at P is defined as the plane
through P perpendicular to the tangent vector. Its equation is
DIFFERENTIAL GEOMETRY
SEC. 26]
63
easily seen to be
(s - r) t = 0
(97)
The third fundamental plane is the rectifying plane through P
perpendicular to the normal n. Its equation is
(s - r) n = 0
(98)
Problems
1. Find the equations of the three fundamental planes for the
curve
x = at,
y=bt2,
z=cta
2. Show that the limiting position of the line of intersection
of two adjacent normal planes is given by (s - r) n = p where
s is the vector to any point on the line.
26. Intrinsic Equations of a Curve. The curvature and torsion
of a curve depend on the point P of the curve and consequently
on the are parameter s. Let is = f(s), r = F(s). These two
equations are called the intrinsic equations of the curve. They
owe their name to the fact that two curves with the same intrinsic
equations are identical except possibly for orientation in space.
Assume two curves with the same intrinsic equations. Let the
trihedrals at a corresponding point P coincide; this can be done
by a rigid motion.
Now
ds
(t1. t2) = t1
,cn2 + xni t2
T (nl n2) = n1 (-Kt - rb2) + n2 (--Kt, -- rbi)
d
Adding, we obtain
s
0
(99)
64
VECTOR AND TENSOR ANALYSIS
(SEC. 27
so that
constant = 3
since at P
tl=t2,
n1=n2,
(100)
b1=b2
Since (100) always maintains its maximum value, we must have
dr, _ dr2
tl = t2, n1 n2, bi = b2 so that
or r1 = r2 locally.
ds
ds
Hence the two curves are identical in a small neighborhood of
P. Since we have assumed analyticity of the curves, they are
identical everywhere.
Problems
1. Show that the intrinsic equations of x = a(9 - sin 8),
y = a(l - cos 8), z = 0 are p2 + s2 = 16a2, 7- = 0, where s is
measured from the top of the are of the cycloid.
2. Show that the intrinsic equation for the catenary
y=a'(ex/a+e-(sla))
2
is ap = s2 + a2, where 8 is measured from the vertex of the
catenary.
Fla. 38.
27. Involutes. Let us consider the space curve r. We construct the tangents to every point of r and define an involute
DIFFERENTIAL GEOMETRY
SEc. 271
65
as any curve which is normal to every tangent of r (see Fig. 38).
From Fig. 39, it is evident that
r,=r+ut
(101)
is the equation of the involute, u unknown. Differentiating
(101), we obtain
dr,
ds,^lt
A
Cdr
d_u \ ds
ds+ u ds+dst
ds,
where s is are length along r and s, is arc length along r'.
(95), (102) becomes
r
ti =
(t+uua+dut)
d
ds
ds,
(102)
Using
(103)
Now t t, = 0 from the definition of
the involute so that
du
1 +=0
and
u=c - s
Fta. 39.
(104)
Therefore r, = r + (c - s)t, and there exists an infinite family
of involutes, one involute for each constant c. The distance
between corresponding involutes remains a constant. An invo-
r
Fia. 40.
lute can be generated by unrolling a taut string of length c which
has been wrapped along the curve. The end point of the string
generates the involute (see Fig. 40). What are some properties
of the involute?
VECTOR AND TENSOR ANALYSIS
66
[SEc. 28
r1=r+(c-s)t
ti = dr,
dr
/
dt
Ids
+
lC
S)
ds ds, -
(C - s)
t]
ds
dsl
ds
K - n
ds1
Hence the tangent to the involute is parallel to the corresponding
normal of the curve. Since ti and n are unit vectors, we must
have (c - s) K d = 1. The curvature of the involute is
l
do ds
dt1
obtained from
- = Kin, ds= ds1
--=
dsl
(-Kt -7-b)
Hence
(c - s)K
+ r2
r
K12
K2
K2(C - 8)2
(105)
28. Evolutes. The curve t'
whose tangents are perpendicular to a given curve is called the
evolute of the curve. The tan-
gent to r' must lie in the plane
of b and n of r since it is perpendicular to t. Consequently
Fla. 41.
rl=r+un+vb
is the equation of the evolute.
tl
=
dr1
= dr
dsl -
{ds
Differentiating, we obtain
d_n
db
d_u
dv
A
+ u ds + v ds + ds n + d8 bl ds,
= It + u(-Kt -rb) +vTn+ds n +d8b]dsl
Now t t1 = 0, which implies I - uK = 0 or u =
dv)
tl=L(-ru+as
, b+( +ds)n
1
= p. Thus
K
d8
Also t1 is parallel to r1 - r = un + vb (see Fig. 41). Therefore
(dv/ds) - UT
(du/ds) + Pr
u
V
DIFFERENTIAL GEOMETRY
SEC. 28]
67
or
T=
uv' - vu'
u2+v2
= dsd- tan-' -vu)
Therefore
=
ra
o
Tds=tan--'U -C
-V
and v = p tan (,p - c) since u = p. Therefore
r1 = r + pn + p tan (,p - c)b
(106)
and again we have a one-parameter family of evolutes to the
curve T.
Problems
1. Show that the unit binormal to the involute is
Kb - Tt
b1 =
(C - S)KK1
2. Show that the torsion of an involute has the value
=
T1
T-K
dS
] [K(K2 + r2)(C - 8)I-1
3. Show that the principal `normal to the evolute is parallel
to the tangent of the curve 1'.
4. Show that the ratio of the torsion of the evolute to its curva-
ture is tan (,p - c).
5. Show that if the principal normals of a curve are binormals
(equal vectors not necessarily coincident) of another curve, then
c(K2 +,r2) = K where c is a constant.
6. On the binormal of a curve of constant torsion T, a point Q
is taken at a constant distance c from the curve. Show that the
binormal to the locus of Q is inclined to the binormal of the given
curve at an angle
tan-'
CT2
K(C2r2 + 1)}
7. Consider two curves which have the same principal normals
(equal vectors not necessarily coincident). Show that the tangents to the two curves are inclined at a constant angle.
VECTOR AND TENSOR ANALYSIS
68
[SEC. 29
29. Spherical Indicatrices
(a) When dealing with a family of unit vectors, it is often
convenient to give them a common origin and then to consider
the locus of their end points.
This locus obviously lies on a
unit sphere. Let us now consider
the spherical indicatrix of the tan-
gent vectors to a curve r = r(s).
The unit tangent vectors are
t(s) = ds- Let r, = t. Then
=
t'
dr,
= dt ds
ds1
ds dal -
ds
ds,
Thus the tangent to the spherical indicatrix P is parallel to the
normal of the curve at the corresponding point. Moreover,
Fla. 42.
1 = K d1' t, = n. Let us now find the curvature K, of the indicatrix.
We obtain
do ds
=K,n,='--=-(-Kt-rn)
ds ds,
as,
dt,
1
K
and
K2 + r2
K12
2
K
(b) The spherical indicatrix of the binormal, r1 = b.
entiating,
dr,
ds
A ds
=
= re
t,=--- ds ds1
ds,
Therefore
rds=1
ds1
t,=n
and
ds,
Differentiating,
dt,
dS1
= x,n1 =
do ds
A dS;
_
1
r
and
K12 =
K2 + r2
T2
`-Kt - rn)
Differ-
SEC. 30J
DIFFERENTIAL. GEOMETRY
69
Problems
1. Show that the torsion of the tangent indicatrix is
T(dK/ds) - K(dr/ds)
T2)
Ti
K(K2 +
2. Show that the torsion of the binormal indicatrix is
Ti
T(dK/ds) - K(dT/ds)
T(K2
r2)
+
3. Find the curvature of the spherical indicatrix of the principal
normal of a given curve.
30. Envelopes. Consider the one-parameter family of surfaces F(x, y, z, c) = 0. Two neighboring surfaces are
F(x, y, z, c) = 0
and F(x, y, z, c + Ac) = 0. These two surfaces will, in general,
intersect in a curve. But these equations are equivalent to the
equations F(x, y, z, c) = 0 and
F(x, y, z, c + Ac) - F(x, y, z, c)
Ac
=0
where Ac # 0. As Ac --> 0, the curve of intersection approaches
a limiting position, called the characteristic curve, given by
F(x, y, z, c) = 0
aF(x, y, z, c)
ac
-0
(107)
Each c determines a characteristic curve. The locus of all
these curves [obtained by eliminating c from (107)] gives us a
surface called the envelope of the one-parameter family. Now
consider two neighboring characteristics
F(x,y,z,c) = 0
aF(x,,z,c)
ac
=0
and
(108)
F(x, y, z, c + Ac) = 0
aF(x, y, z, c + Ac) = 0
which, in general, intersect at a point.
ac
The locus of these points
is the envelope of the characteristics and is called the edge of
70
VECTOR AND TENSOR ANALYSIS
[SEc.31
The edge of regression is given by the three simultaneous equations
F(x, y, z, c) = 0
regression.
aF(x, y, z, c) = 0
ac
(109)
a2F(x,y,z,c) = 0
ace
Example 43. Let us consider the osculating plane at a point P.
From (96) we have [s - r(s)] b(s) = 0. If we let P vary, we
obtain the one-parameter family of osculating planes given by
f(x, y, z, s) = [s - r(s)] b(s) = 0
where s is the parameter and s = xi + yj + A.
Now
of
=
as
dr
ds
A = (s - r)
b + (s -r).ds
of
= 0, we obtain (s - r) n = 0.
as
plane.
This locus is the rectifying
The intersection of f = 0 and as = 0 obviously yields
the tangent lines which are the characteristics.
a If
as2
in, and setting
Now
= -t n + (s - r) (-Kt - A) + (s - r) n da
It is easy
Y to verify
Y that s = r satisfies f =
a
as
=
az
4982
-
= 0, so that
the edge of regression is the original curve r = r(s).
A developable surface, by definition, is the envelope of a oneparameter family of planes. The characteristics are straight
lines, called generators. We have seen that the envelope of the
osculating planes is the locus of the tangent line to the space
curve P.
In general, a developable surface is the tangent surface
of a twisted curve. A contradiction to this is the case of a
cyiinder or cone.
31. Surfaces and Curvilinear Coordinates.
the equations
x = x(u, v)
y = y(u, v)
z = z(u, v)
Let us consider
(110)
DIFFERENTIAL GEOMETRY
SEc. 32]
71
where u and v are parameters ranging over a certain set of values.
If we keep v fixed, the locus of (110) is a space curve. For each
v, one such space curve exists, and if we let v vary, we shall obtain
a locus of space curves which collectively form a surface. We
shall consider those surfaces (110) for which x, y, z have continuous second-order derivatives. Equation (110) may be written
r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k
(111)
where the end point of r generates the surface. The curves
obtained by setting v = constant are called the u curves, and
similarly the v curves are obtained by setting u = constant.
The parameters u and v are called curvilinear coordinates, and
the two curves are called the parametric curves.
32. Length of Arc on a Surface. If we move from the point r
to the point r + dr on the surface, the distance ds is given by
(arudu+-dvN 2
ds2 =
or
2
C_ J
due + 2
9r
o9r
au . av
du dv + (av)2 dv2
or
ds2 = E du2 + 2F du dv + G dv2
(112)
where
E
\12
au/ '
F
ao9r .
u av'
2
G
(Ov)
Equation (112) is called the first fundamental form for the surface
r = r(u, v).
In particular, along the u curve, dv = 0, so that
(ds) = 1'E du
and similarly
(113)
(ds), = VG_ dv
Now
Or
and av are tangent vectors to the u and v curves, so
that the parametric curves form an orthogonal system if and
only
Or
Or
au av
72
VECTOR AND TENSOR ANALYSIS
Example 44.
1Szc. 33
Consider the surface given by
r = r sin 8 cos v i+ r sin 8 sin V j+ r cos 8 k,
r = constant
Differentiating,
ar = r cos 0
cos sP i + r cos 0 sin V j - r sin 8 k
d0
ar
a st
= -r sin6sinpi+rsin8coscpj
and
is
E
a8/ =
c1r
r2,
F=-e
c1r
2
= 0,
G=
r2 Sin2 0
so that ds2 = r2 d82 + r2 sin2 0 dcp2 and the 0-curves are orthogonal to the 9-curves. Of course the surface is a sphere.
33. Surface Curves. By letting u and v be functions of a
single variable t, we obtain
r = r[u(t), v(t)]
(114)
which represents a curve on the surface (111). Along this curve,
(arau
ar dvl
dr = ` du + Wt dt. dr is completely determined when du
dt
av 1
and dv are specified, so that we will use the notation (du, dv) to
specify a given direction on the surface. Now consider another
curve such that ar =
su
au
+ av av, where su and av are the
differential changes of u(t) and v(t) for this new curve. Now
dr or = E du au + F(du av + dv au) + G dv av
(115)
so that two curves are orthogonal if and only if
Edu su +F(duav +dv &u) +Gdv av = 0
or
dv
E+FCa6Vu+au/+Gd
-=0
(116)
If we have a system of curves on the surface given by the differential equation P(u, v) su + Q(u, v) av = 0, the differential equa-
DIFFERENTIAL GEOMETRY
SEC. 341
73
tion for the orthogonal trajectories is given by
E+F,(_P+dv1 - GPdv
Q
``
by
P
bu
Q
since -
dull
Q du
=
0
(117)
Problems
1. Find the envelope and edge of regression of the one-param-
eter family of planes x sin c - y cos c + z tan 8 = c, where c is
the parameter and 0 is a constant.
2. Show that any two v curves on the surface
r = u cos v i + u sin v j + (v + log cos u)k
cut equal segments from all the u curves.
3. Find the envelope and edge of regression of the family of
x2
ellipsoids c2 (a2
+
y
z2
+ c2 = 1 where c is the parameter.
b2
4. If 8 is the angle between the two directions given by
P due + Q du dv + R dv2 = 0
show that tan 0 = H(Q2 - 4PR)}/(ER - FQ + GP), where
H
ar
8
avl.
8u
tvhl
5. Prove at the differential equations of the curves which
bisect the angles between the parametric curves are
VEdu-VGdv=0
and 1/E du + 1i dv - 0.
6. Given the curves uv = constant on the surface r = ui + vj,
find the orthogonal trajectories.
7. Show that the area of a surface is given by
f f (EG
- F2)1 du dv
34. Normal to a Surface. The vectors
or
and av are tangent
to the surface r(u, v) along the u and v curves, respectively.
Consequently,
ar
8r
au x 49V
,
is a vector normal to the surface. Note
VECTOR AND TENSOR ANALYSIS
74
[SEC. 35
that au need not be a unit tangent vector to the u curve since the
parameter u may not represent are length. Since
(ds),, = 1/E du
a necessary and sufficient condition for u to be arc length is that
E = 1. We define the unit normal to the surface as
n _
(Or/au) x (ar/av)
(118)
(ar/au) x (ar/av)
35. The Second Fundamental Form.
Consider all the planes
through a point P of the surface r = r(u, v) which contain the
normal n. These planes intersect the surface in a family of
curves, the normals to the curves being parallel to n. We now
compute the curvature of any one of these curves in the direction
(du, dv). Let ds be length of are along this curve. Now
dr
Or du
Or dv
t=ds=auds+avds
Therefore
d2r
dt
ds2
ds
_
492r
du 2
au2 (ds)
K"n
2
a2r d_u dv
a2r dv 2
+ au av A ds +
av2 (ds)
Or d2u
Or d2v
+
au ds2
+
av ds2
119)
and
(du)2
n)
since n
-
Or
au
(n au2)
n
- = 0.
av
Kn =
ds+ 2 (n au2av) ds ds
)z
Therefore
edue+2fdudv+gdv2
ds2
Kn
edue+2fdudv+gdv2
Edue+2Fdudv+Gdv2
(120)
DIFFERENTIAL GEOMETRY
SEC. 361
75
where we define
a2r
e=n-au2'
a2r
a2r
_n.
auav'
f
(121)
The quantity e due + 2f du dv + g dv2 is called the second fundamental form.
Now consider any curve t on the surface and let its normal be
n, at a point P, the direction of r being (du, dv) at P. Let r' be
Fia. 43.
the normal curve in the same direction (du, dv) with normal n at
P (Fig. 43). We have
r"
K
=n r"K
since n rl" = r" n for two curves with the same (du, dv) [see
(119)].
Therefore
COS0=
-
Kn
K
so that
K = K,, sec 0
(122)
This is Meusnier's theorem.
36. Geometrical Significance of the Second Fundamental
Form. We construct a tangent plane to the surface at the point
r(uo, vo). What is the distance D of a neighboring point
r(uo + Du, vo + Av)
VECTOR AND TENSOR ANALYSIS
76
on the surface, to the plane? It is D = Ar n.
r(uo + Au, vo + AV) = r(uo, vo) +
1
+
2!
from the calculus.
Au2 + 2
Now
ar
+ av
a2r
AV
a2r
+ av2 Av2 +
au av Au AV
Consequently
1/
D=
-
(a2r
ar
Au
au
[SEC. 36
z
z
Av2
except for infinitesimals of higher order.
Thus
2D = e du2 + 2f du dv + g dv2
(123)
Problems
1. For the paraboloid of revolution
r=ucosvi+usinvj+u2k
show that E = 1 + 4u2, F = 0, G = u2, e = 2(1 + 4u)`}, f = 0,
g = 2u2(1 + 4u2)-}, and find the normals to the surface and the
normal curvature for the direction (du, dv).
2. What are the normal curvatures for directions along the
parametric curves?
3. Find the second fundamental form for the sphere
r=rsin0coscpi+rsin6sincpj+rcosOk
r = constant.
4. Show that the curvature K at any point P of the curve of
intersection of two surfaces is given by
K2 Sln2 0 = K12 + K22 - 2K1K2 COS 0
where 9j, K2 are the normal curvatures of the surfaces in the direc-
tion of the curve at P, and 0 is the angle between their normals.
5. Let us make a change of variable u = u(u, v), v = v(u, 1).
Show that E, F, G transform according to the law
E
lz
Caul
+ 2F
au au
+G(av
\au
2
DIFFERENTIAL GEOMETRY
SEC. 37]
F
au au
au av
77
av au
av av
=Eaudo +F (aft do +audo )+Gag do
UP-
(avl 2
au av
au\2
E
Ii + 2F avav
+ G `avl
and that E due + 2F du dv + G dv2 = E due + 2F du do + G W.
Also show that
e=±
(
e
au +
(au)2
au au
au av
g
-±
[e
(au)2
av
au av
(OV\2 1
au au + g au J
au av
av au
au av
au av
au av
+ 2f avav+ g
37. Principal Directions.
(K .E - e) due +
Zf
av av
au CIO]
(av12
av
J
From (120) we have
f) du dv +
g) dv2 = 0
(124)
or
A due + 2B du dv + C dv2 = 0
This quadratic equation has two directions (du, dv), (Su, Sv),
which give the same value for x,,. These two directions will
coincide if the quadratic equation (124) has a double root.
is true if and only if
B2-AC= (K.F'-f)2-
This
0
or
,
2(F2 - EG) +
Moreover, we have
so that
du
gE - 2fF) + (f2 - eg) = 0
(125)
- B and d = - C if B2 - AC = 0,
A
(K. E - e) du + (x F - f) dv = 0
f) du + (xnG - g) dv = 0
6126)
The solutions of (124) give the two directions for a given x,,.
When x is eliminated between (124) and (125), the two directions
coincide and satisfy
(Ef - Fe) due + (Eg - Ge) du dv + (Fg - Gf) dv2 = 0 (127)
VECTOR AND TENSOR ANALYSIS
78
[SEc. 37
The two directions, solutions of (127), are called principal directions and are the only ones with a unique normal curvature, that
is, no other direction can have the same curvature. The normal
curvatures in these two directions are called the principal curvatures at the point. The average of the two principal curvatures
is
HeG+gE-2fF
(128)
2(EG - F2)
which is obtained by taking one-half of the sum of the roots of
(125). The Gaussian curvature K is defined as the product of
the curvatures, that is,
f2eg
K= F2EG
(129)
A line of curvature is a curve whose tangent at any point has a
direction coinciding with a principal direction at that point.
The lines of curvature are obtained by solving the differential
equation (126). The curvature of a line of curvature is not a
principal curvature since the line of curvature need not be a
normal curve.
Example 45. Let us consider the right helicoid
r= ucospi+usine'j+cook
We have
Or
=cosSPi+sinrpj,
au
2
zr
a2G2-0'
8ua
= -u sins
= -sin+Cos(pj,
a2r
Q= -ucosipi - usinrpj
Hence
(ar)2
au-
1,
F=
(Or\2
au - am
0'
G=
am
= u2 + c2
DIFFERENTIAL GEOMETRY
SEC. 371
79
Also
n = (ar/au) x (,9r/app)
= (c sin p i - c cos v j + uk) (C2 +
u2)-}
(ar/au) x (ar/ap)j
82r
a2r
f=n
au ago
= -c(c2 +
u2)-#
2
Equation (125) yields
- (u2 + C2)K,, + C2(C2 + u2)-1 = 0
whence
C
u2+C2
The average curvature is H = 2
the Gaussian curvature is K =
(u2
2
C2)2'
+ C2) =
0, and
The differential
equation (126) for the lines of curvature becomes
u2)-}
-c(c2 +
dug + c(c2 + U2)Id tp2 = 0
so that
d(p = ±
du
= ± log (u + \/u2 + c2) + a
and
(C2 + u2) b
and the lines of curvature are given by
r = u cos [± log (u +
u2 + c2) + aji + u sin -p j + cspk.
Referring to (126) for the two principal directions, we have
dv
du
av
Eg - Ge
+
Fg - Gf
_ Ef - Fe
du au Fg - Gf
au
dv av
(130)
Substituting (130) into (116) we obtain
E - F(Fg-Gf/ +G \F'g -Gf/
so that the principal directions are orthogonal.
0
VECTOR AND TENSOR ANALYSIS
80
[SEC. 38
Now let us choose the principal curves as the parametric lines.
Thus u = constant, v constant are to represent the principal
curves. These two curves satisfy the equation du dv = 0, so
that from (127) we must have
Ef - Fe = 0
Fg-Gf=0
Eg - Ge
0
From these equations we conclude that
f(Eg - Ge) = gfE - feG=Feg - eFg =0
and F(Eg - Ge) = 0, so that f = F = 0. We have shown that
a necessary and sufficient condition that the lines of curvature be
parametric curves is that
f=F=0
(131)
Problems
1. Find the lines of curvature on the surface
x=a(u+v),
y=b(u-v),
z=uv
2. Show that the principal radii of curvature of the right conoid
x = u cos v, y = u sin v, z = {f (v) are given by the roots of
f'=K2
- of"(u2 + fF )iK - (u2 + ft )2 = 0
3. The surface generated by the binormals of the curve r = r(s)
is given by R = r + ub. Show that the Gauss curvature is
K = -.r2/(1 + r2u2)2. Also show that the differential equation
of the lines of curvature is
-T2 du2 -
(K + Kr2u2 + d3 u)
du ds + (1 + r2u2)T M = 0
38. Conjugate Directions. Let P and Q be neighboring points
on a surface. The tangent planes at P and Q will intersect in a
straight line 1. Now let Q approach P along some fixed direction.
The line 1 will approach a limiting position 1'. The directions
PQ and 1' are called conjugate directions.
We now compute the analytical expression for two directions
to be conjugate. Let n be the normal at P and n + do the
DIFFERENTIAL GEOMETRY
SFc. 38]
normal at Q, where dr = PQ =
tion of 1' be given by Sr =
Or
au
Su
Or
an
du +
ar
Sv.
+ av
Or
-
81
dv.
Let the direc-
Since Sr lies in both
planes, we must have Sr n = 0 and Sr (n + dn) = 0. These
two equations imply Sr do = 0, or
(au
Su
+
av
Sv)
Can du +
a dv)
=0
(132)
Expanding, we obtain
far
(_au
-) du bu + I (-av -) Sv du + (auau
Or
anav
Or an
aU) au dvJ
anau
ar an
+
av
av
Sv dv = 0
(133)
Now n au = 0, so that by differenuiating we see that
ar
an
a2r
which implies
an ar
au au
a2r
-n- au,
- = -e
an
Or
an Or
av au
au av
Similarly
-f
an Or
9
so that (133) becomes
e du Su + f (du Sv + dv Au) + g Sv dv = 0
(134)
If the direction (du, dv) is given, there is only one corresponding
conjugate direction (Su, Sv), obtained by solving (134).
Now consider the lines of curvature taken as parametric curves.
Their directions are (du, 0), (0, Sv). Equation (134) is satisfied
by these directions since f = 0 for lines of curvature, so that the
lines of curvature are conjugate directions.
82
VECTOR AND TENSOR ANALYSIS
[SFc. 39
39. Asymptotic Lines. The directions which are self-conjugate
are called asymptotic directions. Those curves whose tangents
are asymptotic directions are called asymptotic lines. If a direcdv
tion is self-conjugate,
=
du
Sv,
Su
so that (134) becomes
e due + 2f du dv + g dv2 = 0
(135)
We see that the asymptotic directions are those for which the
second fundamental form vanishes.
ture rc, vanishes for this direction.
If e = g = 0, f
Moreover, the normal curva-
0, the solution of (135) is u = constant,
v = constant, so that the parametric curves are asymptotic lines
if and only if e = g = 0,f3PK 0.
Example 46. Let us find the lines of curvature and asymptotic
lines of the surface of revolution z = x2 + y2. Let x = u cos v,
y = u sin v, z = u2, and
r=ucosvi+usinvj+u2k
We obtain
Or
=cosvi-{-sinvj+2uk,
ar=
-u sinvi+ucosvj
av
au
azr
8u2
a2r
=2k'
a2r
av2
au av
= -sinvi+cosvj
- -ucosvi - usinvj
n= (ar/au) x (ar/av)
(ar/au) x (ar/av)I
= (-2u2 cos v i - 2u2 sin v j + uk)u-1(1 + 4u2)'I
Therefore
z
=2(1+4u)-},
au2
z
g=n
f=n aua2rav =0
= 2u2(1 +4 U2)-f
2
DIFFERENTIAL GEOMETRY
SEC. 40]
ar
83
ar
Also F = -- - = 0, so that f = F = 0, and from (131) the
au av
parametric curves are the lines of curvature. The asymptotic
lines are given by due + u2 dv2 = 0. These are imaginary, so
that the surface possesses no asymptotic lines.
Problems
1. Show that the asymptotic lines of the hyperboloid
r=acos0seeipi+bsin0sec4, j+ctan 4k
are given by 0 ± ¢ = constant.
2. The parametric equations of the helicoid are
x=ucosv,
y=usinv,
z=cv
Show that the asymptotic lines are the parametric curves, and
that the lines of curvature are u + V 'u2+ c2 c2 = Ae}°. Show
that the principal radii of curvature are ± (u2 + c2)c-1.
3. Prove that, at any point of a surface, the sum of the normal
curvature in conjugate directions is constant.
4. Find the asymptotic lines on the surface z = y sin x.
40. Geodesics. The distance between two points on a surface
(we are allowed to move only on the surface) is given by
8 = I.1 [E (dt/2 + 2F dt dt + G (i)]1 dt
(136)
Among the many curves on the surface that join the two fixed
points will be those that make (136) an extremal. Such curves
are called geodesics. We wish now to determine the geodesics.
To do this, we require the use of the calculus of variations, and
so we say a few words about this important method.
Let us first consider the integral
fQ(xt,yi) (1
(x,, VS)
+ y'dx
(137)
We might ask what must be the function y = y(x) joining the two
points P and Q which will make (137) a minimum. The reader
might be tempted to say, y' = 0 or y = constant, since the integrand is then a minimum. But we find that y = constant will
84
VECTOR AND TENSOR ANALYSIS
[SEC. 40
not, in general, pass through the two fixed points. Hence the
solution to this problem is not trivial. We now formulate a
more general problem: to find
y
y = y(x) such that
j'f(x, y, y') dx
is an extremal.
ry
Ar
}' M
The function
f(x, y,)y ' is given. It is y(x)
and so also y' (x) that are
unknown . Let y = y (x) be
that function which makes
I.-, B
,
(138)
It
(138) an extremal Now 1 t
Y(x, a) = y(x) + arp(x), where
a is arbitrary and independ-
P.x
b
a
ent of x and jp(x) is any function with continuous first
Fzo. 44.
derivative having the property that jp(a) = p(b) = 0 (see Fig. 44). Under our assumption,
J(a) = j,'f(x, Y, ?') dx
is an extremal for a = 0.
dJ
=
da a-0
(139)
dJ
da0 = 0 or
Consequently
Ib\ay,+a-
(140)
since
of
as
of 81'
of 81"
aY as + aY' as
of
if
+
of
C7F
'p,
and for a=0,
of
a> r
of
of
of
ay'
Tay
,
e
We now integrate the right-hand term of (140) by parts and
obtain
P] b
fb
ay
dx
+
Lay
-
Jab
dx ay'/ `p
dx = 0
(141)
DIFFERENTIAL GEOMETRY
SEc. 40]
85
Now (p(a) = p(b) = 0 by construction of v(x), so that
(142)
Now let us assume that
a (-'
ay
ay
dx
(ay')
is continuous.
If
is not identically zero on the interval (a, b), it
will be positive or negative at some point. If it is positive at
x = c, it will be positive in a neighborhood of x = c from continuity (see Sees. 42 and 43). We can construct (p to be positive
on this interval and zero elsewhere. Then
Jb[afd(of)]
ay`p dx > 0
dx
y
so that we have a contradiction to (142). Consequently, the
function of y(x) must satisfy the Euler-Lagrange differential
equation
of
d of
dx Cay'
ay
0
(143)
If f = f(y, y'), we can immediately arrive at an integral of
(143).
Let us consider
f
dx (
y yl
ay
1J, +
a
y
y [Of
_
lay
f-y
aof
y,
yy
of
ay
d
dx ay'
(t\1 = 0
= constant
d (af 1
y/ dx TO
from (143)
(144)
is an integral of (143) if f = f(y, y').
Example 47. To extremalize (137), we have f = (1 + y'2)1,
which is independent of x. From (144),
f - y
of
1
= constant =
a
VECTOR AND TENSOR ANALYSIS
86
[SEC. 40
so that
(1 +y') - Y'(1
J
a
y'2),
and
y'=
and finally
y= ±
av-1x+ 0
(145)
The constants a and fi are determined by noting that the straight
line (145) passes through the two given fixed points.
Example 48.
If f = f Cx',x 2, ... , x",
dt2) ...
dtl,
dtn?
t);
fhfdt is an extremal when the xa(t) satisfy
then
d (ftaf
dt
a
for a = 1, 2,
. .. , n with is =
of
axa = o
X.
(146)
The superscripts are not
powers but labels that enable us to distinguish between the various variables. The formulas (146) are a consequence of the fact
that
fo f dt must be an extremal when x`(1) is allowed to vary
while we keep all other x1 fixed, j = 1, 2, ... , i - 1, i + 1,
,n.
Let us now try to find the differential equations that u(t) and
v(t) must satisfy to make (136) an extremal. We write
s= f"
(E,42 + 2Fuv + Gv2)} dt
and apply (146), where x' = u and x2 = v. We thus obtain
dt \a4/
au - 0
off`
of
dt avl
av
d
o
(147)
(148)
where
f = (E,42+ 2Fuv + Gv2)i = dt'
E = E(u, v), etc.
---
DIFFERENTIAL GEOMETRY
SEc. 401
Now
af
au
Eic + Fv
ice
of
au
f
CIE
au
+ 2uv
OF
au
2f
+ v2
87
aG
au
so that (147) becomes
d Eu + Fvl _ 2t2 + 2"(W/au) + v2(aG/au)
dt
ds/dt /
2 ds/dt
(149)
and if we choose for the parameter t the are length s, then t = a
and dt = 1, so that (149) reduces to
d
ds
(Eic + Fv) = 2( u2 aE + 2uv
OF
+ 62
-n/A
opu
while similarly (148) yields
d (Fu+Gv)
\
(150)
_2(u2av +24vav +v2av/
In Chap. 8 we shall derive by tensor methods a slightly different system of differential equations.
Example 49. Consider the sphere given by
r = a sin 0cos(pi+a sin 0 sin cpj+acosOk
where ds2 = a2 do2 + a2 sin2 0 dp2 so that E = a, F = 0,
G = a2 Sin2 0,
and
OE
80
_
aEE
ac
_
OF
_
OF
a9
av
_
aG
_
ap
8G
0'
aB
_
- a sin 29
2
Hence (150) reduces to
2
ds a ds/
2 ds
sin 2B
d (a sin2 0 ds) = 0
Integrating (151) we have sine 0 LIP = constant.
d
(151)
We can choose
our coordinate system so that the coordinates of the fixed points
VECTOR AND TENSOR ANALYSIS
88
are a, 0, 0 and a, 0, 0. Hence sin' 8
ds
= 0, and
[SEC. 40
ds
= 0, so that
0. Hence the geodesic is the are of the great circle joining
the two fixed points.
Example 50. Let us find y(x) which extremalizes
f y(1 + y")} dx
Since f = y(1 + y")# = f(y, y'), we can apply (144) to obtain
a first integral. We obtain y(1 + y'')t -- y''y(1 + y'')-l = a-1,
and simplifying this expression yields y' = ± (a2y2 - 1)I. A
further integration yields ay = cosh (0 ± ax). These are the
curves (catenoids) which have minimum surfaces of revolution.
Problems
1. Find the geodesics on the ellipsoid of revolution
2
x2+ z2
b2 = 1
+
a2
Hint: Let x = u cos v, z = u sin v.
2. Show that the differential equation of the geodesics for the
right helicoid x = u cos v, y = u sin v, z = cv is
du
TV
+
1
[(u2 + C2)(u2 + C2
h2)1],
h = constant
-h
3. Prove that the geodesics on a right circular cylinder are
helices.
4. Show that the perpendicular from the vertex of a right
circular cone to the tangents of a given geodesic is of constant
length.
5. Find y(x) which extremalizes
f'[(l + y')/y)]} dx.
CHAPTER 4
INTEGRATION
41. Point-set Theory. In geometry and analysis the student
has frequently made use of the concept of a point and of the
notion of a set or collection of elements (objects, points, numbers,
etc.). We shall not define these concepts, but shall take their
notion as intuitive. We may be interested in the points subject
to the condition x2 + y2 < r2. These will be the points interior
to the circle of radius r with center at the origin. We might also
be interested in the rational points of the one-dimensional continuum. For convenience, we shall consider only points of the
real-number axis in what follows. Any set of real numbers will
be called a linear set. The integers form such a set, as do the
rationals and irrationals. All the definitions and theorems
proved for linear sets can easily be extended to any finitedimensional space.
Closed Interval. The set of points { x j satisfying a < x S b
will be called a closed interval. If we omit the end points, that
is, consider those x that satisfy a < x < b, we say that the
interval is open (open at both ends). For example, 0 S x 5 1
is a closed interval, while 0 < x < 1 is an open interval.
Bounded Set. A linear set of points will be said to be bounded
if there exists an open interval containing the set. It must be
emphasized that the ends of the interval are to be finite numbers,
which thus excludes - oo, + oo.
An alternative definition would be the following: A set of
numbers S is bounded if there exists a finite number N such that
-N <x <NforallxinS.
The set of numbers whose squares are less than 3 is certainly
bounded, for if x2 < 3 then obviously -2 < x < 2. However,
the set of numbers whose cubes are less than 3 is unbounded, for
x3 < 3 is at least satisfied by all the negative numbers. This set
is, however, bounded above. By this we mean that there exists
a finite number N such that x < N if x3 < 3. Certainly N = 2
89
90
VECTOR AND TENSOR ANALYSIS
[SEc.41
does the trick. Specifically, a set of elements S is bounded above
if a finite number N exists such that x < N for all x in S. Let the
student frame a definition for sets bounded below.
We shall, in the main, be concerned with sets that contain an
infinite number of distinct points. The rational numbers in the
interval 0 < x < 1 form such a collection. Let the reader
prove that between any two rationals there exists another
rational.
Limit Point. A point P will be called a limit point of a set S
if every open interval containing P contains an infinite number of
distinct elements of S. For example, let S be the set of numbers
(1/2, 1/3, 1/4, . . . , 1/n, . . .). It is easy to verify that any
open interval containing the origin, 0, contains an infinite number of S. In this case the limit point 0 does not belong to S.
It is at once apparent that a set S containing only a finite number
of points cannot have a limit point.
Neighborhood. A neighborhood of a point is any open interval
containing that point.
Interior Point. A point P is said to be an interior point of a
set S if it belongs to S and if a neighborhood N of P exists, every
element of N belonging to S. If S is the set of points 0 < x 5 1,
then 0 and 1 are not interior points of S since every neighborhood
of 0 or 1 contains points that are not in S. However, all other
points of S are interior points.
Boundary Point. A point P is a boundary point of a set S
if every neighborhood of P contains points in S and points not in
S. If S is the set 0 5 x <_ 1, then 0 and 1 are the only boundary
points. A boundary point need not belong to the set. 1 is a
boundary point of the set S for which x > 1, but 1 is not in S
since 1 > 1.
Complement. The complement of a set S is the set of points
not in S. The complement C(S) has a relative meaning, for it
depends on the set T in which S is embedded.
If S, for example,
is the set of numbers -1 < x < 1, then the complement of S
relative to the real axis is the set of points lxl > 1. But the
complement of -1 < x < 1 relative to the set -1 S x S 1 is
the null set (no elements). The complement of the set of nationals relative to the reals is the set of irrationals, and conversely.
Open. Set. A set of points S is said to be an open set (not to be
confused with open interval) if every point of S is an interior
SEC. 411
IXTBGRATION
91
point of S. For example, the set S consisting of points which
satisfy either 0 < x < 1 or 6 < x < 8 is open.
Closed Set. A set containing all its limit points is called a
closed set. For example, the set (0, 1/2, 1/3, 1/4, . . . , 1/n,
.) is closed, since its only limit point is 0, which it contains.
Problems
1. What are the limit points of the set 0 < x < 1? Is the
Open? What are the boundary points?
2. Repeat Prob. 1 with the point x = removed.
3. Show that the set of all boundary points (the boundary)
set closed?
of a set S is closed.
4. Prove that the set of all limit points of a set S is closed.
5. Prove that the complement of an open set is closed, and
conversely.
6. Why is every finite set closed?
7. Prove that the set of points common to two closed sets is
closed. The set of points belonging to both Si and S2 is called
the intersection of Sl and S2, written SI n828. Prove that the set of points which belong to either S, or 82
is open if S, and S2 are open. This set is called the union of S,
and S2, written S, U S29. An infinite union of closed sets is not necessarily closed.
Give an example which verifies this.
10. An infinite intersection of open sets is not necessarily open.
Give an example which verifies this.
Supremum. A number s is said to be the supremum of a set of
points S if
1. x in S implies x 5 s
2. t < s implies an x in S such that x > t
Example 51. Let S be the set of rationale less than 1. Then
1 is the supremum of S, for (1) obviously holds from the definition of S, and if t < 1, it is possible to find a rational r < 1 such
that t < r, so that (2) holds. We give a proof of this statement
in a later paragraph.
Example 52. Let S be the set of rationals whose squares are
to
less than 3, that is, SI x2 < 3. Certainly we expect the
be the supremum of this set. However, we cannot prove this
without postulating the existence of irrationals. We overcome
92
VECTOR AND TENSOR ANALYSIS
[SEC. 41
this by postulating
Every nonempty set of points has a supremum
(152)
Hence the rationals whose squares are less than 3 have a supreIt is obvious that we should define this supremum as the
square root of 3.
The supremum of a set may be + oo as in the case of the set
of all integers, or it may be - oo as in the case of the null set.
The infemum of a set S is the number s such that
1. x in S implies s < x
mum.
2. t > s implies an x in S such that x < t
Example 53. Let a > 0 and consider the sequence a, 2a, 3a,
... , na, ... . If this set is bounded, there exists a finite
supremum s. Hence an integer r exists such that ra > s - (a/2)
so that (r + 1)a > s + (a/2) > s, a contradiction, since na 5 s
for all n. Hence the sequence Ina} is unbounded. This is the
Archimedean ordering postulate.
Example 54. To prove that a rational exists between any two
numbers a, b. Assumed: a > b > 0, so that a - b > 0. From
example 53, an integer q exists such that q(a - b) > 1, or
qa > qb + 1. Also an integer p exists such that p 1 = p > qb.
Choose the smallest p.
Thus p > qb ? p - 1. Hence
qa>gb+1zp>qb,
and a > p/q > b. Q.E.D.
With the aid of (152) we are in a position to prove the wellknown Weierstrass-Bolzano theorem.
"Every infinite bounded set of points S has a limit point."
The
proof proceeds as follows: We construct a new set T. Into T we
place all points which are less than an infinite number of S. T is
not empty since S is bounded below. From (152), T has a
supremum; call it s. We now show that s is a limit point of S.
Consider any neighborhood N of s. The points in N which are
less than s are less than an infinite number of points of S, whereas
those points in N which are greater than s are less than a finite
number of points of S. Hence N contains an infinite number of
S, so that the theorem is proved. We have at the same time
proved that s is the greatest limit point of S. A limit point may
or may not belong to the set.
SEC. 41)
INTEGRATION
93
Problems
1, 2, 3, . . . is
1. The set 1/2, 1/3,. . . , 1/n, . .
unbounded. Does it have any limit points? Does this violate
the Weierstrass-Bolzano theorem?
2. Prove that every bounded monotonic (either decreasing or
increasing) sequence has a unique limit.
3. Prove that lim r" = 0 if Irl < 1. Hint: The sequence
r, r2, . . . , r", . . . is bounded and monotonic decreasing for
r > 0, and r"+1 = rr"
4. Show that if P is a limit point of a set S, we can pick out a
subsequence of 8 which converges to P.
5. Show that (152) implies that every set has an infemum.
6. Show that removing a finite number of elements from a set
cannot affect the limit points.
7. Prove the Weierstrass-Bolzano theorem for a bounded set
of points lying in a two-dimensional plane.
8. Let the sequence of numbers sl, 82, .. . , sn, . . . satisfy
the following criterion: for any e > 0 there exists an integer N
such that Is,,+F - snI < e for n ? N, p > 0. Prove that a unique
limit point exists for the sequence.
.,
9. A set of numbers is said to be countable if they can be
written as a sequence, that is, if the set can be put into one-to-one
correspondence with the positive integers. Show that a countable collection of countable sets is countable. Prove that the
rationals are countable.
10. Show that the set S consisting of x satisfying 0 5 x 5 1
is uncountable by assuming that the set S is countable, the
numbers x being written in decimal form.
Theorem of Nested Sets. Consider an infinite sequence of
nonempty closed and bounded sets S1, S2, . . . , Sn,
. such
There
that Sn contains
that is, S1 M S2 D Sa M
.
exists a point P which belongs to every Si, i = 1, 2, 3, . . . .
The proof is easy. Let P1 be any point of S1, P2 any point of
P,,
S2j etc. Now consider the sequence of points P1, P2,
. . . .
This infinite set belongs to S1 and has a limit point P
which belongs to 8, since Sl is closed. But P is also a limit of
P, Pn+1, ... , so that P belongs to Sn. Hence P is in all Sn,
n = 1, 2, . . . .
... ,
Diameter of a Set. The diameter of a set S is the supremum of
all distances between points of the set. For example, if S is the
VECTOR AND TENSOR ANALYSIS
94
[SEc. 41
set of numbers x which satisfy I < x < 1, 3 < x < 7, the
diameter of S is 7 - = 61. There are pairs of points in S
whose distance apart can be made as close to 6,91 as we please.
Problems
1.
If a set is closed and bounded, the diameter is actually
attained by the set.
2.
Prove this.
If, in the theorem of nested sets, the diameters of the S
approach zero, then P is unique. Prove this.
The Heine-Borel Theorem. Let S be any closed and bounded
set, and let T be any collection of open intervals having the
property that if x is any element of S, then there exists an open
interval 7' of the collection T such that x is contained in T.
The theorem states that there exists a subcollection T' of T
which is finite in number and such that every element x of S is
contained in one of the finite collection of open intervals that
comprise T'.
Before proceeding to the proof we point out that (1) both the
set S and the collection T are given beforehand, since it is no
great feat to pick out a single open interval which completely
covers a bounded set S; (2) S must be closed, for consider the set
S(1, 1/2, 1/3, . . . , 1/n, . . .) and let T consist of the following set of open intervals:
TIIx such that Ix
- 11 < 22
T21x such that Ix
- _f < 32
1
T,,I x such that Ix - 1
n < (n + 1)2
It is very easy to see that we cannot reduce the covering of S
by eliminating any of the given T,,, for there is no overlapping
of these open intervals. Each T,, is required to cover the point
1/n of S contained in it.
Src. 421
INTEGRATION
95
Proof of the theorem: Let S be contained in the interval
-N < x < N.
This is possible since S is bounded. Now divide
this closed interval into two equal intervals (1) -N < x _:5- 0 and
(2) 0 < x < N. Any element x of S will belong to either (1) or
(2).
Now if the theorem is false, it will not be possible to cover
the points of S in both (1) and (2) by a finite number of the
given collection T, so that the points of S in either (1) or (2)
require an infinite covering. Assume that the elements of S in
(1) still require an infinite covering. We subdivide this interval
into two equal parts and repeat the above argument. In this
way we construct a sequence of sets S,
S2
S3
,
such
that each Si is closed and bounded and such that the diameters
of the Si -* 0. From the theorem of nested sets there exists a
unique point P which is contained in each Si. Since P is in S,
one of the open intervals of T, say T, will cover P. rt'his Tp
has a finite nonzero diameter so that eventually one of the Si
will be contained in TD, since the diameters of the Si --> 0. But
by assumption all the elements of this Si require an infinite
number of the { T } to cover them.
This is a direct contradiction
to the fact that a single Tp covers them. Hence our original
assumption is wrong, and the theorem is proved.
42. Uniform Continuity. A real, single-valued function f (x)
is said to be continuous at a point x = c if, given any positive
number e > 0, there exists a positive number a > 0 such that
If(x) - f(c)I < e whenever Ix - cI < S. The a may well depend
on the a and the point x = c. The function f(x) is said to be
continuous over a set of points S if it is continuous at every point
of S.
We now prove that if f (x) is continuous on a closed and
bounded interval, it is uniformly continuous. We define uniform continuity as follows: If, for any e > 0 there exists a S > 0
such that If(XI) - f(X2)1 < e whenever Ix1 - x21 < S, then f(x)
is said to be uniformly continuous. It is important to notice
that 8 is independent of any x on the interval. We make use
of the Heine-Borel theorem to prove uniform continuity.
Choose
an e/2 > 0. Then at every point c of our closed and bounded
set there exists a S(c, a/2) such that If(s) - f (c)I < e/2 for
c - a < x < c + S. Hence every point of S is covered by a
26-neighborhood, and so also by a a-neighborhood. By the
Heine-Borel theorem we can pick out a finite number of these
96
VECTOR AND TENSOR ANALYSIS
[SEc. 43
Let S, be the diameter of the
smallest of this finite collection of 6-neighborhoods. Now consider any two points x, and x2 of S whose distance apart is less
neighborhoods which will cover S.
than bl.
x1.
Let xo be the center of the 6-neighborhood which covers
From continuity lf(xi) - f(xo)I < e/2. But also x2 differs
from xo by less than 26, so that I f (x2) - f (xo) I < e/2. Consequently If(xl) - f(x2)1 < e. Q.E.D.
43. Some Properties of Continuous Functions
(a) Assume f(x) continuous on the closed and bounded interval
a < x < b. As a consequence of uniform continuity, we can
prove that f (x) is bounded. Choose any e > 0 and consider the
corresponding S > 0 such that If(XI) - f(X2)1 < e whenever
-
x21 < S. Now subdivide the interval (a, b) into a finite
number of S-intervals, say n of them. It is easily seen that
Ix1
max If(x) I < If (a) I + ne.
(b) If f (a) < 0 and f (b) > 0, there exists a c such that f (c) = 0,
a < c < b. Let {x} be the set of all points on (a, b) for which
f (x) < 0. The set is bounded and nonvacuous since a belongs
to {x}, for f(a) < 0. The set {x} will have a supremum; call it c.
Assume f (c) > 0. Choose e = f (c) /2. From continuity, a 8 > 0
exists such that If(x) - f (c) I < f(c)12 if Ix cI < 6. Hence
f(x) ? Jf(c) for all x in some neighborhood of x = c. Hence c
is not the supremum of f x 1. Similarly f (c) < 0 is impossible, so
that f (c) = 0.
(c) We prove that f (x) attains its maximum. In (a) we
showed that f (x) was bounded. Let s be the supremum of f (x),
-
a 5 x S b. As a consequence, f (x) <= s for all x on (a, b).
Now consider the set s - e, s - e/2, . . . , s - e/n, ... ,
where e > 0. Since s is a supremum, there exist x1, x2,
x,,, ... such that f (xl) > s - e, f (x2) > s - e/2, ... ,
... ,
Let { xn' } be a subsequence
f xd } will have a limit point c.
of f x,,) which converges to c. Then lim f (x.') z s, since
-s'-+c
e/n -* 0 as n
oo.
But from continuity
lim f(x,') = .f(c).
4C
Hence f (c) = s. Q.E.D.
SEC. 44]
INTEGRATION
97
Problems
1. In the proof of (c) we exclude f(c) > s. Why?
2. If f (x) is continuous on (a, b), show that the set of values
{f(x) }
is closed.
3. If f (x) has a derivative at every point of (a, b), show that
f (x) is continuous on (a, b).
4. If f (x) has a derivative at every point of (a, b), show that a
c exists such that f(c) = 0, a < c < b, when f (a) = f (b) = 0.
5. If f (x) has a derivative at every point of (a, b), show that
a c exists such that f(b) - f(a) _ (b - a)f'(c), a < c < b.
6. Show that if two continuous functions f(x), g(x) exist such
that f (x) = g(x) for the rationals on (a, b), then f (x) = g(x) on
(a,
7. Given the function f (x) = 0 when x is irrational, f (x) = 1/q
when x is rational and equal to p/q (p, q integers and relatively
prime), prove that f (x) is continuous at the irrational points of
(0, 1) and that f(x) is discontinuous at the rational points of
this interval.
44. Cauchy Criterion for Sequences. Let x1, x2, ... ,
xn, ... be a sequence of real numbers. We say that L is the
limit of this sequence, or that the sequence converges to L, if,
given any e > 0, there exists an integer N depending on a such
that IL - xnl < e whenever n > N(e). However, in most cases
we do not know L, so that we need the Cauchy convergence
criterion. This states that a necessary and sufficient condition
that a sequence converge to a limit L is that given any e > 0,
there exists an integer N such that Ixn - xml < e for n >_ N,
m
N. That the condition is necessary is obvious, for
IL - xnl < e/2,
xn,l < e/2 for m, n > N implies Ix,, - xml < e for m, n >_ N.
The proof of the converse is not as trivial. Choose any e/2 > 0.
IL
Then we assume an N exists such that Ix,, -- xml < e/2 for
N, so that the
m, n ? N. Hence Ixn+ < I xNl + (e/2), n
sequence is bounded. We ignore xi, X2,
.
.
.
, xN_, since a finite
number of elements cannot affect a limit point. From the
Weierstrass-Bolzano theorem this infinite bounded set has at
least one limit point L.
Hence, given an a/2, there exists an x.,
with n > N, such that IL - xAi < e/2. But we also have
98
VECTOR AND TENSOR ANALYSIS
[SEC. 45
I xm - x,,l < e/2 for all m, n > N. Hence IL - xml < e for all
m > N. Q.E.D.
Problems
1. Show how the convergence of a series can be transformed
into a problem involving the convergence of a sequence.
2. Show that the Cauchy criterion implies that the nth term
of a convergent series must approach zero as n -- oo.
3. Show that the sequence 1, 1/2, 1/3, . . . , 1/n, . . . converges by applying the Cauchy test.
45. Regular Arcs in the Plane. Consider the set of points
in the two-dimensional plane
such that the set can be repreP2
sented
in
some coordinate
system by x = f (t), y = sp(t),
a S t < 0, where f(t) and ap(t)
are continuous and have continuous first derivatives.
Such
curves are called regular arcs.
A regular curve is a set of
points consisting of a finite
number of regular arcs joined
one after the other (see Fig.
P4
Fia. 45.
45).
PoPI, P1P2, P2t 3, P3P4 are
the regular arcs joined at P1,
P2, P3. Notice that there are at most a finite number of discontinuities of the first derivatives. In Fig. 45 the derivatives
are discontinuous at P1j P2, P3.
46. Jordan Curves. The locus i x = f (t) a < t 5 fl, will be
y = P(t),
called a Jordan curve provided that f (t) and (p(t) are continuous
l curve correspond to two
and that two distinct points on the
distinct values of the parameter t (no multiple points).
A closed Jordan curve is a continuous curve having f(a)
,o(a) = 9(fl) but otherwise no multiple points.
From this we see that a Jordan curve is always "oriented,"
that is, it is always clear which part of the curve lies between two
points on the arc, and which points precede a given point. We
shall be interested in those curves which are rectifiable, or, in
IN TkGRA TION
Sr:c. 47]
99
other words, we shall attempt to assign a definite length to a
given Jordan are or curve.
47. Functions of Bounded Variation. Let f(x) be defined on
the interval a < x < b. Subdivide this interval into a finite
number of parts, say a = xo, x,, . . . x;, . . . , x,-1) x = b.
Now consider the sum
- f(xo)I + I f(x2) - f(xi)1 + .
I f(xi)
. + l f(xn)
- f(xn - 1)I
n
_ i_1
± I f(xi) - f (xt-1)
If the sums of this type for all possible finite subdivisions are
bounded, that is, if
n
s=x
I f(xi) - f(xc-1)I < A < -o
(153)
we say that f(x) is of bounded variation on (a, b).
A finite, monotonic nondecreasing function is always of
bounded variation since
n
n
i=1
If(x=) - f(x=-,)1 = I If(x=) - f(xi-1)) = f(b) - f(a) = A
t=1
An example of a continuous function that is not of bounded
variation is the following:
f(x) = x sin 2x,
0< x 5 1
f (O) = 0
Let us subdivide 0 5 x <_ 1 into the intervals
(n+1) Sx<n,
n = 1, 2, .. . , N. Now f(1 /n) = (1/n) sin (irn/2) so that
N
n
(I
If \n/
f \n }
)I
2
=1 +3
+ 25
+ N2
We cannot bound this sum for all finite N since the series diverges
as N -> w. N was chosen as an odd integer.
VECTOR AND TENSOR ANALYSIS
100
[SEc.48
48. Arc Length. Consider the curve given by x = f(t),
y = c(t), and assume no multiple points. Divide the parameter
t in any manner into n parts, say
a = t0 < 11 < t2 < '
'
' < to = ,8
and consider
Sn =
i=1
1[x(ti)
- x(ti-1)]2 + ly(ti) - y(tti-1)]2J}
This is the length of the
straight-line segments joining
Y
the points x(ti), y(ti), i = 0,
. . . , n (see Fig. 46).
If the set of all such lengths,
obtained by all finite meth-
1) 2,
ods of subdivision, is bounded,
we say that the curve is rectifiable and define the length of
the curve as the supremum of
these lengths.
An important theorem is
----x the following: A necessary
and sufficient condition that
the curve described by
F1o. 46.
x = f(t),
y = 'P(t)
be rectifiable is that f (t) and V(t) be of bounded variation.
Let
n
A=
{=1
[I f (ti) -
B = I IIf(t;) f-l
so that
f(4.-1)12
+ 14p(ti) - p(ti--1) J 2J1
f(4--1)I + Ip(k)
- v(4-1)I I
A==<Bs /2A
Consequently if the curve is rectifiable, f (t) and '(t) are of
bounded variation, and conversely, if f (t) and ap(t) are of bounded
variation, A is bounded, and hence the curve is rectifiable.
If f(t) and V'(t) are continuous, then from the law of the mean,
I f(4) - f (ti--1) I
= I f (ii) (ti - ti-1) I < A I ti - ti-il, where A is the
supremum of If(t)l, a 5 t =< # and 4_1 5 Ti 5 ti.
INTEGRA TION
SEC. 49]
n
n
E If()
i-1
- f(t;-1)I < A 11(4i=1
101
- till = A(# - a)
Similarly, p(t) is of bounded variation, so that the curve is rectifiable. Under these conditions it can easily be shown that the
arc length is given by
s - f8 [(dt)2 + (
p)2J dt
(154)
49. The Riemann Integral. We now develop the theory of
the Riemann integral in connection with line integrals. We
need a curve over which an integration can be performed and a
function to be integrated over this curve. Let r
aSt
x
y=¢(t)
be a rectifiable are. Let f (x, y) be a function continu-
ous at all points of the curve r. Subdivide the parameter t
< t, = 6. Let the coordiinto n parts, a = to < ti < t2 <
and let Osi be the length of are joining
nates of Pi be
Pi-1 to Pi. Since f (x, y) is continuous, it will take on both its
minimum and maximum for each segment Pi_1P;. Multiply each
are length by the maximum value of f (x, y) on this are, say f v,
and form the sum
n
J = A fii (xi, yi) A8i
Let .7 be the infemum of all such sums. Similarly, let K be the
supremum of all sums when the minimum value of f(x, y), say f,,,
is used. If J = K, we say that f (z, y) is Riemann-integrable
over the curve r, and write
J = K = fr f(x, y) ds = LfI(t),
+ * )} dt
whenever So' and J' are continuous.
That J = K for a continuous function defined over a rectifiable
curve is not difficult to prove. For,
J-K=
n
i-1
102
VECTOR AND TENSOR A V.I LYSIS
[SEC. 50
and from the uniform continuity of f(x, y) we can subdivide the
curve r into arcs such that the difference between the maximum
and minimum values of f (x, y) on any are is less than any given
e > 0 so that
JJ
- Kl < e11lAsil < eL
(155)
We leave it as an exercise for the reader to prove that we can make
J as close to J as we please and similarly that the difference
R=R1+R2
(ii)
FIG. 47.
FIG. 48.
between K and K can be made arbitrarily small, for a sufficiently
large number of subdivisions. Since the a of (155) is arbitrary,
we can make the difference between J and K as small as we
please so that J = K since they are fixed numbers.
50. Connected and Simply Connected Regions. A region R
is said to be connected if, given any two points of the region, we
can join them by an arc, every point of the are belonging to the
region R. In Fig. 47, (i) is connected; (ii) is not connected. In
(i), R is the nonshaded region.
INTEGRATION
Sic. 51]
103
If every closed curve of a connected region R can be continuously shrunk to a point of R, we say that the region is simply
connected. In Fig. 48, (i) is connected, but not simply connected; (ii) is simply connected. In (i) the curve r cannot be
continuously shrunk to a point. An analytic expression for
simple connectedness can be set up, but we shall omit this.
51. The Line Integral. Let
f = X(x, y, z)i + Y(x, y, z)j + Z(x, y, z)k
and consider the line integral f ( f ds) ds along a rectifiable
space curve r given by r = r(s).
Since
dsds = dr = dx i + dy j + dz k
we have
Jr (f ds/
ds = fr X dx + Y dy + Z dz
(156)
We use (156) as a means of evaluating the line integral. If
the space curve is given by x = x(t), y = y(t), z = z(t), then
(156) reduces to
ff
dr =
fo'
[x(t)
dt + Y(t) dt + Z(t) dtJ
dt
(157)
In general, the line integral will depend on the path joining
the two end points of integration. If f f dr does not depend
on the curve r joining the end points of integration, we say that f
is a conservative vector field. If f is a force field, we define
fB
f - dr as the work done by f as the unit particle moves from
A
A to B. We shall now work out a few examples for the reader.
Example 55. Let f = x2i + yaj, and let the path of integra-
tion be the parabola y = x2, the integration being performed
We exhibit three methods of solution.
(a) Let x = t, so that y = t2. Thus
from (0, 0) to (1, 1).
f = t2i + t6j,
r = ti + t2j,
dr = (i + 2tj) dt
104
VECTOR AND TENSOR ANALYSIS
[SEc. 52
and
ft t of f
dr = f.' (t2 + 2t7) dt =
(b) Since y = x2 everywhere along r, f = x2i + x6j along r,
and dr = dx i + dy j = (i + 2xj) dx, so that
1
Jr
JO
(x2 + 2x') dx = A
(c)
r (1, 1)
J(o, 0)
((1,
f dr =
1)
J(0, 0)
1+y'1- 7
xa
x2 dx + y$ dy - 3
0
4 (0
12
(c) shows that the integral from (0, 0) to (1, 1) is independent
of the path since x2 dx + y3 dy is a perfect differential, that is,
x2 dx + ys dy = d[(xs/3) + (y4/4)].
Example 56. f = yi - xj, and let the path of integration be
y = x2 from (0, 0) to (1, 1). Then
(1
f(1'1)ydx-xdy=
0,
ro, o)
, o)
f' x2dx-x(2xdx) _ -
Next we compute the integral by moving along the x axis from
x = 0 to x = 1 and then along the line x = 1 from y = 0 to
y = 1. We have
f
f("
(0,0))
( o, o)
1, o)
0, o)
Along the first part of the path, f = -xj and dr = dx i, since
y = dy = 0. Along the second part of the path, f = yi - j,
dr = dy j since x = 1 and dx = 0. Thus
f
(111)
0)
x=1
v
f.--=o -
fv-o
f01-dy=-1
i
The line integral does depend on the path for the vector field
f = yi - xj. We say that f is a nonconservative field.
52. Line Integral (Continued). Let us assume that f can be
written as the gradient of a scalar (p(x, y, z), that is, f = 0p.
Then the line integral JAB f dr is independent of the path of
INTEGRATION
SEC. 521
105
integration from A to B since
JAfdrJAVdrLd(B)(A) (158)
Our final result in (158) depends only on the value of (p(x, y, z)
when evaluated at the points A(xo, yo, zo), B(xi, yi, z,) and in no
way depends on the path of integration. If our path is closed,
then A = B and (p(B) = p(A), so that the line integral around
any closed path vanishes if f = Vip. However, the region for
which f =
must be simply connected. Let us consider the
following example.
Example 57. Let
f
- yi
LL
xi
=x2+y2 -rx2+y2
Then f = VV where p = tan-' (y/x), and if we integrate f over
the unit circle with center at the origin, we have
d
(tan_ixJ)fo2T dO
2r
so that our line integral does not vanish. The region for which
f = V(p is not simply connected since p is not defined at the
origin.
We now prove that if f f dr is independent of the path, then
f is the gradient of a scalar V. Let
P(x, y, z) =
JP(z,y.z)
f
dr
o(xo, YO, zo)
JP(fdr)ds
(159)
Now
(P(x + Ax, y, z) - (P(x, y, z)
Ax
and since the line integral is independent of the path joining P
to Q, we choose the straight line from P to Q as our path of
integration, that is, dr = dx i. Thus
lim
AX-40
cP(x+Ox,y,z) --P(x,y,z) =av
ax
Ax
iim
AX-0
f
z
x+Ax
X (x, y, z) dx
Ox
= X (x, y, z)
106
VECTOR AND TENSOR ANALYSIS
[SEC. 52
We have assumed that f is continuous.
from the calculus.
similarly
-
= Y(x, y, z),
y
Now
az = Z(x, y, z)
so that
f=
a(P
ax
av
i + a7
,
+
a,P
az
k = VV
(160)
If f has continuous derivatives, we can easily conclude whether
f is the gradient of a scalar or not. Assume f = VV, or
(1)
X=-"`P,
(3)Z=
(2)
az
Differentiating (1) with respect toy and (2) with respect to x,
we see that
ax
ay
Similarly
ay
ax
aYaZ
az
ay
ax
az
(161)
azax
This is the condition that V x f = 0. Conversely, assume
V x f = 0. Let
,P(x,y,z) = f:oX(x,y,z)dx+f,0 Y(xo, y, z) dy
+
z,
Z(xo, yo, z) dz
Now
a'P
ax = X (x, y, z)
aq
_ -L.
f..
ay
f
dx+
Y(xo,y,z)
-0Y
ax dx + Y (x,), y, z)
= Y(x, y, z) - Y(xo, y, z) + Y(xo, y, z)
= Y(x, y, z)
(162)
INTEGRATION
Sic. 53]
Similarly
107
Consequently
7.(x, y, z).
f
axi+ayj+azk=Drp
We have proved that a necessary and sufficient condition for f
to be the gradient of a scalar is that.
V xf =0
(163)
If f = V or V x f = 0, f is said to be an irrotational vector.
Example 58.
Let f = 2xyezi + x2ezj + x2yezk.
Vxf=
k
i
j
a
ax
ay
a
az
2xyez
x2ez
x2yeZ
a
Then
=0
and
=
JX
2xy9 dx +
02ez dy +
02. 0 . 9 dz
= x2yeZ
so that f= V (x2yeZ + constant).
Problems
1. Given f = xyi - xj, evaluate f f dr along the curve y = x3
from the origin to the point P(1, 1).
2. Show that if the line integral around every closed path is
zero, that is, if jFf dr = 0, then f = V p.
3. Show that /r dr = 0.
4. Show that the inverse-square force field f = -r/r3 is conservative.
The origin is excepted.
5. f = (y + sin z)i + xj + x cos z k. Show that f is conservative and find rp so that f = V p.
6. Evaluate f x dy - y dx around the unit circle with center
at the origin.
7. If A is a constant vector, show that /A dr = 0,
jrAxdr=0
53. Stokes's Theorem. We begin by considering a surface
of the type encountered in Chap. 3. Now consider a closed
108
VECTOR AND TENSOR ANALYSIS
[SEC. 53
rectifiable curve r that lies on the surface. As we move, along
the curve r, keeping our head in the same direction as the normal
ar
car
x ) we keep track of the area to our left. It is this surface
au av
u curve
that we shall keep in touch
with,
and r
will
be the
boundary of this surface.
We neglect the rest of the
surface r(u, v).
We now con-
sider a mesh of networks on
the surface formed by a collection of parametric curves.
v curve
Fia. 49.
Of course, the boundary r
will not, in general, consist
of arcs of these parametric
curves (see Fig. 49). Consider the mesh A BCD. Let
the surface coordinates of A
be (u, v), so that A(u, v), B(u + du, v),
C(u+du,v+dv)
D(u, v + dv) are the coordinates of A, B, C, D. We also assume
that the parametric curves are rectifiable. Now consider
ABCD f dr
The value of fat A is f(u, v); at B it is f(u + du, v); at D it is
f(u, v + dv).
Now
f (u + du, v) = f (u, v) + dfu
= f(u, v) + (dru V)f
= f(u, v) + (au du v) f
except for infinitesimals of higher order. Similarly
f(u,v+dv)
=f(u,v)+(-dvv)f
INTEGRATION
Six. 53]
109
Hence, but for infinitesimals of higher order,
-auc1r
f+
fi au} dude
av -
(164)
Now
(v x f)
for ar
(au
x av
an
Or
= (° x f) x au
°/
C\au
J
av
av ^ L \av
Hencee
fiABCD f
dr = (v x f) a x
a du dv
v)fI ar
au
(165)
ar
ar
Now d x a du dv = area of sector ABCD in magnitude, and
au av
its direction is along the normal. We define
dd =
Or
Or
aux& du dv
(166)
so that
IABCDfdr= (V
We now sum over the entire network. Interior line integrals
will cancel out in pairs leaving only fir f dr. Also
I
f f (V
(V
a
over surface 8
as the areas approach zero in size. We thus have Stokes's
theorem :
fir f dr = f f (V x f) dd
8
(167)
VECTOR AND TENSOR ANALYSIS
110
[SEC. 54
Comments
1. The reader may well be aware that (165) does not hold for
a mesh that has r as part of its boundary. This is true, but
fortunately we need not worry about the inequality. The line
integrals cancel out no matter what subdivisions we use, and for
a fine network the contributions of those areas next to r contribute little to f f (V x f) dd. The limiting process takes care
of this apparent negligence.
2. We have proved Stokes's theorem for a surface of the type
r(u, v) discussed in Chap. 3. The theorem is easily seen to be
true if we have a finite number of these surfaces connected continuously (edges).
3. Stokes's theorem is also true for conical points, where no
dd can be defined. We just neglect to integrate over a small area
covering this point. Since the area can be made arbitrarily
small, it cannot affect the integral.
4. The reader is referred to the text of Kellogg, "Foundations
of Potential Theory," for a much more rigorous proof of Stokes's
theorem.
54. Examples of Stokes's Theorem
Example 59. Let r be a closed Jordan curve in the x-y plane.
Let f = -yi + xj. Applying Stokes's theorem, we have
ff
ff
S
ax
ay
x
az
-y
0
=2ffdydx=2A
S
or
Area A =-
xdy - ydx
For the ellipse x = a cos t, y = b sin t, dx = -a sin t dt,
dy = b cos t dt
and
A=
f27
ab(cos2 t + sin 2 t) dt = grab
(168)
INTEGRATION
SEc. 541
Example 60.
111
If f has continuous derivatives, then a necessary
and sufficient condition that ff dr = 0 around every closed
path is that V x f = 0.
If V x f = 0, then Cf U. dr = f f (V x f) dd = 0. Cons
versely, assume f t dr = 0 for every closed path. If V x f p4 0,
then V x f 0 at some point P. From continuity, V x f: 0
in some region about P. Choose a small plane surface S in this
region, the normal to the plane being parallel to V x f. Then
f- dr = f f V x f- dd > 0, a contradiction
s
Example 61. We see that an irrotational field is characterized
by any one of the three conditions:
f = VP
Vxf=0
f dr = 0
(i)
(ii)
(iii)
(169)
for every closed path
Any of these conditions implies the other two.
Example 62. Assume f not irrotational. Perhaps a scalar
p(x, y, z) exists such that pf is irrotational, that is, V x (pf) = 0.
If this is so,
pV xf+Vp xf = 0
(170)
and dotting (170) with f, we have, since f VA x f = 0, the
equation
f- (V xf) = 0
(171)
If f = Xi + Yj + Zk, (171) may be written
XYZ
a
a
a
ax
ay
az
=0
(172)
XYZ
In texts on differential equations it is shown that (172) is also
sufficient for p(x, y, z) to exist. We call p(x, y, z) an integrating
factor.
Example 63. Let f = f(x, y, z)a, where a is any constant vector. Applying Stokes's theorem, we have
VECTOR AND TENSOR ANALYSIS
112
ff
[SEC. 54
x a) dd
s
ff
s
a
f
f
f= a
dd x
a
we have
S
ff
[a(V
s
g) - (a V)gl dd
=a- f f (V.g)dd-a f f
dd (summed).
V
There-
fore
xdr = a- f f (ddxv) xg
and
dr x g = f f (dd x V) x g
S
(174)
We notice that in all cases
0 dr * f=
Jf(doxv)*f
(175)
S
The star (*) can denote dot or cross or ordinary multiplication.
In the latter case, f becomes a scalar f.
INTEGRATION
Sec. 541
113
Problems
1. Prove that fdr = 0 from (175).
2. Show that L, 'dr x ri taken around a curve in thex-y plane
is twice the area enclosed by the curve.
3. If f = cos y i + x(1 + sin y)j, find the value of ff - dr
around a circle of radius r in the x-y plane.
4. Prove that fr dr = 0.
5. Prove that ff dd x r =
fir 2 dr.
s
6. Prove that f u Vv dr = - ,f v V u dr.
7. Prove that u Vv - dr = JJvu x Vv dd.
s
8. If a vector is normal to a surface at each point, show that
its curl either is zero or is tangent to the surface at each point.
9. If a vector is zero at each point of a surface, show that
its curl either is zero or is tangent to the surface.
10. Show that fi a x r dr = 2a. Jf dd, if a is constant.
S
11. If fi E dr =
that V x E =
B dd for all closed curves, show
-cat Jf
s
---
C at
12. By Stokes's theorem prove that V x (V(p) = 0.
13. Show that fi dr/r = f f (r/r3) x dd where r = Id.
8
14. Find the vector f such that xy = f(O,(X'o, Y,)
f dr.
o)
15. If f = r/r3, show that V x f = 0 and find the potential yo
such that f = Vsp.
16. Show that the vector f = (- yi + xj)/(x2 + y2) is irrotational and that (f r f dr = 2w, where r is a circle containing the
origin.
Does this contradict Stokes's theorem? Explain.
17. Show that ff v x f dd = 0, where S is a closed surface.
s
18. Let C1 and C2 be two closed curves bounding the surfaces
Show that
S1 and S2.
114
VECTOR AND TENSOR ANALYSIS
k. ki r122 dr1
[SEC. 55
dr2 = -4 fJ dal f f dd2
St
s2
f, fc, r122 dr1 x dr2 = - 2 f f dot x ff dd2
Si
s2
where r12 is the distance between points on the two curves.
55. The Divergence Theorem (Gauss). Let us consider a
region V over which f and V f are continuous. We shall assume
that V is bounded by a finite number of surfaces such that at
each surface there is a well-defined continuous normal. We shall
also assume that f can be integrated over the total surface bounding V. Now, no matter what physical significance f has, if any,
we can always imagine f to be the product of the density and
velocity of some fluid. We have seen previously (Sec. 20) that
the net loss of fluid per unit volume per unit time is given by
V f.
Consequently, the total loss per unit time is given by
JfJ(v.f)dr
(176)
V
Now since f and V f are continuous, there cannot be any point
in the region V at which fluid is being manufactured or destroyed;
that is, no sources or sinks appear. Consequently, the total
loss of fluid must be due to the flow of fluid through the boundary
S of the region V. We might station a great many observers on
the boundary S, let each observer measure the outward flow, and
then sum up each observer's recorded data. At a point on the
surface with normal vector area dd, the component of the velocity
perpendicular to the surface is V N, where N is the unit outward
normal vector. It is at once apparent that pV dd = f dd
represents the outward flow of mass per unit time. Hence the
total loss of mass per unit time is given by
Jff.dd
(177)
Equating (176) and (177), we have the divergence theorem:
JJf.do=fJf(v.f)d.r
8
V
(178)
INTEGRATION
SEC. 55]
115
For a more detailed and rigorous proof, see Kellogg, "Foundations of Potential Theory." We now derive Gauss's theorem by a
different method. Let f be a differentiable vector inside a connected region R with rectifiable surface S. Surround any point
P of R by a small element of volume dr having a surface area AS.
Form the surface integral JJf. dd and consider the limit,
As
lira
f
As
Ar-.0
DT
If this limit exists independent of the approach of Ar to zero, we
define
Jff.dd
div f = lim
As
AT-40
(179)
AT
We can write
tardivf= JJ
(180)
where e -40 as Ar -* 0. If we now subdivide our region into
many elementary volumes, we obtain formulas of the type (180)
for all of these regions. Summing up (180) for all volumes and
then passing to the limit, we have
fffdivfdr__
(181)
Jffdd
Inthe derivation of (181) use has been made of the fact that for
each internal dd there is a -dd, so that all interior surface
integrals cancel in pairs, leaving only the boundary surface S
as a contributing factor. The sum of the ei Ar; vanishes in the
limit, for I Zet Aril 5 ZIel... OT S IEIo,.=V, and if div f is continu-
ous, IeI,. - 0 as 2r --i 0.
By choosing rectangular parallelepipeds and using the method
of Sec. 20, we can show that
fff.do
div f = slim
AS
AT
clu
c3v
aw
ax
+ ay +
49Z
(182)
116
VECTOR AND TENSOR ANALYSIS
[SFC. 55
for f = ui + vj + wk, which corresponds to our original definition of the divergence.
Example 65. Consider a sphere with center at 0 and radius a.
Take f=r=xi+yj+zk,
dd =
(r) dS =
a
(ai)
(
xi+yj+zk)dS
Now V f =3, f dd = (1/a)(x2 + y2 + z2) dS = a dS on the
sphere.
Applying (178), f _f f 3 dr = f f a dS, or 3V = aS, where
S is the surface area of the
sphere. If V is known to be
g 47a8, then S = 4aa2.
Example 6 6. Let f = qr/rs
and let V be a region surround-
ing the origin and let S be its
We cannot apply the
divergence theorem to this resurface.
gion since f is discontinuous at
r = 0. We overcome this difficulty by surrounding the origin
by a small sphere E of radius e
(see Fig. 50). The divergence theorem can be applied to the
connected region V'. The region V' has two boundaries, S and
Fzo. 50.
E.
Applying (178),
ffJ V. () dr
f
8
r ad+J1
E
r"d
(183)
In Example 25 we saw that V (r/r3) = 0. This implies that
(183) reduces to
ff.!.do=
(184)
Now for the sphere 1, dd = -r dS/e, since the outward normal
to the region V' is directed toward the origin and is parallel to
the radius vector. Hence
JJ.dd= fff.dd= ffds=4fq
INTEGRATION
SFe. 551
117
The integral f f f dd is called the flux of
since f f dS =
s
the vector field f over the surface S.
square force f = qr/r3,
fJ
f
S
We have, for an inverse-
dd = 4irq
(185)
Example 67. A vector field f whose flux over every closed
surface vanishes is called a solenoidal vector field.
From (178)
it is easy to verify that V f = 0 for such fields. Hence a
solenoidal vector is characterized by either /f dd = 0 or
0.
Now assume f = V x g, which implies V f = V (V x g) = 0.
Is the converse true? If V f = 0, can we write f as the curl of
some vector g? The answer is "Yes"! We call g the vector
potential of f. This theorem is of importance in electricity
theory, as we shall see later. Notice that g is not uniquely
determined since V x (g + Vp) = V x g. We now show the
existence of g. Let f = Xi + Yj + Zk and assume
g=ai+ij+yk
We wish to find a, $, y such that f = V x g, and hence
x
=---
Y=
Z
ay
a#
ay
az
as
ay
az
ax
as
as
ax
ay
(186)
=---
Now assume a = 0. Then X =
ay
ay
- 8-, Y = - aye Z = as.
8z
ax
Consequently if there is a solution with a = 0, of necessity
$6=
fXzdx+o.(y,z)
y=Now V f = 0 or
ax
o
xos
Ydx+r(y,z)
aY
y
+ az) by assumption.
ax
VECTOR AND TENSOR ANALYSIS
118
f
Hence
ay
aY
JZo
az
dx
,"
+
+
+ ar
- as
ay
az
ay
fx ax
ax
aZ
az /
ar
ay
= X (x, y, z) - X (xo, y, z) +
_
[SEC. 55
aa
az
or
ao
ay -
az
so that (186) is satisfied by a = 0, S = I o Z dx + or (y, z),
jX
y
Y dx by choosing r = 0, o(y, z) = - JZX(xo,y,z) dz.
Hence f =Vxgwhere
g=[jz
Z dx + c(y, z)] j - faVdxk.
In general
g=[ faZdx+v(y,z)]j- foYdxk+Vp
(187)
For example, if f = V(1/r), then V f = 0. Now
X=--
Y=-y,
s,
Z=-raz
where r2 = x2 + y2 + z2. Applying (187)
g=
x
-z dx
y dx
(x2+y2+z2), j+1 (x2+y2+z2)$k+V
,
v arbitrary
and
g = (x2 + y2 + z2)#(y2 +
Example 68. Green's theorem.
and f2 = v Vu and obtain
z2)
We apply (178) to fl = u Vv
f f f V. (u Vv)dr= f f f
R
(-zj + yk) + VSP
ff
R
S
(188)
f f f V V. (v Vu) dr = f f f (v V2u + VV. VU) d7 = Jf v Vu dd
R
S
R
Subtracting, we obtain
f f f (u V2v - v V2u) dr = f f (u Vv - v Vu) dd (189)
R
S
INTEGRATION
SEC. 55]
119
Example 69. Uniqueness theorem. Assume two functions
which satisfy Laplace's equation everywhere inside a region and
which take on the same values over the boundary surface S.
The functions are identical. Let V2(P = 1724, = 0 inside R and
(p ='p on S.
Now from (188) we have
JJJov2odr+JfJvo.vodr=JJove.do
R
R
Define 0 - (p - ¢ so that 1720 = 0 over R and 0=0 on S.
Hence f f f (V0)2 dT = 0, which implies V0 = 0 inside R.
R
Hence 0 = constant = p -'p, and since sc
on S, we must
have cp ='G. We have assumed the existence of Vp, V¢ on S.
Example 70. Another uniqueness theorem. Let f be a vector
whose curl and divergence are known in a simply connected
region R, and whose normal components are given on the surface
S which bounds R. We now prove that f is unique. Let f1 be
another vector such that V f = V f1, V x f = V x f1, and
f dd = f, dd on S. We now construct the vector g = f - f1.
We immediately have that V g = V x g = g dd = 0. In
Sec. 52 we saw that if V x g = 0, then g is the gradient of a scalar
Consequently, V g = V2ip = 0. Applying (188)
cp, g = V<p.
with u = v = (p, we have
If! [V2cc+ (Vcc)21 dr = ff `p vv - dd = f
0
s
R
and hence fff (VV) 2 dr = 0 so that V p - 0 inside R, and
R
g = Vp - 0, so that f = f 1 inside R.
We can also prove that f is uniquely determined if its divergence and curl are known throughout all of space, provided that f
tends to zero like 1/r2 as r--, ao. We duplicate the above
proof and need Jim f p VV dd = 0. If cp tends to zero like
1/r, then V<p tends to zero like 1/r2, and f f cc V(p V. dd tends to
s
zero like 1 /r as r oo.
Example 71. Let f = f(x, y, z)a, where a is constant. Apply
ing (178), we obtain
VECTOR AND TENSOR ANALYSIS
120
a-
[SEC. 55
f ffdo= f f fv.(fa)dr=a f f fVfdr
s
V
V
Hence
fffdd
s
fff
=
7fdr
(190)
V
We leave it to the reader to prove that
JJdo*f = Jff(V*f)dr
s
(191)
V
Problems
1. Prove that f f dd x f = f f f (V x f) dr.
y
s
2. Prove that ff dd = 0 over a closed surface S.
s
c constants, show that
= 4(a + b + c), where S is the surface of a unit
3. If f = axi + byj + czk, a,
Jff.do
b,
s
sphere.
fffdd
4. By defining grad f = Jim
As
gradf =
snow that
OT
4,-+O
afi+
afj+afk
ax
ay
az
fff.dd
5. By defining div f = lim °3
A?--+O
div f =
r2 sin 0 Lar
GT
(r2 sin of,) +
, show that
e (r sin 9 fe) + a
for spherical coordinates.
6. Show that
f
fJ
JfJ (P 0fdr.
(rf,)]
INTEGRATION
SEC. 55]
121
7. If w = V x v, v = V x u, show that
'9
R
v2dr = f f u xv.dd+ f f
S
R
8. Show that
f J f w Vu . Vv dr = f f uw Vv . dd - f if u V . (w Vv) dr
9. If v = Vp and V . v = 0, show that for a closed surface
f J f V2dr = Jf pv.dd.
10. Show that f f IrJ2r . da = 5 fJJ r2 dr.
11. If f = xi - yj + (z2 - 1)k, find the value of f f f . do
over the closed surface bounded by the planes z = 0, z = 1 and
the cylinder x2 + y2 = 1.
12. If f is directed along the normal at each point of the
boundary of a region V, show that f f f (V x f) dr = 0.
V
13. Show that f f r x da = 0 over a closed surface.
S
14. Given f = (xyez + log (z + 1) - sin x)k, find the value of
ffv x f . da over the part of the sphere x2 + y2 + z2 = 1 above
S
the x-y plane.
15. Show that (xi + yj)/(x2 + y2) is solenoidal.
16. If f 1 and f2 are irrotational, show that f 1 x f2 is solenoidal.
17. Find a vector A such that
f =- yzi - zxj + (x2+y2)k = V xA.
18. If r, 0, z are cylindrical coordinates, show that De and
Find the vector potentials.
19. Let S, and S2 be the surface boundaries of two regions
V1 and V2. Let r be the distance between two elementary
V log r are solenoidal vectors.
VECTOR AND TENSOR ANALYSIS
122
Show that
volumes d-r, and dT2 of V1 and V2.
fv, rn:-2 dT,
fsI rm dd1 = -m(mn + 1) fV (IT2
.fs, dii2
and that
1s= dd2
show that V2X =
for V2Y, V2Z.
fs, log r ddl = -
fv,
cIT2
f
dTl
V1
r
V xf ='1i+¢2j+4,ak,f = Xi + Yj + Zk,
20. IfV - f =
Vxf=zi.
[SEC. 56
49 O
-
- a3y -
z2' and find similar expressions
Find a vector f such that V f = 2x + y - 1,
56. Conjugate Functions. Let us consider the two-dimensional vector field w = u(x, y)i + v(x, y)j and an orthogonal
vector field wi = v(x, y)i - u(x, y)j. Obviously w - wl = 0.
What are the conditions on u(x, y), v(x, y) which will make w and
w1 irrotational? From Stokes's theorem
x w dd = ff
ffv
s
s
{ w1 dr = jJv
=
Jf
w dr =
av
au
ax
ay
au
ax-
dydx
av
ay
(192)
dy dx
A necessary and sufficient condition that both w and w1 be
irrotational is that
52).
av
ax
- au = 0 and - au -- av = 0 (see Sec.
ax
ay
a4
This yields
av
ax
_
au
ay
av
au
ay
ax
(193)
The reader who is familiar with complex-variable theory will
immediately recognize (193) as the Cauchy-Riemann equations,
which must be satisfied for the analyticity of the complex func.
tion w = v(x, y) + iu(x, y), i =
INTEGRATION
SEC. 56]
123
On differentiating (193), we obtain
=
V 2U
a2u
axe
+
a2u
2
=
49Y
a2v
V2v=+a
axey =
a2v
2
(194)
and
au av
ax ax
+
au av
ay ay
_
If functions u(x, y), v(x, y) satisfy Eqs. (193) we say that
The importance of such func-
they are harmonic conjugates.
y
v
P (x, yl
Ix
Fia. 51.
tions is due to the fact that they satisfy the two-dimensional
Laplace's equation given by (194). If u satisfies V2u = 0, we
say that u(x, y) is harmonic.
Let us now consider two rectangular cartesian coordinate
systems, the x-y plane and the u-v plane (Fig. 51). Let
r=xi+yj.
Now to every point P(x, y) there corresponds a point Q(u, v)
given by the transformation u = u(x, y), v = v(x, y). Hence the
vector w = u(x, y)i + v(x, y)j corresponds to the vector
r -- xi+yj.
If now P(x, y) traverses a curve C in the x-y plane, Q(u, v) will
trace out a corresponding curve r in the u-v plane.
The curve u(x, y) = constant in the x-y plane transforms into
the straight line u = constant in the u-v plane. Similarly,
124
VECTOR AND TENSOR ANALYSIS
[Ssc. 56
v(x, y) = constant transforms into the straight line v = constant.
The two straight lines are orthogonal. Do the curves
u(x, y) = constant
v(x, y) = constant intersect orthogonally? The answer is "Yes"!
The normal to the curve u(x, y) = constant is the vector
ay
i+u
=au
VU
a
and the normal to the curve v(x, y) = constant is Vv =
av
ax
av
49V
i+
yj
au av
0 from (194).
ax ax + ay ay =
Example 72. Consider the vector field w = 2xyi + (x2 - y2) j.
so that Vu Vv =
Ou av
Here u = 2xy, v = x2 - y2, and
av
= au
ax
ay
av
_
= 2x
au _ _ 2
ay
ax
r
y
so that u and v are conjugate harmonics.
The curves
u = 2xy = constant
and v(x, y) = x2 --- y2 = constant are orthogonal hyperbolas
which transform into the straight lines u = constant, v = constant, in the u-v plane (Fig. 52).
Example 73. Consider
w = ( tan-1
x)
i + J log (x2 + y2)j
Here u(x, y) = tan-1 (y/x), v = I log (x2 + y2), and
au
av
y
ax
ay
x2 + y2
au
av
x
ay
ax
x2 + y2
so that u and v are conjugate harmonics.
SEc. 56]
INTEGRATION
125
Y
I
0'
Fia. 52.
(3)
(4)
------------
Fra. 53.
The circles x2 + y2 = constant transform into the straight
lines v = J log c, while the straight lines y = mx transform into
the straight lines u = tan-' m (Fig. 53).
Example 74. If u(x, y) is given as harmonic, we can find its
conjugate v(x, y). If v(x, y) does exist satisfying (193), then
126
VECTOR AND TENSOR ANALYSIS
dv=axdx+-dy=aydx Now consider the vector field f =
Vxf=-t
au
ay
f&c. 56
au
ax dy
i - au j.
ax
We have that
x + a2 ) k = 0 by our assumption about u(x, y).
y
Hence f is irrotational and so is the gradient of the scalar v,
f = Vv, and
v = fa au
a dx - jyOU dy + c
1!
8x
(195)
As an example, consider u = x2 - y2, which satisfies V2u = 0.
Hence
v=
fox
- 2y dx - f." 2 . 0 dy + c = - 2xy + c
Problems
1. Find the harmonic conjugate of xa - 3xy2, of ex coo y, of
x/(x2 + y2).
2. Show that u(x, y) = sin x cosh y and v(x, y) = cos x sinh y
are conjugate harmonics and that the curves u(x, y) = constant,
v(x, y) = constant are orthogonal. What do the straight lines
y = constant transform into?
3. If u(x, y), v(x, y) are conjugate harmonica, show that the
angle between any two curves in the x-y plane remains invariant
under the transformation u = u(x, y), v = v(x, y), that is, the
transformed curves have the same angle of intersection.
CHAPTER 5
STATIC AND DYNAMIC ELECTRICITY
57. Electrostatic Forces. We assume that the reader
is
familiar with the methods of generating electrostatic charges.
It is found by experiment that the repulsion of two like point
charges is inversely proportional to the square of the distance
between the charges and directly proportional to the product
of their charges. The forces act along the line joining the two
charges. We define the electrostatic unit of charge (e.s.u.) as
that charge which produces a force of one dyne on a like charge
situated one centimeter from it when both are placed in a
vacuum. The electrostatic intensity at a point P is the force
that would act on a unit charge placed at P as a result of the rest
of the charges, provided that the unit test charge does not affect
the original distribution of charges. For a single charge q placed
at the origin of our coordinate system, the electric intensity, or
field, is given by E = (q/r3)r. For many charges the field at P
is given by
E=-
4'a
s
a=1 ra
(196)
ra
where ra represents the vector from P to the charge qa.
We have seen in Example 25 that V V. (r/r3) = 0. Consequently,
V.E=0
(197)
so that the divergence of the electrostatic-field vector is zero at
any point in space where no charge exists. Hence E is solenoidal
except where charges exist, for there E is discontinuous. If the
coordinates of P are t, n, t and the coordinates of qi are xi, yi, zi,
then ri =
I(rS
l
V1
ri
ri
- xi)2 + (,l - y,)2 +
a
(pJ
- z,)2]1, and
_
rr;3
(xi - )i -I- (yi- 7/)i + (zi -t)k
ot
r ri
an
127
r;
VECTOR AND TENSOR ANALYSIS
128
[SEC. 58
so that (196) reduces to
E = -Vp
(198)
n
where P =a-1
I qa/ra. We call w the electrostatic potential.
For any closed path which does not pass through a point charge,
we have f E dr = - ^p dr = - f d(p = 0. Thus E is also
irrotational, and
VxE=O
(199)
We also note that f E dr = cp(P) - p(-o) = tp(P), since
,p(oo) = 0. Hence the work done by the field in taking a unit
charge from P to oo is equal to the potential at P.
58. Gauss's Law. Let S be an imaginary closed surface that
does not intersect any charges. In Example 66 we saw that
J f (qr/r3) dd = 4Tq. This is true for each charge q; inside S.
S
Hence
JS ! La
a-1
qara dd = 41r
raa
a-1
qa
(200)
For a charge outside 8,
Jf.do=JfJv.()dr=0
(201)
since there is no discontinuity in qr/r3, r > 0. Adding (200)
and (201), we obtain Gauss's law,
JJE.dd=4irQ
(202)
s
where Q is the total charge inside S. The theorem in words
is that the total electric flux over any closed surface equals 4w
times the total charge inside the surface.
Example 75. We define a conductor as a body with no electric
field in its interior, for otherwise the "free" electrons would move
and the field would not be static. The charge on a conductor
must reside on the surface, for consider any small volume contained in the conductor and apply Gauss's theorem.
SEC. 58]
STATIC AND DYNAMIC ELECTRICITY
129
f f E dd = 4xq
and since E = 0, we must have q = 0. This is true for arbitrarily small volumes, so that no excess of positive charges over
negative charges exists. Hence the total charge must exist on
the surface of the conductor.
If a body has the property that a charge placed on it continues
to reside where placed in the absence of an external electric field,
Fic. 54.
we call the body an insulator. Actually there is no sharp line
of demarcation between conductors and insulators. Every body
possesses some ability in conducting electrons.
At the surface of a conductor the field is normal to the surface,
for any component of the field tangent to the surface would cause
a flow of current in the conductor, this again being contrary to
the assumption that the field is static (no large-scale motion of
electrons occurring). Such a surface is called an equipotential
surface. The field is everywhere normal to an equipotential
surface, for the vector E = -V(p is normal everywhere to the
surface V(x, y, z) = constant.
Example 76. Consider a uniformly charged hollow sphere E.
We shall show that the field outside the sphere is the same as if
130
VECTOR AND TENSOR ANALYSIS
[SEC. 58
the total charge were concentrated at the center of the sphere
and that in the interior of the
sphere there is no field.
Let P be any point outside
the sphere with spherical coordinates r, 0, cp. Construct an
imaginary sphere through P con(see
centric with the sphere
Fig. 54). From symmetry it is
obvious that the intensity at any
point of the sphere is the same
Fic. 55.
as that at P. Moreover, the
field is radial. Applying Gauss's law, we have f f E dd = 4,rQ,
or
f fEdS=Ef fdS=4irr2E=4aQ
so that
E= r
and
E=
r
(203)
Q
Q
We leave it to the reader to show that E = 0 inside I.
Example 77. Field w thin a parallel-plate condenser.
Consider
two infinite parallel plates with surface densities a and -a.
Fic. 56.
From symmetry the field is normal to the plates. We apply
Gauss's law to the surface in Fig. 55 with unit cross-sectional
area.
f f E do = E = 4ara
(204)
so that the field is uniform.
Example 78. We now determine the field in the neighborhood
of a conductor. We consider the cylindrical pillbox of Fig. 56
and apply Gauss's law to obtain
SEC. 58]
STATIC AND DYNAMIC ELECTRICITY
131
EA = 4raA
or
E = 47ro N
(205)
where a is the charge per unit area and N is the unit normal vector
to the surface of the conductor.
Example 79. Force on the surface of a conductor. We consider
a small area on the surface of the conductor. The field at a
point outside this area is due to (1) charges distributed on the
rest of the conductor (call this field E1), and (2) the field due to
the charge resting on the area in question, say E2 (see Fig. 57).
From Example 78, El + E2 = 41rv. Now the field inside the
Ei+E2
P
Fia. 57.
conductor at the point P' situated symmetrically opposite P is
zero from Example 75. The field at P' is E1 - E2 = 0. Thus
El = 2o per unit charge. For an area dS the force is
dE = (2av) (v dS) = 27rv2 dS
(206)
This force is normal to the surface. A charged soap film thus
tends to expand.
Problems
1. Two hollow concentric spheres have equal and opposite
charges Q and -Q. Find the work done in taking a unit test
charge from the sphere of radius a to the sphere of radius b, b > a.
The outer sphere is negatively charged.
2. Find the field due to any infinite uniformly charged cylinder.
3. Solve Prob. 1 for two infinite concentric cylinders.
VECTOR AND TENSOR ANALYSIS
132
[Sec. 59
4. Let ql, q2, . . . , q be a set of collinear electric charges
residing on the line L. Let C be a circle whose plane is normal
to L and whose center lies on L. Show that the electric flux
n
through this circle is N = 121rga(1 - cos Sa), where Na is the
a-I
angle between L and any line from qa to the circumference of C.
5. Let the line L of Prob. 4 be the x axis, and rotate a line of
force r in the x-y plane about the x axis (see Fig. 58). If no
Y
xa
q3
0
x2
xl
qa
qi
x
I
II
`
I
I
1
1
/
Fia. 58.
charges exist between the planes x = A, x = B, show that the
equation of a line of force is
n
a-1
qa(x - xa)[(x - xa)z + y2]i = constant
6. Point charges +q, -q are placed at the points A, B. The
line of force that leaves A making an angle a with AB meets the
plane that bisects AB at right angles in P. Show that
sin 2 =
59. Poisson's Formula.
E_
sin (+
PA B)
In Sec. 57, we saw that
-v
a-1 r,
STATIC AND DYNAMIC ELECTRICITY
SEC. 591
133
For a continuous distribution of charge density p, we postulate
that the potential is
=
fff
pdr
(207)
where the integration exists over all of space. At any point P
where no charges exist, r > 0, and we need not worry about the
convergence of the integral. Now let us consider what happens
at a point P where charges exist, that is, r = 0. Let us surround
the point P by a small sphere R of radius e. The integral
f f f (p dr/r) exists if p is continuous. We define 9 at P
Y-R
as lim f f f (p dr/r). This limit exists, for using spherical
y-R
coordinates,
If f f "d*T
=I
fo2r fo=
0 pr sin a dr d9 d<p` < MT2E2
where M is the bound of p in the neighborhood of P.
I i l xf p
dT
_ fJf
Rp
V
Thus
I < M? (e2 + el )
where e' is the radius of the sphere R' surrounding P. The
Cauchy criterion holds, so that the limit exists. In much the
same way we can show that
E = f f f Ta dr
(208)
and that at a point P where a charge exists
E(P) = lim f uJ -t dr
r +0
converges.
Now from Gauss's law
ff
8
f f f pdr
v
VECTOR AND TENSOR ANALYSIS
134
[SEC. 59
In order to apply the divergence theorem to the surface integral,
we must be sure that V E is continuous at points where p is
continuous. We assume this to be true, and the reader is referred
to Kellogg's "Foundations of Potential Theory" for the proof
of this.
Thus
JJJ(v.E)dr = 47r f f f p dr
(209)
V
V
Since (209) is true for all volumes, it is easy to see that
V E = 4irp
(210)
Since E _ - Vs,, we have
provided V E and p are continuous.
Poisson's equation
V2,p = -4ap
(211)
and at places where no charges exist, p = 0, so that Laplace's
equation, V2,p = 0, holds.
Example 80. In cylindrical coordinates
V 20
r L or
(r Or + e
r
t90/
+ 8z \r 8z/ J
Consider an infinite cylinder of radius a and charge q per unit
length. At points where no charge exists, we have V2,p = 0.
Moreover, from symmetry, p depends only on r. Thus
r dr = constant = A
vAlogr+B
E_ -Vip= -A r,
r=xi -f- yl
Also 42rc = (Er)r_a = -A/a, so that q = 2,raa = -A/2, and
E=2r
r2
(212)
STATIC AND DYNAMIC ELECTRICITY
SEC. 601
135
Example 81. To prove that the potential is constant inside a
conductor. From Green's formula we have
Jfpvcc.ddJfJ(s7co)2dr+JJJrpv2codr
Inside the conductor no charge exists so that V 2(p = 0.
Moreover,
for any surface inside the conductor, E = -Vgp = 0 so that
Jff (vp)2 dr = 0 for all volumes V inside the conductor.
V
Therefore (VV)2 = 0, and
app
ax
=
app
ay
=
app
az
_ 0, so that p = con-
stant inside the conductor.
Problems
1. Solve Laplace's equation in spherical coordinates assuming
the potential V = V(r).
2. Find the field due to a two-dimensional infinite slab, of
width 2a, uniformly charged. Here we have p = p(x) and must
solve Laplace's equation and Poisson's equation separately for
free space and for the slab, and we must satisfy the boundary
condition for the potential at the edge of the slab. The space
occupied by the slab is given by -a S x < a, - co < y < oo.
3. Solve Laplace's equation for two concentric spheres of radii
a, b, with b > a, with charges q, Q, and find the field.
4. Solve Laplace's equation and find the field due to an infinite
uniformly charged plane.
5. Prove that two-dimensional lines of force also satisfy
Laplace's equation.
6. Show that rp = (A cos nx + B sin nx) (Cell + De-Av) satisfies
x+
a2 $
= 0.
49Y2
7. If Cpl and S02 satisfy Laplace's equation, show that cpl + 4o
and Ipl -- rp2 satisfy Laplace's equation. Does cl92 satisfy
Laplace's equation?
8. If (pi satisfies Laplace's equation and cp2 satisfies Poisson's
equation, show that cpi + 4p2 satisfies Poisson's equation.
60. Dielectrics. If charges reside in a medium other than a
vacuum, it is found that the inverse-square force needs readjustment. That this is reasonable can be seen from the following
136
VECTOR AND TENSOR ANALYSIS
[SEC. 61
considerations. We consider a parallel-plate condenser separated by glass (Fig. 59). Assuming that the molecular structure
/
of glass consists of positive
and negative particles, the
electrons being bound to the
.+++++++++++.....++++
0/0
---
Glass
-----j/
nucleus, we see that the field
due to the oppositely charged
plates might well cause a dis-
Fm. 59
placement of the electrons
away from the negative plate and toward the positive plate.
This tends to weaken the field, so that E = 47rv/x, where x > 1.
x is called the dielectric constant.
It is found experimentally that E -_ (qq'/Kr9)r for charges in a
dielectric. Applying this force, we see that Gauss's law is modified to read
fjE
.d
d=
4 Q
(213)
K
and if x is a constant,
f
f
41rQ
(214)
where D is defined as the displacement vector, D = xE _ -x V.
Poisson's equation becomes V D = -V (K Vg) = 4rp, and for
constantx
4rp
(215)
x
For p = 0 we still have Laplace's equation V2(p = 0.
In the most general case, we have
3
D; = Z x;;E;,
i=i
i = 1, 2, 3
where D = Dli + D2j + Dak, E = E,i + E2j + Eak, and x;; = K;,.
61. Energy of the Electrostatic Field. . Let us bring charges
q,, q2, . . . , q,. from infinity to positions P1, P2, . . . , P,., and
calculate the work done in bringing about this distribution. It
takes no work to bring q, to P,, since there is no field. To bring
q2 to P2, work must be done against the field set up by Q1. This
amount of work is glg2/rl2, where r12 is the distance between Pl
and P2. In bringing q3 to P3, we do work against the separate
STATIC AND DYNAMIC ELECTRICITY
SEC. 61]
137
fields due to q, and q2. This work is gig3/r13 and g2g3/r23.
continue this process and obtain for the total work
=- -
We
n q;q'i
W
(216)
ri,
The J occurs because gig2/r12 occurs twice in the summation
process, once as glg2/r12 and again as g2g1/r21. The quantity W
n
is called the electrostatic energy of the field.
Since (pi =
n
we have W =
gipi.
i=1
;al
q;/rii,
For a continuous distribution of charge,
we replace the summation by an integral, so that
W= jfff pv dr
(217)
Now assume that all the charges are contained in some finite
We have V D = 41rp so that
sphere.
W
fff
fff
8- f f f
Applying the divergence theorem,
W
ff
87 S
ff
v
Now p,is of the order of 1/r for large r, and D is of the order of
1/r2, while do is of the order of r2. We may take our volume
of integration as large as we please, since p = 0 outside a fixed
sphere.
Hence lim f f cpD dd = 0, so that
s
w= 8A f f f (E D) dr
(218)
138
VECTOR AND TENSOR ANALYSIS
[SEC. 62
The energy density is w = (1/8ir)E D. For an isotropic
medium, D= KE and W = (1/87r) Jff KE2 dr.
Example 82. Let us compute the energy if our space contains
a charge q distributed uniformly over the surface of a sphere of
radius a. We have
=a
D = E = qr,r>>
r
and
D=E=O,r<a
The total energy is
22x x
W=s-
ffJ-sin O drdOdcp
q2
2a
62. Discontinuities of D and E at the Boundary of Two Dielectrics. Let S be the surface of discontinuity between two media
with dielectric constant K1 and K2. We apply Gauss's law to a
pillbox with a face in each medium (Fig. 60). Assuming no
charges exist on the surface of
K1
discontinuity, we have
K2
so that
n2=_nl
Si nce nl
D2 n2 = 0.
= - n2, we have
DN, = DN,
(219)
FIG. 60.
We have taken the pillbox very
flat so that the sides contribute a negligible amount to the flux.
Equation (219) states that the normal component of the displacement vector D is continuous across a surface of discontinuity
containing no charges.
We next consider a closed curve r with sides parallel to the
surface of discontinuity and ends negligible in size (Fig. 61).
Since the field is conservative,
ft dr = 0
or
Er1 = Er1
(220)
SEC. 63]
STATIC AND DYNAMIC ELECTRICITY
139
In other words, the tangential component of the electric vector
E is continuous across a surface of discontinuity. Combining
(219) and (220), we have
DN,
DN,
ET,
E,,
K1EN,
K2EN,
ET,
ET,
or
Fio. 62.
for isotropic media.
Hence
tan01_Kl
tan 02
(221)
K2
which is the law of refraction (see Fig. 62).
63. Green's Reciprocity Theorem. Let us consider any distribution of volume and surface charges, the surfaces being conductors. Let p be the volume density and a the surface density.
If p is the potential function for this distribution of charges, then
02p = -4up. We shall make use of the fact that E = -Vp and
that at the surface E. = 4arv, or E . dd = -4uo dS.
A new distribution of charges would yield a new potential
function cp' such that V2sp' = -41ro'. Our problem is to find
VECTOR AND TENSOR ANALYSIS
140
1SEC. 63
a relationship between the fundamental quantities p, o, p of the
old distribution and p', o', of the new distribution. To do so,
we apply Green's formula
JJJ
(vV2V
- ip'V',p)
JJ(cV,' -
v v) dd
S
V
which reduces to
-4v ii (,Pp '-V 'p)dr=4a f f('-rp'o)dS
s
V
or
f if rpp' dr +
ff
' ds = f f f v p dr +
jJ
dS
p
(222)
This is Green's reciprocity theorem. It states that the potential (p of a given distribution when multiplied by the corresponding charge (p', a,) in the new distribution and then summed over
all of the space is equal to the sum of the products of the potentials (pp') in the new distribution by the charges (p, o) in the old
distribution, that is, a reciprocal property prevails.
Example 83. Let a sphere of radius a be grounded, that is, its
potential is zero, and place a charge q at a point P, b units from
the center of the sphere, b > a. The charge q will induce a
charge Q on the sphere. We desire to find Q. We construct a
new distribution as follows: Place a unit charge on the sphere, and
assume no other charges in space. The potential due to this
charged sphere is p' = 1/r. For the sphere we have initially
= 0, Q = ?, and afterward, cp' = 1/a, q' = 1. For the point P
we have initially pp = ?, q = q, and afterward, V = 1/b, q' = 0.
Applying the reciprocity theorem, we have
0.1+vp-0Q'+q
a
b
so that Q = - (alb)q. This is the total charge induced on the
sphere when it is grounded. Note that this method does not
tell us the surface distribution of the induced charge.
.
Problems
1. A conducting sphere of radius a is embedded in the center
of a sphere of radius b and dielectric constant K. The conductor
SFc.64]
STATIC AND DYNAMIC ELECTRICITY
141
is grounded, and a point charge q is placed at a distance r from
its center, r > b > a. Show that the charge induced on the
sphere is Q = -Kabq{r[b + (K - 1)a])-1.
2. A pair of concentric conductors of radii a and b are connected by a wire. A point charge q is detached from the inner
one and moved radially with uniform speed V to the outer one.
Show that the rate of transfer of the induced charge (due to q)
from the inner to the outer sphere is
dQ
dt
= -gab(b - a)-'V(a + Vt)-2
3. A spherical condenser with inner radius a and outer radius b
is filled with two spherical layers of dielectrics Kl and K2, the
boundary between being given by r = J(a + b). If, when both
shells are earthed, a point charge on the dielectric boundary
induces equal charges on the inner and outer shells, show that
K1/K2 = b/a.
4. A conductor has a charge e, and V1, V2 are the potentials
of two equipotential surfaces which completely surround it
(V1 > V2). The space between these two surfaces is now filled
with a dielectric of inductive capacity K. Show that the change
in the energy of the system is - e(V l - V2) (K - 1 )K 1.
5. The inner sphere of a spherical condenser (radii a, b) has a
constant charge E, and the outer conductor is at zero potential.
Under the internal forces, the outer conductor contracts from
radius b to radius b1. Prove that the work done by the electric
forces is E2(b - b1)b-'b1-1.
64. Method of Images. We consider a charge q placed at a
point P(b, 0, 0) and ask if it is possible to find a point Q(z, 0, 0)
such that a certain charge q' at Q will cause the potential over
the sphere x2 + y2 + z2 = a2, a < b, to vanish. The answer is
"Yes"! We proceed as follows : From Fig. 63 we have
82 = z2 + a2 - 2az cos B
t== b2+a2-2abcos6
We choose z so that zb = a2, and call Q(a2/b, 0, 0) the image
point of P(b, 0, 0) with respect to the sphere. Thus
a2
s2 = a (a2 + b2 - 2ab cos 0) =
a2
b2
0
VECTOR AND TENSOR ANALYSIS
142
[SEC. 64
and
The potential at S due to charges q and q' at P and Q is
P=
s,-I- q
t
=-t Cbq'+qJ
and p = 0 if we choose q' _ - (alb)q.
FIG. 63.
The potential at any point R with spherical coordinates r, 0, (p
is
_
(a/b)q
q
(r2 + b2 - 2rb cos 0)#
[r2
+ (a4/b2) - (2a2/b)r cos 0]1
(223)
with = 0 on S and V24 = 0 where no charges exist.
Now let us consider the sphere of Example 83. The function
of (223) satisfies Laplace's equation and is zero on the sphere.
From the uniqueness theorem of Example 69, F of (223) is the
potential function for the problem of Example 83. The radial
field is given by
Er
8
q(r-bcos9)
or
(r2 + b2 - 2rb cos B)'
(a/b)q[r - (a2/b) cos 0]
[r2 + (a4/b2) - (2a2/b)r cos B];
SEC. 651
STATIC AND DYNAMIC ELECTRICITY
143
and the surface distribution is given by
_ (Er)rs _
b2 - a2
q
4ir a(a2 + b2 - 2ab cos 9)'
47r
Problems
1. A charge q is placed at a distance a from an infinite grounded
plane. Find the image point, the field, and the induced surface
density.
2. Two semiinfinite grounded planes intersect at right angles.
A charge q is placed on the bisector of the planes. What distribution of charges is equivalent to this system? Find the field
and the surface distribution induced on the planes.
3. An infinite plate with a hemispherical boss of radius a is
at zero potential under the influence of a point charge q on the
axis of the boss at a distance f from the plate. Find the surface
density at any point of the plate, and show that the charge is
attracted toward the plate with a force
q2
4f2
4g2a3 fa
+
(f' - a')2
65. Conjugate Harmonic Functions. If we are dealing with
a two-dimensional problem in electrostatics, we look for a solution of Laplace's equation V2V = 0. The curves V (x, y) = con-
stant represent the equipotential lines. We know that these
curves are orthogonal to the lines of force, so that the conjugate
function U(x, y) (see Sec. 56) will represent the lines of force.
We know that V2V = V2U = 0.
We now give an example of the use of conjugate harmonic
functions. In Example 73 we saw that
U(x, y) = A tan-' y
x
(224)
V (X, y) = 2 log (x2 + y2)
are conjugate functions satisfying Laplace's equation. If we
take V (x, y) as the potential function, then the equipotentials
are the circles (A/2) log (x2 + y2) = C, or x2 + y2 = e2cie.
144
VECTOR AND TENSOR ANALYSIS
[SEC. 65
Hence the potential due to an infinite charged conducting cylinder
is
log r2 = A log r, r2 = x2 + y2
V (X, y) = 2 log (x2 + y2) =
since A log r satisfies Laplace's2equation and satisfies the bound-
ary condition that V = constant for r = a, the radius of the
charged cylinder. If q is the charge per unit length, then
_ aV
q
(Er)r..a
ar
24ra
41r
41r
r-a
A
4ara
so that A = -2q and V = -2q log r.
1e=Jr)
0
U=U1
U=U0 (0=0)
Frs. 64.
If we choose U(x, y) = A tan-1 (y/x) as our potential function, then the equipotentials U = constant are the straight lines
A tan-1 (y/x) = C, or y = x tan (C/A). As a special case we
may take the straight lines 0 = 0, 0 = v as conducting planes
raised to different potentials (see Fig. 64). The lines of force
are the circles (A/2) log (x2 + y2) = V.
The theory of conjugate functions belongs properly to the
theory of functions of a complex variable. With the aid of the
Schwarz transformation it is possible to find the conjugate functions associated with more difficult problems involving the twodimensional Laplace equation.
Problems
1. By considering Example 72, find the potential function
and lines of force for two semiinfinite planes intersecting at right
angles.
STATIC AND DYNAMIC ELECTRICITY
SEC. 66J
145
2. What physical problems can be solved by the transformation
x = a cosh U cos V, y = a sinh U sin V? Show that
V2U = V2V = O
66. Integration of Laplace's Equation. Let S be the surface
of a region R for which V24p = 0. Let P be any point of R, and
let r be the distance from P to any point of the surface S. We
make use of Green's formula
fJf (,p 02' - ¢ V2V) dr = f f (9 Vi'
DAP)
dd
R
We choose ¢ = 1/r, and this produces a discontinuity inside
R, namely, at P, where r = 0. In order to overcome this difficulty, we proceed as in Example 66. Surround P by a sphere
Z of radius e. Using the fact that V2,P = V24, = 0 inside R'
(R minus the Z sphere), we obtain
0=
f8 f (9 V r
r
V-r) - dd +
f f (p V 1r - r Vp) - dd
(225)
//
B
//
Now V(1/r) = -r/r', and on the sphere X,
r
r8
r
e$
and (1/r) VV dd is of the order eIV(pf, so that by letting e --j 0,
(225) reduces to
v(p) = 4
ffir
VV - sa V
r)
dd
(226)
This remarkable formula states that the value of Sp at any point
P is determined by the value of (p and V(p on the surface S.
Problems
f
1. If P is any point outside the closed surface S, show that
f
V(1/r)] dd = 0, where V2cp = 0 inside S and
r is the distance from P to any point of S.
VECTOR AND TENSOR ANALYSIS
146
a2
2. Let rp satisfy V2cp =
aX2 +
[SEC. 67
2
2 = 0. Let r be the closed
1P
y°
boundary of a simply connected region in the x-y plane. If P
is an interior point of t, show that
a[log (1/r)]
1) app
1
,P(P) - 2x
`0 an
1(`log r an
ds
where use is made of the fact that
ff
(u an -van) da
`
(u V 2V - v V 2U) dA
A
n being the normal to the curve.
3. Let (p be harmonic outside the closed surface S and assume
that ip --> 0 and rI V pl -> 0 as r -' oo. If P is a point outside 8,
show that
P(P) = 4x Ifs (1r V p - ip v
T
dd
where the normal dd is inward on S.
4. Let so be harmonic and regular inside sphere 7. Show that
the value of p at the center of is the average of its values over
the surface of the sphere. Use (226).
67. Solution of Laplace's Equation in Spherical Coordinates.
From Sec. 23, Prob. 1,
+ aB (sin
+
49(p
o
(sin
0
(227)
B
To solve (227), we assume a solution of the form
(228)
V(r, 0, gyp) = R(r)6(0),P(,p)
Substituting (228) into (227) and dividing by V, we obtain
sin 0
d8)
d( r2
dR)
- J +-A (sin O
-+
R dr
1
dr
0 d9
d8
1
d2
-- =
4) sin 0 d(p2
0
STATIC AND DYNAMIC ELECTRICITY
SEc. 67j
147
Consequently
_-
_1 d
dR\
r
R dr
dr
2
d
1
0 sin 0 dB
sin
0
d9
d9
1
d4
- sin 2 0 4, dp2
(229)
The left-hand side of (229) depends only on r, while the righthand side of (229) depends on 0 and gyp. This is possible only
if both quantities are constant, for on differentiating (229) with
respect to r, we obtain
d
d (r2
LR
i
J
= 0. We choose as
the constant of integration c = -n(n + 1), so that
R dr (r2 dR)
n(n + 1)
or
r2
4+24+n(n+1)R=0
(230)
It is easy to integrate (230), and we leave it to the reader to show
that R = Ar" + Br-n-1 is the most general solution of (230).
Returning to (229), we have
1d
'e
= n(n -{- 1) sing 0 --
sin
(sin 0 de)
(231)
d0
41)
Since we have again separated the variables, both sides of
(231) are constant. We choose the constant to be negative,
-m2, m being an integer. This choice guarantees that the solution of
d24
d(p2
-I-mq =0
is single-valued when p is increased by tar.
(232)
The solution of
(232) is 4) = A cos mtp + B sin mcp.
Finally, we obtain that 0(0) satisfies
sin 0 d0 (sino)+[n(n+1)sin2o_m2Je=o (233)
We make a change of variable by letting µ = cos 0,
dµ= -sin0d0
VECTOR AND TENSOR ANALYSIS
148
[SEC. 67
so that (233) becomes
[(1 - µ2)
(1 - µ2)
+ [(1
µ2)n(n + 1) - m2]0
dµ
= 0
(234)
If we assume that V is independent of p (symmetry about the
z axis), we have m = 0, so that (234) becomes
(235)
d
dIA
This is Legendre's differential equation.
By the method of series solution, it can be shown that
0 = P"(14)
1
d"(µ2 - 1)n
2 -n!
dµn
satisfies (235); the P (µ) are called Legendre polynomials.
Two important properties of Legendre polynomials are
the
following:
f 11 P,n(;&)Pn(p) dµ
=0
if m : n
J 11 P.'(µ) dµ = 2n + 1
(236)
(237)
We give a proof of (236). P. and P. satisfy
[(1
dµ
- p!) dµn1 + n(n + 1)Pn = 0
(238)
(239)
Multiplying (238) by P. and (239) by P. and subtracting, we
obtain
P. d- [(1 - 2) a ,, - Pn d [(1
_ u2) d µ"'
+[n(n+1) -m(m+l)PPm=0
SFC. 671
STATIC AND DYNAMIC ELECTRICITY
149
or
-M2)(Pmddn-PndPm)]
µ[(1
+ [n(n + 1) - m(m + 1)]P,aPm = 0 (240;
Integrating between the limits -1 and + 1, we obtain
[n(n + 1) - m(m + 1)J f i P,nPn dµ = 0
and ifm -d n,
1
PmPn dµ = 0
A particular solution of (227) which is independent of gyp, that
is, aV = 0, is given by V(r, 0) = (Anrn + Bnr-' 1)Pn(cos 0).
aip
Now it is easy to show that any sum of solutions of (227) is
also a solution, since (227) is linear in V. Consequently a more
general solution is
V=
n-0
(Anrn + Bn7rn-1)Pn(cos 0)
(241)
provided that the series converges.
If we wish to solve a problem involving V2V = 0 with spherical
boundaries, we try (241) as our solution. If we can find the
constants An, B. so that the boundary conditions are fulfilled,
then (241) will represent the only solution, from our previous
uniqueness theorems involving Laplace's equation.
We list a few Legendre polynomials:
Po(µ) = 1
Pi(µ) = A
P2(p) = . (3µs - 1)
P,(µ) = 4(5Aa - 3µ)
P,(µ) = 9(35."4 - 30u$ + 3)
P.(0) = 0, n odd
P-(0) = (_1)n/x 1-3-5 ... (n - 1)
2.4.6 ... n
Pn(1) = 1
Pn(-µ) _ (-1)"Pn(µ)
(242)
uneven
VECTOR AND TENSOR ANALYSIS
150
[SEC. 68
Problems
1. Prove (237).
2. Solve V2V = 0 for rectangular coordinates by the method
of Sec. 67, assuming V = X(x)Y(y)Z(z).
3. Investigate the solution of VI V = 0 in cylindrical coordinates.
68. Applications
Example 84. A dielectric sphere of radius a is placed in a uni-
form field Eo = Eok. We calculate the field inside the sphere.
The potential due to the uniform field is p = -Eoz = -Eor cos 0.
There will be an additional potential due to the presence of the
dielectric sphere. Assume it to be of the form ArPI = Ar cos 0
inside the sphere and Br-2PI ° Br-2 cos 0 outside the sphere.
We cannot have a term of the type Cr-2 cos 0 inside the sphere,
for at the origin we would have an infinite field caused by the
presence of the dielectric. Similarly, if a term of the type
Dr cos a occurred outside the sphere, we would have an infinite
field at infinity due to the presence of the sphere. If we let VI
be the potential inside and V11 the potential outside the sphere, we
have
VI = --Eor cos 0 + Ar cos 0
(243)
Vu = --Eor cos B + B cos 0
r2
Notice that VI and Vn are special cases of (241). We have two
unknown constants, A, B, and two boundary conditions,
VI= Vu at r=a
or
DN, = D.Y.
(see Sec. 62).
a
I
K
=
II
a
at r = a
(244)
From (243) and (244) we obtain
A=K-'Eo,
x+2
B_aaK-1Eo
K+ 2
so that
VI = I
-'V+2 Eor cos 0 = -
2 Eoz
K
+
(245)
STATIC AND DYNAMIC ELECTRICITY
SEC. 681
151
We see that the field inside the dielectric sphere is
E_ -vVI=K+2Eo
and E is uniform of intensity less than E0 since K > 1.
the sphere
VII = -Eor cos B +
K - 1 _ Eo
cos 8
K + 2 r2
Outside
(246)
The radial field outside the sphere is given by
aVII=Eocos0+2K +
E,
ocos9
2aa
For a given r the maximum E, is found at 0 = 0.
Example 85. A conducting sphere of radius a and
charge Q is surrounded by a
spherical dielectric layer up
to r = b (Fig. 65). Let us
calculate the potential distribution.
From
spherical
symmetry V = V (r), so that
we try
FlG. 65.
The boundary conditions are
VI= VIIatr=b
(i)
aVI
_
aVII
at r = b
Or = Or
J2rJT (a
D dd = K
(11)
(iii)
Q=
41r
ff
s
47
T a2 sin 0 d9 dp
J
From (i) Alb = (B/b) + C; from (ii) -A/b2 = -KB/b2; from
(iii) Q = (K/4x) (B/a2) f f dS = KB.
Hence
VECTOR AND TENSOR ANALYSIS
152
VI = Q'
VII =
Q +90C
1
(SEC. 68
(247)
K
Example 86. A conducting sphere of radius a and charge Q is
placed in a uniform field. We calculate the potential and the
distribution of charge on the sphere. We assume a solution
V = -Eor cos 0 +
B
r
The boundary condition is
Q = 4v
Jf - avlr-a dS
so that
Q
=
1
4
,
f2v
f-
(E cos 8 + a a2 sin 8 d8 dsp = B
and
V= -Eorcos8+Qr
For the charge distribution
aV
a-
cIr
4"
4I(E0cos8+Q
Example 87. Consider a charge q placed at A (b, 0, 0). Let
us compute the potential at any point P(r, 8, gyp) (see Fig. 66).
The potential at P is
4
V=
= q(r2 + b$ - 2rb cos
8)_;
There are two cases to consider:
(a) r < b. Let µ = cos 8, x = r/b, so that
(1 - 2µx
V=
b
+2)-}
Now (1 - 2µx + x=)-} can be expanded in a Maclaurin series in
powers of x, yielding
SEC. 68]
STATIC AND DYNAMIC ELECTRICITY
_
m
V=b
n= b
n0
I
153
m
()fl
n0
P.(--)
( 248)
The proof is omitted here that the Pn(p) are actually the Legendre
polynomials. However, we might expect this, since V satisfies
Laplace's equation and P,,(µ)rn is a solution of VI V = 0.
z
Fia. 66.
(b) r > b. In this case
V = q I Pn6u)
rn=0
/b n
(249)
r
Notice that each term is of the form Pn(µ)r-n-1, which satisfies
Laplace's equation.
Example 88. A point charge +q is placed at a distance b
from the center of two concentric, earthed, conducting spheres
of radii a and c, a < b < c. We find the potential at a point P
for a <r <b.
(b/r)'P,,(cos 0) due to the charge q;
For r > b, V = (q/r)
0
and for r < b, V = (q/b) 2 (r/b)'Pn(cos 0). Moreover, we have
0
VECTOR AND TENSOR ANALYSIS
154
[SEC. 68
an induced potential of the form
V = q I (Anrn + Bnr n-1)Pn(cos 0)
(250)
0
which is due to the spheres, the An and Bn undetermined as yet.
Hence
For r > b:
Vl = q I [Anrn + (Bn + bn)r-n-1]Pn(cos 6)
0
(251)
For r < b:
[(A,, +
V2 = q
b-n-I)rn +
Bnr-n-1]Pn(cos 0
0
The boundary conditions are
V1=0atr=c
(i)
(ii)
(252)
V2 = O at r = a
These yield the equations
Ancn + (Bu + bn)c-n-1 = 0
(An + b-n-1)an + Bna n-1 = 0
(i)
(11)
so that
a2n+1 (C2n+l
Bn T bn+l (a2n+l
-- b2n+l)'
C2n+1)'
A. + b-(n+1) _
b-n-1(C2n+1
-
b2n+1)
a2n+1 - C2n+l
Hence
j
m
V2(P) = q
n
-0
b2n+1 - C2n+1
bn+l(a2n+l - C2n+1)
(rn
a2n+1
- rn+1 Pn(cos 0)
(253)
Problems
1. Show that the force acting on the sphere of Example 86 is
F = kQEok.
2. A charge q is placed at a distance c from the center of a
spherical hollow of radius a in an infinite dielectric of constant K.
Show that the force acting on the charge is
STATIC AND DYNAMIC ELECTRICITY
SEC. 691
155
n(n + 1)
2n+1
,410n+K(n+ 1)(a/
(K - 1)g2''
c2
3. A point charge q is placed a distance c from the center of an
earthed conducting sphere of radius a, on which a dielectric
layer of outer radius b and constant K exists. Show that the
potential of this layer is
m
V
=
(2n + 1 )b2n+1(rn - a2n+lr-n-1)
q
-c n c cn { [(K +1)n + 1]b 2n+1 + (n +1)(K -1)a 2n+1 } Pn(cos e)
0
4. Show that the potential inside a dielectric shell of internal
and external radii a and b, placed in a uniform field of strength E,
is
V=
9KE
9K - 2(1 - K2)[(b/a)3 - 1]
r cos 0
5. The walls of an earthed rectangular conducting tube of
infinite length are given by x = 0, x = a, y = 0, y = b. A point
charge is placed at x = xo, y = yo, z = zo inside the tube. Show
that the potential is given by
V = 8q
I (m2a2 +
nil m-1
n2b2)-e-a-'b-'(mta'+n$t)i*(s-ss)
to
sin
n1rx0
a
sin
nrx
a
sin
mTryo
b
sin
miry
b
69. Integration of Poisson's Equation. Instead of assuming
that p is harmonic, let us consider that 1p satisfies V2co = -47rp.
By applying Green's formula as in Sec. 66, we immediately obtain
,P(P) = Jff rdr+Jf (rVp
- jP 0
(254)
If we make the further assumption that rp is of the order of 1/r
for large r and that jVjpj - 1/r2, we see that by pushing S out to
infinity the surface integral will tend to zero. Our assumption is
valid, for if we assume the charge distribution to be bounded by
some sphere, then at large distances the potential will be of the
order of I /r, since we may consider all the charges as essentially
156
VECTOR AND TENSOR ANALYSIS
[SEc. 70
concentrated at a point. Thus
V(P) = JJf e dr
(255)
70. Decomposition of a Vector into the Sum of Solenoidal and
Irrotational Vectors. In Example 70, we saw that if Ifl tends to
zero like 1/r2 as r --> oo, then f as uniquely determined by its
curl and divergence.
We now proceed to write f as the sum of irrotational and
solenoidal vectors. Let
f dr
W(P) = JfJ
(256)
where r is the distance from P to the element of integration dr.
If we write f = fli + f j + f k, W = WA + W,,j + Wek, then
f
ffrldr
W2 = f f f
f2
dr
(257)
M
W3=fffr$dr
00
We assume that the components of f are such that the integrals
of (257) converge and that I Wl - 1/r, I V W,,l '' 1/r2, n = 1, 2, 3.
From Sec. 69, V2W. = -4xfn, so that
V2W = -4,rf
(258)
From (256)
(259)
VxW= JJJf xVdr
40
Now
V x(V xW) =
--V2W
SEc. 711
STATIC AND DYNAMIC ELECTRICITY
157
so that
f
=4-V
-4
X
X
and hence
f = V x A + Vcp
260)
where
A=4-VxW, =
Problems
1. Show that (256) is a special case of (254).
2. Find an expression for (p(P) if V2(p = -41rp inside S and if
P is on the surface S.
3. f = yzi + xzj + (xy - xz)k. Express f as the sum of an
irrotational and a solenoidal vector.
71. Dipoles. Let us consider two neighboring charges -q and
+q situated at P(x, y, z) and Q(x + dx, y, z). The potential at
the origin 0(0, 0, 0) due to -q is -q/r, and that due to +q is
q/(r + dr), where r = (x2 + y2 + Z2)I and
r+dr= [(x + dx)2 + y2 + Z2]1
The potential at 0(0, 0, 0) due to both charges is
q
r + dr
q
r
g2dr
r2
Now dr = x dx/r, so that rp - qx dx/r3. If we now let q -p co
and dx -+ 0 in such a way that q dx remains finite, we have formed
what is known as a dipole. Let r be the position vector from the
origin to the dipole, and let M = q dr, where dr is the vector from
the negative charge to the positive charge; !drl = dx. We have
(M r)/r8 = (qr dr)/ra = (qx dx)/r3, so that
M is called the strength or moment of the dipole.
than one dipole, the potential at a point P is given by
For more
VECTOR AND TENSOR ANALYSIS
158
i= i
[SEC. 72
(Mi.ri)
ri3
where r; is the vector from P to the dipole having strength M.
Example 89. The field strength due to a dipole is E _ --Vv
so that
r
(262)
E = V rs )
rs
ra
-
-
Example 90.
Potential energy of a dipole in a field of potential V.
Let (p, be the potential at the charge q and c2 the potential at
the charge -q. The energy of the dipole is
L
app
ds = M
W = spiq + IP2(-q) = q(spl - IP2) = g
as
ao
where ds is the distance between the charges. Now
d,p = ds V p
so that W = M as VV = M V(p.
72. Electric Polarization. Let us consider a volume filled
with dipoles. The potential due to any single dipole is given by
(261). If we let P be the dipole moment per unit volume, that is,
P = lim (AM/or), then the total potential due to the dipoles is
A"o
= I f f r ra- dr
(263)
Now V V. (P/r) = (1/r)V P - [(P r)/r3]. The reason that we
have taken V(1/r) t= r/r3 instead of -r/r3 is that
r = [( - x)2 + (n - y)2 + (J
- z)2]'
and V performs the differentiations with respect to x, y, z, the
coordinates of the point P at which V is being evaluated. The
coordinates t, ot, t belong to the region R and are the variables of
integration, and r = (E - x)i + (n - y)j + (r - z)k.
(263) becomes
`P=fffv.(!)dr
R
fff
R
Hence
SEC. 72]
STATIC AND DYNAMIC ELECTRICITY
159
and
(p -
ff!.dd_ fff_!d
r
(264)
by applying the divergence theorem.
Example 91. Let us find the electric intensity at the center
of a uniformly polarized sphere. Here P = Pok, so that V P = 0
inside R. Hence (264) becomes
sp(x, y, z) =
Pok dd
ff
x)2 + (71 - y)2 + ( - z) 2]i
S
and
E=
fj
Po(k
dd)[(E - x)i + (n
- y)j + (1' - z)k]
[(E-x)2+(7] -y)2+(J -z)2]$
S
E(0,0,0)
f
f
ni + 'k)Po(k dd)
(265)
Now for points on the sphere,
52 +' 72+-2 = a2,
and letting t = a sin B cos -o, n = a sin 0 sin tp,
is easily seen that (265) reduces to
E(0, 0, 0) = -frPok
a cos 0, it
(266)
E is independent of the radius of the sphere. By superimposing
(concentrically) a sphere with an equal but negative polarization, we see that the field at the center of a uniformly polarized
shell is zero.
Problems
1. Prove (262).
2. Prove (266).
3. If M1 and M2 are the vector moments of two dipoles at A
and B, and if r is the vector from A to B, show that the energy
- 3(M1 r)(M2 r)r-6.
of the system is W = M1
4. The dipole-moment density is given by P = r over a sphere
of radius a. Calculate the field at the center of the sphere.
Mgr-a
160
VECTOR AND TENSOR ANALYSIS
[SEc. 73
73. Magnetostatics. The same laws that have held for electrostatics are true for magnetostatics with the exception that
OZcp.. = 0 always, since we cannot isolate a magnetic charge.
We make the following correspondences, since all the laws of
electrostatics were derived on the assumption of the inversesquare force law, which applies equally well for stationary
magnets.
EH
Magnetostatics
Electrostatics
q
qm
D ---) B (magnetic induction)
K <---- u (permeability)
D = KE F----- B = µH
(267)
0
74. Solid Angle. Let r be the position vector from a point P
N to a surface of area dS and unit normal N, that
is, do = N dS. We define the solid angle subtended at P by the surface dS to be (see Fig.
67)
dct =
r3
P
Fia. 67.
The total solid angle of a surface is
J'r.do
r
S
3
Example 92.
(268)
Let S be a sphere and P the origin so that
12(P)=
4rdS=4u
r
fJr
Example 93. The magnetic dipole is the exact analogue of the
electric dipole. We consider a magnetic shell, that is, a thin
sheet magnetized uniformly in a direction normal to its surface
(Fig. 68). Let $ be the magnetic moment per unit area and
STATIC AND DYNAMIC ELECTRICITY
SEC. 75]
161
assume S = constant. The potential at P is given by
sff r,MdS=13 f f rras=On
Now let P and Q be opposite points on
the negative and positive sides of the
surface S.
We have
,p(Q) = -P(4ir - 11)
so that the work done in taking a unit
positive pole from a point P on the negative side of the shell to a point Q on the
positive side of the shell is given by
P
Fur. 68.
W = f," H H. dr = - f,' Vp . dr = *(P) - F(Q)
=6U+$(4,r- 9)
W = 4x-8
(269)
75. Moving Charges, or Currents. If two conductors at different potentials are joined together by a metal wire, it is found that
certain phenomena occur (heating of the wire, magnetic field), so
that one is led to believe that a flow of charge is taking place.
Let v be the velocity of the charge and p the density of charge.
We define current density by j = pv. The total charge passing
through a surface per unit time is given by
ffpv.do_ Jfj.dd
Now the total charge inside a closed surface S is Q = f f f p dr.
R
If there are no sources or sinks inside S, then the loss of charge
per unit time is given by - aQ = -
ffpv.dd=-
f f at dr.
R
dr
Thus
162
VECTOR AND TENSOR ANALYSIS
[SEC. 76
Applying the divergence theorem, we have
0
(270)
or
at
=
0
Equations (270) are the statement of conservation of electric
charge.
We define a steady state as one for which p is independ-
ent of the time, a = 0, which implies V j = 0.
It has been found by experiment that if E is the electric field,
then
j = XE = -A VM
(271)
where X is the conductivity of the metal. This is Ohm's law.
a
For the general case, j, = I Xa,gE#, and the simplest case,
69=1
X = constant, so that V2(p = 0 for the steady state.
We now compute the work done on a charge q as it moves
The
from a point of potential p, to one of potential 02, Ip2 > 92.
energy at (pi is qpl and at V2 is q(p2. The loss in energy is
W = (1v1 - Iv2)q
This loss in electrical energy does not go into mechanical energy,
since the flow is assumed steady. Hence the electrical energy
is converted into heat, Q = (,p1 - p2)q. The power loss is
P = dt
= (1P1 - (P2) dq,
and since 01 - 02 = RJ (another form
of Ohm's law, where R is resistance and J current), we have
P=RJ2
(272)
76. Magnetic Effect of Currents (Oersted). Experiments
show that electric currents produce magnetic fields. The
mathematical expression for the magnetic field is given by
dH =
Jr x dr
(273)
cra
STATIC AND DYNAMIC ELECTRICITY
SEC. 76]
163
where r is the vector from the point P; P is the point at which we
calculate the magnetic field dH due to the line current J in that
portion of the wire dr, and c
is a constant, the ratio of the
electrostatic to the electromagnetic unit of charge (see
Fig. 69).
Q (tn'r)
Biot and Savart
established this law for
straight-line currents.
For a closed path
Jrxdr
H=
(274)
P (x,y, z)
cra
Fra. 69.
- x)2 + (n - y)2 + ( - z)2];, and V(1/r) = r/ra,
+ j a + k az. Hence
where V = i
clx
Now r = [(E
y
H=1fiJVI
c
r
x(Jdr
r
c
since V does not operate on dr and J is a constant. Thus
H = V x A, where A = (1/c) J dr/r = (11c) f f f i dr/r is
integrated over all space containing currents.
Now V A = fif V (j/r) dr
f f (j/r)
s
dd, so that if all
currents lie within a given sphere, we may push the boundary
of R to infinity, since nothing new will be added to the integral
yielding A. But when S is expanded to a great distance, j = 0
on S, so that that V A = 0. Also
V x H = -VIA. Now since A = (1/c)
f f f j dr/r, or
0
A. = (1 /C)
f f f is dr/r, A. =
As
(1 /c)
f f f j dr/r,
is dr
r
fff
164
VECTOR AND TENSOR ANALYSIS
we have from Sec. 69 that V2A = -- (47r/c)j.
[SEC. 76
Thus
VxH=47j
c
(275)
Example 94. The work done in taking a unit magnetic pole
around a closed path r in a magnetic field due to electric currents
is
ff
c
S
ff
s
For an electric current J in a wire that loops r, we have
(276)
Example 95. The magnetic field at a point P, r units away
from an infinite straight-line wire carrying a current J, is obtained
by use of Example 94.
H dr = H(2irr) =
4a
c
J
so that
H=
27
cr
We compute the dimensions of c//. Now
Example 96.
f. = qaq,'/Kr2, and f. = qqm'/µr2 so that
[M][L]
[g612
[q+x]2
[712
[K][L]2
[µ][L]2
and
[J]
[dq,/dt]
[c]
[c]
[g]
[c][T]
[M14[L][K]}
[c][T]2
F rom (276) ,
Work
Unit pole
=
H
,
dr
c
so that
[M][L]2 _ [J]
[gm][T]2
= 4'rJ
[c]
SEc. 77]
STATIC AND DYNAMIC ELECTRICITY
165
and
[Ml}[L];
[M]}[Ll'[K]i
[7'][µl}
[c][TJ2
yielding
L/i
[TL]
We see that c// has the dimensions of speed. We shall soon
see the significance of this.
(2)
(1)
Fca. 70.
77. Mutual Induction and Action of Two Circuits.
Consider
two closed circuits with currents J, and J2 (Fig. 70). The
magnetic field at 0 due to J, is H, = V x A, where
A,=J, r dr
c
(1)r
We define the mutual inductance of the two circuits as the mag-
netic flux through the surface B due to a unit current in (1).
This is
IM=
f f L da= f f
a
a
1(2) A1. dr = c
1(2)
(1(1)
drl)
dr2
Hence
M=
dr, dr,
1
C
f(2)J(1)
r
(277)
The current element J2 dr2 seta up a magnetic field, so that
from Newton's third law of action and reaction, any magnetic
166
VECTOR AND TENSOR ANALYSIS
[SEC. 77
field will act on J2 dr2 with an equal and opposite force. Thus
df = J 2 drz x H, =
J,cJ2
(L) dr, x V r
x dr2
(278)
and integrating over (2) we obtain
f-J,J2f drzxf
(2)
C
OZ
(1)
r
xdr,
Now
dr2 x (TI 1 x
r
dr,) = V 1 (dr, dr2) - (dr2 "V 1) dr,
r
r
and J 2) [dr2 - v(1/r)] dr, =
f=
I2)
d(1/r) dr, = 0, so that
Jc
2
J (2)1(1)
(C'
1) (dr,
dr2)
This is the force of loop (1) on loop (2).
(279)
It is equal and
opposite to the force of loop (2) on loop (1), this being immediately deducible from (279) when we keep in mind that
v-=-V1
1
r2,
r12
In (279), r = r2,.
Example 97. We find the force per unit length between two
long straight parallel wires carrying currents J, and J2. We use
(278) and the result of Example 95. We have H, = (2J,/cd)i
at right angles to the plane containing the wires. Hence
df=Jzdr2x 2J,
di.
=
2J,J2
dr2xi
and the force per unit length is F = 2J,J2/cd. If the currents
are parallel, F is an attractive force; if the currents are opposite,
F is a repulsive force.
Problems
1. From (278) show that f = J2 f f dd2 x (V X H,).
a
2. Find the force between an infinite straight-line wire carrying
a current J, and a square loop of side a with current J2, the
SEC. 791
167
STATIC AND DYNAMIC ELECTRICITY
extended plane of the loop containing the straight-line wire, and
the shortest distance from the wire to the loop being d.
3. A current J flows around a circle of radius a, and a current
J' flows in a very long straight wire in the same plane. Show
that the mutual attraction is 4irJJ'/c(sec a - 1), where 2a is
the angle subtended by the circle at the nearest point of the
straight wire.
4. Show that A = (J/c) f f dd x V (1 /r) for a current J in a
S
closed loop bounding the area S.
For a small circular loop, show
that A = (M x r/r$), where r is very much larger than the radius
of the loop and is the vector to the center of the circle, and where
M=c f f dd
78. Law of Induction (Faraday). It has been found by experiment that a changing magnetic field produces an electromotive
force in a circuit. If B is the magnetic inductance, the flux
through a surface S with boundary curve r is given by f f B dd.
S
The law of induction states that
-ca f f
Applying Stokes's theorem, we have
_ - =VxE
C at
(280)
The time rate of change of magnetic inductance is proportional
to the curl of the electric field. Equation (280) is a generalization of V x E = 0, which is true for the electrostatic case in which
B = 0 and for the steady state for which atB = 0.
79. Maxwell's Equations. Up to the present we have, for an
electrostatic field, V x E = 0, V D = 4,rp and, for stationary
currents, V x H = (41r/c)j, V- B = 0.
VECTOR AND TENSOR ANALYSIS
168
Now V x E_- 1CaB
at
[SEC. 79
a generalization of V x E = 0.
is
Maxwell looked for a generalization of V x H = (41r/c)j. He
decided to retain the two laws: (1) V D = 4rp as the definition
of charge, and (2) V j +
at
= 0 as the law of conservation of
charge.
Let us assume
VxH=4w(j+x)
(281)
C
We take the divergence
as a generalization of V x H = (47r/c) j.
of (281) and obtain
(282)
so that
= - V j = at = Oar at (V . D)
V
aD
0
so that
We can choose Z =
VxH= w- (i
+--aD)
(283)
C
We call -
aD
t
the displacement current.
We rewrite Maxwell's equations
V D = 4irp
(284)
V X
4c \1 + 41r aD1
t
SEC. 801
STATIC AND DYNAMIC ELECTRICITY
169
We have in addition the equation
f=p(E+1vxBJ
(v)
\\
(285)
C
where f is the force on a charge p with velocity v moving in an
electric field E and magnetic inductance B. This result follows
from Sec. 77.
Problems
1. Show that the equations of motion of a particle of mass m
and charge e moving between the plates of a parallel-plate condenser producing a constant field E and subjected to a constant
magnetic field H parallel to the plates are
md-
= Be -
dy
d
dx
= He
ML
dt
dtz
Given that
d
dt
= x = y = 0 when t = 0, show that
z = (E/wH) (1 - cos wt), y = (E/(X) (wt - sin wt), where
He
w = -
m
2. From (iii) of (284) show that V - aB = 0.
3. From (i) and (iv) of (284) show that
D) -
at
4. Write down Maxwell's equations for a vacuum where
j=p=O,D=E,B=H.
80. Solution of Maxwell's Equations for "Electrically" Free
Space. We have p = j = 0 and c, u are constants. Equations
(284) become
VECTOR AND TENSOR ANALYSIS
170
[SEC. 80
V E=0
(i)
VH=0
vxE=
(ii)
aH
(286)
K aE
vxHcat
We take the curl of (iii) and obtain
V x (V x E) = V(V E) - V2E _
or
V2E =
µK a2E
C2 at2
(287)
by making use of (i) and (iv). Similarly
V2H -
JLK a2H
C2
8t2
(287a)
Equation (287) represents a three-dimensional vector wave equation. To illustrate, consider a wave traveling down the x axis
with velocity V and possessing the wave profile y = f(x) at
t = 0. At any time t it is easy to see that y = f(x - Vt). From
y = f(x - Vt) we have
a2-y
= f"(x - Vt) and
at= = V2f"(x - V0
so that
a2y
ax2
1 a2y
V2 at2
(288)
Equation (287) represents three such equations, and µK/c2 plays
the same role as 1/V2, so that c/V has the dimensions of a
velocity (see Example 96).
c = 3 X 1010 by experiment.
z
Example 98.
We solve the wave equation V2f = Y2 at2 in
spherical coordinates where f = f(r, t).
STATIC AND DYNAMIC ELECTRICITY
SEC. 80J
171
Vf = f'(r, t) Vr = f' r
V2f
=
V.(rr/
3f+rf" = rf V - r+V(
)'r
f,
Our wave equation is
2 of
19 2f
1 a2f
V28t2
are+rar
(289)
Now let u(r, t) = rf(r, t); then
af_1au
Or
r Or
u
02fi0 2u2au
r2
are
r are
r2 Or
2
+
r2
and substituting into (289), we obtain
a2u
V2 at2
82u
are
of which the most general solution is
u(r, t) = g(r - Vt) + h(r + Vt)
and so
.f(r, t) = r [g(r - Vt) + h(r + Vt)]
(290)
is the most general solution of (289).
Let us now try to determine a solution of Maxwell's equations
for the case E = E(x, t), H = H(x, t).
Now V E = 0, so that
which implies
aE=(x, t)
8E.(x, t)
ax
= 0.
ax
+
aEy(x, t)
ay
+
8Eg(x, t)
az
0,
We are not interested in a uniform
field in the x direction, so we choose E. = 0. Hence
E = EE(x, t)j + Es(x, t)k
and similarly
H = Hy(x, t) j + H.(x, t)k
172
VECTOR AND TENSOR ANALYSIS
[SEC. 80
Now we use Eq. (iii) of (286), V x E _ - c axt , so that
i
j
k
a
a
a
ax
ay
az
0
Ey
E.
AaH, i
LaH,
at c at
c
k
or
(i)
OE.
,.8H4
ax
c
at
aE,
aH,
ax
at
(291)
Similarly, on using (iv) of (286), we obtain
(i)
aH,
K aE
8x
c at
aH,
K aE,
ax
c at
(292)
The four unknowns are E., E,, H,,, H,, which must satisfy (291)
and (292). If we choose H = E, = 0, we see that (i) of (291)
and (ii) of (292) are satisfied. Differentiating (ii) of (291) with
respect to x and (i) of (292) with respect to t, we obtain
a2Ey
axe
UK 61%
C2 at2
(293)
We leave it to the reader to show that
82H.
ax2
Ax a2H,
c2
at=
(294)
'These equations are of the type represented by (288). Hence
a solution to Maxwell's equations is
E = [E,,(') (x - Vt) + Eyes) (x + Vt)Jj
H = [H,(')(x - Vt) + H,{2>(x + Vt)Jk
(295)
where V = c(AK)"}.
Both waves are transverse waves, that is, they travel down
the x axis but have components perpendicular to the x axis.
SEC. 801
STATIC AND DYNAMIC ELECTRICITY
173
Also note that E H = 0, so that E and H are always at right
angles to each other.
By letting H. = Er, = 0, we can obtain another solution,
E = E.(x, t)k, H = H (x, t)j. These two solutions are called
the two states of polarization, the electric vector being always
oriented 90° with the magnetic vector.
Example 99. We compute the energy density.
2
w,n=
_
_
We _
BH
2
and w = w, +
2
µH.2
2
µH2
2 = 2 =w.
=2
2w, = 2 ,. = KE 2, and for both waves
w, = K(E,,2 + E.2). We have here used the fact that
(see Prob. 1).
Example 100. Maxwell's equations in a homogeneous conducting medium are
V.E=4p
K
V.H = 0
OH
7E
c at
VxH=4c1uE+4v
a/
Assume a periodic solution of the form
z)e-;'e
E = Eo(x, y,
H = Ho(x, y,
z)e-'"e
Substituting into (iv), we obtain
e--;tee V
x Ho =
4 1t oe.-;'eE0 -
iKW
or
VxHo
4a/
=-lo-Eo
2Kw
C
4r
(296)
VECTOR AND TENSOR ANALYSIS
174
[SEC. 81
This equation is the same as that which occurs for "electrically"
free space with a complex dielectric coefficient.
Problems
1. By letting E = f(x - Vt), Hz = F(x -- Vt), V = c/,
show that H. = VKIIA E.
2. Derive (287a).
a2y
=
3. Letr=x- Vt, s =x+ Vt, and show thataX2
1 a2y
V2 at2
2
reduces to -ay = 0. Integrate this equation and show that the
Or as
general solution of (288) is y = f(x - Vt) + F(x + Vt), where
f and F are arbitrary functions.
4. Prove that Maxwell's equations for insulators (a = 0) are
v x H=
KaE
C at
V x E_- c a
and
(297)
5. Show that the solution of (297) can be expressed in terms
of a single vector V, the Hertzian vector, where
aV
2
H=-vx at
and V satisfies V2V =at2
c.
6. Prove
thatE=-VxeH=
at
c
-V(V.W)+Kµa
2W
C2 at2
isa
192W
solution of (297), provided that W satisfies V2W = e
at2
7. Derive (294).
8. Look up a proof of the laws of reflection and refraction.
9. By considering (i) and (iv) of Example 100, show that
P=
Poe-4x'`i`
81. Poynting's Theorem. Our starting point is Maxwell's
equations. Dot Eq. (iii) of (284) with H and Eq. (iv) with E
and subtract, obtaining
c(H-V xE -
xH) = -H aB
at
ID
at
(298)
STATIC AND DYNAMIC ELECTRICITY
SEc. 821
Now from (218)
aw,
at
1
E
4ir
aD
aw,,
at
at
_
175
H aB
4r
at
1
Let us write j = jo + jc where jo represents the galvanic current
and j. = pv, the conduction current. Now
dr
awmec,,,,a;c,,
awM
at
at
and from Sec. 75 it is easy to prove that E ja is Joule's power
loss =
aQ
Moreover, H V x E - E V x H = V (E x H), so
that we rewrite (298) as
c V (E X H) = -41r
aw,
aw,
at
+
at
aQl
awm
+
at
+
at
(299)
Integrating over a volume R and applying the divergence
theorem, we obtain
f if aQ dr + 4r f f E x H- dd
f f at dr
J
(300)
where w is the total energy density.
We define s = (c/4x)E x H as Poynting's vector. Equation
(300) states that to determine the time rate of energy loss in a
given volume V, we may find the flux through the boundary
surface of the vector s = (c/4T)E x H and add to this the rate
of generation of heat within the volume. It is natural to
interpret Poynting's vector as the density of energy flow.
Problems
1. Find the value of E and H on the surface of an infinite
cylindrical wire carrying a current. Show that Poynting's vector
represents a flow of energy into the wire, and show that this flow
is just enough to supply the energy which appears as heat.
2. Find the Poynting vector around a uniformly charged
sphere placed in a uniform magnetic field.
3. If E of Sec. 80 is sinusoidal, E = Eo sin co(x - Vt)k, find
the energy density after finding the magnetic wave H.
82. Lorentz's Electron Theory. For charges moving with
velocity v, j = pv, and Maxwell's equations become
VECTOR AND TENSOR ANALYSIS
176
[SEc. 82
4irp
(i)
(ii)
(iii)
(301)
(iv)
V
xH
= 4c
D)
+
(pv
These equations are due to Lorentz. From (ii) we can write
B = V x (Ao + Vx) = V x A. Substitute this value of B into
C
(iii) and obtain V x E
v x Af or its equivalent
vx(E+catA/
0
I
Thus E + A is irrotational, so that E + 1
-vgp.
A
Let D= KE, B = pH, and substitute into (iv). We have
I
µ
V x (V xA) =
-
r
Ipv+ 4a(
z
at2A-vaC1 SO
t
c
and since V x (V x A) = --V2A + V(V A), we obtain
z
v2A
cl-p
v + V j,
c2 at2
(302)
where
vA+c2 at
Now
= -K V2(p -
K(--AKa2S0/
C
at
c
Kp, 12,p
4rp
_
14
c2 812
K
so that
_
vg
cat
at2
(303)
SEC. 821
STATIC AND DYNAMIC ELECTRICITY
177
Equations (302) and (303) would be very much simplified if
VA+
we could make
KA a
c- at
__ O.
This is called the equa-
tion of gauge invariance.
Let us see if this is possible.
i aA° =
Now B = V x A0 and E
-v4p° so that
+ C at
l aA
I aAo
cat - vp
E = -cat where A = A0 + V. Thus
c1
(BA
aA0
at
at
1a
-cat °x =
v((P°
c°)
and
_
Iax _
C at
1 a2x
C ate
(p + constant
asoo
aso
at
at
Now we desire
a'
c2 at
ao
C
z
at
C
at21
or
2
vzx
a°
-v A0
cz atz
cz
(304)
The right-hand side of (304) is a known function of x, y, x, t.
This equation is called the inhomogeneous wave equation, and
if the equation of gauge invariance is to hold, we must be able to
solve it. If we can solve it, the Lorentz equations will reduce
to four inhomogeneous wave equations and so will also be solvable. They are
v2A-c2
a2t2
=
v25p-=-= K a2c
C2 at 2
4cµP,v,
(305)
4w
K
P
VECTOR AND TENSOR ANALYSIS
178
(SEC. 83
Problems
1. For the Lorentz transformations (see Prob. 11, Sec. 24),
show that
a2,p
825,
1 a2p
c2at2
a24,
a2tp
a2,p
a2(p
1 a2V
c2ai2
the D'Alembertian.
We call
2. Consider the four-dimensional vector C; = (A1, A2, A3, -q'),
2
i = 1, 2, 3, 4, the A; satisfying V2A = C2 at2' while p satisfies
2
V 2v = -
with H = V x A, and E=
4,
-cat -
Let
X, = x) x2 = y, x3 = z, x' = ct, and show that
_
F"
Show that
0
_ aC;
8C;
C axi
- H.
Hr
axf
-HY
0
Hx
E.
EY
H, - Es
- Hz - Es
0
E.
--E,
0
axr,s = 0, i = 1, 2, 3, 4, yields V x H =
-1
V E = 0, Fit = -F;;, i or j = 4, otherwise Fu = F;1.
at
and
Also
show that
aFj-
aF.#
axy
+
axp
3. If P;;
+
= °
ax.
aH
yields V x E _ 4
aF#,
a, P, y = 1,2,3,4
and V H = 0.
4
axa axs
c
formations P12 = -171 =
show that for the Lorentz trans-
Ht
ll -
(y2/C2)]l
Complete the ma-
trix P 1.
83. Retarded Potentials. Kirchhoff's Solution of
72
1
V2
2
av
- 47rF(x, y, z, t)
STATIC AND DYNAMIC ELECTRICITY
SEC. 83]
179
To find the solution (p(P, t) of the inhomogeneous wave equation at t = 0, we surround the point P by a small sphere of
Fia. 71.
radius e, and let S be the surface of a region R containing P (see
Fig. 71). We apply Green's formula to this region.
fJ f
dT = f f
V2,p
R
Vp - V#) dd
B
+ f f (# V-p - p Vu')
dd
(306)
9
We choose for ¢ a solution of 'V24, - V2 a
= 0. We know
that 4, = f(r + Vt)/r is one such solution, where f is arbitrary.
Equation (306) now becomes
V2f
If
(4,,,2,p
t_V
actd-r-4'rfff FOd-r= f f
R
.. .
B
+ fJ
a
.
(307)
VECTOR AND TENSOR ANALYSIS
180
[SEc. 83
Equation (307) is true for all values of t so that we may integrate
(307) with respect to t between limits t = tl and t = t2. We
obtain
III
at
1:2l
at
j:2
dT - 4
dt j f f F4, dT
f dt (if ..
rr
=
.+
Now on E, ¢ _ (1/E)f( + Vt) and
v o da = -
j
f
ar,._ = 1 [f(E + Vt)
E
(308)
+ Vt)] dS
2
so that (308) reduces to
12
V2
f
If [f(r + Vt) &p
at
r
- 4w f if
+ 91
Vf'(r + Vt)1t2
r
fiFf(r+
t)dt dr = f,t: dt
Ef'(E + Vt)
dr
t,
f(E + Vt) R]
`
f
E
[f( Vt)
1
, da .+ f f
JJ
E2
...}
(309)
s
Let us now return to a consideration of f(r + Vt). Since f
is arbitrary, let us choose f ° 0 for jr + VtI > 6, with the addias
f(r + Vt) d(r + Vt) = 1, where 6 is
tional restriction that
arbitrary for the moment. Notice that f = 0 for Ir + Vti > a.
Now let us choose t2 > 0 and tl negatively large, so that for
all values of r in the region R, jr + Vt[ > a. Hence
f
[f(r + Vt) &p
r
at
L
_
_
Vf'(r + Vt)
r
Itt,
since Ir + V121 > a, I r + V4J > S.
Moreover
ft
F f(r
+ Vt) dt = 1 fttip f(r + Vt) d(r + Vt)
=Vj
da F
f (x) dx
(310)
for a fixed r. Now if S is chosen very small, the value of (310)
reduces to approximately
STATIC AND DYNAMIC ELECTRICITY
SBc. 831
1
r z=o
(F)
V
IL f(x)
dx =
181
1
V
(r
t_-r/v
Hence the left-hand side of (3' 09) reduces to
4V
(311)
()t_-r/vdr
Now considering the right-hand side of (309), we see that
lim
e-.0
ff
E
rf(E + Vt) Dtp
L
E
(pf'(E + Vt)
+
R] dd = 0
E
since dd is of the order E2, and f, f , tp, Dtp are bounded for a fixed a.
We also have that
lim -
t:
jI
dt f f
since f f dS =
'pE,
f(E + Vt)
. dS = -
1
v 47rcp(P)
(312)
and for small 6,
E
Finally,
fs f i:'
dtlf(r+ Vt)D
(rf'
Dr] dd
2
r2
L
ii fl'
= s
dt{f(r+Vt)Dp-Vr].dd
r
r
f
- f f j,t dt r f'(Dr . dd) = ff f .t dt
[f(r + Vt)
r
+ -f Dr] . dd + f f j i s rV f at dt (Dr dd)
(313)
s
on integrating by parts and noticing that f} = 0. Finally the
right-hand side of (313) becomes equivalent to
V sf f(rD t - -r/V -,p t--r/V
1
r
1
t3(o
rV at t- -r/V
(314)
VECTOR AND TENSOR ANALYSIS
182
[SEC. 83
Combining (311), (312), and (314), we obtain
V(P)
` 1 JR J
F=
dr
-r/ v
1
- i fJ
(v , ~
S
V r + rV at
V
r)
dd
(315)
t= -r/'V
Now let S recede to infinity and assume that (p,
when
evaluated at t = -r/V on the surface S, have the value zero
until a definite time T. For large r, t = -r/V is negative and
so is always less than T. Hence the surface element vanishes,
and
V
(P)=
JJJ
t- -r/V
di
-
(316)
The solutions to (305) are thus seen to be
A (P,
t) =
f fJ
W
4
pv
Ie
r t-(r/V)
,v(P,t)'JLI x t - (r/V)
whe re V =
Finally,
dr
317)
dr
B=VxA
l aA
E=-'cat-V
(318)
The physical interpretation of these results is simple. The
values of the magnetic and electric intensities at any particular
point P at any instant t are, in general, determined not by the
state of the rest of the field (p, v) at that particular instant, but
by its previous history. The effects at P, due to elements at a
distance r from P, depend on the state of the element at a previous time t -- (r/V). This is just the difference in time required
for the waves to travel from the element to P with the velocity
V = c/ V AK, hence the name retarded potential. Had we considered the function f(r - Vt), we should have obtained a solution depending on the advanced potentials. Physically this is
impossible, since future events cannot affect past eventsl
SEC. 831
STATIC AND DYNAMIC ELECTRICITY
183
Problem
1. A short length of wire carries an alternating current,
j = pv = Io (sin wt)k, -1/2 5 z 5 1/2.
(a) At distances far removed from the wire, show that
A=
j-o1
cr
sin w
(t
- rc//J k
and that in spherical coordinates
A,.
=jot sin wit - r1 cos 0
\ c/l
/ Isin0
Ae= - Iotsinwltc//
cr
cr
A, =O
(b) Show that H, = He = 0, and that
Hw =
cr sin 0
cos w Ct - c) + r sin w
IC
t - -) 1
(c) Find p from the equation of gauge invariance, and then
E,, E8, E from E + c A - V o.
CHAPTER 6
MECHANICS
84. Kinematics of a Particle. We shall describe the motion
of a particle relative to a cartesian coordinate system. The
motion of any particle is known when r = x(t)i + y(t)j + z(t)k
is known, where t is the time. We have seen that the velocity
and acceleration, relative to this frame of reference, will be given
by
v' dt i +dt
a=
d2y
d2x
dtz 1
+
dt
j
dtz
k
d zz
+
dt2
k
The velocity may also be given by v = vt, where v is the speed
and t is the unit tangent vector to the curve r = r(t). Differentiating, we obtain
a=dt
dtt+vdsdt
by making use of (95).
=dtt+Kv2n
(319)
Analyzing (319), we see that the accelera-
tion of the particle can be resolved into two components: a
tangential acceleration of magnitude dt, and a normal accelera-
This latter acceleration is called
centripetal acceleration and is due to the fact that the velocity
vector is changing direction, and so we expect the curvature to
play a role here.
For a particle moving in a plane, we have seen in Sec. 17,
Example 18, that the acceleration may be given by
tion of magnitude v2rc = v2/p.
2r
2l
- r ` de]R -}- r d (2 de P
= [ dt
r
a
184
MECHANICS
SEC. 84J
Example 101.
185
Let us assume that a particle moves in a plane
and that its a cceleration is only radial. In this case we must
have r d [ r2 de1 = 0, and integrating, +r2 dte = h = constant.
From the calculus we know
that the sectoral area is given
by dA = }r2 dB (see Fig. 72).
Thus dA = constant, so that
equal areas are swept out in
equal intervals of time.
Example 102. For a particle
Fie. 72.
moving around a circle r = b
with constant angular speed coo =
d
we have dt = 0 and
(r2wo) = 0, so that a = -bwo2R.
Example 103. To find the tangential and normal components
of the acceleration if the velocity and acceleration are known.
v=vt,
and
a v = vat
Also
so that
as =
v
axv=va,nxt= -vanb
and
an = I$)V,i°I
Problems
1. A particle moves in a plane with no radial acceleration and
constant angular speed wo. Show that r = Ae"o' + Be at.
2. A particle moves according to the law
r = cos t i + sin t j + t2k
Find the tangential and normal components of the acceleration.
VECTOR AND TENSOR ANALYST °
186
[SEc. 85
3. A particle describes the circle r = a cos 0 with constant
Show that the acceleration is constant in magnitude and
directed toward the center of the circle.
4. A particle P moves in a plane with constant angular speed
w about 0. If the rate of increase of its acceleration is parallel
speed.
d2r
to OP, prove that ate = 4rw2
5. If the tangential and normal components of the acceleration
of a particle moving in a plane are constant, show that the
particle describes a spiral.
85. Motion about a Fixed Axis. In Sec. 10, Example 12, we
saw that the velocity is given by v = w x r. Differentiating, we
obtain
dw
dr
xr
a=wxa+dt
a=wxv+axr
(320)
Since v = w x r, we
where a is the angular acceleration --'
at
have also
wl
a=wx(wxr)+axr
_ (w - r)w - w2r + a x r
If we take the origin on the
line of w in the plane of the
motion, then w is perpendicular
tororor = 0, so that
a= -w2r+axr
a x r is the tangential acceleraFm. 73.
tion, and w x (w x r) is the
centripetal acceleration.
If we assume that a particle P is rotating about two intersecting
lines simultaneously, with angular velocities cal, w2 (Fig. 73), we
can choose our origin at the point of intersection so that
vi = wl x r,
and the total velocity is
v2 = 632 x r
V = V1 + V2 = (wi + (02) x r
MECHANICS
SFC. 86]
187
A particle on a spinning top that is also precessing experiences
such motion.
86. Relative Motion. Let A and B be two particles traversing
curves r, and r2 (Fig. 74). r, and r2 are the vectors from a point
0 to A and B, respectively.
r2=r+rl
(321)
Definition: dt is the relative velocity of B with respect to A,
written V4(B).
Fia. 74.
Differentiating (321), we have
dr2
dr
dt - dt
dri
+
dt
or
Vo(B) = VA(B) + Vo(A)
(322)
More generally, we have
Vo(A) = V4,(A) + VA,(A1) + VA.(A,) + ..
.
+
Vo(A.)
It is important to note that V4(B) _ -VB(A).
Example 104. A man walks eastward at 3 miles per hour, and
the wind appears to come from the north. He then decreases
his speed to 1 mile per hour and notices that the wind comes
from the northwest (Fig. 75). What is the velocity of the wind?
We have
V0(W) = VM(W) + V0(M)
G(ground)
VECTOR AND TENSOR ANALYSIS
188
ISEc. 86
In the first case
VM(W) = -kj,
V0(M) = 3i
so that
VG(JV) = -kj + 3i
In the second case,
VM(W) = h(i - j),
VG(M) = i
so that
V0(W) = h(i - j) + i = (h + 1)i - hj,
and
3=h+1,
-k= -h,
VG(W) =3i-2j
miles per hour, and its direction
The speed of the wind is
makes an angle of tan-' I with the south line.
N
x.
E
S
Fia. 75.
Fzo. 76.
Example 105. To find the relative motion of two particles
moving with the same speed v, one of which describes a circle of
radius a while the other moves along the diameter (Fig. 76). We
have
P=acos0i-}-asin0j,
adze=v
Q = (a - vt)i
This assumes that both particles started together.
T -dQ=(-a sin0dO+v)i+acos0doi
VQ(P) = v(1 - sin 0)i + v cos 6 j
MECHANICS
Sec. 87]
189
The relative speed is
IVQ(P)I = [v2(1
- sin
0)2 + v2 cos2 B]} = 2}v(1 - sin B)}
Maximum IVQ(P)I occurs at 0 = 3x/2, minimum at 0 = x/2.
Problems
1. A man traveling east at 8 miles per hour finds that the wind
seems to blow from the north. On doubling his speed, he finds
that it appears to come from the northeast. Find the velocity
of the wind.
2. A, B, C are on a straight line, B midway between A and C.
It then takes A 4 minutes to catch C, and B catches C in 6 minutes. How long does it take A to catch up to B?
3. An airplane has a true course west and an air speed of
200 miles per hour. The wind speed is 50 miles per hour from
1300. Find the heading and ground speed of the plane.
87. Dynamics of a Particle. Up to the present, nothing has
been said of the forces that produce or cause the motion of a
particle. Experiment shows that for a particle to acquire an
acceleration relative to certain types of reference frames, there
must be a force acting on the particle. The types of forces
encountered most frequently are (1) mechanical (push, pull), (2)
gravitational, (3) electrical, (4) magnetic, (5) electromagnetic.
We shall be chiefly concerned with forces of the types (1) and
(2). For the present we shall assume Newton's laws of motion
hold for motion relative to the earth. Afterward we shall modify
this. Newton's laws are:
(a) A particle free from the action of forces will remain fixed
or will continue to move in a straight line with constant speed.
(b) Force is proportional to time rate of change of momentum,
that is, f = dt (mv). In general, m = constant, so that
The factor m is found by experiment to be an invariant for a given
particle and is called the mass of the particle. In the theory of
relativity, m is not a constant. my is called the momentum.
(c) If A exerts a force on B, then B exerts an equal and opposite
force on A. This is the law of action and reaction: fAB = -fha
VECTOR AND TENSOR ANALYSIS
190
[SEC. 88
By a particle we mean a finite mass occupying a point in our
Euclidean space. This is a purely mathematical concept, and
physically we mean a mass occupying negligible volume as compared to the distance between masses. For example, the earth
and sun may be thought of as particles in comparison to their
distance apart, to a first approximation.
88. Equations of Motion for a Particle.
Newton's second law
may be written f = m dt = ma. We postulate that the forces
acting on a particle behave as vectors. This is an experimental
fact. Hence if fl, f2j . . . , f act on m, its acceleration is given
by
a
1
(fl+f2+....+fn)_
m
1I
mti_1
We may also write f = m d
dt2
,
f,_ mff
where r is the position vector from
the origin of our coordinate system to the particle. If the particle
d
is at rest or is moving with constant velocity, then
dt2
= 0, and
so f = 0, and conversely. Hence a necessary and sufficient condition that a particle be in static equilibrium is that the vector
sum of the forces acting on it be zero.
A standard body is taken as the unit mass (pound mass). A
poundal is the force required to accelerate a one-pound mass one
foot per second per second. The mass of any other body can be
compared with the unit mass by comparing the weights (force of
f12/
m2 gravit y at mean sea leve l ) o f th e two
objects. This assumes the equivalence
of gravitational mass and inertial mass.
Example 106. Newton's law of gravi-
tation for two particles is that every
pair of particles in the universe exerts
Fzo. 77.
a mutual attraction with a force directed
along the line joining the particles, the
magnitude of the force being inversely
proportional to the square of the distance between them
and directly proportional to the product of their masses.
f12 = (Gmim2/r2)R (see Fig. 77). G is a universal constant.
Let
191
MECHANICS
SEC. 88]
the mass of the sun be M and that of the earth be m. We
shall assume that the sun is fixed at the origin of a given coordinate system (Fig. 78). The force act-
ing on the earth due to the sun is
f = - (Gm.M/r3)r
From the second law
GmM
rs r
dv
dt
d2r
m dt2
so that
dv
GM
dt
r3
M
(323)
Fia. 78.
Now
d
v)
dt (r x
dv
= r x dt
and hence
d
dt(rxv)=rx(-GMr) =0
This implies
r x v = h = constant vector
or
rxa =
dr
(324)
h
Since Ir x drl = twice sectoral area, we have 2 dA = Ihl, or equal
areas are swept out in equal intervals of time. This is Kepler's
first law of planetary motion. Moreover, r I r x dt ] = r h = 0,
so that r remains perpendicular to the fixed vector h, and the
motion is planar. Now
-GMrxh= -GMrx(rxv)
a xh
from (324), and
d
d
(v x h) = d x h, so that
(v x h)
GM
r x (r x v)
(325)
VECTOR AND TENSOR ANALYSIS
192
Now r = rR, where R is a unit vector.
SEC. 88
Hence
R
v
so that (325) becomes
\
d(vxh)=-GMrx (rxrR)
/
=
-GMRx(RxdR)
-GM K R .
ddR
t/
R - R2 dRJ
= GM ddR
(326)
since R is a unit vector.
Integrating (326), we obtain
vxh=GMR+k
and
h' = GMr + rk cos (R, k)
(327)
Thus
h2/GM
r
(328)
1 + (k/GM) cos (r, k)
We choose the direction of the constant vector k as the polar
axis, so that
r
h'/GM
1 + (k/GM) cos 8
(329)
This is the polar equation of a conic section. For the planets
these conic sections are closed curves, so that we obtain Kepler's
second law, which states that the orbits of the planets are ellipses
with the sun at one of the foci.
Let us now write the ellipse in the form
S
r
1+ e cos 9
where e=
GM' p
k
MECHANICS
SEC. 88]
193
The curve ci osses the polar axis at 0 = 0, 0 = it so that the length
of the major axis is
ep + ep
2, - 1+e
1-e
2p
2h2
1-e2
GM(1-e2
For an ellipse, b2 = a2 - c2 = a2 - e2a2, orb = a(l - ,2)1.
area of the ellipse is A = Tab = ,ra2(1 - e2)1, and since
The
dA
= Jh,
dt
the period for one complete revolution is
T
2A
2ara2(1 - e2)i
Zral
- e2)1 = G1Mi
- h - a1GiM1(1
Thus
_
T2
47r2
0
GM =
constant, for all planets
(330)
This is Kepler's third law, which states that the squares
of the periods of revolution of the planets are proportional to the
cubes of the mean distances from the sun.
Problems
1. A particle of mass m is attracted toward the origin with the
force f = - (k2m/r6)r. If it starts from the point (a, 0) with the
speed vo = k/21a2 perpendicular to the x axis, show that the path
is given by r = a cos 0.
2. A bead of mass m slides along a smooth rod which is rotating
with constant angular speed w, the rod always lying in a horizontal plane. Find the reaction between bead and rod.
3. A particle of mass m is attracted toward the origin with a
force - (mk2/r3)R._ If it starts from the point (a, 0) with velocity
vo > k/a perpendicular to the x axis, show that the equation of
the path is
(OW k2)i
r = a sec C
0
-
avo
I
4. In a uniform gravitational field (earth), a 16-pound shot
leaves the putter's fingers 7 feet from the ground. At what angle
should the shot leave to attain a maximum horizontal distance?
194
VECTOR AND TENSOR ANALYSIS
[SEC. 89
5. Assume a comet starts from infinity at rest and is attracted
toward the sun. Let ro be its least distance to the sun. Show
that the motion of the comet is given by r = 2ro/(1 + cos 0).
89. System of Particles. Let us consider a system consisting
of a finite number of particles moving under the action of various
forces. A given particle will be under the influence of two types
of forces: (1) internal forces, that is, forces due to the interaction
of the particle with the other particles of the system, and (2) all
other forces acting on the particle, said forces being called
external forces.
If r; is the position vector to the particle of mass m,, then we
shall designate f,(e) as the sum of the external forces acting on the
jth particle, and f,(i) as the sum of the internal forces acting on
this particle. Newton's second law becomes for this particle
f1( + f1( = m; d2r'
(331)
Unfortunately, we do not know, in general, f;('), so that we shall
not try to find the motion of each particle but shall look rather
for the motion of the system as a whole. Since Eq. (331) is
true for each j, we can sum up j for all the particles. This yields
n
J=1
f'c6> +
d 2r,
f'(i)
;=1
1-1
m' dt2
From Newton's third law we know that for every internal force
n
there is an equal and opposite reaction, so that
This leaves
A
f ®= 0.
n
n
l m,
f,
dt2
(332)
We now define a new vector, called the center-of-mass vector,
by the equation
n
Im,r,
n
ra =1' =
(333)
n
;11
7-1
MECHANICS
SEC. 89)
195
The end point of r. is called the center of mass of the system.
It is a geometric property and depends only on the position of
the particles. Differentiating (333) twice with respect to time,
we obtain
Md2r` _
dt2
n
md?r,
' dt2
;=1
so that (332) becomes
n
f
f.(e)=M
2_1°
(334)
z
)ml
Equation (334) states that the center of mass of the system
accelerates as if the total mass were concentrated there and
all the external forces acted at that point.
Fia. 79.
Example 107. If our system is composed of two particles in
free space and if they are originally at rest, then the center of
mass will always remain at rest, since f = 0 so that d2° = 0, and
r. = constant satisfies the equation of motion and the initial condition
0.
For the earth and sun we may choose the center
of mass as the origin of our coordinate system (Fig. 79). The
equations of motion for earth and sun are
m
d2r1
dt2
r2)2f
= - (rlGmMR
M d2r2
dt2
_ (rlGmMR
- 12)2
Since r. = 0, we have mr, + Mr2 = 0, and
d2r1
d12
_
-GM
rl
[1 + (m/M))2 rig
196
VECTOR AND TENSOR ANALYSIS
[SEc. 91
This shows that m is attracted toward the center of mass by an
inverse-square force.
The results of Example 106 hold by replac-
ing M by M[1 + (m/M)1-2.
Problems
1. Show that the center of mass is independent of the origin
of our coordinate system.
2. Particles of masses 1, 2, 3, 4, 5, 6, 7, 8 are placed at the
corners of a unit cube. Find the center of mass.
3. Find the center of mass of a uniform hemisphere.
4. Find the force of attraction of a hemisphere on another
hemisphere, the two hemispheres forming a full sphere.
90. Momentum and Angular Momentum. The momentum of
a particle of mass m and velocity v is defined as M = mv. The
total momentum of a system of particles is given by M = j m;v1.
j-1
We have at once that
dM
dt
n
I mj
j-1
dv1
dt =
n
I f; (e) = f
(335)
j-1
We emphasize again that the mass of each particle is assumed
constant throughout the motion.
The vector quantity r x my is defined as the angular momentum, or moment of momentum, of the particle about the origin 0.
f
The total angular momentum is given
by
n
H = E r; x mjvj
(336)
j-1
91. Torque, or Force Moment. Let
Fla. 80.
f be a force acting in a given direction
and let r be any vector from the origin
whose end point lies on the line of
action of the force (see Fig. 80). The vector quantity r x f
is defined as the force moment, or torque, of f about 0. For a
system of forces,
n
L = I r, x f,
j-1
(337)
MECHANICS
SEC. 921
197
We immediately ask if the torque is different if we use a differ-
ent vector ri to the line of action of f.
The answer is in the
negative, for
(r, - r) x f = 0
since ri - r is parallel to f. Hence rl x f = r x f.
What of the torque due to two
equal and opposite forces both
acting along the same line? It
is zero, for
r1xf+r1x(-f)
=r1x(f-f)=0
Two equal and opposite forces
with different lines of action
constitute a couple (see Fig. 81).
Let ri be a vector to f and r2 a
vector to -f. The torque due
Fio. 81.
to this couple is
L = r1xf+r2x(-f)
= (r1 - r2) x f
The couple depends only on f and on any vector from the line of
action of -f to the line of action of f.
Problems
1. Show that if the resultant of a system of forces is zero, the
total torque about one point is the same as that about any other
point.
2. Show that the torques about two different points are equal,
provided that the resultant of the forces is parallel to the vector
joining the two origins.
3. Show that any set of forces acting on a body can be
replaced by a single force, acting at an arbitrary point, plus a
Prove this first for a single force.
4. Prove that the torque due to internal forces vanishes.
suitable couple.
92. A Theorem Relating Angular Momentum with Torque.
We are now in a position to prove that the time rate of change
of angular momentum is equal to the sum of the external torques
for a system of particles.
198
VECTOR AND TENSOR ANALYSIS
[SEC. 93
Since
n
H = Ir3xmjv
r, x
j=1
j=1
dri
m'dt
we have on differentiating
dH _
dt
n
Z r, x m,
dt2
n
I
j-1 r' x
aii =
dt
n
d2r,
dri
dri
+ j=1 dt x m' dt
(f1" + f, (') )
n
I r; x
j=1
f;(e)
=L
(338)
93. Moment of Momentum (Continued). It is occasionally
more useful to choose a moving point Q as the origin of our
FIG. 82.
coordinate system.
space. We define
Let 0 be a fixed point and Q any point in
n
dri
HQa = I (r, -- rQ) x m;
di
j-1
(339)
The superscript a stands for absolute momentum, that is, the
velocity of m; is taken relative to 0, whereas the subscript Q
stands for the fact that the lever arm is measured from Q to the
particle m; (Fig. 82).
Differentiating (339), we obtain
MECHANICS
SEc. 941
dHQa
_
dt
drQl
(dri _
drQ
dl
dr,
dt /
+ L (r. -
dt
dr1
n
j-1
m;r,, so that M dt =
m1
dr,
-, and
l1
j=1
j-1
dt2
(r1 - rQ) x (f.(e) + f .(i))
xLm'dt+I
n
Now Mr =
rQ) x "ni
j=1
n
j=1
d2r;
nn`
X 7n,
(dt
j=1
199
n
f,(i)=0,
j=1
r1 x f;(') = 0 from Sec. 91, so that
dHQa
M
dt =
drQ
I
dF
n
(r, - rQ)
dt x dt + j-1
x fj(e)
or
dtQa
LQ (e)
-M
Qx
dt
dtc
(340)
We can simplify (340) under three conditions:
1. Q at rest, so that
drQ
dt
=0
2. Center of mass at rest, dt` = 0
3. Velocity of Q is parallel to velocity of center of mass,
drQ
dre
dt
dt
0
In all three cases
dHe
= LQ(e)
dt
(341)
In particular, if LQ(e) = 0, then He - constant, and this is the
law of conservation of angular momentum.
94. Moment of Relative Momentum about Q. In Sec. 93 we
assumed that the absolute velocity of each particle was known.
It is often more convenient to calculate the velocity of each
particle relative to Q.
This is
dr,
dt
- drQdt
We now define rela-
VECTOR AND TENSOR ANALYSIS
200
[SEC. 94
tive moment of momentum about Q as
n
HQr =
j-1
(rj - rQ) x mj
(rj - rQ)
dl
(342)
Differentiating,
dHQ*
dt
j-1
=
(d2rj
(r' - rQ) x m'
(r7 - rQ) x
dt2
dl2
d2rQ
`
(fj(ei
/
d2rQ
f3(i))
n
x
dt2
I m, (rj - rQ)
j-1
We see that
dtiQr
a,
LQ(' +
'rQ
d12
n
x I mj (rj - rQ)
(343)
js1
Under what conditions does dd Q' = Lo
d2rQ
dt2
?
We need
n
x I mj(rj - rQ) = 0
or
M
d2rQ
dt2
x (r, - rQ) = 0
(344)
Now (344) holds if
1. rQ = rQ, or Q is at the center of mass.
2. Q moves with constant velocity,
dQ
l = 0.
2
3. r. - rQ is parallel to
2
dt'
Problems
n
1. Show that
jet
a rQ
m4(r1 - rQ) x dt2 = M(r. - rQ) x
d2rQ
dt2
2. A system of particles lies in a plane, and each particle
remains at a fixed distance from a point 0 in this plane, each
MECHANICS
SEC. 95]
201
particle rotating about 0 with angular velocity w.
n
Show that
Ho = Iw, where I = I m;r;2, and show that Lo = I at
j=1
3. A hoop rolls down an inclined plane. What point can be
taken as Q so that the equation of motion (343) would be
simplified?
95. Kinetic Energy. We define the kinetic energy of a particle
of mass m and velocity v as T = jmv v.
C
M
For a system of particles,
r.-r
n
T
n
1
2 mv,2
f=1
fdrs\ 2
1
2 m'
.1-1
(345)
dt 1
P;
`r
r
Now let r, be the vector to the center of 0
mass C (Fig. 83). It is obvious that
Fia. 83.
r;=rr+(r,-r-)
so that
dr;
dt
dr; dr;
_ (dro)2
di dt
dt
_ dr,
dt
+
d
dt
(r; - r-)
(r, - rc) -{+2 drd
dt dt
I d.
L dt
(r; - rc) ]
2
Hence
1
(dr12 dr"
T=2M
dtJ +dt
d
n
2
+ jn
Now Mr. =
1
2 m'
n
(346)
n
r,, so that
m;r,
j-1
r`)]
[dt (r' -
j-1
m1 alt (r; - r-) = 0,
;-1
and (346) reduces to
T=IM
()2
-}-
-1 2
m; [-d
dt
(r; - r.)]2
a)(347)
1
This proves that the kinetic energy of a system of particles is
equal to the kinetic energy of a particle having the total mass
202
VECTOR AND TENSOR ANALYSIS
[SEC. 96
of the system and moving with the center of mass, plus the
kinetic energy of the particles in their motion relative to the
center of mass.
96. Work. If a particle moves along a curve r with velocity v
under the action of a force f, we define the work done by this
force as
W=
fr f t
fr f
f acts at right angles to the path, no work is done.
If the field is conservative, f = -V(p, the work done in taking
the particle from a point A to a point B is independent of the
path (see Sec. 52).
Now
dvi = fj(a) + fi(.)
dt
dvi
= fi(e) . V1. +
mivi .
dt
m,
Vi
and integrating and summing over all particles,
n
J`o
mivi
dt' dt =
La
i:1
fi(a)
. vi dt +
fee:' fi('i
. v; dt
or
n
i=1
lmi[vi2(ti) - vi2(to)l = W(e)
W°
(349)
This is the principle of work and energy. The change in the
kinetic energy of a system of particles is equal to the total work
done by both the external and internal forces.
If the particles always remain at a constant distance apart,
(ri - rk)2 = constant, the internal forces do no work. Let r,
and r2 be the position vectors of two particles whose distance
apart remains constant, and let f and -f be the internal forces
of one particle on the other and conversely. Now
(r, - r2) (r, - r2) = constant
MECHANICS
Sec. 97]
203
so that
(r, - r2)
dr,
di
- dr2dt/
0
(350)
Also
W(o =
f
ff
f is parallel to r, - r2, Ave have f = a(r, - r2) and
f (v, - v2) = a(r, - r2) (v, - v2) = 0
from (350). Thus W(° = 0.
Problems
1. A system of particles has an angular velocity w.
n
T=
i-i
Show that
Jm;lw x r,J2.
2. If to of Prob. 1 has a constant direction, show that T = }Iw2,
n
where I =
md;2, d; being the shortest distance from m; to
line of w.
3. Show that dT = w L, by using the fact that T =
4-mv;2
and that v, = to x r;.
4. Show that the kinetic energy of a system of rotating particles is constant if the system is subjected to no torques. What
if L is perpendicular to w?
5. A particle falls from infinity to the earth. Show that it
strikes the earth with a speed of approximately 7.0 miles per
Use the principle of work and energy.
97. Rigid Bodies. By a rigid body we mean a system of
particles such that the relative distances between pairs of points
remain constant during the discussion of our problem. Actually
no such systems exist, but for practical purposes there do exist
such rigid bodies, at least to a first approximation. Moreover,
the rigid body may not consist of a finite number of particles, but
rather will have a continuous distribution, at least to the unaided
eye. We postulate that we can subdivide the body into a great
many small parts so that we can apply our laws of motion for
particles to this system, this postulate implying that we can use
second.
VECTOR AND TENSOR ANALYSIS
204
the integral calculus.
the following form:
[SEC. 98
Our laws of motion as derived above take
T = f f f -pv2 dr,
p = density
R
ff pr dr
f f f p dr
R
dr,
( 351)
2
J
fR f f(e)
dt2
H = f f f prxvdr
R
-it ff r x f(e) dr
where f(.) is the external force per unit volume.
98. Kinematics of a Rigid Body. Let 0 be a point of a rigid
body for which 0 happens to be fixed.
It is easy to prove that the velocity
f V (P1 = yr
P
of any other point P of the body must
be perpendicular to the line joining 0
to P, for if r is the position vector
from 0 to P, we have r - r = constant throughout the motion so that
0.
Q.E.D.
r dt =
We next prove that if two points
of a rigid body are fixed, then all
other particles of the body are rotating around the line joining these two
points.
Let A and B be the fixed
points and P any other point of the
body.
From above we have
so that P is always moving perpendicular to the plane ABP.
Moreover, since the body is rigid, the shortest distance from P
to the line A B remains constant, so that P moves in a circle
Sec. 981
MECHANICS
around AB (Fig. 84).
could be written
205
We saw in Sec. 10 that the velocity of P
VP =wxrp
Is w the same for all particles? Yes! Assume Q is rotating
about AB with angular velocity w,, so that vQ = to, x rQ. Now
(rP - rQ)2 = constant, so that
(rP - rQ) (vp - vQ) = 0
or
(rP - rQ) ((a x rp - w, x rQ) = 0
Thus
x rP -
xrQ = 0
and
rQ x rp
(w, - w) = 0
We leave it to the reader to conclude that w, = w.
VA
B
1
FIG 85.'
If one point of a rigid body is fixed, we cannot, in general, hope
to find a fixed line about which the body is rotating. However,
there does exist a moving line passing through the fixed point so
that at any instant the body is actually rotating around this line.
The proof proceeds as follows: Let 0 be the fixed point of our
rigid body and let r` be the position vector to a point A. From
above we know that the velocity of A, VA, is perpendicular to
rA. Construct the plane through 0 and A perpendicular to VA
(Fig. 85). Now choose a point B not in the plane. We also
have that vB rB = 0, so that we can construct the plane through
0 and B perpendicular to vB. Both planes pass through 0, so
206
VECTOR AND TENSOR ANALYSIS
[SEC. 98
that their line of intersection, 1, passes through 0. Now consider any point C on this line. We have vc rc = 0. Moreover,
(rc - rA) - (rc - rA) ° constant, so that
(re-rA).(vc-VA) =0
and (rc - rA) VC = 0, since vA is perpendicular to (rc - rA).
Similarly (re - rB) vc = 0. Hence the projections of vc in
three directions which are nonplanar are zero.
This means that
FIG. 86.
Vc - 0, so that we have two fixed points at this particular
instant. Hence from the previous paragraph the motion is that
of a rotation about the line 1. If w is the angular-velocity vector,
then v; = w x r;, where r; is the vector from 0 to the jth particle.
Now let us consider the most general type of motion of a rigid
-r represent a fixed coordinate system in space,
body. Let
and let 0-x-y-z represent a coordinate system fixed in the rigid
body (see Fig. 86). Let ,o; and r; represent the vectors from 0'
and 0 to the jth particle, and let a be the vector from 0' to 0.
We have ei = a + r,, and differentiating,
dLD;
dt
- da
dt
+
dr,
dt
MECHANICS
SEC. 98]
207
Now l represents the velocity of Pi relative to 0.
This means
0 is fixed as far as Pi is concerned, and from above we know that
dri
dt - w
Thus
x ri.
_dei =
da
dt
dt
v'
(352)
+wxr;
that is, the most general type of motion of a rigid body is that of
a translation
A
dt
plus a rotation w x r;.
We next ask the following question: If we change our origin
from 0 to, say, 0" does w change? (Fig. 87.) The answer is
"No"! Let b be the vector
0
from 0' to 0". Then
v;
rf
=t=' + w,xr;
But
and
db
da
dt
dt
ri = (a
+w x (b - a)
-- b) + ri
Fio. 87.
Thus
vi = d +w x(b - a)+wi x (a -b)+w, xri
(353)
Subtracting (352) from (353), we obtain
(w - wi) x (b-a)+(w,-w) xr;=0
or
(w - wl) x (b-a-r,)=
xr;"=0
0 and not parallel to the vector
w - wl, at any particular instant. Hence w, ° w.
We can certainly choose an r;"
Problems
1. Show that if r, and r2 are two position vectors from the
origin of the moving system of coordinates to two points in the
rigid body, then r, dt2 + r2 - dtl = 0.
VECTOR AND TENSOR ANALYSIS
208
[SEC. 99
2. A plane body is moving in its own plane. Find the point
in the body which is instantaneously at rest.
3. Show that the most general motion of a rigid body is
a translation plus a rotation
about a line parallel to the
translation.
99. Relative Time Rate of
Change of Vectors. Let S be
an y vector measured in
the
moving system of coordinates
(Fig. 88).
S = S i + Sj + S,k
Fin. 88.
(354)
To find out how S changes with time as measured by an observer
at 0', we differentiate (354),
dS
dS=.
dt = dt 1+
dS
.
dt'+
dk
dS,
di
_
dj
dt k+S= dt+Sydt+S'dt
(355)
We do not keep i, j, k fixed since i, j, k suffer motions relative to
0'. But we do know that dt is the velocity of a point one unit
along the x axis, relative to 0. Hence
dk = w x k.
at
dS
d
= w x is dt = w x it
Hence (355) becomes
dSs
dt - dt
i -}
dS
dt
dS,
j + d k + w x (3j +
S1t)
and
dS
dt
DS
S
dt + w x
(356)
where DS represents the time rate of change of S relative to the
moving frame, for S= is measured in the moving frame and so d _
t
is the time rate of change of S. as measured by an observer in the
moving frame.
MECHANICS
SEC. 1001
209
Intuitively, we expected the result of (356), for not only does S
change relative to 0, but to this change we must add the change
in S because of the rotating frame. The reader might well ask,
What of the motion of 0 itself? Will not this motion have to be
considered? The answer is "No," for a translation of 0 only
pulls S along, that is, S does not change length or direction if 0
is translated. It is the motion of S relative to the frame O-x-y-z
and the rotation about 0 that produce changes in S.
Problems
1. Show that d dl
2. For a pure translation show that
dS
DS
dt =
dt
3. From (356) show that di = w x i.
Let P be any point in space and let 9 and r
be the position vectors to P
100. Velocity.
from 0' and 0, respectively
(see Fig. 89). Obviously
a + r, so that
_ dp _ A dr
v
dt
dt
+
dt
Now r is a vector measured in
the O-x-y-z system, so that
(356) applies to r. This yields
dr _ Dr
+ to x r and
7t
dt
v
dt
This result is expected.
t
-dt
FIG. 89.
xr+Dr
(357)
A is the drag velocity of P, co x r is
the velocity due to the rotation of the 0-x-y-z frame, and
Dr
is
the velocity of P relative to the 0-x-y-z frame. The vector sum
is the velocity of P relative to the frame 0'--i-r.
VECTOR AND TENSOR ANALYSIS
210
(SEC. J 01
101. Acceleration. In Sec. 100 we saw that
v=ac
A dt+wxr+ Dr
dt
To find the acceleration, we differentiate (357) and obtain
dv_d2p_d2a
dt
dt2
dt2
d
+ dt (wxr) +
d(Drl
dt
dt
(358)
We apply (356) to w x r and obtain
d
(wxr) = w x (wxr) + D (w x r)
Similarly
d
dldtl
d2p
dt 2
_
d2a
dt2
+ w x (wxr) +
D
xdt+d`dtl
Dr D2r
do
x r + 2w x
dt
dt + dt2
(359)
If P were fixed relative
Dr = D2r
to the moving frame, we would have
0 and consedt2 =
dt
quently P would still suffer the acceleration
Let us analyze each term of (359).
d2a
dt2
&
+wx(wxr)+dt xr
This vector sum is appropriately called the drag acceleration
of the particle. Now let us analyze each term of the drag acceleration. If the moving frame were not rotating, we would have
2
0, and the drag acceleration reduces to the single term dta
This is the translational acceleration of 0 relative to 0'. Now in
Sec. 84 we saw that w x (w x r) represented the centripetal accel-
eration due to rotation and d x r represented the tangential
component of acceleration due to the angular-acceleration vector
MECHANICS
SEC. 1011
d We easily explain the term
relative to the O-x-y-z frame.
D2r
211
as the acceleration of P
What, then, of the term 2w x dr?
This term is called the Coriolis acceleration, named after its discoverer. We do not try to give a geometrical or physical reason
for its existence. Suffice to say, it occurs in Eq. (359) and must
be considered when we discuss the motion of bodies moving over
the earth's surface. Notice that the term disappears for par-
ticles at rest relative to the moving frame, for then dt = 0.
It
also does not exist for nonrotating frames.
Now Newton's second law states that force is proportional
to the acceleration when the mass of the particle remains constant. It is found that the frame of reference for which this law
holds best is that of the so-called "fixed stars." We call such a
frame of reference an inertial frame. Any other coordinate
system moving relative to an inertial frame with constant velocity
D
d2
is also an inertial frame, since from (359) we have d e =
because w = Of
66
1
0,
dt2r
da
d1
constant, dta = 0.
dt =
Let us now consider the motion of a particle relative to the
earth. If f is the vector sum of the external forces (real forces,
2
that is, gravitation, push, pull, etc.), then d p = m, and (359)
becomes
D2r
d2a
Dr
d4,3
f
dt2 - w x (w x r) - d x r - 2w x dt + m (360)
dt2
This is the differential equation of motion for a particle of mass
m with external force f applied to it.
Example 108. Let us consider the earth as our rotating frame.
The quantity w x ((a x r) is small, since jwl
27r/86,164 rad/sec,
and for a particle near the earth's surface, lrl , (4,000)(5,280)
feet.
Also
d2
dt
- 0 over a short time; da ,= 0 over a short time;
212
VECTOR AND TENSOR ANALYSIS
[SEC. 101
so that (360) becomes
D2r
-2w x dt +
dt2
m
(361)
Now consider a freely falling body starting from a point P at
rest relative to the earth. Let the z axis be taken as the line
joining the center of the earth
to P, and let the x axis be
taken perpendicular to the z
axis in the eastward direction.
We shall denote the latitude
of the place by A, assuming
A > 0. The equation of motion in the eastward direction
is given by
yd1x
_
t2
S
(fm/, = 0.
tion it is
(see Fig. 90).
dj
dt
, + Cfm
/
has no component eastward, so
We do not know
-gtk +
x
Now f (force of attraction)
Fio. 90.
that
-2 (w
at
but to a first approxima-
Moreover, w = w sin A k + w cos A j
i.
Hence (w x
at)
d2x
dt' =
0
= -wgt cos A, and
2wgt cos A
(362)
If the particle remains in the vicinity of latitude A, we can keep A
constant, so that on integrating (362), we obtain
dx
dt
X=
=wgt2cosA
3ts cos A
(363)
(363) is to a first approximation the eastward deflection of a
shot if it is dropped in the Northern Hemisphere. If h is the
MECHANICS
SEc. 1011
213
distance the shot falls, then h = Jgt2 approximately, so that
2
x = 3 wh
cos X
2gh1}
( J
Problems
1. Show that the winds in the Northern Hemisphere have a
horizontal deflecting Coriolis acceleration 2wv sin X at right
angles to v.
2. A body is thrown vertically upward. Show that it strikes
the ground 'jwh cos X (2h/g)1 to the west.
3. Choose the x axis east, y axis south, z axis along the plumb
line, and show that the equations of motion for a freely falling
body are
d2x
W
+2wsinUdt z- 2wcos0dty=0
dt +2w cos0dy =0
dtz-g-2w sin0dx =0
2
where 0 is the colatitude.
CO
Fio. 91.
4. Using the coordinate system of Prob. 3, let us consider the
motion of the Foucault pendulum (see Fig. 91).
Let ii, i2, i$ be the unit tangent vectors to the spherical curves
r, 8, p. We leave it to the reader to show that the acceleration
VECTOR AND TENSOR ANALYSIS
214
[SEC. 101
along the is vector is 2 cos e .06 + sin 0 0 when the string is of
unit length.
The two external forces are mgk along the z axis and
the tension in the string, T = - Tr = - Tit. We wish to find
the component of these forces along the i3 direction. T has no
component in the is direction. Now k is = 0, so that mgk has
no component along the is direction. Finally, we must compute
Dr
The velocity vector is
the is component of -2w x
Dr
dt
_ Bit + sin 6 rpis
Also w = w(- cos A j - sin X k), so that we must find the relationship between i i2, is and i, j, k.
Now
r = i, = sin0cosci+sin Bsin cpj+cos0k
ail
i2
cos 0 cos v i + cos 0 sin
= aB =
_
ail
1
sin B ai
1s
sin 0 k
- sin sp i + coo sp j
Thus
i = (i il)i, + (i i2)i2 + (i
i3)is
= sin 0 cos Tp it + cos 0 cos jP is -- sin rp is
j = sin 0 sin Sp i, + cos B sin 'P i2 + C0803
k = cos 0 it - sin B i2
11
is
i2
-2wx-=2w cos A sin 0 sin P
+ sin A cos 0
0
-sinAsin0
6
sine c
and
(-
r
2w x -D
dt
= #(sin A sin 0 sin ip + sin A cos 0)
Equation (361) yields
2 cos 006+ sin 8;p = 2w(B sin A sin B sin rp + # sin A cos 0)
(364)
MECHANICS
SEC. 102]
For small oscillations, sin 0
215
0, and (364) reduces to
, = w sin X
(365)
Hence the pendulum rotates about the vertical in the clockwise sense when viewed from the point of suspension with an
angular speed w sin X. At latitude 30° the time for one complete
oscillation is 48 hours.
5. Find the equation of motion by considering the i2 components of (361) for the Foucault pendulum.
102. Motion of a Rigid Body with One Point Fixed. The
motion of a rigid body with one point fixed will depend on the
forces acting on the body. Let O-x-y-z be a coordinate system
fixed in the moving body, and let O-- be the coordinate system
fixed in space. 0 is the fixed point of the body. In Sec. 94 we
saw that
-Or
= Lo.
Now Ho* = f if r x p dt dr. We can
replace dt by to x r (w unknown). Thus
H0? = f f f pr x (w x r) dr
R
= fJf p[r2w - (r
w)r] dr
(366)
R
Let
w=w=i+wyj+w
r=xi+yj+zk
so that
r2w - (r w)r = (x2 + y2 + z2) (w i + wyj + W k)
+ (xwx + ywy + zws) (xi + yj + zk)
= [(y2 + z2)wz - xywy - xzws]i
+ [-xyw. + (22 + x2)wy - yzw=lj
+ [-xzwz - yzwy -- (x2 + y2)w]k
We thus obtain
H0? = i[w= f f f p(y2 + z2) dr - w f f f pxy dr - wz f f f pxz dr]
+ j[ - w=f f f pxy dr + wyf f f p(z2 + x2) dr - w: f f f pyz dr]
+ k[ -w=f f f pxz dr - wyf f f pyz dr + w=f f f p(x2 + y2) dr] (367)
216
VECTOR AND TENSOR ANALYSIS
[SEC. 102
The quantities
A = f f f p(y2 + Z2) dr
B=fffp(z2+x2)dr
C = fffp(x2 + y2) dr
D = f f fpyzdr
E = f f fpzxdr
(368)
F = f f f pxy dr
are independent of the motion and are constants of the body.
That they are independent of the motion is seen from the fact
that for a particle with coordinates x, y, z, the scalars x, y, z
remain invariant because the O-x-y-z frame is fixed in the body.
The quantities A, B, C are the moments of inertia about the
x, y, z axes, and D, E, F are called the products of inertia. We
assume the student has studied these integrals in the integral
calculus.
Now from Sec. 99 we have
Lo =
dHor
dt
_
DHor
+ w x Hor so that
dt
DRO,
dt + to x Hor
Hence
L=i+Lvj+LA=i(A
+
ds-F
+j(-F
w
+ k \-E
dt
dtE
dts/
+Bdty-Ddt;/
- D dt + C dta/
i
k
wy
wt
Aws - Fwv - Ewz,
WV
-Fws+Bwv - Dw=,
-Ews - Dwv+Ccos
(369)
In the special case when the axes are so chosen that the
products of inertia vanish (see Sec. 107), we have Euler's celebrated equations of motion :
MECHANICS
SEc. 1031
217.
L. = A d[ + (C - B)wYwz
L = B
dwY
(370)
+ (A - C)w,wZ
dt
L. = C
dw,
dt
+ (B
A)wzwy
103. Applications. If no torques are applied to the body of
Sec. 102, Euler's equations reduce to
A
(i)
dwz
dt-
+ (C
B)wYw, = 0
B dt +(A-C)w,wx=0
(371)
C d,+(B-A}wZwY=0
Multiplying (i), (ii), (iii) by
w,, respectively, and adding,
we obtain
Aws
dws
dt
+ Bw dwY + Cw, dwa = 0
dt
dt
Integrating yields
Aws2 + Bw,,2 + Cw,2 = constant
(372)
This is one of the integrals of the motion. We obtain another
integral by multiplying (i), (ii), (iii) by Aw,z, Bw,,, Cw,, and adding.
This yields
AZ w,
dws
dt
+
BZwy'dw'
dt
+
C2w,
dw,
dt
=
0
so that
A 2w12 + B2wV2 + C2w,2 =constant
(373)
If originally the motion was that of a rotation of angular
velocity w about a principal axis (x axis), then initially
VECTOR AND TENSOR ANALYSIS
218
[SEc. 103
wz(0) = coo
0
(374)
ws(0) = 0
and we notice that (371) and the boundary condition (374) are
satisfied by
wt(t) = -coo
0
wZ(t) = 0
so that the motion continues to be one of constant angular
velocity about the x axis. Here we have used a theorem on the
uniqueness of solutions for a system of differential equations.
Now suppose the body to be rotating this way and then
slightly disturbed, so that now the body has acquired the very
We can neglect ww: as compared
Euler's equations now become
small angular velocities w,,, wt.
to
and wswo.
B dty + (A - C)wswo = 0
Cdt
+(B0
(375)
wo = constant
Differentiating the first equation of (375) with respect to time
and eliminating d s' we obtain
B
z
dtzy +
(- C)((A - B) w02w = 0
(376)
If A is greater than B and C or smaller than B and C, then
(A - C) (A - B)
a2 =
C
> 0, and the solution to (376) is
wy = L cos (at + a)
Also ws =
aBL sin (at + a)
by replacing w,, in (375).
wo(A - C)
SEC. 1041
MECHANICS
219
Problems
1. Solve the free body with A = B for wzj co,, co,.
2. A disk (B = C) rotates about its x axis (perpendicular to
the plane of the disk) with constant angular speed wo. A constant torque Lo is applied constantly in the y direction. Find
w and W..
3. Show that a necessary and sufficient condition that a rigid
body be in static equilibrium is that the sum of the external
forces and external torques vanish.
4. A sphere rotates about its fixed center. If the only forces
acting on the sphere are applied at the center, show that the initial
motion continues.
5. In Prob. 2 a constant torque Lo is also applied in the z
direction.
Find w,, and ws.
104. Euler's Angular Coordinates. More complicated problems can be solved by use of Euler's angular coordinates. Let
O-x'-y'-z' be a cartesian coordinate system fixed in space, and
let O-x-y-z be fixed in the moving body (Fig. 92).
The x-y plane will intersect the x'-y' plane in a line, called the
nodal line N. Let 0 be the angle between the z and z' axes,
L' the angle between the x' and N axes, and (p the angle between
the nodal line and the x axis. The positive directions of these
angles are indicated in the figure.
The three angles &, 0, rp completely specify the configuration
of the body. Now d represents the rotation of the O-z'-N-T'
frame relative to the O-x'-y'-z' frame; de represents the rotation
of the O-z-N-T frame relative to the O-z'-N-T' frame, and finally,
d(P
represents the rotation of the O-x-y-z frame relative to the
O-z-N-T frame. Therefore
+ dO + dt gives us the angular
velocity of the O-x-y-z framed*
relative to the fixed O-x'-y'-z' frame,
and
d1 + dO + d
(377)
220
VECTOR AND TENSOR ANALYSIS
[SEC. 104
The three angular velocities are not mutually perpendicular.
We now define i, j, k, i', j', k', N, T', T as unit vectors along the
x, y, z, x', y', z', N, T', T axes, respectively. Thus
w =
d
k
wzi + w,j + wLk
wz'i' + wy j' + wZ k'
Fia. 92.
Now it is easy to verify that
i = cos rp N + sin cp T
j = - sinpN+cosjpT
i' = cos 4, N - sin ' T'
(378)
MECHANICS
SEC. 1051
221
sin 4, N+cos#T'
sin psin B
cosspsin 0
sin
so that
ki
ws
at
at
at
sin p sin 0 d + cos s dB
day
do
at
dp
at
w = cos (p sin 8 --- - sin (p day
W: = cos B d +
dt
Rewriting this, we have
dt
wy =
d4,
sin 0 sin (P +
sin B cos
wI = d cos 8 +Also
wi=wss+w2+(O2=
cos <p
- dt8 sin So
(379)
d
dd
(d)2
do
d2
(dB 2
+
+Cdt
+ 2 cos 8
dcp d#
dt dt
(380)
For the fixed frame
a
w=-
wy
do
dcp
sin>Gt - cos0sin8dt
(381)
d +cos6d
105. Motion of a Free Top about a Fixed Point. Let us
assume that no torques exist and that the top is symmetric
(A = B). Euler's equations become
VECTOR AND TENSOR ANALYSIS
222
(i)
A
(ii)
A d' + (A - C)cvxwZ = 0
[SEC. 106
(382)
Cdts=O
(iii)
Integrating (iii), we obtain wz = w, = constant.
and add to (i). We obtain
by i =
Multiply
A dt (w= + iwv) + (C - A)wo(wy - iwz) = 0
or
A
d (co., + iwy) = iwo(C - A)(wx + i
)
Integrating,
wz + 2Wy =
ae<'(C-A)/ALot
so that
co. = a cos at
wy = a sin at
(383)
where a = [(C - A)/A]wo and a is a constant of integration.
Now w2 = ws2 + Wye + ws2 = a2 + wp2 = constant, so that the
magnitude of the angular velocity remains constant during the
motion.
Moreover,
fixed space.
d
= 0, so that H is a constant vector in
We choose the z' axis for the direction of H. Now
H = Awzi + Bw,j + Cwk
= Aa cos at i + Aa sin at j + Cwok
(384)
This shows that H rotates around the z axis (of the body) with
constant angular speed a = [(C - A)/A]wo, and since H is fixed
in space, it is the z axis of the body which is rotating about the
fixed z' axis with constant angular speed -a = [(A - C)/A]wo.
Also H k = I HI cos 0 = Cwo, so that 9 is a constant since
I HI = constant. We say that the top precesses about the z' axis.
106. The Top (Continued). We have assumed above that the
weight of the top or gyroscope was negligible, or that the gyroscope was balanced, that is, suspended with its center of mass
at the point of support, so that no torques were produced. We
MECHANICS
SEC. 106]
223
shall now assume that the center of mass, while still located on
the axis of symmetry, is not at the point of support. We now
have the following situation (Fig. 93) :
L = 1kx(.-Wk')
= WI sin ON
The three components of the torque are
Lz = WI sin 0 cos 'v
L" = - Wl sin 0 sin p
L. = 0
ZI
Fro. 93.
Euler's equations become
Wl sin 0 cos p = A
- WI sin 8 sin (p = A
dw=
dt
dwdt"
+ (C - A)w"ws
+ (A - C)wws
(385)
0=CdsforA=B
Multiplying Eqs. (385) by wx, w", Co., respectively, and adding, we obtain
Hence w: = coo.
2
d (Awx2 + Bw"2 + Cw,2) = Wl sin 0(wy cos
w" sin (p)
(386)
224
VECTOR AND TENSOR ANALYSIS
[SEC. 106
From (379) we have wz cos P - w sin cp = de, so that (386)
becomes
1 d (A w=2 +
d8
B
2 dt
dt
and integrating
Cwz2 = -2W1 cos 0 + k
Awx2 +
or, again using (379),
(dO)2
(j)2
= a - a cos B
sin2 B +
(387)
a and a are constants.
Now since Le = L k' = 0) we have HZ- = constant. Also
H = AwJ + Bw j + Cwsk, so that
9+Cwocos0
sin
= constant
Replacing wr and w by their equals from (379), we have
A
sin2 9
sine,p + d sin (p sin B cos
+ di cos2 V sin2 0
- dte cos V sin (p sin 9) + Cwo cos 8 = constant
(388)
or A d sin2 9 + Cwo cos 9 = constant = He.
Let 16 = H,,/A, b = Cwo/A, so that (388) becomes
d'
-bcos9
dt - sin2 9
(389)
From (379)
d4'
w, = wo =
cos B + LIP
dt
(390)
dt
Using (389), (387) becomes
ll2
(
sin9 s 9/
x
+(de = a - a cos 0
(391)
MECHANICS
SEC. 107]
Let z
cos 0, so that dt = - sin 6
225
-, and
2
( - bz)2 +
Cdt
= (a - az) (1 - z2)
Hence
t = fos [(a - az) (1 - z2) - (g - bz) 2]-} dz
(392)
This integral belongs to the class of elliptic integrals. If we can
integrate and find z, then we shall know
9-bz
dip
dt - 1 - z2'
dt
d4,
= wa
d'y
- dt
The reader should look up a complete discussion of elliptic
integrals in the literature.
Fm. 94.
107. Inertia Tensor. The moment of inertia of a rigid body
about a line through the origin may be computed as follows.
Let the line L be given by the unit vector ro = li + mj + nk,
and let r be the vector from 0 to any point P in the body,
r = xi + yj + zk
(see Fig. 94).
The shortest distance from P to L is given by
226
VECTOR AND TENSOR ANALYSIS
[SEC. 107
D2 = r2 -- (r r0)2
_ (x2 + y2 + z2) - (lx + my + nz)2
(12 + m2 + n2)(x2 + y2 + z2) - (lx + my + nz)2
_ 1l2(y2
=
+ z2) + m2(z2 + x2) + n2(x2 + y2) - 2mnyz
- 2lnzx - 2lmxy
Thus
I = f f fpD2dzdydx
= Ale + Bm2 + Cn2 - 2mnD - 2n1E - 2lmF
Let us replace 1, m, n by the variables x, y, z, and let us consider
the surface
,p(x, y, z) = Axe + By2 + Cz2 --- 2Dyz - 2Ezx - 2Fxy
= 1 (393)
A line L through the origin is given by the equation x = It,
y = mt, z = nt. This line intersects the ellipsoid rp(x, y, z) = 1
for t satisfying
(Al2 + Bm2 + Cn2 - 2Dmn - 2n1E - 2lmF)t2 = 1
or t2 = 1/I. The distance from the origin to this point of intersection is given by
d = (1212 + m2t2 + n2t2)} = t =
I-f
so that
(394)
We know that a rotation of axes will keep I fixed, for the line
and the body will be similarly situated after the rotation. We
now attempt to simplify the equation of the quadric surface
,p(x, y, z) = 1. First, let us find a point P on this surface at
which the normal will be parallel to the radius vector to this
point. The normal to the surface is given by Vip, so that we
desire r parallel to Vv, which yields the equations
Ax - Ez - Fy
By - Dz - Fx
Cz - Dy - Ex
x
y
z
(395)
Any orthogonal transformation (Example 8) will preserve the
form of (393) and (395) with x, y, z replaced by x', y', z' and
A, B, . . . , F replaced by A', B', .. . , P. Now choose the
227
MECHANICS
SEC. 107]
z' axis through P so that x' = 0, y' = 0, z' _
satisfy (395).
This yields - E'/O = - D'/0 = C', which means that
E'=D'=0
and (393) reduces to
A'x'' + B'y'' + C'z'' - 2F'x'y' = 1
The rotation
(396)
x"=x'cos0-y'sin0
y"= x' sin0+y'cos0
z" = z'
with tan 20 = F'/(B' - A') reduces (396) to
A"z" + D''y.,, + cf/z,F= = 1
(397)
This is the canonical form desired. We have thus proved the
important theorem that a quadratic form of the type (393) can
always be reduced to a sum of squares of the form (397) by a
rotation of axes. In the proof we made the assumption that
there was a point P such that r is parallel to V o, which yielded
(395). We could have arrived at Eqs. (395) by asking at what
point on the sphere x2 + y2 + z2 = 1 is (p(x, y, z) a maximum.
Since p(x, y, z) is continuous on the compact set
x2+y2+z2 = 1
such a point always exists. Equations (395) are then easily
deduced by Lagrange's method of multipliers.
We can arrange the constants of inertia into a square matrix
-E
I = -F
B -D
-E -D
(398)
C
The elements of the matrix (an array of elements) are called the
components of I. Under a proper rotation we have shown that
we can write
A" 0
I= 0
0
B"
.
0
0
0
C"
(399)
228
VECTOR AND TENSOR ANALYSIS
[SEC. 107
In general, under an orthogonal transformation, I will become
- E'
-F1
A'
I = -F'
-D'
B'
-D'
- E'
(400)
C'
and the components of I in (400) will be related to the components of I in (398) according to a certain law. We shall see in
Chap. 8 that I is a tensor and so is called the inertia tensor.
Referring back to (367), we may write
A -F -E
H _ -F B -D
wZ
Hz
ws
H=
-E -D
wb
C
(401)
from the definition of multiplication of matrices, where
w = w1 + wyJ + w
If we write (398) as
111
121
I31
1112
122
I32
113
123
I3
and
Ho' = H,i + H2J + H3k
w=w1i + w2j + w3k, then (367) may be written
3
H, = I, I,-way
j= 1, 2, 3
(402)
aa1
which is equivalent to the matrix form (401).
Problems
1. Find the moments and products of inertia for a uniform
cube, taking the cube edges as axes.
2. Show that the moment of inertia of a body about any line
is equal to its moment of inertia about a parallel line through the
center of mass, plus the product of the total mass and the square
of the distance from the line to the center of mass.
3. Find the angular-momentum vector of a thin rectangular
sheet rotating about one of its diagonals with constant angular
speed we.
229
MECHANICS
SEC. 1071
4. If
3
Hp =
3
Ipawa,
Hp = I Ipacoa
a-1
a=1
3
lip = I ap'Ha,
3
coo =
a-1
apawa,
3 = 1, 2, 3
a-1
for arbitrary wa, show that
3
3
1 Ipaaa = I Ia°apa,
a-1
6, a = 1, 2, 3
amt
5. Let us consider the form
I = x2 + 9y2 + 18z2 - 2xy - 2xz + 18yz
We may write
I = (x2 - 2xy - 2xz) + 9y2 + 18yz + 18z2
(x-y-z)2+8y2+16yz+1722
y - z)2 + 8(y + z)2 + 9z2
X2 +Y2 + Z2
(x
where X = x - y - z, Y = V"8- (y + z), Z = 3z, a set of linear
transformations from x, y, z to X, Y, Z.
This method may be employed to reduce any quadratic form
to normal form. However, the linear transformations may not
be a rotation of axes. Reduce I to normal form by a rotation of
axes.
CHAPTER 7
HYDRODYNAMICS AND ELASTICITY
108. Pressure. The science of hydrodynamics deals with the
motion of fluids. We shall be interested in liquids and gases, a
liquid or gas being defined as a collection of molecules, which,
when studied macroscopically, appear to be continuous in structure. A liquid differs from a
z
solid in that the liquid will
yield to any shearing stress,
however small, if the stress
is continued long enough.
All liquids are compressible
to a slight extent, but for
many purposes it is simpler
to consider the liquid as being
incompressible. We shall
also be highly interested in
FIG. 95.
These are
perfect fluids.
liquids which possess no shearing stresses.
We now show that the pressure is the same in all directions
for a perfect fluid. Let us consider the motion of the tetrahedron ORST (see Fig. 95). The face ORT has a force acting
on it, since it is in contact with other parts of the liquid. Under
the above assumption, this force acts normal to the face. Call it
Af,,. If we divide Af" by the area of the face ORT, AA,,, we
obtain the pressure on this face, P,, =
f"
AAy
The limit of this
quotient is called the pressure in the direction normal to the face
ORT. The y component of the pressure on the face RST is
P cos l4. Let f" be the y component of the external force per
unit volume, and let p be the density of the fluid.
of motion in the y direction is given by
The equation
+f"oT=d
=
230
p A zr
y
dt°
(403)
HYDRODYNAMICS AND ELASTICITY
SEC. 109]
since
dt
(p Ar) =
dm
dt
231
= 0. Now AA = AA,. cos 0, so that (403)
becomes
(Pv - Pn) + fv A
As A --- 0, we have
AA
oAv dt
Oz
Cp
(404)
i
0, so that if we assume f,,, d
dt2
p
finite, we must have P = P.. Similarly, P = P. = P. = p.
Since the normal n for the tetrahedron can be chosen arbitrarily,
the pressure is the same in all directions and p is a point function,
p = p(x, y, z, t). We leave it to the student to prove that at the
boundary of two perfect fluids the pressure is continuous.
109. The Equation of Continuity. Consider a surface S bounding a simply connected region lying entirely inside the liquid.
Let p be the density of the fluid, so that the total mass of the fluid
inside S is given by
p(x,y,z,t)d-r
M=
ff
R
Differentiating with respect to time and remembering that x, y, z
are variables of integration, we obtain
M
dd
=
f J fat dr
(405)
Now there are only three ways in which the mass of the fluid
inside S can change: (1) fluid may be entering or leaving the
surface.
The contribution due to this effect is JJ vp dd.
s
(2) matter may be created (source), or (3) matter may be
destroyed (sink). Let 4,(x, y, z, t) be the amount of matter
created or destroyed per unit volume. For a source, 4, > 0, and
for a sink, 0 < 0. The net gain of fluid is therefore
ffJ1dT_ jfpv.dd
(406)
Equating (405) and (406) and applying the divergence theorem,
we obtain
232
VECTOR AND TENSOR ANALYSIS
ap + V (Pv) = #(x, y, z, t)
[SEC. 109
(407)
This is the equation of continuity. For no source and sink,
(407) reduces to
0
at + V - (Pv) =
(408)
If furthermore the liquid is incompressible, p = constant,
aP
at
= 0, and (408) becomes
Vv=0
(409)
If the motion is irrotational, that is, if f v dr = 0, then
V=
so that the equation of continuity for an incompressible
fluid possessing no sources and sinks and having irrotational
motion is given by
V2V = 0
(410)
We call p the velocity potential. We solve Laplace's equation
for rp, then compute the velocity from v = Vrp.
Problems
1. If the velocity of a fluid is radial, u = u(r, t), show that the
equation of continuity is
ap
at
LP
+ u ar +
P a
r2 ar
(r2u)
Solve this equation for an incompressible fluid, if '(r, t) = 1/r2.
z
_2xyz
x z - 1<)zj+
2. Showthaty=
(x2 + y2)2
i+ (x2 + y2)2
x2 + y2 kiss
possible motion for an incompressible perfect fluid. Is this
motion irrotational?
3. Prove that, if the normal velocity is zero at every point
of the boundary of a liquid occupying a simply connected region,
and moving irrotationally, rp is constant throughout the interior
of that region.
Sic. 110]
233
HYDRODYNAMICS AND ELASTICITY
4. Prove that if v is constant over the boundary of any simply
connected region, then (p has the same constant value throughout
the interior.
5. Express (407) in cylindrical coordinates, spherical coordinates, rectangular coordinates.
110. Equations of Motion for a Perfect Fluid. Let us consider
the motion of a fluid inside a simply connected region of volume
V and boundary S. The forces acting on this volume are
(1) external forces (gravity, etc.), say, f per unit mass; (2)
pressure thrust on the surface, - p dd, since dd points outward.
The total force acting on V is
F = fff pfdr - f f pdd
=
f f f (pf - Vp)dr
The linear momentum of V is
M = J117 pvdr
and the time rate of change of linear momentum is
dMd
=
fffpvdr
I tf
f pdT + f f f v(pdr)
since the volume V changes with time. However, p dr is the
mass of the volume dT, and this remains constant throughout
the motion, so that
d
(p dT) = 0.
Since F =
dd , we obtain
fff(pf-vp)drfffpdr
This equation is true for all V, so that
pf - VP=p dv
d
I
or
dt
f - p VP
This is Euler's equation of motion.
(411)
234
VECTOR AND TENSOR ANALYSIS
From (76) we have that
dvav
dt
at
SEC. 111
+ (v V)v, so that an alterna-
tive form of (411) is
(412)
Vp
Also from Eq. (9) of Sec. 22, Vv2 = 2v x (V x v) + 2(v - V)v, so
that (412) becomes
av
at
1
+1VV2-Vx(VxV)=f-PVp
2
(413)
111. Equations of Motion for an Incompressible Fluid under
the Action of a Conservative Field. If the external field is con-
servative, f = -Vx, so that f - (1/p) Vp = -V[x + (p/p)] if
p = constan t. Hence (413) becomes
(IV
\
-vx(Vxv)=-V(x+P+2v2)
(414)
-
We consider two special cases:
(a) Irrotational motion. v = V\p and V x v = 0, so that (414)
becomes
at
= -v l x + + 2 v2).
\ P
(b) Steady motion. a = 0, so that (414) becomes
l
1
vX(Vxv)=VlX++1v2
P
For this case we immediately have that
v.[V(x+p+1v2)]=0
p
2
Hence V[x + (p/p) + +v2] is normal everywhere to the velocity
field v. Thus v is parallel to the surface X + (p/p) + j v2 =
constant. The curve drawn in the fluid so that its tangents
are parallel to the velocity vectors at corresponding points is
called a streamline. We have proved that for an incompressible
perfect fluid, which moves under the action of conservative
SEC. 111]
HYDRODYNAMICS AND ELASTICITY
235
forces and whose motion is steady, the expression X + (p/P) +
-v2 rernains constant along a streamline. This is the general
form of Bernoulli's theorem. If X remains essentially constant,
then an increase of velocity demands a decrease of pressure, and
conversely.
1.
Problems
If the motion of a perfect incompressible fluid is both steady
and irrotational, show that x + (p/p) + V2 = constant.
2. If the fluid is at rest, dt = 0. Show that V x (pf) = 0,
and hence that f V x f = 0. This is a necessary condition for
equilibrium of a fluid. Why must pf be the gradient of a scalar
if equilibrium is to be possible?
3. If a liquid rotates like a rigid body with constant angular
velocity w = wk and if gravity is the only external force, prove
that p/p = 1w2r2 - gz + constant, where r is the distance from
the z axis.
4. Write (411) in rectangular, cylindrical, and spherical
coordinates.
5. A liquid is in equilibrium under the action of an external
force f = (y + z)i + (z + x)j + (x + y)k. Find the surfaces of
equal pressure.
6. If the motion of the fluid is referred to a moving frame of
reference which rotates with angular velocity w and has translational velocity u, show that the equation of motion is
dw
Dr D2r
du
f--Vp=dt+dt
Xr+wx (wxr)+2wxdt+
dt2
1
P
and that the equation of continuity is
at
(P Rdt-)
r
For a simply connected region R with
boundary S, the kinetic energy of R is
7. The energy equation.
T=ifffpv2dr
R
Let the surface S move so that it always contains all the original
mass of R. Show that
236
VECTOR AND TENSOR ANALYSIS
dT
= ffJ
R
dt
-dv2
dt
= If!
=
[SEc. 112
v.f!Vpdr
f f f v.fdr- f f
f f f pdt(dr)
S
R
(415)
R
Analyze each term of (415).
8. For irrotational flow show that
at =
p(p),
- (x + P + 2 v2) + C(t),
z
f-D(t).
-f- 2 + x +
t
112. The General Motion of a Fluid. Let us consider the
and if p =
velocities of the particles occupying an element of volume of a
Let P be a point of the
volume or region, and let vp
represent the velocity of the
fluid.
fluid at P (Fig. 96). The veloc-
ity at a nearby point Q is
vQ = vp + dvp
= Vp + (dr . V)vp
from (75).
(416)
By (dr V)vp we
mean that after differentiation,
the partial derivatives of v are
calculated at P. We now replace dr by r for convenience,
so that r = xi .+ y} + zk if we
consider P as the origin and x,
Fio. 96.
y, z large in comparison with x2, y2, z2, zy, etc.
now becomes vQ = vp + (r V)vp. Now
Equation (416)
r x (V x w) +
(417)
from (9), (10), (12) of Sec. 22.
Now let
w = (r V)vp = x av + y av + z avl
al p
ay p
azlp
(418)
SEC:. 1121
and hence
HYDRODYNAMICS AND ELASTICITY
-- P+yayay
ax av
ax ax
ay av
237
az avI
P
+z azazP
=w
We did not differentiate the
(' ayP
axP
I
IV P
' since they have
been evaluated at P and so are constants for the moment.
using (417), we obtain
Thus,
w = j V(r w) + J(V x w) x r
Moreover, v
(418)], and
(419)
=Vp+w, so thatV xv =V xw = (Vxv)p [see
VQ = VP+ -(V x v)P
(420)
It is easy to verify that r w is a quadratic form, that is,
Axe+Bye+Cz2+2Dyz+2Ezx+2Fxy
and so by a rotation (Sec. 107), we can write
and
IV(r w) = axi + byj + czk
We may now write (420) as
VQ = VP + w x r + (axi + byj + czk)
(421)
where w = J(V x v)p.
Let us analyze (421), which states that the velocity of Q is the
sum of three parts:
1. The velocity vp of P, which corresponds to a translation of
the element.
2. w x r represents the velocity due to a rotation about a line
through P with angular velocity J(V x v) p.
3. axi + byj + czk represents a velocity relative to P with
components ax, by, cz, respectively, along the x, y, z axes.
The first two are rigid-body motions; they could still take place
if the fluid were a solid. The third term shows that particles at
238
VECTOR AND TENSOR ANALYSIS
(SEc. 113
different distances from P move at different rates relative to P. If
we consider a sphere surrounding P, the spherical element is trans-
lated, rotated, and stretched in the directions of the principal
axes by amounts proportional to a, b, c. Hence the sphere is
deformed into an ellipsoid. This third motion is called a pure
strain and takes place only when a substance is deformable.
Each point of the fluid will have the three principal directions
associated with it. Unfortunately, these directions are not the
same at all points, so that no single coordinate system will suffice
for the complete fluid.
The most general motion of a fluid is that described above and
is independent of the coordinate system used to describe the
motion. It is therefore an intrinsic property of the fluid.
113. Vortex Motion. If at each point of a curve the tangent
vector is parallel to the vector w = J(V x v), we say that the
dx
dy
wz
wy
curve is a vortex line. This implies that dx =
=
dz
where
w,
dx, dy, dz are the components of the tangent vector and
w=wi+wyJ+w L
The integration of this system of differential equations yields
the vortex lines. The vortex lines may change as time goes on,
since, in general, w will depend on the time.
Let us now calculate the circulation around any closed curve
in the fluid.
C=
ff (V xv) dd
(422)
s
r
If V x v = 0, then C = 0. This is true while we keep the curve
r fixed in space. Let us now find out how the circulation
changes with time if we let the particles which comprise P move
according to the motion of the fluid. As time goes on, assuming
continuity of flow, the closed curve will remain closed.
Now
e=
(423}
r
r'
where s is are length along the particular curve v, at some time t.
At an instant later the curve t' has moved to a new position given
by the curve r". The velocity of the particles over this path is
239
HYDRODYNAMICS AND ELASTICITY
SFC. 1141
slightly different from that over r, and, moreover, the unit
tangents ds have changed.
The parameter s is still a variable of
integration and has nothing to do with the time. Therefore
dt
dt
ds
ds + $6 v
Er- ds
= T dt . ds
+
v.
d
dt Pdsl
ds \dt/
ds
ds
(424)
Euler's equation of motion (411) for a conservative field,
f = -VX
is
dv
dt
1
= -VX - Vp = -VV, where V = x + f dp/p.
There-
1
fore
2I
2
dt
_
--0d(V-J2) =0
(425)
We have arrived at a theorem by Lord Kelvin that the circulation around a closed curve composed of a given set of particles
remains constant if the field is conservative, provided that the density
p is a function only of the pressure p.
If we now consider a closed curve
lying on a tube made up of vortex
lines, but not encircling the tube (see
Fig. 97), then
C=
ff
8
s i nce dd i s normal to V x v.
Fr om
Fio. 97.
Kelvin's theorem, C = 0 for all time,
so that the curve r always lies on the vortex tube.
114. Applications
Example 109. Let us consider the steady irrotational motion
of an incompressible fluid when a sphere moves through the fluid
VECTOR AND TENSOR ANALYSIS
240
ISEC. 114
with constant velocity. Let the center of the sphere travel along
the z axis with velocity vo. We choose the center of the sphere
as the origin of our coordinate system. From Sec. 110, Prob. 6,
we have
f - pop=
at
and
0
Hence
Dr
=
Now at points on the surface
so that V2p = 0.
of the sphere we must have CYT)radWLY = 0, so that
(a?Y-a = 0.
Or
We look for a solution of Laplace's equation satisfying this
boundary condition, so that we try
'P =
(see Sec. 67).
(Ar +
!
r/J
cos 8
(426)
We need
B
-CA-
a3
cosB=0
so that B = a2A/2. Moreover, at infinity we expect the velocity of the fluid to be zero, so that the velocity relative to the
sphere should be -vo. Hence
aVs
az z
.= A =
- vo
and
s
P = -vo (r + 2r2
cos 0
(427)
The velocity of the fluid relative to the sphere is given by v = Vsp
and the velocity of the fluid is v = Vp + vok.
Example 110. Let us consider a fluid resting on a horizontal
surface (x-y plane) and take z vertical. Let us assume a transverse wave traveling in the x direction. For an incompressible
Sec. 114]
HYDRODYNAMICS AND ELASTICITY
241
fluid
z+az?
°2`P-ax
0
2
(428)
We assume a solution of the form p = A (z)e
aLT(X-'i)t.
Sub-
stituting into (428), we obtain
2w
e
tZ_voi f
41x2
- 2 A (z) +
d2A
dx2
I=0
so that
d2A
_
41r2
(429)
\2 A
dz2
The solution to (429) is A = Ape<zr° + Boe (2r/1`)', and a real
solution to (428) is
(x - vt)]
(Aoe(2r/X)z + Boe-(zrn*)z) cos
(430)
L
The fluid has no vertical velocity at the bottom of the plane on
which it rests, so that v. =
= 0 at z = 0. This yields
L(P
az
A 0 = B0, so that
= Ao(e(2r')= + ec2r/l.>=) cos
I2r
(x - Vi)
J
= 20 cosh (-- z)
cos
[
xr (x - vt)]
(431)
From Prob. 8, Sec. 110, we have
-(x++2v2)+C(t)
at
and for a gravitational potential, x = gz, so that
at
=
- (gz +
p
+
2
v2)
+ C(t)
(432)
We now assume that the waves are restricted to small amplitudes and velocities, so that we neglect Jv2. Moreover, at the
VECTOR AND TENSOR ANALYSIS
242
[SEC. 114
surface, p, the atmospheric pressure, is essentially constant, so
that dp = 0. Differentiating (432), we obtain
a2(
_
dC
az
.9 at
ate
+
(433)
dl
and again at the surface at = vZ = az , so that (433) becomes
dC
a_p
a2V
(434)
g az + dt
ate
Substituting (431) into (434), we obtain
2
zcosLA (x - Vt)
-v2 a2 A0 cosh
g
2w-
z cos
I2r
(x - vt) ] + dt
(435)
In order for C to be dependent only on t, we must have the
coefficient of cos [(27r/A) (x - vt) ]identically zero in (435).
Ao
(- v
227r2
cosh
27r
z+
g
sink
27r
z=0
Hence
(435a)
or
v2 =
g tanh - z
In deep water z/X is large so that tank
z
1, and the
velocity of the wave is v = (Ag/2,r)*.
Problems
1. Show that for steady motion of an incompressible fluid
under the action of conservative forces, (v O)w - (w V)v = 0,
wherew = D x v.
2. Show that dt (Pl = \P of v for a conservative system.
SEc. 1151
HYDRODYNAMICS AND ELASTICITY
243
3. If C is the circulation around any closed circuit moving
with the fluid, prove that
dC
=
p d (1) if the field is con-
servative and if the pressure depends only on the density.
4. Show that v = 2axyi + a(x2 - y2)j is a possible velocity
of an incompressible fluid.
5. Verify that the velocity potential (p = A[r + (a2/r)] cos 0
represents a stream motion past a fixed circular cylinder.
115. Small Displacements. Strain Tensor. In the absence
of external forces, a solid body remains in equilibrium and the
forces between the various particles of the solid are in equilibrium
because of the configuration of the particles. If external forces
are added, the particles (atoms, molecules) tend to redistribute
themselves so that equilibrium will occur again. Here we are
interested in the kinematic relationship between the old positions
r
p°
of equilibrium and the new. We
shall assume that the deformations are small and continuous.
We expect, from Sec. 112, that
so
po
r
p
Is
p
Fia. 98 .
in the neighborhood of a given point Po, the remaining points will
be rotated about Po and will suffer a pure strain relative to Po.
Let r be the position vector of P relative to Po, and let s be the
displacement vector suffered by P, and so the displacement suffered by Po (Fig. 98). Then
S = so + ds = so + (r
V)so
(436)
Let s = u(x, y, z, t)i + v(x, y, z, t)j + w(x, y, z, t)k. Since we
will be dealing with static conditions,
s = u(x, y, z)i + v(x, y, z)j + w(x, y, z)k
From (420),
S = so+J(V xs)P,
(437)
where
w=x-asax
as
PO
as
+yay Pc +zoz
PO
since s = v At.
We are interested in the position of P after the deformation
(now P') relative to the new position of Po (now Po ). This is
244
VECTOR AND TENSOR ANALYSIS
[SEC. 115
the vector r' = r `}- s - so, or
r' = r+4(D xs)po x r +
(438)
Since J(V x s)P, x r represents a rigid-body rotation about P0,
we ignore this nondeformation term and so are interested in
r + 'I'p(r w). Now
auj
aw
a-l
+2 D Cx tax +xyax +xz at
av
au
awl
I r
+ yz D xy
1
Po
+2
y Po
P.
+
y2
ay PO
0
ay Po
aw
+ 2D
zx azlPo
+ zy a P.
az Po
and
r+
1
aul
x Ci
x
+
(au
avl
x (aw au
ax! + 2 \ax + az J i
au
_y
+ 2 Cay +
ax
av
av
_z (aw
2y+
+yCl+ay
x (aw
y aw
au
av
az)+2Cay+az
8v
aw 1
az
Jk
(439)
The partial derivatives are evaluated at the point Po.
Let us now consider the matrix
au
i + ax
lIsr'II =
au
av
2 ay
+ Ox
1
+
+ ax/
i aw aul i law
2 Cay
2 C 8x
+
y
azl 2 `ay +
(440)
av
az
The nine components of this matrix form the strain tensor. If
we write r = x'i + x2j + xak and r' = y'i + y2j + yak
(see
245
HYDRODYNAMICS AND ELASTICITY
SEC. 115)
Example 8), then r' = r + I V(r w) may be written
i = 1, 2,3
yi = sfix' + S2 'X2 + 83ix3,
or
3
y' =
(441)
Sa'xa
a-1
We shall see in Chap. 8 that since r and r' are vectors, then, of
necessity, the s/ are the components of a tensor. Notice that
Sii = s; , so that the tensor is symmetric.
The ellipsoid which has the equation
(
aul
avl
(
1 + ax)xr+ 1 +ay y2+
aw
(awl
1
ay +aavlxy
+ az)z2+
au
aw
av
+ (ay + az
(au
yz + (ax
+
az
zx = 1
(442)
is called the strain ellipsoid. From Sec. 107 we know that we
can reduce the ellipsoid to the form
Ax" + By'2 + Cz'z = 1
by a proper rotation. The strain tensor becomes entirely
diagonal,
A
0
0
0 B 0
0 C
0
In the directions of the new x', y', z' axes, the deformation is a
pure translation, and these directions are called the principal
directions of the strain ellipsoid.
Let us now compute the change in the unit vectors, neglecting
the rotation term. The unit vector i has the components
(1, 0, 0), so that from (438) and (439)
i
(
au
1
au
av
- r1 = \1 + ax) 1 + 2 (ay +
1
aw
ax 1 + 2
ax
(LU)
By neglecting higher terms such as
_
Iril
I
1+
au
ax
21
+
az/ k
u-
y
I
.
a_ul
Similarly j --> r2, and lr2l = 1
we have
+
avl
ay
; k - rs,
246
VECTOR AND TENSOR ANALYSIS
and fr31 = 1 +
awl
[SEC. 116
The angle between r1 and r2 is given by
.
I
cos B =
au
+r,jjr2+
av
v ay + ax'
.
The terms of the strain tensor are now fully understood. The
volume of the parallelepiped formed by r1, r2, r3 is
au
av
r2xrs = 1+ ax + ay +
V
so that
V
_
V
_ au
aw
az
aw
av
(443)
ax+8y +az
V
The left-hand side of (443) is independent of the coordinate
system, so that V s is an invariant.
Finally, we see that the deformation tensor due to the tensor
- V (r w) has the components
1
2\ay+8x/ 2(ax+ az)
ax
au
8v'\
2 (ay
aw
1
+ ax/
2 \ ax
av
ay
au
1 (au
+
az
2 \ay + az
+
axi/1I
av
aui
2 (axl
au
av
2 (ay
+ az
1
aw
az
(444)
where
u1 =u,
u2=v,
u3=w,
x1=x
x2=y
x3=z
116. The Stress Tensor. Corresponding to any strain in the
body must be an impressed force which produces this strain.
Let us consider a cube with faces perpendicular to the coordinate
axes. In Sec. 108 we assumed no shearing stresses, but now we
consider all forces possible between two neighboring surfaces.
SF:C. 116]
247
HYDRODYNAMICS AND ELASTICITY
Let us consider the face ABCD (Fig. 99). It is in immediate
contact with other particles of the body. As a consequence, the
resultant force tx on the face ABCD can he decomposed into three
forces: txx, tyx, tzx, where txx is the component of tx in the x direc-
tion, t, is the component of tx in the y direction, and tzz is the
,
z
y
Fia. 99.
component of tt in the z direction. We have similar results for
the other two faces and so obtain the matrix
tzz
tzy
tzz
tyz
tyy
tyz
tzz
tzy
G.
(445)
These are the components of the stress tensor.
By considering a tetrahedron as in Sec. 108, we immediately
see that if dd is the vectoral area of the slant face, then the components of the force f on this face are
f, = tzx dsz + tzy d s + tzz dsz
fy = tyx dsz + tyy dsy + tyz dsz
fz = tzz ds, + tzy dsy + tzz dsz
where dsz = i dd, dsy = j dd, dsz = k dd.
(446)
248
VECTOR AND TENSOR ANALYSIS
[SEC. 117
We immediately see that
tyx =
Of.,
fx,
tx
asv
as.,
3
and that fi = I tia dsa, where f, = f,, f2 = f,,, f3 = f., t12 = t,,,
a=1
.
.
.
.
We shall see later that this explains why the ti; are
called the components of a
dv
tensor.
Let us now consider the
resultant force acting on a
volume V with boundary 8
z
(see Fig. 100).
(446)
Wehavefrom
fz = txx dsx + txv ds + tzx dst
so that
Y
fz _
F. =
JJtxzdSx
+ t,, ds, + tx, dss
=
!s J t.dd
Fio. 100.
where t = tz, +
Applying the divergence theorem, we obtain
Fz
- JIf Jr
V
acz=
acZy
ax
+
4-9t"
ay
+
az
dT
t,, k.
(447)
with similar expressions for F,,, F.
By letting V --p 0, we have that the x component of the force
per unit volume must be Vax- +
qty"
y
+
tz
117. Relationship between the Strain and Stress Tensors. In
the neighborhood of a point P in our region, let us choose the
three principal directions of the stress tensor for the axes of our
cartesian coordinate system. If we assume that the region is
isotropic (only contractions and extensions exist), a cube with
faces normal to the principal directions will suffer distortions only
along the principal axes. Hence the principal directions of the
249
HYDRODYNAMICS AND ELASTICITY
SEC. 1171
strain ellipsoid will coincide with those of the stress ellipsoid.
this coordinate system
el
ie17JJ =
0
0
0
It,
0
0
0
0
0
t2
In
0
0
(448)
0
0
Our fundamental postulate relating the shear components with
those of the stress will be Hooke's law, which states that every
tension produces an extension in the direction of the tension and
is proportional to it. We let E (Young's modulus) be the factor
of proportionality. Experiments also show that extensions in
fibers produce transverse contractions. The constant for this
phenomenon is called Poisson's ratio a. We thus obtain for the
relative elongations of the cube in the three principal directions
the following:
1 -SET
a
el =E;ti --(t2+t3)
_
E
E
1
0,
or
t2 -
e2
1
e3 =
E
E(t1+t2+t3)
(t3 + tl) =
1 r
t2 -
E (t1 + t2 + t3)
t3 "-
(t1 + t2 + t3)
(449)
1-r-a-
a
3 - - (11 + t2)
The formulas for el, e2, e3 apply only in the immediate neighbor-
hood of a point P. Since points far removed from P will have
different stress ellipsoids, the principal directions will vary from
point to point. Hence no single coordinate system will exist
that would enable the stress and strain components to be related
by the simple law of (449). Let us therefore transform the
components of the stress and strain tensors so that they may be
referred to a single coordinate system. The reader should read
Chap. 8 to understand what follows. If he desires not to break
the continuity of the present paragraph, he may take formula
(456) with a grain of salt, at least for the present. Example 8,
Probs. 21 and 22 of Sec. 11, and Prob. 21 of Sec. 15 will aid the
reader in what follows.
If x1, x2, x3 are the coordinates above, and if we change to a
new coordinate system 11'. T2, 23 where
3
x' = I a'xx,
a-1
i = 1, 2, 3
(450)
VECTOR AND TENSOR ANALYSIS
250
(SEC. 117
then the transformation (450) is said to be linear. Notice that
the origin (0, 0, 0) remains invariant.. If, furthermore, we desire
distance to be preserved, we must have
3
3
(xi)
(.Ti), =
In Chap. 8 we shall easily show that this requires
=0ifij
3
a=1
= 1if ij
a.,aaia = S=i
(451)
Equation (451) is the requirement that (450) be a rotation of axes.
Moreover, since we are dealing with tensors, we shall see that the
components of the strain tensor in the x'-x2-x3 coordinate system
are related to the components in the x'-a 2-x3 system by the
following rule:
3
3
e;i = I I a,aa1 ea#
(452)
'6=1 a=1
If we now let i = j and sum on i, we obtain
`3
3
3
3
L, ei, =i=1I I0=1
Z a=1
aiaMeag
:=1
3
3
3
Saisea# = I eaa
a-1
0-1 a-1
so that
e11 + 922+933 = e11 + e22 + e33 = el + e2 + e3
(453)
This is an invariant obtained from the strain tensor [see (443)].
A similar expression is obtained for the stress tensor; namely,
that
111+122+133 = t11 +t22+133 = t1+t2+t3
Equations (449) may now be written as
el -1+0
E t1+
e2=1
(454)
e3=1Eor
ts+
HYDRODYN.4.1IICS AND ELASTICITY
SEC. 117]
251
where 4, is the invariant
E (tl + t2 + t3)
-
(III + 122 + 133)
From Eq. (452) we have
3
eij =
3
E aiaa,8eas
B=1 a-1
and since eap = 0 unless a = 6 [see (448)], we obtain
3
3
1 aiaaJaeaa = I aiaajaea
a=1
a=1
3
a
-l
aiaaja (1
1+ a
F,
E
to
+)
3
3
I aiaajala + Y' 1 aiaaja
a-1
a=1
and
1 + o
eij = E
3
3
of + 4,5ij
(455)
3
since 1j = I I aiaajalas = I aiaaala.
6=1 a -l
a=1
Equation (455) is the relationship between the components of
the strain and stress tensors when referred to a single coordinate
system. We have
Q111
ell =
- E (tll + 122 + 133)
1
1
922 =
E
+Q_
E
122 -
oT
-
(111 + t22 + 138)
O
e33 =
1
1
912 =
+ 133 - E, (111 + E22 + 133)
+v_
E
112
1+0.
923 =
E
l23
1 +Q_
981 =
E
131
(456)
252
VECTOR AND TENSOR ANALYSIS
[SFC. 11
Solving Eels. (456) for the t;; and removing the bars, we obtain
t11 =
1+
[ell +
01
1
(ell + e22 + e33)
2au
E
o
aw
av
(au
=i+o, axl-2vax+ay+az
=
tss
=
E rav
au
v
J
av awt22
1+aLay1-2v (ax+ay+az)]
E taw
v
1+aaz+1-2QV*s
E
t12=t21=
t23 = t32 =
=
t31 - t13 =
E
(457)
(au
av
2(1+o) ay+ax
aw
E
av
2(1 + v) \az + ay
E
(aw au
2(1+Q)ax +az
Equation (447) now becomes
E
F=
[d?u
-}- v
ax2
+
(a2u
1 - 20 \49x2
E
+
_
1
2(1+o)
a2w
+ ax ay +
ax az
a2u
2(1 + u) aye
E
-2(1+a)IV
a2v
.
a (au
a2w
a2v
a2u
ay ax
+ az ax + az2
av
aw
1
zu+1 -2aaxax+ay+ az J
[vu+ 1 1
The forces per unit volume in the y and z directions are
Fy
2(1+a)[Dzv+1 12vay(V - s)I
E
{V2W+
a
1
I-2oaz
(0 s)
so that
f=
+
2(1
[V2S
r)
V(V s)]
+ 1 1 2cr
(458)
If we let R = RJ + R j + R2k be the external body force per
unit volume, p the density of the medium, then Newton's second
253
HYDRODYNAMICS AND ELASTICITY
SEc. 117]
law of motion yields
R+ 2(1 +o,)[V2S+-Lv(V.S)]
=
P
For the case R = 0, (459) reduces to
E
2(1 + a) [v2s
+
1
- 2a °(°
2
s)
=p
at2
(460)
In Sec. 70 we saw that a vector could be written as the sum of
a solenoidal and an irrotational vector. Let s= s1 + s2j where
V s, = 0 and V x s2 = 0. Since (460) is linear in s, we can
consider it as satisfied by s, and s2. This yields
E
V2s1_- p a2S,
2(1 + a)
ate
and
(461)
2(1
E
+a,)
1
[V2s2
+ 1 -2a °(° . s2)
a2s2
= p ate
However,
V X (V X 62) = V(V S2) - V 'S2 = 0
so that
E(1 T a)
(1 +a)(1 - 2a)
V2S2
= pa2s2
at2
(462)
In Sec. 80, we saw that (461) leads to a transverse wave moving with speed Vt = E/2(1 + a)p.
Equation (462) is also a wave equation, but the wave is not
transverse. Let us assume that the wave is traveling along the
x axis. Then
s2=s2(X-Vt)
= u(x - Vt)i + v(x - Vt)j + w(x - Vt)k
=0
V X S2 =
axJ+-k=0
clx
254
VECTOR AND TENSOR ANALYSIS
[SEC. 117
so that w and v are independent of x and therefore are independent
of x - Vt. We are not interested in constant displacements, so
that s2 = u(x - Vt)i, and the di,,placement of s-2 is parallel to
the direction of propagation of the wave.
longitudinal. The speed of the wave is
The wave is therefore
L'(1 - o.)
V,=
P(1 + v)(1
2v)
In general, both types of waves are produced, this result being
useful in the study of earthquakes.
Problems
1. Derive (451).
2. If P, f--, f3 are the components of a vector for a cartesian
coordinate system, prove that the components P, P, j'3 of this
vector in a new cartesian coordinate system are related to the old
3
components by the rule J' =
i = 1, 2, 3, using the
n=i
coordinate transformation (450).
3. If the body forces are negligible and if the medium is in a
state of equilibrium, show that V2S +
1 -1
V(V s) = 0.
4. If the strain of Prob. 3 is radial, that is, if s = s(r)r, find the
differential equation satisfied by s(r).
5. Assuming a = 0 for a long thin bar, find the velocity of
propagation of the longitudinal waves.
6. If µ = E/2(1 + v) (modulus of rigidity) and
Ea
A=
(1 + a) (1 - 2Q)
show that Eq. (459) becomes
49
R+AVss+
x
7. Why do we use ats instead of
's
Pats
x
in in (459)?
8. A coaxial cable is made by filling the space between a solid
core of radius a and a concentric cylindrical shell of internal
radius b with rubber. If the core is displaced a small distance
Ssc. 118]
HYDRODYNAMICS AND ELASTICITY
255
axially, find the displacement in the rubber. Assume that end
effects, gravity, and the distortion of the metal can be neglected.
118. Navier-Stokes Equation. We are now in a position to
derive the equations of motion of a viscous fluid. In the case of
nonviscous fluids, we assumed no friction between adjacent layers
of fluid. As a result of friction (viscosity), rapidly moving layers
tend to drag along the slower layers of fluid, and, conversely, the
slower layers tend to retard the motion of the faster layers. It
is found by experiment that the force of viscosity is directly
proportional to the common area A of the two layers and to the
gradient of the velocity normal to the flow. If the fluid is moving in the x-y plane with speed v, then the viscous force is
since av is the gradient of the speed normal to the direction of flow.
17 is called the coefficient of viscosity.
We shall let Pt; be the stress tensor and u;; the strain tensor
for the fluid analogous to t; and e;; of the previous paragraphs.
We have
_
E
au
2(1 + v) ay
P12
Ov
+
ax
where u, v, w are the components of the velocity vector v (see
Sec. 112). For a fluid moving in the y direction with a gradient
in the x direction, we have u = 0 and au = 0, so that
ay
_
P12
Hence the term
E
2(1
E
av _ av
2(1 + v) ax = '' ax
must be replaced by q.
or)
In addition to the stress components due to viscosity, we must
add the stress components due to the pressure field, which we
assume to be
-p
I
0
0
0
0
-p
0
0
-p
VECTOR AND TENSOR ANALYSIS
256
[SEC. 118
The equations of (457) become
(463)
P;i = 2rio;i + X(911 + 022 + r33)aii - pa+i
where X is undetermined as yet.
We see that
Now let i = j and sum on j.
P11 + P22 + P33 = (2rj + 3X)(ail + 022 + 033) - 3p
We know that P11 + P22 + P33 is an invariant and that for the
static case P11 + P22 + P33 = -3p. Consequently we choose
2, + 3X = 0, so that (463) becomes
p;, = P;, = 2+10;1 - I V(0-11 + 022 + 033) - pa+i
(464)
Moreover, the velocity vector is given by
for small velocities.
v = u11 + 1627 + u3k
and 0;i =
1
au;
2 C axi
auiax:
+
' so that div v = V v = 011 + 022 + 033-
To obtain the equations of motion, we note that from (447)
ap11
fl -
ax,
3
c3
= jG
=1
49p12
+
axe
x1 = x
where x2 = y
49p13
+ axi
x3=z
apli
ax
3
and in general f;
I apt'
Hence
-i
du;
pF; + f1 = p dt
becomes
3
pF. +
ap;i
axi
du:
(465)
P dt
j =1
where F; is the external force per unit mass.
From (464)
a0;i
2 a(diy v)
ap
ap;i
17
= 2,i
axi - 3
axi
a`i - axi a+i
axi
_ 'q a au; au) 2 a(V v)
+1
axi
3
= axi axi + ax'
-
-
ap
a; - axi
257
HYDRODYNAMICS AND ELASTICITY
SEC. 1181
and
3
api;
3
a
au;
au;
2
+ axi/
1-1 ax?
3'
v)
a (V
ap
ax,
The equations of motion (465) are
dui
p dt
- = pFi + v
I
3
J
1
a aui
ax' Cax'
ap
2 a(V v)
+ au;
-7
ax'
axi
axi
3
(466)
or
pat
= pf +,q V2v - - V(V v) - Vp
(467)
For an incompressible fluid V v = 0, and
pd
Along with (467) we have the equation of continuity
at
=0
Problems
1. Derive (467) from (466).
2. Consider the steady flow of an incompressible fluid through
a small cylindrical tube of radius a in a nonexternal field. Let
v = vk and show that p = p(z) and +1 V2v =
ap
Show that the
boundary conditions are v = 0 when r = a, and v = v(r),
r2 = x2 + y2, and that
= r dr (r dr) Hence show that
v = (A/4i7)(r2 - a2), where A is a constant and LP = A.
3. Consider a sphere moving with constant velocity vok (along
the z axis) in an infinite mass of incompressible fluid. Choose
the center of the sphere as the origin of our coordinate system.
Show that the equation of motion is p aE = q V2v - Vp and that
the boundary conditions are v = 0 for r = a, v = -vok at r = oo
258
VECTOR AND TENSOR ANALYSIS
[SEC. 118
and that for steady motion a = 0 for any quantity 4, associated
with the motion. Moreover V v = 0. We shall assume that v
and the partial derivatives of v are small. Show that this implies
(i)
Vp = I V2v
Hence prove that V2p = 0.
Now let v = -V(p + wlk and show
that
awi
=v4o
z
49Z
Assuming p = -,1 V2p and V2w1 = 0, show that (i) and (ii)
will be satisfied.
Let w, = 3voa/2r, rp = coz - (voa3z/4r3) + (3voaz/4r),
P=
3,gvoaz
2r3
and show that
V2w1 = 0,
V2p = 0,
p = -n V2rp
v= -V(p+wik=0forr=a
v = -vok for r = co
4. Solve for the steady motion of an incompressible viscous
fluid between two parallel plates, one of the plates fixed, the
other moving at a constant velocity, the distance between the
plates remaining constant.
5. Find the steady motion of an incompressible, viscous fluid
surrounding a sphere rotating about a diameter with constant
angular velocity.
No external forces exist.
CHAPTER 8
TENSOR ANALYSIS AND RIEMANNIAN GEOMETRY
We shall be interested in sums
119. Summation Notation.
of the type
S = alx, + a2x2 + .
. + anxn
.
(468)
We can shorten the writing of (468) and write
n
S = Z aixi
(469)
i=i
Now it will be much more convenient to replace the subscripts
of the quantities x:, x2,
.
. .
, x by superscripts, x', x2,
. . . , X11.
The superscripts do not stand for powers but are labels that
allow us to distinguish between the various x's. Our sum S now
becomes
n
S=
(470)
aixi
We can get rid of the summation sign and write
(471)
S = aixi
where the repeated index i is to be summed from I to n. This
notation is due to Einstein.
Whenever a letter appears once as a subscript and once as a
superscript, we shall mean that a summation is to occur on this
letter. If we are dealing with n dimensions, we shall sum from
1 to n. The index of summation is a dummy index since the
final result is independent of the letter used. We can write
S = a;xi = a;x' = a,,xa = as0.
Example 111. If f = f(x', x2,
calculus
259
,
xn), we have from the
0,60
VECTOR AND TENSOR ANALYSIS
. + -CIx"dx"
df = ax dxl + ax dx2 +
_
= of
[SEC. 120
of
dx'
ax'
dxa
axa
;.nd
df
afdJxa
dt - axa dt
The index a occurs both as
Hence we first sum on a, say from
Example 112. Let S = gasxax#.
a subscript and superscript.
1 to 3. This yields
S = 9iax'x8 + 92ax'x' + g38xaxe
Now each term of S has the repeated index # summed, say, from
I to 3. Hence
S = g11x'x1 + 912x'x2 + 913x'x3 + 921x2x' + g22x2x2 + 923X2x3
+ 931x3x' + g32x3x2 + 933x8x3
and S = gaoxaxa represents the double sum
3
3
S = Z I g-Oxaxs
0-1 a-1
We also notice that the gap can be thought of as elements of a
square matrix
gii 912 913
921
922
923
931
932
933
120. The Kronecker Deltas. We define the Kronecker o to
be equal to zero if i - j and to equal one if i = j:
a;=0,ipj
1,i=j
We notice that Si = 52 _ 1
1
b_ = 03 =
= SA = 1,
=
r
+1 = u
(472)
TENSOR ANALYSIS
SEC. 120)
If xI, x2,
261
... , x" are n independent variables, then
ax'
axl
= S,
i
for if
axe = 1, and if i
j, there is no change in the vari-
able x' if we change x1 since they are independent variables, so
that
ax!
ax1
= 0.
Example 113. Let S = aaxa.
as
-=
ax"
Then
as ga"
cis
= as ax , and
ax
Now Sµ = 0 except when a = µ, so that on summing on a we
a(axa)
obtain
= a".
ax"
Let S = apxax# = 0 for all values of the vari. . , x^.
We show that a,, + a;; = 0. First
differentiate S with respect to x' and obtain
Example 114.
ables x', x2, .
T
axa
axp
as =
aa$xa
xa = 0
+
a,
ax'
ax'
axi
= aaflxaaf + aa$67x8 = 0
= aaixa + a;se = 0
Now differentiate with respect to xi so that
028
axe
ax'
- a.v + aiob = 0
and
a,,+a;,
0
We define the generalized Kronecker delta
as follows:
The superscripts and subscripts can have any value from 1 to n.
If at least two superscripts or at least two subscripts have the
same value, or if the subscripts are not the same set of numbers
as the superscripts, then we define the generalized Kronecker
delta to be zero. If all the superscripts and subscripts are separately distinct, and the subscripts are the same set of numbers
Qs the superscripts, the delta has the value of + 1, or --1, accord-
262
VECTOR AND TENSOR ANALYSIS
[SEC. 120
ing to whether it requires an even or odd number of permutations
to arrange the superscripts in the same order as the subscripts.
For example, 6x23 = 1, 6213 - - 1, 61231
23 = 1, 6163 = 0,
a1438 - -11 a123
=
0221
= 6312
323
0,
61213
=0
It is convenient to define
eilh ... i.. =
(473)
and
Eiji, ... i. =
Problems
1. Write in full aaxa=bi,a,i= 1, 2, 3.
2. If aae7xax8xY = 0, show that ai;k + aki; + a; + aJik +
akii + aiki = 0.
3i,ti:'412 ...n
i.,i,...i.
6;,i:...;..
3. Show that 612..,,
i. =
4. If yi = aaxa, zi = by, show that zi = bas;x#.
5. Prove that
bas = are - a8r
baBratYaapi = arst + airs
+ astr - sari - aisr - arts
6. Show that the determinant I at at can be written
2
a1
a11
a2'
atz
a2
2
a2
= Etiiaia? = etiia1
2
and that
at
a2
2
z
a1
8
a1
1
a3
at
z
a2
a3
8
a$$
a2
7. Prove that ei1i, ... ie =
8. Show that aij
= Eiika;a;ak = eiikai4a8
e'li2 ' ' i*.
ax axi
9. If yi = yi(xl, x2) ... , xe), i = 1, 2, . . . , n, show that
ayi ax"
ai assuming the existence of the derivatives. Also
a; =
axa y
i
show that ayi
-
axe
e
yi
as = a?. Show that
a2yi
axa ax¢
axa 8x8 ayi
8yk
ayi
492xa
+ Oxa ayi ayk - 0
TENSOR ANALYSIS
SEC. 1211
10. If
= cp(x', x 2 .
. .
.
. .
n),
xi = xi(y', y2'
i = 1,2,
.
yn)
,
= xi(y)
n, and if
<P = 0(y1, y2,
show that
263
.
.
,
.
y=
yn)
= p[xl(y), x2(y),
a
Given 9,, =
aoi
arpi
ayayi -
(ax# _
C) p
acs
.
.
.
,
xn(y)]
, show that
axe, axil
axa) ay, ayl
121. Determinants. We define the determinant lal by the
equation
a1
at
2
a1
a2
at
at
an
2
2
a2
an
i,a1a2
n
a1
n
n
a2
l all
an
an
=
Erik ... i"ai,a?2
.
.
.
(474)
a"n
The reader should note that this definition agrees with the
definition for the special case of second- and third-order determinants which he has encountered in elementary algebra. The
definition of a determinant as given by (474) shows that it consists of a sum of terms, nn in number. Of these, n! are, in general, different from zero. Each term consists of a product of
elements, one element from each row and column. The sign
.
of the term a a'a
a depends on whether it takes an even
or odd permutation to regroup i1i2
in into 12 . . n.
Since i1 and i2 are dummy indices, we can interchange them
so that
fail = Ei,i,... ina,'a2
.
.
.
ann = ei,i, ... inal=a2
.
. , ann
An interchange of the subscripts 1 and 2, however, will mean
that an extra permutation will be needed on i2iii3
in. This
changes the sign of the determinant. Hence interchanging two
columns (or rows) changes the sign of the determinant. As an
VECTOR AND TENSOR ANALYSIS
264
[SEc. 121
immediate corollary, we have Ia;I = 0 if two rows (or columns)
are identical.
Let us now examine the sum
bill:... i.
= Ei,i, ...
a.
(475)
If jl, j2, . . . , jn take on the values 1, 2, . . . , n, respectively,
we know that (475) reduces to (474). If the ji, j2,
. , j,.
take on the values 1, 2, . . . , n, but not respectively, then we
have interchanged the rows. An even permutation reduces (475)
to Ia;I, and an odd permutation of the i's reduces (475) to -Ia 1.
If two of the j's have the same value, (475) is zero, since if two
rows of a determinant are alike it has zero value. Hence
..
Es, ...,.a`;<1
. .. a;-
(475a)
i=
and
t,44 ...
i
ai.
(475b)
ai Ehj, ... i.
Example 115. We now derive the law of multiplication for
two determinants of the same order.
I ail
IasIEC,;,...
Ibil
bri
i.bi'b2
Ei,i, ... i.a is =
= Ei,i, ...
We have
.
. .
s,
a.,b
(ai,bi,)
i.(a1j1,bi')
s, 2
1
b2
.
.
.
{
,. x )
. b%
I ciI
where
abi
(476)
We now derive an expansion of a determinant
in terms of the cofactors of the elements. We have
Example 116.
I41 = Ei,i,...i.aia8
a; (Ei,i,... j.a2
. an
.
.
.
.
.
a'*)
= aIAQ
AQ is called the cofactor of a,.
where Ao =
In general,
a'0-b5-Pa'-#-K +,. .
= apAP.
. an)
(f4 not summed)
Hence
aiAa = IaIB;
(477)
TENSOR ANALYSIS
SEC. 121]
265
Also
aaA7 = IaI
(477a)
Example 117. Let us consider the n linear equations
a, i = 1, 2,
yi = aQxa,
.. , n
(478)
Multiplying by Af, we obtain
A°y' =a' Asxa
so that summing on i, we have from (477)
A"y' = Ialaxa = Ialxx
If IaI 0 0, then
A°y'
_ y'(cofactor of a' in IaI)
IaI -
Example 118.
(479)
IaI
Let yi = y'(x', x2,
. ,
x"), i = 1, 2,
. . .
, n.
i
In the calculus it is shown that if ayi) 96 0 at a point P and if the
partial derivatives are continuous, we can solve for the x' in
terms of the y's, that is, xi = xi(y', y2, ... , y") in a neighborhood of P.
Now we have identically
ayi
8i
ayi
ay' axa
axa ayi
Forming the determinant of both sides,
I O;I
from (476).
ayi axa
ay
- lax- ayi
ax
Hence
axi
ayl
The determinant I I
is called the Jacobian of the y's with respect
to the x's. We have shown that
VECTOR AND TENSOR ANALYSIS
266
J \xl,y2,
', x2,
. . .
xn
,,yry/Jfy',X2,
1, y2, .
.
.
y
xn) -
[SEc. 121
1
(480)
Example 119. If the elements of the (leterminant lal are functions of the variables x', x22, . . . , x", we leave it to the student
to prove that
_ Aa0 aao
alai
(481)
axµ
axp
As a special case, suppose a =
ay
i
, where y' = y'(x', x2,
, xn).
49x1
a yi axa
Let us consider j to be fixed for the moment.
ax- ay,
yt
a , Xi = ax -, so that Yi = aQXa. If
Let Yi = 491,
ax
y
Now a`t =
0
l al
ax11
then
Ya(cotactor of a#" in lal)
lal
from (479), or
a
8x8
S; cofactor of aye in
ay
ax
cix1
1
(cofactor of Lyi in
ay
8x
ax
and so
1
(cofactor of
a?
in
Applying this to (481), we have
ay
ay 8x8
ax
TX 8y1
267
7'EXSOR ANALYSIS
Sic. 121
a,y
a
ay axa
8x
a2ya
ax, ays ax), axa
axµ
or
(482)
la
a log
axa
laxI
ft"
a2ya
aya axµ
axa
We shall make use of this result later.
Example 120. Let yi = y'(xl, x2,
,
ay
xn),
We
3-1 0.
ax
wish to prove that
a2xµ
ayk aye
axa ax'6 axA
-
a2yi
ayi ayk ay` axa axa
ayi axa
Now S' = ax- -,a so
yi that upon differentiating with respect to
yk, we obtain
ayi
0 - axa
axa
a2xa
ayk aye
a2yi
axa
(483)
+ ay; axa axa ayk
1+
Multiplying both sides of (483) by ayi and summing on i, we
obtain
axa axa axµ
a2xµ
o = ayk ayay'
+
As a special case, if y = f (x),
d 2x
dy 2
ayk
ayi
&2yi
Q.E.D.
axa axa
= - (dx3d2y
dy dx2
Problems
i
1. What is the cofactor of each term of
z
i
al
a2
a3
2
2
2
a1
a2
a3
asl
a32
a33
2. Prove (481), (477).
3. If JAI is the derminant of the cofactors of the determinant
IaI, show that I:4I = IaJ"-'.
VECTOR AND TENSOR ANALYSIS
268
[SEC. 122
4. If a = a, show that A; = A. Is laI = jail in all cases?
5. Find an expression for
aaxa
ayi ayi ayk
6. If z' = z'(y1, y2, ... , yn), y' = yt(x1,
that J(z/y)J(y/x) = J(z/x).
7. If 9i; = gap
_
a ia2i-, show that I#I = IgI
a2{
x2,
.
. .
, x"), show
2
I
_I
axa
_
8. If u' = ua axa, Vi = V. clt.
-, show that uaVU = uaVa, where
xi = x{(x1 x2
xn).
ax{
ax i
9. If u{ = ua , show that u{ = ua
ax"
axa a
ax- 00
10. If gi; = gaa
, show that gi; = 9a$
a2i
axi axi
a { , ui = ua axi
azi
, show that u{ = ua
11. If u{ = ua
axa
axa
axa
82i
12. Apply (476) for the product of two third-order determinants.
13. If A; = B,-s ayi ax , show that A{ = B, and that IAI = IBI,
A;A; = B )$B,!.
14. If X is a root of the equation Iai; -- Xbifl = 0, show that X
is also a root of Id;; - Ail = 0 provided that B;; = aaa
!ii =
baa
axa axa ax
a axi
a
axa axa
aji
0.
122. Arithmetic, or Vector, n-space. In the vector analysis
studied in the previous chapters, we set up a coordinate system
with three independent variables x, y, z. We chose three mutually perpendicular vectors i, j, k, and all other vectors could be
written as a linear combination of these three vectors. Any
vector could have been represented by the number triple (x, y, z),
where we imply that (x, y, z) -= xi + yj + zk. The unit vectors
could have been represented by (1, 0, 0), (0, 1, 0), and (0, 0, 1).
A system of mathematics could have been derived solely by
defining relationships and operations for these triplets, and we
need never have introduced a geometrical picture of a vector.
For example, two triplets (a, b, c), (a, #, y) are defined to be equal
TENSOR ANALYSIS
SEC. 1221
269
if and only if a = a, b = P, c = y. We define the scalar product
as (a, b, c) (a, f4, -y) = as + bfi + cy. The vector product, differentiation, etc., can easily be defined. Addition of triples is
defined by (a,b,c)+(a,#,y)=(a+a,b+$,c+y). If A
is a real number, then A(a, b, c) is defined as (Aa, Ab, Ac). The
set of all triples obeying the rules
(i)
(ii)
(a, b, c)+(a,0,y) _ (a+a,b+i,c+y)
(484)
A(a, b, c) = (Aa, Ab, Ac)
is called a three-dimensional vector space, or the arithmetic space
of three dimensions.
It is easy to generalize all this to obtain the arithmetic n-space.
Elements of this space are of the form (x', x2, . . . , x"), the x'
taken as real. In particular, the unit or basic vectors are
(0, 0, . . .
(1,0,0, . . . ) 0), (0,1,0, . . . ,0),
... ,
We shall designate V. as the arithmetic n-space.
By a space of n dimensions we mean any set of objects which
can be put in one-to-one reciprocal correspondence with the
arithmetic n-space. We call the correspondence a coordinate
system. The one-to-one correspondence between the elements
or points of the n-space and the arithmetic n-space can be chosen
in many ways, and, in general, the choice depends on the nature
of the physical problem.
Let the point P correspond to the n-tuple (x', x2,
We now consider the n equations
y' = y'(x', x2,
... , x"),
i = 1, 2,
.
.
.
.
. ,
x^).
.
,n
.
, n (486)
(485)
and assume that we can solve for the x', so that
x' = x'(y', y2,
... , y"), i = 1, 2,
.
We assume (485) and (486) are single-valued. It is at once
obvious that the point P can be put into correspondence with
the n-tuple (y', y2,
y"). The n-space of which P is an
element is also in one-to-one correspondence with the set of
(y', y2, . .. , y"), so that we have a new coordinate system.
The point P has not changed, but we have a new method for
attaching numbers to the points. We call (485) a transforma-
... ,
tion of coordinates.
270
VECTOR AND TENSOR ANALYSIS
[SEC. 123
123. Contravariant Vectors. We consider the arithmetic
n-space and define a space curve in this V by
= 1, 2, ..
x` = x`(t),
at<_/i
n (487)
Note the immediate generalization from the space curve x = x(t),
y = y(t), z = z(t): In our new notation x = x', y = x2, z = x3.
We We remember that
dx d dz
,
are the components of a tangent
dt dt' dt
vector to this curve. Generalizing, we define a tangent vector
to the space curve (487) as having the components
d.ri
'j = 1, 2,
dt
. .
.
,n
(488)
Now let us consider an allowable (one-to-one and single-valued)
coordinate transformation, of the type (485). We immediately
have that
y' = y'(x1, x2,
. . . ,
x") = y`tx'(t), x2(t),
. .
.
, x"(t)1 = y'(t)
as the equation of our space curve for observers using the y
The components of a tangent vector to the
same space curve (remember the points of the curve have not
changed; only the labels attached to these points have changed)
are given by
coordinate system.
dyi
i = 1, 2,
. .
.
,n
(489)
dt
Certainly the x coordinate system is no more important than
the y coordinate system. We cannot say that dt is the tangent
s
vector any more than we can say dt is the tangent vector. If
we considered all allowable coordinate transformations, we would
obtain the whole class of tangent elements, each element claiming
to be the tangent vector for that particular coordinate system.
It is the abstract collection of all these elements that is said to
be the tangent vector. We now ask what relationship exists
between the components of the tangent vector in the x coordinate
system and the components of the tangent vector in the y coordi-
TENSOI? ANALYSIS
SF(,. 123]
nate system.
271
We can easily answer this question, for
We also notice that
dys
0y1 dxa
dt
axa dt
d = ay- dydls
dx1
(490)
We leave it as an exercise
that this result follows from (490) as well as from (486).
We now make the following generalization: Any set of numbers
. , n, which transform accordAi(xe, x2, . . . , xn), i = 1, 2,
ing to the law
..
_
AL(P, 22,
,
xn) = Aa(xl, x2,
. .
,
.
x')
axa
(491)
axa
under the coordinate transformation x= = 21(x', x2, . . . , xn),
are said to be the components of a contravariant vector. The
vector is not just the set of components in one coordinate system
but is rather the abstract quantity which is represented in each
coordinate system x by the set of components Ai(x).
We immediately see that the law of transformation for a
contravariant vector is transitive. Let
A'=Aaa2i,
OX*
.Ai=Aai
a2a
Then
A' _ A8 ax a = Aa ax8
a28
ax _Aaax
axa ate
axa
which proves our statement.
If the components of a contravariant vector are known in one
coordinate system, then the components are known in all other
allowable coordinate systems by (491). A coordinate transformation does not give a new vector; it merely changes the
components of the same vector. We thus say that a contra
variant vector is an invariant under a coordinate transformation.
An object of any sort which is not changed by transformations of
coordinates is called an invariant.
Example 121. Let X, Y, Z be the components of a contravariant vector in a Euclidean space, for an orthogonal coordinate
VECTOR AND TENSOR ANALYSIS
272
[SEC. 123
system, and let ds2 = dx2 + dy2 + dz2. The components of this
vector in a polar coordinate system -,re
R=X-ar+Yaary
0
ae
ae
X ax + Y ay
ar
+Zaz= cos 0X+ sin 0Y
CIO
_
+ Z az^
-sine X + cos e Y
Z=Xa +Yaz+Zaz=Z
r
r
y
where r = (x2 + y2)4, 0 = tan-' (y/x), z = z.
The components R, 0, Z are not the projections of the vector
A = Xi + Yj + Zk on the r, 9, z directions. However, if the
0-component is given the dimensions of a length by multiplying
by r, we obtain Or = - sin 0 X + cos 9 Y, which is the projection of A in the 9-direction. We multiplied by r because r d8
is arc length along the 0-curve. R, 0, Z are the vector components of the vector A in the r-e-z coordinate system, whereas
R, re, Z are the physical components of the same vector.
Problems
1. If A'(x), B'(x) are components of two contravariant vectors, show that C'"(x) = A'(x)B'(x) transforms according to the
ax' ar
law C*' = Cap axa ate, where C" =
2. Show that if the components of a contravariant vector
vanish in one coordinate system, they vanish in all coordinate
systems. What can be said of two contravariant vectors whose
components are equal in one coordinate system?
3. Show that the sum and difference of two contravariant
vectors of order n is another contravariant vector.
4. If X, Y, Z are the components of a contravariant vector in
an orthogonal coordinate system, find the components'in a spherical coordinate system. By what must the 0 and Sp components
be multiplied so that we can obtain the projections of the vector
on the 0- and (p-directions?
5. If A' = Aa a ' show that A' = Aa
'
a.
TENSOR ANALYSIS
SEC. 125]
6. Referring to Prob. 1, show that Cs,
N
7. If Ai = ax1 Aa
273
Caa
ax` ax'
axa al$
N
i
axa, show that Ati =
i
l-A-
I
az
124. Covariant Vectors. We consider the scalar point function So = co(x', x2, . . . , x"), and form the n-tuple
(492)
Now under a coordinate transformation
av
ay'
a(p
axa
(493)
axa ay'
so that the elements of the n-tuple l
1 y2
related to the elements of (492) by (493).
,
y"J are
We say that the axi
are the components of a covariant vector, called the gradient of gyp.
More generally, if
J i=Aaa'
L'
(494)
the A; are said to be the components of a covariant vector. The
remarks of Sec. 123 apply here. What is the difference between
a contravariant and a covariant vector? It is the law of transformation! The reader is asked to compare (494) with (491).
We might ask why it was that no such distinction was made in
the elementary vector analysis. We shall answer this question
in a later paragraph.
125. Scalar Product of Two Vectors. Let A'(x) and &(x)
be the components of a contravariant and a covariant vector.
We form the scalar AaBa. What is the form of AaBa if we make
a coordinate transformation?
Now
ax a
Aa=A8axe'
Baaxa
c12'
VECTOR AND TENSOR ANALYSIS
274
[SEc. 126
so that
AaBa = AFB,
AaBa = A"B,
axa axQ
axp axa
a2s
= Asfl = AaBa
Hence AaBa is a scalar invariant under a coordinate transformation. The product (AaBa) is called the scalar, or dot,
product, or inner product, of the two vectors.
Problems
1. If Ai and Bi are components of two covariant vectors, show
that Ci; = A;B, transforms according to the law Oil = Cap
axa axe
axa ax?
2. Show that C = AB; transforms according to the law
axp
C"i
- C8 V axa
.
a
3. If Ai = A. axi show that A, = Aa
axi
4. If p and J, are scalar invariants, show that
grad (vO) = c grad P + 0 grad cp
grad [F(op)] = F'(,p) grad rp
5. If A'Bi is a scalar invariant for all contravariant vectors
A', show that Bi is a covariant vector.
126. Tensors. The contravariant and covariant vectors
defined above are special cases of differential invariants called
tensors.
The components of the tensor are of the form Ta4b .. b;,
where the indices a,, a2,
the integers 1, 2,
.
. .
, a,
b1, b2,
. . . ,
b, run through
... , n, and the components transform accord-
ing to the rule
ax N
1 b,b,...
ax
alas
TaW*A:...,q,
anal
. .
axa* axle,
. aXa
alb,
. .
. axig
a-.
(495)
TENSOR ANALYSIS
SEC. 126]
We call the exponent N of the Jacobian
tensor field.
275
1092X1
the weight of the
If N = 0, we say that the tensor field is absolute;
otherwise the tensor field is relative of weight N. A tensor
density occurs for N = 1. The vectors of Sees. 123 and 124 are
absolute tensors of order 1. The tensor of (495) is said to be
contravariant of order r and covariant of order s. If s = 0, the
tensor is purely contravariant, and if r = 0, purely covariant;
otherwise it is called a mixed tensor.
Two tensors are said to be of the same kind if the tensors have
the same number of covariant indices and the same number of
contravariant indices and are of the same weight. We can construct further tensors as follows:
(a) The sum of two tensors of the same kind is a tensor of this
kind. The proof is obvious, for if
a...b _ axN
Tc ...
d - ax
b=
A7c.
d
a
... 8xrOxa
axd axa
To ... r age
ax°
laXlN S-..
""T a
axr axa
.
. .
.
alb
ax"
.
.
atd axa
.
a
ax0
then
Ua...b
c...d
Fa...b)
c...d
c...d
_ Ua...
c...z
ax N
ax'
axb
a2
(b) The product of two tensors is a tensor. We show this
for a special case. Let
Tba =
a axQ a2"
alt T s axlaxa
Ox 3
19i =
a2
02i
Sc or,
so that
= ax s
(T bs')
,ax
ax8 a2a all
(TWO)
a26 ax* ax"
The new tensor is of weight N + N' = 3 + 2 = 5.
(c) Contraction.
Consider the absolute tensor
Atk
-
a ax0 axy axi
AOY
a2' axk axa
276
VECTOR AND TENSOR ANALYSIS
[SEC. 126
Replace k by i and sum. We obtain
A,i - Aay
= Aay
a
= Apa
43X'9 axy axi
axi axi axa
M ax,
y
13x#
a
= Aay
axi sa
ax axa
axa
axi
so that Ali are the components of an absolute contravariant
vector. In general, we equate a certain covariant index to a
contravariant index, sum on the repeated indices, and obtain
a new tensor. We call this process a contraction.
(d) Quotient law. We illustrate the quotient law as follows:
Assume that AiB,k is a tensor for all contravariant vectors Ai.
We prove that B,k is a tensor, for
AiE1 = AQBay
= A'Bay
-
CIO axy 492i
x6
axi ax; axa
axa axy
ax axk
or
l
axa axy
Ai ` $,k - Bay axp avk
=0
Since A' is arbitrary, we must have f k = Bay
result.
Example 122.
tensor, for
a2 ax
the desired
The Kronecker delta, S, is a mixed absolute
=
axi axa
' aga axi'
Example 123.
axa axk
axi axa = 80
a
axa axs
= asa
If Ai and Bi are the components of a contra-
variant and a covariant vector, then C = AiB, are the components of a mixed tensor, for
s
Ai = Aa ax ,
axa
axa
B' _ B
a ali
-
277
TENSOR ANALYSIS
SEC. 1261
CO that
AaB#
C
axa axe =
Cs
axa Ox'
Let gij be the components of a covariant tensor
Example 124.
ax-,9x#
Taking determinants and applying
so that gi, = gap - =
axt ox'
Example 115 twice, we obtain
lax
ax
I#I} = IgIl
or
Igl =1g1 l ax
a2
Now if Ai are the components of an absolute contravariant vector, then A{ = Aa ax Y so that
ax
B, = IgI}A' =
B.
ax
aax
IgI'Aa
Of
axa
Ba2i
ax-
are the components of a vector density. This
method affords a means of changing absolute tensors into relative
tensors.
Example 125. Assume gap dxa dxO an invariant, that is,
so that Bi °
IgI#Ai
9aP d7a d2O = gp dxa dxfi
a
Now d2a = axu dxµ, so that
gap
axa aaxu ax,
dxu dx = g,,, dxu dx'
or
(9aO
a0
°1°
^ 9u
axu ax
dxu dx' = 0
(496)
If we assume gap = gaa, then since (496) is identically zero for
arbitrary dxi, we must have (see Example 114)
278
VECTOR AND TENSOR ANALYSIS
ISEC. 126
axa axs
9µY = 9a$
(496a)
axµ axY
or the gµ, are components of a covariant tensor of rank 2.
Example 126. If the components of a tensor are zero in one
coordinate system, it follows from the law of transformation
(495) that the components are zero in all coordinate systems.
This is an important result.
Example 127. Outer product of two vectors. Let A; and B; be
the components of two covariant vectors, so that
C;; = A;B; = AaEp
axa ax#
axti axe =
C°
axa a28
axti axe
The C:; are the components of a covariant tensor of second order,
the outer product of Ai and B.
Example 128. By the same reasoning as in Example 127, we
have that Cc; = A;B; - A;B, are the components of a covariant
tensor of the second order. Notice that C,, is skew-symmetric,
for C;; = -C;;. For a three-dimensional space
11C1;II =
0
A1B8 - AaBI
A2B3 - A3B2
-(A2B3 - A3B2)
0
A1B2 - A2B1
0
-(A1B2 - A2B1)
-(A1B8 - A3B1)
The nonvanishing terms are similar to the components of the
vector cross product.
Problems
1. If A = Aao
axa ax0
ai
axe, show that A;; = Aaa
axa axa
ax' axe
2. Show that A,, can be written as the sum of a symmetric
and a skew-symmetric component.
3. If A are the components of an absolute mixed tensor, show
that A; is a scalar invariant.
4. If Aa# are the components of an absolute covariant tensor,
and if Aa#A$,, = 8 , show that the Aal are the components of an
absolute contravariant tensor. The two tensors are said to be
reciprocal.
SEC. 1271
TENSOR ANALYSIS
279
5. If Al, and A1, are reciprocal symmetric tensors, and if ui
are components of a covariant vector, show that A,;uiul = Aiiu,u;,
where ui = Ai°u,,.
6. Let Ail and Bit be symmetric tensors and let ui, vi be components of contravariant vectors satisfying
(A,,-KBi;)ui=0 i,j= 1, 2,
(A,, -K'Bii)v' = 0,
.
. .
n
KX K
AiiutuWhy
i
is K
Prove that Ai,uiv' = Bi,uiv' = 0, and that K = BiJuiui
an invariant?
7. From the relative tensor A of weight N, derive a relative
scalar of weight N.
8. If Ami is a mixed tensor of weight N, show that Amw is a
mixed tensor of weight N.
9. Show that the cofactors of the determinant lair, are the
components of a relative tensor of weight 2 if ail is an absolute
covariant tensor.
10. If Ai are the components of an absolute contravariant
vector, show that axe are not the components of a mixed tensor.
127. The Line Element. In the Euclidean space of three
dimensions we have assumed that
ds2 = dx2 + dye + dz2
In the Euclidean n-space we have
ds2 = (dx')2 + (dx2)2 + . .. + (dxn)2
= bo dx° dxa
If we apply a transformation of coordinates
xi = xi(21, 22,
.
.
.
,
2' )
we have that dx' = axi d2a, so that (497) takes the form
ds2 = 6,0
ax° axe
dxµ dx
axµ aV
We may write ds2 = vr d2" dz', where
9vr = a°a
axa 8x#
a_
n
ax° ax°
ax+' any
(497)
VECTOR AND TENSOR ANALYSIS
280
[SEc. 127
Thus the most general form for the line element (ds)2 for a
Euclidean space is the quadratic form
ds2 = gas dxa dx5
(498)
The gap are the components of the metric tensor (see Example
125).
The quadratic differential form (498) is called a Rie-
mannian metric. Any space characterized by such a metric is
called a Riemannian space. It does not follow that there exists
a coordinate transformation which reduces (498) to a sum of
squares.
If there is a coordinate transformation
xi = xi(y', y2,
.
.
.
,
y")
such that ds2 = Say dya dy5, we say that the Riemannian space is
Euclidean. The y's will be called the components of a Euclidean
coordinate system. Notice that gas = Sap. Any coordinate system for which the g;; are constants is called a Cartesian coordinate
system.
We can choose the metric tensor symmetric, for
gii = Ij(gv + gig) + (gii - gii)
and the terms $(gt, - gii) dx' dx' contribute nothing to the sum
ds2. The terms J(g,i + gii) are symmetric in i and j.
Example 129. In a three-dimensional Euclidean space
ds2 = tax 1)2 + (dx2)2 + (dx3)2 for an orthogonal coordinate sys-
tem, so that
1
0
0
0 0
1
0
0
1
Let
x' = r sin 0 cos p = y' sin y2 cos y$
x2 = r sin 0 sin rp = y' sin y2 sin y2
x3=rcos9=y'cosy2
Now
axa 8x5
9ti(r, g, w) = gas ay; ayi
ax' ax'
axe axe
OxI 49x8
- ay: ayi + ay: ayi + ay' ayi
TENSOR ANALYSIS
SEC. 127]
281
Hence
gll = (sin y2 cos y3)2 + (sin y2 sin y3)2 + (cos y2)2
Similarly
933 = (y')2(sin y2)2,
922 = (y') 2,
gi,=0 for i5j
so that
ds2 = (dy')2 + (y')2(dy2)2 + (y' sin y2)2(dy3)2
= dr2 + r2 do2 + r2 sin2 0
is the line element in spherical coordinates. Since the g's are
not constants, a spherical coordinate system is not a Cartesian
coordinate system.
Example 130. We define gii as the reciprocal tensor to gi,,
that is, gi.2 gai = d,f (see Prob. 4, Sec. 126).
The gii are the signed
minors of the g;i divided by the determinant of the gi;. For
spherical coordinates in a Euclidean space
1
0
0
r2
0
0
0
0
r2 sin2 B
1
0
0
0
2
r2
0
0
0
1
r2 sin2 9
Example 131. We define the length L of a vector Ai in a
Riemannian space by the quadratic form
L2 = g pA°A#
(499)
The associated vector of Ai is the covariant vector
i = gia
It is easily seen that Ai = gipAp, so that
L2 = gapg'
g"'A"A.
We see that a vector and its associate have the same length.
If L2 = 1, the vector is a unit vector.
Example 132. Angle between two vectors.
unit vectors.
Let Ai and B, be
We define the cosine of the angle between these
VECTOR AND TENSOR ANALYSIS
282
[SEC. 127
two vectors by
cos 0 = A'Bi = Aig;;Bi = gi,AiB'
= giiA,B1 = gi'A;B,
(500)
If the vectors are not unit vectors,
gi,AiB;
(500 a)
cos 0 =
If gi;A'B' = 0, the vectors are orthogonal. We must show that
Assuming a posiI cos 01 < 1. Consider the vector AA' +
tive definite form, that is, gaaza# > 0 unless z' = 0, we have
1Bi.
gas(XAa + µB")(XAa + kB") > 0
or
y = A2(gaAaAa) + 2Aµ(gasA"B1) +µ2(gaaBaBa) > 0
This is a quadratic form in A2/µ2, so that the discriminant must
be negative, for if it were nonnegative, y would vanish for some
value of A/µ or µ/A. Hence
gaaAaBP <
(gaaAaA8)4(gaaBaBa)}
or ,cos 01 < 1. Moreover, if Ai = kBi, it is easy to see that
cos 0 = ± 1. Hence I cos el < 1.
Example 133. A hypersurface in a Riemannian space is given
by x' = x'(ul, u2). If we keep ul fixed, ul = uo, we obtain the
space curve x' = x'(uo, u2), called the u2 curve. Similarly,
x' = x'(ul, uo) represents a ul curve on the surface. These curves
are called the coordinate curves of the surface.
once that on the surface
2
We have at
axa axa
ds2 = gad dxa dxa = I gap - -dui du'
_
au' aui
ca-1
axa axa
g°a aui au
ds2 = h;; du' du'
where hi, = gaa
axa axa
au'aui
dui du'
(501)
TENSOR ANALYSIS
SEC. 127]
Example 134.
283
Let us consider
The special theory of relativity.
the one-parameter group of transformations
x = ,B(x - Vt)
y=y
(502)
z = 2
xJ
where,6 = [1 - (V2/c2)1-} and V is the parameter. c is the speed
of light. These are the Einstein-Lorentz transformations (see
Prob. 11, Sec. 24). The transformations form a group because
(1) if we set V = 0, we obtain the identity transformations
x = 2, y = y, z = 2, t = t; (2) the inverse transformation exists
since 2 _ #(x + Vt),
y, t = z, t = O[t + (V/c')x], the
inverse transformation obtained by replacing the parameter V
by - V; (3) the result of applying two such transformations
yields a new Lorentz transformation, for if
where
_ [1 - (W'/c')]-;, then
x = (2- Ut)
y=?!
z
t=fi -Ux
c
where
U`
V-I-W
- 1 + (VW/c):'
^ r1 _
`\
U2)-1
c
I
We now assume that (x, y, z, t) represents an event in space
and time as observed by S and that (x, 9, 2, 1) represents the
same event observed by S (see Fig. 101).
VECTOR AND TENSOR ANALYSIS
284
The origin b has :f = y = 2 = 0' so that from (502)
[SEC. 127
dx
dt
= - V'
showing that S moves with a constant speed - V relative to S.
Similarly S moves with speed + V relative to S.
y
.7, 1,q
'P1:Mf,7
14
-V
S
S
X
X
z
z
Fia. 101.
From (502) we see that 0 and 0 coincide at t = 0. At this
instant assume that an event is the sending forth of a light wave.
The results of Prob. 11, Sec. 24 show that
x2 + y2 + z2
t2
- xa _'
_2
12
+
22
-- 4
C
so that the speed of light is the same for both observers. This
is one of the postulates of the special theory of relativity. Starting with this postulate and desiring the group property, we could
have shown that the transformations (502) are the only transformations which keep dx2 + d y2 + dz2 - cz dt2 = 0 an invariant.
Let us now consider a clock fixed in the S frame. We have
x = constant, so that dx = 0, and from (502) dt = 0 dt. Hence
a unit of time as observed by S is not a unit of time as observed
by S because of the factor 0 5-6 1. 8 remarks that S's clock is
running slowly.
The same is true for clocks fixed in the S frame.
TENSOR ANALYSIS
SEC. 1271
285
We choose for the interval of our four-dimensional space the
invariant
ds2 = c2 dt2 - dx2 - d y2 -dz2 = (dx') 2 - (dx') 2 - (dx2) 2
- (dxa) 2
where x' = x, x2 = y, x$ = z, x' = ct. The interval ds2 yields
two types of measurements, length and time, but takes care to
distinguish between them. If we keep a clock fixed in the S
frame, then dx = dy = dz = 0, so that ds2 = c2 dt2, and the
measurement of interval ds is real and proportional to the time
dt.
Now if we keep t fixed, dt = 0, and
ds' = - (dx2 + dye + dz2)
so that ds is a pure imaginary, its absolute value denoting length
as measured by meter sticks in a Euclidean space.
We shall describe the laws of physics by tensor equations, the
components of the tensors subject to the transformations (502).
This will guarantee the invariance of our laws of physics.
The momentum of a particle of mass mo will be defined by
pa = mo d8 . If the speed of the particle is u,
()2 ()2
u2
+
+ \dt/2
as measured by S, then
d82 = c2 dt2 - (dx2 + dye + dz2) = (c2 - u2) dt2
so that
mo
dxa
I
mo
dxa
pa = (C2 - u2)} dt = C [1 - (u2/c2)]} dt
and
p = [1
mo
- (u2/C2)]}
-- m
We define the Minkowski force by the equations
_
f a - C2
d(
dxa
ds m0 da )
I
d (m
[1 - (u2/C2)]; dt
a
dt
'
a = 1, 2, 3, 4
286
VECTOR AND TENSOR ANALYSIS
ISEC. 127
The Minkowski force differs from the Newtonian force by the
factor [1 - (u2/C2))-1. The work done by the Newtonian force
Fa = d- (m ddt
for a displacement dxa is
d
dE _
(mxa) dxa
a-1
I (mxa dxa + dd -4- dxa j
a1
mou du
[1 - (u2/C2)11
and integrating, E = [1 - (u2/c2))-lmoc2 - moc2 = (m - mo)c2,
with E = 0 for u = 0. Expanding [1 - (u2/c2)1-1 in a Maclaurin
series, we have E Jmou2 for (u2/c2) << 1.
The reader is referred to Probs. 1, 2, and 3 of Sec. 82 for the
application of special relativity theory to electromagnetic theory.
Let the reader derive (285) by use of this theory, choosing the
frame S so that at a particular instant the charge p is fixed in this
frame. The force on the charge as measured by S is given by
(285).
Problems
1. For paraboloidal coordinates
x1 = yly2 cos ys
x2 = y1y2 sin ya
x8 = 3[(yl)2 - (y2)2]
show that ds2 = [(y')2 + (y2)21[(dyl)2 + (dy2)21 + (yly2)2(dy$)2.
2. Show that for a hypersurface in a three-dimensional
8
Euclidean space, h1 =
a-1
axa axa
t9lis au?
3. Show that the unit vectors tangent to the ul and u2 curves
are given by
1
ax
18x
1
and
-V//-22aU2
Vh11aul
4. If w is the angle between the coordinate curves, show that
cos w = his/.
TENSOR ANALYSIS
SEc. 127J
287
. . . , x") = constant determines a hypersurface
If dx` is any infinitesimal displacement on the hyper-
5. (p(x', x2,
of a V,,.
Why does this show that the ax
surface, we have a dxa = 0.
ax
are the components of a covariant vector that is normal to the
hypersurface?
6. If (p(x', x2,
. . . ,
x") = constant, show that a unit vector
a aP
gra avp
normal to the surface is given by (g°8
ax, axo
axa
7. Consider the vector with components (dx', 0, 0, .
. . ,
0).
Under a coordinate transformation the components become
1
"
2
Cax' dx', ax' dx',
ax' dx' . Consider the new components
for the vectors with components (0, dx2, . . . , 0), . . . , (0, 0,
. .. ) 0, dx"), and interpret
. . . ,
d2' dxa ... dx"
=
1a1 dx' dx2 .
.
Using the result of Prob. 7, Sec. 121, show that
is an invariant.
. dx"
gl dx' dx2
dx"
We define the volume by
V=ff.
. .
f
dx' dxa .
.
. dx"
8. Show that ds is a unit vector for a V".
9. The surfaces xti = constant, i - 1, 2, ... , n, are called
the coordinate surfaces of Riemannian space. On these surfaces
all variables but one are allowed to vary. This determines
subspaces of dimensions (n - 1). If we let only x' vary, we
obtain a coordinate curve. Show that the unit vectors to the
coordinate curves are given by a; = 1//, i = 1, 2, . . . , n,
and that the angle of intersection between two coordinate curves
is given by
r
cosm;; _
gc,
9g
10. Show that a length observed by S appears to be longer as
observed by S. How does S compare lengths with S?
288
VECTOR AND TENSOR ANALYSIS
[SEC. 128
11. Let jx, jj,, jz, p be the components of a vector as measured
What are the components of the same vector as measured
by S?
by S.
12. Let
d2 x
d 29 dz2
72
die' die' die
be the components of the acceleration of a
2z
2y
dz
particle as measured by S. Find dt2, dt2, dt2 from (502).
128. Geodesics in a Riemannian Space.
If a space curve in a
Riemannian space is given by xi = xi(t), we can compute the
distance between two points of the curve by the formula
s
dxa dxO 4
r
(
- ` i \g°a dt dt dt
(503)
To find the geodesics we extremalize (503) (see Sec. 40). The
differential equations of the geodesics are [see Eq. (146)]
d (of
of
of
ax' -
dt
where f = (go '.#)} =
(504)
Now
Wt
of
ax'
0
1 (aga9 zaxs
2f \ ax'
and
d (af
d (ga;xa + gist
dt \az'
dt \ 2 ds/dt
Fz
2 ds/dt
/
(g.a + g+sio + agxps xaxs + x° ve
2
2(ds/dt)2 dt2
(ga:xa +
d
2
If we choose s for the parameter t, s = t, dt = 1, dtQ = 0, and
use the fact that gii = gii, (504) reduces to
a
1 /agai
g'ax + 2
610
ague
+
axa
agoal/ xa x
o =0
- -axti
(505)
TENSOR ANALYSIS
SEC. 1281
280
Multiplying (505) by gri and summing on i, we obtain
gri
xr
+
`(89-i
.99io
+
2
axi /
axa
or
d2xr
dxa dxP
,
(506)
+ r°a ds ds -
dS2
where
9r r° (.9g.,
+
2 \ ax$
ra$
a9oa
ago
axa
axQ
(507)
The functions r" are called the Christoffel symbols of the second
kind. Equations (506) are the differential equations of the
geodesics or paths.
Example 135. For a Euclidean space using orthogonal coordinates, we have ds2 = (dxl) 2 +
+ (dx") 2, so that gap = a.#
and
ft"
d1
Hence the geodesics are given by ds = 0 or
0.
xr = ars + br, a linear path.
Example 136. Assume that we live in a space for which
dal = (dx')2 + [(x1)2 + c2](dx2)2, the surface of a right helicoid
immersed in a Euclidean three-space. We have
0
1
{1gi?il =
0
0
1
1
0
(x1)2 + c2
(x')2 .+ C2
Thus we have
rill =0,
r21 = r14 = 0,
1
x1
r212
(x') 2 + C2
1 -- -x1,
r22
r2
so that the differential equations of the geodesics on the surface
are
d2x'
ds2 d2x2
d82
2x'
dx2 2
x
1
T8
- 0
dx' dx2
+ (x')2 + C2 d8 ds =
0
290
VECTOR AND TENSOR ANALYSIS
(SEC. 129
Problems
1. Derive the r;k of Example 136.
2. Find the differential equations of the geodesics for the line
element ds2 = (dxl)2 + (sin x')2(dx2)2.
3. Show that rqn = r'a.
4. For a Euclidean space using a cartesian coordinate system,
show that r, = 0.
5. Obtain the Christoffel symbols and the equations of the
geodesics for the surface
xl = ul cos u2
x2 = u' sin 412
x3 = 0
This surface is the plane x3 = 0, and the coordinates are polar
coordinates.
6. From (507) show that axµ = g,sroals + g,arp,,.
7. Obtain the Christoffel symbols for a Euclidean space using
cylindrical coordinates. Set up the equations of the geodesics.
Do the same for spherical coordinates.
8. Write out the explicit form for the Christoffel symbols of
the first kind : { i, jk I = g;,rlk.
9. Let ds2 = E due + 2F du dv + G dv2. Calculate IgJ, g'',
i, j = 1, 2. Write out the rk.
axk axe + a2x° axe
10. If r = r;7 axs
a. ax ax ax ax ax show that r;,y - rry
are the components of a tensor.
129. Law of Transformation for the Christoffel Symbols.
Let the equations of the geodesics be given by
d2xi
ds2
d i J dxk
_
+ r "k ds ds ^ 0
(508)
and
AV
ds2
+
P 2k
d.V d2k
ds ds
_-
°
(5 0 9)
for the two coordinate systems xi, xi in a Riemannian space. We
now find the relationship between the r k and r; . Now
TENSOR ANALYSIS
SEc. 129J
dxi
ds
__
a2' dxa
axa ds
d2x`
and
_
a2xi
291
dxa dxa
axa axa ds ds
ds2
d2x°
9.
+ axa ds2
Substituting into (509), we obtain
&V d2xa
axa ds'
dxa dxa
a22 i
+ ax# axa ds A +
rik axi axk dxa dxa
axe, ax" ds as
0
(510)
°
We multiply (510) by axi and sum on i to obtain
d2x°
dx2
axe axk
+ a22i 49x°l dxa dxa
+ (rijk axa axa axgr
a2i
axa axa a-V ds ds -
0
Comparing with (508), we see that (using the fact that rlk = rk;)
rik
axa axy axi
a2x°
Oxi
aY axe axk axa + axi axk a2°
This is the law of transformation for the rk.
(511)
We note that the
rk are not the components of a tensor, so that the rk may be
zero in one coordinate system but not in all coordinate systems.
Example 137. From (481) we have
algl
8
= Iglgae ax'
and from Prob. 6, Sec. 128,
ax#
= garap + g°arPA
so that
= gas9°ar,, + gang°aFBp
= a;rap + a1r00
=rap+r;p=2rµ
or
C log
axp
_
ra
p
(512)
292
VECTOR AND TENSOR ANALYSIS
[SEC. 129
Example 138. We may arrive at the Christoffel symbols and
their law of transformation by another method. Differentiating
the law of transformation
axe, ax8
g'' = ga$ a2i a2;
with respect to xk, we have
ag;i
agar axY axa ax8
axk - axY a2k
gay
ax0 aaxa
492X#
axa
49.t- __-
j a.Ti
ak a +a2k
I02i x2a2
(513)
If we now subtract (513) from the two equations obtained from
it by cyclic permutations of the indices i, j, k, we obtain
axs axY axi
ti
a
r" =
2 g°' (agk,
axj
a2xa
axi
r - rdy afj a2k axa + 02i axk axa
511a)
where
agi,
+ axle
_
agile
ax"
Example 139. Let us consider a Euclidean space for which
ds2 = (dx 1) 2 + (dx2) 2 + .
-
. + (dx") 2
In this case the r (x) = 0. In any other coordinate system, we
have
a2x° a2i
TV a2k ax°
If the new coordinate system is also Cartesian (the g,, = constants), then r k = 0, or
82x°
aP a2k
=0
x° = alga + b°
(514)
where a', b° are constants of integration.
Hence the coordinate transformation between two cartesian
coordinate systems is linear. If, furthermore, we desire the
distance between two points to be an invariant, we must have
n
n
d2° d2- _
dx° dx° _
ff-1
n
W-1
a:a- dxa dx8
C-1
TENSOR ANALYSIS
SEC. 129]
293
so that
n
I a,-,a; = Sas
(515)
V-1
A linear transformation such that (515) holds is called an orthogonal transformation.
For orthogonal transformations,
'
=g"° ax. axe
a' axe
axa axe
a:' CIO
reduces to S;; =
We multiply both sides by
amt
ax),
and
sum on i, so that
azf
axe
axp
8x'
8x8
is a = #0 axe - iii
= aaa
(516)
Now let us compare the laws of transformation for covariant
and contravariant vectors. We have
A' = Aa
Replacing
621
a2'
_,
axa
A; =
Aa
axa
82'
(517)
by a from (516), we see that
axa
n
a-1
Aaa
(518)
so that orthogonal transformations affect contravariant vectors
in exactly the same way that covariant vectors are affected [compare (517) and (518)]. This is why there was no distinction
made between covariant and contravariant vectors in the
elementary treatment of vectors.
Problems
1. From (511) show that
_
r'k
82xC, of
all To ax- + afj axk axe
a ax8 axy a2'
=ray
VECTOR AND TENSOR ANALYSIS
294
[SEC. 129
2. By differentiating the identity gaga; = S , show that
ag{k
axi =
_gkxr h4 - 9htrh;
3. Derive (513) by performing the permutations.
s0
2 aQ
4. If az.
xk
ax = 0, show that
0.
az axk =
5. If
090 My axi
r;k
a21°
axi
= rPr ax; 021; axa + axi axk at°
show that
r;k =
a axA 8x' axi
6. If xa = xa(ul, u2,
if h;; = gas
axa axs
Taui
r;, a.C axk
. . . ,
axa
+
a2x' axi
axi axk ax°
u*), a = 1, 2,
.
.
.
, n, r < n, and
and if
1
i
{rik)h
= 2 h.,
(ah
auk
ahk,
ah,kl
+ aui _ au°
show that
hair k)k = gap(ra,)o
axa axs axy
axa a2xs
+
gas
aui aui auk
aut aui auk
7. Define gs(x) by the equation gas(x) = µ(x)gas(x). We see
that the metric tensor gas(x) is determined only up to a factor of
multiplication u(x). In this space (conformal) we do not compare lengths at two different points, or, in other words, the unit
of length changes from point to point. Show that
T;r(x) = r;, (x} + rPYaa + PSaY - ga°gstimo
1alogµ
where 'Po = 2 axe
, and ra sy are defined by (507) using
I'sa
gas and gas.
8. Prove that a geodesic of zero length (minimal geodesic)
[that is, xa(s) satisfies
(506)
and
dxa dxs
gas ds A
= 0] remains a
minimal geodesic under a conformal transformation.
295
TENSOR ANALYSIS
SEC. 130J
130. Covariant Differentiation. Let us differentiate the absolute covariant vector given by the transformation
At-Aaaxa
axi
We obtain
aA;
aAa axa axa
axe
axa ax= ax' + Aa axi axi
It is at once apparent that
aA;
axe
a2xa
(519)
are not the components of a
tensor. However, we can construct a tensor by the following
device: From (511a) (see page 292)
ax' axr axa
ax; axe axo
82x,7
+
axa
axi axi ax"
(520)
Multiplying (520) by A. and subtracting from (519), we obtain
-A
&V
aA;
a 1'
_
( axa
-Ar
O
aa
Ox- axa
ax' axe
(521 )
so that if we define
A;;=-Aa1
8x1
49A;
(522)
we have that
A..s = Aa.a
axa axa
axi axi
and A;,, is a covariant tensor of rank 2. The tensor is called the
covariant derivative of A; with respect to xi. The comma will
denote covariant differentiation. For a cartesian coordinate sysaA;
tem, r,,E = 0, so that A;,1 = axe our ordinary derivative.
For a scalar of weight N we have
A=I-
821
A
296
VECTOR AND TENSOR ANALYSIS
(SEC. 130
so that
axlN aA axa
ax 8xa 02'
+N
ax N-1
a
ax
lax
aax'
_ laxe a2xs
ax
and from (482),
Hence
ax ax i ax
aA
axi -
ax N aA axa
latl axa axi
Multiplying r* = n.,
+
ax N axa
N
H
82x0
axs &V C120
a x, axa
A
(523)
2
axa
+
axi
axa Hi axQ
by NA and subtracting
from (523), we have
OA
ax
- NAr; _
N (claAx
a
ax
clx
- NA r-.1, i a
(524)
Hence A,, -= ax - NAr`,, is a relative covariant vector of weight
N.
A.
It is called the covariant derivative of the relative scalar
For a cartesian coordinate system it reduces to the ordinary
derivative.
In general, it can be proved that if Tp,s,::: p; is a relative tensor
of weight N, then
+
,,p
axe
+
TW-a,ry"
-A-
a v,
Ta,a,...a,
Ain
(525)
is a relative tensor of weight N, of covariant order one greater
than Ts,,:::a;, and it is called the covariant derivative of TB,A :::8.
Example 140. We have
ag;;
gii,k = axk - gwl'.k - g+Nr k
so that from Prob. 6, Sec. 128,
g i.k = 0
(526)
TENSOR ANALYSIS
SEC. 130J
297
Example 141. If p is an absolute scalar,
_ gyp, we call
-p,; = ax the gradient of V.
Example 142.
Curl of a vector.
aA;
ant vector. We have A;,; =
axj
aaxA;
- Aar .
axi
aA1
- ax` is a covariant tensor of rank 2.
It is called the curl of the vector Ai.
of the gradient of a scalar, Ai =
curl A; ==
Similarly,
"' - Ar,i
A,.; Hence Ai,5 - A;,; =
Let A; be an absolute covari-
If the A; are the components
a-P,
axi
then
- ax' ax' = o
ax1 ax'
621P
a2ip
so that the curl of a gradient is zero. It can be shown that the
converse holds. If the curl is identically zero, the covariant
vector is the gradient of a scalar.
Example 143.
Intrinsic derivatives.
tensors, we know that Ai.;
dxi
ds
.
is a covariant vector.
the intrinsic derivative of A;.
A;.,
dxi
ds
8A;
as
Since Ai,1 and
dx
are
We call it
We have
aAi dxi
a dxf
a axe ds - A°`r`'
dA;
ds -
ds
dx1
(527)
`oar"ds
and write the intrinsic derivative of A; as
BA;
as
Example 144. The divergence of an absolute contravariant
vector is defined as the contraction of its covariant derivative.
298
VECTOR AND TENSOR ANALYSIS
(SEC. 130
Hence
div A' = A' = ax
Now rQ;
as =
a logx
IgI
div A
-4- AarQ,
from (512), so that
a
1
_
ax-
t
div As =
+
1
Aa
ax.
N/191
aa
a (V f gj Aa)
(528)
In spherical coordinates, we have
1
`'9=
0
0
0
r2
0
0 r2 sin2 0
0
i
= r2 sin 0
so that
div A' =
i
a
r2 sin 8 ar
(r2 sin 0 A,)+
a.
d9
(r2 sin 0 AB)
+ app (r2 sin 0 Ac)
and changing At and A" into physical components having the
dimensions of A (see Example 121), we have
div A' =
1
Car (r2 sin 0 A*) + a9 (r sin 0 A°) + a (rAc) J
r2 sin o
Example 145. The Laplacian of a scalar invariant. If p is a
scalar invariant, -r,; is the gradient of jp, and the div (p,,) is called
the Laplacian of .p.
a
Lapcp=VZ(p=div(g,)=diva'
Thus
v 2(p
1,91
aaa (v' Fg19°°
)
TX
(529)
TENSOR ANALYSIS
Sec. 1301
into a contravariant vector so that we could
We changed
apply (528).
299
aThe associate of
81P
is g«i LIP.
ax'
axe
In spherical coordinates
i
1
0
0
0
r2
0
0
0
r2 sin2 B
0
0
t g' l =
7
I
0
0
r2 sin 9
so that
VF =
a
1 r2 sin B
r2 sin 0 ar \\\
1
{
Example 146.
aF
ar
-1-
a (sin B aF' + a / 1 aF'
00 \
00a l\sin 9 app/)
In Example 144 we defined the divergence of
the vector A' as div A' = A
Ma
+ Aar' - For a Euclidean
Cti
space using cartesian coordinates, the rk = 0, so that
0A2
OA'
div A'=axl+axe+
... + aA^
axn
The quantity A* is a scalar invariant. If we let Ni be the components of the unit normal vector to the surface do, then AaNa
is also an invariant. In cartesian coordinates the divergence
theorem is
JJ
divAdr = ffA.dd = JJA.N&r
In tensor form it becomes
fJfAT = JJA"Nad r
(530)
We can obtain Green's formula by considering the covariant
vectors 4O,; and y ,;. Now let
A, = kp,i - soO,c
300
VECTOR AND TENSOR ANALYSIS
{SEC. 130
The associated vector of A; is A° = g'=A; = g°.i(4,v,s - 4,i).
We easily see that A' = g°i(4y,; -
Now g°'co,,Q is an
invariant and in cartesian coordinates reduces to
a29
+
(ax1)2
a2_
+
(ax2)2
a24
_ Lap (p
(ax3)2
Hence, using (530), we obtain
f if (t' Lap so - v Lap ¢) dr =
f f g°iA;NQ do
S
=
ff
Svf.,)N; dv
S
Let us consider the covariant vector Fa. We
multiply it by the contravariant vector d--° and sum on a to
Example 147.
obtain the invariant F. dxa = F.
ds
ds, which reduces to f dr
in cartesian coordinates. In Example 142 we constructed the curl
of a vector, which turned out to be a tensor of rank 2. We
now construct a vector whose components will also be those of
the curl of a vector. We know that F.,s is a tensor. Now
define 4ah =
1
w
1
if a, ft, y is an even permutation of 1, 2, 3;
'
is an odd permutation of 1, 2, 3;
4°Ar = 0 otherwise. Thus
41st
-,
1
211 -
-L
I
4132 =
1
411! = 0
191
We obtain a new invariant
G" = - 4"0''Fa,F
In cartesian coordinates Fa,P = ate, and
1 = -EarilFR .a
= -j E2111F2 s + 021F3,2
= aFs
aF2
axe - ax=
TENSOR ANALYSIS
SEC. 130]
Hence Stokes's theorem in tensor
and similarly for G2 and G3.
form reads
a
f F.
301
ds ds =
f
do
(531)
Problems
1. By starting with Ai = All
A.1
a2i
Oza
, show that
axi + Aara1
is a mixed tensor.
2. Prove that (g{aAa),, = g{aA'.
3. Prove that (A-B.).1 = AaBa,, + A Ba.
4. Prove that 14i = 0.
5. Use (529) to find the Laplacian of F in cylindrical coordinates.
6. Prove that a;
7. Prove that
0.
1
axa (vflg g`-) + rQgas = 0.
8. As in Example 143, show that the intrinsic derivative
as
ds
+
Aarocp
ds
is a contravariant vector.
9. Show that the intrinsic derivative of a scalar of weight N
is b = ds - NAr;, ds,
so that if A is an absolute constant,
10. Show that (g,#AaAO) j = gapA ;Ap + 9apAaA,'
11. Show that A; , = aA' +
ax°
mixed tensor A.
12. Show that DZ(vo)
13. Show that Aaa =
r;,A' for an absolute
V;k -I- 2 0v - DO -}- J, pz9.
1
0
(VTg_j A") - Abr.
gl
exa
302
VECTOR AND TENSOR ANALYSIS
[Sac. 130
14. If A; = A;(x, t) is a covariant vector, show that
oA;
at
_
8A;
at
+ Ai,J
dx'
dt
av;
_
and hence that the acceleration fc °
at
8vi
at + v ,,v'.
15. Let X. be an arbitrary vector whose covariant derivative
Consider Jf Ta,' XaN,6 da, and apply
s
the divergence theorem to the vector TOXa. Hence show that
vanishes; that is, Xa,$ = 0.
fJ T aaNN da = ff7 T dT.
S
16. Let s; be the displacement vector of any particle from its
position of equilibrium (see Sec. 115). We know that s;, is a
covariant tensor. The relative displacements of the particles
are given by
as; = s,,J dx
_ 4(s,,J + sJ,:) dx' +
s,,.) dx'
Show that the term 4(8;,J - 8J,;) dxJ represents a rotation.
We define the symmetric strain tensor E;J by the equation
Ei; = +(s.,J + ss,;)
The stress tensor T;, is defined by the equations
AF; = TJNJ Ao-
where AF; is the force acting on the element of area Aa with
normal vector Ni (see Sec. 116).
Let f, be the acceleration of the volume dT and F, be the force
per unit mass acting on the mass in question. Show that
f j f pF, dT + f f TJNJ da =
f f f pf, dT
or using contravariant components,
fffa pF'dr+ f sf T*JNJda= fffR pf'dT
TENSOR ANALYSIS
SEC. 1311
303
Now deduce the equations of motion
pFr + T = pfr
If Tn = pgri, show that
pFr - p.igr' = pfr
or
[see (411)]
pFr --- p.r = pfr
The equations of the geodesics
131. Geodesic Coordinates.
are given by
dxi dxk
d2xi
+ r,k
as2
= 0
ds as
where the Ik transform according to the law
ask
-
rasY
49x8 ax" axi
ax axk axa
192xa
axi
+ 491i axk axa
(532)
We ask ourselves the following question : If the I k are different
from zero at a point x` = qi, can we find a coordinate system
such that rk 0 at the corresponding point? The answer is
"Yes"!
Let
(xi - q') +
a and
so that
I
(533)
= 1, and moreover the transforma-
ax1i
a
tion (533) is nonsingular.
q«)(xd - 4s)
The point xi = q' corresponds to the
point x' = 0.
Now differentiating (533) with respect to x', we obtain
axii
a=
+ (ri OP )
because of the symmetry of ri,8.
Q(xa - qa) ate?
I
Hence 6z! = 6f.
Q
Differentiating (534) with respect to xk, we obtain
a2x8
axa axs
a2x'
+
(rae)
a
+
(r10)
a(x0
q°`)
axk a21
axk ax'
axk axi
(534)
VECTOR AND TENSOR ANALYSIS
304
[SEC. 131
so that
ax° axe
a2x'
axk axj
= - (ray) 4 ask
Q
Q
axe
-(raa)Q6-50 _ -k
Substituting into (532), we obtain
(rij k)o = (rfY)Qaj kaa -
(rk)Qaa
= (r k), - (r;k)Q = o,
Q.E.D.
Any system of coordinates for which (r7k)P = 0 at a point
P is called a geodesic coordinate system. In such a system,
the covariant derivative, when evaluated at the origin, becomes
the ordinary derivative evaluated at the origin. For example,
\(Mi
(As')O _
axi )0
+ (ra;)o(Aa)o = ax, )0
since rq; = 0 at the origin.
The covariant derivative of a sum or product of tensors must
obey the same rules that hold for ordinary derivatives of the
calculus, for at any point we can choose geodesic coordinates so
that
A'1
OA'
aBi
ax;
+axi
_ a(Ai + Bi)
ax'
(A'+B');
and A', + B`; - (A' + B'),, is a zero tensor for geodesic coordinates. Hence A + B`i - (Ai + B'),1 is zero in all coordinate
systems, so that
A` + B`; _ (A' + B').;
We leave it as an exercise for the reader to prove that
(A'B;).k = A'kB7 + A'B1,k
Equation (533) yields one geodesic coordinate system.
There
are infinitely many such systems, since we could have added
c,fiy(x) (x° - q') (xa - qP) (xr - q7) to the right-hand side of
(533) and still have obtained (r k)o = 0.
A special type of geodesic coordinate is the following: Let
x' = x'(s) be a geodesic passing through the point P, x' = xo, and
TENSOR ANALYSIS
SEc. 131]
let
xi r
305
Define
X' = Vs
(535)
where s is arc length along the geodesic. Each V determines a
geodesic through P, and s determines a point on this geodesic.
Hence every point in the neighborhood of P has the definite
coordinate xi attached to it. The equations of the geodesics in
this coordinate system are
d2z1
dS2
2
d21 dxk
1
+ r'k ds ds =
0
i
But ds = V and ds = 0, so that
rikl:'k = 0
(536)
Since this equation holds at the point P for all directions t', we
must have r + r 1 = 21 k = 0, so that the x= are geodesic
coordinates. The X' = `s are called Riemannian coordinates.
Example 148. If ' is a unit vector, we have
gaa°
=1
The intrinsic derivative is
a
gas s e + gaak°
since (g.p),1 = 0 (see Example 140).
adx.
gQs°
0.
Usp + -r" $) =
dt1
Q
as
=
0
Hence gaat"
as
0, and
We see that the vector
dx°
is normal to the vector t'.
Problems
1. Show that
dx°, dxgab
ds ds
remains constant along a geodesic.
2. Show that for normal coordinates 2, r;2' k = 0.
VECTOR AND TENSOR ANALYSIS
306
[SEC. 132
3. If s is are length of the curve C, show that the intrinsic
i
derivative of the unit tangent
ds
in the direction of the curve
has the components
pi -
d2xs
+
ds2
.
dx1 dxk
r'k
A
TS
What are the components for a geodesic?
a
4. Prove that bt (X"Ya) = sat Ya + Xa sat.
132. The Curvature Tensor. Let us consider the absolute
contravariant vector Vi. Its covariant derivative yields the
mixed tensor
v`; =
a'
ax; + V ar°`'
On again differentiating covariantly, we obtain
avt.
v',k = axk' + v rak
= a 2V
s
axk 49x1
- viark
+ aV a ri + Va arm, +
axk
'
49xk
(a
a
+ VOr;, rak
Interchanging k and j and subtracting, we have
Vt;k - Vsk; = V"Ba;k
(537)
where
Since V';k - Vik; and Vi are tensors, Va;k must be the components
of a tensor, from the quotient law (Sec. 126). It is called the
curvature tensor. We can obtain two new tensors of the second
order by contraction.
TENSOR ANALYSIS
SEC. 133]
307
Let
Ri'
Ba;
ara
axa
ara
axa + r ara; - r%raa
(539)
This tensor is called the Ricci tensor and plays an important role
in the theory of relativity.
We obtain another tensor by defining
Si; = Be..
ara
aa i
ara
a;
ax1
axi
(540)
Evidently Si; = -8;i. and if we use the fact that
a log "ICI
- ra
1aµ.
ax"
we have that
a2 log
a2 log -VIgl
axi ax1
ax1
=
0
axi
Now Ri; - R;; = Si; = 0, so that the Ricci tensor is symmetric
in its indices. We could have deduced this fact by examining
(539) directly.
The invariant R = gi"Ri; is called the scalar curvature.
133. Riemann-Christoffel Tensor. The tensor
Rhitk = gpaBk
(541)
is called the Riemann-Christoffel, or covariant curvature, tensor.
Let us note the following important result: Assume that the
Riemannian space is Euclidean and that we are dealing with a
cartesian coordinate system. Since r;k(x) = 0, we have from
(538)
B';ki = 0
(542)
in this coordinate system. But if B';k= = 0 in one coordinate
system, the components are zero in all coordinate systems.
Hence if a space is Euclidean, the curvature tensor must vanish.
We shall show later that if B1,-k1 = 0, the space is Euclidean.
308
VECTOR AND TENSOR ANALYSIS
[Sac. 134
If we differentiate (538) and evaluate at the origin of a geodesic
coordinate system, we obtain
Bajk a
a2raj
_
- ax° axk
a2rak
8x° axj
Permuting j, k, a and adding, we have the Bianchi identity
(543)
Bakj.o + Ba0j.k + Baka,1 = 0
134. Euclidean Space. We have seen that if a space is
Euclidean, of necessity Bike = 0. We shall now prove that if
the Bki = 0, the space is Euclidean. Now
axfi axy 00
ayj ayk axa
rik(y) = rsY
a2xa ayi
+ ayj ayk aXa
If there is a coordinate system (x', x2,
r;,.(x) = 0, then
a2xa
. .
.
,
xn) for which
ayi
r;k(y) = ayj ayk axa
(544)
and conversely, if (544) holds, the r;y(x) = 0. Now let us investigate under what conditions (544) may result. We write (544)
as
a2xa
axe
ayj ayk = ayi rk(y1
(545)
which represents a system of second-order differential equations.
Let us define
axa
u7-aycc
(546)
so that (545) becomes
auk = u7Nkly)
ayj
(547)
For each a we have the first-order system of differential equations
given by (546) and (547), which are special cases of the more
general system
TENSOR ANALYSIS
SEC. 1341
azk
ay'
= f (z', z2,
. . .
z", z"+1, y`,
,
yn)
309
k=1, 2,
..,n+I
j=1,2,...,n
(548)
If we let z' = x°, z2 = 211, z3 = 112i
and (547) reduce to (548).
.
.
2
2
We certainly must have aye ayj = ayj
afk
8y1
, Zn+' = un, Eqs. (546)
.
yi, and this implies
_ aft af, av
ay' - ay' + aZM ay'
afk az"
+
az
(549)
or
k
afk
afk
ay'
+ azi, fill
+ fk
u
ay
az"
fj
If the f; are analytic, it can be shown that the integrability conditions (549) are also sufficient that (548) have a solution satisfying the initial conditions z1; = zo at yi = yo. The reader is
referred to advanced texts on differential equations and especially
to the elegant proof found in Gaston Darboux, "Lecons sysOmes
orthogonaux et les coordonnees curvilignes," pp. 325-336,
Gauthier-Villars, Paris, 1910.
The integrability conditions (549), when referred to the system
(546), (547), become
1
jkua
nklua
aj"U'.=0
(550)
The first equation of (550) is satisfied from the symmetry of the
I';k, and the second is satisfied if 19j"n = 0. Hence, if f"kj = 0,
we can solve (545) for z' = x° in terms of y', y2, . . . , y". For
the coordinate system (x', x2, . . , x"), we have T k(x) = 0.
Problems
1. Show that Rhijk = - Rihik = - Rhikj and that
Riijk = Rhikk = 0
2. Show that Rhijk + Rhkij + Rhiki = 0-
3. If Rij = kgii, show that R = nk.
310
VECTOR AND TENSOR ANALYSIS
[SEC. 134
4. For a two-dimensional space for which g12 = 921 = 0,
show that R12 = 0, 811922 = R22911 = R1221 and that
R_
R1221
911922
R:; = JRg.;.
5. Show that B k1 76 0 for a space whose line element is given
by ds2 = (dx')2 + (sin xl)2(dx2)2.
6. Derive (550) from (546), (547), (549).
1 49R
7. If R = gtiaR,, show that (R'),; =
2 ax;.
CHAPTER 9
FURTHER APPLICATIONS OF TENSOR ANALYSIS
i
135. Frenet-Serret Formulas. Let Ai =
ds
be the unit tangent
vector to the space curve xi = xi(t), i = 1, 2, 3, in a Riemannian
space. In Example 148 we saw that the contravariant vector
t
as
is normal to Xi.
Let us define the curvature as K =
0}
g°0 88 aas
°
and the principal unit normal µi by
aai
as
_
=
Kµi
(551)
i
Since pi is a unit vector, we know that aS is normal to pi.
Now
g°0A°,us = 0, so that the intrinsic derivative yields
Sµa
bA°
as
since g°0,; = 0 or a3s
µB = 0
+ g°0
g°'eA°
bs
Hence
0.
aµ0
g°a A°
+ Kg°aa°i8 = 0
as
or
(B68
µs
+ KAS
g°aa°
=0
(552)
since gp°µs = g°aX 'X 8 = 1.
i
Equation (552) shows that Ss + KAi is normal to V, and since
5Fdi
as
and Ai are normal to µi,
aµi
as
+ KAi is also normal to µi.
311
We
VECTOR AND TENSOR ANALYSIS
312
[SEC. 135
define the binormal vi by the equation
-
v'
Sµ + KAt
1
(553)
(68
T
or
aIAi
as
-= -#c? - TY'
(554)
(i)
where ,r is called the torsion and is the magnitude of
Since v` is normal to both X and µ', we have
gaPvax$
=0
(555)
g«avO'µP = 0
By differentiating (555) and using (551), (554), (555), we leave
it to the reader to show that
gaaµa
/ - ava
as
(T/!a
=0
(556)
The vector rµe - as is thus normal to all three vectors V, Af, Pi.
Since we are dealing in three-space, this is possible only if
av
TEIZ
as
= 0, or
avi
=
as
(557)
Writing (551), (554), (557) in full, we have the Frenet-Serret
formulas
&
dxP
ds
ds
KAi
i
d+
µa s = - (Ki + Tv`)
a
ds
+
Pa
ds
= Tµ
(558)
SEC. 1361 FURTHER APPLICATIONS OF TENSOR ANALYSIS 313
= 0,
For a Euclidean space using cartesian coordinates, the r
and (558) reduces to the formulas encountered in Sec. 24.
Problems
1. Derive (556).
2. Using cylindrical coordinates,
ds2 = (dxl)2 + (x')2(dx2)2 + (dxa)2
and for a circle xl = a, x2 = t, x3 = 0. Expand (558) for this
case, and show that K = 1/a, r = 0.
2i
-K2X + U' - KTY'.
3. Show that Ss
=
as
4. Since (558) is true for a Euclidean space using cartesian
coordinates, why would (558) hold for all other coordinate systems in this Euclidean space?
.
2
5. Since 'Xi = d3 , show that
dx
d
a
+ ra' ds
Q
dare the
com
ponents of a contravariant vector.
136. Parallel Displacement of Vectors. Consider an absolute
contravariant vector Ai(xl, x2, . . . , xn) in a cartesian coordinate system. Let us assume that the components Ai are con-
stants. Now
A' = Aa
a2'
A' = Aa
axa
axi
a2a
so that
dA: - _ Aa
822'
axfi
8xs axa (921y
dxy
since dA' = 0. We thus obtain
dA = Ao
axe axa
d,y
ax' axa a2y 492'
a22i
From (511a) (see page 292)
-
a2i
r~"° = a2y a2° axa
a2xa
a2xi
since r, = 0
axO axa
ax8 axa axy axe
;
from (483)
so that
dAi = - A°1;y d2y
(559)
VECTOR AND TENSOR ANALYSIS
314
[SEC. 136
In general, a Riemannian space is not Euclidean. We generalize (559) and define parallelism of a vector field Ai with respect
to a curve C given by x' = x'(s) as follows: We say that Ai is
parallelly displaced with respect to the Riemannian V,, along the
curve C, if
dAi
ds
dx*
°T ds
or
aas=
=0
d-si+r,,A°ds
(560)
We say that the vector Ai suffers a parallel displacement along
the curve. Notice that the intrinsic derivative of A' along the
curve xi(s) vanishes.
d2xi +
dxa dxe = 0,
In particular, for a geodesic we have
ri
°s ds ds
ds2
so that the unit tangent vector
dxi
d suffers a parallel displacement
along the geodesic.
Example 149. Let us consider two unit vectors Ai, Bi, which
undergo parallel displacements along a curve. We have
cos 0 = gapAaB$
and
a(cos 0)
as
_ gas
SAa
as
B$
aBQ
+
gaaAa
as
=0
so that 0 = constant. Hence, if two vectors of constant magnitudes undergo parallel displacements along a given curve, they
are inclined at a constant angle.
Two vectors at a point are said to be parallel if their corresponding components are proportional. If A' is a vector of
constant magnitude, the vector B' = WA', cp = scalar, is parallel
to A'. If A' is also parallel with respect to the V. along a curve
t
xi = xi(s), we have Ss = 0. Now
aBi
as -
0A'
as
dc*
+ ds
`4s
d A'
ds
1 dip
- p ds
r
d(log y)
d8
SEC. 137] FURTHER APPLICATIONS OF TENSOR ANALYSIS 315
We desire B' to be parallel with respect to the V along the curve,
so that a vector B' of variable magnitude must satisfy an equation of the type
a13'
B =
f(s)Bb
(561)
if it is to be parallelly displaced along the curve.
Problems
1. Show that if the vector At of constant magnitude is parallelly displaced along a geodesic, it makes a constant angle with
the geodesic.
2. If a vector A' satisfies (560), show that it is of constant
magnitude.
3. If a vector Bi satisfies (561) along a curve r, by letting
At = #Bi show that it is possible to find ,y so that At suffers a
parallel displacement along IF.
4. Let xi(t), 0 < t < 1, be an infinitesimal closed path. The
change in the components of a contravariant vector on being
parallelly displaced along this closed path is 1 Ai = -jrra,Aa dxs,
from (560). Expand Aa(x), ri (x) in Taylor series about
xo = x4(0), and neglecting infinitesimals of higher order, show that
AA' = }RO,,A'
xy dx# - x0 dxy
where Rte,, is the curvature tensor (see Sees. 132, 133, 134).
137. Parallelism in a Subspace. We start with the RieV,,, ds2 = gab dxs dxs. If we consider the
mannian space,
transformation
xa = xa(ul, u2,
.
. . ,
u"),
m<n
(562)
we see that a point with coordinates ul, u2, . . . , u"' is a point
of V. and also a point of V,,. The converse is not true, for given
the point with coordinates x', x2, . . . , x", there may not exist
U', u2, . . . , U. which satisfy xa = xa(ul, u2, . . , ur), since
m < n. Now
axa 8xs
sub 8u1
ds2 = 94 dxa dxs = gab -
= h;; du' du'
- du' du;
316
VECTOR AND TENSOR ANALYSIS
[SEC. 137
so that the fundamental metric tensor in the subspace, V,,,, is
given by
jE;j =gap
axa M
aui au1
a
Now dxa = aua dui, so that if due are the components of a contra-
variant vector in the Vm, dxa are the components of the same
In general, if ai(u',
vector in the V,,.
.
.
.
,
u»1), i = 1, 2,
. . . ,
m, are the components of a contravariant vector in the V., we
say that
a = 1, 2,
= aui ai
Aa
. .. , n
(563)
are the components of the same vector in the V,,.
Ma
We now find a relationship between
and as, where s is
arc length along the curve ui = ui(s) or the space curve
xa = xa[ui(s)]
Differentiating (563), we have
dAa
ds
axa dai
au' A +
a2xa duy i
a
aui aui Ts
and
6Aa
as
dAa
+ (r *-y
ay
ds
dxy
ds
_
a2xa du'
as
aui ds + aui aui ds
ax,- dai
ax8 axy du1
+ (rayaauui aui ds
Hence
9'a
ax' Ma
auk Ss
dai
_ ;k ds + ai
du'
ds
ax,
9oa
auk SSsa
h`k dst
CIO axy ax'
9a«(ray>°
+ ai
ds
aui au' auk
h`k(r'i)h,
+ 9Qa
a2xa ax°l
aui aui auk
(see Prob. 6, See. 129)
Sr;c. 1371 FURTHER APPLICATIONS OF TENSOR ANALYSIS 317
Hence
ax" SAa
9aa
_
auk as
hlk
I dal
ds
dull
+
(r'i)ha ds
y
J
and
ax" SAa
9°a
auk as
Sat
hck
(564)
as
From (564) we see that if ai is parallelly displaced along
xa[ui(s)], that is, if
as
i
= 0, then Ss = 0. Thus the theorem:
If a curve C lies in a subspace V. of V,,, and a vector field in V.
is parallel along C with respect to V,,, then it is also parallel along
C with respect to V.Problems
1. Prove that if a curve is a geodesic in a
it is a geodesic in
any subspace V. of V,,. Consider the unit tangents to the
geodesics.
2. By considering k fixed, show that auk is a contravariant
vector of V,, tangent to the uk curve, obtained by considering
ul, u2,
.
uk-1, uk+',
.
.
. . .
u'" fixed in the equations
xa = xa(u'
um)
3. If ai is parallel along C with respect to the Vm, show that
Ma
is normal to the space V, that is, normal to the ui curves,
as
i = it 2,
.
.
.
, M.
4. Under a coordinate transformation ui = ui(u', . .. , um),
i = 1, 2, . . . , m, the xa remain invariant. Hence show that
axa
xaa =
au',
agaa
9ad.i - ax° x.f
where the covariant derivatives are performed relative to the
aai
metric hi;, that is, a1, = aa + ak(r k
au'
318
VECTOR AND TENSOR ANALYSIS
5. Show that ga$(x kxa + x {x k) + xaxIxk
[SFC. 138
a
0, where
ax, =
covariant differentiation is with respect to ui and h;;.
6. Show that A ; = x;ai + x°,as, for each a.
138. Generalized Covariant Differentiation. The quantities
aui are contravariant vectors if we consider i fixed; for if
xa = xa(ul
and ya = ya(xl,
. .
u'")
xn), a = 1, 2,
. ,
aya
aui
_
.
.
.
, n, then
aya axa
CIO aui
a
showing that the aui transform like a contravariant vector.
axa
However, if we consider a as fixed, the -- , i = 1, 2,
aui
. .. , m,
are covariant vectors in the V.; for if u' = ui(ul, .. . , u'n),
axa
axa and
we have - = - - We propose to consider tensors of this
aui
our au'
type, Latin indices indicating tensors of the V., and Greek indices
indicating tensors of the V..
Let us consider the tensor A7. We wish to derive a new tensor
which will be a tensor in the V. for Greek indices and a tensor in
V. for Latin indices. We consider a curve C in V. given by
u` = ui(s) and by xa = xa(s) in V,,. Let bi be the components of a
vector field in V. parallel along C with respect to
and let ca
be the components of a vector field in V. parallel along C with
respect to V. We have
dbi
ds
dca
rrkb,
+
dd k
s
=0
(565)
dx#
dsds=0
We now consider the product bicaA;. In V. this product is
an invariant (scalar product) for each i, and in V. it is a scalar
invariant and is a function of are length s along C. Its derivative
is
SEc. 1381 FURTHER APPLICATIONS OF TENSOR ANALYSIS 319
i
ds
(b°c«A;)
bica
dsi +
biAi"
ds
°
since
ds
c°A;
k\
d
= bica d' + A; rag
making use of (565).
+
d3
- Air,k
ds
)
Since b1 and ca are arbitrary vectors, and
d
(b'c«A;) is a scalar invariant, it follows from the quotient
ds
law that
_j + `4'r ,,
dA°
dx#
ds
ds
dull
(566)
- A`r' k ds
is a tensor of the same type as A7. We call it the intrinsic derivative of A; with respect to s.
We may write (566) as
aA°
axe
(auk + A; r«a auk
- Air;)\ duk
ds
k
and since this is a tensor for all directions ds (the directions
k
of C are arbitrary), it follows from the quotient law that
aAuk°
Aff:k =a
axe
+ A;r auk - Air?k
(567)
is the generalized covariant derivative of A7 with respect to the
V,,,.
Problems
a contravariant vector in V,i?
1. Why is A ",
ill
2. Show that
aA"
Aa;:, =
- r#rA
r,A,; ax*
au,
ax?
aut
- rAflk
is a mixed tensor, by considering the scalar invariant b8c'daA';.
a
3. Show that x = x`; = aus, and that
VECTOR AND TENSOR ANALYSIS
320
_
(SEC. 139
a2xa
44 = au' aui - r .xh + r;yx x
= xai + FRxx4.
Show that
ga$xQ4 k = 0, and show by cyclic
permutations that gaaxa;x = 0.
5. The x°i of Prob. 4 arek normal to the vectors 4k, the tangent
vectors to the surface. Hence the x"i are components of a vector
normal to the subspace Vm. If Ni are the components of a unit
normal to Vm, we must have x = b;)Na. We call B = b;i du' dui
the second fundamental form. Show that b;i = gapx" NP. If
the V is a Euclidean V3, gas = as, show that b11 = e, b12 = f,
b22 = g (see Sec. 35) for the subspace r = r(u, v).
139. Riemannian Curvature. Schur's Theorem. Let us consider a point P of a Riemannian space. We associate with P two
independent vectors X,, A2. These vectors determine a pencil
of directions at P, given by
a1Al + a2A2 = aiX;
Every pair of numbers at, a2 determines a direction Ea. Since
the geodesics are second-order differential equations, the point P
and the direction Sa at P determine a unique geodesic. The locus
of all geodesics determined in this manner will yield a surface.
In a Euclidean space the surface will be a plane, since the geodesics are straight lines and two vectors determine a plane.
We now introduce normal coordinates ya with origin at P.
The equations of the geodesics take the form ya = has, where
ka
dya
= C ds p, and the geodesic surface is given by
ya = a'8A1 + a2sX2
= ut?4 + u2A2 = uiX7
(568)
j summed from 1 to 2 and ul = a's, U2 = a2s.
The element of distance on the surface is given by
ds2 = h;i du' dui
(569)
and if ds2 = gab dya dyfi for the V, then
aya ayo
hti
= gap au' aui = gasX A
(569a)
SEC. 139] FURTHER APPLICATIONS OF TENSOR ANALYSIS 321
where the gas represent the components of the fundamental
metric tensor in the system of normal coordinates.
Now let R;;kc be the components of the curvature tensor for
the surface S with coordinates ul, u2. Let us note the following:
The gas of a Riemannian space completely determine the Christoffel symbols r;,, which in turn specify completely the RiemannChristoffel tensor Ras,.a. Once the metric of a surface embedded
in a V. is determined, we can determine the F;k for this surface,
and the R;;kc can then be determined. We need not make any
reference to the embedding space, V,,, to determine the R:;k1
it is apparent that the ht; can be determined without leaving the
surface, so that all results and formulas derived from the h;; are
intrinsic properties of the surface. All we are trying to say is
that ds2 = h21 dus du' is the fundamental metric tensor for a
Riemannian space which happens to be a surface embedded in a
Riemannian V,,. We shall use Latin indices for the space determined by the metric h;; and Greek letters for the V,,.
The indices of R;;k1 take on the values 1 or 2, and from Prob. 1,
Sec. 134, we have that
R1212 = R2121 = -R1221 = -R2112
= R2222
81111 = 1?1122 = R1122 = . .
= 81121 =
. .
(570)
=0
If we make an analytic transformation, uti = uti(ul, u2), i = 1, 2,
then
R;;k1(u) = Rabcd(u)
aua aub au` and
anti au' auk aul
and
au!, aub auc and
aul au2 aa1 au2
au1 au2 2
au1 au2
= R1212ka-lvau2
81212 = Rabcd
au2aul
(571)
by making use of (570). Thus
R 1212 =81212IJ1,LL
(572)
aUa
aub
Moreover, ds2 = h;; du= du', and h;; = hab -- --- so that
aut alai
322
VECTOR AND TENSOR ANALYSIS
IF = I hIJ 2.
[SEC. 139
We rewrite (572) in the form
R1212
R1212
Ih
IhI
(573)
Equation (573) shows that K is an invariant, and it is called the
Gaussian curvature. It is an intrinsic property of the surface.
We now determine an alternative form for K in terms of the
directions Ai and X and the curvature tensor for the V at the
point P. The coordinate transformation between the Christoffel
symbols is given by (see Prob. 6, Sec. 129)
h;1rik(u) = gaprs.,(y)
ay- 8y8 ayT
au' auj auk
+ gaa
ay°
8u1 auk aui
a2yP
which reduces to
har3k = g.pr,,'X AgAk
aya
since
aui -
''
(574)
a2ya
0, from (568).
auk auj
At the point P, rP1(y) = 0 (see Sec. 131), so that h;irik = o
or himhiir;k = r, = 0. Hence the curvature tensor can be
written
R1212(P) = h11Rg12(P)
= h1i
[ar1(P)
au2
from (538) and (541).
From (574), h11r`21(u) =
of Riemannian coordinates
or,
hit
since
ar'22(P)i
au1 J
(575)
XRX , so that at the origin
= gaN
y*
(576)
ag-0 = g,,911.,. + g,ar p = 0 at the origin (Prob. 6, See. 128),
y
and from (569a) auk = X'XOXk agy
= 0.
Y
h1i
12 = gap
ayr
"
Similarly
(577)
Sec. 139] FURTHER APPLICATIONS OF TENSOR ANALYSIS 323
Using (538), (541), (576), (577), it is easy to show that
R1212 = A72A1t2Rapyr
Finally,
Ihl _
hll
h12
h2l
h22
hllh22 - hit
and
aya c3y9
hi, =
h12
h22
gap
c3u1 c3u1
= ga9Xx1
=
gaAX2
so that
Ihi
=
x7x2xip21'(gaagp. - ga,gpa)
(579)
Thus
7
K
aps,X aj X 2'a1
2
(580)
We are now in a position to prove Schur's theorem. If at each
point of a Riemannian space, K is independent of the orientation
(X "j, as), K is constant throughout the space.
It follows at once if K is independent of X,, X7i that
Rap,,, = K(gaagp, - ga,gpe)
and so
Rapa=, = K,N(gaegp, - ga,gp,,)
= K,,(gaagp, -- gaogpa)
Rap,,
Raaµ.e = K,,(ga,gpµ - gw.gp,)
Adding and using Bianchi's identity, (543), we have
K.p(ga,gp - ga,gpa) + K.,(ga,,gsa -- ga,,gp,M) + K,n(ga,gp,, - gawgpr)
=0
Multiplying the above equation by ga° and summing, we have
K, ,(ngp, - gp.) + K.,(gp - ngp,,) + K,,gp,, - K,,,gp, = 0
or
(n - 2)gp,K,,, = (n - 2)K,,gpp
and gp,K,,, = gp K,,, or 5,'.K,,, =
is no arbitrary orientation.
K,,, if n > 2. For n = 2 there
VECTOR AND TENSOR ANALYSIS
324
[SEC. 140
If we choose o = r X u, K, = 0. This is true for all µ since IL
can be chosen arbitrarily from I to n. Hence K = constant
throughout all of space. Such a space is said to be of constant
curvature.
Problems
1. Derive (571).
2. Derive (575).
3. Derive (578).
4. For a V3 for which gii = 0, i 54 j, show that if h, i, j are
unequal,
1
Rii = - Riaai
9hh
1
1
R, = - Raica + - Rhiih
gii
s
gii
1
R=
Riiii
9ii9ii
5. If R; = gaiRii, show that R"a =
l aR
2 axi
6. If Rii = kgii (an Einstein space), show that R = giiRii = nk,
or Rii = (R/n)gii
7. Show that a space of constant curvature K is an Einstein
space and that R = Kn(1 - n).
140. Lagrange's Equations. Let L be any scalar invariant
function of the coordinates q', q2, . .. , qn, their time derivatives 41, 42, ... , 4n, and the time t:
L = L(qi, 4i, t) = L(qi, qi, t)
If we perform a transformation of coordinates,
qa
a = 1, 2,
the
. . .
= ga(gl, q2,
, n, then qa =
aga
aqa
.
.
.
, qn)
so that 4a is a function of
Now
aL 84a
aL
aL aqa
aqi = aqa aqi + a4a aqi
aL aqa
aL a2ga
= aqa aqi + a4a a4i
aq0
41
(581)
SEC. 140] FURTHER APPLICATIONS OF TENSOR ANALYSIS 325
where we consider q' and q' as independent variables in L. Now
also
aL _ aL a4
aqa aq;
aq'
aL aqa
aqa aqr
so that
d
(582)
dt \agi1 - ala a \aq"I + aq q8 aqt
Subtracting (581) from (582), we obtain
d (IL)
dt
/
d 8L,)
L dt \aqa
aqaq-
aq'
which shows that the
covariant vector.
d
- (aqa
aqa
c
qla
are the components of a
For a system of particles, let L = T - V, where T is the
1
1
(ds;12 _
kinetic energy; T
:=1 2
I/
ma
2
s
V (1
xI x1 x1
1
i1
2
s
2
x2, x2f x2f
s
7
xn)
is the potential function, - aV = (F*),. Then
e
aL
dt ax;
_d
_
aL
ax;
-d(magma) _
dt
n
1
8ga0 x,a2
2
ax *
_ m;
'
+ aV
-.
ax;
=max;-(F,)a, ifgaQ=aae
and m,x; - (FT), = 0 for la Euclidean space and Newtonian
ax vanishes in all coordinate
W \-x
systems. We replace the x; by any system of coordinates q',
q2, . . . , qn which completely specify the configuration of the
mechanics.
Hence
system of particles, and Lagrange's equations of motion are
(;) _ aL
d aL
aq=
0, r= 1 ' 2'
'n
(583)
Example 150. In spherical coordinates, a particle has the
square of the velocity v2 = t2 + r262 + r2 sin2 0 rp2, so that
VECTOR AND TENSOR ANALYSIS
326
[SEC. 140
L = T - V = 2 (t2 + r262 + r2 sin2 0 02) - V
-
= m(r92 + r sing 0 #2) - aV
_d
aL
dt
ar
= mr
and one of Lagrange's equations of motion is
mr - m(r62 + r sin 2 9 02) + aV = 0
c1r
Since - aV represents the radial force, the quantity
r - (r62 + r sing 0 02)
must be the radial acceleration.
If no potential function exists, we can modify Lagrange's equa-
tions as follows: We know that T = (m; 2)gapxaO is a scalar
invariant, so that
d (BT)
aT
Qr = dt
aT
axr
(584)
are the components of a covariant vector. In cartesian coordinates, the Q, are the components of the Newtonian force, so that
Q, is the generalized force vector. If f, are the components of the
force vector in a y'-yL -y" coordinate system, then
-
aya
Qr - fa axf
and
Qr dx r =
f
a
49y
a
a
dxr = fa dya
is a scalar invariant. The reader will immediately realize that
fa dya represents the differential of work, dW, so that
aW
Q+
= ax'
(585)
SEC. 1401 FURTHER APPLICATIONS OF TENSOR ANALYSIS 327
, xi-',
We obtain Qi by allowing xi to vary, keeping x', x2,
.
. , x% fixed, calculate the work A Wi done by the forces,
and compute
xi+',
.
d Wi
Qi =..L-lim
-,
O AV
i not summed
Example 151. A particle slides in a frictionless tube which
rotates in a horizontal plane with constant angular speed co.
The only horizontal force is the reaction R of the tube on the
particle.
We have T = (m/2) (j62 + r2A2), so that (584) becomes
mr" - mr02 = Q,
dt (mr2o) = Qe
with Q, = 0, Qe =
R(r de)
do
(586)
= rR. The solution to (586) is
r = Aew' + Be", gR = 2mw dt, since 6 = w.
Problems
1. A particle slides in a frictionless tube which rotates in a
vertical plane with constant angular speed w. Set up the equations of motion.
2. For a rigid body with one point fixed,
T =JA(w;+wy)+jCwa
Using Eulerian angles, show that if Q, =
0, then
C(O + cos 0 +') = R
A¢sin20+Rcos0=S
A#-A,2sinocoso+R,1 sino =Qe
where R, S are constants of integration.
3. If T =
q")gags, show that 2T =
oL(q,
aq
, q^, p,,
4. Define p, =
q* = gr(gl)
.
.
aT
aq-
q°°.
and assuming we can solve for
and
show that the Hamiltonian
328
VECTOR AND TENSOR ANALYSIS
[SEC. 141
II defined by H = pages - L satisfies
II = T + V = It (a constant)
where V = V (q',
.
.
,
q"),
T = aas(q',
.
show that
aH
.
. ,
gn)gags.
Also
pr
a'
all = g.
ap'.
These are Hamilton's equations of motion; the pr are called the
generalized momentum coordinates. Show that they are the
components of a covariant vector.
5. By extremalizing the integral f " L(x1, x', t) dl, show that
Lagrange's equations result.
6. If the action integral
e [(h
A=
f`
dxs
- V )gas '-6
a aA ]
1
dX
is extremalized, show that the result yields
d
dt
aT
IT
aV
- ax' = - ax,
Cair
where T + V = h, the constant of energy. The path of the
particle is the same as the geodesic of a space having the metric
ds2 = 2M(h - V)gas dxa dxs
141. Einstein's Law of Gravitation.
We look for a law of
motion, which will be independent of the coordinate system used,
describing the gravitational field of a single particle. In the
special theory of relativity, the line element for the space-time
coordinates is given by
ds2 = -dx2 - dy2 - dz2 + c2 dt2
-dr2 - r2 do2 - r2 sin2 0 dp2 + c2 d12
(587)
In the space of x, y, z, t, the gas are constants and the space is
SEc. 1411 FURTHER APPLICATIONS OF TENSOR ANALYSIS 329
flat (Euclidean), so that BJk: = 0. For a gravitating particle
we postulate that the Ricci tensor Ri, vanish (see Probs. 5 and 6
of this section).
Since Ri; = R;;, a four-dimensional space yields
n(n + 1)/2 = 10 equations involving the gi; and their deriva1 aR , where
From Prob. 7, Sec. 134, we have R ,i =
tives.
2 ax;
R = gi°R°;, R = g°pRay, and for j = 1, 2, 3, 4 the 10 equations
are essentially reduced to 6 equations.
We assume the line element (due to Schwarzchild) to be of
the form
ds2 = -e"''' dr2 - r2 d82 - r2 sin2 0 dp2 + e'(') dt2 (588)
so that our space is non-Euclidean. We do not include terms
of the form dr do, etc., because we expect our space to be homogeneous and isotropic. We have
911 =
-ea,
-r2,
g22
and
g11 = -e,
g33 = -r2 sin2 0,
i j
gi, = 0,
-r-2'
g22
agke
ago;
(axk
+
ax'
(589)
g33 = - (r2 sin2 6)-1
gii = 0,
g44 = e-',
Now r,k - 21'
9
944 = e'
ag;k
- ax°
iPd j
and since g'° = 0 for
i o e, we have
rik
n
+ ax's - agi)
Caxk
i not summed
If i, j, k are different, then rk = 0. We also see that
rii
_
1
agii
2 g ,i axi
9ii
rik =
gii
kk
2
1
agkk
rkk = - 2 9ii axi
Applying (590), we have
(590)
VECTOR AND TENSOR ANALYSIS
330
1_
r11
1
r22
,tag, _ 1da
1
2g
1
= _
_
=
or
agaa
1 g11
1
raas=
1
ag44
1
dv
or
2
dr
z
1
s
_= -r sine 0 e-A
Or
--g - _
112 = 2
gs2 ag zz
r1aa=
1
1
a9aa
22
29
Or
1
933
2
rt = -219
4
44
4
- sin 0 cos 0
00 =
an..
gsa2
(591)
r
Or
_
-re a
=
g11 a92z
2
1441
2 dr
Or
2
1
[SEC. 141
agsa
ae
r
= cot 0
8g 44
1 dv
Or
2 dr
and all other r;k vanish.
From (539)
ar:
R;; _
ar8,
axe - axa + r
- r°raa
so that
R11 -
a
ar1R
+ art, +
4
ar14
ar
Or
Or
+
rilril + risrs1 + ri3r 1
+ ri4ril - r1i(ri1 + rig + ria + r44)
1(dv12
1
1
r2
r2+2dr2+r2+r2+4`drl
1
d2D
1
1
l da
2r dr
1 da
2r dr
l da dv
4 dr dr
(592)
by making use of (591). Hence Einstein's law R;; = 0 yields
_
R11
1 (_f)2
1 d2v
2 dr2
+
4 Cdr
1 A dv
4 dr dr
1A
r dr -
0
SEC. 141] FURTHER APPLICATIONS OF TENSOR ANALYSIS 331
Similarly
R22 = e-a C 1 -
R33 = sine 8
2
C
1= 0
dr) I-
r (dr
(ddrv
1
1+
L
da
dr I
1 (dv 2
e.-a r_ 1 dzv
R44 =
_
2r
2 dr2
4
- sin2 0 = 0
1 dX dv
+ 4 dr dr
dr
(593)
1 dvl _
r drJ - 0
Dividing R44 by e'-x and adding to R11, we obtain
d +dr =o
or
X + v = constant = co
We desire the form of (588), as r --f oo, to approach that of
(587).
co..
ev
C1
This requires that X and v approach zero as r approaches
Hence \X + v = 0 or X
P.
+ r dr) = 1. Let y =
From R22 = 0 we have
so that dr = 1 dy and
'Y r
yCl+ydr)
or
dy
dr
1-y
r
and
2m
(594)
r
where 2m is a constant of integration.
The equations of the geodesics are
d2x'
a82 +
r;k
dx' dxk
d ds = 0
which yield
d20
2
CdLp)2
dr d_9
Z-2 + 2T E ds A
+
2
r33
Vs
J
=0
332
VECTOR AND TENSOR ANALYSIS
or
d20
+
d,82
a dB
ds
If 0 = 2,
2 dr db
- sin 0 cos 8
r ds ds
a
2
= 0 initially, then 0 =
boundary conditions.
We also obtain
(595)
0
ds
satisfies (595) and the
()2
1'11(ds/2
ds2+
(j2
[SEC. 141
ds+ r44(d- 2 =0
or
dt
ds2
1 da
+ 2 dr
(dr)2
ds
-
-x
re a (thp)2
1 e v-'
+2
dv (dt)2
dr ds/
=0
(596)
=0
(597)
making use of 0 = 7r/2.
Also
d29
ds2
dr dcp
a
2I'19
ds ds
dt dr
d21
ds2
=
+ 2x14 ds ds -
0
d2
or
ds2
0
2drdip
+r
d2t
or
ds2
+
dsds
dv dt dr
dr ds ds
- 0 (598)
Integrating (597) and (598), we obtain
r2
log
dt
d
d(p
A
= h
(599)
dt
+ v = log c
or
c
ds = y
(600)
where h and c are constants of integration. Equation (588)
becomes
ds2 = - 1 dr2 - r2 d(P 2 + y dt2
y
or
1
-
1
y \ds2 - r2
2
\ds2 + y
or
(l
r4 + y
(r
d(p)2
r2
y
1
1
z
SEc. 1411 FURTHER APPLICATIONS OF TENSOR ANALYSIS 333
or
Ch drl 2
2m
h.2
- c2 - 1
r2 dip/ + r2
h2
+ 2m
r
r
r2
and writing it = 1/r, we obtain
()2
au + u2 =
C
z -
h2
1
+ h2 u + 2mu8
and differentiating, we finally obtain
d + u = h2 + 3mu2
d2
(601)
2U
We obtain an approximate solution of (601) in the following
manner: We first neglect the small term 3mu2 = 3m/r2, for large
r.
ad z
The solution of dz + u = h2 is
T =u=-[l+eeos((p-w)j
This is Newton's solu-
where e, w are constants of integration.
tion of planetary motion. We substitute this value of u in the
term 3mu2, and we obtain
m
3m3
6m3
W,p2+u =h2+h;
+ h4
ecos(cp-w)
d2u
33
+ h4 [1 +
23
cos 2((p - co)]
We now neglect certain terms which yield little to our solution
and obtain
dz
6m'3
d2U + u = h2 +
e Cos (,p - w)
From the theory of differential equations the solution of our new
equation is
z
U = h2
+ e cos ((p - w) + h2
= h2 [1 + e cos (rp - co - e)J,
where e = (3m2,'h2)cp and e2 is neglected.
sin ((p - w)
approximately
VECTOR AND TENSOR ANALYSIS
334
(SEC. 141
When the planet moves through one revolution, the advance
of the perihelion is given by 8(w + e) = (3m2/h2) 3 = 6irm2/h2.
When numerical results are given to the constants, it is found
that the discrepancy between observed and calculated results on
the advance of the perihelion of Mercury is removed.
Problems
1. Derive (591).
2. Derive (593).
3. For motion with the speed of light, ds = 0, so that from
(599), h = oo, and (601) becomes
s
d2u + u = 3mu2
(602)
2
Integrate d Y + u = 0, replace this value of u in 3mu2, and
P
obtain an approximate solution of (602) in the form
u = co`p+ 2(coal jp+2sin2(p)
where R is a constant of integration.
Since u = 1/r, x = r coa rp,
y = r sin tp, show that
m x2 +
2y2
x = R + R(x2+y2)1
The term (m/R) (x2 + 2y2)/(x2 + y2)} is the small deviation of the
path of a light ray from the straight line x = R. The asymptotes
are found by taking y large compared with x. Show that they
are x = R + (m/R) (± 2y) and that the angle (in radians)
between the asymptotic lines is approximately 4m/R. This is
twice the predicted value, on the basis of the Newtonian theory,
for the deflection of light as it passes the sun and has been verified during the total eclipse of the sun.
4. If R;; = ag;; is taken for the Einstein law, show that if
y = el, then y = 1 - (2m/r) - }are and
d2U
d2 + U = h2 + 3mu2 - 3 h2 us
5. Assume the following: ds2 = gap dxa dxi, gap = 0 for a p&
agaa
gae
1, 8x4
= 0,
dxa
ds
'
0, a = 1, 2, 3,
dx4
%:W
ds
1,
SEC. 142] FURTHER APPLICATIONS OF TENSOR ANALYSIS 335
41
_ J944+ constant
xl = x, x2 = y, xa = z, x4 = ct. Show that the equations of the
dzi
geodesics reduce to Newton's law of motion die + a ' = 0,
i = 1, 2, 3.
6. With the assumptions of Prob. 5 show that R44 = 0 yields
Laplace's equation V24
= 0.
Tensors.
142. Two-point
The tensors that we have studied
have been functions of one point. Let us now consider the
functions ga,s(xl, x2) which depend on the coordinates of two
points. We now allow independent coordinate transformations
at the two points M1(x;, x;, . . . , x;), M2(x4i x2, xs, . . . ,
If in the new coordinate systems xl, 22 we have
ga.0(
1, 22) = g, (xl, x2)
8xi 8xg2
8-i -
(603)
then the go.,A are the components of a two-point tensor, a covariant
vector relative to Ml and a covariant vector relative to M2.
Indices preceding the comma refer to the point M1, indices
If we keep the coordinates of
M2 fixed, that is, if 2_ = x', then (603) reduces to
following the comma refer to M2.
xs) = gw.a(xl, x2)
dxi
8x
(604)
1
so that relative to Ml, g,,s behaves like a covariant vector. A
similar remark applies at the point M2.
We leave it to the reader to consider the most general type of
two-point tensor fields. We could, indeed, consider a multiple
tensor field depending on a finite number of points. What
difficulties would one encounter for tensors depending on a countable collection of points?
We may consider a two-point tensor field as special one-point
tensors of a 2n-dimensional space subject to a special group of
coordinate transformations.
The scalar invariant
ds2 = ga,s(xl, x2) dxi dxs
(605)
is an immediate generalization of the Riemann line element.
Indeed, when x, and x2 coincide, we obtain the Riemann line
VECTOR AND TENSOR ANALYSIS
336
element.
sdt-
[SEC. 142
Assuming ds2 > 0 for a < t < 0, we can extremalize
fd
fB(ga'odx" dx,
dt
dt
(606)
dt
and obtain a system of differential equations.
z:
1
+
ra+.o,x-°ze1
1
x$ + r:
+
C..'6x1
axe2= 0
a,
xax2 + CO,
(607)
10=0
where
rye, =g ''° a x '
a9°.e
ax`
r,"# = g°'' axe
_ a9°.,)
C`
axiz
=
9"'
;
a9µ..
_
ax1°
ag°.
(608)
axµ IF
dx;
9"'`9µ.i = a;7r
x1 = ds
The unique solutions of (606), xi(s), x4(s), subject to the
initial conditions x1(8o) = ao, X '(so) _ Oo, 1 '(so) = a', xs(so) = o;,
are called dyodesice, or dyopaths.
Problems
1. Derive (607).
2. Show that the CQ,#(x1, X2) are the components of a two-point
tensor, a mixed tensor relative to M1, and a covariant vector
relative to M2.
3. Show that the law of transformation for the linear connec-
tion ri, is
(2 1, x2) = r°";,(xl, x2)
4. Show that rip, = r
,
axi
ax;
'9X'
azi awl ax;
a2xi
axi
+ ax; awl axi
if and only if g.,p =
axa,
where
1
,p,g(x x2) is a scalar relative to M1 and a covariant vector relative to M2. If also r: = r' , show that of necessity
a2
9°`'0
= axi axQ
where 4, is a scalar relative to both M1 and M2.
SEc. 142] FURTHER APPLICATIONS OF TENSOR ANALYSIS 337
5. If
ds2 = -el, dr, dr2 - r,r2eM dpi
e=
-
[1mM2
(m + L
el=(1+rl
ex
=
er dt, dt2
- (mm2M
'
+ r' M)2 r2
1 1 11
M)2 rl J
C1+m/
e2µ-r
show that the two-point tensors
ax
ar:'
1
a
2
+
ro,
a,c
9x°
ax-1
T.a .
a,T
2
vanish identically (m, M are constants). Show that the dyodesics satisfy
rir2 ds1 = he-,,
= he-a
rir2
dt1
ds
__
Cie_
dt2
ds = C2e
d2v
d,P2
+
Mm
v
L
1 + [1
+ (m/M))'hi
M
11 + (m/M)]2h2l
+3Mv2r1+(mMM)J
2
L
provided that Mr2 = mr1, v = 1/r1, hi = (M/m)h. For m << M,
d
M
Mm/hi << 1, we have
+v
+ 3Mv2, the Einstein solution for the motion of an infinitesimal particle moving in the field
of a point gravitational mass M.
REFERENCES
Brand, L. "Vectorial Mechanics," John Wiley & Sons, Inc.,
New York. 1930.
Brillouin, L. "Les Tenseurs," Dover Publications, New York
1946.
Graustein, W. C. "Differential Geometry," The Macmillan
Company, New York. 1935.
Houston, W. V. "Principles of Mathematical Physics,"
McGraw-Hill Book Company, New York. 1934.
Joos, G. "Theoretical Physics," G. E. Stechert & Company,
New York. 1934.
Kellogg, O. D. "Foundations of Potential Theory," John Murray, London. 1929.
McConnell, A. J. "Applications of the Absolute Differential
Calculus," Blackie & Son Ltd., Glasgow. 1931.
Michal, A. D. "Matrix and Tensor Calculus," John Wiley &
Sons, Inc., New York. 1947.
Milne-Thomson, L. M. "Theoretical Hydrodynamics," The
Macmillan Company, New York. 1938.
Page, L. "Introduction to Theoretical Physics," D. Van
Nostrand Company, New York. 1935.
Phillips, H. B. "Vector Analysis," John Wiley & Sons, Inc.,
New York.
1933.
Smythe, W. R. "Static and Dynamic Electricity," McGrawHill Book Company, New York. 1939.
Thomas, T. Y. "Differential Invariants of Generalized Spaces,"
Cambridge University Press, London. 1934.
Tolman, R. C. "Relativity Thermodynamics and Cosmology,"
Oxford University Press, New York. 1934.
Veblen, O. "Invariants of Quadratic Differential Forms,"
Cambridge University Press, London. 1933.
Weatherburn, C. E. "Elementary Vector Analysis," George
Bell & Sons, Ltd., London. 1921.
"Advanced Vector Analysis," George Bell & Sons, Ltd.,
London, 1944.
339
VECTOR AND TENSOR ANALYSIS
3.10
"Differential Geometry," Cambridge University Press,
London.
1927.
"Riemannian Geometry," Cambridge University Press,
London. 1942.
Wilson, W. "Theoretical Physics," vols. I, II, III, Methuen
& Co., Ltd, London. 1931, 1933, 1940.
INDEX
A
Acceleration, angular, 186
centripetal, 30-31, 184
Coriolis, 210-211
linear, 30, 184, 210-211
Action integral, 328
Addition, of tensors, 275
of vectors, 2
Angular momentum, 196-200
Angular velocity, 22
Are length, 71, 100
Archimedean ordering postulate, 92
Arcs, 98
rectifiable, 98, 100
regular, 98
Arithmetic n-space, 268-269
Associated vector, 281
Associative law of vector addition, 3
Asymptotic directions, 81
Asymptotic lines, 81
Average curvature, 78
B
Bernoulli's theorem, 234-235
Bianchi's identity, 308
Binormal, 58, 312
Biot-Savart law, 163
Boundary point, 90
Boundary of a set, 91
Bounded set, 89
Bounded variation, 99
C
Calculus of variations, 83-85
Cartesian coordinate system, 280,
292
Cauchy-Riemann equations, 122
341
Cauchy's criterion for sequences, 97
Cauchy's inequality, 13
Center of mass, 194
Centripetal acceleration, 30-31, 184
Ceva's theorem, 8
Characteristic curves, 69
Charges, 127
moving, 146
Christoffel symbols, 289-293
law of transformation of, 290-291
Circulation, 238
Closed interval, 89
Closed set, 91
Commutative law of vector addition, 3
Complement of a set, 90
Components of a tensor, 274
Components of a vector, 8, 270-271
Conductivity, 162
Conductor, 128
field in neighborhood of, 130-131
force on the surface of, 131
Conformal space, 294
Conjugate directions, 80
Conjugate functions, 122, 143
Connected region, 102-103
Conservation of electric charge, 162
Conservative field, 103
Continuity, 95
equation of, 231-232
uniform, 95
Contraction of a tensor, 275-276
Contravariant tensor, 274-275
Contravariant vector, 270-272
Coordinate curves, 52
Coordinate system, 9, 269
Coordinates, geodesic, 303-305
Riemannian, 305
transformation of, 269
Coriolis acceleration, 210-211
342
VECTOR AND TENSOR ANALYSIS
Couple, 197
Covariant, differentiation, 295-296
generalized differentiation of, 318319
Direction cosines, 13
Directional derivative, 38, 297
Discontinuities of D and E, 138139
Covariant tensor, 274-275
Covariant vector, 273
Curl, 45, 55, 297, 300
of a gradient, 46, 297
Currents, displacement, 168
electric, 161-162
Curvature, average, 78
of a curve, 58, 311
Gaussian, 78, 322
lines of, 78
Riemannian, 307, 320-323
tensor, 306-307
Curve (see Space curve)
Curvilinear coordinates, 50, 70
curl, divergence, gradient, Laplacian in, 54-55
D
D'Alembertian, 178
Del (v), 40
Deflection of light, 334
Deformation tensor, 246
Desargues's theorem, 7
Derivative, covariant, 295-296
intrinsic, 297-298
of a vector, 29
Determinants, 263-267
cofactors of, 264
derivative of, 266
multiplication of, 264
Developable surfaces, 70
Diameter of a set, 93
Dielectrics, 135-136
Differentiation, covariant, 295-296
generalized covariant, 318-319
rules, 32
of vectors, 29
Dipole, 157-158
energy of, 158
field of, 158
magnetic, 160-161
moment of, 157
potential of, 157
Displacement current, 168
Displacement vector, 136
Distributive law, 3, 11, 21
Divergence, 42, 54, 120, 297-298
of a curl, 46
of a gradient, 44
Divergence theorem of Gauss, 114120, 299
Dot, or scalar, product, 10
Dynamics of a particle, 189
Dynamics of a system of particles,
194
Dyodesics, 336
E
Edge of regression, 69-70
Einstein, Albert, law of gravitation,
328-329
space, 324
special theory of relativity, 283286
summation notation, 259
Einstein-Lorentz transformations,
283
Electric field, 127
discontinuity of, 138
polarization of, 158-159
Electromagnetic wave equations,
170-173
Electrostatic dipoles, 157-158
Electrostatic energy, 136-138
Electrostatic field, 127
Electrostatic flux, 128
Electrostatic forces, 127
Electrostatic intensity, 127
Electrostatic polarization, 158
Electrostatic potential, 128
Electrostatic unit of charge, 127
Electrostatics, Gauss's law of, 128
Green's reciprocity theorem of,
139-140
Ellipsoid of inertia, 226
of strain, 245
INDEX
Energy, equation for a fluid, 235-236
of electromagnetic field, 175
of electrostatic field, 136-138
kinetic, 201
Envelopes, 69
Equation, of continuity, 231-232
of gauge invariance, 177
of motion for a fluid, 233-236,
302-303
Equipotential surfaces, 129
Euclidean space, 8, 279, 308-309
Euler's angular coordinates, 219-221
equation of motion, for a fluid, 233,
302-303
for a rigid body, 216-217
Euler-Lagrange equation, 85
Evolutes, 66
343
Gauss, divergence theorem of, 114120
electrostatic law of, 128
Generalized force vector, 326
momentum, 328
Geodesic coordinates, 303-305
Geodesics, 83, 288-289
minimal, 294
Gradient, 36, 120, 273, 297
Gravitation, Einstein's law of, 328329
Newton's law of, 190
Green's formula, 118, 299-300
Green's reciprocity theorem, 139140
Gyroscope, motion of, 222-225
H
F
Faraday's law of induction, 167
Field, 9
conservative vector, 103
nonconservative vector, 104
solenoidal vector, 117
steady, 9
uniform, 10
Fluid, 230
general motion of, 236-238
Force moment, 196-200
Foucault pendulum, 213-215
Frenet-Serret formulas, 60, 311-312
Functions of bounded variation, 99100
conjugate, 122
continuous, 95
properties of, 96
harmonic, 123
Fundamental forms, first, 71
second, 74-75
Fundamental planes, 62-63
normal, 62
osculating, 62
rectifying, 63
G
Gauge invariance, 177
Gauss, curvature, 78, 322
Hamilton's equations of motion, 328
Harmonic conjugates, 123
functions, 123, 143
Heine-Borel theorem, 94
Helix, 60
Hooke's law, 249
Hypersurfaces, 282
I
Images, method of, 141-143
Induction, law of, 167
Inertia, moment of, 216
product of, 216
tensor, 225-228
Inertial frame, 211
Infemum, 92
Inhomogeneous wave equation, 177
solution of, 178-182
Insulator, 129
Integral, line, 101, 103-105
Riemann, 101
Integrating factor, 111
Integration, of Laplace's equation,
145
of Poisson's equation, 155
Interior point, 90
Interval, closed, 89
open, 89
VECTOR AND TENSOR ANALYSIS
344
Intrinsic equations of a curve, 63
Invariant, 271
Involutes, 64
Irrotational motion, 232
Irrotational vectors, 107, 111
J
Jacobian, 49, 265
Jordan curves, 98
K
Kelvin's theorem, 239
Kepler's laws of planetary motion,
191-193
Kinematics, of a particle, 184
of a rigid body, 204-207
Kinetic energy, 201
Kirchhoff's solution of the inhomogeneous wave equation, 178-182
Kronecker delta, 260-262
L
Lagrange's equations, 324-327
Laplace's equation, 123
integration of, 145
solution in spherical coordinates,
146-149
uniqueness theorem, 119
Laplacian, 45, 298-299
in cylindrical coordinates, 55
in spherical coordinates, 56, 299
Law of induction, 167
of refraction, 139
Legendre polynomials, 148-149
Legendre's equation, 148
Limit point, 90
Line, of curvature, 78
element, 279
of Schwarzchild, 329
of force, 132
integral, 101, 103-105
Linear function, 3
set, 89
Liquids, general motion of, 233-234
Lorentz's electron theory, 175-177
transformations, 61, 283
M
Magnetic dipole, 160-161
effect of currents, 162-164
Magnetostatics, 160
Mass of a particle, 189, 285
Maxwell's equations, 167-169
for a homogeneous conducting
medium, 173
solution of, 169-173
Menelaus' theorem, 8
Meusnier's theorem, 75
Minkowski force, 285
Moment of inertia, 216
Momentum, 196
angular, 196-200
generalized, 328
relative angular, 199-200
Motion, in a plane, 33
irrotational, 234
relative, 187-188
steady, 234
vortex, 238-239
Moving charges, 161-162
Mutual induction of two circuits,
165-166
N
Navier-Stokes equation, 255-257
Neighborhood, 90
Newton's law of gravitation, 190
Newton's law of motion, 189, 211
Nonconservative field, 104
Normal acceleration, 184
plane, 62
to a space curve, 58, 311
to a surface, 73, 109
Number triples, 15, 268-269
O
Oersted, magnetic effect of currents,
162-165
Ohm's law, 162
Open interval, 89
Open set, 90
Orthogonal transformation 292-293
Osculating plane, 62
345
INDEX
P
Parallel displacement, 313-315
Parallelism in a subspace, 315-317
Parametric lines or curves, 71
Particle, acceleration of, 30, 210
angular momentum of, 196
dynamics of, 189
kinematics of, 184
mass of, 189, 285
momentum of, 196
Newton's laws of motion for, 189,
211
rotation of, 22
velocity of, 30, 184, 209
Particles, system of, 194
Perihelion of Mercury, 334
Permeability, 160
Planetary motion, 190-193
Point, 89
boundary, 90
interior, 90
limit, 90
neighborhood of, 90
set theory, 89
Poisson's equation, 132-134
integration of, 155
Poisson's ratio, 249
Polarization, 158-159
Potential, of a dipole, 157
electrostatic, 128
vector, 117
velocity, 232
Power, 162
Poynting's theorem, 174-175
Poynting's vector, 175
Pressure, 230
Principal directions, 77-78
Q
Quadratic differential form, 280
Quotient law of tensors, 276
R
Radius of curvature, 24
Recapitulation of differentiation
formulas, 48
Reciprocal tensors, 281
Rectifying plane, 63
Refraction, law of, 139
Regions, connected, 102
simply connected, 102-103
Regular arcs, 98
Relative motion, 187-188
time rate of change of vectors, 208
Resistance, electric, 162
Retarded potentials, 178-182
Ricci tensor, 307
Riemann integral, 101
Riemannian, coordinates, 305
curvature, 307, 320-323
metric, 280
space, 280
geodesics in, 288-289
hypersurface in, 282
Riemann-Christoffel tensor, 307-308
Rigid bodies, 203
motion of, 215-225
S
Scalar, 1
curvature, 307
gradient of, 36, 273, 297
Laplacian of, 45, 298-299
product of vectors, 10, 273-274
Schur's theorem, 323-324
Schwarzchild line element, 329
Second fundamental form, 74-75
geometrical significance, 75
Sequence, 97
Cauchy criterion for convergence
of, 97
Set, 89
bounded, 89
closed, 91
complement of, 90
countable, 93
diameter of, 93
infemum of, 92
limit point of, 90
linear, 89
open, 90
supremum of, 91
theorem of nested, 93
Simply connected region, 102
346
VECTOR AND TENSOR ANALYSIS
Sink, 231
Solenoidal vector, 117
Solid angle, 160
Source, 231
Space, conformal, 294
Space curve, 31
Tensor, components of, 274
contraction of, 275-276
contravariant, 274-275
covariant, 274-275
curvature, 306-307
deformation, 246
inertia, 225-228
mixed, 275
Ricci, 307
Riemann-Christoffel, 307-308
strain, 243-246
stress, 246-248
two-point, 335
weight of, 275
Tensors, 274-278
absolute, 275
addition of, 275
cross product of, 278
outer product of, 278
product of, 275
quotient law of, 276
reciprocal, 281
relative, 275
Theorem, of Ceva, 8
of Desargue, 7
of Menelaus, 8
Top, motion of, 222-225
Torque, 196-200
Torsion of a space curve, 59, 312
Transformation of coordinates, 269
Trihedral, 59
Triple scalar product, 23
Triple vector product, 24
Two-point tensors, 335
arc length of, 100-101
curvature of, 58, 311
intrinsic equations of, 63
Jordan, 98
on a surface, 72
radius of curvature of, 58
tangent to, 31, 58, 311
torsion of, 59, 312
unit binormal of, 59, 312
unit principal normal of, 58, 311
Space of n-dimensions, 268-269
Special theory of relativity, 283-286
Spherical coordinates, 35, 50
indicatrix, 68
Steady field, 9
Stokes's theorem, 107-112, 300-301
Strain, ellipsoid, 245
tensor, 243-246
Streamline, 234
Stress tensor, 246-248
Subtraction of vectors, 3
Summation convention, 259
Superscripts, 14, 259
Supremum, 91-92
Surface, 70
are length on, 71
asymptotic curves on, 81
average curvature of, 78
conjugate directions on, 80
U
curves on, 72
developable, 70
Uniform continuity, 95
first fundamental form of, 71
Uniform vector field, 10
Gauss curvature of, 78
Uniqueness theorems, 119
geodescis of, 83
Unit charge, 127
normal to, 73, 109
principal directions on, 77
V
second fundamental form of, 74
Vector, associated, 281
T
basis, 3, 8
center of mass, 194
Tangent to a space curve, 31, 58, 311
components of, 8-9
Tangential acceleration, 184
conservative field, 108
INDEX
Vector, contravariant, 270-272
covariant, 273
curl of, 45, 55, 297, 300
definition of, 1
differentiation of, 29
displacement, 136
divergence of, 42, 54, 120, 297-298
field, 9
irrotational, 107, 111
length of, 1, 281
operator del (v), 40
physical components of, 272
potential, 117
solenoidal, 117
space, 268-269
sum of a solenoidal and an irrotational vector, 156-157
unit, 1, 281
zero, 1
Vectors, addition of, 2, 275
angle between two, 10, 281-282
differentiation of, 29
equality of, 1
fundamental unit, 8
linear combination of, 3
347
Vectors, parallel, 2, 314
parallel displacement of, 313-315
scalar, or dot, product of, 10, 273274
subtraction of, 3
triple scalar product of, 23-24
triple vector product of, 24-25
vector, or cross, product of, 20-23
Velocity, angular, 22-23
linear, 30, 184, 209
potential, 232
Vortex motion, 238-239
W
Waves, equation of, 170
inhomogeneous equation of, 177
longitudinal, 253-254
transverse, 172, 253
IN'eierstrass-Bolzano theorem, 92
Work, 103, 202
Y
Young's modulus, 249