Appendix
C
Vectors and tensors
C.1
Vectors
A vector x can be represented in a (not necessarily orthogonal) coordinate system
x = x1 ê1 + x2 ê2 + x3 ê3
= xp êp
(C.1)
where êi are unit length basis vectors. They form a base for a cartesian (orthogonal)
coordinate system if all of the following conditions are fulfilled
êi · êj = δij .
(C.2)
The symbol δij is the Kronecker symbol and is defined by
(
1 if i = j,
δij :=
0 if i 6= j.
Kronecker symbo
In Equation (C.1) we have used the summation convention: we sum over all summation conve
indices that appear twice.
We now consider another orthogonal coordinate system K 0 that is rotated with
respect to the original coordinate system K, but has the same origin. The new base
vectors ê0i also fulfill the condition
ê0i · ê0j = δij .
(C.3)
The vector x can then be written in the new base as
x = x01 ê01 + x02 ê02 + x03 ê03
= x0p ê0p .
(C.4)
The vector x stays the same, its representation is different in both coordinate systems
(the components xi and x0i are different).
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Appendix C
Vectors and tensors
K
ê3
K, K’
ê3 = ê′3
ê′2
ê2
ê2
θ
ê′1
θ
ê1
ê1
Figure C.1: The original coordinate system K (solid lines) and the rotated system
K 0 (dashed lines)
Rotation matrix
direction cosine
We now derive the connection between the representations of the vector x in both
coordinate systems. For this we first define the direction cosine
αij := ê0i · êj =
cos(ê0i , êj )
| {z }
direction cosine
(C.5)
The quantity αij is the scalar product of the unit vectors ê0i and êj and therefore also
the cosine of the angle between the vectors ê0i and êj (remember a · b = ab cos θ).
With help of the summation convention (Eq. C.1) we write
x = xp êp ,
(C.6)
an make the scalar product with the unit vector êi
x · êi = xp êp · êi
= xp δpi
= xi
(C.7)
and therefore
xi = x · êi
= x0p ê0p · êi
=
x0p αpi
(Equation C.4)
.
(Equation C.5)
(C.8)
Therefore we have shown that the representations of the vector x in both coordinate
systems K and K 0 are linked by
xi = αpi x0p .
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(C.9)
Appendix C
Vectors and tensors
It can also be shown that the inverse transformation is given by
x0i = αip xp .
(C.10)
We now derive the same rules for the direction cosines αij . First we write
x0i = xp αip
= αjp x0j αip
(Equation C.10)
(Equation C.9)
= αjp αip x0j .
and use x0i = δij x0j so that we obtain δij x0j = αjp αip x0j . This is true for all values of
x0j so that we arrive at
αip αjp = δij .
(C.11)
Similarly it can be shown that
αpi αpj = δij .
It is convenient to write the direction cosines as

α11 α12

[αij ] = α21 α22
α31 α32
(C.12)
a matrix

α13
α23 
α33
(C.13)
Equations (C.11) and (C.12) can then be written more compactly as
αip αjp = δij
or
[αij ][αij ]T = 1 ,
αpi αpj = δij
or
[αij ]T [αij ] = 1 ,
(C.14)
and Equation (C.9) is written as
x = [αij ]T x0 .
(C.15)
The matrix [αij ] is called the rotation matrix. As Equation (C.14) shows, it has rotation matrix
the important property that the transpose of the rotation matrix is identical to its
inverse
[αij ]T = [αij ]−1 .
(C.16)
The inverse matrix [αij ]−1 is defined through
inverse matrix
[αij ][αij ]−1 = [αij ]−1 [αij ] = 1 .
(C.17)
Example We assume that the coordinate system K 0 is rotated by the angle θ with
respect to the coordinate system K (Figure C.1). The components of the rotation
matrix can be obtained by calculating the scalar products êi · ê0j :
·
ê01
ê02
ê03
ê1
cos θ
− sin θ
0
C3
ê2
sin θ
cos θ
0
ê3
0
0
1
Appendix C
Vectors and tensors
For the calculation of the components α21 and α12 in the table we have made use of
the relations
cos(θ − π/2) = sin(θ)
cos(θ + π/2) = − sin(θ)
For example
α21 = cos(ê02 , ê1 ) = cos(−π/2 − θ)
= − sin(θ) .
Therefore the rotation matrix is


cos θ sin θ 0
[αij ] = − sin θ cos θ 0
0
0
1
(C.18)
Next we take a point P which position vector has the coordinates (2,1,3) in the
coordinate system K
P = 2ê1 + ê2 + 3ê3 .
We want to calculate the coordinates of P in the system K 0 . With the relation
(Eq. C.10)
x0i = αip xp
we obtain
x01 = α11 x1 + α12 x2 + α13 x3
= cos θ · 2 + sin θ · 1 + 0 · 3
0
x2 = α21 x1 + α22 x2 + α23 x3
= − sin θ · 2 + cos θ · 1 + 0 · 3
0
x3 = α31 x1 + α32 x2 + α33 x3
= 0 · 2 + 0 · 1 + 1 · 3,
and arrive at
P = (2 cos θ + sin θ)ê01 + (−2 sin θ + cos θ)ê02 + 3ê03 .
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Appendix C
Vectors and tensors
Comma notation
We consider the scalar field f = f (x) (e.g. temperature field), the vector field
fk = fk (x) (e.g. velocity field) and the tensor field fpq = fpq (x) (e.g. stress field). We
introduce the compact comma notation f,i for the derivative of f with respect to comma notation
spatial direction xi
∂f
i = 1, 2, 3.
f,i :=
∂xi
Examples of the comma notation
(I)
(xk fk ),i = fi + xk fk,i
(II) (xk fk ),ij = fi,j + fj,i + xk fk,ij
Proofs:
(I)
(xk fk ),i = xk,i fk + xk fk,i
= δki fk + xk fk,i
= fi + xk fk,i
(II) (xk fk ),ij = (fi + xk fk,i ),j
= fi,j + xk,j fk,i + xk fk,ij
= fi,j + fj,i + xk fk,ij
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Appendix C
C.2
Vectors and tensors
Tensors
First order tensor
A vector a can be written in two different bases êk and ê0k as
a = ap êp
a = a0p ê0p .
and
(C.19)
Further valid expressions are
a0i = αip ap
ai = αpi a0p
tensor of order 1
and
(C.20)
where αij are the components of a rotation matrix. We now define a (cartesian)
tensor of order 1 as a quantity, which is represented by three real numbers that
under the change from the {xi }-system to the {x0i }-system are transformed as
a0i = αip ap .
(C.21)
Second order tensor
The above definition is extendable. We consider two vectors a and b. Analogous to
Equation (C.20) we can write
b0i = αip bp
bi = αpi b0p .
and
(C.22)
We now form the product ai bj and look at its transformation behavior
a0i b0j = (αip ap )(αjq bq )
= αip αjq ap bq
and
ai bj = (αpi a0p )(αqj b0q )
= αpi αqj a0p b0q
(C.23)
These equations yield the relation between ai bj and a0i b0j . We now write the product
ai bj as a new quantity
cij := ai bj
and also
c0ij := a0i b0j .
tensor product
The new quantity [ai bj ] = [cij ] can be written as 3 × 3 matrix, the tensor product
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Appendix C
Vectors and tensors
of the vectors a and b


a1 b 1 a1 b 2 a1 b 3
[ai bj ] = a2 b1 a2 b2 a2 b3 
a3 b 1 a3 b 2 a3 b 3
=
a
⊗ b}
| {z
tensor product
With the above definitions, Equations (C.23) can be written
cij = ai bj = αpi αqj a0p b0q = αpi αqj c0pq
c0ij
=
a0i b0j
and
= αip αjq ap bq = αip αjq cpq .
(C.24)
A tensor of order 2 is defined by this transformation rule, i.e. a tensor A of order 2
with the components [A]ij = Aij always transforms like
A0ij = αip αjq Apq .
(C.25)
Tensor of order n
Definition: Given the 3n numbers a0i1 ,i2 ,··· ,in that transform as
a0i1 ,i2 ,··· ,in = αi1 ,j1 αi2 ,j2 · · · αin ,jn aj1 ,j2 ,··· ,jn
(C.26)
under the change from the cartesian coordinate system xi to x0i . These numbers are
tensor of order n
called cartesian tensor of order n
A tensor is defined by its transformation properties. To test whether a given quantity
is a tensor, the components have to transform according to Equation (C.26).
Example: We take some scalar field φ = φ(x), which could be for example the
temperature field. We define the quantity ai := φ,i (the temperature gradient) in
any coordinate system K. Are the three numbers (a1 , a2 , a3 ) the components of a
tensor?
To answer this question we inquire the transformation properties of ai . Following
the definition of ai , which is valid for all coordinate systems, we have
a0i =
∂φ
,
∂x0i
and with the chain rule
a0i =
∂φ
∂φ ∂xk
∂xk
=
= φ,k 0 .
0
0
∂xi
∂xk ∂xi
∂xi
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Appendix C
Vectors and tensors
Equation shows that
∂xk
= αik ,
∂x0i
such that
a0i = αik φ,k = αik ak .
Comparison with Equation (C.26) shows that ai is indeed a tensor of first order.
Example: We now figure out how the second order tensor A looks in the K 0 -system,
when it has the following form in the K-system


0 γ 0
A = γ 0 0
0 0 1
The K 0 -system is rotated with respect to the K-system about the ê3 axis by an
angle of θ = π/4.
Since A is a second order tensor, we have
a0ij = αik αjl akl
with


cos θ sin θ 0
[αij ] = − sin θ cos θ 0 ,
0
0
1
with θ = π/4. Using the transformation formula we obtain
a011 = α1k α1l akl
= α11 (α11 a11 + α12 a12 + α13 a13 )
+ α12 (α11 a21 + α12 a22 + α13 a23 )
+ α13 (α11 a31 + α12 a32 + α13 a33 )
= cos θ(0 cos θ + γ sin θ + 0)
+ sin θ(γ cos θ + 0 sin θ + 0)
+ 0 (. . . )
1 1
= 2γ cos θ sin θ = 2γ √ √ = γ.
2 2
Therefore we have
a011 = γ.
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Appendix C
Vectors and tensors
Similarly for the next component we get
a012 = α1k α2l akl
= α11 (α21 a11 + α22 a12 + α23 a13 )
+ α12 (α21 a21 + α22 a22 + α23 a23 )
+ α13 (α21 a31 + α22 a32 + α23 a33 )
= cos θ(−0 sin θ + γ cos θ + 0)
+ sin θ(−γ sin θ + 0 cos θ + 0)
+ 0 (. . . )
= γ(cos2 θ − sin2 θ)
= γ(cos2 θ − sin2 θ)|(θ=π/4) = 0 ,
and therefore
a012 = γ.
With this we arrive at the representation

γ
A = 0
0
of tensor A in the K 0 system

0 0
−γ 0 .
0 1
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Appendix C
Vectors and tensors
Invariants
invariant
A quantity which is independent of the orientation of the coordinate system is called
an invariant. This is best explained with some examples.
Example: a and b are vectors with components ai and bj . Show that the scalar
product c = a · b is invariant.
To show this we have to prove that the scalar product is independent of the orientation of the coordinate system, i.e. that c0 = a0i b0i and c = ai bi are the same
c0 = a0i b0i
= αip ap αiq bq
= αip αiq ap bq
= δpq ap bq
= ap bp
= ai bi
=c
Example: cij are the components of a second order tensor. Show that the trace cii
is an invariant.
c0ii = αip αiq cpq
= δpq cpq
= cpp = cii
Example: Show that cik cki is an invariant.
c0ik c0ki = αiq αkp cqp αkr αis crs
= αiq αis αkp αkr cqp crs
= δqs δpr cqp crs
= cqp cpq = cik cki
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Appendix C
Vectors and tensors
The permutation symbol ijk
The permutation symbol ijk has the value zero if at least two of the indices are permutation symb
the same. Otherwise the symbol has the value +1 or -1, depending on whether the
order of the indices is cyclic or anticyclic. Thus we have ε123 = ε231 = ε312 = 1 (cyclic
indices), ε132 = ε213 = ε321 = −1 (anticyclic indices), and εiij = εijj = εjii = 0 (no
summation over repeated indices!) for i and j = 1, 2, 3.


+1, i, j, k in cyclic order
εijk := −1, i, j, k in anticyclic order


0,
two or more indices have the same value
The permutation symbol is also known as the Levi-Civita ε symbol. This symbol is
mainly used to write the vector product in index notation. One valid expression is
εijk = êi · (êj × êk ).
A useful relation between the Kronecker symbol and the permutation symbol is the
δ − ε relation
εijk εkpq = δip δjq − δjp δiq .
(C.27)
Isotropic tensors
An isotropic tensor is a tensor with the same entries in each coordinate system. isotropic tensor
For each isotropic tensor of order n this relation holds
A0ijk... = Aijk... .
The unit tensor δij is an example of an isotropic tensor, since
δij0 = αip αjq δpq
(Eq. C.25)
= αip αjp
= δij
(Eq. C.11) ,
and therefore δij0 = δij .
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Appendix C
Vectors and tensors
We note some relations without proof, that will be useful in further sections.
• Each isotropic second order tensor A with components [A]ij can be written in
the form
[A]ij = α δij
where α is a scalar.
• Each isotropic third order tensor A with components [A]ijk can be written in
the form
[A]ijk = α εijk
where α is a scalar.
• Each isotropic fourth order tensor A with components [A]ijkl can be written
in the form
[A]ijkl = α δij δkl + β δik δjl + γ δil δjk
where α, β and γ are a scalar quantities.
C 12