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A Some Basic Rules of Tensor Calculus
The tensor calculus is a powerful tool for the description of the fundamentals in con-
tinuum mechanics and the derivation of the governing equations for applied prob-
lems. In general, there are two possibilities for the representation of the tensors and
the tensorial equations:
- the direct (symbolic) notation and - the index (component) notation
The direct notation operates with scalars, vectors and tensors as physical objects
defined in the three dimensional space. A vector (first rank tensor) a
a a is considered
as a directed line segment rather than a triple of numbers (coordinates). A second
rank tensor AAA is any finite sum of ordered vector pairs AAA = aaa ⊗⊗⊗ bbb +... + ccc ⊗⊗⊗ ddd. The
scalars, vectors and tensors are handled as invariant (independent from the choice of
the coordinate system) objects. This is the reason for the use of the direct notation
in the modern literature of mechanics and rheology, e.g. [29, 32, 49, 123, 131, 199,
246, 313, 334] among others.
The index notation deals with components or coordinates of vectors and tensors.
For a selected basis, e.g. g
g g
i
, i = 1, 2, 3 one can write
aaa = a
i
ggg
i
, AAA =
a
i
b
j
+... + c
i
d
j
ggg
i
⊗ ggg
j
Here the Einstein’s summation convention is used: in one expression the twice re-
peated indices are summed up from 1 to 3, e.g.
a
k
g
g g
k
3
k= 1
a
k
g
g g
k
, A
ik
b
k
3
k= 1
A
ik
b
k
In the above examples k is a so-called dummy index. Within the index notation the
basic operations with tensors are defined with respect to their coordinates, e. g. the
sum of two vectors is computed as a sum of their coordinates c
i
= a
i
i
. The
introduced basis remains in the background. It must be remembered that a change
of the coordinate system leads to the change of the components of tensors.
In this work we prefer the direct tensor notation over the index one. When solv-
ing applied problems the tensor equations can be “translated” into the language
of matrices for a specified coordinate system. The purpose of this Appendix is to
168 A Some Basic Rules of Tensor Calculus
give a brief guide to notations and rules of the tensor calculus applied through-
out this work. For more comprehensive overviews on tensor calculus we recom-
mend [54, 96, 123, 191, 199, 311, 334]. The calculus of matrices is presented in
[40, 111, 340], for example. Section A.1 provides a brief overview of basic alge-
braic operations with vectors and second rank tensors. Several rules from tensor
analysis are summarized in Sect. A.2. Basic sets of invariants for different groups
of symmetry transformation are presented in Sect. A.3, where a novel approach to
find the functional basis is discussed.
A.1 Basic Operations of Tensor Algebra
A.1.1 Polar and Axial Vectors
A vector in the three-dimensional Euclidean space is defined as a directed line seg-
ment with specified magnitude (scalar) and direction. The magnitude (the length) of
a vector aaa is denoted by |aaa|. Two vectors aaa and bbb are equal if they have the same
direction and the same magnitude. The zero vector 000 has a magnitude equal to zero.
In mechanics two types of vectors can be introduced. The vectors of the first type are
directed line segments. These vectors are associated with translations in the three-
dimensional space. Examples for polar vectors include the force, the displacement,
the velocity, the acceleration, the momentum, etc. The second type is used to char-
acterize spinor motions and related quantities, i.e. the moment, the angular velocity,
the angular momentum, etc. Figure A.1a shows the so-called spin vector a
a a ∗
which
represents a rotation about the given axis. The direction of rotation is specified by
the circular arrow and the “magnitude” of rotation is the corresponding length. For
the given spin vector aaa ∗
the directed line segment aaa is introduced according to the
following rules [334]:
- the vector aaa is placed on the axis of the spin vector,
- the magnitude of aaa is equal to the magnitude of aaa ∗
aaa
∗
aaa
aaa ∗ ∗
aaa
a
a a
a
b c
Figure A.1 Axial vectors. a Spin vector, b axial vector in the right-screw oriented reference
frame, c axial vector in the left-screw oriented reference frame
170 A Some Basic Rules of Tensor Calculus
aaa aaa
b
b
b
bbb
n
n
n
aaa
a
a
a
|a
a
a|
(bbb ··· nnn
aaa
)nnn
aaa
a b
Figure A.3 Scalar product of two vectors. a Angles between two vectors, b unit vector and
projection
Scalar (Dot) Product of two Vectors. For any pair of vectors aaa and bbb a scalar
α is defined by
α = aaa ··· bbb = |aaa||bbb| cos ϕ ,
where ϕ is the angle between the vectors aaa and bbb. As ϕ one can use any of the two
angles between the vectors, Fig. A.3a. The properties of the scalar product are
- aaa ··· bbb = bbb ··· aaa (commutativity), - aaa ··· (bbb + ccc) = aaa ··· bbb + aaa ··· ccc (distributivity)
Two nonzero vectors are said to be orthogonal if their scalar product is zero. The
unit vector directed along the vector aaa is defined by (see Fig. A.3b)
n
n n aaa
aaa
|aaa|
The projection of the vector b
b b onto the vector a
a a is the vector (b
b b ·
· n
n n aaa
)n
n n aaa
, Fig. A.3b.
The length of the projection is |b
b b|| cos ϕ |.
Vector (Cross) Product of Two Vectors. For the ordered pair of vectors aaa and
bbb the vector ccc = aaa × bbb is defined in two following steps [334]:
- the spin vector ccc ∗
is defined in such a way that
- the axis is orthogonal to the plane spanned on aaa and bbb, Fig. A.4a,
- the circular arrow shows the direction of the “shortest” rotation from a
a a to b
b b,
Fig. A.4b,
a a||b
b b| sin ϕ , where ϕ is the angle of the “shortest” rotation from a
a a
to b
b b,
- from the resulting spin vector the directed line segment ccc is constructed according
to one of the rules listed in Subsect. A.1.1.
The properties of the vector product are
aaa × bbb = −bbb × aaa, aaa × (bbb + ccc) = aaa × bbb + aaa × ccc
The type of the vector ccc = aaa × bbb can be established for the known types of the
vectors a
a a and b
b b, [334]. If a
a a and b
b b are polar vectors the result of the cross product
A.1 Basic Operations of Tensor Algebra 171
aaa aaa aaa b
b b bbb b
b b
ϕ
ϕ
ϕ
ccc ∗
c
c c
a b c
Figure A.4 Vector product of two vectors. a Plane spanned on two vectors, b spin vector, c
axial vector in the right-screw oriented reference frame
will be the axial vector. An example is the moment of momentum for a mass point m
defined by r
r r × (m
v
v v), where r
r r is the position of the mass point and v
v v is the velocity
of the mass point. The next example is the formula for the distribution of velocities
in a rigid body vvv = ωωω × rrr. Here the cross product of the axial vector ωωω (angular
velocity) with the polar vector rrr (position vector) results in the polar vector vvv.
The mixed product of three vectors aaa, bbb and ccc is defined by (aaa × bbb) ··· ccc. The result
is a scalar. For the mixed product the following identities are valid
aaa ··· (bbb × ccc) = bbb ··· (ccc × aaa) = ccc ··· (aaa × bbb) (A.1.1)
If the cross product is applied twice, the first operation must be set in parentheses,
e.g., a
a a × (b
b b × c
c c). The result of this operation is a vector. The following relation can
be applied
a
a a × (b
b b × c
c c) = b
b b(a
a a ·
· c
c c) − c
c c(a
a a ·
· b
b b) (A.1.2)
By use of (A.1.1) and (A.1.2) one can calculate
(aaa × bbb) ··· (ccc × ddd) = aaa ··· [bbb × (ccc × ddd)]
= aaa ··· (ccc bbb ··· ddd − ddd bbb ··· ccc)
= a
a a ·
· c
c c b
b b ·
· d
d d − a
a a ·
· d
d d b
b b ·
· c
c c
(A.1.3)
A.1.3 Bases
Any triple of linear independent vectors eee 1
, eee
2
, eee
3
is called basis. A triple of vectors
eee i
is linear independent if and only if eee
1
··· (eee
2
× eee
3
For a given basis eee
i
any vector aaa can be represented as follows
a
a a = a
1
e
e e
1
2
e
e e
2
3
e
e e
3
≡ a
i
e
e e
i
The numbers a
i
are called the coordinates of the vector aaa for the basis eee
i
. In order to
compute the coordinates the dual (reciprocal) basis eee
k
is introduced in such a way
that
e
e e
k
· e
e e
i
= δ
k
i
1, k = i,
0, k 6 = i
A.1 Basic Operations of Tensor Algebra 173
Inner Dot Product. For any two second rank tensors AAA and BBB the inner dot prod-
uct is specified by AAA ··· BBB. The rule and the result of this operation can be explained
in the special case of two dyads, i.e. by setting AAA = aaa ⊗ bbb and BBB = ccc ⊗ ddd
AAA ··· BBB = aaa ⊗ bbb ··· ccc ⊗ ddd = (bbb ··· ccc)aaa ⊗ ddd = α aaa ⊗ ddd, α ≡ bbb ··· ccc
The result of this operation is a second rank tensor. Note that AAA · BBB 6 = BBB · AAA. This
can be again verified for two dyads. The operation can be generalized for two second
rank tensors as follows
AAA ··· BBB =
3
∑
i= 1
aaa
(i)
⊗ bbb
(i)
3
∑
k= 1
ccc
(k)
⊗ ddd
(k)
3
∑
i= 1
3
∑
k= 1
(bbb
(i)
··· ccc
(k)
)aaa
(i)
⊗ ddd
(k)
Transpose of a Second Rank Tensor. The transpose of a second rank tensor
A
A
A is constructed by the following rule
AAA
T
3
∑
i= 1
aaa
(i)
⊗ bbb
(i)
T
3
∑
i= 1
bbb
(i)
⊗ aaa
(i)
Double Inner Dot Product. For any two second rank tensors A
A
A and B
B
B the double
inner dot product is specified by A
A
A ··
·· B
B
B The result of this operation is a scalar. This
operation can be explained for two dyads as follows
AAA ······ BBB = aaa ⊗ bbb ······ ccc ⊗ ddd = (bbb ··· ccc)(aaa ··· ddd)
By analogy to the inner dot product one can generalize this operation for two second
rank tensors. It can be verified that AAA ······ BBB = BBB ······ AAA for second rank tensors AAA and
BBB. For a second rank tensor AAA and for a dyad aaa ⊗ bbb
A
A
A ··
·· a
a a ⊗ b
b b = b
b b ·
· A
A
A ·
· a
a a (A.1.6)
A scalar product of two second rank tensors AAA and BBB is defined by
α = A
A
A ··
·· B
B
B
T
One can verify that
AAA ······ BBB
T
= BBB
T
······ AAA = BBB ······ AAA
T
Dot Products of a Second Rank Tensor and a Vector. The right dot product
of a second rank tensor AAA and a vector ccc is defined by
AAA ··· ccc =
3
∑
i= 1
aaa
(i)
⊗ bbb
(i)
··· ccc =
3
∑
i= 1
(bbb
(i)
··· ccc)aaa
(i)
For a single dyad this operation is
aaa ⊗ bbb ··· ccc = aaa(bbb ··· ccc)
174 A Some Basic Rules of Tensor Calculus
The left dot product is defined by
c
c c ·
· A
A
A = c
c c ·
3
∑
i= 1
a
a a
(i)
⊗ b
b b
(i)
3
∑
i= 1
(c
c c ·
· a
a a
(i)
)b
b b
(i)
The results of these operations are vectors. One can verify that
AAA ··· ccc 6 = ccc ··· AAA, AAA ··· ccc = ccc ··· AAA
T
Cross Products of a Second Rank Tensor and a Vector. The right cross
product of a second rank tensor AAA and a vector ccc is defined by
AAA × ccc =
3
∑
i= 1
aaa
(i)
⊗ bbb
(i)
× ccc =
3
∑
i= 1
aaa
(i)
⊗ (bbb
(i)
× ccc)
The left cross product is defined by
c
c c × A
A
A = c
c c ×
3
∑
i= 1
a
a a
(i)
⊗ b
b b
(i)
3
∑
i= 1
(c
c c × a
a a
(i)
) ⊗ b
b b
(i)
The results of these operations are second rank tensors. It can be shown that
AAA × ccc = −[ccc × AAA
T
]
T
Trace. The trace of a second rank tensor is defined by
tr A
A
A = tr
3
∑
i= 1
a
a a
(i)
⊗ b
b b
(i)
3
∑
i= 1
a
a a
(i)
· b
b b
(i)
By taking the trace of a second rank tensor the dyadic product is replaced by the dot
product. It can be shown that
tr A
A
A = tr A
A
A
T
, tr (A
A
A ·
· B
B
B) = tr (B
B
B ·
· A
A
A) = tr (A
A
A
T
· B
B
B
T
) = A
A
A ··
·· B
B
B
Symmetric Tensors. A second rank tensor is said to be symmetric if it satisfies
the following equality
AAA = AAA
T
An alternative definition of the symmetric tensor can be given as follows. A second
rank tensor is said to be symmetric if for any vector ccc 6 = 000 the following equality is
valid
ccc ··· AAA = AAA · ccc
An important example of a symmetric tensor is the unit or identity tensor III, which
is defined by such a way that for any vector ccc
c
c c ·
· I
I
I = I
I
I ·
· c
c c = c
c c
176 A Some Basic Rules of Tensor Calculus
Linear Transformations of Vectors. A vector valued function of a vector ar-
gument fff (aaa) is called to be linear if fff ( α 1
aaa
1
2
aaa
2
) = α
1
fff (aaa
1
) + α
2
fff (aaa
2
) for any
two vectors aaa 1
and aaa
2
and any two scalars α
1
and α
2
. It can be shown that any linear
vector valued function can be represented by f
f f (a
a a) = A
A
A ·
· a
a a, where A
A
A is a second
rank tensor. In many textbooks, e.g. [32, 293] a second rank tensor A
A
A is defined to
be the linear transformation of the vector space into itself.
Determinant and Inverse of a Second Rank Tensor. Let aaa, bbb and ccc be ar-
bitrary linearly-independent vectors. The determinant of a second rank tensor AAA is
defined by
det A
A
A =
(AAA ··· aaa) ··· [(AAA ··· bbb) × (AAA ··· ccc)]
aaa ··· (bbb × ccc)
The following identities can be verified
det(AAA
T
) = det(AAA), det(AAA ··· BBB) = det(AAA) det(BBB)
The inverse of a second rank tensor AAA
− 1
is introduced as the solution of the follow-
ing equation
AAA
− 1
··· AAA = AAA ··· AAA
− 1
= III
AAA is invertible if and only if det AAA 6 = 0. A tensor AAA with det AAA = 0 is called
singular. Examples of singular tensors are projectors.
Cayley-Hamilton Theorem. Any second rank tensor satisfies the following
equation
AAA
3
− J
1
(AAA)AAA
2
+ J
2
(AAA)AAA − J
3
(AAA)III = 00 0, (A.1.7)
where A
A
A
2
= A
A
A ·
· A
A
A, A
A
A
3
= A
A
A ·
· A
A
A ·
· A
A
A and
J
1
(A
A
A) = tr A
A
A, J
2
(A
A
A) =
[(tr A
A
A)
2
− tr A
A
A
2
],
J
3
(AAA) = det AAA =
(tr AAA)
3
tr AAAtr AAA
2
tr AAA
3
(A.1.8)
The scalar-valued functions J i
(A
A
A) are called principal invariants of the tensor A
A
A.
Coordinates of Second Rank Tensors. Let eee i
be a basis and eee
k
the dual basis.
Any two vectors aaa and bbb can be represented as follows
aaa = a
i
eee
i
= a
j
eee
j
, bbb = b
l
eee
l
= b
m
eee
m
A dyad aaa ⊗ bbb has the following representations
a
a a ⊗ b
b b = a
i
b
j
e
e e
i
⊗ e
e e
j
= a
i
b
j
e
e e
i
⊗ e
e e
j
= a
i
b
j
e
e e
i
⊗ e
e e
j
= a
i
b
j
e
e e
i
⊗ e
e e
j
For the representation of a second rank tensor AAA one of the following four bases can
be used
eee
i
⊗ eee
j
, eee
i
⊗ eee
j
, eee
i
⊗ eee
j
, eee
i
⊗ eee
j
With these bases one can write
A.1 Basic Operations of Tensor Algebra 177
AAA = A
ij
eee
i
⊗ eee
j
= A
ij
eee
i
⊗ eee
j
= A
i∗
∗j
eee
i
⊗ eee
j
= A
∗j
i∗
eee
i
⊗ eee
j
For a selected basis the coordinates of a second rank tensor can be computed as
follows
A
ij
= eee
i
··· AAA · eee
j
, A
ij
= eee
i
··· AAA · eee
j
A
i∗
∗j
= e
e e
i
· A
A
A · e
e e
j
, A
∗j
i∗
= e
e e
i
· A
A
A · e
e e
j
Principal Values and Directions of Symmetric Second Rank Tensors.
Consider a dot product of a second rank tensor AAA and a unit vector nnn. The resulting
vector aaa = AAA ··· nnn differs in general from nnn both by the length and the direction.
However, one can find those unit vectors n
n n, for which A
A
A ·
· n
n n is collinear with n
n n, i.e.
only the length of n
n n is changed. Such vectors can be found from the equation
AAA ··· nnn = λ nnn or (AAA − λ III) ··· nnn = 000 (A.1.9)
The unit vector nnn is called the principal vector and the scalar λ the principal value
of the tensor AAA. Let AAA be a symmetric tensor. In this case the principal values are
real numbers and there exist at least three mutually orthogonal principal vectors.
The principal values can be found as roots of the characteristic polynomial
det(AAA − λ III) = − λ
3
+ J
1
(AAA) λ
2
− J
2
(AAA) λ + J
3
(AAA) = 0
The principal values are specified by λ I
, λ
I I
, λ
I I I
. For known principal values and
principal directions the second rank tensor can be represented as follows (spectral
representation)
AAA = λ
I
nnn
I
⊗ nnn
I
I I
nnn
I I
⊗ nnn
I I
I I I
nnn
I I I
⊗ nnn
I I I
Orthogonal Tensors. A second rank tensor QQQ is said to be orthogonal if it sat-
isfies the equation QQQ
T
··· QQQ = III. If QQQ operates on a vector its length remains un-
changed, i.e. let bbb = QQQ ··· aaa, then
|bbb|
2
= bbb ··· bbb = aaa ··· QQQ
T
··· QQQ ··· aaa = aaa ··· aaa = |aaa|
2
Furthermore, the orthogonal tensor does not change the scalar product of two arbi-
trary vectors. For two vectors aaa and bbb as well as aaa
′
= QQQ ··· aaa and bbb
′
= QQQ ··· bbb one can
calculate
aaa
′
··· bbb
′
= aaa ··· QQQ
T
··· QQQ ··· bbb = aaa ··· bbb
From the definition of the orthogonal tensor follows
Q
Q
Q
T
= Q
Q
Q
− 1
, Q
Q
Q
T
· Q
Q
Q = Q
Q
Q ·
· Q
Q
Q
T
= I
I
I,
det(Q
Q
Q ·
· Q
Q
Q
T
) = (det Q
Q
Q)
2
= det I
I
I = 1 ⇒ det Q
Q
Q = ± 1
Orthogonal tensors with det QQQ = 1 are called proper orthogonal or rotation tensors.
The rotation tensors are widely used in the rigid body dynamics, e.g. [333], and in
the theories of rods, plates and shells, e.g. [25, 32]. Any orthogonal tensor is either
A.2 Elements of Tensor Analysis 179
e
e e 1
eee
2
eee
3
x
2
x
3
x
1
rrr
q
1
q
2
q
3
rrr
1
r
r r 2
r
r r 3
P
q
3
const
Figure A.5 Cartesian and curvilinear coordinates
must be valid. With these assumptions the position vector can be considered as a
function of curvilinear coordinates q
i
, i.e. rrr = rrr(q
1
, q
2
, q
3
). Surfaces q
1
= const,
q
2
= const, and q
3
= const, Fig. A.5, are called coordinate surfaces. For given
fixed values q
2
= q
2
∗
and q
3
= q
3
∗
a curve can be obtained along which only q
1
varies. This curve is called the q
1
-coordinate line, Fig. A.5. Analogously, one can
obtain the q
2
3
-coordinate lines. The partial derivatives of the position vector
with respect the to selected coordinates
rrr
1
∂ rrr
∂ q
1
, rrr
2
∂ rrr
∂ q
2
, rrr
3
∂ rrr
∂ q
3
, rrr
1
··· (rrr
2
× rrr
3
define the tangential vectors to the coordinate lines in a point P, Fig. A.5. The vec-
tors r
r r
i
are used as the local basis in the point P. By use of (A.1.4) the dual basis
r
r r
k
can be introduced. The vector dr
r r connecting the point P with a point P
′
in the
differential neighborhood of P is defined by
dr
r r =
∂ rrr
∂ q
1
dq
1
∂ rrr
∂ q
2
dq
2
∂ rrr
∂ q
3
dq
3
= r
r r
k
dq
k
The square of the arc length of the line element in the differential neighborhood of
P is calculated by
ds
2
= dr
r r ·
· dr
r r = (r
r r
i
dq
i
· (r
r r
k
dq
k
) = g
ik
dq
i
dq
k
where g ik
≡ r
r r
i
· r
r r
k
are the so-called contravariant components of the metric tensor.
With g ik
one can represent the basis vectors rrr
i
by the dual basis vectors rrr
k
as follows
r
r r
i
= (r
r r
i
· r
r r
k
)r
r r
k
= g
ik
r
r r
k
180 A Some Basic Rules of Tensor Calculus
Similarly
rrr
i
= (rrr
i
··· rrr
k
)rrr
k
= g
ik
rrr
k
, g
ik
≡ rrr
i
··· rrr
k
where g
ik
are termed covariant components of the metric tensor. For the selected
bases rrr i
and rrr
k
the second rank unit tensor has the following representations
III = rrr
i
⊗ rrr
i
= rrr
i
⊗ g
ik
rrr
k
= g
ik
rrr
i
⊗ rrr
k
= g
ik
rrr
i
⊗ rrr
k
= rrr
i
⊗ rrr
i
A.2.2 The Hamilton (Nabla) Operator
A scalar field is a function which assigns a scalar to each spatial point P for the
domain of definition. Let us consider a scalar field ϕ (rrr) = ϕ (q
1
, q
2
, q
3
). The total
differential of ϕ by moving from a point P to a point P
′
in the differential neighbor-
hood is
d ϕ =
∂ϕ
∂ q
1
dq
1
∂ϕ
∂ q
2
dq
2
∂ϕ
∂ q
3
dq
3
∂ϕ
∂ q
k
dq
k
Taking into account that dq
k
= drrr ··· rrr
k
d ϕ = dr
r r ·
· r
r r
k
∂ϕ
∂ q
k
= dr
r r ·
∇ ϕ
The vector ∇
∇ ϕ is called the gradient of the scalar field ϕ and the invariant operator
∇ (the Hamilton or nabla operator) is defined by
∇∇∇ = rrr
k
∂ q
k
For a vector field aaa(rrr) one may write
daaa = (drrr ··· rrr
k
∂ a
a a
∂ q
k
= drrr ··· rrr
k
∂ a
a a
∂ q
k
= drrr ··· ∇∇∇ ⊗ aaa = (∇∇∇ ⊗ aaa)
T
··· dddrrr,
∇∇∇ ⊗ aaa = rrr
k
∂ aaa
∂ q
k
The gradient of a vector field is a second rank tensor. The operation ∇∇∇ can be applied
to tensors of any rank. For vectors and tensors the following additional operations
are defined
divaaa ≡ ∇∇∇ ··· aaa = rrr
k
∂ a
a a
∂ q
k
rotaaa ≡ ∇∇∇ × aaa = rrr
k
×
∂ aaa
∂ q
k
The following identities can be verified
∇∇∇ ⊗ rrr = rrr
k
∂ r
r r
∂ q
k
= rrr
k
⊗ rrr
k
= III, ∇∇∇ ··· rrr = 3
182 A Some Basic Rules of Tensor Calculus
- Curl Theorems
∫
V
∇∇∇ × aaa dV =
∫
A(V)
nnn × aaa dA,
∫
V
∇∇∇ × AAA dV =
∫
A(V)
nnn × AAA dA
A.2.4 Scalar-Valued Functions of Vectors and Second
Rank Tensors
Let ψ be a scalar valued function of a vector aaa and a second rank tensor AAA, i.e.
ψ = ψ (aaa, AAA). Introducing a basis eee
i
the function ψ can be represented as follows
ψ (a
a a, A
A
A) = ψ (a
i
e
e e
i
, A
ij
e
e e
i
⊗ e
e e
j
) = ψ (a
i
, A
ij
The partial derivatives of ψ with respect to a
a a and A
A
A are defined according to the
following rule
d ψ =
∂ψ
∂ a
i
da
i
∂ψ
∂ A
ij
dA
ij
= daaa ··· eee
i
∂ψ
∂ a
i
j
⊗ eee
i
∂ψ
∂ A
ij
dA
ij
(A.2.4)
In the coordinates-free form the above rule can be rewritten as follows
d ψ = daaa ···
∂ψ
∂ a
a a
∂ψ
∂ A
A
A
T
= daaa ··· ψ
,aaa
,A
A A
T
(A.2.5)
with
ψ ,aaa
∂ψ
∂ aaa
∂ψ
∂ a
i
e
e e
i
, ψ
,AAA
∂ψ
∂ AAA
∂ψ
∂ A
ij
e
e e
i
⊗ e
e e
j
One can verify that ψ ,aaa
and ψ
,A
A A
are independent from the choice of the basis. One
may prove the following formulae for the derivatives of principal invariants of a
second rank tensor AAA
J
1
(A
A
A)
,AAA
= I
I
I, J
1
(A
A
A
2
,AAA
= 2 A
A
A
T
, J
1
(A
A
A
3
,AAA
= 3 A
A
A
2
T
J
2
(AAA)
,AAA
= J
1
(AAA)III − AAA
T
, (A.2.6)
J
3
(AAA)
,AAA
= AAA
2
T
− J
1
(AAA)AAA
T
+ J
2
(AAA)III = J
3
(AAA)(AAA
T
− 1
A.3 Orthogonal Transformations and Orthogonal
Invariants
An application of the theory of tensor functions is to find a basic set of scalar invari-
ants for a given group of symmetry transformations, such that each invariant relative
A.3 Orthogonal Transformations and Orthogonal Invariants 183
to the same group is expressible as a single-valued function of the basic set. The ba-
sic set of invariants is called functional basis. To obtain a compact representation
for invariants, it is required that the functional basis is irreducible in the sense that
removing any one invariant from the basis will imply that a complete representation
for all the invariants is no longer possible.
Such a problem arises in the formulation of constitutive equations for a given
group of material symmetries. For example, the strain energy density of an elastic
non-polar material is a scalar valued function of the second rank symmetric strain
tensor. In the theory of the Cosserat continuum two strain measures are introduced,
where the first strain measure is the polar tensor while the second one is the axial
tensor, e.g. [108]. The strain energy density of a thin elastic shell is a function of
two second rank tensors and one vector, e.g. [25]. In all cases the problem is to find
a minimum set of functionally independent invariants for the considered tensorial
arguments.
For the theory of tensor functions we refer to [71]. Representations of tensor
functions are reviewed in [280, 330]. An orthogonal transformation of a scalar α , a
vector a
a a and a second rank tensor A
A
A is defined by [25, 332]
α
′
≡ (det Q
Q
Q)
ζ
α , a
a a
′
≡ (det Q
Q
Q)
ζ
Q
Q
Q ·
· a
a a, A
A
A
′
≡ (det Q
Q
Q)
ζ
Q
Q
Q ·
· A
A
A ·
· Q
Q
Q
T
, (A.3.1)
where Q
Q
Q is an orthogonal tensor, i.e. Q
Q
Q ·
· Q
Q
Q
T
= I
I
I, det Q
Q
Q = ± 1 , I
I
I is the second
rank unit tensor, ζ = 0 for absolute (polar) scalars, vectors and tensors and ζ = 1
for axial ones. An example of the axial scalar is the mixed product of three polar
vectors, i.e. α = aaa ··· (bbb × ccc). A typical example of the axial vector is the cross product
of two polar vectors, i.e. ccc = aaa × bbb. An example of the second rank axial tensor
is the skew-symmetric tensor WWW = aaa × III, where aaa is a polar vector. Consider a
group of orthogonal transformations S (e.g., the material symmetry transformations)
characterized by a set of orthogonal tensors QQQ. A scalar-valued function of a second
rank tensor f = f (AAA) is called to be an orthogonal invariant under the group S if
∀QQQ ∈ S : f (AAA
′
) = (det QQQ)
η
f (AAA), (A.3.2)
where η = 0 if values of f are absolute scalars and η = 1 if values of f are axial
scalars.
Any second rank tensor B
B
B can be decomposed into the symmetric and the skew-
symmetric part, i.e. B
B
B = A
A
A + a
a a × I
I
I, where A
A
A is the symmetric tensor and a
a a is the
associated vector. Therefore f (B
B
B) = f (A
A
A, a
a a). If B
B
B is a polar (axial) tensor, then a
a a is
an axial (polar) vector. For the set of second rank tensors and vectors the definition
of an orthogonal invariant (A.3.2) can be generalized as follows
∀QQQ ∈ S : f (AAA
′
1
, AAA
′
2
,... , AAA
′
n
, aaa
′
1
, aaa
′
2
,... , aaa
′
k
= (det Q
Q
Q)
η
f (A
A
A
1
, A
A
A
2
,... A
A
A
n
, a
a a
1
, a
a a 2
,... , a
a a
k
), A
A
A
i
= A
A
A
T
i
(A.3.3)
A.3 Orthogonal Transformations and Orthogonal Invariants 185
Invariants for a Single Second Rank Symmetric Tensor. Consider the
proper orthogonal tensor which represents a rotation about a fixed axis, i.e.
QQQ( ϕ mmm) = mmm ⊗ mmm + cos ϕ (III − mmm ⊗ mmm) + sin ϕ mmm × III, det QQQ( ϕ mmm) = 1,
(A.3.6)
where m
m m is assumed to be a constant unit vector (axis of rotation) and ϕ denotes
the angle of rotation about m
m m. The symmetry transformation defined by this tensor
corresponds to the transverse isotropy, whereby five different cases are possible, e.g.
[299, 331]. Let us find scalar-valued functions of a second rank symmetric tensor A
A
A
satisfying the condition
f (AAA
′
( ϕ )) = f (QQQ( ϕ mmm) ··· AAA ··· QQQ
T
( ϕ mmm)) = f (AAA), AAA
′
( ϕ ) ≡ QQQ( ϕ mmm) ··· AAA ··· QQQ
T
( ϕ mmm)
(A.3.7)
Equation (A.3.7) must be valid for any angle of rotation ϕ. In (A.3.7) only the left-
hand side depends on ϕ. Therefore its derivative with respect to ϕ can be set to zero,
i.e.
d f
d ϕ
dAAA
′
d ϕ
∂ f
∂ AAA
′
T
= 0 (A.3.8)
The derivative of AAA
′
with respect to ϕ can be calculated by the following rules
dA
A
A
′
( ϕ ) = dQ
Q
Q( ϕ m
m m) ·
· A
A
A ·
· Q
Q
Q
T
( ϕ m
m m) + Q
Q
Q( ϕ m
m m) ·
· A
A
A ·
· dQ
Q
Q
T
( ϕ m
m m),
dQ
Q
Q( ϕ m
m m) = m
m m × Q
Q
Q( ϕ m
m m)d ϕ ⇒ dQ
Q
Q
T
( ϕ m
m m) = −Q
Q
Q
T
( ϕ m
m m) × m
m m d ϕ
(A.3.9)
By inserting the above equations into (A.3.8) we obtain
(mmm × AAA − AAA × mmm) ······
∂ f
∂ AAA
T
= 0 (A.3.10)
Equation (A.3.10) is classified in [92] to be the linear homogeneous first order par-
tial differential equation. The characteristic system of (A.3.10) is
dAAA
ds
= (m
m m × A
A
A − A
A
A × m
m m) (A.3.11)
Any system of n linear ordinary differential equations has not more then n − 1
functionally independent integrals [92]. By introducing a basis eee i
the tensor AAA can
be written down in the form AAA = A
ij
eee
i
⊗ eee
j
and (A.3.11) is a system of six ordi-
nary differential equations with respect to the coordinates A
ij
. The five integrals of
(A.3.11) may be written down as follows
g
i
(A
A
A) = c
i
, i = 1, 2,... , 5,
where c i
are integration constants. Any function of the five integrals g
i
is the so-
lution of the partial differential equation (A.3.10). Therefore the five integrals g i
represent the invariants of the symmetric tensor AAA with respect to the symmetry
transformation (A.3.6). The solutions of (A.3.11) are
186 A Some Basic Rules of Tensor Calculus
AAA
k
(s) = QQQ(smmm) ··· AAA
k
0
··· QQQ
T
(smmm), k = 1, 2, 3, (A.3.12)
where A
A
A
0
is the initial condition. In order to find the integrals, the variable s must
be eliminated from (A.3.12). Taking into account the following identities
tr (QQQ ··· AAA
k
··· QQQ
T
) = tr (QQQ
T
··· QQQ ··· AAA
k
) = tr AAA
k
, mmm ··· QQQ(smmm) = mmm,
(Q
Q
Q ·
· a
a a) × (Q
Q
Q ·
· b
b b) = (det Q
Q
Q)Q
Q
Q ·
· (a
a a × b
b b)
(A.3.13)
and using the notation Q
Q
Q
m
≡ Q
Q
Q(sm
m m) the integrals can be found as follows
tr (A
A
A
k
) = tr (A
A
A
k
0
), k = 1, 2, 3,
mmm ··· AAA
l
··· mmm = mmm ··· QQQ
m
··· AAA
l
0
··· QQQ
T
m
··· mmm
= mmm ··· AAA
l
0
··· mmm, l = 1, 2,
mmm ··· AAA
2
··· (mmm × AAA ········· mmm) = mmm ··· QQQ
T
m
··· AAA
2
0
··· QQQ
m
··· (mmm × QQQ
T
m
··· AAA
0
··· QQQ
m
··· mmm)
= m
m m ·
· A
A
A
2
0
· Q
Q
Q
m
[
(Q
Q
Q
T
m
· m
m m) × (Q
Q
Q
T
m
· A
A
A
0
· m
m m)
]
= m
m m ·
· A
A
A
2
0
· (m
m m × A
A
A
0
· m
m m)
(A.3.14)
As a result we can formulate the six invariants of the tensor AAA with respect to the
symmetry transformation (A.3.6) as follows
I
k
= tr (A
A
A
k
), k = 1, 2, 3, I
4
= m
m m ·
· A
A
A ·
· m
m m,
I
5
= m
m m ·
· A
A
A
2
· m
m m, I
6
= m
m m ·
· A
A
A
2
· (m
m m × A
A
A ·
· m
m m)
(A.3.15)
The invariants with respect to various symmetry transformations are discussed in
[79]. For the case of the transverse isotropy six invariants are derived in [79] by the
use of another approach. In this sense our result coincides with the result given
in [79]. However, from our derivations it follows that only five invariants listed
in (A.3.15) are functionally independent. Taking into account that I 6
is the mixed
product of vectors mmm, AAA ··· mmm and AAA
2
··· mmm the relation between the invariants can be
written down as follows
I
2
6
= det
1 I
4
I
5
I
4
I
5
mmm ··· AAA
3
··· mmm
I
5
mmm ··· AAA
3
··· mmm mmm ··· AAA
4
··· mmm
(A.3.16)
One can verify that m
m m ·
· A
A
A
3
· m
m m and m
m m ·
· A
A
A
4
· m
m m are transversely isotropic invari-
ants, too. However, applying the the Cayley-Hamilton theorem (A.1.7) they can be
uniquely expressed by I 1
, I
2
,... I
5
in the following way [54]
mmm ··· AAA
3
··· mmm = J
1
I
5
+ J
2
I
4
+ J
3
mmm ··· AAA
4
··· mmm = (J
2
1
+ J
2
)I
5
+ (J
1
J
2
+ J
3
)I
4
+ J
1
J
3
where J 1
, J
2
and J
3
are the principal invariants of A
A
A defined by (A.1.8). Let us
note that the invariant I 6
cannot be dropped. In order to verify this, it is enough to
consider two different tensors
