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1.2 Algebra of Tensors

19

Every tensor A can be decomposed into its so-called spherical part and its deviatoric part, i.e.

A= al+ devA

or

 

 

 

1

1

1

 

rl• = -trA =-(I: A)= -4-·

u

3

3

34

u •

(1.131)

(l.132)

Any tensor of the fonn al, with a denoting a scalar, is known as a spherical tensor, while elevA is known as a deviator of A, or a deviatoric tensor. The deviatoric op-

erator dev( •) is denoted by the short-hand notation, i.e. dev( •) (1/3)tr( •)I, or dev( • )iJ = (• )ij - (1/3)( • )kk8iJ .

Computing the trace of ( 1.131) we deduce with ( 1.92), ( 1.87)5 , ( 1.22)1 and ( l. I32h

that Aii = 3(Aii/3) + dev Aii. This relation yields the important property

 

tr(devA) = 0 ,

(l.133)

which means that the trace of the deviator of A is always zero.

 

EXERCISES

1.Let u be any vector and e any unit vector. Use eqs. ( 1.125), ( 1.126) and the triple vector product ( 1.47) to show that u can be resolved into components parallel, i.e. 011'and perpendicular, i.e. Uj_, to unit vector e, according to

u = u11 + u.l = (u · e)e +ex (u x e) .

2.Prove eqs. ( 1.57) and ( 1.58) and show that

3.Let A be a tensor with matrix [A] and AT the transpose of A with matrix [A]T. Show that [AT) = [A]T.

4.Show by means of representation (1.59) and properties ( 1.98) and ( J.61) that the

double contraction of the two tensors A and B yields the component form Aii Bii.

5.Starting from eq. (l.100), verify that detA = EijkAi 1Ai2 Ak3 and show property

(1.101).

6. Consider eq. (1.93) 1 and relations (l.103), (l.88) in order to obtain the property

A -·r:A=3 .

(1.134)

20

1 Introduction to Vectors and Tensors

7.Given that S is a symmetric tensor and W is antisymmetric, prove property

(l.117).

8.Two tensors A and B are given in the form of their matrix representations

 

cos(}

sinO

0 l

[A) = -sinO

cosO

0

[

0

0

1

 

Verify that these matrices are orthogonal. Show that [A] describes a rotation and

[B] a reflection. Give a geometrical interpretation.

9. Find the axial vector of the skew tensor W = 1/2{u ® v - v 0 u).

10. Let A be a tensor whose matrix is

[A]= [ t t tl

Find the spherical and the deviatoric parts of A.

1.3 Higher-order Tensors

In the following we discuss higher-order tensors. Any tensor of order (or rank) n may be expressed in the form

Ai 1 i2 ••• in ei 1 © ei2 0 · · · ® ein .

(1.135)

 

(_

As (1.135) shows, a tensor of order n has 3n components Ai1 i 2 ••• i,.,

provided with n

indices i1 , i 2 , ••• , in. In particular, a tensor of order zero has 3° = 1 component with 0 (no) index, which simply is a scalar a (i.e. a single real number). A tensor of order one has 31 = 3 components characterizing any vector v, i.e. 'Vi as indicated in ( 1.23 ). A tensor of order two, for example, A, has 32 = 9 components, i.e. Aij as indicated in ( 1.60).

Tensor of order three. A tensor of order three we distinguish by the notation A,

B, C, ... According to (1.135), A may be expressed as

(l.136)

where Aiik characterizes the 33 = 27 components of A.

1.3 Higher-order Tensors

21

A particular example for a third-order tensor is the so-called triadic product of

three vectors u, v, w, denoted by u ® v ® w, with the properties

 

(u®v)®w=u®v®w,

 

(1.137)

(u ® v 0 w)x = (w · x)u 0 v

,

(1.138)

(u ® v ® w) : (x 0 y) = (v · x)(w · y)u

,

(l.139)

(u®v®w):l=(v·w)u.

 

(1.140)

Hence, by analogy with the procedure which led to eq. (1.62) we may determine the components of A, i.e.

( 1.141)

The double contraction of a third-order tensor A with a second-order tensor B produces a vector. It is denoted by A : B and may be found with representations (l.136), (1.59) and properties (1.139), (1.21) and (l.61) as

A: B = AiikB1m(ei 0 ei ® ek) : (e1 0 em)

=

=

AijkBtm(ej ·eL)(ek ·em)ei = AijkBtm8j18kmei

AijkBjkei .

(1.142)

EXAMPLE 1.4 Let E denote the third-order permutation tensor, which may be expressed as

(1.143)

Here, Cijk = (ei x ei) ·ek are the 33 components of E which we have introduced as the permutation symbol (recall eqs. (1.33) and (1.35)).

Show that the double contraction of E with the dyad of any vectors u and v, i.e. u ® v, gives the cross product of v and u.

Solution. With expressions (E)ijk = €ijk and (u 0 v)1m = 'U(Vm we may write

(1.144)

Using properties (1.139), (1.21) and (1.22)2 , we find that

(l.145)

which is a vector with components €ijk"UfVk. Finally, by means of the permutation

22

1

Introduction to Vectors and Tensors

 

property €ijk

= -€kji and expressions (1.43h, ( 1.26) we obtain

 

 

 

e : (u ® v) = v x u '

(1.146)

which is the desired result.

•

 

 

 

 

 

Tensor of order four.

Any tensor of order four, denoted by A, JR, C, ... , has

34 = 81 components, i.e.

Aiikl, Biikl, Ciikl, .. ., respectively. We may express any

tensor A in terms of the three Cartesian basis vectors, we write

(l.147)

in which Aijkl are the components of A.

One example for a fourth-order tensor is the product of the four vectors u, v, w, x,

denoted by u ® v 0 w ®

x. We have the useful properties:

 

 

(u 0 v) 0 (w 0 x) = u ® v ® w ® x

,

(I. 148)

(u 0 v ®

w ® x) : (y 0 z) = (w . y)(x . z)(u ®

v) .

(1.149)

With these properties we can show that the components Aiikl of A can be expressed as

(1.150)

Hence, the double contraction of A with a second-order tensor B is a second-order tensor, denoted by A : B. With representations (I. I47), (1.59), and properties (1.149), ( 1.21) and (1.61) we find that

A: B = AijklBmn(ei ® ei ®

ek ® e,) : (em® en)

 

= Aijk1Bmn8krn81nei ®

ei

(_

 

= Aijk1Bk1ei ® ei .

 

(1.151)

Following the definitions above and by analogy with (1.53), (1.56) and (1.57), note the important rules

{A® B) : C = A(B : C) =

(B : C)A

,

(1.152)

A : (B ® C) = {A : B)C = C(A : B)

,

(1.153)

(A® B) : (C ® D) = (B: C){A ®

D) = (A 0 D)(B: C) ,

(1.154)

and the property

 

 

 

A(B ® C)D = (AB) 0 (CD) .

 

(1.155)

1.3 Higher-order Tensors

23

EXAMPLE 1.5 Show eq. (l.152) which is the analogue of eq. (l.53).

Solution. Express tensor A, B, C in terms of their basis vectors according to ( 1.59).

With properties (1.148), (1.149), (1.21) and (1.61) we find that

(A® B): C = AijBktCmn(ei ® ej) ® (ek ® e,): (em® en)

=AijBklCmn(ek · em)(e, · en)(ei ® ej)

=AiiBktCmn8km81nei ® ei

= AiiBk1Ck,ei ® ei .

( 1.156)

By using representation ( 1.59)

and definition (1.93) we obtain AiiBk,Cklei ® ei

A{B : C) = (B : C)A, which is

the desired result.

•

 

 

 

A further example for a fourth-order tensor is the tensor product D = A ® B, while

A and Bare second-order tensors with components Aii and Bii, respectively. In index notation we may write Dijkl = AiiBkt·

The unique transpose of a fourth-order tensor A denoted by AT is governed, by

analogy with (I .81 ), by the identity

 

B : AT : C = C : A : B = {A : B) : C

(1.157)

for all second-order tensors B and C. The properties of a fourth-order tensor

 

(A?)T =A ,

( 1.158)

{A®B)T = B®A

( 1.159)

hold. From the identity ( 1.157) we deduce the index relation, namely (AT) iJkl = AkliJ.

We now define fourth-order unit tensors Il and Il so that

A=Il:A, AT=Il:A' ( 1.160)

for any second-order tensor A. These transformations assign to any A the tensor A itself, and its transpose, respectively. The fourth-order unit tensors may be represented by

n= 8ik8jlei

®

ej 0

ek ®

e,

= ei ®

ej ®

ei ®

ej

'

(1.161)

Il = c5ilc5jkei

0

ej 0

ek ®

el

= ei 0

ej 0

Cj 0

ei

,

(l.162)

where (Il)iikl = 8ik8i1 and (Il)ijkt

= 8itc5jk

define the Cartesian components of Il and Il,

respectively. Note that ni= nT.

 

 

 

 

 

 

 

 

 

= oik8il - (1/3)8ii8kl·

24

1 Introduction to Vectors and Tensors

The deviatoric part of a second-order tensor A may be described by means of a fourth-order tensor. From relations ( 1.13 I), ( 1.132h and transformations ( 1. I 60) 1,

(1.152) we conclude that

1

1

~I} : A ,

 

devA =A - 3(1: A)I = {Il -

31

(1.163)

 

 

 

or with the definition of a fourth-order projection tensor P

devA = P: A ,

(1.164)

The components of devA and A are related through the expression elev Aii = Pijkl Akt, with Pijkl

EXERCISE

 

I. Two fourth-order tensors §, W are given in the form

 

1

-

w = -(Il -

JI)

2

 

(observe the similar structure to eq. ( 1.1I2)). Show that

1 {A +A·r ) =§:A ,

~(A - AT) = w : A ,

2

-

which assign to any second-order tensor A its symmetric and skew parts, respectively.

1.4 Eigenvalues, Eigenvectors of Tensors

The scalars Ai characterize eigenvalues (or principal values) of a tensor A if there exist corresponding nonzero normalized eigenvectors Di (or principal directions or principal axes) of A, so that

(i = 1, 2, 3; no summation) .

(1.165)

To identify the eigenvectors of a tensor, we use subsequently a hat on the vector quantity concerned, for example, ft.

Thus, a set of homogeneous algebraic equations for the unknown eigenvalues ,.\,

i = 1, 2, 3, and the unknown eigenvectors ni, 'i = 1, 2, 3, is

 

(A

-

,.\·l)n· -

o

'

(i = 1, 2, 3; no summation) .

(1.166)

 

1

l -

 

Cayley-Hamilton equation

1.4 Eigenvalues, Eigenvectors of Tensors

25

Eigenvalues characterize the physical nature of a tensor. They do not depend on coordinates. For a positive definite symmetric tensor A, all eigenvalues ,.\i are (real and) positive since, using (1.165), we have,.\ =Di·Ani > 0, i = 1, 2, 3. Moreover, the set of eigenvectors of a symmetric tensor A form a mutually orthogonal (orthonormal) basis {ni} (the proof of this statement is omitted).

Principal scalar invariants. For the system ( 1.166) to have solutions ni # o the determinant of the system must vanish. Thus,

(1.167)

where

( 1.168)

This requires that we solve a cubic equation in ,.\, usually written as

( 1.169)

called the characteristic polynomial (or equation) for A, the solutions of which are the eigenvalues Ai, i = 1, 2, 3.

Here, Ii(A), i = 1, 2, 3, are the so-called principal scalar invariants of A. In

terms of A and its principal values,.\, i

= 1, 2, 3, these are given by

 

 

 

11 (A) = Aii = trA = .A1 + ,.\2

+ ,.\3

,

 

 

 

 

 

 

 

(I. I70)

! 2 (A)

1

1 [

(trA)

2

-

tr(A

2 ]

= trA

-1

detA

= -(AiiAji - AiiAii) = -

 

)

 

 

2

2

 

 

 

 

 

 

 

 

 

 

= ,.\ 1.A2 + .A1.A3 + A2A3

,

 

 

 

 

 

 

 

 

(1.171)

/3(A)

= EijkA1iA2jAak = detA = A1A2A3 .

 

 

 

 

 

(l.172)

Sometimes we shall use the alternative notation

 

 

 

 

 

 

 

 

 

 

 

 

 

 

l3(A) =

IIIA

 

 

(1.173)

for the three principal invariants interchangeably.

 

 

 

 

 

 

 

 

A repeated application of tensor A to eq. ( 1.165) yields A0 ni =

.Arni, i = 1, 2, 3,

for any positive integer a. Using this relation and ( 1.169) multiplied by Di, we obtain the well-known

(1.174)

It states that every (second-order) tensor A satisfies its own characteristic equation.

Spectral decomposition of a tensor. Any symmetric tensor A may be represented by its eigenvalues ,.\i, i = 1, 2, 3, and the corresponding eigenvectors of A forming an

26 1 Introduction to Vectors and Tensors

orthonormal basis {Di}· Using the unittensor, by analogy with ( l.65h, i.e. I= Di 0Di, and relations (1.58), (1.165) and (l.55) we obtain an expression which is known as the spectral decomposition (or spectral representation) of A, i.e.

3

 

A =AI= (Aili) 0 Di = L ,.\iDi ®Di .

(1.175)

i=l

The components Aii of tensor A relative to a basis of principal directions follow with ( 1.59) by replacing ei with the three orthonormal basis vectors Di. With

eqs. (l.165) and (1.21) we obtain

 

(j = 1, 2, 3; no summation) ,

( 1.176)

which produces a diagonal matrix [A] in the form

 

 

)q

 

[A)=

~

(l.177)

 

[

 

where the diagonal elements are the eigenvalues of A. This result may be obtained directly from the spectral decomposition (1.175) of A.

For a pair of equal roots, i.e. ,.\ 1 = ,.\2 # ,.\3 ( ,.\ 1 = ,.\2 == ,.\), with one unique eigenvector D3 associated with ,.\3 , we deduce from (1.175) that

(l.178)

where P~3, Pt3 denote projection tensors introduced in (1.125) and (1.126). In (1.178)

the eigenvector D

is perpendicular to a plane spanned by the mutually orthogonal

.

~

3d~

~

eigenvectors n1 an

n2.

 

 

The third case to distinguish, namely for equal roots (,.\ 1 =

,.\2 = ;\3 = ,.\), is

 

 

 

 

(l.179)

where every direction is a principal direction and every set of mutually orthogonal basis denotes principal axes.

EXERCISES

1.Show that the quantities in eqs. ( 1.170)-( 1.172), i.e. / 1, / 2 , / 3 , are indeed principal invariants of a tensor A under an orthogonal transformation of coordinates.

1.4 Eigenvalues, Eigenvectors of Tensors

 

 

27

2. Let a tensor A be given by

 

 

 

A=cr(I-e1®e1)+f3(e1®e2+e2®e1)

,

 

 

where er, {j are scalars and e 17 e2 are orthogonal unit vectors.

 

 

(a) Show that the eigenvalues Ai, i = 1, 2, 3, of A are

 

 

 

,,\3 =

a

(cr2

2) 1/2

2 -

4 + /3

.

(b)Derive the associated normalized eigenvectors Di, i = 1, 2, 3, which in matrix representation read

3.Let a non-symmetric tensor A be given by the tensor product e1 ® e:h with the property e 1 • e2 = 0. Determine the eigenvalues and eigenvectors of tensor A.

4.Let A be a given second-order tensor. Show that the invariants of its deviatoric part, according to ( 1.163), are

fdevA = 0 ,

1

2

,

(l.180)

/IdcvA=-?tr(devA)

 

 

~

 

 

 

IIIdcvA = det.(devA) =

1

 

)3

 

3tr(devA · .

 

5. Use the relations (I. I 70)-( I. I 72), ( 1. I 80)

and

(I. I 63) 1 to show

that I Ic1evA'

I I IdevA may be expressed in terms of IA, I IA and I I IA as

 

JldevA = JJA -

1

2

'

 

3/A

 

1

 

 

2 :J

(1.181)

l/lc1cvA =///A - - /A//A +~IA ·

3

 

 

21

 

Hint: To derive eq. ( 1.181) apply the Cayley-Hamilto11 equation in the form of

(devA) 3 - lc1evA (devA) 2 + IldevA devA - Illc1evA I= 0.

28

1 Introduction to Vectors and Tensors

1.5Transformation Laws for Basis Vectors and Components

Consider two sets of mutually orthogonal basis vectors which share a common origin. They correspond to a 'new' and an 'old' (original) Cartesian coordinate system which we assume to be right-handed characterized by the two sets {ei} and {ei} of basis vectors, respectively. Hence, the new coordinate system may be obtained from the original one by a rotation of the basis vectors ei about their origin (see Figure 1.7). We find from ( 1.10) that the cosine of the angle between ei and ei is represented by Qii = cosO(ei, ei) = ei · ei. Note that the first index on Qii indicates the 'old' components whereas the second index holds for the 'new' components.

'

'

O(c2, c:!)'\

'~

'

'

'~ ' -

~C2

Figure 1.7 Rotation of mutually orthogonal basis vectors ei into ei.

(_

It is worthwhile to mention that vectors and tensors themselves remain invariant upon a change of basis - they are said to be independent of any coordinate system. However, their respective components do depend upon the coordinate system introduced, which is arbitrary. The components change their magnitudes by a rotation

(and/or reflection) of the basis vectors, but are independent of any translation (for the

characterization of a translation, see p. 62).

We now set up the transformation laws for various components of vectors and

tensors under a change of basis. Following the considerations above we have

 

T-

'-1'J3

 

and

ei= Q ei '

(1.182)

i. -

,-,

'

where Q denotes an orthogonal tensor (seep. 16), with components QiJ which are the same in either basis. The components describe the orientation of the two sets of basis

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