Mathematical Methods for Physics and Engineering : A Comprehensive Guide

26.9 ISOTROPIC TENSORS

are independent of the transformationLij. Specifically,δ 11 has the value 1 in

all coordinate frames, whereas for a general second-order tensorTall we know

is that ifT 11 =f 11 (x 1 ,x 2 ,x 3 )thenT 11 ′ =f 11 (x′ 1 ,x′ 2 ,x′ 3 ). Tensors with the former

property are calledisotropic(orinvariant)tensors.

It is important to know the most general form that an isotropic tensor can take,

since the description of the physical properties, e.g. the conductivity, magnetic

susceptibility or tensile strength, of an isotropic medium (i.e. a medium having

the same properties whichever way it is orientated) involves an isotropic tensor.

In the previous section it was shown thatδijandijkare second- and third-order

isotropic tensors; we will now show that, to within a scalar multiple, they are the

only such isotropic tensors.

Let us begin with isotropic second-order tensors. SupposeTijis an isotropic

tensor; then, by definition, foranyrotation of the axes we must have that

Tij=Tij′=LikLjlTkl (26.37)

for each of the nine components.

First consider a rotation of the axes by 2π/3 about the (1, 1 ,1) direction; this

takesOx 1 ,Ox 2 ,Ox 3 intoOx′ 2 ,Ox′ 3 ,Ox′ 1 respectively. For this rotationL 13 =1,

L 21 =1,L 32 = 1 and all otherLij= 0. This requires thatT 11 =T 11 ′ =T 33.

SimilarlyT 12 =T 12 ′ =T 31. Continuing in this way, we find:

(a)T 11 =T 22 =T 33 ;

(b)T 12 =T 23 =T 31 ;

(c)T 21 =T 32 =T 13.

Next, consider a rotation of the axes (from their original position) byπ/ 2

about theOx 3 -axis. In this caseL 12 =−1,L 21 =1,L 33 = 1 and all otherLij=0.

Amongst other relationships, we must have from (26.37) that:

T 13 =(−1)× 1 ×T 23 ;

T 23 =1× 1 ×T 13.

HenceT 13 =T 23 = 0 and therefore, by parts (b) and (c) above, each elementTij=

0 except forT 11 ,T 22 andT 33 , which are all the same. This shows thatTij=λδij.

Show thatλijkis the only isotropic third-order Cartesian tensor.

The general line of attack is as above and so only a minimum of explanation will be given.

Tijk=Tijk′ =LilLjmLknTlmn (in all, there are 27 elements).
Rotate about the (1, 1 ,1) direction: this is equivalent to making subscript permutations
1 → 2 → 3 →1. We find

(a)T 111 =T 222 =T 333 ,

(b)T 112 =T 223 =T 331 (and two similar sets),

(c)T 123 =T 231 =T 312 (and a set involving odd permutations of 1, 2 ,3).