Isotropic tensors

TENSORS
A useful application of (26.30) is in obtaining alternative expressions for vector
quantities that arise from the vector product of a vector product.
Obtain an alternative expression for ∇ × (∇ × v).
As shown in the previous example, ∇ × (∇ × v) can be expressed in tensor form as
[∇ × (∇ × v)]i = ijk klm
∂2 vm
∂xj ∂xl
∂2 vm
= (δil δjm − δim δjl )
∂xj ∂xl
∂2 vi
∂
∂vj
−
=
∂xi ∂xj
∂xj ∂xj
= [∇(∇ · v)]i − ∇2 vi ,
where in the second line we have used the identity (26.30). This result has already
been mentioned in chapter 10 and the reader is referred there for a discussion of its
applicability. By examining the various possibilities, it
generally,
δip
ijk pqr = δjp
δ
kp
is straightforward to verify that, more
δiq
δjq
δkq
δir
δjr
δkr
(26.34)
and it is easily seen that (26.30) is a special case of this result. From (26.34) we
can derive alternative forms of (26.30), for example,
ijk ilm = δjl δkm − δjm δkl .
(26.35)
The pattern of subscripts in these identities is most easily remembered by noting
that the subscripts on the first δ on the RHS are those that immediately follow
(cyclically, if necessary) the common subscript, here i, in each -term on the LHS;
the remaining combinations of j, k, l, m as subscripts in the other δ-terms on the
RHS can then be filled in automatically.
Contracting (26.35) by setting j = l (say) we obtain, since δkk = 3 when using
the summation convention,
ijk ijm = 3δkm − δkm = 2δkm ,
and by contracting once more, setting k = m, we further find that
ijk ijk = 6.
(26.36)
26.9 Isotropic tensors
It will have been noticed that, unlike most of the tensors discussed (except for
scalars), δij and ijk have the property that all their components have values
that are the same whatever rotation of axes is made, i.e. the component values
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26.9 ISOTROPIC TENSORS
are independent of the transformation Lij . Specifically, δ11 has the value 1 in
all coordinate frames, whereas for a general second-order tensor T all we know
= f11 (x1 , x2 , x3 ). Tensors with the former
is that if T11 = f11 (x1 , x2 , x3 ) then T11
property are called isotropic (or invariant) 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 that δij and ijk are 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. Suppose Tij is an isotropic
tensor; then, by definition, for any rotation of the axes we must have that
Tij = Tij = Lik Ljl Tkl
(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
takes Ox1 , Ox2 , Ox3 into Ox2 , Ox3 , Ox1 respectively. For this rotation L13 = 1,
L21 = 1, L32 = 1 and all other Lij = 0. This requires that T11 = T11
= T33 .
= T31 . Continuing in this way, we find:
Similarly T12 = T12
(a) T11 = T22 = T33 ;
(b) T12 = T23 = T31 ;
(c) T21 = T32 = T13 .
Next, consider a rotation of the axes (from their original position) by π/2
about the Ox3 -axis. In this case L12 = −1, L21 = 1, L33 = 1 and all other Lij = 0.
Amongst other relationships, we must have from (26.37) that:
T13 = (−1) × 1 × T23 ;
T23 = 1 × 1 × T13 .
Hence T13 = T23 = 0 and therefore, by parts (b) and (c) above, each element Tij =
0 except for T11 , T22 and T33 , which are all the same. This shows that Tij = λδij .
Show that λijk is 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.
= Lil Ljm Lkn Tlmn
Tijk = Tijk
(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) T111 = T222 = T333 ,
(b) T112 = T223 = T331
(and two similar sets),
(c) T123 = T231 = T312
(and a set involving odd permutations of 1, 2, 3).
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