Mathematical Methods for Physics and Engineering - Matematica.NET

26.9 ISOTROPIC TENSORSare independent of the transformation L ij . Specifically, δ 11 has the value 1 inall coordinate frames, whereas for a general second-order tensor T all we knowis that if T 11 = f 11 (x 1 ,x 2 ,x 3 )thenT 11 ′ = f 11(x ′ 1 ,x′ 2 ,x′ 3 ). Tensors with the formerproperty 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, magneticsusceptibility or tensile strength, of an isotropic medium (i.e. a medium havingthe same properties whichever way it is orientated) involves an isotropic tensor.In the previous section it was shown that δ ij and ɛ ijk are second- and third-orderisotropic tensors; we will now show that, to within a scalar multiple, they are theonly such isotropic tensors.Let us begin with isotropic second-order tensors. Suppose T ij is an isotropictensor; then, by definition, for any rotation of the axes we must have thatT ij = T ij ′ = L ik L jl T kl (26.37)for each of the nine components.First consider a rotation of the axes by 2π/3 about the (1, 1, 1) direction; thistakes Ox 1 , Ox 2 , Ox 3 into Ox ′ 2 , Ox′ 3 , Ox′ 1 respectively. For this rotation L 13 =1,L 21 =1,L 32 = 1 and all other L ij = 0. This requires that T 11 = T 11 ′ = T 33.Similarly T 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) by π/2about the Ox 3 -axis. In this case L 12 = −1, L 21 =1,L 33 = 1 and all other L ij =0.Amongst other relationships, we must have from (26.37) that:T 13 =(−1) × 1 × T 23 ;T 23 =1× 1 × T 13 .Hence T 13 = T 23 = 0 and therefore, by parts (b) and (c) above, each element T ij =0 except for T 11 , T 22 and T 33 , which are all the same. This shows that T ij = λδ ij .◮Show that λɛ ijk 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.T ijk = T ijk ′ = L il L jm L kn T lmn (in all, there are 27 elements).Rotate about the (1, 1, 1) direction: this is equivalent to making subscript permutations1 → 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).945