Kroner

14 Major symmetries of tensor D
Write:
D ijkl = − 1
4πµ (M ijkl + 2κ 1 N ijkl ) (240)
M ijkl =
N ijkl =
∫ π ∫ 2π
0 0
∫ π ∫ 2π
0
0
(δ ij z k z l sinΦ) dΘdΦ (241)
(z i z j z k z l sinΦ) dΘdΦ (242)
It is clear that N satisfies both major and minor symmetries. It remains to
show that M ijkl satisfies major symmetry. Clearly, if i ≠ j, then M ijkl = 0. It
can also be shown that if k ≠ l, then M ijkl = 0. For this purpose, since M ijkl
satisfies minor symmetries, it suffices to show that:
M ij12 = M ij13 = M ij23 = 0 (243)
Putting Eqs. 79,80,81 into Eq. 241 as needed to form M ij12 , M ij13 , M ij23 , we
find that in each case the Θ-integral vanishes.
For M ij12 , the Θ-integral involves:
∫ 2π
0
(cosΘsinΘ)dΘ = 1 4
∫ 2π
For M ij13 , the Θ-integral involves:
∫ 2π
For M ij23 , the Θ-integral involves:
0
∫ 2π
0
0
(sin2Θ)d(2Θ) = (− 1 4 )[cos2Θ]2π 0 = 0 (244)
cosΘdΘ = [sinΘ] 2π
0 = 0 (245)
sinΘdΘ = −[cosΘ] 2π
0 = 0 (246)
From the above, it is clear that the only non-zero values of M ijkl are obtained
only if i = j and k = l. With i = j, M ijkl reduces to the second order quantity:
M kl =
∫ π ∫ 2π
0
0
(z k z l sinΦ) dΘdΦ (247)
With k = l = 1, 2, 3 and putting Eqs. 79,80,81 into Eq. 247 as needed, we find
that:
M 11 =
∫ π ∫ 2π
0
0
sin 2 Φcos 2 ΘsinΦdΘdΦ =
[ Θ
=
2 + 1 ] 2π
4 sin2Θ
0
∫ 2π
0
cos 2 ΘdΘ
∫ π
[
− 1 3 cosΦ(sin2 Φ + 2)
0
sin 3 ΦdΦ (248)
] π
0
= 4 3 π (249)