192 11 Isotropic Tensors
Δμν,λκ,σ τΔμ′ν′,λκ,σ′τ′=5
48
(Δμν,μ′ν′Δστ,σ′τ′+Δμν,σ′τ′Δστ,μ′ν′)−
1
24
Δμν,σ τΔμ′ν′,σ′τ′. (11.39)The double contraction{στ}={σ′τ′}, in the last equation, yields
Δμν,λκ,σ τΔμ′ν′,λκ,σ τ=7
12
Δμν,μ′ν′, (11.40)and consequently
Δμν,λκ,σ τΔμν,λκ,σ τ=35
12
,
which impliesΔμν,λκ,σ τΔμν,λκ,σ τ =Δμν,νκ,κμ, see the last equation of (11.38).
The following contraction of four isotropic tensors yields the same numerical value,
viz.
Δμν,βγ ,μ′ν′Δμ′ν′,αγ ,σ τσ τ,λ,μνεβλα=35
12
. (11.41)
Furthermore, by analogy to (11.22), the tensor defined by (11.36) can also be
expressed by
Δμν,λκ,αβ=5
2
Δ(μνσ,αβτ^3 ) Δλκ,σ τ. (11.42)11.4.2 Tensor Product of Second Rank Tensors
For irreducible second rank tensorsSandA, the expression (11.35) reduces to
(S·A)μν≡ SμλAλν =Δμν,λκ,μ′ν′SλκAμ′ν′. (11.43)Products of the coupling tensorΔμν,λκ,σ τwith symmetric traceless dyadic tensors
constructed from the components of two vectorsaandbare listed next:
aμaλ bλbν ≡Δμν,λκ,σ τ aλaκ bσbτ=a·baμbν−1
3
(
b^2 aμaν+a^2 bμbν)
, (11.44)
aμbλ aλbν ≡Δμν,λκ,σ τ aλbκ aσbτ=− 1
6
a·baμbν+1
4
(
b^2 aμaν+a^2 bμbν