406 Appendix: Exercises...
Δ()μ
1 μ 2 ···μ− 1 λ,μ′ 1 μ′ 2 ···μ′− 1 λ=
2 + 1
2 − 1
Δ(μ−^1 )
1 μ 2 ···μ− 1 ,μ′ 1 μ′ 2 ···μ′− 1.
For=2, it reduces toΔμλ,νλ=^53 δμν.
From the definition of theΔ-tensor follows
Δμλ,νλ=1
2
(δμνδλλ+δμλδνλ)−1
3
δμλδνλ=(
2 −
1
3
)
δμν=5
3
δμν,as expected.
11.2 DetermineΔ(μνλ,μ^3 ) ′ν′λ′(p.186)
Hint: computeΔ(μνλ,μ^3 ) ′ν′λ′,in terms of triple products ofδ-tensors, from(11.15)for
=3.
Use of (11.15)for=3 with (10.6) yields
Δ(μνλ,μ^3 ) ′ν′λ′=1
6
∇μ∇ν∇λrμ′rν′rλ′=
1
6
∇μ∇ν∇λ[
rμ′rν′rλ′−1
5
r^2 (rμ′δν′λ′+rν′δμ′λ′+rλ′δμ′ν′)]
.
Successive application of the differential operators∇μ∇ν∇λon the first term after
the second equality sign of the equation above yields
∇μ∇ν∇λ(rμ′rν′rλ′)=∇μ∇ν(rμ′rν′δλλ′+rμ′rλ′δλν′+rλ′rν′δλμ′)
=∇μ[
(rμ′δνν′+rν′δνμ′)δλλ′
+(rμ′δνλ′+rλ′δνμ′)δλν′+(rλ′δνν′+rν′δνλ′)δλμ′]
=(δμμ′δνν′+δμν′δνμ′)δλλ′
+(δμμ′δνλ′+δμλ′δνμ′)δλν′+(δμλ′δνν′+δμν′δνλ′)δλμ′.Likewise, the application of the same differential operators onr^2 rμ′occurring above
as factor ofδν′λ′yields
∇μ∇ν∇λ(r^2 rμ′)=∇μ∇ν( 2 rλrμ′+r^2 δλμ′)= 2 ∇μ(rμ′δνλ+rλδνμ′+rνδλμ′)
= 2 (δμμ′δνλ+δμλδνμ′+δμνδλμ′).The derivatives ofr^2 rν′andr^2 rλ′which are factors ofδμ′λ′andδμ′λ′are given by
analogous expressions where justμ′is replaced byν′andλ′. All terms forΔ(^3 )put
together, with the numerical factors, leads to