Appendix: Exercises... 407
Δ(μνλ,μ^3 ) ′ν′λ′=
1
6
[
(δμμ′δνν′+δμν′δνμ′)δλλ′+(δμμ′δνλ′+δμλ′δνμ′)δλν′
+(δμλ′δνν′+δμν′δνλ′)δλμ′
]
−
1
15
[
(δνλδμμ′+δμλδνμ′+δμνδλμ′)δν′λ′
+(δνλδμν′+δμλδνν′+δμνδλν′)δμ′λ′
+(δνλδμλ′+δμλδνλ′+δμνδλλ′)δμ′ν′
]
. (A.13)
Notice that allδ-tensors in the bracket[...]behind the factor^16 in (A.13) contain
one primed and one unprimed subscript. On the other hand, the triple products of
δ-tensors behind the factor 151 contain one unit tensor with two primed subscripts,
one with two unprimed ones, and one with mixed subscripts.
The contractionλ′=λyieldsΔμνλ,μ(^3 ) ′ν′λ=^75 Δ(μν,μ^2 ) ′ν′. This is in accord with (11.6).
Exercises Chapter 12
12.1 Verify the Numerical Factor in(12.7)for the Integral over a Triple Product
of Tensors(p.202)
Hint: Putν=λ, κ=σ, τ=μand use the relevant formulae given in Sect.11.4.
The recommended contraction, on the left hand side of (12.7)involves
̂rμ̂rν̂rν̂rκ̂rκ̂rμ =
1
3
̂rμ̂rκ̂rκ̂rμ =
2
9
,
and the subsequent integration still yields^29.
On the other hand, due toΔμν,νκ,κμ=^3512 ,cf.(11.38), the right hand side becomes
8
105
35
12 =
2
9 , just as expected.
12.2 Prove that the Fokker-Planck Equation Implies an Increase of the
Orientational Entropy with Increasing Time(p.212)
Hint: The time change of an orientational average isd〈ψ〉/dt=
∫
∂(ψf)/∂td^2 u.
The time change offln(f/f 0 )is ln(f/f 0 )∂f/∂t+ff−^1 ∂f/∂t. Thus the time change
of the orientational entropysa=−kB
∫
fln(f/f 0 )d^2 uis
d
dt
sa=−kB
∫
ln(f/f 0 )
∂
∂t
fd^2 u=−kBν 0
∫
ln(f/f 0 )LμLμfd^2 u,
where
∫ ∂
∂tfd
(^2) u=0 and (12.41) have been used. HereLμ=εμνλuν∂
∂uλis the
relevant differential operator. An integration by part leads to the expression
d
dt
sa=kBν 0
∫
f−^1 (Lμf)Lμfd^2 u,