Tensors for Physics

202 12 Integral Formulae and Distribution Functions

whereSis symmetric traceless second rank tensor, yields a product of the form

SμνSνλSλκSκμ, in the first case and 3(SμνSνμ)^2 , in the second case. Both terms are

proportional to each other, see (5.52). Thus also the two isotropic tensors of rank 8

are equivalent, apart from a numerical factor. More specifically, one has

Δμ 1 ν 1 ,λκΔμ 2 ν 2 ,κσΔμ 3 ν 3 ,σ τΔμ 4 ν 4 ,τ λ

=

1

6

(Δμ 1 ν 1 ,μ 2 ν 2 Δμ 3 ν 3 ,μ 4 ν 4 +Δμ 1 ν 1 ,μ 3 ν 3 Δμ 2 ν 2 ,μ 4 ν 4 +Δμ 1 ν 1 ,μ 4 ν 4 Δμ 2 ν 2 ,μ 3 ν 3 ).

(12.8)

Thus it suffices to consider one of these isotropic tensor, e.g. the second one. Then

the desired integral is

1

4 π

∫

̂rμ 1 ̂rν 1 ̂rμ 2 ̂rν 2 ̂rμ 3 ̂rν 3 ̂rμ 4 ̂rν 4 d^2 ̂r

=

4

105

1

3

(Δμ 1 ν 1 ,μ 2 ν 2 Δμ 3 ν 3 ,μ 4 ν 4 +Δμ 1 ν 1 ,μ 3 ν 3 Δμ 2 ν 2 ,μ 4 ν 4 +Δμ 1 ν 1 ,μ 4 ν 4 Δμ 2 ν 2 ,μ 3 ν 3 ).

(12.9)

Multiplication of this equation withSμ 1 ν 1 Sμ 2 ν 2 Sμ 3 ν 3 Sμ 4 ν 4 implies

1

4 π

∫ (

Sμν̂rμ̂rν

) 4

d^2 ̂r=

4

105

(SμνSμν)^2. (12.10)

12.1 Exercise: Verify the Numerical Factor in(12.7)for the Integral over
a Triple Product of Tensors
Hint: Putν=λ,κ=σ,τ=μand use the relevant formulae given in Sect.11.4.

12.2 Orientational Distribution Function

12.2.1 Orientational Averages

The orientation of a single molecule or particle with uniaxial symmetry can be

specified by a unit vectoru. When needed, the vectorucan be expressed in terms of

the spherical polar anglesθandφ, just as the unit vector̂r.

The orientation of many of those molecules in a molecular liquid or a nematic

liquid crystal or of rod-like particles in a colloidal solution, is characterized by an

orientational distribution functionf=f(u). With the normalization

∫

f(u)d^2 u= 1 , (12.11)