Tensors for Physics

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12.3 Averages Over Velocity Distributions 217


expression, used for the dynamics of colloidal Brownian particles, is



(

δF
δt

)

FP

=FMΩFP(Φ)=−

1

2

ν 0


∂Vκ

(

FM


∂Vκ

Φ

)

. (12.73)

Hereν 0 >0 is the translational relaxation frequency. For particles with a radiusR
larger than the molecules of the surrounding liquid, the hydrodynamic result for the
Stokes friction force (10.58) applies. Then one hasν 0 =m−^16 πηR. The quantity
ΩFP(Φ), defined by (12.73), is referred to as the Fokker-Planck relaxation operator.
The expansion tensors used here are eigenfunctions of this operator.
The implications of the flow term for the moment equations apply to both the
Boltzmann and the Fokker-Planck equations. After an insertion of the expansion
(12.66) into the flow term of (12.72), the multiplication of the expansion tensors of


rank,e.g.φ(s


′+ 1 )
ν 1 ν 2 ···ν′with the vectorVλyields irreducible tensors of ranks′−1 and
′+1, see Sect.11.5. Multiplication of (12.72) by the functionφ(μs 1 )μ 2 ···μ, subsequent
integration overd^3 Vand the use of the relevant orthogonality relations leads to



∂t

aμ(s) 1 μ 2 ···μ+


s′

c(s|− 1 s′)∇λa(s

′)
λμ 2 ···μ− 1

+


s′

c(s|+ 1 s′)∇λa(s

′)
μ 1 μ 2 ···μ− 1 +...=^0. (12.74)

The terms which stem from the collision term or from the damping term of the
Fokker-Planck equation are indicated by the dots. Due to


〈φ(μs 1 )μ 2 ···μVμφ(s

′)
ν 1 ν 2 ···ν− 1 〉^0 ∼Δ

()
μ 1 μ 2 ···μ,μν 1 ν 2 ···ν− 1 ,

and (11.7), the flow term coefficientsc(.. .)are determined by


c(s|− 1 s′)=


2 c 0 ( 2 + 1 )−^1 〈φμ(s 1 )μ 2 ···μVμφ(s

′)
μ 1 μ 2 ···μ− 1 〉^0. (12.75)

The coefficientc(s|+ 1 s′)can be inferred fromc(− 1 s′|s)=c(s|− 1 s′)and
subsequent replacement→+1 and the interchange ofsands′. The first few of
the coefficientsc(...)are


c( 01 | 1 s′)=c 0 δ 1 ,s′, c( 11 | 0 s′)=c 0 δ 1 ,s′+


2
3

c 0 δ 2 ,s′, c( 11 | 2 s′)=


2 c 0 δ 1 ,s′.

For the Fokker-Planck case, the first two of the moment equations are



∂t

a(^1 )+c 0 ∇μa(μ^1 )= 0 ,


∂t

a(μ^1 )+c 0 ∇μa(^1 )+


2

3

c 0 ∇μa(^2 )+


2 c 0 ∇νa(νμ^1 )+ν 1 a(μ^1 )= 0 ,(12.76)
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