Tensors for Physics

230 12 Integral Formulae and Distribution Functions

withx=exνrνandy=e

y

νrν, whereexandeyare unit vectors parallel to thex-axis

andy-axis, respectively. The unit vector parallel to thez-axis isez. The constant shear

rate is

γ=

∂vx

∂y

. (12.126)

The vorticity and the shear rate tensor, cf. are

ωλ=−

1

2

γezλ,γμν=γ eyνexμ =

1

2

γ(eyνexμ+eyμexν). (12.127)

Here, the first order expression (12.121) implies

δg(^1 )=−τγxyr−^1 g′eq,

and one has

gμν=−τγrg′eqeyνexμ,

in this approximation. The more general ansatz

gμν=g+eyνexμ+g−

1

2

(exνexμ−eyνeyμ)+g 0 ezνezμ (12.128)

contains all terms which obey the plane Couette symmetry, viz. which are invariant

when bothexandeyare replaced by−exand−ey. The remaining two terms of the 5

components of the irreducible second rank tensor which, however, do not have this

symmetry, are proportional toezνexμ andezνeyμ.

The quantitiesg+,g−andg 0 are functions ofr. In first order in the shear rateγ,

one hasg+=−τγr−^1 g′eqandg−=g 0 =0. In second order,g−=γτg+,g 0 = 0

is found. The calculation is deferred to the next exercise.

12.4 Exercise: Pair Correlation Distorted by a Couette Flow
Compute the functionsg+,g−andg 0 in first and second order in the shear rateγ,in
steady state, from the plane Couette version of the kinetic equation (12.120)

γy

∂

∂x

δg+τ−^1 δg=−γy

∂

∂x

geq.

Hint:usey^2 =(x^2 +y^2 )/ 2 −(x^2 −y^2 )/2 andx^2 +y^2 =r^2 −z^2 = 2 r^2 / 3 −(z^2 −r^2 / 3 ),
furthermore decomposex^2 y^2 =exμexνeyλeyκrμrνrλrκinto its parts associated with
tensors of ranks= 0 , 2 ,4 with the help of (9.6). Compareg−withg+.