Tensors for Physics

408 Appendix: Exercises...

where the integrand is positive, furthermoreν 0 >0, and thus

d

dt

sa≥ 0

holds true.

12.3 Second Order Contributions of the Kinetic Heat Flux and Friction

Pressure Tensor to the Entropy(p.221)

The ‘non-equilibrium’ entropy, per particle, associated with the velocity distribution

function f=fM( 1 +Φ),is given by s=−kB〈ln(f/fM)〉=−kB〈( 1 +Φ)ln( 1 +

Φ)〉M,where fMis the local Maxwell distribution andΦis the deviation of f from

fM.By analogy with(12.39),the contribution up to second order in rhe deviation is

s=−kB 21 〈Φ^2 〉M.

Determine the second order contributions to the entropy associated with heat flux

and the symmetric traceless pressure tensor.

The quantityΦto be used here is, cf. (12.96),

Φ=〈φμ〉φμ+〈φμν〉φμν,

whereφμandφμνare the expansion tensors pertaining to the kinetic contributions

to the heat flux and the friction pressure tensor, see (12.93) and (12.94). Due to

the orthogonality of the expansion tensors, the expression for the entropy becomes

s=−kB 21 (〈φμ〉〈φμ〉+〈φμν〉〈φμν〉). The dimensionless moments are related to the

heat flux vectorqμkinand the pressure tensorpμνkinvia (12.93) and (12.94). In terms

of these quantities, the desired contribution to the entropy is

s=−kB

1

2

(nkBT)−^2

[

2

5

m

kBT

qμkinqμkin+

1

2

pkinμν pkinμν

]

.

Compared with thermal equilibrium, the non-equilibrium state has a smaller entropy

and thus a higher order.

12.4 Pair Correlation Distorted by a Plane Couette Flow(p.230)

Compute the functions g+,g−and g 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: Use y^2 =(x^2 +y^2 )/ 2 −(x^2 −y^2 )/ 2 and x^2 +y^2 =r^2 −z^2 = 2 r^2 / 3 −(z^2 −
r^2 / 3 ),furthermore decompose x^2 y^2 =exμexνeyλeyκrμrνrλrκinto its parts associated
with tensors of ranks= 0 , 2 , 4 with the help of(9.6).Compare g−with g+.
The first order result is

δg(^1 )=−γτxy r−^1 geq′,