8.3: Weakly Inhomogeneous Gas
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Consider a gas which is only weakly out of equilibrium. We follow the treatment in Lifshitz and Pitaevskii, §6. As the gas is only slightly out of equilibrium, we seek a solution to the Boltzmann equation of the form f=f0+δf, where f0 is describes a local equilibrium. Recall that such a distribution function is annihilated by the collision term in the Boltzmann equation but not by the streaming term, hence a correction δf must be added in order to obtain a solution.
The most general form of local equilibrium is described by the distribution f0(r,Γ)=Cexp(μ−ε(Γ)+V⋅pkBT) , where μ=μ(r,t), T=T(r,t), and V=V(r,t) vary in both space and time. Note that df0=(dμ+p⋅dV+(ε−μ−V⋅p)dTT−dε)(−∂f0∂ε)=(1ndp+p⋅dV+(ε−h)dTT−dε)(−∂f0∂ε) where we have assumed V=0 on average, and used dμ=(∂μ∂T)∗pdT+(∂μ∂p)∗Tdp=−sdT+1ndp , where s is the entropy per particle and n is the number density. We have further written h=μ+Ts, which is the enthalpy per particle. Here, c∗p is the heat capacity per particle at constant pressure5. Finally, note that when f0 is the Maxwell-Boltzmann distribution, we have −∂f0∂ε=f0k∗BT .
The Boltzmann equation is written (∂∂t+pm⋅∂∂r+F⋅∂∂p)(f0+δf)=(∂f∂t)†coll . The RHS of this equation must be of order δf because the local equilibrium distribution f0 is annihilated by the collision integral. We therefore wish to evaluate one of the contributions to the LHS of this equation, ∂f0∂t+pm⋅∂f0∂r+F⋅∂f0∂p=(−∂f0∂ε){1n∂p∂t+ε−hT∂T∂t+mv⋅[(v⋅∇)V]+v⋅(m∂V∂t+1n∇p)+ε−hTv⋅∇T−F⋅v} . To simplify this, first note that Newton’s laws applied to an ideal fluid give ρ˙V=−∇p, where ρ=mn is the mass density. Corrections to this result, e.g. viscosity and nonlinearity in V, are of higher order.
Next, continuity for particle number means ˙n+∇⋅(nV)=0. We assume V is zero on average and that all derivatives are small, hence ∇⋅(nV)=V⋅∇n+n∇⋅V≈n∇⋅V. Thus, ∂lnn∂t=∂lnp∂t−∂lnT∂t=−∇⋅V , where we have invoked the ideal gas law n=p/kBT above.
Next, we invoke conservation of entropy. If s is the entropy per particle, then ns is the entropy per unit volume, in which case we have the continuity equation ∂(ns)∂t+∇⋅(nsV)=n(∂s∂t+V⋅∇s)+s(∂n∂t+∇⋅(nV))=0 . The second bracketed term on the RHS vanishes because of particle continuity, leaving us with ˙s+V⋅∇s≈˙s=0 (since V=0 on average, and any gradient is first order in smallness). Now thermodynamics says ds=(∂s∂T)∗pdT+(∂s∂p)∗Tdp=c∗pTdT−kBpdp , since T(∂s∂T)†p=c∗p and (∂s∂p)†T=(∂v∂T)†p, where v=V/N. Thus, c∗pkB∂lnT∂t−∂lnp∂t=0 . We now have in eqns. [ptea] and [pteb] two equations in the two unknowns ∂lnT∂t and ∂lnp∂t, yielding ∂lnT∂t=−kBc∗V∇⋅V∂lnp∂t=−c∗pc∗V∇⋅V . Thus Equation [LHSA] becomes ∂f0∂t+pm⋅∂f0∂r+F⋅∂f0∂p=(−∂f0∂ε){ε(Γ)−hTv⋅∇T+mv∗αv∗βQ∗αβ+h−Tc∗p−ε(Γ)c∗V/k∗B∇⋅V−F⋅v} , where Q∗αβ=12(∂V∗α∂x∗β+∂V∗β∂x∗α) .
Therefore, the Boltzmann equation takes the form {ε(Γ)−hTv⋅∇T+mv∗αv∗βQ∗αβ−ε(Γ)−h+Tc∗pc∗V/k∗B∇⋅V−F⋅v}f0kBT+∂δf∂t=(∂f∂t)†coll . Notice we have dropped the terms v⋅∂δf∂r and F⋅∂δf∂p, since δf must already be first order in smallness, and both the ∂∂r operator as well as F add a second order of smallness, which is negligible. Typically ∂δf∂t is nonzero if the applied force F(t) is time-dependent. We use the convention of summing over repeated indices. Note that δ∗αβQ∗αβ=Q∗αα=∇⋅V. For ideal gases in which only translational and rotational degrees of freedom are excited, h=c∗pT.