Transition from local equilibrium to global equilibrium in Boltzmann's equation

Question

My question is about the evolution of a system from local equilibrium to global equilibrium. The system is described by a Boltzmann transport equation $$ \dfrac{\partial f}{\partial t}+\mathbf{v}\cdot\dfrac{\partial f}{\partial \mathbf{r}}+\left ( \dfrac{\mathbf{F}}{\hbar} \right )\cdot\dfrac{\partial f}{\partial \mathbf{k}}=I(f) $$ My application is electrons in a solid, but it could be any system to which the Boltzmann equation applies. The global equilibrium distribution $$ f(\mathbf{k})=\dfrac{1}{\exp (\frac{\epsilon(\mathbf{k})-\mu}{k_BT})+1} $$ satisfies the equation if we recall that it cancels the collision integral (right-hand side of Boltzmann's equation) and that $\mathbf{v}=d\epsilon(\mathbf{k})/d\mathbf{k}$ (Again, I am using a Fermi-Dirac distribution because it is the appropriate one for my electrons, but for the sake of my question, this could also be a system that obeys Maxwell-Boltzmann's statistics).

Note that any distribution that has the functional form of a Fermi-Dirac function with respect to $\mathbf{k}$ will cancel the collision integral. One such function would be $$ f(\mathbf{r},\mathbf{k},t)=\dfrac{1}{\exp (\frac{\epsilon(\mathbf{k})-\mu(\mathbf{r},t)}{k_BT(\mathbf{r},t)})+1} $$ which corresponds to a local equilibrium at time $t$. This distribution is not a solution of the entire Boltzmann equation, however, because the right-hand side is zero but the left-hand side (streaming terms) is not.

The above means (I guess) that if I start the system with any local Fermi-Dirac distribution, the Boltzmann equation would drive it towards the global equilibrium distribution. How do I show this? For example, I tried to solve the simpler case in which the temperature is already uniform, so only the chemical potential is a function of position and time. In that case, if I assume that the system evolves by preserving the functional form of a local distribution, so that the collision integral is zero at all times, I end up with a equation $$ \frac{\partial \mu}{\partial t}=\mathbf{v}\cdot \frac{\partial \mu}{\partial \mathbf{r}} $$ but any function $\mu(\mathbf{r}+\mathbf{v}t)$ would satisfy it. This would displace the chemical potential gradient, not reduce it...

Answer 1

Perhaps you could try perturbation theory.

First off, I'll replace your collision integral $I(f)$, with a dimensionalized form ${1 \over \tau}I(f)$, where $\tau$ is the equilibration time, which is assumed to be small.

For further convenience, let me replace the operator on the left-hand-side of the Boltzmann equation with ${d \over dt}$, so the Boltzmann equation becomes $$ {d f\over dt} = {1 \over \tau} I(f). $$ Now, expand $f$ to first-order in $\tau$ (i.e., $f = f_0 + \tau f_1$). The Boltzmann equation is then $$ \tau {d f_0\over dt} = I(f_0)+\tau I'(f_0) f_1 + \mathcal{O}(\tau^2). $$ To zeroth-order, we find that the zeroth-order distribution function $f_0$ is a local solution to the collision operator $$ I(f_0)=0, $$ which in your case, the solution would be the Fermi-Dirac distribution you mentioned.

To first-order in $\tau$, we find $$ I'(f_0) f_1= {d f_0\over dt}, $$ which defines an operator equation for $f_1$. Usually, inversion of the perturbed collision operator is not a simple task, but, for the sake of simplicity, I'll denote the inverse of the operator on the l.h.s. as $ I'(f_0)^{-1}$. The solution for $f_1$ takes the form $$ f_1= I'(f_0)^{-1} {d f_0\over dt}. $$ The right-hand-side of this equation depends on the deviation of $f_0$ from global equilibrium (i.e., nonzero ${d f_0 \over dt}$) while the left-hand-side determines the thermodynamics fluxes.

The general interpretation is that deviations in the zeroth-order distribution from global equilibrium give rise to thermodynamic fluxes that seek to bring the system closer to global equilibrium.

One can go on to talk about Onsager reciprocal relations and the second law of thermodynamics, but I'll leave it here for now.