Temperature relaxation from kinetic equation

Question

Suppose we have the kinetic equation $$ \frac{\partial f}{\partial t}=-\frac{f-f_0}\tau $$ for the electron distribution function in momentum space $f(\mathbf{k},t)$. Here $\tau$ is the relaxation time, $$ f_0(\mathbf{k})=\frac1{e^{[\hbar^2k^2/2m-\mu_0]/T_0}+1} $$ is the equilibrium Fermi-Dirac distribution at room temperature $T_0$.

Is it meaningful to assume that $f(\mathbf{k},t)$ is always of the Fermi-Dirac form $$ f(\mathbf{k},t)=\frac1{e^{[\hbar^2k^2/2m-\mu(t)]/T(t)}+1} $$ with the time-varying temperature $T(t)$ and chemical potential $\mu(t)$ subject to some equations $$ \frac{dT}{dt}=A(T,\mu),\qquad\frac{d\mu}{dt}=B(T,\mu)? $$ Of course, $d\mu/dt$ can be found from the particle number conservation $\int d\mathbf{k}\:f(\mathbf{k},t)=\mathrm{const}$, but what about the temperature? Can we approximate its time evolution by something like $$ \frac{dT}{dt}=-\frac{T-T_0}\tau? $$

Answer 1

There is no reason to make an Ansatz, for this is a first order differential equation. It is inhomogeneous however, but that can be solved by simply taking the time derivative of the whole equation. You’ll get: $\frac{\partial^2 f}{\partial t^2} = -\frac{\partial f}{\partial t}/τ$. This you can solve for $\frac{\partial f}{\partial t}$: namely $\frac{\partial f}{\partial t} = α e^{-t/τ}$, with $α$ some constant that could still depend on $\textbf{k}$. Integrating this and knowing that $f(\textbf{k}, t)$ should become $f_0$ if t goes to infinity, we find: $f(\textbf{k}, t) = - α τ e^{-t/τ} + f_0$. So $-α τ + f_0$ should be equal to $f(\textbf{k}, 0)$. Substitute this in our solution for $f(\textbf{k}, t)$: $f(\textbf{k}, t) = f(\textbf{k}, 0) e^{-t/τ} + f_0(1-e^{-t/τ})$. That last term is definitely not of Fermi-Dirac form.

So short answer: no, you can’t assume $f(\textbf{k}, t)$ is always of Fermi-Dirac form.

But yes, you can assume the temperature goes like that approximately, since what you wrote down is exactly Newton’s law of cooling.