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 $$ \frac{dT}{dt}=-\frac{T-T_0}\tau? $$


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.

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    $\begingroup$ Thank you! But what I don't understand: if this Ansatz of Fermi-Dirac distribution with exponentially decaying temperature conforms to experiments (Newton's law of cooling) then why it cannot be accepted as, at least, approximate solution of the kinetic equation? Perhaps it turns out to be quite accurate if the temperature difference $T(0)-T_0$ is not very high? $\endgroup$ – Alexey Sokolik Jul 26 '18 at 16:37
  • $\begingroup$ I would say yes: it all depends on f(k,0), which would have approximately a Fermi-Dirac shape, if the initial conditions are not too far from thermal equilibrium, which is true when T(0) - T_0 is not very high $\endgroup$ – Antaios Jul 28 '18 at 16:52

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