Relaxation of the Boltzmann transport equation

Question

My professor in kinetic gas theory said that when considering the Boltzmann Transport Equation (BCE) $$ \partial_tf + \frac{\vec{p}}{m}\cdot\nabla_{\vec{q}}f + \vec{F}\cdot\nabla_{\vec{p}}f = (\partial_tf)_{Coll} $$ Over long periods of time, the system tends to relax, which makes the distribution $f$ homogenous ($\nabla_{\vec{q}}f$ = 0) and time independent ($\partial_tf = 0$). This means that the system tends to return to equilibrium, which makes sense to me. However, my prof. said that the momentum term does not relax. I.e. $\nabla_{\vec{p}}f \neq 0$ even if the system is in equilibrium.

Why is that so? I would have thought that for a system in equilibrium the particles should have similar velocities and thus similar momentums to have a homogenous distribution of energy. Moreover, if we're only considering particle collisions as interaction term the momentum should remain constant.

Answer 1

If uniform distribution is promised as equilibrium, it means the zero external force $\mathbf{F}$, which means that the $\nabla_\mathbf{p}f$ can have any value. without violating the equation.

As an example, let us consider Maxwell-Boltzmann distribution: $$ f(\mathbf{p},\mathbf{q})=Z^{-1}e^{-\beta\left(\frac{\mathbf{p}^2}{2m}+U(\mathbf{q})\right)},\text{ where }\beta=\frac{1}{k_BT}. $$ This distribution turns to zero the collision integral (how exactly depends on the form of the integral used, see, e.g., these notes suggested by @Quillo), it is independent on time, i.e., $\partial_t f=0$, whereas the gradients are: $$ \nabla_\mathbf{q}f=-\beta f \nabla_\mathbf{q}U = \beta \mathbf{F} f,\\ \nabla_\mathbf{p}f=-\beta\frac{\mathbf{p}}{m}f, $$ that is the sum of the two gradient term is zero. The density distribution under this distribution is not uniform, unless the potential is a constant, in which the distribution becomes Maxwell distribution $$ f(\mathbf{p})=Z^{-1}e^{-\beta\frac{\mathbf{p}^2}{2m}}, $$ that gives a non-zero gradient $\nabla_\mathbf{p}f=-\beta\frac{\mathbf{p}}{m}f$, but still satisfies the Boltzmann equation, since now $\mathbf{F}=-\nabla_\mathbf{q}U=0$.

Answer 2

Having $f$ independent of $t$ means that the distribution is similar at different times; having $f$ independent on $\vec{q}$ means that particles at different positions have similar distribution.

Similarly, having $f$ independent of $\vec{p}$ means that particles with different momenta distribute similarly. But we know that this is not the case at equilibrium. Higher energy states are less probable as $p \propto \exp(-E/k_B T)$. Also the density of states is momentum dependent, meaning that different $\vec{p}$ have more or less states available to populate. Having $\nabla_\vec{p} f = 0$ means that all these features are irrelevant, and a particle has the same probability to be in each value of $\vec{p}$ just as likely. This is not the case in equilibrium.