# Relaxation of the Boltzmann transport equation

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.

• Commented Jun 28, 2023 at 23:31
• "Relaxed $f$" just means $\partial_t f=0$. Depending on the system at hand, this time-independent $f$ can depend on $p$ and/or $q$. Also, the relaxed $f$ may not be unique (therefore, the exact relaxed $f$ may depend on the initial conditions). When $F=0$ and the system is spatially homogeneous, the relaxed state is $0=\partial_t f |_{coll}$, and this is realized by some $f(p)$ (Max.-Boltz. distribution for classical binary elastic collisions). In a collisionless gas, any $f(p)$ could be a "relaxed" distribution (the one that is realized depends on the initial preparation). Commented Jul 31, 2023 at 15:05

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 $$\mathbf{F}=-\nabla_\mathbf{q}U=0$$.
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.