Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2017 / S7 / 3

Hamilton's equations define phase-space velocity $({\dot{q}}_i,{\dot{p}}_i)=(H_{p_i},-H_{q_i})$. Its divergence isThus infinitesimal phase-space volumes are preserved, assuming sufficient differentiability. This is Liouville's theorem; spatial density alone need not be incompressible because the statement concerns the combined coordinate-momentum space.

To match the source's order define $[a,b]={\sum }_i(a_{p_i}{b}_{q_i}-a_{q_i}{b}_{p_i})$. The chain rule and Hamilton's equations then giveAntisymmetry implies $[H,H]=0$, hence $dH/dt=H_t$, which vanishes for a time-independent Hamiltonian. With the alternative common convention $\{a,b\}={\sum }_i(a_{q_i}{b}_{p_i}-a_{p_i}{b}_{q_i})$, the same law is $dg/dt=g_t+\{g,H\}$; mixing the two orders would reverse the physical time evolution.

Conservation of probability obeys ${\partial }_{t}f+{\nabla }_{q,p}\cdot (f\mathbf{v})=0$. Expand the divergence and use incompressibility to obtain ${\partial }_{t}f+\mathbf{v}\cdot {\nabla }_{q,p}f=0$, or $\boxed{df/dt=0}$. This is fine-grained transport for Hamiltonian trajectories without sources or collisions. Coarse graining can obscure this constancy without contradicting it.

For $f=F(H)$ with no explicit time dependence, ${\partial }_{t}f=0$ and $[H,F(H)]=F^{\prime }(H)[H,H]=0$. It therefore solves the stationary transport equation. Normalizability and the accessible energy region still have to be checked for a proposed probability distribution.

Axisymmetry supplies conserved $p_{\theta }$ in addition to conserved energy. Consequently $\boxed{f=F(H,p_{\theta })}$ is a more general stationary solution, since both arguments are constant on every trajectory. Additional independent integrals may permit still more general distributions. This is a sufficient construction, not a claim that every stationary distribution must have exactly these two arguments.