Liouville's theorem when density depends only on constants of the motion

Question

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Reif's Fundamentals of Statistical and Thermal Physics, pages 627-628, presents Liouville's theorem. I did not understand the punchline. Starting with Hamilton's equations, they derive

$$\frac{\partial\rho}{\partial t}=-\sum_{i=1}^f\left(\frac{\partial\rho}{\partial q_i}\dot{q_i}+\frac{\partial\rho}{\partial p_i}\dot{p_i}\right)$$

where $\rho$ is the density of systems in phase space, f is the degrees of freedom, and q and p are the canonical coordinates.

and from there

$\frac{\partial\rho}{\partial t}=-\sum\left(\frac{\partial\rho}{\partial q_i}\frac{\partial H}{\partial p_i}-\frac{\partial\rho}{\partial p_i}\frac{\partial H}{\partial q_i}\right).$

I was fine with that. Then they go on to say

"Suppose that at any given time... the systems are uniformly distributed over all of phase space. Or, more generally, suppose that $\rho$ is at time $t$ only a function of the energy $E$ of the system, this energy being a constant of the motion.

Then

$\frac{\partial \rho}{\partial q_i}=\frac{\partial\rho}{\partial E}\frac{\partial E}{\partial q_i}=0$ and $\frac{\partial \rho}{\partial p_i}=\frac{\partial \rho}{\partial E}\frac{\partial E}{\partial p_i}=0$

This question was answered, and it was determined that the $=0$ parts were in error. However, Reif goes on to state, without motivation, that $\frac{\partial \rho}{\partial t}=0$ whenever $\rho$ at any given time t depends only on constants of the motion. I am unable to figure out why this is true.

Answer 1

Intuitively, if $\rho$ is a function of the constants of motion, its level sets are also level sets of the constants of motion. They are therefore invariant by the phase flow, and since due to Liouville's equation $\rho$ is transported by the phase flow, it is time independent.

Actually, you don't need $\rho$ to depend on constants of motion at all time $t$, you only need the condition at one instant. Liouville's equation will guarantee that it will be at all time (past and future).

More formally, let $A_1,...,A_n$ be functions of phase space, constants of motion, and $\rho = f(A_1,...,A_n)$ at a certain instant, which I can take to be the origin of time, with $f$ a certain function. Writing $\phi_t$ the flow from $0$ to $t$, you have from Liouville's equation: $\rho_t = \rho(\phi_t^{-1})$. Since by assumption $A_k(\phi_t) = A_k$, you have $\rho_t = \rho$.

In particular, if $H$ is time-independent, then taking $\rho$ a function of $H$ gives a stationary ensemble since $H$ is a constant of motion. This is the case of the usual ensembles of SM, including the microcanonical ensemble: $\rho = \delta(H-E)/W$ and the canonical ensemble: $\rho=e^{-\beta H}/Z$.

Hope this helps.

Answer 2

Explanation without mentioning explicitly Hamiltonian flows.

If $\rho$ is a function of the canonical coordinates through $m$ constants of motion $\alpha_k(\{q_i,p_i\})$, $ ( k = 1,\dots,m)$, $$ \frac{\partial\rho}{\partial t}=-\sum_i\left(\frac{\partial\rho}{\partial q_i}\frac{\partial H}{\partial p_i}-\frac{\partial\rho}{\partial p_i}\frac{\partial H}{\partial q_i}\right)=-\sum_i \sum_k\left(\frac{\partial\rho}{\partial \alpha_k}\frac{\partial\alpha_k}{\partial q_i}\frac{\partial H}{\partial p_i}-\frac{\partial\rho}{\partial \alpha_k}\frac{\partial\alpha_k}{\partial p_i}\frac{\partial H}{\partial q_i}\right)\\= -\sum_k \frac{\partial\rho}{\partial \alpha_k}\sum_i\left(\frac{\partial\alpha_k}{\partial q_i}\frac{\partial H}{\partial p_i}-\frac{\partial\alpha_k}{\partial p_i}\frac{\partial H}{\partial q_i}\right) $$ but, for each $k$, the inner sum is the Poisson bracket $\{\alpha_k,H\}$ which vanishes for every constant of motion $\alpha_k$ (remember that for every function $f(\{q_i,p_i,t)$, $\frac{{\mathrm d}f}{{\mathrm d}t}=\{f,H\}+\frac{\partial{f}}{\partial{t}}$).