Reif's Fundamentals of Statistical and Thermal Physics, pages 627-628, presents Liouville's theorem. I do 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'm fine with that. Then they go on to say

"Suppose that at any give 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.


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

I am unable to understand why this follows from the italicized supposition. For example, couldn't we have $E=\frac{p^2}{2m}+\frac{1}{2}kq^2$ and $\frac{\partial E}{\partial q}\ne0$?

  • $\begingroup$ That explains what I guess is the main point, that $\frac{\partial\rho}{\partial t}=0$ but I'm still wondering if the statement $\frac{\partial \rho}{\partial q_i}=\frac{\partial\rho}{\partial E}\frac{\partial E}{\partial q_i}=0$ is in error or, if it is correct, why? $\endgroup$
    – Anna Naden
    Jul 12, 2022 at 15:26
  • 1
    $\begingroup$ No, it's not correct. Because, in general E can depend on q, as you noted. $\endgroup$
    – hft
    Jul 12, 2022 at 15:53

3 Answers 3


If $\rho(H)$ is a function of the energy only (the Hamiltonian), then $$ \frac{\partial \rho}{\partial q_i}=\frac{\partial \rho}{\partial H}\frac{\partial H}{\partial q_i}, \frac{\partial \rho}{\partial p_i}=\frac{\partial \rho}{\partial H}\frac{\partial H}{\partial p_i}. $$ Then $$ \frac{\partial \rho}{\partial t}=-\sum_i\left[\frac{\partial \rho}{\partial H}\frac{\partial H}{\partial q_i}\frac{\partial H}{\partial p_i}- \frac{\partial \rho}{\partial H}\frac{\partial H}{\partial p_i}\frac{\partial H}{\partial q_i}\right]=0, $$ because the two terms in the brackets are identical.


The above time derivative calculated for $\rho$ is actually a total derivative and not a partial derivative.


As I write and prove in [1] and rewrite in [2], the punchline has more to do with the total derivative than the partial derivative....

[Liouville's theorem] Consider a Hamiltonian system characterised by the Hamiltonian $ H\!\left(t,\boldsymbol {q},\boldsymbol {p} \right)$, where the state of the system, $\mathbf{r}$, is described by generalized coordinates $\boldsymbol {q}$ and $\boldsymbol {p}$, which correspond to $N$ generalized postions and $N$ generalized momenta, respectively. Let the trajectory, $\mathbf{r}(t)$, be the solution of the intial-value problem given by Hamilton's equations together with the initial condition. Consider the Hamiltonian system's joint distribution function $f\!\left(t,\mathbf{r}(t)\right)$. Allowing that (i) the flux of the number of particles per unit phase-space volume is a continuously differentiable vector field defined on each and every neighborhood of the phase-space volume, and allow that a velocity field describes the relevant flow, (ii) the quantity of particles per unit phase-space volume is a conserved quantity that cannot be created nor destroyed in any neighborhood of the phase space volume and (iii) the Hamiltonian has continuous second partial derivatives, then the total time derivative of the joint distribution function is identically zero. In other words $$ \frac{df\!\left(t,\mathbf{r}(t)\right)}{dt} =0. $$

[1] A proof of Liouville’s theorem

[2] What does Liouville's Theorem actually mean?


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