In Kardar's book on statistical physics it is claimed that Liouville's theorem gives support for the common assumption that the points in phase space compatible with the hamiltonian are all equally probable. As far as I can gather the reasoning goes as follows: in equilibrium $$\partial \rho/\partial t = -\{\rho, H\}=0$$ and this is compatible with the assumption that $\rho$ is purely a function of $H$ since this yields $$\{\rho, H\}=\rho'(H)\{H, H\}=0.$$ My questions are:

  1. Are all functions $\rho$ that satisfy $\{\rho, H\}=0$ necessarily of the form $\rho(H)$?

  2. If not, the PDFs of the form $\rho = \rho(H)$ are only a subset of the (possibly much larger) set of possible PDFs that satisfy $\{\rho, H\}=0$ and we seem to be no better off than we were before invoking Liouville's theorem.

    Are there additional facts/assumptions that together with the requirement $\{\rho, H\}=0$ make $\rho = \rho(H)$ a well founded assumption?

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    $\begingroup$ Near duplicate. In a multidimensional phase space invariant hypersurfaces are different, in general. Behold a superintegrable system! $\endgroup$ Commented Dec 31, 2021 at 22:27

1 Answer 1


Before I answer your question I would like to make this comment about the Liouville equation: it is a necessary condition on $\rho$, but it is not sufficient. This means that the correct $\rho$ must satisfy the Liouville equation, but not every $\rho$ that satisfies the Liouville equation is the correct probability density of microstate.

Turning to your first question, the answer is, no, $\rho(H)$ is not the only form that satisfies the Liouville equation. But the fact that $\rho(H)$ is an acceptable solution is important because it ties $\rho$ to a mechanical observable, the energy of the system. The premise of thermodynamics is that the macroscopic state is defined by a small number of variables, $(E,V,N)$, for example. A theory that would give $\rho=\rho(\Gamma)$ as a function of the microstate $\Gamma$ is of no use if $\Gamma$ is not something that we can observe and control experimentally.

Gibbs noticed that $\rho(H)$ is an acceptable solution and set out to guess mathematical forms of $\rho(H)$ that made thermodynamic sense. His first guess was the canonical distribution $\rho = c e^{-H/\Theta}$. Gibbs explained his reasons for choosing the exponential form as follows (Gibbs uses $P$ for what we have called $\rho$):

The distribution represented by \begin{equation}\tag{90} \eta=\log P = \frac{\psi-\epsilon}{\Theta}, \end{equation} or \begin{equation}\tag{91} P = e^{\dfrac{\psi-\epsilon}{\Theta}}, \end{equation} where $\Theta$ and $\psi$ are constants, and $\Theta$ positive, seems to represent the most simple case conceivable, since it has the property that when the system consists of parts with separate energies, the laws of the distribution in phase of the separate parts are of the same nature, a property which enormously simplifies the discussion, and is the foundation of extremely important relations to thermodynamics.

"Seems to represent the most simple case conceivable" is a modest way of saying "a stroke of genius"!


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