Typically in physics (at least the way I learned mechanics), this is derived using the multi-dimensional divergence theorem on the $2N$-dimensional phase space i.e.

$0=\partial_t \rho + \sum\limits_{i=1}^N \left(\frac{\partial(\rho \dot{q_i})}{q_i} + \frac{\partial(\rho \dot{p_i})}{p_i} \right)$,

and the terms in brackets are simplified using Hamiltonian equation of motion to obtain Liouville's Theorem:

$\frac{\mathrm{d}\rho}{\mathrm{d}t} = \frac{\partial \rho}{\partial t} + \sum\limits_i \left(\frac{\partial \rho}{\partial q_i}\dot{q_i} + \frac{\partial \rho}{\partial p_i}\dot{p_i}\right)=0$.

I am wondering if it is possible to derive this without assuming "physics," i.e. Hamiltonian equation of motion, but instead from maximum (differential) entropy principle.

Suppose $\rho=\rho(\vec{p},\vec{q};t)$ is a phase-space distribution with maximum differential entropy $\mathcal{H}[\rho] = \int -\rho\log{\rho}\, \mathrm{d}\vec{p}\,\mathrm{d}\vec{q}$.

Since the entropy is maximized, we can write:

$\frac{\mathrm{d} \mathcal{H}}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}\int -\rho\log{\rho} \, \mathrm{d}\vec{p}\,\mathrm{d}\vec{q} = \int\mathrm{d}\vec{p}\,\mathrm{d}\vec{q} \, (-\log\rho - 1) \frac{\mathrm{d}\rho}{\mathrm{d}t} = 0$

However, does this necessarily imply $\frac{\mathrm{d}\rho}{\mathrm{d}t} =0$ ? Or, is there an alternative approach to this problem? One way, perhaps, is to show that the differential entropy $\mathcal{H}[\rho]=\int -\rho\log{\rho}\, \mathrm{d}\vec{p}\,\mathrm{d}\vec{q}$ is (up to a constant) equivalent to Boltzmann's definition of entropy $S = k_B \log\left|\Gamma\right|$ (again, without assuming physics).

(I am basically trying to strip as much physics off from statistical mechanics as possible.)

  • 1
    $\begingroup$ Maybe interesting: physics.stackexchange.com/q/334691/75518 $\endgroup$
    – image357
    Commented Jul 4, 2017 at 18:37
  • $\begingroup$ It's very interesting you are saying "I am basically trying to strip as much physics off from statistical mechanics as possible". It's possible to strip off physics but to do it you need to forget about Liouvillee (too much physics there already). You could take a look at springer.com/gp/book/9783030041489. $\endgroup$
    – Themis
    Commented Aug 22, 2019 at 17:27
  • $\begingroup$ The condition $\frac{\mathrm{d} \mathcal{H}}{\mathrm{d}t} = 0$ is not maximization of $\mathcal{H}$, but constancy of $\mathcal{H}$ in time. That's a different thing. Any density that is a function of time-independent Hamiltonian $\rho(H(q,p))$ is constant in time at any point of the phase space, and thus for such function, $\mathcal{H}[\rho]$ is constant in time. Thus $\rho$ with $\mathcal{H}[\rho]$ constant in time need not be maximizing the information entropy; most of them aren't. $\endgroup$ Commented Feb 26 at 1:19

1 Answer 1


The condition $\mathrm d\mathcal H/\mathrm d t=0$ does not maximize entropy. Entropy maximization is to be done not with respect to time but with respect to $\rho$. The derivative that must be set to zero is $$ \frac{\mathrm d\mathcal H}{\mathrm d \rho} = 0 $$ subject to $$ \int \rho\, \mathrm{d}\vec{p}\,\mathrm{d}\vec{q} = 1 $$ along with any other constraints that define the macroscopic state.


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