Suppose we have a classical statistical problem with canonical coordinates $\vec{q} = (q_1, q_2, \dots, q_n)$ and $\vec{p} = (p_1, p_2, \dots, p_n)$ such that they fulfill the usual Poisson brackets:
\begin{align}
\{ q_i, p_j \} & = \delta_{i,j} \\
\{ q_i, q_j \} & = 0 \\
\{ p_i, p_j \} & = 0 \\
\end{align}
One can show that the time evolution of every dynamical quantity $f(\vec{q}, \vec{p}, t)$ is given by
$$ \frac{d f}{d t} = \frac{\partial f}{\partial t} + \{ f, H \} = \frac{\partial f}{\partial t} + \sum_{i=1}^n \frac{\partial f}{\partial q_i} \dot{q_i} + \frac{\partial f}{\partial p_i} \dot{p_i} = \frac{\partial f}{\partial t} + (\nabla f) \cdot \vec{v} = \frac{\partial f}{\partial t} + \nabla (f \cdot \vec{v}) $$
with $\nabla = (\frac{\partial}{\partial q_1}, \dots, \frac{\partial}{\partial q_n}, \frac{\partial}{\partial p_1}, \dots, \frac{\partial}{\partial p_n})$, $\vec{v} = (\dot{q_1}, \dots, \dot{q_n}, \dot{p_1}, \dots, \dot{p_n})$ and $H$ the Hamiltonian of the system.
Liouville's theorem states that:
$$ \int d^n p ~ ~ d^n q = \int d^n p' ~ ~ d^n q' $$
if $(\vec{q}~', \vec{p}~')$ and $(\vec{q}, \vec{p})$ are both canonical coordinates, e.g. related by a canonical transformation.
So the phase space volume is a constant between systems which are described by canonical coordinates.
Now consider the phase space density $\varrho(\vec{q}, \vec{p}, t)$ which is the density of dynamically allowed trajectories at a given point $(\vec{q}, \vec{p})$ in phase space for a given instance of time $t$.
Liouville's equation reads:
$$\frac{d \varrho}{d t} = 0$$
which (together with the equation for $f$) says that $\varrho$ is a (locally) conserved density in phase space. Because $\varrho \ge 0$ one can conclude that there are no sources of trajectories in phase space: trajectories do not start, end or cross.
Usually the Liouville equation is derived from Liouville's theorem. However I haven't seen such a derivation for which at some point it wasn't assumed that $\varrho$ is a (locally) conserved density. Hence, that reasoning is circular.
Do you know a non-circular derivation of Liouville's equation or is it indeed an axiom of classical statistical mechanics?
Best Answer
Why There is a Need for a Further Axiom
To derive Liouville's equation, you indeed need another axiom further to your assumptions. Something like: "there is no nett creation or destruction of any particle of any species throughout the particle system state evolution". The easiest way to understand the need for this axiom is to cite a system wherefor Liouville's equation cannot hold, even though particles undergo dynamical evolution described by Hamilton's equations throughout their lifetimes: a system of particles undergoing a far-from-equilibrium chemical reaction. In such a system, reactant particle species are consumed by the reaction, and disappear from phase space. Reaction product particles appear in phase space in their place. Moreover, chemical energy is converted to kinetic energy (or contrariwise), so that a product species will "suddenly" appear at a different point in phase space from the one where the correspondingly consumed reactant species particles vanished. Liouville's equations would conceptually be replaced by a coupled system of equations, one for each species $j$, of the form:
$$\frac{\partial\,\rho_j(X,\,t)}{\partial\,t} = \{H,\,\rho_j\} + \sum\limits_k \int_\mathcal{P} M_{j\,k}(X,\,X^\prime)\,\rho_k(X^\prime,\,t)\,\mathrm{d}\Gamma^\prime$$
where the integral is over all phase space $\mathcal{P}$, $\Gamma$ is the measure defined by the volume form and the kernel $M_{j\,k}$ expresses detailed stochimetric balance between the chemically reacting species as well as other physical principles such as conservation of energy, momentum and strict increase with time of entropy. Note that I said "nett" creation or destruction: Liouville's equation would work if the reaction were at equilibrium.
Complete Axioms
The following axioms (1. and 2. are equivalent to yours) will get you Liouville's equation:
From Complete Axioms to Liouville's Equation
From these axioms, the chain of inference you need is as follows:
Circularity of Other Proofs
Ultimately, I don't believe that proofs of Liouville's equation grounded on the divergence theorem are different from the above: I think that they are tacitly introducing Axiom 3 as "obvious" (even though I hope I have shown at the beginning of my answer that it doesn't always hold) and then the continuity equation and incompressible flows are simply an expression of this tacitly assumed axiom. So I don't think that these "proofs" are circular, just somewhat badly written in making use of tacit assumptions.
Summary
User Image sums all this up nicely (I was perhaps too brainfried to make the last step):
and indeed, in the presence of the other two, my axiom 3 is logically equivalent to Liouville's equation. My version is perhaps more physically transparent, but open to interpretation, and so the assertion of Liouville's equation as an axiom is perhaps more succinct and precise. So the answer to the title question is that Liouville's Equation must indeed be added as an axiom, and, in the presence of Axioms 1 and 2, it has the meaning that particle number of all species is conserved.