To obtain the result $\frac{\text d \rho }{\text d t}=0$ you need two facts: the first is that the hamiltonian flow preserves the volume of phase space. The second fact is the conservation of probability, that is, the probability that the system is found in a volume $U$ at time $t=0$ equals the probability of finding it within $\Phi _t U$ at time $t$, where $\Phi _t$ denotes the hamiltonian flow. This is a direct consequence of the deterministic nature of classical mechanics: the two propositions “$(p(0),q(0))\in U$” and “$(p(t),q(t))\in \Phi _t U$” are equivalent.
Using conservation of probability, for an arbitrary volume $U$ we can write an equation: $$\int _U \rho(p,q,0) \text d p \text d q=\int _{\Phi _t U} \rho(p,q,t)\text dp \text d q .$$
By Jacobi's theorem: $$\int _{\Phi_t U} \rho (p,q,t)\text d p \text d q=\int _U\rho (\Phi _t (p,q),t)\text J_{\Phi _t}d p \text d q.$$
The Jacobian $J_{\Phi _t}=1$, because the flow preserves volumes. It follows that: $$\int _U \rho (p,q,0)\text d p \text d q =\int _U \rho (\Phi _t (p,q),t)\text d p \text d q,$$
and, since the volume $U$ was arbitrary, $\rho (p,q,0)=\rho (\Phi _t (p,q),t)$, or $\text d\rho /\text d t=0$.
Introduction
Let us define the density of particles of species $s$ in a volume element, $d\mathbf{x} \ d\mathbf{v}$, at a fixed time, $t$, centered at $(\mathbf{x}, \mathbf{v})$ as the quantity $f_{s}(\mathbf{x},\mathbf{v},t)$. I assume this function is non-negative, contains a finite amount of matter, and it exists in the space of positive times and $\mathbb{R}^{3}$ and $\mathbb{R}_{\mathbf{v}}^{3}$, where $\mathbb{R}_{\mathbf{v}}^{3}$ is the space of all possible 3-vector velocities. Then one can see that there are two ways to interpret $f$: (1) it can be an approximation of the true phase space density of a gas (large scale compared to inter-particle separations); or (2) it can reflect our ignorance of the true positions and velocities of the particles in the system. The first interpretation is deterministic while the second is probabilistic. The latter was used implicitly by Boltzmann. Let us assume that $f_{s}(\mathbf{x},\mathbf{v},t)$ $\rightarrow$ $\langle f \rangle + \delta f$, where $\langle f \rangle$ is an ensemble average of $f_{s}$ and I have dropped the subcript out of laziness.
Liouville's Equation
I know that $\langle f \rangle$ satisfies Liouville's equation, or more appropriately, $\partial \langle f \rangle$/$\partial t = 0$. In general, the equation of motion states:
$$
\begin{equation}
\frac{ \partial f }{ \partial t } = f \left[ \left( \frac{ \partial }{ \partial \textbf{q} } \frac{ d\textbf{q} }{ dt } \right) + \left( \frac{ \partial }{ \partial \textbf{p} } \frac{ d\textbf{p} }{ dt } \right) \right] + \left[ \frac{ d\textbf{q} }{ dt } \cdot \frac{ \partial f }{ \partial \textbf{q} } + \frac{ d\textbf{p} }{ dt } \cdot \frac{ \partial f }{ \partial \textbf{p} } \right] \tag{1}
\end{equation}
$$
where I have defined the canonical phase space of $(\mathbf{q}, \mathbf{p})$. If I simplify the terms dA/dt to $\dot{A}$ and let $\boldsymbol{\Gamma} = (\mathbf{q}, \mathbf{p})$, then I find:
$$
\begin{align}
\frac{ \partial f }{ \partial t } & = - f \frac{ \partial }{ \partial \boldsymbol{\Gamma} } \cdot \dot{\boldsymbol{\Gamma}} - \dot{\boldsymbol{\Gamma}} \cdot \frac{ \partial f }{ \partial \boldsymbol{\Gamma} } \tag{2a} \\
& = - \frac{ \partial }{ \partial \boldsymbol{\Gamma} } \cdot \left( \dot{\boldsymbol{\Gamma}} f \right) \tag{2b}
\end{align}
$$
where one can see that the last form looks like the continuity equation. If I define the total time derivative as:
$$
\begin{equation}
\frac{ d }{ dt } = \frac{ \partial }{ \partial t } + \dot{\boldsymbol{\Gamma}} \cdot \frac{ \partial }{ \partial \boldsymbol{\Gamma} } \tag{3}
\end{equation}
$$
then I can show that the time rate of change of the distribution function is given by:
$$
\begin{align}
\frac{ d f }{ dt } & = \frac{ \partial f }{ \partial t } + \dot{\boldsymbol{\Gamma}} \cdot \frac{ \partial f }{ \partial \boldsymbol{\Gamma} } \tag{4a} \\
& = - \left[ f \frac{ \partial }{ \partial \boldsymbol{\Gamma} } \cdot \dot{\boldsymbol{\Gamma}} + \dot{\boldsymbol{\Gamma}} \cdot \frac{ \partial f }{ \partial \boldsymbol{\Gamma} } \right] + \dot{\boldsymbol{\Gamma}} \cdot \frac{ \partial f }{ \partial \boldsymbol{\Gamma} } \tag{4b} \\
& = - f \frac{ \partial }{ \partial \boldsymbol{\Gamma} } \cdot \dot{\boldsymbol{\Gamma}} \tag{4c} \\
& \equiv - f \Lambda\left( \boldsymbol{\Gamma} \right) \tag{4d}
\end{align}
$$
where $\Lambda \left( \boldsymbol{\Gamma} \right)$ is called the phase space compression factor. Note that Equations 4a through 4d are different forms of Liouville's equation, which have been obtained without reference to the equations of motion and they do not require the existence of a Hamiltonian. I can rewrite Equation 4d in the following form:
$$
\begin{equation}
\frac{ d }{ dt } \ln \lvert f \rvert = - \Lambda\left( \boldsymbol{\Gamma} \right) \tag{5}
\end{equation}
$$
Relation to Hamiltonian
Most readers might not recognize Equations 4d and 5 as Liouville's equation because one usually derives it from a Hamiltonian. If the equations of motion can be generated from a Hamiltonian, then $\Lambda \left( \boldsymbol{\Gamma} \right) = 0$, even in the presence of external fields that act to drive the system away from equilibrium. Note that the existence of a Hamiltonian is a sufficient, but not necessary condition for $\Lambda \left( \boldsymbol{\Gamma} \right) = 0$. For incompressible phase space, I recover the simple form of Liouville's equation:
$$
\begin{equation}
\frac{ d f }{ dt } = 0
\end{equation}
$$
However, Liouville's theorem can be violated by any of the following:
- sources or sinks of particles;
- existence of collisional, dissipative, or other forces causing $\nabla{\scriptstyle_{ \mathbf{v} }} \cdot \mathbf{F} \neq 0$;
- boundaries which lead to particle trapping or exclusion, so that only parts of a distribution can be mapped from one point to another;
- spatial inhomogeneities that lead to velocity filtering (e.g., $\mathbf{E} \times \mathbf{B}$-drifts that prevent particles with smaller velocities from reaching the location they would have reached had they not drifted); and
- temporal variability at source or elsewhere which leads to non-simultaneous observation of oppositely-directed trajectories.
Source of Irreversibility
Irreversibility is somewhat of a conundrum because it arises largely due to our choice of boundary conditions, smoothing assumptions (e.g., coarse graining or mean field theory), and limits. For instance, if I assume a velocity distribution of particles can be represented by a continuous model function, the use of a continuous distribution function inserts irreversibility into the equation. One can argue that this is splitting hairs because it is obvious that irreversibility exists in nature. However, I think it is important because your question points at a deeper issue.
If I assumed perfectly elastic binary particle collisions and ignore quantum uncertainties, one could, in principle, follow the trajectories of all particles in a system forward and backward in time. There would be no irreversibility in this model, if I had strong enough computers. However, binary particle collisions are not truly elastic, so our assumption of elasticity has created a loss of information.
Another subtle point is that Boltzmann a priori defined his, now famous, H-theorem such that time would increase in the correct direction (i.e., positive time). He did not originally relate the H-theorem to entropy, that interpretation came later (I believe with Gibbs, but someone correct me if I am wrong here).
The point is that the concepts of irreversibility and entropy are coupled, but not necessarily through direct means. I am inclined to think that the irreversibility to which you refer arises from our methods of solving the math necessary for modeling dynamical statistical systems.
References
- Evans, D.J. "On the entropy of nonequilibrium states," J. Statistical Phys. 57, pp. 745-758, doi:10.1007/BF01022830, 1989.
- Evans, D.J., and G. Morriss Statistical Mechanics of Nonequilibrium Liquids, 1st edition, Academic Press, London, 1990.
- Evans, D.J., E.G.D. Cohen, and G.P. Morriss "Viscosity of a simple fluid from its maximal Lyapunov exponents," Phys. Rev. A 42, pp. 5990–5997, doi:10.1103/PhysRevA.42.5990, 1990.
- Evans, D.J., and D.J. Searles "Equilibrium microstates which generate second law violating steady states," Phys. Rev. E 50, pp. 1645–1648, doi:10.1103/PhysRevE.50.1645, 1994.
- Gressman, P.T., and R.M. Strain "Global classical solutions of the Boltzmann equation with long-range interactions," Proc. Nat. Acad. Sci. USA 107, pp. 5744–5749, doi:10.1073/pnas.1001185107, 2010.
- Hoover, W. (Ed.) Molecular Dynamics, Lecture Notes in Physics, Berlin Springer Verlag, Vol. 258, 1986.
- Paschmann, G., and P.W. Daly "Analysis Methods for Multi-Spacecraft Data," ISSI Sci. Rep. Ser. 1, ESA/ISSI, Vol. 1. ISBN:1608-280X, 1998.
- Villani, C., Chapter 2 A review of mathematical topics in collisional kinetic theory, pp. 71–74, North-Holland, Washington, D.C., doi:10.1016/S1874-5792(02)80004-0, 2002.
- Villani, C. "Entropy production and convergence to equilibrium for the Boltzmann equation," in XIVTH International Congress on Mathematical Physics, Edited by J.-C. Zambrini, pp. 130–144, doi:10.1142/9789812704016_0011, 2006.
Best Answer
To answer the response in the comment, yes this is merely a coincidence. The Hamiltonian mechanics theorem is a "dynamical statement". It says if you have a Hamiltonian $H$ and look at the integral flow $\Phi_{H,t}$, then this flow leaves the symplectic form invariant $\Phi_{H,t}^*\omega=\omega$ for all $t$. It is dynamical in the sense that it talks about the Hamiltonian $H$ itself, its vector field $X_H:=\omega^{\sharp}(dH)$ and the corresponding integral flow. Meanwhile, the symplectic form $\omega$ talks about the geometry of phase space. So, this version of Liouville's theorem says roughly that the geometry of phase space is unaffected by the dynamics.
The one about complex analysis is just an emphasis of the rigid behavior of holomorphic (or even harmonic) functions. There are so many theorems in math/physics which deal with some quantity being constant/ being conserved under certain actions. Some have relationships with one another, but not always (and this isn't one of those cases). The fact that bounded entire functions are constant is a special case of the following result (whose proof is very similar to the usual proof of Liouville' theorem).
This says if $f$ satisfies some kind of bound, then $f$ has to be a polynomial of not-too large degree. In particular if you take $0\leq \alpha<1$, you can deduce $f$ is constant (the special case $\alpha=0$ corresponds to the usual Liouville theorem). As you can see this theorem is more of a restriction on how crazy holomorphic functions can be (namely, not very crazy at all... compared to smooth functions on $\Bbb{R}$ which can have crazy behavior).