Classical phase density

Set

 context $\langle \mathcal M, H\rangle$ … classical Hamiltonian system definiendum ${\hat\rho} \in \mathrm{it}$ postulate $\langle \mathcal M, H\rangle$ … Hamiltonian system range $\Gamma_{\mathcal M} \equiv \mathcal M\times T\mathcal M$ postulate $\hat\rho: \Gamma_{\mathcal M} \times \mathbb R \to \mathbb R_+$ range $\hat\rho:: \hat\rho({\bf q},{\bf p},t)$ postulate $\frac{\partial}{\partial t}{\hat\rho} = - \nabla ({\hat\rho} \cdot X_H )$
todo: Total derivative for the 'Continuity equation' (last postulate)
todo: Hamiltonian vector field

Discussion

For all initial values $\pi(0)\in\Gamma_{\mathcal M}$, the solutions of the Hamiltonian equations of motion follow the Hamiltonian flow $X_H$. Because phase trajectories can't intersect (the Hamiltonian equations are first order in time), a given sub volume $\Sigma$ of $\Gamma_{\mathcal M}$ flows along $X_H$ with only smooth distortion of its boundary $\partial \Sigma$. Morally, the phase density counts the number of system points in any given subset of the phase volume: If we specify such a volume $\Sigma_{t_0}\subset \Gamma_{\mathcal M}$ where the index denotes some point in time, then $\int_{\Sigma_{t_0}}\hat\rho({\bf q},{\bf p},t_0)=\int_{\Sigma_{t_1}}\hat\rho({\bf q},{\bf p},t_1)$. The phase density doesn't literally count ensemble points, as there are be infinitely many. So $\hat\rho$ is assigned any initial value $\hat\rho({\bf q},{\bf p},0)$ which is soon factored out in a normalization, see Classical probability density function.

We denote the measure in $\Gamma_{\mathcal M}$ simply by $\mathrm d\Gamma$.

Using the Hamiltonian equations, we can pull out $X_H$ and get the Liouville equations:

Theorems

Liouville equation

$\left(\frac{\mathrm \partial}{\mathrm \partial t}+X_H\cdot\nabla\right){\hat\rho}=0$

which can also be written as

$\frac{\mathrm d}{\mathrm dt}{\hat\rho}(\pi(t),t)=0$

where $\pi$ is the solution of the Hamiltonian equations.