# Current density in phase space

$\newcommand{\dd}{{\rm d}}$ I have a question which arises from looking at the impact free Boltzmann equation.

Let $(\vec{x},\vec{v})$ be a vector in our phase space $\Gamma^N = \mathbb{R}^{6N}$. The dynamics of a state are determined by the distribution function $f(\vec{x}, \vec{v}, t)$. Where $f(\vec{x}, \vec{v}, t) \dd^3x\, \dd^3v$ is the amount of particles at time $t$ in the volume element $\dd^3x\, \dd^3v$.

To derive the impact free Boltzmann equation we simply have to equate the time derivative of the volume-integral of $f$ to the flow of particles out of that volume (the amount of particles going out of a certain phase space volume determine how the state goes on in time).

This means: $$\int_V \frac{\partial}{\partial t} f(\vec{x}, \vec{v}, t) \dd V = - \int_V \nabla_{\vec{x}, \vec{v}}((\vec{v},\vec{a})f(\vec{x}, \vec{v}, t))\dd V$$

This is where my question arises. Why is the right side of the equation the flow of particles out of $\dd V$? $(\vec{v},\vec{a})$ is the time derivative of $(\vec{x}, \vec{v})$, but I still don't see it, can somebody give me some pointers what I have to read about to get an intuitive and mathematical feeling for why this is right?

• Any fluid mechanics text should help (Kundu & Cohen, Fox & MacDonald, etc) with understanding the problem. Jul 21, 2013 at 17:03
• @KyleKanos which chapter in Kundu & Cohen esp.?
– user17574
Jul 21, 2013 at 17:44
• Chapter 4: Conservation Laws should be the one. Jul 21, 2013 at 17:52
• Just at a glance, it looks like you are performing integrals over a 6-dimensional phase space, i.e., the single-particle phase space. This is appropriate for the Boltzmann equation but is not appropriate for the 6N dimensional phase space of the full system. Mar 18, 2016 at 7:13

## 1 Answer

This is just a continuity equation. The fact that it is in phase space should not matter--it is just flow in a higher-dimensional space (6*N*).

In general, a density ρ changes in time in proportion to the efflux rate, which is the divergence of the current, j=v ρ. Specifically, the number of "particles" out of a volume decreases in time as $${\partial \over \partial t} \int dV \rho =-\int_S d\vec{S} \cdot \vec{v} ~\rho = -\int dV ~~ \nabla\cdot (\vec{v} \rho).$$ where the subscripted integral is a closed surface integral of that volume, with the surface element turned into a vector perpendicular to that surface. So the particle efflux is the flow of ρ out of the volume, per unit time--only the component of the velocity perpendicular to the surface pushes them out of the volume. The second expression is just application of Gauss's law.

Since this works for every volume and surface, the differential version of that eqn is the conventional continuity diff eqn, $${\partial \over \partial t} \rho + \nabla\cdot (\vec{v} \rho)=0.$$

Also take a look at 172813. Transcribing all this to phase space is straightforward. Recalling that f behaves like an incompressible fluid, by Hamilton's eqns, this continuity equation reduces to Liouville's theorem.