# 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?

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Any fluid mechanics text should help (Kundu & Cohen, Fox & MacDonald, etc) with understanding the problem. – Kyle Kanos Jul 21 '13 at 17:03
@KyleKanos which chapter in Kundu & Cohen esp.? – user17574 Jul 21 '13 at 17:44
Chapter 4: Conservation Laws should be the one. – Kyle Kanos Jul 21 '13 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. – Nanite Mar 18 at 7:13

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, $${\partial \over \partial t} \rho + \nabla\cdot (\vec{v} \rho)=0.$$