In the derivation of the Boltzmann equation (link to Wikipedia) for a collisionless gas it is assumed that: $$ f\left( \vec{r} + \frac{ \vec{p} }{m} \Delta t, \ \vec{p} + \vec{F} \Delta t, \ t + \Delta t \right) = f\left( \vec{r}, \vec{p}, t \right) $$

What is the reasoning for this expression.

  • $\begingroup$ Hint: In the absence of collisions all particles are independent. $\endgroup$
    – WoofDoggy
    Commented Dec 9, 2015 at 15:50
  • $\begingroup$ (so the phase space density changes just due to the movement of the particles in phase space). $\endgroup$ Commented Dec 9, 2015 at 20:04

1 Answer 1


What is the reasoning for this expression

This is another way of expressing a specific form of Liouville's theorem given by: $$ \frac{d \ f\left( \mathbf{r}, \mathbf{p}, t \right)}{d t} = 0 $$ where $f\left( \mathbf{r}, \mathbf{p}, t \right)$ is the phase space density. It is another way of saying that the phase space density is conserved along trajectories in phase space [i.e., a trajectory is a curve in $\left( \mathbf{r}, \mathbf{p} \right)$ space] or that phase space is incompressible for this system.

It simply means that $f\left( \mathbf{r}, \mathbf{p}, t \right)$ does not change.

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