1
$\begingroup$

Basically, the mathematical statement of Liouville's theorem is:

$$\frac{\partial \rho }{\partial t}= -\sum_{i}\left(\frac{\partial \rho}{\partial q_i}\,\dot{q_i}+\frac{\partial\rho}{\partial p_i}\,\dot p_i\right)$$

While I could comprehend the derivation which is nicely done in Reif's Fundamentals of Statistical and Thermal Physics, I could not get what this theorem actually wants to imply.

The Wikipedia article mentions:

It asserts that the phase-space distribution function is constant along the trajectories of the system [...]

What does this mean?

What does the word trajectory mean in the present context?

Is $\rho$ not a function of time?

Can anyone please clarify what that quoted line actually means?

$\endgroup$
5
$\begingroup$

$$\frac{\partial \rho }{\partial t}= -\sum_{i}\left(\frac{\partial \rho}{\partial q_i}\,\dot{q_i}+\frac{\partial\rho}{\partial p_i}\,\dot p_i\right)$$

This means that if we have a function of $t, p, q$ namely $\rho(t,\vec p,\vec q)$ and we have a trajectory that is a curve in $(p,q)$ space, namely $q_i(t), p_i(t), i=1\ldots N,$ then:

$$ \frac{\mathrm d}{\mathrm dt} \rho(t, \vec p(t), \vec q(t)) =\frac{\partial \rho }{\partial t}+ \sum_{i}\left(\frac{\partial \rho}{\partial q_i}\,\dot{q_i}+\frac{\partial\rho}{\partial p_i}\,\dot p_i\right) $$

How if $\rho$ is constant along trajectories, then LHS is 0 and the equation you have written follows directory.

So:

  • a trajectory is any curve in 2N dimensional space described in $q_i$ and $p_i$ coordinates
  • $\rho$ is a function of both time and $\vec q$ and $\vec p$
  • whole concept is just an application of a chain rule.
$\endgroup$
  • $\begingroup$ Thanks for answering; I was confused for the relation was proved taking a volume in the phase space and that the number of systems in the ensemble are constant. After the relation, the author without talking about the volume any more, just said $\rho$ is constant along trajectory. I mean he was talking about systems of ensemble in that volume and then he turned to trajectory. That's what I couldn't understood. Also, if the time-derivative of a function is zero, wouldn't that function be independent of time? $\endgroup$ – user36790 Feb 18 '16 at 2:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy