4
$\begingroup$

I was wondering if there is any relationship between equilibrium in Stat Mechanics and the phase space density of a system? This does not seem to be completely independent, as Entropy is maximized in equilibrium and this quantity is definitely related to the phase space density somehow.

Intuitively I would say that $\frac{\partial \rho}{\partial t}=0$ in equilibrium, but since in a microcanonical ensemble, the uniform distribution is the most probably one, I also feel that the derivative with respect to space and momenta coordinates should vanish, is this true?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

The condition you wrote, namely that the partial derivative of the phase density with respect to time vanishes, is a standard one placed on phase densities that describe equilibrium systems. See, for example, page 29 of Eric D'Hoker's statistical mechanics lecture notes which can be found here:

https://www.pa.ucla.edu/faculty-websites/dhoker-lecture-notes.html

On the other hand, the vanishing of the partial derivatives with respect to phase space coordinates is generally not a condition placed on equilibrium ensembles. The microcanonical distribution is special in this regard.

For example, the phase density for the canonical ensemble is \begin{align} \rho(p,q,t) = \frac{1}{Z(\beta)}e^{-\beta H(p,q)} \end{align} which is manifestly non-constant over phase space provided $H(p,q)$ is non-constant.

Improve this answer
$\endgroup$
5
  • $\begingroup$ thanks for these great lecture notes. you said that the microcanonical ensemble is special in this regard, but in your lecture notes it sounds as this is only true for the so-called uniform ensemble? $\endgroup$
    – Xin Wang
    Commented Nov 5, 2014 at 0:39
  • 1
    $\begingroup$ @XinWang Sure thing. When a system is described by the microcanonical ensemble, it has a fixed energy (by definition). This means that if phase space has dimension $n$, then the state of the system is restricted to lie on an $n-1$ dimensional hypersurface of constant energy. Moreover, the probability density assigned to this hypersurface is a uniform density; all states with the prescribed energy are equally likely. $\endgroup$
    – joshphysics
    Commented Nov 5, 2014 at 1:35
  • $\begingroup$ but how does that mean that $\frac{\partial \rho}{\partial q} =0$ and similarly for the momentum? $\endgroup$
    – Xin Wang
    Commented Nov 5, 2014 at 8:27
  • $\begingroup$ @XinWang Since the system is restricted to lie on a surface of constant energy, it's phase space should effectively be treated as that surface. Suppose that $\xi^i$ are coordinates on that surface, then since in the microcanonical ensemble the probability density is uniform, the derivatives of the phase density with respect to these coordinates will be zero; $\partial\rho/\partial \xi^i = 0$. Note that this is an idealization. In practice one often picks the system not to have exactly fixed energy, but energy fixed in some range of values. $\endgroup$
    – joshphysics
    Commented Nov 5, 2014 at 16:53
  • $\begingroup$ @joshphysics How is the equilibrium condition $\frac{\partial\rho}{\partial t}=0$ is enforced by the ergodic hypothesis? $\endgroup$
    – quirkyquark
    Commented Apr 27, 2019 at 10:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.