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  1. Let $\mu_1=(q,p)$ and $\mu_2=(q',p')$ be two accessible microstates (points in phase space) corresponding to some macrostate $M$.
  2. Let us even assume that the trajectory of a system visits every accessible microstate if one waits long enough.

For a general Hamiltonian $H$, $$\left|\left(\frac{\partial H(q,p)}{\partial p},-\frac{\partial H(q,p)}{\partial q}\right)\right|= \left|\frac{d\mu_1}{dt}\right| \neq \left|\frac{d\mu_2}{dt}\right|=\left|\left(\frac{\partial H(q',p')}{\partial p},-\frac{\partial H(q',p')}{\partial q}\right)\right|$$ which means that the system may flow through $\mu_1$ at a different `speed' than through $\mu_2$. Therefore, if we take a microscopic snapshot of the system at a random point in time, the probability of finding it at $\mu_1$ is not necessarily equal to the probability of finding it at $\mu_2$, unless the Hamiltonian $H(q,p)$ has the special property that makes $\left|\frac{d\mu_1}{dt}\right| = \left|\frac{d\mu_2}{dt}\right|$.

How can we justify the assumption that the probability of finding the system in an accessible microstate is independent of the microstate, without showing that the Hamiltonian has the special property required by the assumption? I'm all for hiding our ignorance behind a democratic assumption. But doing so without regard to the structure of the Hamiltonian is troubling. Am I missing something? For example, can it be shown that the speeds at $\mu_1$ and $\mu_2$ are the same if $H=E_0$ constraint is imposed?

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  • $\begingroup$ what do the $\left|\right|$ symbols mean here? My sense is that they mean cardinality. Please advise. $\endgroup$
    – Michael Levy
    Commented Jan 26 at 18:53

2 Answers 2

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The Hamiltonian does have such a special property, but it's not the one you mention: it's the conservation of phase space volume, which is the basis for Liouville's Theorem. Take a look at these lecture notes which show that the distribution function is constant along a trajectory in phase space.

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  • $\begingroup$ Could you expand a little? E.g., state Liouville theorem and show it's equivalent to OP's 'special property' $\endgroup$
    – innisfree
    Commented Jun 10, 2017 at 5:18
  • $\begingroup$ 1. One does not need any special property of the Hamiltonian to establish Louiville's Theorem. 2. How does Louiville's Theorem address this question? I'm afraid I don't see the relevance. $\endgroup$
    – Coriolis1
    Commented Jun 10, 2017 at 20:54
  • $\begingroup$ Separately, the discussion in the link you've posted assumes $\frac{\partial \rho}{\partial t} =0$. I don't see a basis for that assumption either. $\endgroup$
    – Coriolis1
    Commented Jun 10, 2017 at 21:04
  • $\begingroup$ If I understand you correctly there are two separate questions here. The first is where the assumption of equal probability of microstates in the microcanonical ensemble comes from. This is typically motivated by Liouville's theorem, as described in the links (Tuckerman's book on Statistical Mechanics also has a very nice description). The other question is why this assumption is not equivalent to requiring "equal speed" through all phase space points along a trajectory. I'm not sure I understand this concept -- what units would $|(\dot{q}, \dot{p})|$ have? $\endgroup$
    – user8153
    Commented Jun 11, 2017 at 9:38
  • $\begingroup$ Louiville's Theorem + ergodicity assumption imply that the phase space density is a constant over the surface of accessible microstates. That does not imply that every microstate is equally probable. Every microstate would be equally probable only if the system spends equal amount of time in every microstate it visits. $\endgroup$
    – Coriolis1
    Commented Jun 12, 2017 at 18:59
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That the speeds differ only means that one system is slower than the other, not that the equal probability assumption is violated. The slower system will take a longer time to explore all the microstates, but it will explore all with the same probability (imagine watching the same system running at normal speed versus watching it in a slow motion movie).

Regarding the assumption itself, it is an assumption that works well enough. There lots of discussions about how justifiable it is. You can start here.

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  • $\begingroup$ We are talking about just one system, not two. At time $t$, the system is at $\mu_1$, and at time $t'$ it is at $\mu_2$ in the phase space. So, respectfully, I don't think your comment addresses the issue. $\endgroup$
    – Coriolis1
    Commented Jun 9, 2017 at 20:14
  • $\begingroup$ You are right, I missintrpreted the question. So your argument is that the system spends more time in some states, and thAt will make them moe likely? $\endgroup$
    – user126422
    Commented Jun 9, 2017 at 20:38
  • $\begingroup$ Yes. In extreme cases, if the Hamiltonian has a critical point $\left(\frac{\partial H}{\partial q} = \frac{\partial H}{\partial p} =0\right)$, the system can get stuck there making the critical point a significantly more probable microstate, and leading to extreme violation of the postulate. $\endgroup$
    – Coriolis1
    Commented Jun 9, 2017 at 22:21

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