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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?

Coriolis1
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2 Answers2

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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.

user8153
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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.