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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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  • Could you expand a little? E.g., state Liouville theorem and show it's equivalent to OP's 'special property' – innisfree Jun 10 '17 at 05:18
  • 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.
    • – Coriolis1 Jun 10 '17 at 20:54
  • 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. – Coriolis1 Jun 10 '17 at 21:04
  • 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? – user8153 Jun 11 '17 at 09:38
  • 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. – Coriolis1 Jun 12 '17 at 18:59
  • To give an analogy consider traffic on a road. Cars stop for an extended period at a red light and then speed to the next traffic light. If we take a snapshot of the traffic at a random time, we are more likely to see cars at a traffic light than in the region between traffic lights. Louiville's Theorem is analogous to saying that the number of vehicles you can place on a stretch of 100 m of the road is the same near traffic lights as in the region between traffic lights. That is the road has constant width. It does not imply we are equally likely to see a car at any point on the road. – Coriolis1 Jun 12 '17 at 19:02
  • Your observation about the units of $|(\dot q, \dot p)|$ is well taken. There are hidden powers of $h, c$ and $G$ to make the units of $\dot q$ and $\dot p$ the same. The notion of phase space velocity is used even in the link you have provided and it is sloppy there also in the sense that it is also expressed in natural units. – Coriolis1 Jun 12 '17 at 19:09
  • I want to reiterate my earlier comment that I don't see a basis for the assumption of stationarity of the ensemble ($\frac{\partial \rho}{\partial t} = 0$). It is one of those statements that appear very benign, but it is not necessarily so because it is needed to establish Louiville's Theorem and I don't understand the basis for it. – Coriolis1 Jun 12 '17 at 19:14
  • I might be wrong, but I believe that the lecture notes talk about the phase space velocity $(\dot{q},\dot{p})$, where there is no problem with the units (it's ok if the two components of this vector have different dimensions). But if I understand you correctly you would like to convert this velocity to a speed, i.e., compute its magnitude, and I don't see a way to do that since $\sqrt{\dot{q}^2 + \dot{p}^2}$ doesn't work. – user8153 Jun 12 '17 at 21:32
  • The construction you suggest can be applied to the harmonic oscillator, $H(q,p) = k q^2/2 + p^2/2m$. There the dynamical equations are $\dot{q}=p/m$ and $\dot{p}=-k x$, and one can define a "velocity" $\vec{v} = (\sqrt{m/2}\dot{x}, 1/\sqrt{2 k}\dot{p})$ such that the "speed" $v = \sqrt{\vec{v}^2} = \sqrt{p^2/2m + k q^2/2 }= \sqrt{E}$. For this system one can define a conserved "speed" along a trajectory, and its conservation is that of energy. I don't see how one can generalize that to arbitrary systems though (I think the Liouville equation might be that generalization). – user8153 Jun 13 '17 at 06:50
  • Assuming that $q$ has dimensions of length, one could scale it with a constant factor as $s=q \left(\frac{c^3}{G}\right)$, where $c$ is the velocity of light in vacuum and $G$, the gravitational constant. $s$ has dimensions of momentum as does $p$. $\sqrt{\dot s^2 + \dot p^2}$ is well-defined. If $q$ is a generalized coordinate and does not have the dimension of length, one can again scale $q$ using $h^\alpha c^\beta G^\gamma$ ($h$ is Planck's constant) with appropriate values of $\alpha, \beta, \gamma$ to make the dimensions of the scaled $q$ agree with the dimensions of $p$. – Coriolis1 Jun 13 '17 at 07:27
  • If electromagnetism is involved then, in addition to $h, c, G$, we would need an additional constant such as $\epsilon_0$, the electric permittivity of vacuum, for scaling. – Coriolis1 Jun 13 '17 at 07:33