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This question is about statistical mechanics:

Why does it make sense to postulate, that in thermal equilibrium all micro-states with fixed U within a micro-canonical ensemble are equally probable?

When I was a student, I focused mainly on the mathematical derivation, and it was clear for me, that this is the condition for maximizing the "degree of uncertainty", so we end up with entropy $$ S = k_B\cdot \ln W $$

with $W$ the number of possible micro-states and the famous Boltzmann result.

Now, years after, when I reviewed my old textbooks, this is not plausible anymore. I could easily image a system made up of two "bins" A,B, each capable to hold 0, 1, or 2 portions of (same) energy of amount 1.

The total energy of the system is 2.

The micro-states are defined by the tuple (a,b), a,b denoting number of portions within A,B, respectively.

In each round those two bins exchange energy portions with probabilities according to this scheme:

Enter image description here

Then, after some while, it is much more likely to find the system in state (2,0) as in one of the others. Even when I replace $p=0$ by $p=10^{-6}$ most of the time the system would remain in (2,0).

So in this case, the probabilities are not the same for each micro-state.

Are such systems excluded by some subtle physical reason which I couldn't identify yet?

My textbook, on the other hand, doesn't make assumptions, how systems are constructed physically.

Something must be wrong...does it mean, thermal equilibrium is not defined in those cases? But what else is required to justify the assertion?

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    $\begingroup$ I think the issue here is what is causing these various transition probabilities? It seems like you would need an external mechanism to do this. $\endgroup$ – BioPhysicist May 20 at 14:32
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    $\begingroup$ I think you are missing something. Just because you assume something and then find a system that doesn't follow that assumption doesn't meant there is an issue. It just means that the conclusions from the assumption are not valid in this system. $\endgroup$ – BioPhysicist May 20 at 14:49
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    $\begingroup$ Your mechanism is not time reversible, it cannot be in equilibrium $\endgroup$ – Wolphram jonny May 20 at 14:49
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    $\begingroup$ @Wolphramjonny the underlying mechanics doesn't need to be time reversible to get statistical mechanics. This Markov chain does have an attractor which seems problematic, but the point stands that if you wrote down an arbitrary Markov chain you would still get this issue that the stationary distribution is not uniform. $\endgroup$ – jacob1729 May 20 at 14:54
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    $\begingroup$ @michael I think with your model with an absorbing state then it should not thermalise. But my intuition is that if you had a large number $N\to \infty$ of states and a Markov model imposed on them, that you should get thermal behaviour for almost all possible Markov models. In your case, that I think means that your $p$ should be above some threshold that maybe goes like $N^{-1}$, but $N=3$ is too low to declare to be infinity. $\endgroup$ – jacob1729 May 20 at 15:25
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I think this is a good question and it gets at what statistical mechanics is actually doing when it assigns probabilities to states. In particular, it is not giving you the correct probability to be in any given state. This is even more clear in the fully deterministic case, where the pdf to be in state $x$ at time $t$ is given by $p(x,t) = \delta(x-r(t))$ where $r(t)$ is the solution to the equations of motion of the system.

Your example is similar. You consider a non-deterministic system (a Markov chain) for transitioning between states in some micro-canonical energy window. Standard statistical mechanics says to assign a uniform probability distribution over these states in the absence of other knowledge. However, you have decided to fully specify the dynamics by assigning transition probabiities $T_{ab}=\mathbb{P}(a\rightarrow b)$, where $\mathbb{P}(a \rightarrow b)$ are the probabilities of transitioning from state $a$ to state $b$. This is a well studied problem and there is some steady-state probability distribution that cou can obtain from the matrix $T_{ab}$ corresponding to on of it's (left) eigenvectors. How to do this is not important, the point is that it's typically non-uniform as you observed. How do we square this with statistical mechanics?

Well, one way to deal with this is to require 'detailed balance'. That is, demand that $T_{ab}$ is symmetric. This makes the eigenvector with eigenvalue $1$ a uniform distribution so gives us microcanonical results. I think this is unphysical - systems like yours presumably do exist and I'd expect them to reach thermal equilibrium so long as they are large. My interpretation of what is going on is that for a large number of states, you will not be able to extract useful information out of the Markov model, much like you can't solve (or even use the solution of) the classical deterministic ODEs. Thus, statistical mechanics by maximising entropy is giving you the most reliable estimate you can get without solving the system. In the thermodynamic limit, you'd expect that this becomes strongly peaked in the sense that almost any Markov matrix $T_{ab}$ you can write would agree with the microcanonical prediction for any observable quantity $\langle O \rangle = \sum_a p_a O_a$.

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  • $\begingroup$ Interesting, bit I have to sleep overnight to digest it ;-) $\endgroup$ – michael May 20 at 15:22
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The basic desideratum when you want to determine a suitable probability measure to describe your isolated system at equilibrium is that it must describe a system at equilibrium!

Let's go back to the standard case of a classical Hamiltonian system. The reason the microcanonical measure is a plausible candidate to describe the equilibrium state of the system is that it is left invariant by the dynamics. Were it not the case, it would be inconsistent to use it to describe a system at equilibrium, as starting the system according to this measure, its distribution would change over time (hence, it would not be be at equilibrium). Of course, there are many other invariant measures. This is one of the reasons that the foundational problem in equilibrium statistical mechanics is still wide open.

In the case of a finite-state, irreducible Markov chain, as you propose, it is trivial to determine the set of all probability measures left invariant by the dynamics: there is only one, the stationary measure. In your case, it is clear that the stationary measure is not uniform, hence that there is no hope to describe its equilibrium state using the uniform measure (and, of course, we knew that, since the proper measure to use is the stationary measure!).

(It is easy to characterize reversible Markov chains that leave the uniform measure invariant. In this case, it is necessary and sufficient that the transition probability be symmetric: $p(i\to j) = p(j\to i)$ for all $i,j$.)

So, my point is that a (trivial) necessary condition for the microcanonical ensemble (=uniform measure) to apply to a given system is that it be left invariant by the dynamics, which is not the case in your example.

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  • $\begingroup$ Does this mean that the OP's system cannot be described by a Hamiltonian (or a Lagrangian)? $\endgroup$ – ratsalad May 20 at 23:38
  • $\begingroup$ Well, his system is stochastic, so this would in any case require coarse-graining of the phase space or similar procedure. But yes, a Hamiltonian flow must preserve the microcanonical measure, so either his underlying microscopic dynamics is not Hamiltonian, or his coarse-graining is not suitable (and thus what he calls microstates after coarse-graining does not fairly account for the true microstates). $\endgroup$ – Yvan Velenik May 21 at 7:20
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I think that this is not a statistical system, in the sense that we study in statistical mechanics. Or, alternatively, this is not a system in equilibrium.

One of the basic ideas when we describe a system in statistical mechanics is that the system is macroscopic, with quantities that we can define as extensive and intensive, and these quantities are homogeneous across the system. That is, if we cut up the system into $N$ separate subsystems that are still large (i.e. macroscopic themselves), each of the copies will retain the intensive quantities, and have $1/N$ of the extensive quantities. Your system clearly violates this assumption.

By the way, I would say that "all micro-states with fixed $U$ within are equally probable" is the definition of the micro-canonical ensemble.

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  • $\begingroup$ In my book a micro-canonical ensemble is defined as such, where U is fixed and not subject to averaging. From the principle of maximizing sum of p*ln(p) it is derived that a uniform distribution of p is required. No pre-requisits are assumed that this must be possible at all. So does it mean, that in my case, thermal equilibrium is in fact not defined? Why is it defined for an ideal gas? There must be some kind of "required property" as system must have, to be in thermal equilibrium. $\endgroup$ – michael May 20 at 15:01
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    $\begingroup$ yes, I would say that your system is not in thermal equilibrium, similar to a case where we put two systems with different temperatures and internal energy next to each other, without any coupling between them. We cannot lump them together and declare them to be a single system in thermal equilibrium, as all the statistical quantities are not well defined. $\endgroup$ – user245141 May 20 at 15:10
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    $\begingroup$ a more interesting (maybe) example which is challenging to describe within the tools of statistical mechanics is a system with infinitely long ranged interactions. The energy of each subset is affected by the existence of other subsets, which means that $U$ is not extensive. For thermal equilibrium we want short-ranged interactions (for example, in an ideal gas, which is completely non interacting), and then we can say that each (large, macroscopic) subsystem is independent of the others $\endgroup$ – user245141 May 20 at 15:13
  • $\begingroup$ But when I replace p=0 with p=0,01 the picture would not change significantly. However, the sub-systems are still coupled and can exchange energy. Where is the limit for p ? $\endgroup$ – michael May 20 at 15:14
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    $\begingroup$ @michael The limit for p depends on your tolerance for inaccuracy in the predictions of statistical mechanics. The higher p is, the more accurate statistical mechanics will be in predicting the behavior of this system. You might have to set p to be pretty high to get a reasonable amount of accuracy from statistical mechanics. $\endgroup$ – probably_someone May 20 at 17:50
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The usual explanation is that a physical system is expected to have a symmetric transition probability, i.e., "detailed balance", as noted by Yvan Velenik and jacob1729. Fundamentally this comes from PT invariance (I suggest this is the "subtle physical reason" you seek), which holds in most practical systems. If a system is microscopically irreversible as in your example, that would fall outside standard statistical mechanics.

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