1
$\begingroup$

Suppose I have a gas enclosed in a thermally insulated box, and so I suppose, this is an example of a system in a micro-canonical ensemble. Now, I want to understand which microstates of the system could be said to correspond to the equilibrium situation. Consider the following cases:

(1) The gas is uniformly distributed in the box, and the distribution of the speeds of the particles follows the standard Maxwell-Boltzmann distribution. I think everybody would agree that in this case the gas is in equilibrium.

(2) Initially, the gas is placed in one of the corners of the box, and the speeds of the particles are assigned randomly such that the speed distribution is different from the Maxwell-Boltzmann distribution. Again, I think, that everybody would agree that the gas is not in equilibrium.

(3) Initially, the gas is placed in one of the corners of the box but the speeds are assigned to the particles such that their distribution follows the Maxwell-Boltzmann distribution. It is not clear to me what to call this situation.

Since the total energy of the gas is not going to change, in the phase-space, each microstate of the gas occupies a point on a constant energy surface, and hence the three situations outlined above do lie on this surface. However, the first one seems to be clearly that of equilibrium, the second one clearly to be of non-equilibrium, and the third completely ambiguous. Is the third one that of equilibrium or not? Also, if all these microstates lie on the same energy surface, how does one find out which ones correspond to the equilibrium situation or which ones don't? I think that in the third case the center of mass of the system will move till it eventually settles to the center of the box, and so one could define equilibrium based on two things: the motion of the center of mass, and the speed distribution of the particles. But I am not sure.

Improve this question
$\endgroup$
2
  • $\begingroup$ If you put all the gas in one corner of the box they are certainly not going to stay in that corner of the box and so can't be said to be in any sort of equilibrium $\endgroup$
    – By Symmetry
    Jul 6 '19 at 10:10
  • $\begingroup$ @BySymmetry: Exactly. So how does one classify the microstates based on whether they correspond to the equilibrium or non-equilibrium when the energy is same for all of them? $\endgroup$
    – Peaceful
    Jul 6 '19 at 10:37
2
$\begingroup$

In the microcanonical ensemble all microstates with given energy can be realized at equilibrium. But it is overwhelmingly unlikely (as in: practically impossible for a large system) to observe one in which the density of particle is not uniformly distributed over the box and the velocities are not distributed according to the Maxwell-Boltzmann distribution.

Statistical mechanics is all about probability measures on phase space, not individual configurations. However, if you really want to discuss the latter, then there is no loss in restricting to a subset in which all the nice things you want occur, as the latter subset has measure essentially 1.

This is the same as the following: throw $N=10^{23}$ times a fair coin and record the result. You get a long string of H(ead) and T(ail). All strings are possible (and have the same probability of occurring). However, you will never (in practice) observe strings in which the fraction of H deviates substantially from 1/2, or a string in which all the H are at the beginning, or strings in which there is a run of H of length much longer than log N, etc.

You may decide to define a notion of "typical" strings, that possess the properties you expect to observe in practice. This would be the same as trying to define a notion a "equilibrium configurations". But observe that it is not unambiguously defined: you can add whatever additional property you want as long as you keep the probability of the subset thus defined close to 1.

Improve this answer
$\endgroup$
8
  • $\begingroup$ This does clear up several things for me. However, what about the macroscopic properties for such "abnormal" configurations? For example, is temperature or pressure defined for the configuration in which gas is in one of the corners? $\endgroup$
    – Peaceful
    Jul 7 '19 at 9:31
  • $\begingroup$ @Peaceful : temperature, at least, is not. Temperature is a statistical property: it is not a property of microstates, but of macrostates. To some extent, the same is true of pressure. Now, what you could do, I guess, is the following: if you only know that the gas occupies one of the corners (but do not specify the explicit microstate), then you can look at the marginal measure on all microstates satisfying the constraint. For this set of microstates (but not the microstates themselves), you might be able to associate well-defined temperature and pressure. $\endgroup$
    – Yvan Velenik
    Jul 7 '19 at 9:45
  • $\begingroup$ We do talk about "microstates corresponding to a particular value of temperature". Don't we? $\endgroup$
    – Peaceful
    Jul 8 '19 at 9:57
  • $\begingroup$ Not really. You can talk about microstates corresponding to a particular value of the energy, for example, because the energy is a function defined on the phase space and can thus be used to partition the latter into "cells". But there is no function associated to the temperature, so you cannot use it to partition phase space. You can also see that from the fact that fixing the temperature means that you are considering the (grand)canonical ensemble and not the microcanonical one, and under the Gibbs measure associated to the (grand)canonical ensemble, all microstates are possible. $\endgroup$
    – Yvan Velenik
    Jul 8 '19 at 11:19
  • $\begingroup$ Hmm. But when the gas reaches the equilibrium state (This I anyway don't understand, and was part of the question I asked), how can we define temperature for it then? $\endgroup$
    – Peaceful
    Jul 10 '19 at 4:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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