6

From wikipedia:

In physics and thermodynamics, the ergodic hypothesis1 says that, over long periods of time, the time spent by a system in some region of the phase space of microstates with the same energy is proportional to the volume of this region, i.e., that all accessible microstates are equiprobable over a long period of time.

So if I understood it right, given enough time the system will move through all possible states. However, from thermodynamics we know that state of equilibrium is in a sense the "final state" in which system will get once and won't move to other states after that.

Aren't these two things in contradiction? If ergodic hypothesis is true then wouldn't that mean that system which is already in state of equilibrium will spontaneously move out of equilibrium into some other state (after enough time has passed)?

valerio
  • 16,231
matori82
  • 883
  • 1
    The statistical fluctuation of systems at equilibrium to slightly out of equilibrium states and back again is well understood and well measured. Has been for decades, though it is a difficult experiment to do on even slightly macroscopic samples. You should think of this as refining what is meant by equilibrium. – dmckee --- ex-moderator kitten Jul 26 '16 at 00:04
  • I’ve seen Leonard Suskind go over that on a number of occasions. Many of his talks and classes are on youtube. – JDługosz Jul 26 '16 at 05:30

4 Answers4

15

You have to be careful to distinguish between microstates and macrostates. Thermodynamic equilibrium is a macrostate which consists of a mixture of all possible microstates of energy $E$ weighted by a Boltzmann weight $e^{- \beta E} / Z$. A state in macroscopic thermal equilibrium can be thought of as "moving through phase space" ergodically (i.e. the microstate is constantly changing, but the fraction of time spent in each microstate is fixed to the Boltzmann weight).

David Z
  • 76,371
tparker
  • 47,418
  • Are you saying that ergodic hypothesis applies only to the microstates that correspond to the one macrostate of equilibrium? I mean, this part of the quote from the first post "all accessible microstates are equiprobable over a long period of time." applies only for microstates that give just that one macrostate? In other words, once system is in equilibrium it will move through all microstates that belong only to that macrostate but it won't move to some other microstate that would correspond to other non-equilibrium macrostate? – matori82 Jul 26 '16 at 00:16
  • 3
    @matori82 No, all the system microstates contribute to the thermal equilibrium macrostate, it's just that the high-energy ones don't contribute very much at low temperature. The statement "all accessible microstates are equiprobable over a long period of time" is confusing and misleading, because it seems to contradict the Boltzmann weighting. What they mean is that all accessible microstates of the system and the thermal bath together are equiprobable. When you restrict your attention to only considering the system microstates, this reduces to the Boltzmann weighting. – tparker Jul 26 '16 at 00:21
  • Sorry, still trying to wrap my head around these concepts. Can you explain this in more simple words? Eg. without Boltzmann weighting as I am still not very familiar with it. I want to understand what is the basic idea behind introducing this hypothesis if it can be explained in simple words without equations. Maybe some simple example would help me get the intuition for the concept. – matori82 Jul 26 '16 at 02:23
  • @matori82 Thermodynamic equilibrium does not mean that the system is in a particular microstate - it refers to a particular probability distribution over microstates: the simplest possible one, in which every microstate with the same energy is assigned the same probability. This probability distribution does not change over time (hence the "equilibrium"), even though the specific system microstate is constantly changing. – tparker Jul 26 '16 at 04:12
  • @tparker Consider an ideal gas that freely expands into a container of bigger volume. The total energy remains the same but does ergodic hypothesis states that at some time the gas may have the configuration of the inital state (of smaller volume)? – Antonios Sarikas Jun 05 '20 at 17:11
  • @AntoniosSarikas Yes, absolutely. In fact, this is guaranteed by the Poincare recurrence theorem. But for a gas with many particles, it will spend an infinitesimally tiny fraction of the time in such a smaller state, because such a state represents an incredibly tiny fraction of all the possible states. – tparker Jun 06 '20 at 04:11
1

As far as I understand, the answer is: not only ergodicity but Poincare reccurence theorem "contradicts a bit" with the second law of thermodynamics.

The point is that actually time that every ergodic system (for instance, Boltzmann billiard, as Sinai proved) is in some measurable part of full phase space of system is proportional to the phase volume of this part. But due to some kind of large numbers law almost all of the phase space belongs to the parameters of the maximum entropy state.

So if you remove the boundary between two parts of the half-empty volume, system will periodically return in half-empty state (without boundary now), but the phase volume of this family of states is quite small (in fact, devastatingly tiny), so the proportion of time system belongs to this volume is devastatingly tiny too.

If you want, you can set up computer experiment with 1,2,3,...,10 molecules to see the character of rapid decreasing of the phase volume of small entropy (half-empty rectangle, for instance) while the number of molecules increases.

So in fact entropy is not increasing, it achieves its maximum possible value just when you "open the door" for it, and it persists for very long time interval (transcendentally huge, yes).

Sorry for terrible English.

polytheme
  • 11
  • Consider an ideal gas that freely expands into a container of bigger volume. The total energy remains the same but does ergodic hypothesis states that at some time the gas may have the configuration of the inital state (of smaller volume)? – Antonios Sarikas Jun 05 '20 at 17:22
1

To be concrete lets picture a box with an ideal mono-atomic gas in equilibrium and containing a constant energy. Let's not consider the deviations from the mean values of the atom momenta, so all the atoms have the same mean value of the momentum (I know this isn't realistic, but it is a good approximation of the situation). There are a lot of distributions of the momenta and positions of the atoms in phase space corresponding to the same, macroscopic equilibrium state of the gas. And the bigger the volume, the more of these distributions are possible (the distributions for which the positions of the atoms are, say, all concentrated in one corner of the box, or the momenta are separated into one high momentum part and a low momentum part are not considered, and I already stated that the momenta of the atoms are to be considered equal).

So you don't have to worry that the atoms in the box suddenly show a non-equilibrium sign (like all the atoms residing in one corner: for this to take place you have to wait for a much very much longer time than the time mentioned in the quote).

There is a nice metaphor for ergodicity: Imagine a lonely man taking a random walk, from the entrance to the exit, in the park each day. You can draw his path for many days after each other. This will give (more or less) the same result if you draw the random paths of a lot of people taking a random walk in the park one one single day.

Deschele Schilder
  • 1
0

Ergodic hypothesis is not contradiction to notion of equilibrium. In fact, it is the pillar of equilibrium statistical physics. The measurable quantities(like pressure, temperature,..) which are evaluated at the equilibrium assume that ergodic hypothesis is valid at equilibrium. One starts with finding the partition function(Z) $$ Z = \sum_{\epsilon} e^{-\beta \epsilon} $$ and later on one finds the observable quantities (here we assumed discrete energy levels and we have taken Canonical Ensemble). Here we have summed over all the possible values of energy. In other words, ergodic hypothesis can also be stated as "at equilibrium, time average is equivalent to ensemble average" which we have used while finding the canonical ensemble partition function. We have summed over all the values of microstates as ergodic hypothesis states that " all accessible microstates are equiprobable at equilibrium ".

Hemanshu Dua
  • 1