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I am not physicist and this question may seem trivial. But I understand that in the equilibrium the magnitudes such as temperature or volumen do not vary. Is the same for entropy? My logical says that it should not, and here is why:

Entropy is proportional to the number of microstates of a macrostate In the equilibrium, a system is one of the macrostates with more microstates -> But not necessarily in the one with more of them. The system may visit macrostates with a slightly different number of microstates If this is the case, entropy would be varying (although maybe not too much, but it would be)

Am I wrong here or the reasoning is good?

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  • $\begingroup$ There are different (equivalent) formulations of the second law of thermodynamics. Planck has stated the following version: "Every process occurring in nature proceeds in the sense in which the sum of the entropies of all bodies taking part in the process is increased. In the limit, i.e. for reversible processes, the sum of the entropies remains unchanged.". The important insight here is, that the entropy increase for reversible processes is zero. A system in equilibrium is the most trivial reversible system (the reversed process is identical to the original one). $\endgroup$ – CuriousOne Aug 28 '14 at 19:57
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Equilibrium is defined in the original notions of thermodynamics as the asymptotic static state. I.e., by this definition no macroscopical quantity varies in equilibrium.

Statistical physics however tells us that the system varies in a certain "random walk" around all the possible states and never stops. We just cannot distinguish most of these states macroscopically. But once the system enters the microstate which is one of the overwhelmingly dominant set of microstates which we observe as "equilibrium", it is very likely it will choose another of the "equilibrium" microstates in it's next random step. So we observe it to stay there and to be in equilibrium without change.

But does entropy change? Entropy is strictly a property of the macrostate, i.e. the rough observed state - it is just a measure of microstates giving the same macrostate. There is a single macrostate called "equilibrium". So entropy in a fixed macrostate cannot vary by definition. In this sense the answer is a very strict no.

But as mentioned in the second paragraph, the system never stops it's random walk. So the system actually does fluctuate around equilibrium even macroscopically. Since it does reach a different macrostate, it will necessarilly vary in entropy. Why? Because the equilibrium is a local maximum in entropy in the macrostate space, so any adjoint macrostate will strictly have a different entropy. Hence, we can even say what the direction of the variations will be - the entropy will always shortly fluctuate into values slightly smaller than in equilibrium.


EDIT: In this discussion, I assume that we are able to observe distinct fluctuations in the macroscopic parameters such as internal energy and volume, and entropy is then defined using the phase space volume $\Omega_c$ (number of microstates) constrained by the immediate values of the macroscopic parameters: $$S = k_B \log \Omega_c$$ But we do not observe internal energy directly, so we can think about an isolated system with a fixed volume $V$, start out of equilibrium and observe fluctuations in pressure. There we would observe exactly the mentioned effects.

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  • $\begingroup$ Even in case of a statistical interpretation, entropy is not a momentary quantity, but defined as an average over ALL microstates of the system (or whatever ensemble we chose for a particular application). As such, it is a constant for a system in equilibrium, that characterizes all relevant states of the system, and not the actual state that it is in at any given moment. $\endgroup$ – CuriousOne Aug 28 '14 at 21:03
  • $\begingroup$ Well, by the axiomatic statistical definition you are right. But consider the following: macroscopical parameters fluctuate, we do not know what kind of ensemble this is. But they will always fluctuate around maximal entropy in the macroscopic parameter space no matter what constraint is put on them. Any process of reaching equilibrium in any case when seen only in the macroscopic parameters will be seen as a convergence to a formally maximal entropy of the system with the given constraints. I guess I should clarify this in the answer. $\endgroup$ – Void Aug 28 '14 at 22:10
  • $\begingroup$ I just realized I let myself get confused. Even by statistical definition there would be no problem with the argument. In both the canonical and grandcanonical ensemble the entropy is not sharp and not defined by the averaging over all states - the free energy and grand potential are. In any good sense entropy is defined exactly as I have used it in the question in canonical and grandcanonical ensembles. $\endgroup$ – Void Aug 28 '14 at 22:46
  • $\begingroup$ I think you are confusing thermodynamics with non-equilibrium thermodynamics, where one TRIES to model these quantities as dynamic variables on finite volume elements, using ad-hoc assumptions (usually linear functions) about heat flow, mixing etc., but those assumptions are are much harder to justify than even the ensemble averaging assumptions of statistical mechanics. Truth to be told, as the problems with fully developed turbulence and phase transitions show, non-equilibrium thermodynamics walks an extremely narrow band of potential usefulness. $\endgroup$ – CuriousOne Aug 28 '14 at 23:43
  • $\begingroup$ Then please tell me how do you think entropy is defined in a canonical and grandcanonical ensemble. $\endgroup$ – Void Aug 29 '14 at 7:24
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I am not entirely happy with the given answer so I will provide mine. First of all there is not just a single entropy to talk about, even at equilibrium. So it is misleading to talk about a single entropy.

To give an example of what I mean, let us consider an ideal gas in a box with fixed energy $E$, number of particles $N$ and volume $V$.

What equilibrium statistical mechanics tells us is that the probability distribution of the micro states of the ideal gas in that box does not vary with time. Incidentally, one can also associate a non-varying entropy to that probability distribution which is the total entropy of the system. It basically tells you how much you know nothing about the exact state of the gas once it has equilibrated.

Now, because you know nothing about the gas besides that it has fixed $(E,N,V)$ does not mean you cannot ask any question about its macroscopic behaviour at equilibrium. For instance, most people imagine that when a gas is at equilibrium then it fills uniformly whatever container it is trapped in. But because all micro states are equally likely, that is not necessarily so.

To see this, let us split the box in two equal parts $V_1 = V_2 = V/2$. We can ask ourselves what is the entropy of having a gas in a macroscopic configuration with $N_1$ particles in $V_1$ and $N-N_1$ in $V_2$. By definition, this entropy will be $S(N_1) = k_B \ln \Omega(N_1)$ where \begin{equation} \Omega(N_1) = \frac{N!}{N_1!(N-N_1)!}\left(\frac{V}{2}\right)^{N}E^{3N/2-1} \end{equation} Now, within the Stirling approximation, this expression will read: \begin{equation} S(N_1) = k_B \left\lbrace -N_1 \ln N_1 - (N-N_1)\ln (N-N_1) + constant \right\rbrace \end{equation} If you plot this function you get the following graph:

Configuration entropy associated to having $N_1$ particles in a half of the box (up to an additive constant and in units of $k_B$).

It is easy to see that this graph has a maximum at $N_1 = N_2 = N/2$. So what we find is that, in a gas at equilibrium, having the same number of particles in both halves of the containing box is the macroscopic configuration with the highest configurational entropy. However, all the other configurations with $N_1 \neq N/2$ are also possible it is just that their odds for happening goes as $e^{S(N_1)-S(N)}$ and this function is very strongly peaked at $N_1 = N/2$.

Overall, for any macroscopic configuration other than that of the imposed constraints themselves, the entropy can always vary at equilibrium and will do so by fluctuating about the most probable value.

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