Gibbs entropy, Clausius' entropy and irreversibility

Question

I have a bunch of doubts and confusions on the concept of entropy which have been bothering me for a while now. The most important ones are of a more technical nature, arisen from the reading of this and Jaynes' paper "Gibbs vs Boltzmann entropies". Although they seem like great texts, they leave me confused when they argue that the Gibbs entropy remains constant and that this is precisely what one needs to prove in the Second Law. I quote, from the first link:

we return to the specific case of a gas of N particles, this time confined to one side of a box containing a removable partition. We suppose that the initial state is such that we can descnbe it using the canonical probability distribution. From our earlier discussion we can then say that the Gibbs entropy SG is maximized and equal to the experimental entropy SE.

We now suppose that the partition is opened and the atoms occupy the whole box. We wait until the state variables stop changing, so in that sense the system is in equilibrium and a new experimental entropy S'E can be defined. Also, all the motions of the gas molecules are Hamiltonian, so that the Gibbs entropy S'G has not changed: S'G = SG

The probability distribution of the N particles is no longer the canonical one, however, because of the (very subtle!) correlations it contains reflecting the fact that the molecules were originally on one side of the partition. This means that the Gibbs entropy S'G is now in general less than the maximum attainable for the new values of the state variables, which is in turn equal to the new experimental entropy. So

SE = SG = S'G ≤ S'E

Well, I can say they totally lost me here. After the expansion, the system reaches an equilibrium. We shouldn't care about how we reached that equilibrium, so I would think that maximizing $S_G= -K_B \int \rho \log \rho d \mu$ -with $\rho$ the joint probability distribution of the position and momentum of the $N$ particles and with the restriction that the average energy is, say, $U$- should give again the probability distribution $\rho$ in the new equilibrium. So, I defintely don't understand at all why they say that "The probability distribution of the N particles is no longer the canonical one," neither do I understand the statement "because of the (very subtle!) correlations it contains reflecting the fact that the molecules were originally on one side of the partition," since I believe it is not important at all how we reached the equilibrium! If this was the case, why could we prove that in the first case (when all the molecules are in one side of the box) the Gibbs entropy coincided with the "experimental" one (I, by the way, don't know exactly what they mean by "experimental" entropy; do they mean Clausius'?). Didn't Jaynes prove that Gibbs entropy coincide with Clausius'? (I got that from his paper, at least). But how can Gibbs entropy remain constant if the entropy "MUST" increase?

Jaynes, in his paper, writes things like $(S_G)_{2}-(S_G)_{1},$ so that changes in the Gibbs entropy must be something meaninful, despite being so clear that the hamiltonian evolution leaves $\rho$ the same.

Well, I guess it's being really hard to explain accurately the nature of my confusions, but hopefully someone around here has struggled with similar issues and can give an enligthening clarification.

Thank you very much in advance.

EXPANDING THE ORIGINAL QUESTION: Let me follow the notation from Jaynes' paper, which I linked above. If a let the gas go through a free adiabatic expansion, since the evolution is hamiltonian, it is clear that $\frac{dW_N}{dt}$ obeys Liouville equation, but since $\{ e^{-\beta H}, H \}=0,$ it is clear that the $W_N$ remains constant and thus so do the Gibbs entropy $S_G.$ However, at the end of section IV in Jaynes' paper, he states

If the time-developed distribution function $W_N(t^{\prime}$...

And then I don't know what's going on anymore..! Does the $W_N$ change in time or not?

And, if in a thermal equilibrium we do not use the Canonical Ensemble, which ensemble are we using instead? How is the distribution function? What is the mathematical expresion of the new macroscopic restrictions we should add when maximizing the entropy functional to derive that new probability distribution?

Answer 1

You say:

We shouldn't care about how we reached that equilibrium

In fact the entire point of the example is that we do. I shall try to explain why.

Jaynes and Gull both work in the framework of Bayesian inference (I can recommend the introductory text: http://www.amazon.co.uk/Data-Analysis-A-Bayesian-Tutorial/dp/0198568320. The title may seem unrelated to thermodynamics, but is actually at the heart of science itself). One important concept they use is the Principle of Maximum Entropy. It states essentially what you said in your question:

so I would think that maximizing $ S_G=−K_B∫\rho \log\rho\: \mathrm d\mu $ - with ρ the joint probability distribution of the position and momentum of the N particles and with the restriction that the average energy is, say, U - should give again the probability distribution ρ in the new equilibrium.

This yields the canonical ensemble. Now the assumption behind this is that there is no prior information that we have about the system. But in the case of the box we DO have prior information about the system, namely that the particles used to be on the right of the box. So the particles are in a different ensemble. This ensemble could probably be calculated, but it would be complicated and is not necessary for the sake of his argument.

About your questions regarding the "experimental entropy": What they mean by that is the entropy that is defined by:

$$ \Delta S_E = \int_\rm{reversible} \frac{\partial Q}{T} $$

So yes, I guess that is what people also call "Clausius' entropy". What Jaynes proved is that $ S_G $ and $ S_E $ (choosing the right constant offset for $ S_E $) coincide for the canonical ensemble. So in the argument above we have noted that the ensemble changes from canonical to something else. The new ensemble will have same $ S_G $ (since it was shown to be constant under the equations of motion) but $ S_E $ will have increased. This is then a restatement of the second law of thermodynamics (for systems starting in the canonical ensemble. I have struggled to find a prove for systems starting in other ensembles, would be happy if someone could link me.)

So that should also answer your question about how $ S_G $ staying constant is consistent with the second law.

Answer 2

There is great confusion about the Liouville equation and what it represents –or doesn't.

It represents a mechanical system that that undergoes mechanical processes that are strictly reversible, that is, processes that are isentropic.

It does not (in fact, it cannot) represent a thermodynamic system that exchanges heat, heat is not a concept that mechanics understands. It cannot represent the evolution of a system whose entropy changes. And it does not/cannot represent relaxation to equilibrium.

Many have argued, incorrectly, that since the Gibbs entropy, $-\int \rho(\Gamma)\log\rho(\Gamma)) d\Gamma$, remains unchanged in the Liouville equation, this must mean that the Gibbs entropy is wrong, or incomplete, or not exactly the Clausius entropy and so on. This argument blames Liouville's inadequacies on Gibbs.

The Liouville equation is inadequate because it applies deterministic mechanics, and as we know, thermodynamics requires statistical mechanics. The determinism that plagues Liouville arises from the fact that it treats the boundaries of the system as deterministic. As Gibbs put it:

The forces are supposed to be determined for every system by the same law, being functions of the coordinates of the system $q_1,\cdots q_n$, either alone or with the coordinates $a_1$, $a_2$, etc. of certain external bodies. It is not necessary that they should be derivable from a force-function. The external coordinates $a_1$, $a_2$, etc. may vary with the time, but at any given time have fixed values. In this they differ from the internal coordinates $q_1,\cdots q_n$, which at the same time have different values in the different systems considered. (Gibbs, Elementary Principles in Statistical Mechanics, p. 5).

The "external coordinates" represent interactions with the surroundings and are treated as deterministic functions of time. Of course the surroundings are made of particles whose motion depends on degrees of freedom that are not included in the Liouville equation. This is the approximation in the Liouville treatment that robs it of the ability to exhibit thermodynamic behavior.

To return to Jaynes's view, it is our ignorance about the precise state of the surroundings that increases entropy: a real system evolves differently from what Liouville predicts because we are ignorant of information that is necessary to predict the future state. We end up with more microstates than we thought we had at $t=0$, because we did not know the number of microstates available in the surroundings of the system. These microstates "seep" into the system in the form of entropy generation.