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This question is essentially a duplicate of Gibbs Paradox - why should the change in entropy be zero?. The question concerns the following situation: I have some gas of identical particles and they are in a box which has been partitioned into two halves by a removable divider. Now the question is "If I remove the divider, why should the change in entropy be 0?" The standard answer is given in the above-linked-to question. The main idea of the answer is that the particles are supposed to be considered indistinguishible. If you treat the particles this way, you find that there is no change in entropy when the partition is removed.

I understand this answer, and I suppose it should work for identical atoms, but you can easily imagine a situation where you have a collection of objects which are distinguishible, say some nanoparticles with differing numbers of constituent atoms.

Furthermore, you should be able to treat a gas of identical particles as classical distinguishable bodies, and still get the right answer from statistical mechanics. I would say this is a good check of understanding of statistical mechanics.

If this isn't enough motivation to come up with an alternate resolution to the paradox, consider this paradox, which requires essentially the same resolution that the gibbs paradox does: Imagine two systems initially in thermal contact and in thermal equilibrium. Now suppose they are separated. Here we might think the entropy will decrease in this process because in the initial configuration the energy of both systems were allowed to change as long as their sum remained constant. After separation, the systems' energies are fixed at some value. Clearly the set of final states is a subset of the set of initial states, so the entropy has decreased. This is the analogy of the gibbs paradox where instead of the systems exchanging particle number, they are exchanging energy. I would expect it to have essentially the same resolution.

So my question is, "Why should the change in entropy be zero, even in the particles are distinguishible?"

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" Imagine two systems initially in thermal contact and in thermal equilibrium. Now suppose they are separated.... Clearly the set of final states is a subset of the set of initial states, so the entropy has decreased." If the systems are only in thermal contact and keep their particles, thermodynamic entropy does not depend on whether there is thermal contact or not; this entropy is additive. This is because there is no change in macroscopic variables due to thermal contact - volume, energy, number of particles stay the same for both systems. – Ján Lalinský Mar 9 '14 at 23:59
Please read a paper called "the Gibbs Paradox" by Edwin Jaynes. It is available online, but I can't easily post a link because posting from a phone. It resolves the paradox in a very nice way, and from the arguments in your question I think you will like it. – Nathaniel Mar 10 '14 at 1:52
I second the recommendation for Jaynes' beautiful paper . It can be found here: . – Yvan Velenik Mar 10 '14 at 7:56
@YvanVelenik I recommend all of the papers in the folder Jaynes was a truly remarkable philosopher. – WetSavannaAnimal aka Rod Vance Mar 10 '14 at 21:48

2 Answers 2

up vote 7 down vote accepted

So my question is, "Why should the change in entropy be zero, even if the particles are distinguishable?"

In statistical physics, entropy can be defined in many different ways.

One possibility is to define it as log of the accessible phase space, given the macroscopic constraints (volume). Such entropy is not a homogeneous function of energy, volume and the number of particles. If the wall is removed, this entropy increases. See the paper by Veerstegh and Dieks:

Another way is to define it as log of the accessible phase space divided by the number of permutations of the particles. This entropy is a homogeneous function of energy, volume and the number of particles. If the wall is removed, this entropy stays the same.

Both these definitions are valid, and lead to different concepts of entropy. There is no "the correct" entropy. However, for convenience it is often preferable to use the second definition. The entropy of two interacting sub-systems in equilibrium is then sum of their corresponding entropies and removal of the wall does not change the entropy, which are very practical conventions expected already from classical thermodynamics.

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Thank you for the paper reference. I read it and I think its explanation is mostly satisfying. To sum it up for other people, he basically makes a distinction between stat mech entropy and thermodynamic entropy. He says that the thermodynamic entropy doesn't change when you add or remove the partition, but the stat mech entropy, defined as usual by the log of the number of accessible states, does change. However, he says no predictions from such a stat mech theory would be different from the usual predictions. – NowIGetToLearnWhatAHeadIs Mar 10 '14 at 1:56
So, for example, you can't make a perpetual motion machine or anything even though the stat mech entropy does decrease. He then says that since it doesn't make a difference, you might as well include an N! in the definition of the stat mech entropy so that the stat mech entropy and the thermodynamic entropy will be really the same thing. Here the motivation is not that the particles are indistinghishable, but that it is convenient to have the two notions of entropy to be the same. – NowIGetToLearnWhatAHeadIs Mar 10 '14 at 1:56

It is true that there is no correct a priori way of deriving the "correct" entropy as the notion of "correct" will depend upon what we mean by this word.

However, if statistical mechanics introduces a quantity that has the same name as the most important quantity of thermodynamics, I suspect it has, at the very least, to carry a meaning very close to the original object. In particular, in a mixing problem, the entropy change has to be related to the amount of work to provide to separate back two species into different compartments.

It is perfectly valid to say, as Veerstegh and Dieks do, that consistency of statistical mechanics is not lost if one "counts" correctly the number of states that matter for the macro-observables even if the statistical entropy is shown to increase but this seems to me a very convoluted way of solving the problem. It is even more so that this correct "weighting" of the probabilities relies on exactly the same argument as the one that consists in diving the entropy by N!.

Even more problematic is that the logarithm of the partition functions thus defined loses both its interpretation as being the generating functional of the cumulants of the distribution and its interpretation as being the right thermodynamic potential in the statistical ensemble under study.

In a recent paper, we have proposed to look at a mixing problem where the quantum indistinguishability argument is certainly failing, namely, the problem of mixing between two fully polydisperse systems with possibly different compositions.

We found that even in such a case, an N! was appearing naturally in the expression of the entropy (or free energy) of a polydisperse system together with something that we termed the composition entropy of the system.

Incidentally, we found that upon mixing two polydisperse systems, the entropy difference actually measures a mathematical distance (as measured by the so called Jensen-Shannon entropy) between the probability distributions characterizing the compositions in each of the two compartments. If the compositions are the same, the distance between them is zero (by definition of a metric) and there cannot be any entropy change upon mixing while, if the compositions differ then this mathematical distance increases continuously until it reaches an upper bound that is $\ln 2$ corresponding to a genuine binary mixture.

At the end, although we use a different "philosophical approach" to the problem, we reach essentially the same conclusion as Veerstegh and Dieks i.e. that there is no need for indistinguishability and that furthermore distinguishability is much more ubiquitous in Nature than its nemesis indistinguishability and so it is not satisfactory to rely on an argument solely based on quantum indistinguishability.

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