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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 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 at 1:52
    
I second the recommendation for Jaynes' beautiful paper . It can be found here: bayes.wustl.edu/etj/articles/gibbs.paradox.pdf . –  Yvan Velenik Mar 10 at 7:56
    
@YvanVelenik I recommend all of the papers in the folder bayes.wustl.edu/etj/node1.html Jaynes was a truly remarkable philosopher. –  WetSavannaAnimal aka Rod Vance Mar 10 at 21:48

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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: http://arxiv.org/abs/1012.4111

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 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 at 1:56

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