In the standard derivation of the Sackur-Tetrode equation, the accounting for the indistinguishability of ideal gas molecules adds an extra factor of $N!$ in the partition function. This is usually approximated by Stirling's approximation.

Let us assume that the volume of the box is very large so that the energy spacing is very small, so we can replace the sum in the partition function with a Gaussian integral. Then the entropy of a monoatomic gas without the large $N$ assumption is exactly $$S= Nk\left[\log(n_Q V)+\frac{3}{2}\right]-k \log N!,$$ where $n_Q=\left(2\pi m k T/h\right)^{3/2}$ is an intensive quantity.

We can expand the Stirling series, $$S= Nk\left[\log(n_Q V)+\frac{3}{2}\right]-k \left(N \log N-N +\log\sqrt{2\pi N}+\mathcal O\left(\frac{1}{N}\right)\right).$$ The normal resolution of the Gibbs paradox is given by the truncation of the entropy at the leading order, $$S= Nk\left[\log(n_Q)+\log \frac V N+\frac{5}{2}\right]+ k \log\sqrt{2\pi N}+\mathcal O\left(\frac{1}{N}\right),$$ for which the term in the square brackets is extensive as one scales $N$ and $V$ simultaneously. It is said that this is how indistinguishability resolves the Gibbs paradox, so that entropy remains extensive. However, it is manifest that the subleading corrections do not scale properly.

What happens to the smaller terms at finite $N$? Does this mean the Gibbs paradox isn't fully resolved, or don't we have extensive entropy? The more physical question might be, if we did an experiment with extremely dilute gases where $N$ is small, can we detect a non-extensivity? If not, where does this calculation break down?

  • $\begingroup$ I don't quite get your question. Are you asking if there is a relation like the Sackur-Tetrode equation at small $N$? $\endgroup$
    – WwW
    Oct 9, 2020 at 16:41
  • $\begingroup$ No, the first equation above is already the exact analogue for Sackur-Tetrode for small $N$. The issue is that according to this formula entropy of the gas is not extensive, e.g. if you double $N$ and $V$, $S$ doesn't double. In the standard textbook derivations of Sackur-Tetrode, only the leading terms of the Stirling approximation are kept, which are indeed extensive. This approximation is used to explain the Gibbs paradox. The explanation does not seem to hold at small $N$ though. $\endgroup$
    – NoWayHaze
    Oct 9, 2020 at 17:37

1 Answer 1


The answer is given in the third remark at the end of Section 3 of my paper "Demonstration and resolution of the Gibbs paradox of the first kind" Eur. J. Phys. 35 (2014) 015023 (freely available at arXiv).

In short, let's assume you combine two subsystems S1 and S2, each with N indistinguishable particles, by removing a partition between them. As a result, you get a new System S with 2N particles. The entropy of S is a little bit larger than the sum of the entropies of S1 and S2, because, after the removal of the partition, there is an uncertainty about how many particles are in each of the two subvolumes. (For example, there could be N+1 particles in the first subvolume and N-1 in the second. Before the removal of the partition there were, per definition, exactly N particles in each subvolume.) For this reason, the entropy of an ideal gas of indistinguishable particles (as a function of T, V and N) is only approximately extensive, but not exactly.

  • $\begingroup$ I see, you're basically claiming that the difference is the entropy for each half mixing into the other half. The precise difference is $k \log (2^N)-k\log \binom{2N}{N}$ which you say goes to zero by Stirling's. Thanks, this resolves the first subleading correction to the entropy. However, you still invoked Stirling's so your argument doesn't account for the sub-subleading corrections, i.e. the difference is still not precisely zero for small $N$. For example when $N=1$, the difference is $k\log 2$. So my question isn't fully resolved. $\endgroup$
    – NoWayHaze
    Oct 12, 2020 at 19:54
  • $\begingroup$ If we calculate the Shannon entropy of the binomial distribution of size $2N$, $\Delta S=-k \sum p \log p = -k\sum_m \binom{2N}{N}2^{-2N} \log \left(\binom{2N}{N} 2^{-2N}\right)$, the dominant term is indeed the entropy difference we are seeking. So the "surprise" of finding the $2N$ system split perfectly into $N+N$ gives is the entropy, so is the claim then the residual difference $k\log 2^{2N}-k\log \binom{2N}{N}$ is the surprise of finding the system in other configurations like $(N+1)+(N-1)$ etc.? $\endgroup$
    – NoWayHaze
    Oct 12, 2020 at 20:19
  • $\begingroup$ Oh, this is right. I think the difference is just the conditional entropy of finding the system in such a $N+N$ state. $\endgroup$
    – NoWayHaze
    Oct 12, 2020 at 20:22
  • $\begingroup$ Your last (third) comment suggests that you got it and that your first and second comment are no longer relevant. Nevertheless, regarding these two obsolete comments, I want to remark that the term ${2N}\choose{N}$ only appears in expressions concerning distinguishable particles, such as in my cited paper (where "distinguishable" means that interchanging two particles leads to a new microstate). However, your question refers to indistinguishable particles. $\endgroup$
    – HjP
    Oct 13, 2020 at 19:22
  • $\begingroup$ Yes, I agree your paper uses distinguishable particles, but the residual difference I was referring to is still has a $\binom{2N}{N}$ for indistinguishable particles. Taking the first equation of my original post as a given, $S(2N,2V)-2 S(N,V)= 2Nk \log 2 - k \log (2N)! + 2 k \log N! = k \log 2^{2N} - k \log \binom{2N}{N}$. $\endgroup$
    – NoWayHaze
    Oct 14, 2020 at 2:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.