# Small $N$ ideal gas entropy and extensive entropy: Finite $N$ Sackur-Tetrode and Gibbs Paradox

In the standard derivation for Sackur-Tetrode, 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. The entropy of an monoatomic gas without the large $$N$$ assumption is exactly (We are also assuming 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 an gaussian integral.): $$S= Nk\left[\log(n_Q V)+\frac{3}{2}\right]-k \log N!$$ 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 Gibbs paradox isn't fully resolved, or we don't 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?

• I don't quite get your question. Are you asking if there is a relation like the Sackur-Tetrode equation at small $N$?
– WwW
Oct 9, 2020 at 16:41
• 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. Oct 9, 2020 at 17:37

• 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. Oct 12, 2020 at 19:54
• 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.? Oct 12, 2020 at 20:19
• Oh, this is right. I think the difference is just the conditional entropy of finding the system in such a $N+N$ state. Oct 12, 2020 at 20:22
• 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.
• 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, $S(2N,2V)-2 S(N,V)= 2Nk \log 2 - k \log (2N)! + 2 k \log N! = k \log 2^{2N} - k \log \binom{2N}{N}$. Oct 14, 2020 at 2:27