The Gibbs paradox

The result of the previous section is correct for particles that all have their own identities, and indeed it was originally obtained by Gibbs in this form. However, he quickly realised that expressed in this way a number of physical properties of a gas were coming out wrong and so Eq.  is not quite right. The issue is the same as the one discussed Sec.  for identical particles and can be traced to their de-localisation and their equal access to the whole volume $V$. This implies that having particle 1 in one single particle microstate and particle 2 in another, or the other way round, should not be counted as two separate microstates for the whole gas, because there is no way to distinguish between them. One could argue that in classical physics it is possible, at least in principle, to follow the motion of each particle and so one should be able to tell which molecule is in which position and which velocity once an initial configuration has been observed. But if one were simply presented with one particular snapshot configuration at a later time it would be impossible to tell which molecule is which and any permutation of those molecules would produce an identical configuration.^4.2 The probability in Eq.  must take this issue into account: if the molecules are all in different single particle microstates there are $N!$ permutations of them that leave the microstate of the whole gas unchanged. Each of these terms needs to counted only once, which means that the correct partition faction is:

$$Z = \frac{1}{N!}(Z_1)^N.$$ (4.8)

This issue is known as the Gibbs paradox, and is discussed with various level of emphasis by different authors, with disparate takes on what counts as distinguishable in classical mechanics, at least in principles. However, they eventually all agree (and must do so to get the correct physics) that the $N!$ factor is required (see also discussion in Sec. ).