I have the following problem.

Given a finite system, let us say a box of side $L$ containing $N$ "ideal gas" non-interacting particles, properties such as the entropy can be defined, in this case, $S = - \log ((L^3)^N/N!)$ (I am neglecting the momentum contribution, just configurational entropy).

In the thermodynamic limit, as $L \to \infty$ and $N \to \infty$ such that the density $ \rho = \frac{L^3}{N}$ is constant, all works perfectly.

But how to handle a system where $N$ is constant, and $L \to \infty$ ?

Take the simplest case, $N = 1$. The entropy of one particle in an infinite box diverges due to the volume term in the logarithm, while the density $\to 0$.

This seems absurd to me: a syste with "infinite entropy" and nihil density. Furthermore, if anybody mentioned a zero density case, can this refer to an arbitrary, yet finite number of particles in an "infinite" volume?

I must be missing some point.

  • $\begingroup$ How did you come up with that formula for entropy? It doesn't look right. Entropy is an extensive quantity $\endgroup$ – dedekindCuttage Feb 24 at 14:41
  • $\begingroup$ Ok I will address your point by editing the question $\endgroup$ – Smerdjakov Feb 24 at 16:10

$N=1$ is not a thermodynamic system, but, indeed, if $N$ is large but fixed and we take the volume to infinity then the entropy per particle $S/N$ will diverge. This is simply configurational entropy, so there is no question this result is reliable. Note that

1) The entropy per particle diverges very slowly. In order to increase $S/N$ by 10$k_B$ we have to increase the volume by $e^{10}$.

2) A very dilute gas equilibrates very slowly. To reach the correct equilibrium entropy from a generic initial state will take a verly long time (the equilibration time diverges at least as a power of the volume).

  • $\begingroup$ thanks for your reply...Could you please clarify what you meant by, "this is simply configurational entropy, so there is no question this result is reliable"? $\endgroup$ – Smerdjakov Feb 24 at 17:29
  • $\begingroup$ $S=k_B\log(\Omega)$ as engraved in Boltzmann's tomb stone, and clearly $\Omega\sim V^N$ $\endgroup$ – Thomas Feb 24 at 17:39
  • $\begingroup$ Ok I just wanted to check your statement meant, "the expression is so straightforward that there are no diubts about its correctness", and not " usingvobly the configurational entropy makes the physical representation so poor that rhis result is only formal at best", if you see what i mean... $\endgroup$ – Smerdjakov Feb 24 at 17:51
  • $\begingroup$ Yes, that's what I had in mind. $\endgroup$ – Thomas Feb 25 at 3:11

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