# An issue with thermodynamic properties of infinite systems

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.

• How did you come up with that formula for entropy? It doesn't look right. Entropy is an extensive quantity – dedekindCuttage Feb 24 '19 at 14:41
• Ok I will address your point by editing the question – Smerdjakov Feb 24 '19 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 $$e^{10}$$.
• $S=k_B\log(\Omega)$ as engraved in Boltzmann's tomb stone, and clearly $\Omega\sim V^N$ – Thomas Feb 24 '19 at 17:39