For atoms that are fixed in a crystal with the following assumptions:

  1. There are quantised energy levels with energy $E_i$ for each atom.
  2. Each state has a distinct energy $E_i$.

Such that..

$$N = \sum_i n_i$$ $$U = \sum_i E_i n_i$$

Where $N$ and $U$ are the total number of particles and total energy in the system respectively.

Why is $\Omega$ then..

$$\Omega = \frac{N!}{\prod_i n_i!}$$

I feel it must be derived from the combinations formula but I don't know how exactly. It's $N$ choose what and why?

  • $\begingroup$ Are you sure this is for distinguishable particles? $\endgroup$ – nasu Apr 8 at 16:30

Consider that we randomly mix up all N particles, but consider fixed partitions that divide the particles into groups of $n_i$.

There are N! ways of mixing the particles up. But, if we mix up particles within a group, this does not correspond to a different macrostate.

So we overcount. By how much? For each group, there are $n_i!$ ways to mix the group's members, yielding a different microstate.

So we've overcounted by $\prod_i n_i!$. This explains the denominator.

  • $\begingroup$ Just for extra-extra clarity, why does mixing the particles up within a group not correspond to a different macrostate? $\endgroup$ – sangstar Apr 14 at 16:58
  • $\begingroup$ Because particles (or quanta here) are indistinguishable. $\endgroup$ – thedoctar Apr 15 at 6:42

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