1
$\begingroup$

I am taking a thermodynamics course and we have talked about stat mech and the number of possible combinations of $N$ indistinguishable particles given degeneracy $g$.

We stated that for the distinguishable case, there would be g^(N) possible configurations. Cool, that makes sense. Then to consider the case of indistinguishable particles, we just divide out by the N! equivalent arrangements. Word, totally on board... BUT the resulting form:

(g^(N))/N!

gives me some grief in that you can arrive at non-integer solutions. For example g = 3 , N = 2. Can anybody shed some light on the issue?

$\endgroup$
  • $\begingroup$ Furthermore, I reckon we are considering bosons in this instance. A distribution of fermions would be much more intuitive, as I believe it'd just be g P n ( i.e. g! / N! (g-N)! ) $\endgroup$ – Sam.Robison Apr 24 at 16:29
  • $\begingroup$ For Bosons, it is $^{n+g-1} C_{g-1}$. It is basically an arrangement of g-1 partitions and n particles arranged without repetitions. $\endgroup$ – exp ikx Apr 24 at 16:35
  • $\begingroup$ @Sam.Robison, exp ikx is absolutely right, in your notation it will be $(N+g-1)!/(N!(g-1)!)$. This is an exact formula that goes into an approximate $g^N/N!$ valid when $N<<g$ $\endgroup$ – Aleksey Druggist Apr 26 at 13:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.