I was reading Statistical Physics by F. Mandl. And in Chapter 2.5 pg.61 he derived the general definition of entropy(gibbs entropy forumla). But there is a step that I do not understand. This is the flow of thought:

For a macrostate consisting of an ensemble of a large number, $v$, of identical systems. The number of systems being in state $r$ is $v_r$. Therefore there is the constraint: $$ \sum^{N}_{i}v_i=v $$ In total there are $N$ number of states

Mandl goes on to state that the statistical weight $\Omega_v$ of the ensemble when $v_1$ systems are in state 1, $v_2$ systems are in state 2, and so on and so forth all the way to $v_N$ systems are in state $N$..is given by: $$ \Omega_v = \frac{v!}{v_{1}!v_{2}!...v_{N}!} $$

This is what I do not understand. If the statistical weight of the systems is the number of ways this distribution of states can be realised, then shouldn't $\Omega_v = \Omega_{v_1}\Omega_{v_2}...\Omega_{v_N}$? i.e. $\Omega_v$ should be given by the following formula: $$\Omega_v=\frac{(v!)^N}{(v_{1}!v_{2}!...v_{N}!)(v-v_{1})!(v-v_{2})!...(v-v_{N})!}$$ since $$\Omega_{v_i} = {v \choose v_i} = \frac{v!}{v_i!(v-v_i)!}$$ Furthermore, even by approximating $v_i << v$, i.e. $N$ is very large, the approximated $\Omega_v$ does not converge to the one written to mandl, instead it converges to $\Omega_v = \frac{v!}{v_1!v_2!...v_N!\cdot v}$.

Please help me for I might have made wrong assumptions about the workings of the systems. Any advice is greatly appreciated!


The $\Omega_v$ are multinomial coefficients. They are a direct extension of the binomial coefficients ${{v}\choose{v_1}}$.

The binomial coefficients count the number of ways to put $v_1$ and $(v-v_1)$ out of $v$ objects into two distinct bins (states).

The multinomials count the number of ways to put $v_1, v_2, \dots v_N$ out of $v$ object ($\sum v_i = v$) into $N$ distinct bins (states). That is exactly what you want to do here.

Here is a more constructive way to see this:

  1. Given $v$ systems. How many ways are there to chose $v_1$ of them to be in state 1? $$ \Omega_{1}={{v}\choose{v_1}}$$
  2. For the remaining $v-v_1$, how many ways are there to pick $v_2$ of them to be in state 2? $$ \Omega_2={{v-v_1}\choose{v_2}} $$
  3. Iterate until $v_N$. $$\Omega_v = \prod_i \Omega_i = \frac{v!(v-v_1)!(v-v_1-v_2)!\cdots}{v_1!(v-v_1!)\cdot v_2(v-v1-v2)!\cdots} = \frac{v!}{v_1!v_2!\cdots v_N!} $$
  • $\begingroup$ @Tian Very good. If you find my answer satisfactory, please accept it by clicking on the check-mark. $\endgroup$
    – Nephente
    Feb 12 '17 at 10:56

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.