2
$\begingroup$

I know that the Stirling approximation states that

$$\ln(x!) \approx x\ln x - x$$

However, in some derivations, this is also applied to what looks like a sum of factorial terms. For example, here, which is similar to the derivation in other places, one states that for $W=\frac{N !}{\sum_{i}^{r} n_{i} !}$, we have

$$\ln W=N \ln N-N-\sum_{i}^{r} n_{i} \ln n_{i}-n_{i}$$

How does one show the Stirling approximation for the denominator term of $W$?

$\endgroup$
  • 1
    $\begingroup$ Hint $\ln(a\cdot b)=\ln(a)+\ln(b)$ and $\ln(\frac{a}{b})=\ln(a)-\ln(b)$ $\endgroup$ – Peter Jul 29 at 18:35
  • $\begingroup$ In the linked paper, they write : the number of microstates is given by the following equation $W=\frac {N!}{n_0! \,n_1!\,n_2! \cdots}$ $\endgroup$ – Claude Leibovici Jul 30 at 5:16
5
$\begingroup$

The formula is wrong.

If the denominator had product instead of sum, the formula would be correct.

It doesn't, so it isn't.

$\endgroup$

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.