1
$\begingroup$

I got an expression for free energy F in a book as: $$F(N_1) = \frac{N_1N_2}{N}W-kT(\log N!-\log N_1!-\log N_2!),\tag1\label1$$ where $N=N_1+N_2$. As what the book later introduced, the derivative of $\eqref1$ is: $$F'(N_1)=\frac{N_1-N_2}{N}W-kT\log\frac{N_1}{N_2}.\tag2\label2$$

Can anyone tell me the steps needed to derive $\eqref2$ from $\eqref1$?

$\endgroup$
1
$\begingroup$

You need to use Stirling's Approximation: $$\ln n!=n\ln n-n$$

Then $$F(N_1) = \frac{N_1N_2}{N}W-kT(\log N!-\log N_1!-\log N_2!)$$ becomes $$F(N_1) = \frac{N_1N_2}{N}W-kT(N\log N-N-N_1\log N_1+N_1-N_2\log N_2+N_2)$$ $$\Rightarrow F(N_1) = \frac{N_1N_2}{N}W-kT(N\log N-N_1-N_2-N_1\log N_1+N_1-N_2\log N_2+N_2)$$ $$\Rightarrow F(N_1) = \frac{N_1N_2}{N}W-kT(N\log N-N_1\log N_1-N_2\log N_2)$$

See if this helps.

$\endgroup$

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.