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I am trying to derive eq. (2.15) on page 34 of Birger Bergersen's and Michael Plischke's textbook on Equilibrium Statistical Mechanics second edition.

We have equation $S(E,V,N)=k_B \log \Omega(E,V,N)$. $$(2.14) \ \ \ \ \ \ \Omega(E,V,N) = \frac{V^N}{h^{3N}N!} \frac{(2\pi m E)^{3N/2}}{(3N/2-1)!}\frac{\delta E}{E}$$

Now, they say that, if $|\log(\delta E/E)|\ll N$, we find, using Stirling's approximation, $\log N! \approx N\log N-N$,

$$S(E,V,N) \approx Nk_B\log V/N + \frac{3N}{2}k_B\log(\frac{4\pi mE}{3Nh^2})+5Nk_B/2$$

Now, I did the calculation, but I get something else:

$$S(E,V,N)\approx k_B(N\ln V - N\ln N+N)+k_B\bigg[ \frac{3N}{2}\ln\frac{2\pi m E}{h^2}-(3N/2-1)\ln(3N/2-1)-(3N/2-1) \bigg]$$

But how to continue from here, I assume because $N\gg 1$ that $\ln(3N/2-1)\approx \ln(3N/2)$,and $3N/2\gg 1$ but then I get a term with $-Nk_B/2$ and not as it's written in the textbook with $5Nk_B/2$.

Where did I go wrong here?

Edit: I forgot to say that I neglected $\ln (\delta E/E)$, since it's much less than $N$.

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closed as off-topic by Kyle Kanos, stafusa, JamalS, Jon Custer, JMac Oct 10 '17 at 18:01

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Firstly, I think you have a typo in the expression you derived. The extreme last term, $\left(\frac{3N}{2}-1\right)$ should have a positive sign, and not negative. Other than that, you are on the right track.

You should get, $$S(E,V,N)\approx k_B(N\ln V-N\ln N +N)+ k_B\left[\frac{3N}{2}\ln\frac{2\pi mE}{\hbar^2}-\left(\frac{3N}{2}-1\right)\ln\left(\frac{3N}{2}-1\right)+\left(\frac{3N}{2}-1\right)\right]$$ You can immediately see that, $$k_B(N\ln V-N\ln N +N)=k_B\ln\frac{V}{N}+Nk_B$$ The term in square brackets can be approximated as, $$k_B\left[\frac{3N}{2}\ln\frac{2\pi mE}{\hbar^2}-\frac{3N}{2}\ln\frac{3N}{2}+\frac{3N}{2}\right]=\frac{3Nk_B}{2}\ln\frac{4\pi mE}{3N\hbar^2}+\frac{3Nk_B}{2}$$ Adding everything up, you get the required expression mentioned in the book.

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