# A derivation in problem 3.1 of Problems on Statistical Mechanics.(by Dalvit, et al)

My question is regarding the following derivation: First the problem itself as stated in the book:

Find the entropy $S(E,V,N)$ of an ideal gas of $N$ classical monatomic particles, with a fixed total energy $E$, contained in a $d$ dimensional box of volume $V$. Deduce the equation of state of this gas, assuming that $N$ is very large.

They write on page 41 (in the solution to problem 3.1) that the entropy is given by: $$S(E,V,N;\Delta E)=k\ln\bigg(\bigg( \Omega(E,V,N;\Delta E\bigg)/(h^{dN}N!)\bigg)$$

To calculate $\Omega$ we express it as $\Omega = \Delta E \partial \nu /\partial E$.

Now they derived the equality: $$\nu(E,V,N) = \frac{V^N (2\pi m E)^{dN/2}}{(dN/2)\Gamma(dN/2)}$$

On page 42 they get $$S(E,V,N;\Delta E) = k\bigg\{ N \ln \bigg( \frac{V(2\pi m E)^{d/2}}{h^d}\bigg) -\ln \bigg[ \Gamma(dN/2) \bigg] - \ln(N!) + \ln(\Delta E/E) \bigg\}$$

Which I concur to.

But then comes the approximation for large $N$ with Stirling approximation, and I don't see how do they approximate $\ln ( \Gamma(dN/2))$, is $dN/2$ for large $N$ an integer or non-integer? if it's an integer then $\Gamma(dN/2) = (dN/2)!$ and we can use Stirling's approximation, but if not then I don't see how to approximate $\ln(\Gamma(dN/2))$.

In the end they receive the following approximation for $N \to \infty$: $$S(E,V,N) \approx Nk\bigg\{ \ln\bigg[ \frac{V}{N}\bigg(\frac{4\pi m E}{dNh^2}\bigg)^{d/2}\bigg] +\frac{d+2}{2}\bigg\}$$

I don't see how did they get this expression.

Anyone care to elaborate on the calculations here?

Thanks.

There's no upper bound on rigour. I hope this suffices.

Suppose first than $Nd/2$ is an integer. Then $\Gamma(Nd/2)=(Nd/2-1)!$ and we can use Stirling's approximation. If $M$ is large, then $M! \approx M\log{M}-M+O(1)$. Focusing on the factorial terms, we have: $$\log⁡Γ(Nd/2)+\log⁡N!=$$ $$=(Nd/2-1)\log⁡(Nd/2-1)-(Nd/2-1)+N \log⁡N-N+O(1)=$$ $$=Nd/2 \log⁡(Nd/2)-Nd/2+N \log⁡N-N+O(\log⁡N )=$$ $$=N\log⁡{\left[N\left(\frac{Nd}{2}\right)^{d/2}\right]}- \frac{N (d+2)}{2}+O(\log⁡N)$$

Substituting into the formula for $S$ above, pulling out a factor of $N$ and doing some algebra inside the logarithms we have: $$S = k N \left[\log{\left(\frac{V}{N} \left(\frac{4\pi m E}{N h^2d}\right)^{d/2}\right)}+ \frac{d+2}{2} +\frac{1}{N} \log\frac{\Delta E}{E} + O\left(\frac{\log{N}}{N}\right)\right]$$ The only term that's left to deal with is the one involving $\Delta E$. The result we need follows in the most general case from something like the Fluctuation-Dissipation Theorem. I don't know how Landau deals with it. However, we can make an example by computing the ratio in the canonical ensemble (it doesn't actually matter which ensemble we use, as all ensembles agree in the thermodynamic limit). The (average) energy of a system is given by $$\langle E \rangle=-\frac{\partial}{\partial \beta} \log{Z}$$ where $\beta =1/kT$ and $Z$ is the partition function. For a multiparticle gas $Z \sim Z_1^N$ so $E\sim N$. Then the fluctuation is given by: $$\Delta E^2 = \langle E^2 \rangle-\langle E \rangle^2=-\frac{\partial \langle E \rangle}{\partial \beta}$$ This is enough for what we need, as it means that $\Delta E^2 \sim E \sim N$, which then means that $\Delta E / E \sim 1/\sqrt{N}$. We therefore obtain: $$\log{\frac{\Delta E}{E}} = O(\log{N})$$ and the result you need follows, since we can then absorb this term into the big $O$.

The only thing that's left to do is the case when $Nd/2$ is not an integer. But if it's not an integer, it must be a half-integer. On such points the Gamma function has a closed formula: $$\Gamma\left(n+\frac{1}{2}\right)=\frac{(2n-1)!!}{2^n}\sqrt{\pi}$$ We want to see how big a mistake we are making by replacing it with $\Gamma(n+1)=n!$. This paper (page 5) provides a useful approximation: $$\log{(2n-1)!!}=n\log{\frac{2n}{e}}+\frac{1}{2}\log{\frac{2n-1/12}{n+1/24}}+O\left(\frac{1}{n^3}\right)$$ Therefore: $$\log{\Gamma\left(n+\frac{1}{2}\right)} = n\log{\frac{2n}{e}}+\frac{1}{2}\log{\frac{2n-1/12}{n+1/24}}-n\log{2}+O(1)=$$ $$=n\log{n}-n+O(1)$$ since the logarithm of the ratio is $O(1)$ for large $n$. Therefore, using Stirling, the error that we make is: $$\Gamma(n+1/2)-\Gamma(n+1)=O(1)$$ By making this change we still get the same formula, since the effect is substituting $Nd+1$ in place of $Nd$. After applying Stirling's the error the we make in removing the $1$ is $O(\log{N})$.

• Miao. I just noticed the question is 8 months old! Sorry for bringing it back to life. I guess necromancy is not appreciated – John Donne Nov 20 '17 at 23:35
• It's appreciated nonetheless. – MathematicalPhysicist Nov 21 '17 at 16:03

The Stirling approximation is valid for a large positive argument of gamma-function (https://en.wikipedia.org/wiki/Stirling%27s_approximation#Stirling.27s_formula_for_the_gamma_function), not just factorial, so it does not matter if the argument is integer or not.

• I still would like someone to give a rigorous derivation of this last approximation. I don't see how to derive it. – MathematicalPhysicist Feb 25 '17 at 10:07
• I mean the approximation of $S(E,V,N)$ in my post. – MathematicalPhysicist Feb 25 '17 at 10:13
• @MathematicalPhysicist: I don't think it's difficult, but I am not sure what is the relationship between S(E,V,N;$\Delta$E) and S(E,V,N) – akhmeteli Feb 25 '17 at 10:27
• akhmeteli, it's written on page 41 that: "Since $\Delta E$ is small, we have $\Omega(E,V,N;\Delta E) = \Delta E \Sigma(E,V,N)$ where $\Sigma(E,V,N)$ is a measure of the area of a constant energy surface"; but I don't see the relation either... :-( – MathematicalPhysicist Feb 25 '17 at 10:37