Why is the entropy calculated differently in the microcanonical ensemble?

[LagrangeEuler]

Homework Statement

Ideal gas which consist of $N$ identical particles which moving free inside volume $V$ where all collisions between particles and walls of container are absolute elastic. Calculate phase volume $\Gamma$, entropy $S$, temperature $T$ and pressure of gas.

Relevant Equations

Hamiltonian
$$H=\sum^{3N}_{i=1}\frac{p_i^2}{2m}$$
$$\Gamma(E,N,V)=\int_{H(p,q) \leq E}\frac{dpdq}{h^{3N}N!}$$

I have a problem to understand why this problem is microcanonical ensemble problem? And why entropy is calculated as
$$S(E,N,V)=\ln \Gamma(E,N,V)$$
When in microcanonical ensemble we spoke about energies between $E$ and $E+\Delta E$.

 

[Lord Jestocost]

$\Gamma(E,N,V)$ is defined as the phase volume which contains all points in $\Gamma$ space with energy lower than or equal to $E$. Regarding a microcanonical ensemble in the energy range $[E,E + \delta E]$, the corresponding volume is given by $\omega(E,N,V)=\Gamma(E+\delta E,N,V)-\Gamma(E,N,V)$.
For the entropy, one has then $S(E,N,V)=k_Bln(\omega(E,N,V))$.

For very large $N$, one finds that the following definitions for the entropy are identical up to terms of order $lnN$ and constants:
$S_{\omega}=k_Bln(\omega(E))$ and $S_{\Gamma}=k_Bln(\Gamma(E))$

Have a look, for example, at chapter 2.2 of Statistical Physics by Manfred Sigrist: https://www.e-booksdirectory.com/details.php?ebook=6060

 

[LagrangeEuler]

Yes undestand what you want to say. But is not than the case that entropy
$$S=k_Bln(\omega(E))=k_Bln(\Gamma(E+\delta E))$$
Because in large number of dimension almost all volume of sphere is concentrated around the surface.

 

[Lord Jestocost]

$$\omega(E,N,V)=\Gamma(E+\delta E,N,V)-\Gamma(E,N,V) =\frac {\partial \Gamma(E,N,V)} {\partial E} \delta E$$

 

[LagrangeEuler]

Yes I understand what is $\omega$. My question is why
$$S=k_Bln(\omega(E))=k_Bln(\Gamma(E+\delta E))$$?

 

[dRic2]

Because in large number of dimension almost all volume of sphere is concentrated around the surface.

You already have the answer.

If I understand the notation correctly, then for large $N$: $\omega(E) \approx \Gamma(E)$

 

[LagrangeEuler]

I am confused there. Because if I have that n large number of dimension almost all volume of sphere is concentrated around the surface. That means to my mind that if energy changes from $E-\Delta E$ to $E$ that I can use approximation that we are talking about.

 

[dRic2]

Sorry, I'm not understanding your problem.

 

[Lord Jestocost]

Yes I understand what is $\omega$. My question is why
$$S=k_Bln(\omega(E))=k_Bln(\Gamma(E+\delta E))$$?

You overcomplicate the issue:
$$S=k_Bln(\omega(E))\approx k_Bln(\Gamma(E))$$
holds for a microcanonical ensemble with energies between $E$ and $E+\delta E$ in case of very large $N$. Have a look at equations (2.43) and (2.44) in chapter 2.2 of Statistical Physics by Manfred Sigrist.