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

I have a problem to understand why this problem is microcanonical ensemble problem? And why entropy is calculated as
[tex]S(E,N,V)=\ln \Gamma(E,N,V)[/tex]
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
[tex]S=k_Bln(\omega(E))=k_Bln(\Gamma(E+\delta E))[/tex]
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
[tex]S=k_Bln(\omega(E))=k_Bln(\Gamma(E+\delta E))[/tex]?

 

[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
[tex]S=k_Bln(\omega(E))=k_Bln(\Gamma(E+\delta E))[/tex]?

You overcomplicate the issue:
[tex]S=k_Bln(\omega(E))\approx k_Bln(\Gamma(E))[/tex]
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.