Prove the 3 definitions of entropy are equivalent (stat. mechanics)

[Tosh5457]

Homework Statement

$$S(E,V) = kln(\Gamma(E) )\\<br /> S(E,V) = kln(\omega(E) )\\<br /> S(E,V) = kln(\Sigma(E) )\\$$

S entropy, k Boltzmann's constant. Prove these 3 are equivalent up to an additive constant.

Homework Equations

$$\Gamma(E) = \int_{E<H<E+\Delta}^{'}dpdq\\<br /> \Gamma(E)=\omega\Delta \\<br /> \Delta << E\\<br /> <br /> \Sigma(E) = \int_{H<E}^{'}dpdq\\<br /> \omega = \frac{\partial \Sigma}{\partial E}\\$$

H is the system's Hamiltonian and E is an arbitrary energy. These are integrations over all the p and q's, I wrote them like that to abbreviate.

The Attempt at a Solution

Using the 1st definition I can get to the 2nd one, but I can't reach at sigma's definition.
$$<br /> kln(\Gamma(E)) = kln(\omega\Delta) = kln(\omega) + kln(\Delta)\\<br /> ln(\Delta) << ln(\omega) => S = kln(\omega)\\$$

 

[Greg Bernhardt]

I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?

 

[bloby]

I would say $\Gamma (E)=\Sigma (E+\Delta) -\Sigma (E)=\Delta \frac{\partial \Sigma}{\partial E}+...$