Explanation of definition of entropy I am trying to understand the definition of the entropy.
The lecture notes I use define it as
$$ S(E,V) = k_B \ln(\Gamma(E,V)) $$
with
$$\Gamma(E) = \int_{E < H(p,q) < E + \Delta} d^{3N}p\ d^{3N}q\ \ \rho (p,q) $$
where  $\rho$ is the distribution of $p$ and $q$ in the 6N-dimensional $\Gamma$ space of $p,q$ of a system with N elements.
Now I have 3 questions: 


*

*The $V$ in the definition, is it the volume, which corresponds to N?

*What is the $\Delta$ ?

*What do the integral boundaries mean? We look at all the Hamilton Functions with the Energry between $E$ and $E+\Delta$? But shouldn't the Hamilton Function for a certain system be unique? So why look at different energies? And what does it mean to integrate over Hamilton functions like that?

 A: Okay, let me try to answer your questions one by one:


*

*This is correct. The V corresponds to the volume which in which the N particles are observed.

*The Delta means a finite element of energy. As mentioned below: it is an arbitrary, finite energy, which you later want to decrease until it is indefinite.

*Here you should keep in mind, what q,p,Г and H really mean. Г is the volume of the phase space which is occupied by the system with N particles which has the energy H(p,q). q & q are here 3N-dimensional coordinates of all relevant particles. If you compare this case with a normal 2D phase space you will easier see why you need to integrate like this: by observing a simple 1D-pendulum and its 2D phase space (which is a circle) you can get a good example. By allowing more energies and not only one, you will get an annulus.

*In case you're wondering why you integrate system between this boundaries: You're just doing this to subtract this integral over less-energy-states to get a finite differential of the phase space volume. By making Delta indefinite, you can write it as an derivative.
I hope this helps you to get a better understanding of this definition.
