Undergrad Do Simple 2D Ising Models Have Constant Density of States?

[rabbed]

Do simple 2D Ising models have constant density of states?
How is it calculated?

 

[phinds]

What research have you done so far on this? What have you found out?

 

[rabbed]

I just learned about density of states and the Boltzmann factor.
If the density of states does not depend on the energy/is constant, we can just use the Boltzmann factor to calculate the probability of a particle being in a state of certain energy. And with the Ising model only the BF is used, right?
I googled and found some people finding the DOS using some algorithm, but no calculations or justifications for why it would be constant.
Pretty new to the Ising model also, just thought the DOS would be a fundamental thing to know when deriving it

 

[rabbed]

Okay, I found an explanation for the Ising model..

Next question - The MB energy distribution is: MB_PDF(E) = 2*sqrt(E/pi) * 1/(kB*T)^(3/2) * e^(-E/(kB*T))
How do I arrive at the density of states which hides inside the expression 2*sqrt(E/pi) * 1/(kB*T)^(3/2) ?
I've only seen DOS that have "h" in them.. I want it to contain only E, pi, kB and T..
This is how far I've gotten (using a momentum vector):
V = 4*pi*p^3/3
dV = 4*pi*p^2*dp
dV = 4*pi*(2*m*E)*sqrt(m/(2*E))*dE (since p = sqrt(2*m*E) and dp = sqrt(m/(2*E))*dE)
dV = 2*pi*(2*m)^(3/2)*sqrt(E)*dE

How do I get rid of the m and how do I get in kB and T?

 

[rabbed]

Since no one cares, I might as well ask Another question:
Is there a DOS-expression D for the ideal gas which will both fit into MB_PDF(E, T) = D * e^(-E/(kB*T)) / Z (where Z normalizes the distribution)
as well as giving an extensive entropy S = kB*ln(D) ?
It should be the same quantity, right?

 

[phinds]

Since no one cares, I might as well ask Another question:
Is there a DOS-expression D for the ideal gas which will both fit into MB_PDF(E, T) = D * e^(-E/(kB*T)) / Z (where Z normalizes the distribution)
as well as giving an extensive entropy S = kB*ln(D) ?
It should be the same quantity, right?

Sorry to see you're not getting any help, but starting a new question in the same thread is a VERY bad idea. I'd suggest that you delete it here and start a new thread.