# Density of states

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

phinds
Gold Member
2021 Award
What research have you done so far on this? What have you found out?

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

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?

Last edited:
Since noone 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