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

AI Thread Summary
The discussion centers on whether simple 2D Ising models have a constant density of states (DOS) and how this can be calculated. Participants explore the implications of a constant DOS on the use of the Boltzmann factor for determining the probability of particle states. There is a request for clarification on deriving the DOS from the Maxwell-Boltzmann energy distribution, specifically how to express it solely in terms of energy, pi, kB, and temperature without including mass. Additionally, a question arises about finding a DOS expression for an ideal gas that aligns with the Maxwell-Boltzmann probability distribution and also provides an extensive entropy expression. The conversation highlights the complexities of these calculations and the need for clearer guidance on the topic.
rabbed
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Do simple 2D Ising models have constant density of states?
How is it calculated?
 
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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 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?
 
rabbed said:
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.
 
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