# Density of states in the ideal gas

• I
The MB energy distribution is: MB_PDF(E, T) = 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?

On the same theme:

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?

DrClaude
Mentor
How do I arrive at the density of states which hides inside the expression
You can't. The best you can do is arrive at the density of states divided by the partition function.

I've only seen DOS that have "h" in them.. I want it to contain only E, pi, kB and T..
That doesn't make sense. The density of states depends on Planck's constant, so there is no way to write the DOS without it. Even Gibbs, way before Planck, figured out that there was a parameter ##h## that needed to be included in the partition function. It cancels out if you take the DOS divided by Z, see my comment above.

dextercioby
Thanks, looks like an interesting discussion that I will read.

So the way i've understood this is: Boltzmann's counting of states didn't give the same (extensive) entropy as the entropy of Clausius. Then Gibbs fixed this (Gibbs paradox) by considering indistinguishable particles?

This document seem to say that Boltzmann counting does give extensive entropy after all: https://www.hindawi.com/journals/jther/2016/9137926/

Does Gibbs give 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) ?
Should it be the same quantity in both these places and in that case, does any h-parameters cancel out in the entropy expression also?

Just trying to sort things out.. :)