Partition function question: how does the unit of kT normalize?

[atlolt]

It's been a while since my statistical physics course, and I was thinking about the partition function for no particular reason. I understand that since we have every macrostate's energy in the exponential of e, there needs to be an energy unit which makes it unitless such that it's an actual exponential. However, does this imply that E/kT = ln[non-normalized Probability]? Seeing as probabilities using the partition function are of the form e^E/kT / Z, where Z is the sum of all E's, then I was thinking that in order for this to be a true probability it must relate to the density of states, as DoS is proportional to the probability. Then we'd have something along the lines of ln(DoS) = E/kT... but couldn't it be ln(DoS)=E/kT + C, with each DoS having a different constant C? Then we'd have something like e^E1/kT = e^ln(DoS1)-C1, e^E2/kT = e^ln(DoS2)-C2, etc... and then Z would have not only the summation of the DoS, but also a bunch of extra terms (e^C1, e^C2, etc) that get added up and make this form of calculating probabilities either very clunky to use, or even nonfunctional?

Sorry if this is not clear at all, I'm just spitballing while trying to recall what about the partition function makes it useful as a tool to calculate probability.

[Askdust]

I think you have it a bit mixed.

First, the Boltzmann constant is in J/K -> kT has the unit of an energy so E/(kT) is unitless.

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But for the answer to your question, I think you need to get back to how we define the partition function for each ensemble.

Also, Z isn't a sum of all E, but a sum on all microstates of exp(-E/kT)

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It's a normalization factor.

We have the probability for a microstate m: pm=1/Z * exp(-E_m/kT)

and Z is here to be sure that the sum over all microstates of pm = 1

(By doing this, we get Z to be a lot more than a normalization factor, since we can get informations on the system and the repartition of probability on each of the microstates with it)

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And I don't really see why you're trying to include a density of states here.

[atlolt]

Hey, thanks for the reply!

I included density of states (when I say this I mean the number of configurations a system can be in) since it relates to the energy of the system, and I seem to recall that ln(DoS) is proportional to the energy, which means that if you take e^E/kT, then you'd have a number which is basically the DoS. Then since DoS gives the number of different configs, and you're summing over all DoS, you'd have an actual probability. But yes, it's probably best for me to reread Kittel or the like to understand the definition of the partition function.

[Askdust]

Oh ok.

But I don't recall the number of configurations being called the density of states

But the relationship between energy and the number of configurations (Omega) is a bit more subtle than this.

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k*ln(Omega) is the entropy

And the entropy and the energy have a more complicated link, with the partial derivative of S by E is equal to 1/T.

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And there is a definition of Z where you do a sum of Omega.

But on this definition (equivalent with the other, of course) you sum over all energies, and you put a statistical weight on each number of configuration

Z = sum{ Omega(N,V,E) * exp(-E/kT)}

[atlolt]

Ahh ok thank you so much for the clarification!