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Grothor comments on Entropy and Temperature - Less Wrong Discussion
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Comments (96)
I'l follow suit with the previous spoiler warning.
SPOILER ALERT .
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I took a bit different approach from the others that have solved this, or maybe you'd just say I quit early once I thought I'd shown the thing I thought you were trying to show:
If we write entropy in terms of the number of particles, N and the fraction of them that are excited: α ≡ E/(Nε) , and take the derivative with respect to α, we get:
dS/dα = N log [(1-α)/α]
Or if that N is bothering you (since temperature is usually an intensive property), we can just write:
T = 1/(dS/dE) = E / log[(1-α)/α]
This will give us zero temperature for all excited or no excited particles (which makes sense, because you know exactly where you are in phase space), and it blows up at half particles are excited. This means that there is no reservoir hot enough to get from α < .5 to α = .5 .
I posted some plots in the comment tree rooted by DanielFilan. I don't know what you used as the equation for entropy, but your final answer isn't right. You're right that temperature should be intensive, but the second equation you wrote for it is still extensive, because E is extensive :p
You're right. That should be ε, not E. I did the extra few steps to substitute α = E/(Nε) back in, and solve for E, to recover DanielFilan's (corrected) result:
E = Nε / (exp(ε/T) + 1)
I used S = log[N choose M], where M is the number of excited particles (so M = αN). Then I used Stirling's approximation as you suggested, and differentiated with respect to α.
Good show!