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They state that the chemical potential in a canonical ensemble is given by:

$$\mu = -kT \frac{\partial{\ln Z(N,V,T)}}{\partial{N}} \tag{1}$$

But if I use the definition of chemical partial (which I assume to be general):

$$\mu = -T\frac{\partial{S}}{\partial{N}}$$ And I substitute $S(N,V,T) = k\ln Z+kT\frac{\partial{\ln Z}}{\partial{T}} $ I get almost (1), but with an extra term. I have no clue why this term should be zero.

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You need to pay attention to what is being kept fixed when you take the partial derivative. In your first and second definitions, the things that are kept fixed are different. Therefore, the second formula should use the entropy from the microcanonical ensemble and the first one the free energy in the canonical one. – gatsu May 20 '13 at 12:59
Ah, I see what you mean. Is it possible though to derive the first formula from the second? Or are they defining the chemical potential in different ways for different ensembles? – yarnamc May 20 '13 at 14:41
$\mu = -T\left(\partial S/ \partial N \right)_{E,V}$ and $\mu = \left(\partial F/ \partial N \right)_{T,V}$ are two equivalent definitions but you can't really derive one from the other explicitely. That is because when varying $F$ with respect to $N$ at fixed $T$, the internal energy varies as well and one needs then to account for this variation; that's what does the derivative of the log of $Z$ in your formula. Formally however, you can relate the two definitions via properties of Legendre transforms. – gatsu May 20 '13 at 21:06
Aha thank you, that solved the problem. – yarnamc May 20 '13 at 21:10

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