Why does the cross derivative of the partition function disappear here?

Question

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):

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

GiorgioP · Answer

I have found this old  question without a final answer.
Actually, it is quite simple to show the equivalence of the two formulas.

Indeed, from 
$$
dE = T dS -pdV + mu dN,
$$
we see that, at constant $V$ and $T$, this equation implies:

$$
mu =
-T left.frac{partial{S}}{partial{N}}right|_{T,V} + left.frac{partial{E}}{partial{N}}right|_{T,V}=left.frac{partial{F}}{partial{N}}right|_{T,V},
$$
while at constant $V$ and $E$:
$$
mu =- Tleft.frac{partial{S}}{partial{N}}right|_{E,V},
$$
then, the cancellation of the extra term follows.

Why does the cross derivative of the partition function disappear here?

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