Partial derivatives in Statistical Mechanics

Question

In Statistcal Mechanics, in the grand canonical ensemble, the "fugacity" $z$ is introduced and it is defined as

$z = e^{\frac{\mu}{kT}}$

(of course $\mu$ id the chimical potential and K the Boltzsmann's constant).

We can also define the grand-partition function,

$q(z,V,T) = ln \left( \Sigma_{N_s=0}^{+\infty}z^{N_s}Q_{N_s}(V,T) \right)$

where $N_s$ is a fixed number of particle in an element of our ensemble and $Q(V,T)$ is the canonical partition function.

It can be shown that in the grand-canonical description, the mean energy, $U$ is equal to

$U= kT^2 \left[ \frac{\partial}{\partial T} q(z,V,T)\right]_{z,V}$.

Well, later, in Pathria's book is written the following equation, but, to me, it isn't obvious:

$\left(\frac{\partial U}{\partial T}\right)_{z,V}= \left(\frac{\partial U}{\partial T}\right)_{N,V} + \left(\frac{\partial U}{\partial N}\right)_{T,V} \left(\frac{\partial N}{\partial T}\right)_{z,V} .$

Can anyone help me please?

Answer 1

This is a very complicated way to write the chain rule (https://en.wikipedia.org/wiki/Chain_rule).

The idea is that here, $z$ is a function of $N, T$ and V : $z = z(N,T,V)$ . In general, $U$ is a function of $N, T, V$ : $U = U(T,N,V)$. However, one of those three variables can be replaced by $z$. So one has :

$$U(N, T, V) = U (z(N,T,V), T, V) = U(z, T, V)$$

In mathematics, $U(N, T, V)$ and $U(z, T, V)$ would be two different functions. However, as they represent the same physical quantity, physicists denote it the same. In order to know the arguments of the function, one uses parenthesis :

$$\left(\frac{\partial U}{\partial T}\right)_{z,V}$$ means the partial derivative of U with respect to T when U is expressed as $U(z, T, V)$.

Finally, the chain rule gives that

$$\left(\frac{\partial U}{\partial T}\right)_{z,V}= \left(\frac{\partial U}{\partial N}\right)_{T,V} \left(\frac{\partial N}{\partial T}\right)_{z,V} + \left(\frac{\partial U}{\partial T}\right)_{N,V} \left(\frac{\partial T}{\partial T}\right)_{z,V} + \left(\frac{\partial U}{\partial V}\right)_{N,T} \left(\frac{\partial V}{\partial T}\right)_{z,V} $$

And as $\left(\frac{\partial V}{\partial T}\right)_{z,V} = 0$ and $\left(\frac{\partial T}{\partial T}\right)_{z,V} = 1$, one reaches

$$\left(\frac{\partial U}{\partial T}\right)_{z,V}= \left(\frac{\partial U}{\partial T}\right)_{N,V} + \left(\frac{\partial U}{\partial N}\right)_{T,V} \left(\frac{\partial N}{\partial T}\right)_{z,V}$$.

If needed I can be more rigorous and introduce properly the substitution but this is tedious and not very helpful I think.