Many thanks to Bart Andrews for this contribution!

### Question

- Show that a relation of the kind \(f(x, y, z) = 0\) between the three quantities x, y, and z implies the relation \[\left(\frac{\partial x}{\partial y}\right)_z \left(\frac{\partial y}{\partial z}\right)_x \left(\frac{\partial z}{\partial x}\right)_y = -1\] between the partial derivatives.
- The grandcanonical partition sum Z
_{G}is a function of the three parameters β, V, and μ. All other state variables are derived from Z_{G}and its partial derivatives. Therefore, there must be a relation \(f(N, μ, V, T) = 0\) for the four quantities N, μ, V, and T. Consider this equation for constant T and show that\[ \left( \frac{\partial {\langle N \rangle}}{\partial \mu} \right)_{V,T}=-{\langle N \rangle}\frac{\left(\frac{\partial {\langle N \rangle}}{\partial V}\right)_{\mu,T}}{\left(\frac{\partial {\langle N \rangle} \mu}{\partial V}\right)_{{\langle N \rangle},T}} \]

### Solution

This is once again a two part question. It makes sense, however, to do the first part first on this occasion. The first part asks you to derive the cyclic relation for partial derivatives. So let us first write down the total differential for \(dz\) and then move along the curve of constant \(z\).

\[dz = \bigg( {\partial z \over \partial x} \bigg)_y dx + \bigg( {\partial z \over \partial y} \bigg)_x dy \stackrel{!}{=} 0 \]

It follows that the total derivative for \(y\) is given by…

\[ dy = \bigg( {\partial y \over \partial x} \bigg)_z dx \]

…since \(dz=0\). Now if we substitute this expression back into the total derivative for \(z\) we arrive at the cyclic relation (assuming the reciprocity relation).

\begin{align}

0 & = \bigg( {\partial z \over \partial x} \bigg)_y dx + \bigg( {\partial z \over \partial y} \bigg)_x \bigg( {\partial y \over \partial x} \bigg)_z dx \nonumber \\

0 & = \Bigg( \bigg( {\partial z \over \partial x} \bigg)_y + \bigg( {\partial z \over \partial y} \bigg)_x \left( \partial y \over \partial x \right)_z \Bigg) dx \;\;\;\;\;\;\;\; \forall x \nonumber \\

\bigg( {\partial z \over \partial x} \bigg)_y & = – \bigg( {\partial z \over \partial y} \bigg)_x \bigg( {\partial y \over \partial x} \bigg)_z \nonumber \\

\bigg( {\partial x \over \partial y} \bigg)_z \bigg( {\partial y \over \partial z} \bigg)_x \bigg( {\partial z \over \partial x} \bigg)_y & = -1 \; \blacksquare \nonumber

\end{align}

The next part of the question asks you to consider a specific case in statistical mechanics. The principle is same, it is just that the equation has physical meaning. The question asks you to start with \(f( \langle N \rangle , \mu , V , T)=0\). However, \(T\) is not a variable in this case, it is a constant, and so we know that \(f( \langle N \rangle , \mu , V)=0\). By analogy we can write down the cyclic relation.

\[ \bigg( {\partial {\langle N \rangle} \over \partial \mu} \bigg)_{V,T} \bigg( {\partial \mu \over \partial V} \bigg)_{\langle N \rangle, T} \bigg( {\partial V \over \partial {\langle N \rangle}} \bigg)_{\mu,T} = -1 \]

After some algebraic manipulation with the assumption of the reciprocity relation, we arrive at the desired result.

\begin{eqnarray}

\bigg( {\partial {\langle N \rangle} \over \partial \mu} \bigg)_{V,T} {\langle N \rangle} \bigg( {\partial \mu \over \partial V} \bigg)_{\langle N \rangle, T} \bigg( {\partial V \over \partial {\langle N \rangle}} \bigg)_{\mu,T} & & = -{\langle N \rangle} \nonumber \\

\bigg( {\partial {\langle N \rangle} \over \partial \mu} \bigg)_{V,T} \bigg( {\partial {{\langle N \rangle} \mu} \over \partial V} \bigg)_{\langle N \rangle, T} \bigg( {\partial V \over \partial {\langle N \rangle}} \bigg)_{\mu,T} & & = -{\langle N \rangle} \nonumber \\

\bigg( {\partial {\langle N \rangle} \over \partial \mu} \bigg)_{V,T} \bigg( {\partial {{\langle N \rangle} \mu} \over \partial V} \bigg)_{\langle N \rangle, T} & & = -{\langle N \rangle} {\bigg( {\partial {\langle N \rangle} \over \partial V} \bigg)_{\mu,T}}\nonumber \\

\bigg( {\partial {\langle N \rangle} \over \partial \mu} \bigg)_{V,T} & & = -{\langle N \rangle} \frac{{\left( {\partial {\langle N \rangle} \over \partial V} \right)_{\mu,T}}}{\left( {\partial {{\langle N \rangle} \mu} \over \partial V} \right)_{\langle N \rangle, T}} \; \blacksquare \nonumber

\end{eqnarray}