Does Sachdev's Maxwell relation always hold?

Question

In the paper Bekenstein-Hawking Entropy and Strange Metals, the following Maxwell relation is given (equation 40) $$\left(\frac{\partial S}{\partial Q}\right)_T = - \left(\frac{\partial \mu}{\partial T}\right)_Q\tag{40}$$ where $S$ is the entropy per particle, $Q$ is the fermionic number density $Q = \frac{1}{N}\sum _{i=1}^N\langle \hat{c}_i^\dagger \hat{c}_i\rangle$, $T$ is the temperature and $\mu$ is the chemical potential.

The Maxwell relations in the Grand canonical ensemble, should be derivable from the differential $$d\Omega = - S dT - Q d \mu - p d V.$$

Given the Sachdev's identity however, it would imply the existence of some differential $$dz = S dQ - \mu dT.$$

Which I cannot see as a general relation, so what am I missing? There is a chance that this relation only holds as one is taking $T\to 0$. Then, I would still like to know how one shows this in general.

Answer 1

Consider the Legendre transform $G=\Omega+\mu Q$ then $$dG=-SdT+\mu dQ-pdV$$ The first derivatives are $$S=-\left({\partial G\over\partial T}\right)_{Q},\quad\quad \mu=\left({\partial G\over\partial Q}\right)_{T}$$ The equality $${\partial^2G\over\partial T\partial Q} ={\partial^2G\over\partial Q\partial T}$$ leads to $$-\left({\partial S\over\partial Q}\right)_{T} =\left({\partial \mu\over\partial T}\right)_{Q}$$ as expected.