Free energy density equals pressure

We are referring here to the grand canonical ensemble where all independent variables are taken to be intensive, except the volume. The independent variables are volume, temperature, and all chemical potentials.

Now $\log Z$ is an extensive quantity, so it will be simply proportional to the volume. Thus $F := - T \log Z =: V \, f(T, \mu)$ . We can now compute the pressure using $p = - dF/dV = -F/V$ . The relation $p = -f(T,\mu)$ may be called the equation of state of the system under consideration.

The extensive quantities conjugate to $T$ and $\mu$ are computed via the derivatives $S = - dF/dT$ and $N = - dF/d\mu$ . The relation $E - \mu N = -\partial_\beta \log Z = \partial_\beta (\beta F)$ implies $E - \mu N = F + TS$ and thus $E - \mu N - TS = - pV$ .

One common form that appears in the literature is $\epsilon + p - \mu \, n = T s$ . This is a useful form, because the combination $\epsilon+p = \epsilon - f$ shows up in various places.

If we do not have any conserved particle number, say in a gas of photons, we do not have any chemical potential. We get $\epsilon + p = Ts$ , which also follows (in this instance) from conformality of the photon gas.

About Raghu Mahajan

Postdoctoral research fellow in theoretical physics, studying quantum gravity.

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