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While teaching statistical mechanics, and describing the common thermodynamic ensembles (microcanonical, canonical, grand canonical), I usually give a line on why there can be no $(\mu, P, T)$ thermodynamic ensemble ($\mu$ being the chemical potential, $P$ being the pressure, and $T$ being the temperature). My usual explanation is a handwaving that all of the control parameters would be intensive, which leaves all the extensive conjugated parameters unbounded, and you can't write your sums and integrals anymore.

However, I have always felt uneasy because:

  1. I never was able to properly convince myself where exactly the problem arose in a formal derivation of the ensemble (I am a chemist and as such, my learning of statistical mechanics didn't dwell on formal derivations).

  2. I know that the $(\mu, P, T)$ can actually be used for example for numerical simulations, if you rely on limited sampling to keep you out of harm's way (see, for example, F. A. Escobedo, J. Chem. Phys., 1998). (Be sure to call it a pseudo-ensemble if you want to publish it, though.)

So, I would like to ask how to address point #1 above: how can you properly demonstrate how chaos would arise from the equations of statistical mechanisms if one defines a $(\mu, P, T)$ ensemble?

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What does 'P' stand for? If you mean pressure - 'p' (lowercase) is a more common notation (in general, except for the temperature - lowercase for intensive variables, while uppercase for extensive). – Piotr Migdal Sep 15 '11 at 14:07
@Piot: I edited to clarify, $P$ is pressure. I've seen it both lowercase and uppercase depending on textbooks. – F'x Sep 15 '11 at 14:09
up vote 23 down vote accepted

If you have only one species of particles then working with $(\mu,p,T)$ ensemble does not make sense, as its thermodynamic potential is $0$.

$$U = TS -pV + \mu N,$$ so the Legendre transformation in all of its variables (i.e. $S-T$, $V-(-p)$ and $N-\mu$) $$U[T,p,\mu] = U - TS + pV - \mu N$$ is always zero.

The fact is called Gibbs-Duhem relation, i.e. $$0 = d(U[T,p,\mu]) = -S dT + V dp - N d\mu.$$

However, if you have more species of particles, you can work with a thermodynamic potential as long as you have at least one extensive variable (e.g. number of one species of particles).

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