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$(\mu,P,T)$ pseudo-ensemble: why is it not a proper thermodynamic ensemble?

+ 23 like - 0 dislike

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?

This post has been migrated from (A51.SE)

asked Sep 14, 2011 in Theoretical Physics by F'x (175 points) [ revision history ]
retagged May 12, 2014 by dimension10
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).

This post has been migrated from (A51.SE)
@Piot: I edited to clarify, $P$ is pressure. I've seen it both lowercase and uppercase depending on textbooks.

This post has been migrated from (A51.SE)

The simplest answer is that all three variables are intensive. 

2 Answers

+ 17 like - 0 dislike

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).

This post has been migrated from (A51.SE)
answered Sep 15, 2011 by Piotr Migdal (1,255 points) [ no revision ]
+ 1 like - 0 dislike

A thermodynamic setting in which $T,P,\mu$ figure as the basic variables, and everything else is derived from it, can be found in http://www.mat.univie.ac.at/~neum/ms/phenTherm.pdf

The key is to define in place of the thermodynamic potential a function that vanishes identically in equilibrium, whose derivatives are, up to a scale, the standard extensive thermodynamic variables. 

The setting is very close to the grand canonical ensemble. 

answered Apr 19, 2014 by Arnold Neumaier (11,775 points) [ no revision ]

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