39
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ 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). $\endgroup$ Sep 15, 2011 at 14:07
  • 2
    $\begingroup$ @Piot: I edited to clarify, $P$ is pressure. I've seen it both lowercase and uppercase depending on textbooks. $\endgroup$
    – F'x
    Sep 15, 2011 at 14:09

1 Answer 1

25
$\begingroup$

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).

$\endgroup$
3
  • $\begingroup$ The Legendre transform is zero because it is implicitly assumed the thermodynamic limit. If one computes the Laplace transform from the NPT ensemble with respect to N or that of the grandcanonical ensemble with respect to V, one finds a non-zero thermodynamic potential. $\endgroup$
    – neophyte
    Aug 1, 2018 at 20:19
  • 1
    $\begingroup$ Furthermore, I don't see any physical reason to exclude a system that exchanges energy, volume and particles with its surrounding made of the same species of particles: for instance, a gas confined in a elastic and porous box. Such a physical system at thermal equilibrium would be described by the $\mu PT$ ensemble, following Jaynes' derivation of the Gibbs measure or generalizing the derivation of the grandcanonical ensemble from the canonical ensemble to derive the $\mu PT$ ensemble from the grandcanonical. What else would model this physical system? What is wrong in the above derivations? $\endgroup$
    – neophyte
    Aug 1, 2018 at 20:52
  • 1
    $\begingroup$ What you described is not really a system, is it? It is not bounded in any meaningful way... How would you distinguish between what is and what is not a part of the system in your description? $\endgroup$ Apr 22, 2022 at 14:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.