I have recently learned about ensembles in statistical mechanics, and I've seen multiple applications and interpretations of the EVN (microcanonical), TVN (canonical), $\mu$VT (grand canonical) and NPT (isothermal isobaric) ensembles, I've also learned that the $\mu$PT ensemble doesn't exist. What happens with the $\mu$VE ensemble? is it possible to postulate it? can there be a system that exchanges particles with its surroundings but its energy is fixed? I find it hard to come up with a physical way to do this, but there seems to be not much problem with the mathematics.

A quick search in google scholar brought up only one page mentioning this kind of ensemble, a doctoral thesis about molecular simulation based around it. Does this mean that its possible to formulate it but not used very much?

A similar question could be posed about the $\mu$PE ensemble, I haven't found a single publication referring to it.

  • $\begingroup$ $\downarrow$ is this the kind of argument you were looking for? $\endgroup$ – Phonon Sep 2 '15 at 12:35
  • $\begingroup$ I guess so, I'm not enteriley convinced, mostly on the grounds that I z $\endgroup$ – Ignacio Sep 2 '15 at 14:06
  • $\begingroup$ Sorry, I had some problems with my last comment =P. Also, while writing it I realized that your answer IS actually what I was looking for. According to you it seems like you can have E, V, $\mu /T$ as the natural variables of a potential, would you say that this potentials are not usually used because these are akward variables to control? $\endgroup$ – Ignacio Sep 2 '15 at 14:21
  • $\begingroup$ Indeed, specially maintaining $\mu/T$ experimentally can be very delicate, same goes for $E,$ compared to say $N,V,T.$ $\endgroup$ – Phonon Sep 2 '15 at 14:33

I haven't given this enough thought yet, but at a first glance, I would say no, a potential having $\mu, V, E$ as natural variables would not be a valid one. One possible attempt to obtain such a thermodynamic potential $Q$ that is a natural function of $\mu,V,E$, would be a Legendre transform of the entropy $S(E,V,N).$ We have:

\begin{align} dS = \frac{1}{T}dE+\frac{p}{T}dV-\frac{\mu}{T}dN \tag{1} \end{align}

Let's introduce a new variable for $\mu/T,$ say $\gamma.$ Then if you try to write $Q$ as a Legendre transform of $S:$ \begin{align} Q &= S+\gamma N \\ dQ &= dS + d(\gamma N)\\ &= \frac{1}{T}dE+\frac{p}{T}dV+d\gamma N \end{align} Meaning at best you can have a potential $Q(E,V,\mu/T)$ that is a natural function of the internal energy, volume and the ratio of chemical potential with temperature. Entropy $S$ is a valid thermodynamic potential, thus a valid Legendre transform of $S$ would be one that contains no more or less information than $S.$ So if $Q(E,V,\mu/T)$ contains the same information as $S,$ surely a potential $X(E,V,\mu)$ cannot.

As for your last question, I guess you could say that if the above argument is true, then $X(E,V,\mu)$ cannot exist nor can its Legendre transforms, i.e. $Y(\mu,P,E)=X-pV$ cannot be a valid thermodynamic potential either.


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