Is the $\mu VE$ ensemble possible to formulate?

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.

• $\downarrow$ is this the kind of argument you were looking for? Sep 2 '15 at 12:35
• I guess so, I'm not enteriley convinced, mostly on the grounds that I z Sep 2 '15 at 14:06
• 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? Sep 2 '15 at 14:21
• Indeed, specially maintaining $\mu/T$ experimentally can be very delicate, same goes for $E,$ compared to say $N,V,T.$ 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.

• Going back to this question after five years I realized that it is trivial to create a thermodynamic potential that is a legendre transform of the energy and is a function of the variables $(S, V, \mu)$, this potential would be $\Phi = U - \mu T$, It should then be possible to encode all information of the system there, right? Also, I know that legendre transforms of the entropy are more naturally related to ensembles, but I never saw a definite proof that if a given entropy transofrm doesn't exist then its associated ensamble doesn't either... Mar 3 '20 at 22:53
• I still think your answer is pretty good, that is probably the reason the ensemble doesn't exist, I just found it interesting that I missed this fact at that time. Mar 3 '20 at 22:56

Starting from the entropy expression

$$dS=\frac{1}{T}dU+\frac{p}{T}dV-\frac{\mu}{T}dN$$

we can introduce a new variable $$L=U-\mu N$$, with $$dL=dU-Nd\mu -\mu dN$$. Insertion gives:

$$dS=\frac{1}{T}dL+\frac{N}{T}d\mu +\frac{p}{T}dV$$

This means we can define an ensemble with the variables $$L=U-\mu N$$, $$\mu$$ and $$V$$. Most importantly, this is an example of an adiabatic system (only exchange of matter, but no exchange of heat with the environment). This can be seen, e.g. by realizing that when keeping these 3 variables constant:

$$dL=0=dU-\mu dN\Rightarrow dU=\mu dN$$

This means that energy exchange is only possible via transfer of matter, there is no possibility to merely exchange heat. 