# Generallized Canonical Ensemble - Isobaric Ensemble

I am trying to understand the way generalized canonical ensembles like the pressure ensemble are derived from the standard canonical ensemble.

In the derivation for the standard form, one defines a system $S$ and a reservoir $R$. With a total microcanonical Hamiltonian: $$H(X)=H_S(X)+H_R(X)$$ My question is what do we do to put in the volume exchange? What is the basic idea of going from pure energy exchange with a reservoir to different additional things like volume.

My guess is to just add it as an energy term that is not part of the Hamiltonian?

So then: $$H(X)=H_S(X)-V p+H_R(X)$$ or in general with $y$ being an intensive and $x$ being an intensive variable: $$H(X)=H_S(X)+x y +H_R(X)$$

I would be glad about a correction of my guess or a verification, of course.

• really no one? I think I may have found the answer. If I answer myself will this thread be deleted at some point? – pindakaas Dec 2 '15 at 8:33

The Hamiltonians $H_S$ and $H_R$ both implicitly depend on their respective volumes (or confining potential strength). To allow volume exchange between the two systems, you simply impose the constraint $V_R = V_{tot}-V_S$. The joint Hamiltonian is always given by $H_S+H_R$.
You can check that in mechanical equilibrium, $\partial_{V_S} H_S+\partial_{V_S}H_R=p_S-p_R=0$ implies that $p_S=p_R$.
If the reservoir is so large that $p_R$ changes negligibly as $V_S$ changes, then it is reasonable to work in terms of the intensive quantities $p_R$, $T_R$, $\mu_R$, etc, and we have
$$H_R=p_R\Delta V_S+\dots \; ,$$ where $\dots$ stands for terms that have only a very weak volume dependence. Hence, $$H=H_S+p_R\Delta V_S+\dots \; ,$$ where $\dots$ represents terms that do not influence the Hamiltonian of the system. In mechanical equilibrium, we have $$p_S=\partial_{V_S}H_S\Big|_{\text{entropy}}(V_S,\dots)=p_R \; ,$$ so that $$H=H_S(V_S(p_S),\dots)+p_S\Delta V_S \;.$$ This gives rise to the enthalpy expression, similar to the one you wrote down (up to a sign). Note, however, that free energies always involve a coupling between an extensive quantity and an intensive quantity (coordinate and derivative of energy), so that the full expression is extensive.