Small problem that has been bugging me for a while and I can't seem to demonstrate the validity of $(4)$ below. Starting with enthalpy,
$$H(S,P)-U(S,V)=PV$$
Recognizing the Legendre transform here,
$$\left ( \frac{\partial U}{\partial V} \right )_S=-P = \left ( \frac{\partial \langle E_i \rangle}{\partial V}\right )_S\tag{1}$$
And since,
$$F(V,T) = U(S,V) – TS$$
$$dF = dU – TdS – SdT$$
Then using the fundamantal thermodynamic equation (leaving out changing particles),
$$dU = TdS – PdV$$
Therefore,
$$dF = – PdV -SdT$$
And,
$$\left ( \frac{\partial F}{\partial V} \right )_T = -P \tag{2}$$
We are effectively holding $S$ and $T$ constant, so no heat exchange can occur here.
If we evaluate $(2)$ as it relates to the partition function $Z$ we get, (from Pathria, Statistical Mechanics 3rd Edition, page 51 equation $11$),
$$F=-kT\ln Z$$
$$\frac{\partial F}{\partial V}=-kT\frac{\partial\ln Z}{\partial V}=\frac{-kT}{Z}\frac{\partial Z}{\partial V} = \frac{-kT}{Z}\frac{\partial }{\partial V}\sum_i \exp(-E_i/kT) $$
$$= \frac{-kT}{Z}\sum_i \frac{\partial }{\partial V} \exp(-E_i/kT) = \frac{-kT}{Z}\sum_i (-1/kT)\exp(-E_i/kT)\frac{\partial E_i}{\partial V} $$
$$= \frac{1}{Z}\sum_i \exp(-E_i/kT)\frac{\partial E_i}{\partial V} = \sum_i \rho_i \frac{\partial E_i}{\partial V} $$
$$\left ( \frac{\partial F}{\partial V} \right )_T=\left ( \left \langle \frac{\partial E_i}{\partial V}\right \rangle \right )_T = -P \tag{3}$$
Putting together $(1)$, $(2)$ and $(3)$ it seems that:
$$
\left( \frac{\partial \langle E_i \rangle}{\partial V} \right )_S=\left ( \left \langle \frac{\partial E_i}{\partial V}\right \rangle \right )_T \tag{4}$$
Evaluating the left hand side:
$$\left ( \frac{\partial}{\partial V}\frac{1}{Z}\sum_{i} E_i \exp(-E_i/kT) \right)_S \tag{5}$$
Is there an obvious simplification I am missing? We know the internal energy is a function of volume. Is the partition function a function of volume? (I thought yes).
If someone can elucidate this for me I would appreciate it. Hope the notation is clear, $E_i$ is the energy of a microstate.
$(2)$ is also true based on the Legendre Transform of the Gibbs free energy $G-F=PV$. So we know it is true, I am more interested in $(4)$ since it should also hold.
Best Answer
Not sure if I fully understand your exact question, but I think it is possible to shed a bit of light on the right-hand side of Eq. (4).
Given your derivation of Eq. (3), you are assuming that the possible microstates stay the same when you change the system volume: just their energy changes. In other words, Eq. (3) only holds if the number of microstates -- or their density in phase space -- stays constant as you change $V$, only their energy. This means that a the volume change you consider cannot correspond to a rescaling of the particle coordinates, just a change in the volume boundaries.
For example, consider a gas or liquid whose volume is changed by moving a piston. The only effect this has on the energy of any given microstate is via the interaction between the particles and the piston: the kinetic energy and the interactions between the particles do not change and their contribution to $\partial E_i/\partial V$ is zero. Hence, the right-hand side of Eq. (4) is simply: $$\left\langle \frac{\partial U_{piston}}{\partial V} \right\rangle = \left\langle \frac{\partial U_{piston}}{\partial x} \frac{\partial x}{\partial V} \right\rangle = \left\langle -f_{piston} / A_{piston} \right\rangle = -P_{piston}, $$ where $U_{piston}$ is the energy arising from interactions of particles with the piston, $x$ is the position of the piston, $f_{piston}$ is the total force of the particles in the system exerted on the piston, $A_{piston}$ is the piston surface area, and $P_{piston}$ is the average pressure the particles exert on the piston.
So, the right-hand side of Eq. (4) can simply be interpreted as the (negative of the) pressure experienced by the piston, or equivalently by the walls of the system. This indeed also corresponds to the bulk pressure of the system.
So I think the equation indeed holds, but the sum notation you use here can be a bit confusing in a continuous system. If we were considering a bulk system without explicit walls (e.g. periodic boundary conditions), Eq. (2) would still hold. We could also still consider the effect of volume moves on the free energy, like you do in the derivation of (3), but then typically a volume change would correspond to a rescaling of the system, rather than moving a piston. Such a rescaling changes both the number of accessible microstates and the spacing between particles (and hence the pair interactions). In that case, Eq. (3) would pick up an "ideal gas" contribution from the change in the number of microstates. As mentioned by user3725600 in a comment, a bulk ideal gas is a good example of where this would come into play.