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One can derive all the numerous thermodynamic potentials (Helmholtz, Gibbs, Grand, Enthalpy) by Legendre transformations, but I'm interested in seeing each from Stat Mech instead (ie taking the log of a partition function). I can do this easily for all but the Enthalpy, which has me stumped.

Easy example: Deriving Helmholtz

The Helmholtz energy can be derived by putting our system $S$ in thermal contact with a heat bath $B$ at temperature $T$ and maximizing the total entropy:

$$S_\mathrm{tot}/k_b=\log\sum_{E_S}\Omega_B(E_\mathrm{tot}-E_S)\Omega_S(E_S)$$

If the heat bath is big enough to maintain constant temperature $1/T=\left(\frac{dS_B}{dE_B}\right)_{V_B}$, then we can say $S_B\approx -E_S/T+S_0$ for small $E_S$. Then $\Omega_B\approx \Omega_0 e^{-\beta E_S}$ where $S_0=k_b\log\Omega_0$, $\beta=1/k_bT$, so

$$S_\mathrm{tot}/k_b=\log\sum_{E_S}\Omega_0e^{-\beta E_S}\Omega_S(E_S)=S_0/k_b+\log Z_S$$ where $Z_S=\sum_{E_S}e^{-\beta E_S}\Omega_S(E_S)$. So maximizing the total entropy is just maximizing $Z_S$. If we define the Helmholtz Free Energy $A_S$ as $-\beta A_S=\log Z_S$, and use the Shannon entropy $S/k_b=-\sum p\log p $, we see

$$S_S/k_b=-\sum \frac{\Omega_S(E_S)e^{-\beta E_S}}{Z}\log \frac{\Omega_S(E_S)e^{-\beta E_S}}{Z}$$ $$S_S/k_b=\frac{\beta}{Z}\sum \Omega_S(E_S)E_S e^{-\beta E_S} + \log Z$$ $$S_S/k_b=-\frac{\beta}{Z}\partial_\beta Z + \log Z$$ $$S_S/k_b=\beta \langle E_S\rangle -\beta A_S$$ $$A_S=\langle E_S\rangle -TS_S$$

The other thermodynamic potentials at fixed $T$ are similarly easy to derive.

But now try deriving Enthalpy

The same procedure does not work for enthalpy, because,

$$S_\mathrm{tot}/k_b=\log\sum_{V_S}\Omega_B(V_\mathrm{tot}-V_S,E_{B0}+pV_S)\Omega_S(V_S,E_{S0}-pV_S)$$

...if the bath is big enough to maintain a constant temperature, then its total entropy is constant as a function of $V_S$. That is, if $V_S$ increases, the bath entropy decreases due to less volume, but the bath energy increases by the same amount due to increased energy. So, to first-order, $\Omega_B$ is constant and the total entropy splits into a sum of both individual subsystem entropies.

Is there a way to derive the enthalpy from stat mech considerations, as there is for the other potentials, rather than by Legendre transforming the energy?

By "derive the enthalpy", I mean "derive that the quantity which should be minimized in equilibrium is given by $H=\langle E \rangle+p\langle V \rangle$."

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In case you want to compute the ensemble.

The ensemble you are looking for is called the Isoenthalpic-isobaric Ensemble. The paper I read to learn about it is cited in Wikipedia, Andersen, H. C. Journal of Chemical Physics 72, 2384-2393 (1980). Unfortunately, if you aren't at a university, you might not have access or want to pay, so let me summarize.

First off, we want to reproduce thermodynamics, \begin{align} {\displaystyle dH(S,p)=T\,dS+V\,dp.} \end{align}

Lets consider, \begin{align} \Gamma(N, P, H) &= C^{-1}\int dV~d^{3N}q~d^{3N}p~ \delta(\mathcal{H}(q, p; V) + PV - H) \\ &= \frac{1}{i2\pi} C^{-1} \int d\beta~e^{\beta H} \int dV~d^{3N}q~d^{3N}p~ e^{-\beta(\mathcal{H}(q, p; V) + PV)} \\ &= \frac{1}{i2\pi} C^{-1} \int d\beta~e^{\beta H} \mathcal{Z}_G(\beta, P, N) \\ \end{align} where $ C^{-1} $ just gets rid of units.

Then, \begin{align} \frac{\partial}{\partial P} k_B \log \Gamma(N, P, H) &= k_B \Gamma^{-1} \left[ \frac{1}{i2\pi} C^{-1} \int d\beta~e^{\beta H} \int dV~d^{3N}q~d^{3N}p~ (-\beta V)~e^{-\beta(\mathcal{H}(q, p; V) + PV)} \right]\\ &= - k_B \langle \beta V \rangle = -\left\langle \frac{V}{T} \right\rangle \end{align}

So we would naturally want to call,

$$ S = k_B \log \Gamma(N, P, H) $$

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You may want to find the expectation value of V

$\langle V \rangle = \frac{1}{\beta}\frac{\partial \log Z}{\partial p}$

And then just use the normal definition of Enthalpy with respect to the internal energy

$\langle H \rangle = \langle E \rangle + p \langle V \rangle $

If we maximize entropy with respect to another constraint, this time on the expected pressure;

$\langle p \rangle = \sum_{i} P_i p_i$

Then our probability distribution will have the form of

$P_i = \frac{1}{Z}e^{-\beta E_i - \lambda p_i}$

And so the condition on maximum entropy is given by $S_{max} = -k \log(P)$ which give us

$S_{max} = k\log(Z) + k\beta E_i + k\lambda p_i$

Which upon rearrangement gives for the energy

$E_i = \frac{S}{k \beta} - \frac{1}{\beta}\log(Z) - \frac{\lambda p_i}{\beta}$

We also know from the thermodynamic equations that $\langle V \rangle = \langle -\frac{\partial E_i}{\partial p}\rangle$ which gives us

$\langle V \rangle = \frac{\lambda}{\beta}$

And so, $\lambda = \beta \langle V \rangle$. This then tells us that the partition functino is equal to $Z = e^{-\beta H}$ where H is the enthalpy. Taking the log of the partition function from maximizing entropy we have

$E_i + p_i\langle V \rangle = H$

Let me know if I made any mistakes in the logic here.

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  • $\begingroup$ Thank you, N. I realize my question was ambiguous. What I'd like to do is derive that the last expression you wrote is the quantity which gets minimized in equilibrium. I updated the question to reflect this. $\endgroup$ – Sam Bader Mar 15 '16 at 2:14
  • $\begingroup$ Hey Sam, I made some edits, let me know if this jives at all $\endgroup$ – N. Carrara Mar 16 '16 at 14:43

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