Helmholtz Free Energy, Partition Function

Question

I'm trying to develop some basic intuition here, so this comes mostly as a jumble of commentary/questions. Hope its acceptable.

Helmholtz Free Energy: $A = -{\beta ^{-1}}lnZ$. I find this statement to be incredible profound. Granted, I found it yesterday.

Suppose my system has one energy state with no degeneracy. $Z = e^{-\beta E_1}$, then $A = E_1$, which I suppose says if the system consists of one particle, all its internal energy is available for work. That's nice.

Now, if we introduce some degeneracy $\gamma$, we get $Z = \gamma e^{-\beta E_1}$, and so $A = E_1 - \beta ^{-1}ln \gamma$, and we have clearly lost some of our free energy to the degeneracy (ie. to the fact that there are multiple microstates for our given macrostate, and so we have limited information about the actual configuration of the system, which is free to explore its micro states, limiting the energy we can get from it). So that's nice too.

We can go further by introducing more energies, so $Z = \Sigma \gamma_i e^{-\beta E_i}$, but nice analysis is confounded by my inability to deal coherently with sums in a logarithm. Though I managed to show that $A$ for such a multi-state system is strictly less than $\Sigma [E_i-\beta ^{-1}ln\gamma _i]$, ie. less than the sum of the free energies for independent systems of a given energy $E_i$ and degeneracy $\gamma_i$ . This result, however, requires $E_i > 0$, which I take for granted, but makes plenty sense.

Now, what does it mean for A to be negative? Perhaps more importantly, how does one simply go about obtaining work from a system with some A (a practical question)? Or, perhaps even more importantly, is it this requirement that there be a second final state, seemingly of lower free energy, that makes $A$ itself not so significant, but rather $\Delta A$? And if so, what happens to the intuition about a system with only one state having exactly its energy as free-energy?

Your insights on these and related matters pertaining to legendary $Z$ and its relation to $A$, as well as pointers on where my thinking may be flawed or enlightened, are much appreciated.

Answer 1

Now, what does it mean for A to be negative?

It isn't negative in general. The mistake you're doing is to assume that the logarithm is positive. But $\ln Z$ may be both positive and negative depending on whether $Z$ is greater than one or smaller than one. Both options are possible because $Z=\sum \exp(-E_i/kT)$ and if $E_i$ is smaller than $kT$ (and smaller by an extra margin that compensates the degeneracy), you will get $Z\lt 1$ and $A\gt 0$. After all, the quantity $A$ has the same additive shift ambiguity as any form of energy (any thermodynamic potential), $E\to E+\Delta E$ doesn't change anything about non-gravitational physics, so whether $A$ is smaller or greater than a particular ad hoc threshold (although it's denoted "zero") is a physically inconsequential question.

Perhaps more importantly, how does one simply go about obtaining work from a system with some $A$ (a practical question)?

There's obviously no "mechanical recipe" to construct clever and/or efficient engines. The value of $-A$ nevertheless encodes the maximum amount of useful work that may be extracted from the system whose Helmholtz free energy is $A$. If one wants not to lose energy, one tries to make the individual stages of the engine as reversible as possible, like in the Carnot cycle. So one should better be sure that the heat is transferred between bodies of (almost) the same temperature etc.

Or, perhaps even more importantly, is it this requirement that there be a second final state, seemingly of lower free energy, that makes $A$ itself not so significant, but rather $\Delta A$?

The claim that $-A$ measures the maximum amount of useful work that may be extracted holds assuming that $E=0$ is the minimum non-thermal energy where one can get. So yes, the additive shift ambiguity is always there and the maximum extraction of the useful work really means that we are interested in the value of $\Delta A$, the changes of energy or thermodynamic potentials, and not the values themselves. The final state's having $E=0$ is what de facto identifies $E_{initial}$ and $-\Delta E$ and similarly for other quantities.

And if so, what happens to the intuition about a system with only one state having exactly its energy as free-energy?

If there's only one state, there is no interesting thermodynamics or statistical physics. The state has energy $E=A$ and the maximum amount of useful work that may be extracted is whatever the mechanical properties of the state allow. If it's not specified otherwise, the assumption is that $E=0$ which also means $A=0$ in this case is the minimum value one can get to.