Consider the definition of Helmholtz free energy (assume canonical ensemble and classical particles):

$$ A = - k_B T \ln Z $$

Now we know that at any temperature above absolute zero, $k_B T$ term is positive. Furthermore, based on the definition of canonical partition function (ignoring multiplicity, and assuming discrete microstate energies to keep things simple):

$$ Z = \sum_i \exp\left(-\frac{E_i}{k_B T}\right) $$

For the first term in the sum, energy is zero ($E_i = 0$), therefore the term equals 1. For higher energies, the terms each equal less than 1, but positive. Therefore, the sum is always greater than 1. This means $\ln Z$ is always greater than zero.

If $k_B T$ is always positive, and $\ln Z$ is positive, should $A$ be negative?

The only way $A$ can be positive is when $\ln Z$ is negative, meaning $Z$ less than 1. That is, if the multiplicity/degeneracy ($g_j$) of the system is such that $E_j = 0$ term doesn't contribute (and ideally many states between 0 and $k_B T$, or even up to a few $k_B T$, don't contribute). $g_j$ is used the following way,

$$ Z = \sum_j g_j \exp\left(-\frac{E_j}{k_B T}\right) $$

(I replaced the index $i$ with $j$ to indicate that now we're talking about energy levels, and not microstates).

Does that mean that, for $A$ to be positive, it is necessary, but not sufficient, that the multiplicity/degeneracy of the zero-energy microstate be zero?


Taking into account degeneracy does not help to make $Z<0$. The expression $$ \sum_i \mathrm{exp}\left( -\frac{E_i}{k_BT} \right) $$ contains a sum over the states, whose degeneracy in energy does not matter in this sum. The problem is that you make two independent wrong assumptions:

  1. that the lowest energy is zero;
  2. that the lowest energy does not depend on thermodynamic variables.

As far as point 1, it is well known that energy is defined only within an arbitrary constant. If all the energies of the states are shifted by adding the same constant $E_0$ to the energy of each state, the partition function gets a factor $\mathrm{exp}\left(-E_0/k_BT\right)$ which results in a free energy $A^{\prime}=A+E_0$. And, of course, $E_0$ may be positive or negative.

Point 2 is a little more subtle. It is true that, at fixed volume $V$ and number of particles $N$, the energies $E_i$ are limited below. But such a lower limit does depend on $V$ and $N$. Therefore, if, for a given $V_0$ and $N_0$, one takes the lower bound of the energy as reference energy, and consequently the lowest energy at that $V_0$ and $N_0$ is zero, this is not implying that free energy must be always negative. Taking into account that with this choice $A(T=0,V_0,N_0)=0$, the simultaneous validity of the third principle of thermodynamics ($S(T=0,V,N)=0$ for all $V$ and $N$ ), and the concavity of $A$ as a function of $T$, imply $A(T,V_0,N_0)<0)$ but also that for positive pressures there will be an increasing interval of temperatures where $A>0$, for $V<V_0$.


Yes $A$ (sometimes termed $F$) can be negative, although both options possible. See accepted Stack answer here

Anyhow, with $k_B=1$

$A= -T ln⁡Z= U – TS$

$U-TS=-W$ so $W=-A$

The negative of $A$ is the useful free work $W$ obtainable. Note that Helmholtz free energy $A$ is an equilibrium concept.

Have a look also here and GSU

  • 1
    $\begingroup$ Answer by @LubošMotl you are referring contains a mistake, although the main point (the additive shift ambiguity of energy) is ok. If the lowest energy appearing in the sum over the states is zero, the is no way to make $Z<1$ whatever is the temperature. $\endgroup$
    – GiorgioP
    Feb 2 '20 at 8:46
  • $\begingroup$ Well I suggest you add that comment under that accepted answer, so @LubosM can respond, since it will no doubt be noticed more than here lol. Personally, my original edit of my answer had A as zero or negative. Maybe I should ask can A be positive! $\endgroup$ Feb 2 '20 at 9:01

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