# Is thermodynamic free energy and potential energy the same thing?

The equation for free energy $F$ and potential energy $E_{pot}$ are: $$F=U-TS \\ E_{pot} = E_{tot} -E_{kin}$$ But the temperature $T$ is proportional to the average kinetic energy of a system. So the equations are really similar in case the entropy is constant.

In addition the potential energy is sometimes defined as capacity of a system to do work. On the other side free energy is defined as the amount of work that a thermodynamic system can perform. Isn't that the same thing?

No, because $F$, in fact, has a statistical meaning, $F$, or rather $e^{- \frac{F}{kT}}$, is a sum along different possible energies configurations :

$Z = e^{- \frac{F}{kT}} = \sum_i e^{- \frac{E_i}{kT}}$

So you cannot compare with one-particle (non statistical) physical quantities (for instance the energy of the particle would have only one value).

What is possible is analogies between quantum field theories ans statistical theories, because in quantum field theories, you are summing on all possible paths.

• Can one say that thermodynamic free energy is the same as potential energy in the limit when the number of particles N -> 1? Aug 15 '14 at 9:58
• I think the meaningful interpretation of thermodynamics is statistical. Thermodynamics is not interesting if you consider only one (classical) particle. Aug 15 '14 at 11:16

Thermodynamic free energy is a subset of potential energy. There are at least two types of thermodynamic free energy - Gibbs free energy and Helmholtz free energy. Both describe the potential energy of a system under certain conditions. For Gibbs free energy, the conditions are constant temperature and pressure - useful for biochemists - because the two are usually more or less constant in the human body. Helmholtz free energy specifies different conditions - just constant temperature (and not constant pressure). This makes the use of Helmholtz free energy better for processes that do not occur at constant pressure (i.e. bomb engineering).

This is a good question, but the answer is that the energy equations of thermodynamics and dynamics now cannot correspond to one by one, completely, in that the theoretical structures of the two theories are different. Since the internal energy

\begin{align}U=TS+Yx+\sum_j\mu_jN_j-pV.\end{align}

Helmholtz free energy

\begin{align}F=U-TS=Yx+\sum_j\mu_jN_j-pV.\end{align}

Here $Yx$ is the potential energy of the system, $\sum_j\mu_jN_j$ involve the structure of matter and phases, $-pV$ is the thermodynamic potential, denotes the free energy that can be obtain from heat conversion. $pV$ involves the thermal motion within the system.

Your speculation is interesting, but it is totally misleading, I am afraid. Do not go deep into that.