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? | |||
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).
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Your speculation is interesting, but it is totally misleading, I am afraid. Do not go deep into that.
Your additional remark is rather confused because of terminology: 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? There, it is "the difference in free energy" and not the "free energy" (like potential, free energy is defined up to an additive constant). And the "thermodynamics system" needs to be in contact with a heat bath in order for your statement to be correct. You can think of free energy as a mechanical potential like gravitation, and work can be extracted from the system by changing the free energy. The difference is: the free energy change includes the heat exchanged with the heat bath. Note: be very careful with terminology and always read statements of thermodynamics in full, otherwise you will be lost. | |||
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. | |||
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. | |||||||||
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