Legendre transformation and thermodynamics

Question

I have a question about Legendre transformation.

Imagine that we have a function depending on $x$.

$$ df = \frac{df}{dx} dx = u(x) dx$$

We assume that this quantity has to be minimized (for example $f$ could be $F$, the Helmoltz function (where we only vary the volume for example).

So : $$ dF(V)=\frac{dF}{dV} dV = P(V) dV $$

We are interested in studying the equilibrium in function of the pressure (we apply a constant pressure on our system and we want to see how the volume will be).

As $$ d(PV)=PdV+VdP $$

We can define $G=F-PV$ such as $$ dG=-VdP$$

And here is my question:

I want to transform a function $F$ depending only on the volume to a function $G$ depending only on the pressure.

I know from my knowledge on thermodynamics that $G$ has to be minimized at equilibrium.

But can it be proved mathematically from the fact that $F$ must be minimized at equilibrium in regards of $V$, so $G$ has to be minimized in regards of $P$? I didn't succeed to do it.

Or it can't be proved mathematically : we have to admit that $G$ has to be minimized by using law of thermodynamics (it is not a consequence of Legendre transformation).

(by the way if someone can help me on this thread I would be grateful :D Why is the the differential of Helmoltz free energy dT dependent? )

Answer 1

You have to remember that the statements that "Helmholtz Free Energy is minimized at equilbrium" and "Gibbs Free Energy is minimized at equilbrium" are referring to two different notions of minimizing. In the former case, you minimize at fixed volume; in the latter case, you minimize at fixed pressure.

Once you understand this then the equivalence of the two statements is easy to see. Imagine for generality that in addition to pressure/volume your system has another parameter $\alpha$. (For example, $\alpha$ could be the ionization fraction of water molecules.) Then you have $$ dF = -P dV + \kappa d\alpha$$ where $\kappa = (\partial F)/(\partial \alpha)_V$. This implies that $$dG = V dP + \kappa d\alpha.$$ Now you see that the statement "Helmholtz free energy is minimized at fixed volume (dV =0)" and "Gibbs Free energy is minimized at fixed pressure (dP=0)" are both equivalent to $\kappa=0$.