# Legendre transformation and thermodynamics

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? )

• arxiv.org/pdf/0806.1147.pdf you might want to read a bit about the mathematical properties of Legendre transforms. I don't find my old course scripts from stat mech, but I think most of what you say can be shown. The linked article might be a good start. – Sanya Sep 23 '16 at 18:33
• Please refer Thermodynamics by Callen. There is a chapter on thermodynamic potentials where such proofs are given. – Deep Sep 24 '16 at 4:56

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 $\kappa=0$.