# Why the convexity of $E(S,V,N)$ implies that at thermodynamic equilibrium the energy is minimal?

I do understand that in the case of mechanics for example, if there is a soccer ball inside a "well", at mechanical equilibrium, the ball will be at the bottom of the well.

(source: utk.edu)

This is a consequence, of the fact that the potential has a minimum there. By definition, the force (derived from this potential) acting on the ball will make the ball go at the bottom for small fluctuations around equilibrium position.

But in the case of $$E(S,V,N)$$, I don't see where is the "force" that will keep E at the minimum for small fluctuations of $$S,V$$ and $$N$$.

Actually, the following section of my course does the following to define equilibrium:

$$\frac{\partial E}{\partial S} =T$$

$$\frac{\partial E}{\partial V} =0$$

$$\frac{\partial E}{\partial N} =0$$

• I'm confused. When the free energy has two minima with energies $F_+>F_-$ then the system at $F_+$ is metastable in my dictionary and I can compute the probability for it to transition into the deeper minimum, i.e. a phase transition. I'm very intrigued by what you say regarding 'linearising' the concave region between $F_+$ and $F_-$ and how this corresponds to 'being in a phase transition'. Can you please elaborate or provide a reference? Thank you Commented Mar 4, 2021 at 20:00