# Gibbs free energy of superconducting and normal phases of a metal

Why do the Gibbs free energies of a superconducting and a normally conducting phase must be equal at the transition point?

I'm following Mehran Kardan's Statistical Physics course on MIT OCW. Working through a problem set I got stuck, caved in and looked up the solution to a particular problem, but this gave rise to the following question.

Suppose you are given a metal with a normal and a superconducting phase. I can calculate the Gibbs free energy $G$ of each individual phase as $$G=E-TS-BM.$$ This is not an issue since I had previously determined S and E.

According to the Gibbs-Duhem relation: $$E=TS+BM+\mu N$$ so we deduce $$G=E-TS-BM=\mu N,$$

for each individual phase.

According to the solutions: "At the transition point, the chemical potentials (and hence the Gibbs free energies) must be equal, leading to [...]", and the desired result follows easily from equating $G$ for each phase and solving the resulting equation. It is assumed $B=0$ at this point.

The notes previously showed (using the Helmholtz free energy $F$) that for systems in equilibrium between two phases the chemical potentials must be the same; how does the equality of the $G$ follow from this if no information about $N$ is given? or more generally: why do the Gibbs energies of both phases have to be equal at the transition point?

Link to solutions: http://li.mit.edu/Archive/CourseWork/Ju_Li/MITCourses/8.333/1997/tests/midSolution.pdf (Problem 4, point (d))

• Are you asking the general question on why the Gibbs free energies are equal at a (any) phase transition, or specifically why they are for the normal-to-superconducting transition? – Jon Custer Jan 3 '17 at 14:14
• I would prefer the general answer, yes. Why are the Gibbs free energies equal at a phase transition, from a purely thermodynamic perspective. – alonso s Jan 4 '17 at 8:11