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))