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 course: https://ocw.mit.edu/courses/physics/8-333-statistical-mechanics-i-statistical-mechanics-of-particles-fall-2013/index.htm

Link to assignment: https://ocw.mit.edu/courses/physics/8-333-statistical-mechanics-i-statistical-mechanics-of-particles-fall-2013/assignments/MIT8_333F13_pset1.pdf

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

  • $\begingroup$ 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? $\endgroup$ – Jon Custer Jan 3 '17 at 14:14
  • $\begingroup$ I would prefer the general answer, yes. Why are the Gibbs free energies equal at a phase transition, from a purely thermodynamic perspective. $\endgroup$ – alonso s Jan 4 '17 at 8:11

In the superconductivity field you can find your answers in the Ginzburg-Landau fenomenological theory [1].

In general, every time that you describe your statistical state in function of one (or more) order parameter, you have all termodinamic quantities in function of it, and so you can find divergences points in function of the order parameter, called critical point, that identify the phase transition of the system.

[1] Quantum Theory of Many-Particle System - A.L. Fetter, J.D. Walecka pag. 430-439


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