What kind of free energy do we use for a superconductor in a magnetic field?

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Deja Navarro · Accepted Answer

Ah, this is a classic and rather tragic issue in most statistical-mechanics books. I won't talk about superconductivity, but the general case of magnetic systems. The problem is thermodynamics as it is, was formulated for fluid systems for which pressure and volume were observables and could be manipulated. In that case, say I calculated the partition function of a given microscopic Hamiltonian of the system, then

Z = T r e^{- β H}

and

F = - k_{B} T \ln Z

is the Helmholtz free energy of the system as

(T, V)

are kept constant (It is an (N,V,T) ensemble if we include the chemical potential

μ

). Anyway, if we impose upon the system an external force, aka, pressure, then the ensemble changes to a

(p, T)

one and we instead have the Gibbs potential to work with. So,

Z [p] = T r e^{- β [H + \int p V]}

and

G (p, T) = - k_{B} T \ln Z [p]

. Now the analogue of pressure and volume in magnetic systems is total net magnetization (M) and the external field (H). They form a conjugate pair of generalized extensive variable and generalized intensive force. So, the correspondence is

M \leftrightarrow V

and

H \leftrightarrow - p

Z [H] = T r e^{- β [H - \int M H]}

Hence, in the presence of an external field, we actually have the Gibbs free energy

G (H, T)

instead of the Helmholtz free energy

F (M, T)

, which will only be used in the null field case.
Most books gloss over this and continue to call the potential the Helmholtz free energy in all cases for magnetic systems and almost always get away with it. It doesn't change most of the calculations but it can be confusing conceptually, as in your case.

What kind of free energy do we use for a superconductor in a magnetic field?

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