# Magnetic field interaction energy term

I am trying to derive the Ginzburg-Landau free energy for superconductors, and I follow most of it apart from the introduction on the term containing the magnetic field.

This is the bit I don't understandt, in my notes:

A final part in the free energy is the relevant magnetic field energy density $B_M^2/2\mu_0$, where $B_M=B-B_E$ is due to currents in the superconductor and $B_e$ is due to external sources. (Note that when the material is introduced the total field energy density changes from $B_E^2/2\mu_0$ to $B^2/2\mu_0$, but the part $B_EB_M/\mu_0$ is taken up by the external sources (Waldram, Ch. 6).)

So finally we arrive at the Ginzburg-Landau free energy density:

$$f=\alpha|\psi|^2 + \frac\beta2 |\psi|^4 +\frac{1}{2m} \left| (-i\hbar \nabla +2eA)\psi \right|^2 + \frac{1}{2\mu_0} (B-B_E)^2$$

With just an external field the energy density should be $\frac{1}{2\mu_0}B_E^2$. After the introduction of the material, the total field is now $\mathbf{B} = \mathbf{B_E} + \mathbf{B_M}$ and the energy density is $\frac{1}{2\mu_0}B^2 = \frac{1}{2\mu_0}(\mathbf{B_E}+\mathbf{B_M})^2$.

So the energy added to the system should be their difference, i.e. $$\frac{1}{2\mu_0}(\mathbf{B_E}+\mathbf{B_M})^2 - \frac{1}{2\mu_0}B_E^2 = \frac{1}{2\mu_0}B_M^2 + \frac{1}{\mu_0}\mathbf{B_M}\cdot\mathbf{B_E}.$$

Questions:

1. Why should I not include the $\frac{1}{\mu_0}\mathbf{B_M}\cdot\mathbf{B_E}$ term, like it is said in the notes? I did have a look on Waldram but still did not get it.

2. Is $\mathbf{B} = \mathbf{B_E} + \mathbf{B_M}$ the same thing as $\mathbf{B} = \mu_0(\mathbf{H} + \mathbf{M})?$

3. Is the vector potential $\mathbf{A}$ given by $\mathbf{B} = \nabla \times \mathbf{A}$, i.e. the total field? Or just $\mathbf{B_M} = \nabla \times \mathbf{A}$?

• I have a feeling it is just a matter of poor choice of vocabulary. To me $B$ is the field induced by the supercurrents, which excactly cancels out $B_E$ in the superconducting state. In the bulk, you thus have $B_M \neq 0$ in the normal state and $B_M = 0$ in the superconducting state. Does it make sense ? – Dimitri Apr 21 '16 at 8:43

I always find problems involving the correct thermodynamic form of magnetic field work and energy to be tricky precisely because it is not always easy to decide whether you should include the work done by or on the external power supply. To escape this worry, I take the view that the free energy functional should be such that its variation gives the correct equations, including Maxwell. If you write ${\bf B}=\nabla\times {\bf A}$ and vary ${\bf A}$ (and ignore any integrated out boundary terms) you get from your expression for $F[{\bf A},\psi]$ $$0=\frac{\delta F}{\delta {\bf A}}= \nabla \times \frac 1{\mu_0} (\nabla \times {\bf A} - {\bf B}_{\rm ext})-2e {\bf J}_{\rm int},$$ where $${\bf J}_{\rm int} = \frac{\hbar}{2mi} \left(\psi^*\left(\nabla-\frac{2e}{\hbar}{\bf A}\right)\psi - \psi \left(\nabla+\frac{2e}{\hbar}{\bf A}\right)\psi^*\right)$$ is the number-current of charge $2e$ superconducting Cooper pairs. As $$\nabla\times\frac 1 {\mu_0} {\bf B}_{\rm ext} = {\bf J}_{\rm ext}$$ is the external current, the $\delta {\bf A}$ variation seems to give the correct Maxwell $\nabla\times {\bf H}={\bf J}$ equation, ${\bf J}$ including both internal and external currents. Consequently the given free energy expression is the correct one.
1) The omitted term has to do with the work being done by the power supply that mantains the external current against the induction back EMF when the field ${\bf B}_{\rm int}$ changes. The omission is analogous to adding $PV$ to the Helmholtz free energy $F=U-TS$ to get the Gibbs free energy $G=U-TS+PV$ when we have a system at constant pressure rather than at constant volume.
2)and 3) The field ${\bf B}= \nabla\times {\bf A}$ is the total field. I don't think it necessary to introduce "${\bf M}$" which would be defined by $\nabla \times {\bf M} = 2e {\bf J}_{\rm int}$ so that setting ${\bf B}=\mu_0(\tilde {\bf H}+{\bf M})$ would give an equation $\nabla \times \tilde {\bf H}= J_{\rm ext}$ that conceals the currents due to the superconductor. This is what "${\bf H}$" is designed to do in a magnet (conceal the bound currents), but it seems inappropriate here, so my original ${\bf H}$ (with no tilde) is just ${\bf B}/\mu_0$ where ${\bf B}$ is the total field.