The london model

+We assume that the current flows without dissipation and has the form

+

+\begin{equation} \mathbf{j}_s = n_s e \mathbf{v}_s \end{equation}

+

+

+whence the velocity of superconducting electrons is

+

+\begin{equation} \mathbf{v}_s = \frac{\mathbf{j}}{n_s e} \end{equation}

+

+where n_s is their density. The total free energy is a sum of the kinetic energy of superconducting electrons and the magnetic energy

+

+\begin{equation} F = \frac{1}{2} n_s m v_s^2 + \frac{1}{8\pi} \int dV \, h^2 \end{equation}

+

+Here h is the “microscopic” magnetic field. Its average over a large area in the sample gives the magnetic induction B. Using the Maxwell equation

+

+\begin{equation} \mathbf{j}_s = \frac{c}{4\pi} \nabla \times \mathbf{h} \end{equation}

+

+

+we transform this to the following form

+

+\begin{equation} F = \int dV \left( \frac{h^2}{8\pi} + \lambda_L^2 \left( \nabla \times \mathbf{h} \right)^2 \right) \end{equation}

+

+

+where

+

+\begin{equation} \lambda_L = \sqrt{\frac{m c^2}{4\pi n_s e^2}} \end{equation}

+

+is called the London penetration depth. In equilibrium, the free energy is minimal with respect to the distribution of the magnetic field. Variation with respect to h gives

+

+\begin{equation} \frac{1}{4\pi} \delta F = \int dV \left( \mathbf{h} \cdot \delta \mathbf{h} + \lambda_L^2 (\nabla \times \mathbf{h}) \cdot (\nabla \times \delta \mathbf{h}) \right) \end{equation}

+

+

+Using the identity

+

+\begin{equation} \text{div} [\mathbf{b} \times \mathbf{a}] = \mathbf{a} \cdot [\nabla \times \mathbf{b}] - \mathbf{b} \cdot [\nabla \times \mathbf{a}] \end{equation}

+

+and putting \mathbf{a} = \nabla \times \mathbf{h}, \mathbf{b} = \delta \mathbf{h}, looking for a free energy minimum and omitting the surface term we obtain the London equation:

+

+\begin{equation} \mathbf{h} + \lambda_L^2 \nabla \times \nabla \times \mathbf{h} = 0 \end{equation}

+

+

+Since

+

+\begin{equation} 2 \nabla \times \nabla \times \mathbf{h} = \nabla \, \text{div} \, \mathbf{h} - \nabla \mathbf{h} \end{equation}

+

+and \text{div} \, \mathbf{h} = 0, we find

+

+\begin{equation} \nabla^2 \mathbf{h} - \lambda_L^2 \nabla \mathbf{h} = 0 \end{equation}

+