# Electric field potential gradient in superconductivity

The following equation is equivalent to the London equations for superconductivity iff $\mathbf{\nabla} \phi = \mathbf{0}$:

$\mathbf{J} = - \dfrac{\mathbf{A}}{\Lambda c}$

where $\mathbf{J}$ is the current density, $\mathbf{A}$ is the vector potential, and $\Lambda$ is a constant.

In the steady state $\dot{\mathbf{A}} = \dot{\mathbf{J}} = \mathbf{0}$, this means $\mathbf{E} = \nabla \phi = \mathbf{0}$ which I suppose is an expression of perfect conductivity? What is the physical interpretation of $\mathbf{\nabla} \phi = \mathbf{0}$ outside of the steady state when $\mathbf{E} = \nabla \phi - \dfrac{1}{c}\dfrac{\partial \mathbf{A}}{\partial t}$?

No free charges, i.e. no electric monopoles. This is not surprising, considering that holes and electrons are bound in cooper pairs. This is because the London equation $$\mathbf{J}=-\frac{\mathbf{A}}{\Lambda c}$$ is true only in the Coulomb gauge, where $\nabla\cdot\mathbf{A}=0$. After all, $\mathbf{J}$, as a measurable physical quantity, cannot be gauge invariant. From $$\mathbf{E}=\nabla\phi -\frac{1}{c}\frac{\partial A}{\partial t}$$ we get $$\nabla \cdot \mathbf{E}=0 \Rightarrow \rho = 0,$$ $\rho$ being the electric charge density.
• Thank you. But why does $\nabla \phi = 0$ imply that there are no free charges? – oweydd Mar 30 '17 at 7:43
• Also don't the Cooper pairs act like free charges with a charge $\pm 2 e$? – oweydd Mar 30 '17 at 13:01
• I'm slowly starting to grasp it, but I still don't understand why $\rho$ should equal zero? – oweydd Mar 30 '17 at 16:56
• For a homogeneous superconductor, $\nabla \cdot \mathbf{E}$ equal zero would imply that $\nabla\cdot \mathbf{D}$ is also zero, which by means of Gauss's law, $\nabla \cdot \mathbf{D} = \rho,$ gives $\rho = 0$ – zap Mar 30 '17 at 19:29