Past exam of the theoretical physics major option course of the University of Oxford / 2015 / 8 / c / Solution

Keep the index convention of b: $A^{\mu }=(A_0,\mathbf{A})$, so $A_{\mu }=(A_0,-\mathbf{A})$ and the spatial gauge covariant derivative is $\nabla +ie\mathbf{A}$. The physical three-current is the spatial part of $J^{\mu }$:Equivalently, writing $\phi =R{e}^{i\theta }$ gives $\mathbf{j}=-2e{R}^2(\nabla \theta +e\mathbf{A})$. Under the gauge transformation in a, $\theta \mapsto \theta +\alpha$ and $\mathbf{A}\mapsto \mathbf{A}-\nabla \alpha /e$, so this expression is manifestly invariant. Switching to a convention with the opposite spatial gauge sign reverses the phase-gradient term, not the screening term proportional to $e^2$.

For static fields the spatial Ampère-Maxwell equation is $\nabla \times \mathbf{B}=\mathbf{j}$, while $\nabla \cdot \mathbf{B}=0$. Taking its curl yields the general magnetic-field equationNo uniform-condensate approximation has yet been used. For a varying scalar amplitude the gradient terms must be kept; simply replacing $|\phi {|}^2$ by a constant before deriving the current would conceal their origin. The homogeneous-condensate limit supplies the London screening equation in d.