ass2: Add comma, answer the gauge equality criteria

superconductivity_assignment2_kvkempen.tex

@@ -28,6 +28,7 @@
 \usepackage{hyperref}
 \usepackage{float}
 \usepackage{mathtools}
+\usepackage{amsmath}
 
 \newcommand{\pfrac}[2]{\frac{\partial #1}{\partial #2}}
 
@@ -135,7 +136,7 @@ This gives us our final expression for $B(z)$:
 	\end{cases}.
 \]
 
-The supercurrent follows from the Maxwell-Amp\`ere law considering that there are no other currents, and we look at a current steady over time ($\pfrac{\vec{E}}{t} = 0$):
+The supercurrent follows from the Maxwell-Amp\`ere law, considering that there are no other currents, and we look at a current steady over time ($\pfrac{\vec{E}}{t} = 0$):
 \[
 	\nabla\times\vec{B}(z) = \mu_0\vec{J_s}
 \]
@@ -164,6 +165,26 @@ as $A_y \perp \hat{y}$, giving zero partial derivative.
 In our case, indeed the rigid gauge choice gives the criterium for the London gauge ($\nabla \cdot \vec{A} = 0$).
 \end{enumerate}
 
+In the rigid gauge, the order parameter $\psi$ is constant in space and time.
+To then also have that $\nabla \cdot \vec{A} = 0$, follows from the expression for the supercurrent as we saw earlier.
+
+Reversely, assume that $\nabla \cdot \vec{A} = 0$, and look at what conditions need to be met in order to imply rigid gauge.
+Again, we look at the expression for the supercurrent as function of $\theta$ and $\vec{A}$,
+\begin{align*}
+	\vec{J_s} &= -\frac{2e\hbar n_s}{m}(\nabla\theta + \frac{2e\vec{A}}{\hbar}) \\
+	\iff \frac{2e}{\hbar}\vec{A} &= -\frac{m}{2e\hbar n_s} \vec{J_s} - \nabla\theta,
+\end{align*}
+and take the divergence,
+\[
+	\frac{2e}{\hbar}\nabla \cdot \vec{A} = -\frac{m}{2e\hbar n_s} \nabla \cdot \vec{J_s} - \Delta\theta = 0
+	\implies \Delta\theta = -\frac{m}{2e\hbar n_s} \nabla \cdot \vec{J_s}.
+\]
+This is what only the London gauge implies.
+But when is then the rigid gauge applied by this?
+This is the case for $\nabla \cdot \vec{J_s}$, or, in words, when there is no conservation of supercurrent.
+If this is not the case (if the divergence is non-zero), there is conversion between normal current and supercurrent.
+This result seems to Waldram's conclusion in \cite[p. 24--26]{waldram}.
+
 \section{Type II superconductors and the vortex lattice}
 
 \section{Currents inside type-II superconducting cylinder}