ass2: Do assignment 4 as far as possible

superconductivity_assignment2_kvkempen.tex

@@ -77,7 +77,92 @@ For the specific heat, we find
 	\end{cases}.
 \]
 There is thus a discontinuity in $C(T)$ at $T = T_c$ with size $\Delta C(T) = \frac{a^2}{\beta}T_c$.
+
 \section{Type-I superconducting foil}
+\begin{enumerate}
+\item
+The screening equation is given as
+\[
+	\nabla^2\vec{B} = \frac{\vec{B}}{\lambda}.
+\]
+For easy of calculation, we will use cartesian coordinates,
+and put the external magnetic field $B_E$ along the $x$ axis:
+$\vec{B_E} = B_E \hat{x}$.
+A foil with thickness $a$ we put parallel to the $xy$ plane with the middle of the thickness at $z = 0$ such that the foil fills $-\frac{a}{2} < z < \frac{a}{2}$.
+Due to symmetry in the $xy$ plane of the system, the field inside the foil can only depend on $z$ coordinates.
+So we define the magnitude of the field $|\vec{B}| = B(z)$. 
+
+Using the screening equation, we look for a solution.
+\[
+	\nabla^2\vec{B} = \nabla(\nabla\cdot\vec{B}) - \nabla\times(\nabla\times\vec{B})
+\]
+$\vec{B}$ is divergenceless, so we are left with the latter term.
+Next, we take the curls writing $B_i$ for the $i$th component of $\vec{B}$,
+and realize that we only have $z$ dependence, and $B_y = 0 = B_z$.
+\[
+	-\nabla\times(\nabla\times\vec{B}) = -\nabla\times(\pfrac{B_x}{z} \hat{y}) = -(-\pfrac{^2B_x}{z^2} \hat{x})
+\]
+Rewriting yields
+\[
+	\vec{B} = \lambda \pfrac{^2B_x}{z^2}\hat{x}.
+\]
+For this we know the general solution:
+\[
+	\vec{B} = B_0 \left[ C \cdot e^{z/\lambda} + D \cdot e^{-z/\lambda} \right],
+\]
+with constants $B_0$, $C$, and $D$.
+
+Now we can apply two boundary conditions to find the solution inside the material.
+
+First, due to mirror symmetry in $z$, we require $B(z) = B(-z)$, giving that $C = D$,
+thus we contract the constants as $B'_0 = CB_0 = DB_0$.
+This allows us to write the exponents into $cosh$ form.
+\[
+	B(z) = B'_0 \left[ e^{z/\lambda} + e^{-z/\lambda} \right] = B'_0 \cosh{\frac{z}{\lambda}}
+\]
+
+Second, just outside the foil, at $z = \pm \frac{a}{2}$, the field must be $B_E$, and the field should be continuous across the boundary:
+\[
+	B_E = B(\frac{a}{2}) = B'_0 \cosh{\frac{a}{2\lambda}} \iff B'_0 = \frac{B_E}{\cosh{\frac{a}{2\lambda}}}
+\]
+
+This gives us our final expression for $B(z)$:
+\[
+	B(z) = 
+	\begin{cases}
+		B_E\frac{1}{\cosh{\frac{a}{2\lambda}}}\cosh{\frac{z}{\lambda}} & |z| \leq \frac{a}{2} \\
+		B_E & |z| \geq \frac{a}{2}
+	\end{cases}.
+\]
+
+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 ($\pfrac{\vec{E}}{t} = 0$):
+\[
+	\nabla\times\vec{B}(z) = \mu_0\vec{J_s}
+\]
+Reordering and calculating the curl gives:
+\[
+	\vec{J_s} = \frac{1}{\mu_0} \nabla \times (\pfrac{B(z)}{z}\hat{x}) = \frac{B_E}{\mu_0 \lambda \cosh{\frac{a}{2\lambda}}} \sinh{\frac{z}{\lambda}} \hat{y}
+\]
+
+\item
+From the derivation of the Ginzburg-Landau theory, we get the following expression for the supercurrent $\vec{J_s}$:
+\[
+	\vec{J_s} = -\frac{2e\bar{h}n_s}{m}(\nabla\theta + \frac{2e\vec{A}}{\bar{h}})
+\]
+Using the rigid gauge, we set $\theta = 0$.
+Next, we can equate the previously found supercurrent for our foil to the Ginzburg-Landau found one and reorder to find $\vec{A}$:
+\[
+	\vec{A} = \frac{-B_E m \sinh{\frac{z}{\lambda}}}{4\lambda\mu_0 e^2 n_s \cosh{\frac{a}{2\lambda}}} \hat{y}
+\]
+
+\item
+\[
+	\nabla \cdot \vec{A} = \pfrac{A_x}{x} + \pfrac{A_y}{y} + \pfrac{A_z}{z} = \pfrac{0}{x} + \pfrac{A_y}{y} + \pfrac{0}{z} = 0,
+\]
+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}
 
 \section{Type II superconductors and the vortex lattice}