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\title{Superconductivity - Assignment 2}
\author{
	Kees van Kempen (s4853628)\\
	\texttt{k.vankempen@student.science.ru.nl}
}
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\begin{document}

\section{Temperature dependence in Landau model}
In the Landau model, free energy is given as function of order parameter $\psi$ and temperature $T$ as
\[
	\mathcal{F} = a(T - T_c) \psi^2 + \frac{\beta}{2}\psi^4.
\]

The equilibrium state as function of temperature $T$ is the state of minimal free energy with respect to the order parameter $\psi(T)$.
This point we call $F_0(T)$ with order parameter $\psi_0(T)$.
For this, we will take the derivative of $F$ with respect to $\psi$ and equate it to zero.

\[
	0 = \pfrac{\mathcal{F}}{\psi} = \pfrac{}{\psi} \left[ a(T-T_c)\psi^2 \right] = 2a(T-T_c)\psi + 2\beta\psi^3
\]

Extreme points are found at $\psi = 0$ and $\psi = \pm\sqrt{\frac{-a}{\beta}(T-T_c)}$.

For $T \geq T_c$, $\psi_0(T \geq T_c) = 0$ gives the minimum, i.e. $\mathcal{F}_0(T \geq T_c) = 0$.

For $T \leq T_c$, $\psi_0(T \leq T_c) = \sqrt{\frac{-a}{\beta}(T-T_c)}$ is the minimum,
giving free energy
\[
	\mathcal{F}_0(T \leq T_c) = \frac{-a^2}{\beta}(T-T_c)^2 + \frac{a^2}{2\beta}(T-T_c)^2 = \frac{-a^2}{2\beta}(T-T_c)^2 \leq \mathcal{F}_0(T \geq T_c)
\]
where we chose the positive of the $\pm$ as the order parameter is understood to increase from finite at the phase transition.
\todo{Is this a reasonable statement? It actually does not really matter that much as mostly $\psi^2$ is used, but the physical meaning is totally different. It implies some kind of symmetry, too. It seems that also \cite{abrikosov} mentions this.}
	
For the specific heat, we find
\[
	C(T) = -T\pfrac{^2\mathcal{F}}{T^2} =
	\begin{cases}
		0 & T > T_c \\
		\frac{a^2}{\beta}T & T < T_c
	\end{cases}.
\]
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 $\vec{J_s}$:
\[
	\vec{J_s} = -\frac{2e\hbar n_s}{m}(\nabla\theta + \frac{2e\vec{A}}{\hbar})
\]
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 ($\nabla \cdot \vec{A} = 0$).

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}.

\item
We apply a gauge transformation as follows.
\begin{align}
	\chi(\vec{r}, t) &= \frac{-\hbar}{2e}(\omega t - \vec{k} \cdot \vec{r}) \\
	\vec{A} &\to \vec{A} + \nabla\chi = \vec{A} + \frac{\hbar}{2e} \vec{k} \\
	\phi &\to \phi - \pfrac{\chi}{t} = \phi + \frac{\hbar}{2e} \omega
\end{align}

\todo{Do I really need to put in the previously found $\vec{A}$?}
\end{enumerate}

\section{Type II superconductors and the vortex lattice}
\begin{em}
The nobel prize lecture by Abriskosov \cite{abrikosov} was really interesting.
The start was a good recap of the breakthroughs relevant to conventional superconductivity,\footnote{Why is that every story on superconductivity includes KGB captivity?}
but in pages 61--63, the theory is worked through a little quickly.
I might reread it some times.
\end{em}

\section{Currents inside type-II superconducting cylinder}

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%\appendix

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