superconductivity/superconductivity_assignment2_kvkempen.tex

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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.
	
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 ($\nabla \cdot \vec{A} = 0$).
\end{enumerate}

\section{Type II superconductors and the vortex lattice}

\section{Currents inside type-II superconducting cylinder}

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%\appendix

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