235 lines
9.0 KiB
TeX
Executable File
235 lines
9.0 KiB
TeX
Executable File
\documentclass[a4paper, 11pt]{article}
|
|
\usepackage[utf8]{inputenc}
|
|
|
|
\usepackage[
|
|
a4paper,
|
|
headheight = 20pt,
|
|
margin = 1in,
|
|
tmargin = \dimexpr 1in - 10pt \relax
|
|
]{geometry}
|
|
|
|
\usepackage{fancyhdr} % for headers and footers
|
|
\usepackage{graphicx} % for including figures
|
|
\usepackage{booktabs} % for professional tables
|
|
\setlength{\headheight}{14pt}
|
|
|
|
|
|
\fancypagestyle{plain}{
|
|
\fancyhf{}
|
|
\fancyhead[L]{\sffamily Radboud University Nijmegen}
|
|
\fancyhead[R]{\sffamily Superconductivity (NWI-NM117), Q3+Q4}
|
|
\fancyfoot[R]{\sffamily\bfseries\thepage}
|
|
\renewcommand{\headrulewidth}{0.5pt}
|
|
\renewcommand{\footrulewidth}{0.5pt}
|
|
}
|
|
\pagestyle{fancy}
|
|
|
|
\usepackage{siunitx}
|
|
\usepackage{hyperref}
|
|
\usepackage{float}
|
|
\usepackage{mathtools}
|
|
\usepackage{amsmath}
|
|
\usepackage{todonotes}
|
|
\setuptodonotes{inline}
|
|
|
|
\newcommand{\pfrac}[2]{\frac{\partial #1}{\partial #2}}
|
|
|
|
\title{Superconductivity - Assignment 2}
|
|
\author{
|
|
Kees van Kempen (s4853628)\\
|
|
\texttt{k.vankempen@student.science.ru.nl}
|
|
}
|
|
\AtBeginDocument{\maketitle}
|
|
|
|
% Start from 4
|
|
\setcounter{section}{3}
|
|
|
|
\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}
|
|
|
|
|
|
|
|
\todo{The essay so far is just a draft. Choosing a topic was hard.}
|
|
|
|
\section{Currents inside type-II superconducting cylinder}
|
|
For $B_{c1} < B_E < B_{c2}$, the cylinder of type-II superconductor material is in the mixed state.
|
|
In the mixed or vortex state, superconductors let through a number of finite flux quanta $\Phi_0$.
|
|
Some small regions of the material are not superconducting, but in the normal state.
|
|
Flux passes through these regions in multiples of $\Phi_0$, but usually just one $\Phi_0$ per region,
|
|
and a supercurrent is generated to expel the field from the rest of the material.
|
|
This supercurrent moves around these region in a vortex-like shape.
|
|
|
|
Please see the figure below for a beautiful drawing.
|
|
It was not specified what the direction of $\vec{B_E}$ was with respect to the cylinder orientation, so I chose what I thought was most reasonable as an example.
|
|
|
|
\begin{figure}
|
|
\centering
|
|
\includegraphics[width=.8\textwidth]{cylinder-vortex-state.png}
|
|
\end{figure}
|
|
|
|
\bibliographystyle{vancouver}
|
|
\bibliography{references.bib}
|
|
|
|
%\appendix
|
|
|
|
\end{document}
|