ass2: The essay is hard.

.gitignore

@@ -44,3 +44,6 @@
 # Assignment 1
 sc_elements.pdf
 superconductivity_assignment1_kvkempen.pdf
+
+# Assignment 2
+superconductivity_assignment2_kvkempen.pdf

makefile

@@ -3,3 +3,6 @@
 ass1:
 	./sc_elements.py
 	latexmk -xelatex superconductivity_assignment1_kvkempen
+
+ass2:
+	latexmk -xelatex superconductivity_assignment2_kvkempen

references.bib

@@ -59,3 +59,15 @@ Electrical conductivity or specific conductance is the reciprocal of electrical
 	editor = {Lide, David R.},
 	year = {2003},
 }
+
+@report{abrikosov,
+	title = {Type {II} superconductors and the vortex lattice},
+	url = {https://www.nobelprize.org/prizes/physics/2003/abrikosov/lecture/},
+	abstract = {The Nobel Prize in Physics 2003 was awarded jointly to Alexei A. Abrikosov, Vitaly L. Ginzburg and Anthony J. Leggett "for pioneering contributions to the theory of superconductors and superfluids".},
+	language = {en-US},
+	urldate = {2022-02-23},
+	institution = {The Nobel Foundation},
+	author = {Abrikosov, Alexei A.},
+	year = {2003},
+	pages = {29--67},
+}

superconductivity_assignment2_kvkempen.tex

@@ -0,0 +1,262 @@
+\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}
+In 2003, Alexei Abrikosov was one of the winners of the Nobel Prize in Physics ``for pioneering contributions to the theory of superconductors and superfluids''.
+For this occasion, he gave a lecture called ``Type II superconductors and the vortex lattice''\cite{abrikosov}
+explaining the discoveries that led to the understanding of conventional superconductors.
+To get started, let me first explain what superconductors are.
+
+% Begin copy from philosophy
+Superconductors are characterized by perfect diamagnetism and zero resistance.
+Perfect diamagnetism is the ability by superconductors to have a net zero magnetic field inside.
+If you apply an external magnetic field, this thus means that a superconductor will let a current flow on its inside to generate a field to counteract this external field $\vec{H}$.
+This generated current is called a supercurrent.
+This is, however, a phase of the material.
+Superconductors only have these properties below a certain temperature, its critical temperature $T_c$, and can only expel a maximum external magnetic field, its critical magnetic field $B_c(T)$, which is a function of the temperature.
+
+The class of superconductors we have a model for, is the class of conventional superconductors, which are explained by a theory called BCS (and some extensions).
+In this class, there are two types, called type-I and type-II superconductors.
+% End copy from philosophy
+
+In type-I superconductors, there is only one phase in which the superconductor material exhibits perfect diamagnetism:
+when the externally applied magnetic field $H < B_c(T)$ and $T < T_c$.
+
+In type-II superconductors, there are two phases distinct from the normal conducting state.
+One is the superconducting state which behaves as in type-I superconductors, with critical field $B_{c1}(T)$.
+This state is reached for $T < T_c$ and $B_E < B_{c1}(T)$.
+The other state is a mixed state that allows some flux to pass through the material.
+This passing through is done by creating normally conducting channels throughout the material where a fixed amount of flux can pass through.
+This fixed amount is a multiple of the flux quantum $\Phi_0$.
+The material generates current around these channels cancelling the field on the inside of the superconducting part of the material.
+
+---
+
+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 it 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.
+
+
+
+\todo{The essay so far is just a draft. Choosing a topic was hard. As we are to aim at bachelor students not knowing sc, I thought a proper introduction was appropriate.}
+
+\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}