ass3: Wrap up the set and hand it in

Hoi Roos,

In de bijlage vind je de pdf van mijn uitwerkingen en de code voor taak
12 (die ook in de appendix staat). Door tijdgebrek ben ik niet overal
aan toegekomen. Het was erg veel werk.

Met vriendelijke groet,
Kees van Kempen

.gitignore

@@ -47,3 +47,6 @@ superconductivity_assignment1_kvkempen.pdf
 
 # Assignment 2
 superconductivity_assignment2_kvkempen.pdf
+
+# Assignment 3
+superconductivity_assignment3_kvkempen.pdf

ass3-12-a-weak-junction.py

@@ -0,0 +1,60 @@
+#!/bin/env python3
+
+import numpy as np
+from scipy.integrate import odeint
+from matplotlib import pyplot as plt
+import pandas as pd
+
+def phi_dot(phi, t, I_DC, I_RF):
+    R        = 10.e-3 #Ohm
+    I_J      = 1.e-3 #A
+    omega_RF = 2*np.pi*.96e9 #rad/s
+    hbar     = 1.0545718e-34 #m^2kg/s
+    e        = 1.60217662e-19 #C
+
+    return 2*e*R/hbar*( I_DC + I_RF*np.cos(omega_RF*t) - I_J*(np.sin(phi)) )
+    # I attempted to solve it without the constants, as I suspected overflows
+    # were occurring. The next line did not improve the result.
+    #return R*( I_DC + I_RF*np.sin(omega_RF*t) - I_J*(np.sin(phi)) )
+
+# We need an initial value to phi
+phi_0 = 0
+
+# Let's try it for a lot of periods
+N_points = 10000
+t = np.linspace(0, 100, N_points)
+
+hbar     = 1.0545718e-34 #m^2kg/s
+e        = 1.60217662e-19 #C
+
+df = pd.DataFrame(columns=['I_DC','I_RF','V_DC_bar'])
+
+# For testing:
+#phi = odeint(phi_dot, phi_0, t, (.5e-3, .5e-3))[:, 0]
+#for I_DC in [1e-4, .5e-3, 1.e-3, 1.5e-3, 2.e-3, 2.5e-3]:
+
+for I_DC in np.arange(0, 1e-3, 1e-5):
+    for I_RF in [0., .5e-3, 2.e-3]:
+        # The individual solutions for phi do seem sane, at least, the ones
+        # I inspected.
+        phi = odeint(phi_dot, phi_0, t, (I_DC, I_RF))
+        # I initially thought to average over the tail to look at the asymptotic behaviour.
+        #N_asymp = N_points//2
+        #V_DC_bar = hbar/(2*e)*np.mean(phi[N_asymp:]/t[N_asymp:])
+        # Then I choose to just take the last point to see if that gave better results.
+        V_DC_bar = hbar/(2*e)*phi[-1]/t[-1]
+        print("For I_DC =", I_DC, "\t I_RF = ", I_RF, "\twe find V_DC_bar =", V_DC_bar)
+        df = df.append({'I_DC': I_DC, 'I_RF': I_RF, 'V_DC_bar': V_DC_bar}, ignore_index = True)
+
+## Plotting the thing
+plt.figure()
+
+plt.xlabel("$\\overline{V_{DC}}$")
+plt.ylabel("$I_{DC}$")
+
+for I_RF in df.I_RF.unique():
+    x, y = df[df.I_RF == I_RF][["V_DC_bar", "I_DC"]].to_numpy().T
+    plt.plot(x, y, label="$I_{RF} = " + str(I_RF) + "$")
+
+plt.legend()
+plt.show()

makefile

@@ -6,3 +6,6 @@ ass1:
 
 ass2:
 	latexmk -xelatex superconductivity_assignment2_kvkempen
+
+ass3:
+	latexmk -xelatex superconductivity_assignment3_kvkempen

references.bib

@@ -71,3 +71,40 @@ Electrical conductivity or specific conductance is the reciprocal of electrical
 	year = {2003},
 	pages = {29--67},
 }
+
+@article{dai-synthesis-1995,
+	title = {Synthesis and neutron powder diffraction study of the superconductor {HgBa2Ca2Cu3O8} + δ by {Tl} substitution},
+	volume = {243},
+	issn = {0921-4534},
+	url = {https://www.sciencedirect.com/science/article/pii/0921453494024618},
+	doi = {10.1016/0921-4534(94)02461-8},
+	abstract = {Substitution of Tl for Hg was performed in the Hg based 1223 phase HgBa2Ca2Cu3O8 + δ (Tc = 135 K), resulting in an increase of the superconducting transition temperature to 138 K for samples with a nominal composition of Hg0.8Tl0.2Ba2Ca2Cu3O8 + δ. The crystal structure of this solid solution has been investigated by neutron powder diffraction techniques at room temperature and at 10 K. The compound has the same crystal as Hg-1223 with the space group symmetry P4/mmm and lattice parameters a = 3.8489(1), c = 15.816(1) Å. Rietveld analysis results indicate that Hg is partially replaced by Tl, and the oxygen content, δ, is 0.33. The lattice-parameter changes resulting from the Tl substitution are too small to account for the Tc change by mimicking the effect of pressure. No phase transition occurs down to 10 K.},
+	language = {en},
+	number = {3},
+	urldate = {2022-03-07},
+	journal = {Physica C: Superconductivity},
+	author = {Dai, P. and Chakoumakos, B. C. and Sun, G. F. and Wong, K. W. and Xin, Y. and Lu, D. F.},
+	month = mar,
+	year = {1995},
+	pages = {201--206},
+}
+
+@article{chubukov,
+author = {Chubukov, Andrey and Pines, David and Schmalian, Jörg},
+year = {2002},
+month = {02},
+pages = {51},
+title = {A Spin Fluctuation Model for D-wave Superconductivity}
+}
+
+@misc{ray-2016, title={Master's thesis: Structural investigation of La(2-x)Sr(x)CuO(4+y) - Following staging as a function of temperature}, url={https://figshare.com/articles/thesis/Structural_investigation_of_La_2_x_Sr_x_CuO_4_y_Following_staging_as_a_function_of_temperature/2075680/2}, DOI={10.6084/m9.figshare.2075680.v2}, abstractNote={A thesis submitted to the Niels Bohr Institute at the Faculty of Science at the University of Copenhagen, Denmark, in partial fulfilment of the requirements for the degree of Master of Science in physics. Submission date was November 19, 2015, and the defence was held on November 30, 2015, where the degree was also awarded.
+
+The cuprate La2-xSrxCuO4+y – a high-temperature superconductor – was discovered almost three decades ago. However the mechanisms behind the superconductivity in the material for different doping values x and y are still not fully understood. A small part of this large puzzle is added to the pile with this thesis, where results on the structure for several different samples are presented.
+
+The emphasis in this thesis is on a specific superstructure thought to be connected to the ordering of interstitial oxygen, known from the isostructural compound La2NiO4+y as staging. Four single crystal samples with different co-doping values are investigated by the use of both X-rays and neutrons.
+
+Staging is observed for all four samples at low temperatures with X-ray measurements. The sample with strontium doping x = 0.00 shows several coexisting staging levels with staging numbers between 2 and 8, with the highest contribution from a staging level between 4 and 5. The co-doped samples show increasing staging number with increasing x. It is found that the staging belongs to a structural phase assumed in space group Fmmm, while the unstaged fraction of the samples are in the Bmab space group. These two structural phases are found to have significantly different lengths of the long crystal axis for the two low x samples, in the order of a fraction of a percent, while the two higher x samples had a difference of only a small fraction of a permille.
+
+The temperature dependent phase transitions for both the Bmab structure and the staging reflections are investigated between 5 and 300 K. The critical exponents for the Bmab reflections are found to be significantly lower than results from similar materials in literature, although with transition temperatures consistent with literature for comparable sample compositions. It is found that the critical exponents for the staging reflections increase for increasing doping while the transition temperatures decrease, both consistent with results on the isostructural La2NiO4+y.
+
+Results from previous neutron measurements are found to be consistent with the X-ray measurements in this work, and measured reciprocal space maps from this work show a large variety of other superstructure reflections which will be interesting to investigate in the future.}, publisher={figshare}, author={Ray, Pia Jensen}, year={2016}, month={Feb} } 

superconductivity_assignment2_kvkempen.tex

@@ -31,6 +31,7 @@
 \usepackage{amsmath}
 \usepackage{todonotes}
 \setuptodonotes{inline}
+\usepackage{mhchem}
 
 \newcommand{\pfrac}[2]{\frac{\partial #1}{\partial #2}}
 
@@ -57,21 +58,19 @@ 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.
 
 \[
-	0 = \pfrac{\mathcal{F}}{\psi} = \pfrac{}{\psi} \left[ a(T-T_c)\psi^2 \right] = 2a(T-T_c)\psi + 2\beta\psi^3
+	0 = \pfrac{\mathcal{F}}{\psi} = \pfrac{}{\psi} \left[ a(T-T_c)\psi^2 + \frac{\beta}{2}\psi^4 \right] = 2a(T-T_c)\psi + 2\beta\psi^3
 \]
 
 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,
+For $T \leq T_c$, $\psi_0(T \leq T_c) = \pm\sqrt{\frac{-a}{\beta}(T-T_c)}$ is the minimum,
 giving free energy
 \[
-	\mathcal{F}_0(T \leq T_c) = \frac{-a^2}{\beta}(T-T_c)^2 + \frac{a^2}{2\beta}(T-T_c)^2 = \frac{-a^2}{2\beta}(T-T_c)^2 \leq \mathcal{F}_0(T \geq T_c)
+	\mathcal{F}_0(T \leq T_c) = \frac{-a^2}{\beta}(T-T_c)^2 + \frac{a^2}{2\beta}(T-T_c)^2 = \frac{-a^2}{2\beta}(T-T_c)^2 \leq \mathcal{F}_0(T \geq T_c).
 \]
-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
 \[
 	C(T) = -T\pfrac{^2\mathcal{F}}{T^2} =
@@ -81,13 +80,20 @@ For the specific heat, we find
 	\end{cases}.
 \]
 There is thus a discontinuity in $C(T)$ at $T = T_c$ with size $\Delta C(T) = \frac{a^2}{\beta}T_c$.
+See the following sketches of the $T$-dependence of the derived quantities.
+
+\begin{figure}[H]
+	\centering
+	\includegraphics[width=.9\textwidth]{ass2-landau-theory-T.pdf}
+	\caption{For the Landau theory, we find the drawn temperature dependences for equilibrium values of $\mathcal{F}_0$, $\psi_0$ and $C$. Note the minus sign for the free energy $\mathcal{F}_0$. At $T = 0$, there are  $y$ axis intersections for all three quantities, namely a minimum $\mathcal{F}_0(0) = \frac{-a^2}{2\beta}T_c^2$, $\psi_0(0) = \pm\sqrt{\frac{-a}{b}T_c}$, and $C(0) = 0$, which I forgot to indicate in the sketches. Do also note that there thus is an intersection in the $\mathcal{F}_0(T)$ curve at $T = 0$, although the drawing may look asymptotic.}
+\end{figure}
 
 \section{Type-I superconducting foil}
 \begin{enumerate}
 \item
 The screening equation is given as
 \[
-	\nabla^2\vec{B} = \frac{\vec{B}}{\lambda}.
+	\nabla^2\vec{B} = \frac{\vec{B}}{\lambda^2}.
 \]
 For easy of calculation, we will use cartesian coordinates,
 and put the external magnetic field $B_E$ along the $x$ axis:
@@ -98,17 +104,15 @@ So we define the magnitude of the field $|\vec{B}| = B(z)$.
 
 Using the screening equation, we look for a solution.
 \[
-	\nabla^2\vec{B} = \nabla(\nabla\cdot\vec{B}) - \nabla\times(\nabla\times\vec{B})
+	\nabla^2\vec{B} = \left[ \pfrac{^2}{x^2} + \pfrac{^2}{y^2} + \pfrac{^2}{z^2} \right] \vec{B}
 \]
-$\vec{B}$ 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 $B_y = 0 = B_z$.
+we realize that we only have $z$ dependence, and $B_y = 0 = B_z$.
 \[
-	-\nabla\times(\nabla\times\vec{B}) = -\nabla\times(\pfrac{B_x}{z} \hat{y}) = -(-\pfrac{^2B_x}{z^2} \hat{x})
+	\nabla^2\vec{B} = \pfrac{^2B_x}{z^2} \hat{x}
 \]
 Rewriting yields
 \[
-	\vec{B} = \lambda \pfrac{^2B_x}{z^2}\hat{x}.
+	\vec{B} = \lambda^2 \pfrac{^2B_x}{z^2}\hat{x}.
 \]
 For this we know the general solution:
 \[
@@ -158,12 +162,16 @@ Next, we can equate the previously found supercurrent for our foil to the Ginzbu
 \[
 	\vec{A} = \frac{-B_E m \sinh{\frac{z}{\lambda}}}{4\lambda\mu_0 e^2 n_s \cosh{\frac{a}{2\lambda}}} \hat{y}
 \]
+This we can rewrite using $\lambda = \sqrt{\frac{m}{\mu_0 n_s e^2}}$ for the London penetration depth as
+\[
+	\vec{A} = \frac{-B_E \lambda \sinh{\frac{z}{\lambda}}}{4 \cosh{\frac{a}{2\lambda}}} \hat{y}.
+\]
 
 \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.
+as $A_y$ is independent of $y$.
 
 In our case, indeed the rigid gauge choice gives the criterium for the London gauge ($\nabla \cdot \vec{A} = 0$).
 
@@ -181,11 +189,12 @@ 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}.
+This is what the London gauge implies.
+Now the question is under what circumstances the rigid gauge follows from the London gauge.
+This is the case for $\nabla \cdot \vec{J_s} = 0$, or in words, when there is conservation of superelectrons.
+If this is not the case (if the divergence is non-zero), there is conversion between normal electrons and superelectrons.
+This would take place if the temperature is lowered, as more superelectrons allow for a larger supercurrent, thus a larger critical magnetic field.
+This result seems to agree with Waldram's conclusion in \cite[p. 24--26]{waldram}.
 
 \item
 We apply a gauge transformation as follows.
@@ -195,7 +204,6 @@ We apply a gauge transformation as follows.
 	\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}
@@ -209,10 +217,12 @@ 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.
+Superconductivity 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 zero resistance property follows from the perfect diagmagnetism.
+It is impossible for the material to let these supercurrents flow indefinitely with resistance, as heat would be generated.
 
-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).
+The class of superconductors we have a model for, is the class of conventional superconductors.
 In this class, there are two types, called type-I and type-II superconductors.
 % End copy from philosophy
 
@@ -225,7 +235,53 @@ 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 material generates current around these channels in accordance to the Maxwell-Amp\`ere law, conforming to the let through magnetic field inside the vortex and cancelling the field on the outside the vortex.
+
+There are lots of applications for both the perfect diagmagnetism and the zero resistivity.
+There is even a Wiki about them: \url{https://en.wikipedia.org/wiki/Technological\_applications\_of\_superconductivity}.
+What is most notable about these applications, is that maintaining a temperature below the critical temperature is the biggest challenge.
+A real breakthrough for superconductivity would be the discovery of room-temperature superconductors at atmospheric pressure, or materials close to that.
+Currently, the highest $T_c$ material we know is carbonaceous sulfur hydride (\ce{CH8S}) with $T_c = \SI{15}{\degreeCelsius}$ but at a pressure of a whopping $\SI{267}{\giga\pascal}$.
+At atmospheric pressure, the highest $T_c$ material known is a cuprate \cite{dai-synthesis-1995} \ce{HgBa2Ca2Cu3O_{8+\delta}} at $T_c = \SI{135}{\kelvin}$.
+The quest for this breakthrough is intensely researched, although most is experimental.
+The clue is that most of the high $T_c$ materials that are being discovered, are unconventional superconductors.
+As there is no theory for them (yet), the search is mostly educational guessing.
+By trying to find patterns in the previously high $T_c$ materials, similar materials are studied to see if they also exhibit superconductivity.
+One of the patterns is that superconductivity in cuprates is high $T_c$.
+We'll focus on these materials in the following.
+
+Currently, most hopeful candidates are cuprates.
+These materials are made of layers of copper oxides (\ce{CuO2}) alternated with layers of other metal oxides.
+The copper oxide layers are the superconductive layers, and the other metal oxides are used as charge reservoirs, doping electrons (or holes) into the copper oxide layers.
+Due to the geometry of these materials, there is anisotropy in the resistivity of the material.
+Parallel to the layers, superconductivity takes place in the copper oxide layers.
+Perpendicular to the layers, this is not the case.
+
+The behaviour of the material can be tuned by tuning the doping, thus the other metal oxides as mentioned before.
+A typical phase diagram as function of the doping can be seen in figure \ref{fig:cuprate-phase}.
+The material can be steered from being antiferrimagnetic to superconductive by increasing doping.
+
+\begin{figure}
+	\centering
+	\includegraphics{cuprate-phase.pdf}
+	\caption{For high $T_c$ superconducting cuprates, a typical phase diagram as function of doping looks like this.\cite{chubukov} }
+	\label{fig:cuprate-phase}
+\end{figure}
+
+As can be seen, there is an optimal doping fraction for achieving highest $T_c$.
+Aiming for this doping yields the desired material.
+
+Now the question is what direction to search for.
+The timeline in figure \ref{fig:timeline} might give a direction for the most promising types of cuprates to look into.
+It could be, however, that other types have higher $T_c$.
+A lot of creativity is therefore needed to find them.
+
+\begin{figure}
+	\centering
+	\includegraphics[width=\textwidth]{Timeline-of-Superconductivity-from-1900-to-2015.pdf}
+	\caption{The last century, a lot of research has been done in the direction of cuprate superconductivity. Pia Jensen Ray made this overview for his master thesis.\cite{ray-2016} The different paths are different types of cuprates. Please see his thesis for the meaning of the labels. On the right side, an idea of the temperature is givin by comparing it to common cooling agents.}
+	\label{fig:timeline}
+\end{figure}
 
 ---
 
@@ -234,24 +290,25 @@ The start was a good recap of the breakthroughs relevant to conventional superco
 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.
+These flux allowing regions are called vortices, due to their shape and direction of current flow.
+Vortices look like channels (or tubes), and supercurrents move around these channels in a spiraling fashion.
+One can visualize this as current through a coil such that on the inside of the coil, the field is in one direction perpendicular to it, and on the outside it is the opposite direction.
+The current direction is governed by the Maxwell-Amp\`ere equation.
+In this case, the current is such that the field inside the cylinder but outside these channels is counteracted.
 
 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}
+	\includegraphics[width=.6\textwidth]{SchermafbeeldingKees-vortex-by-fleur-ahlers.png}
+	\caption{The direction of $\vec{J_s}$ is such that a magnetic field is generated to counteract and even expel the external field outside the vortices inside the material. Around the vortices, that means that the supercurrents run anti-clockwise. The field is then along $\vec{B_E}$ inside the vortices, but along $-\vec{B_E}$ outside the vortices but inside the material. Around the outside border of the cylinder, however, $\vec{J_s}$ runs clockwise and again cancels $\vec{B_E}$ on the inside of the material.}
 \end{figure}
 
 \bibliographystyle{vancouver}

superconductivity_assignment3_kvkempen.tex

@@ -0,0 +1,176 @@
+\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}
+\usepackage{mhchem}
+\usepackage{listings}
+
+\newcommand{\pfrac}[2]{\frac{\partial #1}{\partial #2}}
+
+\title{Superconductivity - Assignment 3}
+\author{
+	Kees van Kempen (s4853628)\\
+	\texttt{k.vankempen@student.science.ru.nl}
+}
+\AtBeginDocument{\maketitle}
+
+% Start from 8
+\setcounter{section}{7}
+
+\begin{document}
+
+\section{\ce{Nb3Sn} cylinder}
+Consider a cylinder of \ce{Nb3Sb}.
+From lecture 4, we have the following properties for \ce{Nb3Sn}:
+$T_c = \SI{18.2}{\kelvin}$,
+$\xi = \SI{3.6}{\nano\meter}$,
+$\lambda = \SI{124}{\nano\meter}$,
+$\kappa = \frac{\lambda}{\xi} = 34 > \frac{1}{\sqrt{2}}$,
+which means we are indeed dealing with a type-II superconductor.
+As $B_{c1} < B_E < B_{c2}$, the cylinder is in the vortex state.
+From the previous set of assignments, we know what the currents in the cylinder look like.
+From free energy considerations, we have found in lecture 4 that for type-II superconductors, it is favorable to allow flux quanta inside the superconductor in this vortex state.
+In this derivation, the contribution of one flux quantum is considered, but the consideration holds for many vortices, until they start to interact and repel eachother.
+At that point, the vortex-vortex interaction orders the vortices in a lattice.
+When the vortex cores start to overlap, there are no superconducting regions left, thus the material enters the normal conducting state.\footnote{I wanted to paint a complete picture althought it is not needed to answer the question.}
+Minimizing the free energy over the flux shows the energy is lowered for determined thresholds $B_{c1} < B_E < B_{c2}$.
+
+Let's start with the result from said free energy considerations.
+The average field inside the cylinder is given by the following self-consisting equation as
+\[
+	B = B_E - \frac{\phi_0}{8\pi\lambda^2}\ln{\frac{\phi_0}{4\exp{(1)}\xi^2B}}.
+\]
+Plugging in the values for \ce{Nb3Sn}, $B_E = \SI{1}{\tesla}$, and $\phi_0 = \SI{2.0678}{\weber}$, $B$ is found as $B = \SI{.986}{\tesla} \approx B_E$ by intersection.
+%https://www.wolframalpha.com/input?i=B+%3D+1+-+%282.0678*10%5E-15%29%2F%288*pi*%28124*10%5E-9%29%5E2%29+*+ln%282.0678*10%5E-15%2F%284*exp%281%29*%283.65*10%5E-9%29%5E2+*+B%29%29
+This is in the range as provided in the assignment ($B = \SI[separate-uncertainty]{.981\pm.019}{\tesla}$).
+
+To investigate the inhomogenity of the field inside the cylinder, we look at the gradient $\nabla B$ inside the material.
+As we assume a vortex lattice that fully fills the cross section of the cylinder,
+and we assume that the fields due to each vortex die out quickly enough to not overlap,
+it suffices to calculate the gradient over just one vortex.
+These assumptions coincide with slide 15 of lecture 4, from which I took figure \ref{fig:lec4-vortexlattice}.
+
+\begin{figure}[H]
+	\centering
+	\label{fig:lec4-vortexlattice}
+	\includegraphics[width=.4\textwidth]{lec4-vortexlattice.png}
+	\caption{The vortices are arranged in a lattice to maximize their distance, as this lowers their repulsive interaction and thus the energy.}
+\end{figure}
+
+On slide 19 from the same week, we find an expression $B(r)$ for the field at distance $r$ from the vortex core as
+\[
+	B = \frac{\phi_0}{2\pi\lambda^2} K_0(r/\lambda) = B_0 K_0(r/\lambda),
+\]
+where $K_0$ is the modified Bessel function of the second kind.
+For small $r$ (i.e. $r << \lambda$), we can approximate this and find that
+\[
+	K_0 \propto - \ln{(r/\lambda)},
+\]
+and notice a singularity at $r = 0$.
+For the gradient we thus find
+\[
+	\nabla B \propto \pfrac{K_0}{r}(r/\lambda) \propto \pfrac{-\ln{(r/\lambda)}}{r} = -\lambda/r.
+\]
+The size of the supercurrent density has the same relation, $J_S \propto 1/r$.
+
+\section{Superconducting wire}
+\textbf{(a)}
+The voltage $U = \SI{1.5e-5}{\volt}$ across the wire of length $\ell = \SI{.08}{\meter}$ induces a current $J_t$. % through the resistive wire with unknown resistivity $\rho$ according to Ohm's law.
+Due to the presence of the magnetic field $B = \SI{5}{\tesla}$, if the vortices move with velocity $v_L$, a Lorentz force $f_L$ per vortex acts on the vortices.
+This results in a power input $P_L = f_Lv_L = J_tBv_L$ per vortex.
+%$\epsilon = Bv_L$
+This power should come from the current induced by the voltage, thus $P_L = \epsilon J_t = \frac{U}{\ell}J_t$.
+Equating these expressions and rewriting yields
+\[
+	v_L = \frac{U}{B\ell} = \SI{3.75e5}{\meter\per\second}.
+	% https://www.wolframalpha.com/input?i=1.5*10%5E-5+%2F+%285*+.08%29
+\]
+
+\textbf{(b)}
+The vortices are aranged in a lattice with separation $r_{sep} = \sqrt{\frac{\Phi_0}{B}}$.
+They move along the wire with velocity $v_L$ as determined above.
+The expected frequency is then given by their velocity over the separation, as that is the period of the changing fields due to the vortices:
+\[
+	f = \frac{v_L}{r_{sep}} = \frac{U}{B\ell}\sqrt{\frac{B}{\Phi_0}} = \frac{U}{\ell\sqrt{B\Phi_0}} = \SI{1.84}{\kilo\hertz},
+	% https://www.wolframalpha.com/input?i=1.5*10%5E-5+%2F+%28.08%29+%2Fsqrt%285+*+2.067*10%5E%28-15%29%29
+\]
+where we used that $\Phi_0 = \SI{2.067e-15}{\volt\second}$.
+This is very close to what is written in the assignment, but not precisely the same, so maybe I used a different value for $\Phi_0$.
+
+\section{Fine type-II superconducting wire}
+
+\section{Critical currents}
+\textbf{(a)}
+Silsbee's rule states that the supercurrents through the wire must not generate magnetic fields in excess of $B_c$ at the surface of the wire.
+We assume that the supercurrent is maximal at the surface with a maximum value of $J_{max}$, and that the supercurrent decays linearly from the surface to zero at a penetration depth $\lambda$ deep.
+We thus find a relation for the supercurrent as function of the cylindrical radius $r$ as
+\[
+	J_s(r) = \frac{J_{max}}{\lambda} \left[ r - R + \lambda \right].
+\]
+
+Now we can use the Maxwell-Amp\`ere law to find this value for $J_{max}$.
+\[
+	\oint \vec{B}\cdot d\vec{\ell} = \mu_r\mu_0\iint\vec{J_s}\cdot d\vec{S}
+\]
+Using $\vec{B} = \vec{B_c}$, $\mu_r = 1$ as we're calculating the field outside the sc, $J_s = J_s(r)$, the area over which $d\vec{S}$ runs to be the small ring from $r = R - \lambda$ to $r = R$, and the path along which $d\vec{\ell}$ runs to be the loop $2\pi R$ along the surface of the wire.
+This gives us
+\[
+	2\pi R B_c = \mu_0 \int_{\phi = 0}^{2\pi}\int_{r=R-\lambda}^R \frac{J_{max}}{\lambda}\left[ r - R + \lambda \right] rdr d\phi.
+\]
+Solving the integral over $\phi$ results in
+\[
+	R B_c = \mu_0 \int_{r=R-\lambda}^R \frac{J_{max}}{\lambda}\left[ r - R + \lambda \right] rdr = \frac{\mu_0J_{max}}{\lambda} \left[ \frac{r^3}{3} - \frac{(R + \lambda)r^2}{2} \right]_{r = R-\lambda}^R.
+\]
+Solving for $J_{max}$, this yields the beautiful expression
+\[
+	J_{max} = \frac{6B_c\lambda R}{\mu \left[ 4\lambda^3 - 9\lambda^2R + 3\lambda R^2- 3\lambda R + 3R^3 -3R^2 \right]}.
+	% https://www.wolframalpha.com/input?i=R*B+%3D+m*x%2Fl*%28%28R%5E3-%28R-l%29%5E3%29%2F3+-+%28R%2Bl%29*%28R+-+%28R-l%29%5E2%29%2F2%29
+\]
+
+\textbf{(b)}
+
+
+\section{A weak junction}
+See the code in appendix \ref{appendix:program-task-12}.
+
+It unfortunately does not seem to produce any useful results.
+In the code, I left many comments as it is mostly in a debugging state.
+
+\bibliographystyle{vancouver}
+\bibliography{references.bib}
+
+\appendix
+\section{Program to task 12}
+\label{appendix:program-task-12}
+\lstinputlisting[language=python,breaklines=true]{ass3-12-a-weak-junction.py}
+
+\end{document}