ass4: Scan my solutions

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

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}