Prof. B. Stritzker\\
WS 2007/08\\
\vspace{8pt}
- {\Large\bf Tutorial 2}
+ {\Large\bf Tutorial 2 - proposed solutions}
\end{center}
\section{Phonons 1}
$M_1\ddot{u}_s=C(v_s+v_{s-1}-2u_s)$\\
$M_2\ddot{v}_s=C(u_{s+1}+u_s-2v_s)$
\item Ansatz:\\
- $u_s=u\exp{i(ska-\omega t)}$\\
- $v_s=v\exp{i(ska-\omega t)}$
+ $u_s=u\exp(i(ska-\omega t))$\\
+ $v_s=v\exp(i(ska-\omega t))$
\item Solution of the equation system:\\
- $-\omega^2M_1u=Cv[1+\exp(-ika)]-2Cu$\\
- $-\omega^2M_2v=Cu[\exp(ika)+1]-2Cv$\\
+ $-\omega^2M_1u\exp(i(ska-\omega t))=
+ C\exp(-i\omega t)[v\exp(iska)+v\exp(i(s-1)ka)-2u\exp(iska)]$\\
+ $\Rightarrow -\omega^2M_1u=Cv(1+\exp(-ika))-2Cu$\\
+ $-\omega^2M_2v\exp(i(ska-\omega t))=
+ C\exp(-i\omega t)[u\exp(i(s+1)ka)+u\exp(iska)-2v\exp(iska)]$\\
+ $\Rightarrow -\omega^2M_2v=Cu[\exp(ika)+1]-2Cv$\\
Non trivial solution only if determinant of coefficients
$u$ and $v$ is zero.\\
$\Rightarrow
-C[1+\exp(ika)] & 2C-M_2\omega^2
\end{array}
\right|=0$\\
+ $\Rightarrow
+ 4C^2+M_1M_2\omega^4-2C\omega^2(M_2+M_1)-
+ \underbrace{C^2(1+\exp(ika))(1+\exp(-ika))}_{
+ C^2(\underbrace{1+1+\exp(ika)+\exp(-ika)}_{
+ 2+2\cos(ka)=2(1+\cos(ka))})}$\\
$\Rightarrow
M_1M_2\omega^4-2C(M_1+M_2)\omega^2+2C^2(1-\cos(ka))=0$
\end{itemize}
--- /dev/null
+\pdfoutput=0
+\documentclass[a4paper,11pt]{article}
+\usepackage[activate]{pdfcprot}
+\usepackage{verbatim}
+\usepackage{a4}
+\usepackage{a4wide}
+\usepackage[german]{babel}
+\usepackage[latin1]{inputenc}
+\usepackage[T1]{fontenc}
+\usepackage{amsmath}
+\usepackage{ae}
+\usepackage{aecompl}
+\usepackage[dvips]{graphicx}
+\graphicspath{{./img/}}
+\usepackage{color}
+\usepackage{pstricks}
+\usepackage{pst-node}
+\usepackage{rotating}
+
+\setlength{\headheight}{0mm} \setlength{\headsep}{0mm}
+\setlength{\topskip}{-10mm} \setlength{\textwidth}{17cm}
+\setlength{\oddsidemargin}{-10mm}
+\setlength{\evensidemargin}{-10mm} \setlength{\topmargin}{-1cm}
+\setlength{\textheight}{26cm} \setlength{\headsep}{0cm}
+
+\renewcommand{\labelenumi}{(\alph{enumi})}
+
+\begin{document}
+
+% header
+\begin{center}
+ {\LARGE {\bf Materials Physics I}\\}
+ \vspace{8pt}
+ Prof. B. Stritzker\\
+ WS 2007/08\\
+ \vspace{8pt}
+ {\Large\bf Tutorial 4}
+\end{center}
+
+\section{Hall effect and magnetoresistance}
+The Hall effect refers to the potential difference (Hall voltage)
+on the opposite sides of an electrical conductor
+through which an electric current is flowing,
+created by a magnetic field applied perpendicular to the current.
+Edwin Hall discovered this effect in 1879.
+
+Consider the following scenario:
+An electric field $E_x$ is applied to a wire extending in $x$-direction
+and a current density $j_x$ is flowing in that wire.
+There is a magnetic field $B$ pointing in the positive $z$-direction.
+Electrons are deflected in the negative $y$-direction
+due to the Lorentz force $F_L=-evB$
+until they run against the sides of the wire.
+An electric field $E_y$ builds up opposing the Lorentz force
+and thus preventing further electron accumulation at the sides.
+The two quantities of interest are:
+\begin{itemize}
+ \item the magnetoresistance
+ \[
+ \rho(B) = \frac{E_x}{j_x} \textrm{ and}
+ \]
+ \item the Hall coefficient
+ \[
+ R_H(B) = \frac{E_y}{j_xB} \textrm{ .}
+ \]
+\end{itemize}
+In this tutorial the treatment of the Hall problem is based on a simple
+Drude model analysis.
+\\\\
+First of all the effect of individual electron collisions can be expressed
+by a frictional damping term into the equation of motion for the momentum
+per electron.
+
+\begin{enumerate}
+ \item Recall the Drude model.
+ Given the momentum per electron $p(t)$ at time t
+ calculate the momentum per electron $p(t+dt)$
+ an infinitesimal time $dt$ later.
+ {\bf Hint:} What is the probability of an electron taken at random at
+ time $t$ to not suffer a collision before time $t+dt$?
+ If not experiencing a collision it simply evolves under the influence
+ of the force $f(t)$.
+ Combine contributions of the order of $(dt)^2$ to the term
+ $O(dt)^2$.
+ \item Write down the equation of motion for the momentum per electron
+ by dividing the above result by $dt$
+ and taking the limit $dt\rightarrow 0$.
+ \item Sketch a schematic view of Hall's experiment.
+ \item Find an expression for the Hall coefficient.
+ {\bf Hint:} Insert an appropriate force into the equation of motion
+ for the momentum per electron.
+ Consider the steady state and acquire the equations
+ for the $x$ and $y$ component of the vector equation.
+ To find an expression for the Hall coefficient use the second equation
+ and the fact that there must not be transverse current $j_y$
+ while determining the Hall field.
+\end{enumerate}
+
+\end{document}
--- /dev/null
+\pdfoutput=0
+\documentclass[a4paper,11pt]{article}
+\usepackage[activate]{pdfcprot}
+\usepackage{verbatim}
+\usepackage{a4}
+\usepackage{a4wide}
+\usepackage[german]{babel}
+\usepackage[latin1]{inputenc}
+\usepackage[T1]{fontenc}
+\usepackage{amsmath}
+\usepackage{ae}
+\usepackage{aecompl}
+\usepackage[dvips]{graphicx}
+\graphicspath{{./img/}}
+\usepackage{color}
+\usepackage{pstricks}
+\usepackage{pst-node}
+\usepackage{rotating}
+
+\setlength{\headheight}{0mm} \setlength{\headsep}{0mm}
+\setlength{\topskip}{-10mm} \setlength{\textwidth}{17cm}
+\setlength{\oddsidemargin}{-10mm}
+\setlength{\evensidemargin}{-10mm} \setlength{\topmargin}{-1cm}
+\setlength{\textheight}{26cm} \setlength{\headsep}{0cm}
+
+\renewcommand{\labelenumi}{(\alph{enumi})}
+
+\begin{document}
+
+% header
+\begin{center}
+ {\LARGE {\bf Materials Physics I}\\}
+ \vspace{8pt}
+ Prof. B. Stritzker\\
+ WS 2007/08\\
+ \vspace{8pt}
+ {\Large\bf Tutorial 4 - proposed solutions}
+\end{center}
+
+\section{Hall effect and magnetoresistance}
+\begin{enumerate}
+ \item \begin{itemize}
+ \item probability: $1-\frac{dt}{\tau}$
+ \item momentum contribution of non-colliding electrons:
+ $f(t)dt+O(dt)^2$
+ \item momentum per electron at time $t+dt$:\\
+ \[
+ p(t+dt)=\left(1-\frac{dt}{\tau}\right)
+ \left[p(t)+f(t)dt+O(dt)^2\right]
+ =p(t)-\frac{dt}{\tau}p(t)+f(t)+O(dt)^2
+ \]
+ \end{itemize}
+ \item \[
+ p(t+dt)-p(t)=-\frac{dt}{\tau}p(t)+f(t)dt+O(dt)^2
+ \]
+ \[
+ \frac{p(t+dt)-p(t)}{dt}=-\frac{p(t)}{\tau}+f(t)+\frac{O(dt)^2}{dt}
+ \]
+ \[
+ \stackrel{dt\rightarrow 0}{\Rightarrow} \quad
+ \frac{dp(t)}{dt}=-\frac{p(t)}{\tau}+f(t)
+ \]
+ \item
+\end{enumerate}
+
+\end{document}
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