\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 2}
\end{center}

\section{Phonons 1}
\begin{enumerate}
 \item \begin{itemize}
        \item $r_i=r_{i0}+u_i$\\
	      $\rho=r_2-r_1=r_{20}+u_2-r_{10}-u_1=(r_{20}-r_{10})+(u_2-u_1)
	           =\rho_0+\sigma$
	\item $\Phi-\Phi_0=\frac{D}{2}(\rho-\rho_0)^2
	                  =\frac{D}{2}(\rho^2+\rho_0^2-2\rho_0\rho)$\\
	      $\rho^2=\rho_0^2+\sigma^2+2\rho_0\sigma$ 
	      $\Rightarrow$ $\rho=\sqrt{\rho_0^2+\sigma^2+2\rho_0\sigma}$\\
              $\Rightarrow$ $\Phi-\Phi_0=\frac{D}{2}
	                     [2\rho_0^2+\sigma^2+2\rho_0\sigma-
			      2\rho_0\sqrt{\rho_0^2+\sigma^2+2\rho_0\sigma}]$
       \end{itemize}
 \item $\sigma \parallel \rho_0$:
       \begin{enumerate}
        \item \begin{flushleft}
	       \includegraphics[height=6cm]{elongation_p01.eps}
	       \includegraphics[height=6cm]{elongation_p02.eps}
	       \includegraphics[height=6cm]{elongation_p03.eps}
              \end{flushleft}
        \item $\sigma = \sigma_{\parallel}$:\\
	      $\rho_0 \sigma_{\parallel} = |\rho_0| |\sigma_{\parallel}|$\\
	      $\Phi-\Phi_0=\frac{D}{2}\left(2\rho_0^2+\sigma_{\parallel}^2+
	                   2\rho_0\sigma_{\parallel}-
		           2\rho_0\sqrt{(\rho_0+\sigma_{\parallel})^2}\right)
			  =\frac{D}{2}\sigma_{\parallel}^2$
       \end{enumerate}
 \item $\sigma \perp \sigma_0$:
       \begin{enumerate}
        \item \begin{flushleft}
	       \includegraphics[height=5.3cm]{elongation_n01.eps}
	       \includegraphics[height=5.3cm]{elongation_n02.eps}
	       \includegraphics[height=5.3cm]{elongation_n03.eps}
              \end{flushleft}
        \item $\sigma=\sigma_{\perp}$:\\
	      $\sigma_{\perp} \rho_0 = 0$\\
	      $\Phi-\Phi_0=\frac{D}{2}\left[2\rho_0^2+\sigma_{\perp}^2-
		           2\rho_0\sqrt{\rho_0^2+\sigma_{\perp}^2}\right]$

	\item $\sigma_{\perp} = \alpha \rho_0$, $\alpha \ll 1$\\
	      $\sqrt{\rho_0^2+\sigma_{\perp}^2}=
	       \sqrt{\rho_0^2+\alpha^2\rho_0^2}=
	       \rho_0\sqrt{1+\alpha^2}=
	       \rho_0(1+\frac{\alpha^2}{2}-\frac{\alpha^4}{8}+\ldots)$\\
	      $\Rightarrow \Phi-\Phi_0=
	       \frac{D}{2}\left[\rho_0^2\left(2+\alpha^2-
	       2(1+\frac{\alpha^2}{2}-\frac{\alpha^4}{8}+\ldots)\right)\right]=
	       \frac{D}{2}\left[\rho_0^2(\frac{\alpha^4}{4}+\ldots)\right]$\\
	      $\Rightarrow \Phi-\Phi_0\stackrel{\alpha\ll 1}{=}
	       \frac{D}{2}\rho_0^2\frac{\alpha^4}{4}=
	       \frac{D}{2}\sigma_{\perp}^2\frac{\alpha^2}{4}$
	\item $\sigma_{\parallel}$, $\sigma_{\perp} \ll \rho_0$\\
	      $\Rightarrow$ potential contribution of $\sigma_{\perp}$
	      compared to contribution of $\sigma_{\parallel}$
	      negligible small.
       \end{enumerate}
 \item \begin{itemize}
        \item As long as the displacements and thus the elongation is small
	      compared to the equilibrium state the change in the potential
	      due to the perpendicular elongation is negligible small.
        \item Regarding a possible existence of perpendicular elongation
	      the model of the linear chain is unproblematic.
        \item In a real crystal couplings in other directions exist.
	      These can only be neglected if they are small compared to the
	      coupling of the considered direction.
       \end{itemize}
\end{enumerate}

\section{Phonons 2}
\begin{enumerate}
\item Derive the dispersion relation for a linear chain with two different
      alternating types of atoms.
\item Discuss the two solutions for $\omega^2$.
\end{enumerate}

\end{document}
