added 1_02s, first excercise only!

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+\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}