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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})}
+\renewcommand{\labelenumii}{\arabic{enumii})}
+\renewcommand{\labelenumiii}{\roman{enumiii})}
+
+\begin{document}
+
+% header
+\begin{center}
+ {\LARGE {\bf Materials Physics II}\\}
+ \vspace{8pt}
+ Prof. B. Stritzker\\
+ SS 2008\\
+ \vspace{8pt}
+ {\Large\bf Tutorial 3 - proposed solutions}
+\end{center}
+
+\vspace{8pt}
+
+\section{Specific heat in the classical theory of the harmonic crystal -\\
+         The law of Dulong and Petit}
+
+\begin{enumerate}
+ \item Energy:
+       \begin{eqnarray}
+       w&=&-\frac{1}{V}\frac{\partial}{\partial \beta}
+       ln \int d\Gamma \exp(-\beta H)
+       =-\frac{1}{V}\frac{1}{\int d\Gamma \exp(-\beta H)}
+       \frac{\partial}{\partial \beta} \int d\Gamma \exp(-\beta H)\nonumber\\
+       &=&-\frac{1}{V}\frac{1}{\int d\Gamma \exp(-\beta H)}
+       \int d\Gamma \frac{\partial}{\partial \beta} \exp(-\beta H)\nonumber\\
+       &=&-\frac{1}{V}\frac{1}{\int d\Gamma \exp(-\beta H)}
+       \int d\Gamma \exp(-\beta H) (-H) \qquad \textrm{ q.e.d.} \nonumber
+       \end{eqnarray}
+ \item Potential energy:
+       \[
+       U=\frac{1}{2}\sum_{{\bf RR'}}\Phi({\bf r}({\bf R})-{\bf r}({\bf R'}))
+        =\frac{1}{2}\sum_{{\bf RR'}}
+         \Phi({\bf R}-{\bf R'}+{\bf u}({\bf R})-{\bf u}({\bf R'}))
+       \]
+       Using Taylor and
+       $U_{\text{eq}}=\frac{1}{2}\sum_{{\bf R R'}} \Phi({\bf R}-{\bf R'})$:
+       \[
+       U=U_{\text{eq}}+
+         \frac{1}{2}\sum_{{\bf RR'}}({\bf u}({\bf R})-{\bf u}({\bf R'}))
+        \nabla\Phi({\bf R}-{\bf R'})+
+        \frac{1}{4}\sum_{{\bf RR'}}
+        [({\bf u}({\bf R})-{\bf u}({\bf R'})) \nabla]^2
+        \Phi({\bf R}-{\bf R'}) + \mathcal{O}(u^3)
+       \]
+       Linear term:\\
+       The coefficient of ${\bf u}({\bf R})$ is
+       $\sum_{\bf R'}\nabla\Phi({\bf R}-{\bf R'})$
+       which is minus the force excerted on atom ${\bf R}$
+       by all other atoms in equlibrium positions.
+       There is no net force on any atom in equlibrium.
+       The linear term is zero.\\\\
+       Harmonic term:\\
+       $(a\nabla)^2 \Phi=
+        a\nabla a\nabla \Phi=
+       a\nabla \sum_u a_u \frac{\partial\Phi}{\partial r_u}=
+       \sum_v \frac{\partial \sum_u a_u
+       \frac{\partial\Phi}{\partial r_u}}{\partial r_v} a_v=
+       \sum_{uv}\frac{\partial}{\partial r_v} a_u
+       \frac{\partial \Phi}{\partial r_u} a_v=
+       \sum_{uv}a_u \frac{\partial^2\Phi}{\partial r_u \partial r_v} a_v$\\
+       \[\Rightarrow
+       U_{\text{harm}}=\frac{1}{4}\sum_{\stackrel{{\bf R R'}}{\mu,v=x,y,z}}
+       [u_{\mu}({\bf R})-u_{\mu}({\bf R'})]\Phi_{\mu v}({\bf R}-{\bf R'})
+       [u_v({\bf R})-u_v({\bf R'})],
+       \quad \Phi_{\mu v}({\bf r})=
+        \frac{\partial^2 \Phi({\bf r})}{\partial r_{\mu}\partial r_v}.
+       \]
+ \item Change of variables:
+       \[
+       {\bf u}({\bf R})=\beta^{-1/2}\bar{{\bf u}}({\bf R}), \qquad
+       {\bf P}({\bf R})=\beta^{-1/2}\bar{{\bf P}}({\bf R})
+       \]
+       \[
+       \Rightarrow
+       d{\bf u}({\bf R})=\beta^{-3/2}d\bar{{\bf u}}({\bf R}), \qquad
+       d{\bf P}({\bf R})=\beta^{-3/2}d\bar{{\bf P}}({\bf R}), \qquad
+       \]
+       Kinetic energy contribution:
+       \[
+       H_{\text{kin}}=\frac{{\bf P}({\bf R})^2}{2M}
+       \]
+       Integral:
+       \[
+       \int d\Gamma \exp(-\beta H)=
+       \int d\Gamma \exp\left[-\beta\left(\sum \frac{{\bf P}({\bf R})^2}{2M}+
+       U_{\text{eq}} + U_{\text{harm}}\right)\right]
+       \]
+
+\end{enumerate}
+
+\section{Specific heat in the quantum theory of the harmonic crystal -\\
+         The Debye model}
+
+As found in exercise 1, the specific heat of a classical harmonic crystal
+is not depending on temeprature.
+However, as temperature drops below room temperature
+the specific heat of all solids is decreasing as $T^3$ in insulators
+and $AT+BT^3$ in metals.
+This can be explained in a quantum theory of the specific heat of
+a harmonic crystal, in which the energy density $w$ is given by
+\[
+w=\frac{1}{V}\frac{\sum_i E_i \exp(-\beta E_i)}{\sum_i \exp(-\beta E_i)}.
+\]
+\begin{enumerate}
+ \item Show that the energy density can be rewritten to read:
+       \[
+   w=-\frac{1}{V}\frac{\partial}{\partial \beta} ln \sum_i \exp(-\beta E_i).
+       \]
+ \item Evaluate the expression of the energy density.
+       {\bf Hint:}
+       The energy levels of a harmonic crystal of N ions
+       can be regarded as 3N independent oscillators,
+       whose frequencies are those of the 3N classical normal modes.
+       The contribution to the total energy of a particular normal mode
+       with angular frequency $\omega_s({\bf k})$ 
+       ($s$: branch, ${\bf k}$: wave vector) is given by
+       $(n_{{\bf k}s} + \frac{1}{2})\hbar\omega_s({\bf k})$ with the
+       excitation number $n_{{\bf k}s}$ being restricted to integers greater
+       or equal zero.
+       The total energy is given by the sum over the energies of the individual
+       normal modes.
+       Use the totals formula of the geometric series to expcitly calculate
+       the sum of the exponential functions.
+ \item Separate the above result into a term vanishing as $T$ goes to zero and
+       a second term giving the energy of the zero-point vibrations of the
+       normal modes.
+ \item Write down an expression for the specific heat.
+       Consider a large crystal and thus replace the sum over the discrete
+       wave vectors with an integral.
+ \item Debye replaced all branches of the vibrational spectrum with three
+       branches, each of them obeying the dispersion relation
+       $w=ck$.
+       Additionally the integral is cut-off at a radius $k_{\text{D}}$
+       to have a total amount of N allowed wave vectors.
+       Determine $k_{\text{D}}$.
+       Evaluate the simplified integral and introduce the
+       Debye frequency $\omega_{\text{D}}=k_{\text{D}}c$
+       and the Debye temperature $\Theta_{\text{D}}$ which is given by
+       $k_{\text{B}}\Theta_{\text{D}}=\hbar\omega_{\text{D}}$.
+       Write down the resulting expression for the specific heat.
+\end{enumerate}
+
+\end{document}