From 99bdd766a8ef8502e677e534d5d0387444ab5253 Mon Sep 17 00:00:00 2001 From: hackbard Date: Tue, 3 Jun 2008 18:05:25 +0200 Subject: [PATCH] added initial 2_03s file --- solid_state_physics/tutorial/2_03s.tex | 173 +++++++++++++++++++++++++ 1 file changed, 173 insertions(+) create mode 100644 solid_state_physics/tutorial/2_03s.tex diff --git a/solid_state_physics/tutorial/2_03s.tex b/solid_state_physics/tutorial/2_03s.tex new file mode 100644 index 0000000..a8eab9d --- /dev/null +++ b/solid_state_physics/tutorial/2_03s.tex @@ -0,0 +1,173 @@ +\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} -- 2.20.1