--- /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})}
+\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 4 - proposed solutions}
+\end{center}
+
+\vspace{4pt}
+
+\section{Legendre transformation and Maxwell relations}
+
+\begin{enumerate}
+ \item Legendre transformation:
+ \begin{eqnarray}
+ dg &=& df - \sum_{i=r+1}^{n} d(u_ix_i)\nonumber\\
+ &=& df - \sum_{i=r+1}^{n} (x_idu_i + u_idx_i)\nonumber\\
+ &=& \sum_{i=1}^r u_idx_i - \sum_{i=r+1}^n x_idu_i\nonumber
+ \end{eqnarray}
+ \[
+ \Rightarrow g=g(x_1,\ldots,x_r,u_{r+1},\ldots,u_n)
+ \]
+ \item Use $T=\left.\frac{\partial E}{\partial S}\right|_V$ and
+ $-p=\left.\frac{\partial E}{\partial V}\right|_S$.\\
+ Start with internal energy $E=E(S,V)$:
+ \[
+ \Rightarrow dE=\frac{\partial E}{\partial S}dS +
+ \frac{\partial E}{\partial V}dV =
+ TdS - pdV
+ \]
+ Enthalpy $H=E+pV$:
+ \[
+ \Rightarrow dH=dE+Vdp+pdV=TdS-pdV+Vdp+pdV=TdS+Vdp
+ \]
+ \[
+ \Rightarrow
+ \left.\frac{\partial H}{\partial S}\right|_p=T \textrm{ and }
+ \left.\frac{\partial H}{\partial p}\right|_S=V
+ \]
+ Helmholtz free energy $F=E-TS$:
+ \[
+ \Rightarrow dF=dE-SdT-TdS=TdS-pdV-SdT-TdS=-pdV-SdT
+ \]
+ \[
+ \Rightarrow
+ \left.\frac{\partial F}{\partial V}\right|_T=-p \textrm{ and }
+ \left.\frac{\partial F}{\partial T}\right|_V=-S
+ \]
+ Gibbs free energy $G=H-TS=E+pV-TS$:
+ \[
+ \Rightarrow dG=dH-SdT-TdS=TdS+Vdp-SdT-TdS=Vdp-SdT
+ \]
+ \[
+ \Rightarrow
+ \left.\frac{\partial G}{\partial p}\right|_T=V \textrm{ and }
+ \left.\frac{\partial G}{\partial T}\right|_p=-S
+ \]
+ \item Maxwell relations:\\
+ Enthalpy: $dH=TdS+Vdp$
+ \[
+ \frac{\partial}{\partial S}
+ \left(\left.\frac{\partial H}{\partial p}\right|_S\right)_p=
+ \frac{\partial}{\partial p}
+ \left(\left.\frac{\partial H}{\partial S}\right|_p\right)_S
+ \Rightarrow
+ \left.\frac{\partial V}{\partial S}\right|_p=
+ \left.\frac{\partial T}{\partial p}\right|_S
+ \]
+ Helmholtz free energy: $dF=-pdV-SdT$
+ \[
+ \frac{\partial}{\partial V}
+ \left(\left.\frac{\partial F}{\partial T}\right|_V\right)_T=
+ \frac{\partial}{\partial T}
+ \left(\left.\frac{\partial F}{\partial V}\right|_T\right)_V
+ \Rightarrow
+ \left.-\frac{\partial S}{\partial V}\right|_T=
+ \left.-\frac{\partial p}{\partial T}\right|_V
+ \]
+ \item For a thermodynamic potential $\Phi(X,Y)$ the following identity
+ expressing the permutability of derivatives holds:
+ \[
+ \frac{\partial^2 \Phi}{\partial X \partial Y} =
+ \frac{\partial^2 \Phi}{\partial Y \partial X}
+ \]
+ Derive the Maxwell relations by taking the mixed derivatives of the
+ potentials in (b) with respect to the variables they depend on.
+ Exchange the sequence of derivation and use the identities gained in (b).
+\end{enumerate}
+
+\section{Thermal expansion of solids}
+
+It is well known that solids change their length $L$ and volume $V$ respectively
+if there is a change in temperature $T$ or in pressure $p$ of the system.
+The following exercise shows that
+thermal expansion cannot be described by rigorously harmonic crystals.
+
+\begin{enumerate}
+ \item The coefficient of thermal expansion of a solid is given by
+ $\alpha_L=\frac{1}{L}\left.\frac{\partial L}{\partial T}\right|_p$.
+ Show that the coefficient of thermal expansion of the volume
+ $\alpha_V=\frac{1}{V}\left.\frac{\partial V}{\partial T}\right|_p$
+ equals $3\alpha_L$ for isotropic materials.
+ \item Find an expression for the pressure as a function of the free energy
+ $F=E-TS$.
+ Rewrite this equation to express the pressure entirely in terms of
+ the internal energy $E$.
+ Evaluate the pressure by using the harmonic form of the internal energy.
+ {\bf Hint:}
+ Step 2 introduced an integral over the temperature $T'$.
+ Change the integration variable $T'$ to $x=\hbar\omega_s({\bf k})/T'$.
+ Use integration by parts with respect to $x$.
+ \item The normal mode frequencies of a rigorously harmonic crystal
+ are unaffected by a change in volume.
+ What does this imply for the pressure
+ (Which variables does the pressure depend on)?
+ Draw conclusions for the coefficient of thermal expansion.
+ \item Find an expression for $C_p-C_V$ in terms of temperature $T$,
+ volume $V$, the coefficient of thermal expansion $\alpha_V$ and
+ the inverse bulk modulus (isothermal compressibility)
+ $\frac{1}{B}=-\frac{1}{V}\left.\frac{\partial V}{\partial p}\right|_T$.\\
+ $C_p=\left.\frac{\partial E}{\partial T}\right|_p$ is the heat capacity
+ for constant pressure and
+ $C_V=\left.\frac{\partial E}{\partial T}\right|_V$ is the heat capacity
+ for constant volume.
+\end{enumerate}
+
+\end{document}
projects / lectures / latex.git / blobdiff