2 \documentclass[a4paper,11pt]{article}
3 \usepackage[activate]{pdfcprot}
7 \usepackage[german]{babel}
8 \usepackage[latin1]{inputenc}
9 \usepackage[T1]{fontenc}
13 \usepackage[dvips]{graphicx}
14 \graphicspath{{./img/}}
20 \setlength{\headheight}{0mm} \setlength{\headsep}{0mm}
21 \setlength{\topskip}{-10mm} \setlength{\textwidth}{17cm}
22 \setlength{\oddsidemargin}{-10mm}
23 \setlength{\evensidemargin}{-10mm} \setlength{\topmargin}{-1cm}
24 \setlength{\textheight}{26cm} \setlength{\headsep}{0cm}
26 \renewcommand{\labelenumi}{(\alph{enumi})}
27 \renewcommand{\labelenumii}{\arabic{enumii})}
28 \renewcommand{\labelenumiii}{\roman{enumiii})}
34 {\LARGE {\bf Materials Physics II}\\}
39 {\Large\bf Tutorial 4 - proposed solutions}
44 \section{Legendre transformation and Maxwell relations}
47 \item Legendre transformation:
49 dg &=& df - \sum_{i=r+1}^{n} d(u_ix_i)\nonumber\\
50 &=& df - \sum_{i=r+1}^{n} (x_idu_i + u_idx_i)\nonumber\\
51 &=& \sum_{i=1}^r u_idx_i - \sum_{i=r+1}^n x_idu_i\nonumber
54 \Rightarrow g=g(x_1,\ldots,x_r,u_{r+1},\ldots,u_n)
56 \item Use $T=\left.\frac{\partial E}{\partial S}\right|_V$ and
57 $-p=\left.\frac{\partial E}{\partial V}\right|_S$.\\
58 Start with internal energy $E=E(S,V)$:
60 \Rightarrow dE=\frac{\partial E}{\partial S}dS +
61 \frac{\partial E}{\partial V}dV =
66 \Rightarrow dH=dE+Vdp+pdV=TdS-pdV+Vdp+pdV=TdS+Vdp
70 \left.\frac{\partial H}{\partial S}\right|_p=T \textrm{ and }
71 \left.\frac{\partial H}{\partial p}\right|_S=V
73 Helmholtz free energy $F=E-TS$:
75 \Rightarrow dF=dE-SdT-TdS=TdS-pdV-SdT-TdS=-pdV-SdT
79 \left.\frac{\partial F}{\partial V}\right|_T=-p \textrm{ and }
80 \left.\frac{\partial F}{\partial T}\right|_V=-S
82 Gibbs free energy $G=H-TS=E+pV-TS$:
84 \Rightarrow dG=dH-SdT-TdS=TdS+Vdp-SdT-TdS=Vdp-SdT
88 \left.\frac{\partial G}{\partial p}\right|_T=V \textrm{ and }
89 \left.\frac{\partial G}{\partial T}\right|_p=-S
91 \item Maxwell relations:\\
92 Internal energy: $dE=TdS-pdV$
94 \frac{\partial}{\partial S}
95 \left(\left.\frac{\partial E}{\partial V}\right|_S\right)_V=
96 \frac{\partial}{\partial V}
97 \left(\left.\frac{\partial E}{\partial S}\right|_V\right)_S
99 \left.-\frac{\partial p}{\partial S}\right|_V=
100 \left.\frac{\partial T}{\partial V}\right|_S
102 Enthalpy: $dH=TdS+Vdp$
104 \frac{\partial}{\partial S}
105 \left(\left.\frac{\partial H}{\partial p}\right|_S\right)_p=
106 \frac{\partial}{\partial p}
107 \left(\left.\frac{\partial H}{\partial S}\right|_p\right)_S
109 \left.\frac{\partial V}{\partial S}\right|_p=
110 \left.\frac{\partial T}{\partial p}\right|_S
112 Helmholtz free energy: $dF=-pdV-SdT$
114 \frac{\partial}{\partial V}
115 \left(\left.\frac{\partial F}{\partial T}\right|_V\right)_T=
116 \frac{\partial}{\partial T}
117 \left(\left.\frac{\partial F}{\partial V}\right|_T\right)_V
119 \left.-\frac{\partial S}{\partial V}\right|_T=
120 \left.-\frac{\partial p}{\partial T}\right|_V
122 Gibbs free energy: $dG=Vdp-SdT$
124 \frac{\partial}{\partial p}
125 \left(\left.\frac{\partial G}{\partial T}\right|_p\right)_T=
126 \frac{\partial}{\partial T}
127 \left(\left.\frac{\partial G}{\partial p}\right|_T\right)_p
129 \left.-\frac{\partial S}{\partial p}\right|_T=
130 \left.\frac{\partial V}{\partial T}\right|_p
134 \section{Thermal expansion of solids}
137 \item Coefficients of thermal expansion:\\
138 Consider a cube with side lengthes $L_1,L_2,L_3$.
139 Isotropic material: $\frac{1}{L_1}\frac{\partial L_1}{\partial T}=
140 \frac{1}{L_2}\frac{\partial L_2}{\partial T}=
141 \frac{1}{L_3}\frac{\partial L_3}{\partial T}=
144 \alpha_V&=&\frac{1}{V}\frac{\partial V}{\partial T}=
145 \frac{1}{L_1L_2L_3}\frac{\partial}{\partial T}(L_1L_2L_3)=
146 \frac{1}{L_1L_2L_3}\left(L_2L_3\frac{\partial L_1}{\partial T}+
147 L_1L_3\frac{\partial L_2}{\partial T}+
148 L_1L_2\frac{\partial L_3}{\partial T}\right)
150 &=&\frac{1}{L_1}\frac{\partial L_1}{\partial T}+
151 \frac{1}{L_2}\frac{\partial L_2}{\partial T}+
152 \frac{1}{L_3}\frac{\partial L_3}{\partial T}=3\alpha_L\nonumber
155 dF=-pdV-SdT \Rightarrow p=-\left.\frac{\partial}{\partial V}\right|T
158 dE=TdS-pdV \Rightarrow
160 Find an expression for the pressure as a function of the free energy
162 Rewrite this equation to express the pressure entirely in terms of
163 the internal energy $E$.
164 Evaluate the pressure by using the harmonic form of the internal energy.
166 Step 2 introduced an integral over the temperature $T'$.
167 Change the integration variable $T'$ to $x=\hbar\omega_s({\bf k})/T'$.
168 Use integration by parts with respect to $x$.
169 \item The normal mode frequencies of a rigorously harmonic crystal
170 are unaffected by a change in volume.
171 What does this imply for the pressure
172 (Which variables does the pressure depend on)?
173 Draw conclusions for the coefficient of thermal expansion.
174 \item Find an expression for $C_p-C_V$ in terms of temperature $T$,
175 volume $V$, the coefficient of thermal expansion $\alpha_V$ and
176 the inverse bulk modulus (isothermal compressibility)
177 $\frac{1}{B}=-\frac{1}{V}\left.\frac{\partial V}{\partial p}\right|_T$.\\
178 $C_p=\left.\frac{\partial E}{\partial T}\right|_p$ is the heat capacity
179 for constant pressure and
180 $C_V=\left.\frac{\partial E}{\partial T}\right|_V$ is the heat capacity