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2 \documentclass[a4paper,11pt]{article}
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10 \usepackage{amsmath}
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15 \usepackage{color}
16 \usepackage{pstricks}
17 \usepackage{pst-node}
18 \usepackage{rotating}
19 \usepackage{eepic}
20
21 \setlength{\headheight}{0mm} \setlength{\headsep}{0mm}
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26
27 \renewcommand{\labelenumi}{(\alph{enumi})}
28
29 \begin{document}
30
31 % header
32 \begin{center}
33  {\LARGE {\bf Materials Physics I}\\}
34  \vspace{8pt}
35  Prof. B. Stritzker\\
36  WS 2007/08\\
37  \vspace{8pt}
38  {\Large\bf Tutorial 2 - proposed solutions}
39 \end{center}
40
41 \section{Phonons 1}
42 \begin{enumerate}
43  \item \begin{itemize}
44         \item $r_i=r_{i0}+u_i$\\
45               $\rho=r_2-r_1=r_{20}+u_2-r_{10}-u_1=(r_{20}-r_{10})+(u_2-u_1)
46                    =\rho_0+\sigma$
47         \item $\Phi-\Phi_0=\frac{D}{2}(\rho-\rho_0)^2
48                           =\frac{D}{2}(\rho^2+\rho_0^2-2\rho_0\rho)$\\
49               $\rho^2=\rho_0^2+\sigma^2+2\rho_0\sigma$ 
50               $\Rightarrow$ $\rho=\sqrt{\rho_0^2+\sigma^2+2\rho_0\sigma}$\\
51               $\Rightarrow$ $\Phi-\Phi_0=\frac{D}{2}
52                              [2\rho_0^2+\sigma^2+2\rho_0\sigma-
53                               2\rho_0\sqrt{\rho_0^2+\sigma^2+2\rho_0\sigma}]$
54        \end{itemize}
55  \item $\sigma \parallel \rho_0$:
56        \begin{enumerate}
57         \item \begin{flushleft}
58                \includegraphics[height=6cm]{elongation_p01.eps}
59                \includegraphics[height=6cm]{elongation_p02.eps}
60                \includegraphics[height=6cm]{elongation_p03.eps}
61               \end{flushleft}
62         \item $\sigma = \sigma_{\parallel}$:\\
63               $\rho_0 \sigma_{\parallel} = |\rho_0| |\sigma_{\parallel}|$\\
64               $\Phi-\Phi_0=\frac{D}{2}\left(2\rho_0^2+\sigma_{\parallel}^2+
65                            2\rho_0\sigma_{\parallel}-
66                            2\rho_0\sqrt{(\rho_0+\sigma_{\parallel})^2}\right)
67                           =\frac{D}{2}\sigma_{\parallel}^2$
68        \end{enumerate}
69  \item $\sigma \perp \sigma_0$:
70        \begin{enumerate}
71         \item \begin{flushleft}
72                \includegraphics[height=5.3cm]{elongation_n01.eps}
73                \includegraphics[height=5.3cm]{elongation_n02.eps}
74                \includegraphics[height=5.3cm]{elongation_n03.eps}
75               \end{flushleft}
76         \item $\sigma=\sigma_{\perp}$:\\
77               $\sigma_{\perp} \rho_0 = 0$\\
78               $\Phi-\Phi_0=\frac{D}{2}\left[2\rho_0^2+\sigma_{\perp}^2-
79                            2\rho_0\sqrt{\rho_0^2+\sigma_{\perp}^2}\right]$
80
81         \item $\sigma_{\perp} = \alpha \rho_0$, $\alpha \ll 1$\\
82               $\sqrt{\rho_0^2+\sigma_{\perp}^2}=
83                \sqrt{\rho_0^2+\alpha^2\rho_0^2}=
84                \rho_0\sqrt{1+\alpha^2}\stackrel{Taylor}{=}
85                \rho_0(1+\frac{\alpha^2}{2}-\frac{\alpha^4}{8}+\ldots)$\\
86               $\Rightarrow \Phi-\Phi_0=
87                \frac{D}{2}\left[\rho_0^2\left(2+\alpha^2-
88                2(1+\frac{\alpha^2}{2}-\frac{\alpha^4}{8}+\ldots)\right)\right]=
89                \frac{D}{2}\left[\rho_0^2(\frac{\alpha^4}{4}+\ldots)\right]$\\
90               $\Rightarrow \Phi-\Phi_0\stackrel{\alpha\ll 1}{=}
91                \frac{D}{2}\rho_0^2\frac{\alpha^4}{4}=
92                \frac{D}{2}\sigma_{\perp}^2\frac{\alpha^2}{4}$
93         \item $\sigma_{\parallel}$, $\sigma_{\perp} \ll \rho_0$\\
94               $\Rightarrow$ potential contribution of $\sigma_{\perp}$
95               compared to contribution of $\sigma_{\parallel}$
96               negligible small.
97        \end{enumerate}
98  \item \begin{itemize}
99         \item As long as the displacements and thus the elongation is small
100               compared to the equilibrium state the change in the potential
101               due to the perpendicular elongation is negligible small.
102         \item Regarding a possible existence of perpendicular elongation
103               the model of the linear chain is unproblematic.
104         \item In a real crystal couplings in other directions exist.
105               These can only be neglected if they are small compared to the
106               coupling of the considered direction.
107        \end{itemize}
108 \end{enumerate}
109
110 \section{Phonons 2}
111 \begin{enumerate}
112 \item \begin{itemize}
113        \item Convention:\\
114              Atom type 1: $M_1$, $u_s$ (elongation of atom $s$ of type 1)\\
115              Atom type 2: $M_2$, $v_s$ (elongation of atom $s$ of type 2)\\
116              Lattice constant: $a$, Spring constant: $C$
117        \item Equations of motion:\\
118              $M_1\ddot{u}_s=C(v_s+v_{s-1}-2u_s)$\\ 
119              $M_2\ddot{v}_s=C(u_{s+1}+u_s-2v_s)$
120        \item Ansatz:\\
121              $u_s=u\exp(i(ska-\omega t))$\\
122              $v_s=v\exp(i(ska-\omega t))$
123        \item Solution of the equation system:\\
124              $-\omega^2M_1u\exp(i(ska-\omega t))=
125              C\exp(-i\omega t)[v\exp(iska)+v\exp(i(s-1)ka)-2u\exp(iska)]$\\
126              $\Rightarrow -\omega^2M_1u=Cv(1+\exp(-ika))-2Cu$\\
127              $-\omega^2M_2v\exp(i(ska-\omega t))=
128              C\exp(-i\omega t)[u\exp(i(s+1)ka)+u\exp(iska)-2v\exp(iska)]$\\
129              $\Rightarrow -\omega^2M_2v=Cu[\exp(ika)+1]-2Cv$\\
130              Non trivial solution only if determinant of coefficients
131              $u$ and $v$ is zero.\\
132              $\Rightarrow
133               \left|
134               \begin{array}{cc}
135               2C-M_1\omega^2 & -C[1+\exp(-ika)]\\
136               -C[1+\exp(ika)] & 2C-M_2\omega^2
137               \end{array}
138               \right|=0$\\
139              $\Rightarrow
140               4C^2+M_1M_2\omega^4-2C\omega^2(M_2+M_1)-
141               \underbrace{C^2(1+\exp(ika))(1+\exp(-ika))}_{
142               C^2(\underbrace{1+1+\exp(ika)+\exp(-ika)}_{
143                               2+2\cos(ka)=2(1+\cos(ka))})}$\\
144              $\Rightarrow
145               M_1M_2\omega^4-2C(M_1+M_2)\omega^2+2C^2(1-\cos(ka))=0$
146       \end{itemize}
147 \newpage
148 \item \begin{eqnarray}
149       \omega^2&=&C\left(\frac{2C(M_1+M_2)}{2M_1M_2}\right)\pm
150                  \sqrt{\frac{4C^2(M_1+M_2)^2}{4M_1^2M_2^2}-
151                        \frac{2C^2(1-cos(ka))}{M_1M_2}} \nonumber \\
152               &=&C\left(\frac{1}{M_1}+\frac{1}{M_2}\right)\pm
153                  \sqrt{C^2\frac{(M_1+M_2)^2}{M_1^2M_2^2}-
154                        \frac{1}{M_1M_2}2C^2(1-cos(ka))} \nonumber \\
155               &=&C\left(\frac{1}{M_1}+\frac{1}{M_2}\right)
156                  \stackrel{{\color{red}+}}{{\color{blue}-}}
157                  C\sqrt{\left(\frac{1}{M_1}+\frac{1}{M_2}\right)^2-
158                         \frac{2(1-\cos(ka))}{M_1M_2}} \nonumber
159       \end{eqnarray}
160       \begin{figure}[!h]
161
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165 \footnotesize
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167 \color{black}
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186 \color{black}
187 \put(2550,1300){\makebox(0,0)[r]{$\sqrt{\frac{2C}{M_2}}$}}
188 \put(2550,800){\makebox(0,0)[r]{$\sqrt{\frac{2C}{M_1}}$}}
189 \put(2350,-10){\makebox(0,0)[r]{$\frac{\pi}{a}$}}
190 \put(1500,-30){\makebox(0,0)[r]{$k$}}
191 \put(700,-10){\makebox(0,0)[r]{$0$}}
192 \put(650,1500){\makebox(0,0)[r]{$\sqrt{2C(\frac{1}{M_1}+\frac{1}{M_2})}$}}
193 \put(600,800){\makebox(0,0)[r]{$\omega$}}
194 \put(1800,1000){\makebox(0,0)[r]{$M_1>M_2$}}
195 \put(1989,1636){\makebox(0,0)[r]{optical branch}}
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211
212       \end{figure}
213       \begin{itemize}
214        \item $ka\ll 1$:\\
215              $\rightarrow \cos(ka)\approx 1-\frac{1}{2}k^2a^2$ (Taylor)\\
216              $\Rightarrow$\\
217              $\sqrt{(\frac{1}{M_1}+\frac{1}{M_2})^2-
218               \frac{k^2a^2}{M_1M_2}}=$
219              $(\frac{1}{M_1}+\frac{1}{M_2})
220               \sqrt{1-\frac{k^2a^2}{M_1M_2(1/M_1+1/M_2)^2}}
221               \stackrel{Taylor}{\approx}
222               (\frac{1}{M_1}+\frac{1}{M_2})
223               (1-\frac{1}{2}\frac{k^2a^2}{M_1M_2(1/M_1+1/M_2)^2})$\\
224              $\omega \approx \sqrt{C(\frac{1}{M_1}+\frac{1}{M_2})}
225               \sqrt{1\pm (1-\frac{1}{2}\frac{k^2a^2}{M_1M_2(1/M_1+1/M_2)^2})}$\\
226              $\stackrel{{\color{red}+}}{\rightarrow}
227               \sqrt{C(\frac{1}{M_1}+\frac{1}{M_2})}
228               \sqrt{2-\frac{1}{2}\frac{k^2a^2}{M_1M_2(1/M_1+1/M_2)^2}}
229               \stackrel{Taylor}{\approx}
230               \sqrt{C(\frac{1}{M_1}+\frac{1}{M_2})}\sqrt{2}
231               (1-\frac{1}{2}\frac{1}{4}\frac{k^2a^2}{M_1M_2(1/M_1+1/M_2)^2})$\\
232              $\stackrel{{\color{blue}-}}{\rightarrow}
233               \sqrt{C(\frac{1}{M_1}+\frac{1}{M_2})}
234               \sqrt{\frac{1}{2}\frac{k^2a^2}{M_1M_2(1/M_1+1/M_2)^2}}=
235               \sqrt{C(\frac{1}{M_1}+\frac{1}{M_2})}
236               \sqrt{\frac{1}{2}\frac{1}{M_1M_2(1/M_1+1/M_2)^2}}ka$\\
237              {\color{red}Optical branch}: $\omega\stackrel{ka\ll 1}{\approx}
238                               \sqrt{2C\left(\frac{1}{M_1}+
239                                             \frac{1}{M_2}\right)}$\\
240              {\color{blue}Acoustic branch}: $\omega\stackrel{ka\ll 1}{\approx}
241                                \sqrt{\frac{C/2}{M_1+M_2}}ka$\\
242        \item $k=0$:\\
243              $\rightarrow u/v = - M_2/M_1$ (out of phase)\\
244        \item $k=\pi/a$\\
245              $\rightarrow \omega^2=2C/M_2,2C/M_1$
246       \end{itemize}
247 \end{enumerate}
248
249 \end{document}