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28 \begin{document}
29
30 % header
31 \begin{center}
32  {\LARGE {\bf Materials Physics I}\\}
33  \vspace{8pt}
34  Prof. B. Stritzker\\
35  WS 2007/08\\
36  \vspace{8pt}
37  {\Large\bf Tutorial 2 - proposed solutions}
38 \end{center}
39
40 \section{Phonons 1}
41 \begin{enumerate}
42  \item \begin{itemize}
43         \item $r_i=r_{i0}+u_i$\\
44               $\rho=r_2-r_1=r_{20}+u_2-r_{10}-u_1=(r_{20}-r_{10})+(u_2-u_1)
45                    =\rho_0+\sigma$
46         \item $\Phi-\Phi_0=\frac{D}{2}(\rho-\rho_0)^2
47                           =\frac{D}{2}(\rho^2+\rho_0^2-2\rho_0\rho)$\\
48               $\rho^2=\rho_0^2+\sigma^2+2\rho_0\sigma$ 
49               $\Rightarrow$ $\rho=\sqrt{\rho_0^2+\sigma^2+2\rho_0\sigma}$\\
50               $\Rightarrow$ $\Phi-\Phi_0=\frac{D}{2}
51                              [2\rho_0^2+\sigma^2+2\rho_0\sigma-
52                               2\rho_0\sqrt{\rho_0^2+\sigma^2+2\rho_0\sigma}]$
53        \end{itemize}
54  \item $\sigma \parallel \rho_0$:
55        \begin{enumerate}
56         \item \begin{flushleft}
57                \includegraphics[height=6cm]{elongation_p01.eps}
58                \includegraphics[height=6cm]{elongation_p02.eps}
59                \includegraphics[height=6cm]{elongation_p03.eps}
60               \end{flushleft}
61         \item $\sigma = \sigma_{\parallel}$:\\
62               $\rho_0 \sigma_{\parallel} = |\rho_0| |\sigma_{\parallel}|$\\
63               $\Phi-\Phi_0=\frac{D}{2}\left(2\rho_0^2+\sigma_{\parallel}^2+
64                            2\rho_0\sigma_{\parallel}-
65                            2\rho_0\sqrt{(\rho_0+\sigma_{\parallel})^2}\right)
66                           =\frac{D}{2}\sigma_{\parallel}^2$
67        \end{enumerate}
68  \item $\sigma \perp \sigma_0$:
69        \begin{enumerate}
70         \item \begin{flushleft}
71                \includegraphics[height=5.3cm]{elongation_n01.eps}
72                \includegraphics[height=5.3cm]{elongation_n02.eps}
73                \includegraphics[height=5.3cm]{elongation_n03.eps}
74               \end{flushleft}
75         \item $\sigma=\sigma_{\perp}$:\\
76               $\sigma_{\perp} \rho_0 = 0$\\
77               $\Phi-\Phi_0=\frac{D}{2}\left[2\rho_0^2+\sigma_{\perp}^2-
78                            2\rho_0\sqrt{\rho_0^2+\sigma_{\perp}^2}\right]$
79
80         \item $\sigma_{\perp} = \alpha \rho_0$, $\alpha \ll 1$\\
81               $\sqrt{\rho_0^2+\sigma_{\perp}^2}=
82                \sqrt{\rho_0^2+\alpha^2\rho_0^2}=
83                \rho_0\sqrt{1+\alpha^2}\stackrel{Taylor}{=}
84                \rho_0(1+\frac{\alpha^2}{2}-\frac{\alpha^4}{8}+\ldots)$\\
85               $\Rightarrow \Phi-\Phi_0=
86                \frac{D}{2}\left[\rho_0^2\left(2+\alpha^2-
87                2(1+\frac{\alpha^2}{2}-\frac{\alpha^4}{8}+\ldots)\right)\right]=
88                \frac{D}{2}\left[\rho_0^2(\frac{\alpha^4}{4}+\ldots)\right]$\\
89               $\Rightarrow \Phi-\Phi_0\stackrel{\alpha\ll 1}{=}
90                \frac{D}{2}\rho_0^2\frac{\alpha^4}{4}=
91                \frac{D}{2}\sigma_{\perp}^2\frac{\alpha^2}{4}$
92         \item $\sigma_{\parallel}$, $\sigma_{\perp} \ll \rho_0$\\
93               $\Rightarrow$ potential contribution of $\sigma_{\perp}$
94               compared to contribution of $\sigma_{\parallel}$
95               negligible small.
96        \end{enumerate}
97  \item \begin{itemize}
98         \item As long as the displacements and thus the elongation is small
99               compared to the equilibrium state the change in the potential
100               due to the perpendicular elongation is negligible small.
101         \item Regarding a possible existence of perpendicular elongation
102               the model of the linear chain is unproblematic.
103         \item In a real crystal couplings in other directions exist.
104               These can only be neglected if they are small compared to the
105               coupling of the considered direction.
106        \end{itemize}
107 \end{enumerate}
108
109 \section{Phonons 2}
110 \begin{enumerate}
111 \item \begin{itemize}
112        \item Convention:\\
113              Atom type 1: $M_1$, $u_s$ (elongation of atom $s$ of type 1)\\
114              Atom type 2: $M_2$, $v_s$ (elongation of atom $s$ of type 2)\\
115              Lattice constant: $a$, Spring constant: $C$
116        \item Equations of motion:\\
117              $M_1\ddot{u}_s=C(v_s+v_{s-1}-2u_s)$\\ 
118              $M_2\ddot{v}_s=C(u_{s+1}+u_s-2v_s)$
119        \item Ansatz:\\
120              $u_s=u\exp(i(ska-\omega t))$\\
121              $v_s=v\exp(i(ska-\omega t))$
122        \item Solution of the equation system:\\
123              $-\omega^2M_1u\exp(i(ska-\omega t))=
124              C\exp(-i\omega t)[v\exp(iska)+v\exp(i(s-1)ka)-2u\exp(iska)]$\\
125              $\Rightarrow -\omega^2M_1u=Cv(1+\exp(-ika))-2Cu$\\
126              $-\omega^2M_2v\exp(i(ska-\omega t))=
127              C\exp(-i\omega t)[u\exp(i(s+1)ka)+u\exp(iska)-2v\exp(iska)]$\\
128              $\Rightarrow -\omega^2M_2v=Cu[\exp(ika)+1]-2Cv$\\
129              Non trivial solution only if determinant of coefficients
130              $u$ and $v$ is zero.\\
131              $\Rightarrow
132               \left|
133               \begin{array}{cc}
134               2C-M_1\omega^2 & -C[1+\exp(-ika)]\\
135               -C[1+\exp(ika)] & 2C-M_2\omega^2
136               \end{array}
137               \right|=0$\\
138              $\Rightarrow
139               4C^2+M_1M_2\omega^4-2C\omega^2(M_2+M_1)-
140               \underbrace{C^2(1+\exp(ika))(1+\exp(-ika))}_{
141               C^2(\underbrace{1+1+\exp(ika)+\exp(-ika)}_{
142                               2+2\cos(ka)=2(1+\cos(ka))})}$\\
143              $\Rightarrow
144               M_1M_2\omega^4-2C(M_1+M_2)\omega^2+2C^2(1-\cos(ka))=0$
145       \end{itemize}
146 \item \begin{eqnarray}
147       \omega^2&=&C\left(\frac{2C(M_1+M_2)}{2M_1M_2}\right)\pm
148                  \sqrt{\frac{4C^2(M_1+M_2)^2}{4M_1^2M_2^2}-
149                        \frac{2C^2(1-cos(ka))}{M_1M_2}} \nonumber \\
150               &=&C\left(\frac{1}{M_1}+\frac{1}{M_2}\right)\pm
151                  \sqrt{C^2\frac{(M_1+M_2)^2}{M_1^2M_2^2}-
152                        \frac{1}{M_1M_2}2C^2(1-cos(ka))} \nonumber \\
153               &=&C\left(\frac{1}{M_1}+\frac{1}{M_2}\right)\pm
154                  C\sqrt{\left(\frac{1}{M_1}+\frac{1}{M_2}\right)^2-
155                         \frac{2(1-\cos(ka))}{M_1M_2}} \nonumber
156       \end{eqnarray}
157       \begin{itemize}
158        \item $ka\ll 1$:\\
159              $\rightarrow \cos(ka)\approx 1-\frac{1}{2}k^2a^2$ (Taylor)\\
160              Optical branch: $\omega^2\approx
161                               2C\left(\frac{1}{M_1}+\frac{1}{M_2}\right)$\\
162              Acoustic branch: $\omega^2\approx
163                                \frac{C/2}{M_1+M_2}k^2a^2$\\
164        \item $k=0$:\\
165              Optical branch: $u/v = - M_2/M_1$ (out of phase)\\
166        \item $k=\pm \pi/a$:\\
167              $\rightarrow \omega^2=2C/M_2,2C/M_1$
168       \end{itemize}
169 \end{enumerate}
170
171 \end{document}