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\begin{document}

% header
\begin{center}
 {\LARGE {\bf Materials Physics I}\\}
 \vspace{8pt}
 Prof. B. Stritzker\\
 WS 2007/08\\
 \vspace{8pt}
 {\Large\bf Tutorial 2 - proposed solutions}
\end{center}

\section{Phonons 1}
\begin{enumerate}
 \item \begin{itemize}
        \item $r_i=r_{i0}+u_i$\\
	      $\rho=r_2-r_1=r_{20}+u_2-r_{10}-u_1=(r_{20}-r_{10})+(u_2-u_1)
	           =\rho_0+\sigma$
	\item $\Phi-\Phi_0=\frac{D}{2}(\rho-\rho_0)^2
	                  =\frac{D}{2}(\rho^2+\rho_0^2-2\rho_0\rho)$\\
	      $\rho^2=\rho_0^2+\sigma^2+2\rho_0\sigma$ 
	      $\Rightarrow$ $\rho=\sqrt{\rho_0^2+\sigma^2+2\rho_0\sigma}$\\
              $\Rightarrow$ $\Phi-\Phi_0=\frac{D}{2}
	                     [2\rho_0^2+\sigma^2+2\rho_0\sigma-
			      2\rho_0\sqrt{\rho_0^2+\sigma^2+2\rho_0\sigma}]$
       \end{itemize}
 \item $\sigma \parallel \rho_0$:
       \begin{enumerate}
        \item \begin{flushleft}
	       \includegraphics[height=6cm]{elongation_p01.eps}
	       \includegraphics[height=6cm]{elongation_p02.eps}
	       \includegraphics[height=6cm]{elongation_p03.eps}
              \end{flushleft}
        \item $\sigma = \sigma_{\parallel}$:\\
	      $\rho_0 \sigma_{\parallel} = |\rho_0| |\sigma_{\parallel}|$\\
	      $\Phi-\Phi_0=\frac{D}{2}\left(2\rho_0^2+\sigma_{\parallel}^2+
	                   2\rho_0\sigma_{\parallel}-
		           2\rho_0\sqrt{(\rho_0+\sigma_{\parallel})^2}\right)
			  =\frac{D}{2}\sigma_{\parallel}^2$
       \end{enumerate}
 \item $\sigma \perp \sigma_0$:
       \begin{enumerate}
        \item \begin{flushleft}
	       \includegraphics[height=5.3cm]{elongation_n01.eps}
	       \includegraphics[height=5.3cm]{elongation_n02.eps}
	       \includegraphics[height=5.3cm]{elongation_n03.eps}
              \end{flushleft}
        \item $\sigma=\sigma_{\perp}$:\\
	      $\sigma_{\perp} \rho_0 = 0$\\
	      $\Phi-\Phi_0=\frac{D}{2}\left[2\rho_0^2+\sigma_{\perp}^2-
		           2\rho_0\sqrt{\rho_0^2+\sigma_{\perp}^2}\right]$

	\item $\sigma_{\perp} = \alpha \rho_0$, $\alpha \ll 1$\\
	      $\sqrt{\rho_0^2+\sigma_{\perp}^2}=
	       \sqrt{\rho_0^2+\alpha^2\rho_0^2}=
	       \rho_0\sqrt{1+\alpha^2}\stackrel{Taylor}{=}
	       \rho_0(1+\frac{\alpha^2}{2}-\frac{\alpha^4}{8}+\ldots)$\\
	      $\Rightarrow \Phi-\Phi_0=
	       \frac{D}{2}\left[\rho_0^2\left(2+\alpha^2-
	       2(1+\frac{\alpha^2}{2}-\frac{\alpha^4}{8}+\ldots)\right)\right]=
	       \frac{D}{2}\left[\rho_0^2(\frac{\alpha^4}{4}+\ldots)\right]$\\
	      $\Rightarrow \Phi-\Phi_0\stackrel{\alpha\ll 1}{=}
	       \frac{D}{2}\rho_0^2\frac{\alpha^4}{4}=
	       \frac{D}{2}\sigma_{\perp}^2\frac{\alpha^2}{4}$
	\item $\sigma_{\parallel}$, $\sigma_{\perp} \ll \rho_0$\\
	      $\Rightarrow$ potential contribution of $\sigma_{\perp}$
	      compared to contribution of $\sigma_{\parallel}$
	      negligible small.
       \end{enumerate}
 \item \begin{itemize}
        \item As long as the displacements and thus the elongation is small
	      compared to the equilibrium state the change in the potential
	      due to the perpendicular elongation is negligible small.
        \item Regarding a possible existence of perpendicular elongation
	      the model of the linear chain is unproblematic.
        \item In a real crystal couplings in other directions exist.
	      These can only be neglected if they are small compared to the
	      coupling of the considered direction.
       \end{itemize}
\end{enumerate}

\section{Phonons 2}
\begin{enumerate}
\item \begin{itemize}
       \item Convention:\\
             Atom type 1: $M_1$, $u_s$ (elongation of atom $s$ of type 1)\\
	     Atom type 2: $M_2$, $v_s$ (elongation of atom $s$ of type 2)\\
	     Lattice constant: $a$, Spring constant: $C$
       \item Equations of motion:\\
             $M_1\ddot{u}_s=C(v_s+v_{s-1}-2u_s)$\\ 
             $M_2\ddot{v}_s=C(u_{s+1}+u_s-2v_s)$
       \item Ansatz:\\
             $u_s=u\exp(i(ska-\omega t))$\\
	     $v_s=v\exp(i(ska-\omega t))$
       \item Solution of the equation system:\\
             $-\omega^2M_1u\exp(i(ska-\omega t))=
	     C\exp(-i\omega t)[v\exp(iska)+v\exp(i(s-1)ka)-2u\exp(iska)]$\\
	     $\Rightarrow -\omega^2M_1u=Cv(1+\exp(-ika))-2Cu$\\
	     $-\omega^2M_2v\exp(i(ska-\omega t))=
	     C\exp(-i\omega t)[u\exp(i(s+1)ka)+u\exp(iska)-2v\exp(iska)]$\\
             $\Rightarrow -\omega^2M_2v=Cu[\exp(ika)+1]-2Cv$\\
	     Non trivial solution only if determinant of coefficients
	     $u$ and $v$ is zero.\\
	     $\Rightarrow
	      \left|
              \begin{array}{cc}
              2C-M_1\omega^2 & -C[1+\exp(-ika)]\\
	      -C[1+\exp(ika)] & 2C-M_2\omega^2
	      \end{array}
	      \right|=0$\\
	     $\Rightarrow
	      4C^2+M_1M_2\omega^4-2C\omega^2(M_2+M_1)-
	      \underbrace{C^2(1+\exp(ika))(1+\exp(-ika))}_{
	      C^2(\underbrace{1+1+\exp(ika)+\exp(-ika)}_{
	                      2+2\cos(ka)=2(1+\cos(ka))})}$\\
	     $\Rightarrow
	      M_1M_2\omega^4-2C(M_1+M_2)\omega^2+2C^2(1-\cos(ka))=0$
      \end{itemize}
\newpage
\item \begin{eqnarray}
      \omega^2&=&C\left(\frac{2C(M_1+M_2)}{2M_1M_2}\right)\pm
                 \sqrt{\frac{4C^2(M_1+M_2)^2}{4M_1^2M_2^2}-
		       \frac{2C^2(1-cos(ka))}{M_1M_2}} \nonumber \\
              &=&C\left(\frac{1}{M_1}+\frac{1}{M_2}\right)\pm
                 \sqrt{C^2\frac{(M_1+M_2)^2}{M_1^2M_2^2}-
	               \frac{1}{M_1M_2}2C^2(1-cos(ka))} \nonumber \\
              &=&C\left(\frac{1}{M_1}+\frac{1}{M_2}\right)
	         \stackrel{{\color{red}+}}{{\color{blue}-}}
                 C\sqrt{\left(\frac{1}{M_1}+\frac{1}{M_2}\right)^2-
	                \frac{2(1-\cos(ka))}{M_1M_2}} \nonumber
      \end{eqnarray}
      \begin{figure}[!h]

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      \end{figure}
      \begin{itemize}
       \item $ka\ll 1$:\\
             $\rightarrow \cos(ka)\approx 1-\frac{1}{2}k^2a^2$ (Taylor)\\
	     $\Rightarrow$\\
	     $\sqrt{(\frac{1}{M_1}+\frac{1}{M_2})^2-
	      \frac{k^2a^2}{M_1M_2}}=$
	     $(\frac{1}{M_1}+\frac{1}{M_2})
	      \sqrt{1-\frac{k^2a^2}{M_1M_2(1/M_1+1/M_2)^2}}
	      \stackrel{Taylor}{\approx}
	      (\frac{1}{M_1}+\frac{1}{M_2})
	      (1-\frac{1}{2}\frac{k^2a^2}{M_1M_2(1/M_1+1/M_2)^2})$\\
	     $\omega \approx \sqrt{C(\frac{1}{M_1}+\frac{1}{M_2})}
	      \sqrt{1\pm (1-\frac{1}{2}\frac{k^2a^2}{M_1M_2(1/M_1+1/M_2)^2})}$\\
	     $\stackrel{{\color{red}+}}{\rightarrow}
	      \sqrt{C(\frac{1}{M_1}+\frac{1}{M_2})}
	      \sqrt{2-\frac{1}{2}\frac{k^2a^2}{M_1M_2(1/M_1+1/M_2)^2}}
	      \stackrel{Taylor}{\approx}
	      \sqrt{C(\frac{1}{M_1}+\frac{1}{M_2})}\sqrt{2}
	      (1-\frac{1}{2}\frac{1}{4}\frac{k^2a^2}{M_1M_2(1/M_1+1/M_2)^2})$\\
	     $\stackrel{{\color{blue}-}}{\rightarrow}
	      \sqrt{C(\frac{1}{M_1}+\frac{1}{M_2})}
	      \sqrt{\frac{1}{2}\frac{k^2a^2}{M_1M_2(1/M_1+1/M_2)^2}}=
	      \sqrt{C(\frac{1}{M_1}+\frac{1}{M_2})}
	      \sqrt{\frac{1}{2}\frac{1}{M_1M_2(1/M_1+1/M_2)^2}}ka$\\
	     {\color{red}Optical branch}: $\omega\stackrel{ka\ll 1}{\approx}
	                      \sqrt{2C\left(\frac{1}{M_1}+
			                    \frac{1}{M_2}\right)}$\\
	     {\color{blue}Acoustic branch}: $\omega\stackrel{ka\ll 1}{\approx}
	                       \sqrt{\frac{C/2}{M_1+M_2}}ka$\\
       \item $k=0$:\\
             $\rightarrow u/v = - M_2/M_1$ (out of phase)\\
       \item $k=\pi/a$\\
	     $\rightarrow \omega^2=2C/M_2,2C/M_1$
      \end{itemize}
\end{enumerate}

\end{document}
