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32 {\LARGE {\bf Materials Physics I}\\}
37 {\Large\bf Tutorial 2}
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)
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 $\Rightarrow$ $\Phi-\Phi_0=\frac{D}{2}
51 [2\rho_0^2+\sigma^2+2\rho_0\sigma-
52 2\rho_0\sqrt{\rho_0^2+\sigma^2+2\rho_0\sigma}]$
54 \item $\sigma \parallel \rho_0$:
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}
61 \item $\sigma = \sigma_{\parallel}$:\\
62 $\rho_0 \sigma_{\parallel} = |\rho_0| |\sigma_{\parallel}|$\\
63 $\Phi-\Phi_0=\frac{D}{2}\left(2\rho_0^2+\sigma_{\parallel}^2+
64 2\rho_0\sigma_{\parallel}-
65 2\rho_0\sqrt{(\rho_0+\sigma_{\parallel})^2}\right)
66 =\frac{D}{2}\sigma_{\parallel}^2$
68 \item $\sigma \perp \sigma_0$:
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}
75 \item $\sigma=\sigma_{\perp}$:\\
76 $\sigma_{\perp} \rho_0 = 0$\\
77 $\Phi-\Phi_0=\frac{D}{2}\left[2\rho_0^2+\sigma_{\perp}^2-
78 2\rho_0\sqrt{\rho_0^2+\sigma_{\perp}^2}\right]$
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 $\sigma_{\parallel}$
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.
111 \item \begin{itemize}
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 $M_2\ddot{v}_s=C(u_{s+1}+u_s-2v_s)$
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 $-\omega^2M_1u=Cv[1+\exp(-ika)]-2Cu$\\
124 $-\omega^2M_2v=Cu[\exp(ika)+1]-2Cv$\\
125 Non trivial solution only if determinant of coefficients
126 $u$ and $v$ is zero.\\
130 2C-M_1\omega^2 & -C[1+\exp(-ika)]\\
131 -C[1+\exp(ika)] & 2C-M_2\omega^2
135 M_1M_2\omega^4-2C(M_1+M_2)\omega^2+2C^2(1-\cos(ka))=0$
138 \omega^2=C\left(\frac{1}{M_1}+\frac{1}{M_2}\right)\pm
139 C\sqrt{\left(\frac{1}{M_1}+\frac{1}{M_2}\right)^2-
140 \frac{2(1-\cos(ka))}{M_1M_2}}
144 $\rightarrow \cos(ka)\approx 1-\frac{1}{2}k^2a^2$\\
145 Optical branch: $\omega^2\approx
146 2C\left(\frac{1}{M_1}+\frac{1}{M_2}\right)$\\
147 Acoustic branch: $\omega^2\approx
148 \frac{C/2}{M_1+M_2}k^2a^2$\\
150 Optical branch: $u/v = - M_2/M_1$ (out of phase)\\
151 \item $k=\pm \pi/a$:\\
152 $\rightarrow \omega^2=2C/M_2,2C/M_1$