Exercises
7. Take $CO_2$ as an example of linear triatomic molecule, that the central atom is $M$, the other two are $m$. For the y direction, $$
T=\frac{1}{2}m\left( \dot{y}_{1}^{2}+\dot{y}_{3}^{2} \right) +\frac{1}{2}M\dot{y}_{2}^{2},
$$ $$
V=\frac{k}{2}\left( y_{1}^{2}+2y_{2}^{2}+y_{3}^{2}-2y_1y_2-2y_2y_3 \right).
$$ The z direction is the same, we have, $$
A_{11}=k, \,m_{11}=m
$$ $$
A_{22}=2k, \,m_{22}=M
$$ $$
A_{33}=k, \,m_{33}=m
$$ $$
A_{12}=A_{21}=-k,\, m_{12}=m_{21}=0
$$ $$
A_{13}=A_{31}=0, \,m_{13}=m_{31}=0
$$ $$
A_{23}=A_{32}=-k, \,m_{23}=m_{32}=0
$$ Therefore the secular equation is, $$
\left| \begin{matrix}
k-mw^2& -k& 0\\
-k& 2k-Mw^2& -k\\
0& -k& k-mw^2\\
\end{matrix} \right|=0.
$$ That is, $$
w^2\left( k-mw^2 \right) \left[ Mmw^2-\left( 2m+M \right) k \right] =0.
$$ The eigenfrequencies are, $$
\omega _1=0
$$ $$
\omega _2=\sqrt{\frac{k}{m}}
$$ $$
\omega _3=\sqrt{\frac{\left( 2m+M \right) k}{Mm}}.
$$ The zero frequency shows that the redundancy of variables, because one of the $y_1,\,y_2,\,y_3$ can be reduced by the conservation of momentum of the mass center. It shows the symmetry that the mass center can be translated arbitrarily. The other two frequencies shows two modes that two $m$ move along the same or opposite direction respectively.
13. The equations of motion are $$
m\ddot{x}_1=-kx_1-k\left( x_1-x_2 \right) -\frac{q^2}{4\pi \varepsilon _0a^2}\left( 1+\frac{2\left( x_1-x_2 \right)}{a} \right),
$$ $$
m\ddot{x}_2=-kx_2-k\left( x_2-x_1 \right) +\frac{q^2}{4\pi \varepsilon _0a^2}\left( 1+\frac{2\left( x_1-x_2 \right)}{a} \right).
$$ That is $$
m\ddot{x}_1+\left( 2k+\frac{q^2}{2\pi \varepsilon _0a^3} \right) x_1-\left( k+\frac{q^2}{2\pi \varepsilon _0a^3} \right) x_2=-\frac{q^2}{4\pi \varepsilon _0a^2},
$$ $$
m\ddot{x}_2+\left( 2k+\frac{q^2}{2\pi \varepsilon _0a^3} \right) x_2-\left( k+\frac{q^2}{2\pi \varepsilon _0a^3} \right) x_1=\frac{q^2}{4\pi \varepsilon _0a^2}.
$$ We solve the homogeneous first, $$
m\ddot{x}_1+\left( 2k+\frac{q^2}{2\pi \varepsilon _0a^3} \right) x_1-\left( k+\frac{q^2}{2\pi \varepsilon _0a^3} \right) x_2=0,
$$ $$
m\ddot{x}_2+\left( 2k+\frac{q^2}{2\pi \varepsilon _0a^3} \right) x_2-\left( k+\frac{q^2}{2\pi \varepsilon _0a^3} \right) x_1=0.
$$ We attempt a solution of the form, $$
x_{10}=B_1e^{iwt},
$$ $$
x_{20}=B_2e^{iwt}.
$$ Then, $$
\left( -mw^2+2k+\frac{q^2}{2\pi \varepsilon _0a^3} \right) B_1-\left( k+\frac{q^2}{2\pi \varepsilon _0a^3} \right) B_2=0
$$ $$
\left( -mw^2+2k+\frac{q^2}{2\pi \varepsilon _0a^3} \right) B_2-\left( k+\frac{q^2}{2\pi \varepsilon _0a^3} \right) B_1=0.
$$ Therfore the secular equation is $$
\left| \begin{matrix}
-mw^2+2k+\frac{q^2}{2\pi \varepsilon _0a^3}& -\left( k+\frac{q^2}{2\pi \varepsilon _0a^3} \right)\\
-\left( k+\frac{q^2}{2\pi \varepsilon _0a^3} \right)& -mw^2+2k+\frac{q^2}{2\pi \varepsilon _0a^3}\\
\end{matrix} \right|=0.
$$
14. We denote the charge on $C_1$, $C_2$ and $C_3$ is $q_1$, $q_2$ and $q_1+q_2$ respectively. The Lagrangian is $$
L=\frac{1}{2}L_1\dot{q}_{1}^{2}+\frac{1}{2}L_2\dot{q}_{2}^{2}+\frac{1}{2}L_3\left( \dot{q}_1+\dot{q}_2 \right) ^2-\frac{q_{1}^{2}}{2C_1}-\frac{q_{2}^{2}}{2C_2}-\frac{\left( q_1+q_2 \right) ^2}{2C_3}.
$$ Then $$
U=\frac{q_{1}^{2}}{2C_1}+\frac{q_{2}^{2}}{2C_2}+\frac{\left( q_1+q_2 \right) ^2}{2C_3},
$$ $$
T=\frac{1}{2}L_1\dot{q}_{1}^{2}+\frac{1}{2}L_2\dot{q}_{2}^{2}+\frac{1}{2}L_3\left( \dot{q}_1+\dot{q}_2 \right) ^2.
$$ Therefore $$
A_{11}=\frac{\partial ^2U}{\partial q_{1}^{2}}=\frac{1}{C_1}+\frac{1}{C_3}
$$ $$
A_{12}=A_{21}=\frac{\partial ^2U}{\partial q_1\partial q_2}=\frac{1}{C_3}
$$ $$
A_{22}=\frac{\partial ^2U}{\partial q_{2}^{2}}=\frac{1}{C_2}+\frac{1}{C_3}.
$$ And $$
m_{11}=L_1+L_3
$$ $$
m_{12}=m_{21}=L_3
$$ $$
m_{22}=L_2+L_3.
$$ Therefore we need to solve, $$
\left| \begin{matrix}
\frac{1}{C_1}+\frac{1}{C_3}-w^2\left( L_1+L_3 \right)& \frac{1}{C_3}-w^2L_3\\
\frac{1}{C_3}-w^2L_3& \frac{1}{C_2}+\frac{1}{C_3}-w^2\left( L_2+L_3 \right)\\
\end{matrix} \right|=0
$$ Therefore the eigenfrequencies satisfy, $$
w^4\left( L_1L_2+L_1L_3+L_2L_3 \right) -w^2\left( \frac{L_2+L_3}{C_1}+\frac{L_1+L_3}{C_2}+\frac{L_1+L_2}{C_3} \right) +\left( \frac{1}{C_1C_2}+\frac{1}{C_1C_3}+\frac{1}{C_2C_3} \right) =0.
$$
15. For a single channel, the normal coordinate satisfy, $$
\ddot{\zeta}_i=-w_{i}^{2}\zeta +\frac{1}{\sqrt{2\pi}}G_i\left( w \right) e^{-iwt}.
$$ The solution should have the form, $$
\zeta _i=Ae^{iwt}+Be^{-iwt}.
$$ Thus, $$
-w^2\left( Ae^{iwt}+Be^{-iwt} \right) =-w_{i}^{2}\left( Ae^{iwt}+Be^{-iwt} \right) +\frac{1}{\sqrt{2\pi}}G_i\left( w \right) e^{-iwt}
$$ Therefore $$
A=0,\,B=\frac{1}{\sqrt{2\pi}}\frac{G_i\left( w \right)}{\left( w_{i}^{2}-w^2 \right)}
$$ Integrating each channel, we have $$
\zeta _i=\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty}{\frac{G_i\left( w \right)}{\left( w_{i}^{2}-w^2 \right)}}e^{-iwt}dw.
$$ Considering the dissipation, $$
\mathcal{F} =\frac{1}{2}k_i\dot{\zeta}_{i}^{2}.
$$ By the Lagrange’s equation, $$
\frac{d}{dt}\frac{\partial L}{\partial \dot{\zeta}_i}-\frac{\partial L}{\partial \zeta _i}+\frac{\partial \mathcal{F}}{\partial \dot{\zeta}_i}=Q_i,
$$ we have, $$
\ddot{\zeta}_i+k_i\dot{\zeta}_i+w_{i}^{2}\zeta _i=\frac{1}{\sqrt{2\pi}}G_i\left( w \right) e^{-iwt}.
$$ Again let $$
\zeta _i=Ae^{iwt}+Be^{-iwt},
$$ we can derive, $$
\left( w_{i}^{2}-w^2 \right) \left( Ae^{iwt}+Be^{-iwt} \right) +\left( iwk_iAe^{iwt}-iwk_iBe^{-iwt} \right) =\frac{1}{\sqrt{2\pi}}G_i\left( w \right) e^{-iwt}.
$$ Thus $$
A=0,\,B=\frac{1}{\sqrt{2\pi}}\frac{G_i\left( w \right) \left( w_{i}^{2}-w^2-iw\mathcal{F} \right)}{\left( w_{i}^{2}-w^2 \right) ^2+w^2\mathcal{F} ^2}
$$ Therefore, integrating all the channels, $$
\zeta _i=\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty}{\frac{G_i\left( w \right) \left( w_{i}^{2}-w^2-iw\mathcal{F} \right)}{\left( w_{i}^{2}-w^2 \right) ^2+w^2\mathcal{F} ^2}}e^{-iwt}dw
$$