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Transverse vibration of a triatomic molecule ABA which consists of two identical atom A and one another atom B such as $\mathrm{CO}_2$ is depicted in page 24 Sec. 24 Landau Mechanics.

The transverse displacements $y_{1}$, $y_{2}$, $y_{3}$of the atoms are, according to (24.1) and (24.2), related by $m_{A}\left(y_{1}+y_{2}\right)+m_{B}y_{2}=0$, $y_{1}=y_{3}$(a symmetrical bending of the molecule; Fig. 28c). The potential energy of this vibration can be written as $\frac{1}{2}k_{2}l^{2}\delta^{2}$ where $\delta$ is the deviation of the angle ABA from the value $\pi$, given in terms of the displacements by $\delta=\left[\left(y_{1}-y_{2}\right)+\left(y_{3}-y_{2}\right)\right]/l$. Expressing $y_{1}$, $y_{2}$, $y_{3}$ in terms of $\delta$, we obtain the Lagrangian of the transverse motion:

$L =\frac{1}{2}m_{A}\left(\dot{y}_{1}^{2}+\dot{y}_{3}^{2}\right)+\frac{1}{2}m_{B}\dot{y}_{2}^{2}-\frac{1}{2}k_{2}l^{2}\delta^{2}$

where $l$ is the equilibrium length between A-B bonding.

Here, I can derive the kinetic energy terms very easily. However, I cannot understand how the potential energy term can be expressed with $\delta$.

For first A-B bond, the displacement of boding length is given by $\Delta = \sqrt{l^2 + (y_1 - y_2)^2} - l$ therefore, the potential engergy $U_1$ with a Hooke's constant $k_2$ is given by,

$U_{1}=\frac{1}{2}k_{2}\left[\sqrt{l^{2}+\left(y_{1}-y_{2}\right)^{2}}-l\right]^{2}=\frac{1}{2}k_{2}l^{2}\left[\sqrt{1+\left(y_{1}-y_{2}\right)^{2}/l^{2}}-1\right]^{2}\simeq\frac{1}{8}k_{2}\frac{\left(y_{1}-y_{2}\right)^{4}}{l^{2}}$

with very small displacement ($|y_1 - y_2| \ll l$).

Likewise, I can get the second potential energy of B-A bond and I can write the total potential energy:

$U=\frac{1}{8}k_{2}\frac{\left(y_{1}-y_{2}\right)^{4}}{l^{2}}+\frac{1}{8}k_{2}\frac{\left(y_{3}-y_{2}\right)^{4}}{l^{2}}$

I cannot find any connection between this result and $\delta$-expressed one.

Any advice would be helpful.

Thank you.

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