# Transverse vibration of triatomic molecule from Landau Mechanics Sec. 24

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.