# Lagrangian of the Stretching Mode Vibration of the Acetylene Molecule

Actually, this is part of a homework question in my classical mechanics class. The question requires me to derive the eigenfrequencies of the acetylene molecule's bending and stretching modes under the harmonic approximation.

For the molecule $$\rm H-C\equiv C-H$$, with masses of the atoms $$m_{\rm H}$$ and $$m_{\rm C}$$, stiffness of the bonds $$k_{CC}$$ and $$k_{HC}$$, and equilibrium lengths of the bonds $$l_{CC}$$ and $$l_{CH}$$. Enumerating the atoms from left to right, we should have the Lagrangian for stretching mode to be $$L = \frac{m_H}{2}(\dot{x_1}^2 + \dot{x_4}^2) + \frac{m_C}{2}(\dot{x_2}^2 + \dot{x_3}^2) - \frac{k_{HC}}{2}[(x_4 - x_3 - l_{HC})^2 + (x_2 - x_1 - l_{HC})^2] - \frac{k_{CC}}{2}[(x_3 - x_2 - l_{CC})^2].$$

Naturally, we require that the total momentum and angular momentum be zero to remove translations and rotations of the molecule as a whole.

And part of the hint for the question provided by my professor is as follows:

Simplify the Lagrangian by exploring symmetry. Think carefully about the good choice of generalized coordinates, recall symmetric/antisymmetric modes.

I know that from the vanishing momentum assumption, we have $$m_H(u_1 + u_4) + m_C(u_2 + u_3) = 0,$$ where $$u_i$$ is the derivation of atom $$i$$ from its equilibrium position. But how can I simplify the Lagrangian from this and the symmetry of the molecule? Any help is greatly appreciated! :)

Center-of-mass of H: $$q_1 = \dfrac{x_1 + x_4}{2}$$.
Center-of-mass of C: $$q_2 = \dfrac{x_2 + x_3}{2}$$.
Correspondingly, since we want four generalized coordinates, so we also need two more, which naturally come from the COMs above: $$q_3 = \dfrac{x_2 - x_3}{2},\quad q_4 = \dfrac{x_1 - x_4}{2}$$