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! :)


Speaking of symmetry and anti-symmetry, you might want to consider using the center of mass of H and C separately as a "generalized" coordinates:

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: $$ q_3 = \dfrac{x_2 - x_3}{2},\quad q_4 = \dfrac{x_1 - x_4}{2} $$

With this transformation, you can simplify the kinetic terms and harmonic potential terms.


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