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! :)


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.