# In counting degrees of freedom of a linear molecule, why is rotation about the axis not counted?

I was reading about the equipartition theorem and I got the following quotations from my books:

A diatomic molecule like oxygen can rotate about two different axes. But rotation about the axis down the length of the molecule doesn't count. - Daniel V. Schröder's Thermal Physics.

A diatomic molecule can rotate like a top only about axes perpendicular to the line connecting the atoms but not about that line itself. - Resnick, Halliday, Walker s' Fundamentals of Physics.

Why is it so? Doesn't the rotation take place that way?

The energy levels of a diatomic molecule are $E = 2B, 6B, 12B$ and so on, where $B$ is:

$$B = \frac{\hbar^2}{2I}$$

Most of the mass of the molecule is in the nuclei, so when calculating the moment of inertia $I$ we can ignore the electrons and just use the nuclei. But the size of the nuclei is around $10^{-5}$ times smaller than the bond length. This means the moment of inertia around an axis along the bond is going to be about $10^{10}$ smaller than the moment of inertia around an axis normal to the bond. Therefore the energy level spacings will be around $10^{10}$ times bigger along the bond than normal to it.

In principle we can still excite rotations about the axis along the bond, but you'd need huge energies to do it.

• I'm not sure if it's ok to be asking a further question on a post from 3 years ago, but better than asking a duplicate I guess.... My first question: From your post, I see that the moment of intertia is small along the axis, so indeed the magentic dipole is very large, which would mean a large energy stored if the dipole is in a magnetic field. But why would we assume that the molecule is in a magnetic field in the first place? If it is not, then no energy would be stored in these rotational modes anyway.
– Meep
Apr 17 '18 at 16:29
• My second question: why are we considering the energy stored in rotational modes in terms of magnetism at all? If we consider the molecule simply as two neutral balls on a light stick, then energy is stored in its rotation. It's not clear to my why we do not talk about this energy.
– Meep
Apr 17 '18 at 16:33
• Follow up to your comment (which is related to my second question), ten the energy of the rotational modes has nothing to do with the magenic moment. In fact, if I consider classical rigid body dynamics, the length of the inertia ellipsoid on a w axis is proportional to $sqrt{\frac{1}{I}}$ so a small moment of inertia means that the moleule can have a very rapid rotation about that axis for a given kinetic energy...
– Meep
Apr 17 '18 at 16:36
• I thought the energies were refering to dipole moments (I must have gotten confused with the B)
– Meep
Apr 17 '18 at 16:38
• I see now that I was mistaken with regards to the magnetic field, however this does not explain the discrepancy with the inertia ellipsoid method. Has this to do with this being a classical approach? But still, but the correspondence principle the smallest moment of inertia should correspond to highest/lowest energies in both the classical and quantum cases...
– Meep
Apr 17 '18 at 16:42

Just an addition to John Rennie's answer. The equipartition theorem can only be derived in classical statistical physics. In quantum statistics it is not correct. For each degree of freedom there is a characteristic temperature below which the quantum effects are significant. This temperature is very high for rotation around the axis of the molecule; I guess it is much higher than the temperature at which the molecule dissociates, so this degree of freedom is "frozen" as long as the molecule exists and can be ignored.