How do the energy eigenvalues of rotational degrees of freedom in statistical mechanics come about?

Question

I want to understand the hierarchy different degrees of freedom of a mechanical system. Specifically, I want to understand which subsystems equibrilate faster and why. This question comes up:

Why is the rotational temperature $T_\text{rot}\propto\frac{\hbar^2}{I}$ invesely proportional to mass?

Shouldn't that energy increse, like in the case of the Maxwell-Boltzmann distribution, where it is proportional to the kinetic energy (translational degrees of freedom)?

The positive energy-powers are in the Planks constants $\hbar$ and essentially comes from the angular momentum (operator), which makes the particle dependend quantity $[I]\propto mass$. Probably, the answer becomes clear, once one has a general picture of the coupling of the system to it's rotational degrees of freedom.

Answer 1

I don't know if this is what OP is asking(v1), but the rotational temperature $T_{\rm rot}$ times Boltzmann's constant $k_B$ denotes a typical energy $k_BT_{\rm rot}$ involved in the difference $\Delta E$ between two neighboring rotational energy levels of, say, a molecule.

This fact follows from the classical formula $\frac{{\bf L}^2}{2I}$ for rotational energy and the fact that angular momentum is quantized in units of $\hbar$.

Thus a molecule with a big moment of inertia $I$ adjusts more easily its rotational state with the surroundings (as compared to a molecule with small moment of inertia).