# Why does symmetry reduce heat capacity?

For monatomic gasses, the heat capacity approaches $$3R/2$$ because there are three degrees of freedom (DOF) of translational motion. For diatomic gasses it approaches $$7R/2$$ because the symmetry about its axis cancels out one rotational DOF, but it also has two vibrational modes. And for complex molecules you get up the $$(3 + DOF_{Rotational} + 2DOF_{Vibrational})R/2$$

What I don't understand is why rotational symmetry means heat cannot be "stored" in that degree of freedom. Just because a disc is symmetrical doesn't mean we can't use it as a flywheel. Just because a football is symmetrical doesn't mean that its rotational energy won't get used up as its path through the air is bent.

My first thought was maybe the answer is "Because quantum", but that's ridiculous because one of the first operators you learn about is the orbital angular momentum operator and you apply that to atoms, so you can definitely have angular momentum being stored in an atom. It also reminded me that a lot of atoms aren't even spherically symmetrical because the electron shells extend out in bulbs and things.

Why can't molecules store heat in their axes of rotational symmetry?

• I noticed this degrees of freedom of a diatomic molecule question has been asked before. However, I find this new answer by Sean E. Lake better than the answer to the question of the link. Aug 14, 2022 at 15:56
• Incidentally: the actual vibrational motion is a single degree of freedom, yet vibrational motion of a diatomic molecule contributes in approximation an amount of $2R/2$ to the heat capacity. Where does the doubling come from? For simplicity assume the vibration is harmonic oscillation. As we know: in the case of harmonic oscillation the time average of potential energy is the same amount as the time average of kinetic energy. See also: on physicsforums: degrees of freedom of a diatomic molecule Aug 14, 2022 at 18:07
• Yeah I glossed over the vibration point because I understood that one. It is interesting though. Aug 14, 2022 at 19:18

Your understanding is incorrect. The symmetry about the axis for diatomic molecules doesn't cancel that rotational degree of freedom. The system has translation symmetries, and if symmetry canceled degrees of freedoms out, you'd have zero heat capacity.

To understand what's going on, you have to look at the energy written in terms of the quantum mechanically relevant variables: $$E = \frac{P^2}{2M} + \frac{1}{2}\vec{L} \cdot I^{-1} \vec{L},$$ where $$P$$ is the momentum of the center of mass, $$M$$ is the total mass, $$\vec{L}$$ is the angular momentum about the center of mass, and $$I$$ is the moment of inertia tensor. For a diatomic molecule the moment of inertia for two of the rotations is quite large - about $$M\ell^2 / 8$$, where $$M$$ is the total mass of the molecule and $$\ell$$ is the total distance from nucleus to nucleus. The third moment of inertia is quite small, though, of order $$M$$ times the size of the nucleus squared. Because you divide by the moment of inertia tensor, that means that the energy needed to excite rotations along that axis is very high.

For comparison, rotations around the long axes take of order $$1\,\mathrm{meV}$$ of energy. Rotations around the small axis take $$10\,\mathrm{MeV}$$ (i.e., 10 billion times as much energy)! Every atom I'm familiar with would fully ionize long before this degree of freedom unfroze.

To calculate this I assumed the nucleus has a mass of one proton, the molecule is about 1 ångström, and the nucleus is about 1 femtometer. You then just have to assume that the minimum non-zero angular momentum is $$\hbar$$, and you can calculate the answer yourself.

• Thank you so much. Whenever I asked my lecturer I just got trite responses about how "the molecule doesn't look different when you turn it that way" and I couldn't make any sense of that logic. This makes far more sense. Aug 14, 2022 at 13:31
• It might be worth adding that there is some "because quantum" in this. In classical thermodynamics a degree of freedom will contribute to the heat capacity no matter how much energy it takes to activate, and no matter what temperature. Only quantum thermodynamics explains why some degrees of freedom don't "turn on" at lower temperatures as you have explained. Aug 14, 2022 at 19:04
• Yes this is a good point. Its just that in my course it wasn't made clear that the rotational modes about the "thin" axis were just stupendously high energy compared to vibrational ones. It was presented as though they didn't "count" or maybe even didn't exist because the molecule wouldn't "look different" when rotated about that axis. Which was an incorrect explanation. Aug 14, 2022 at 19:21
• @Disgusting That explanation is not completely incorrect, it's just an approximation. If you clasically assume nuclei as point charges or spherically symmetric charged particles and an averaged-out electron cloud, there's no way any external electric field could exert a torque around the axis of symmetry. Only when you consider electrons as orbiting around the nuclei, you can start thinking about exciting them by an oscillating field resonant with their "orbital period", which is going to require similarly high frequencies/energies. Aug 15, 2022 at 11:34
• @TooTea That's only true for static electric fields. The moment you get dynamic fields, like in light, the field has a curl. A curling electric field will classically exert a torque on any electric charges, regardless of their symmetry. For the ideal plain wave the torque will average out over a cycle, but it's not hard to imagine a wave packet and moving atom could combine to give a net torque. Even for a point-like particle, classically the torque is zero, but so is the moment of inertia. No, the only way this works is this quantum mechanics. Aug 16, 2022 at 1:10