# Rotational degrees of freedom I don't get it.

## Main Question or Discussion Point

"Rotational degrees of freedom"...I don't get it.

How can you determine how many rotational degrees of freedom an object or system has? For example, how many rotational degrees of freedom does a diatomic molecule have, and why? I've never been good at figuring this out.

## Answers and Replies

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Molecule:

in standard kinetic gas theory
1 atom in molecule 0 degrees
2 atoms in molecule 2 degrees
3 - infinity atoms in molecule 3 degrees

The only exception is many atoms on a line those also have only 2 degrees. I don't know of a molecule with 1 degree of rotational freedom.

The idea is, that you cannot rotate something with zero diameter.

The idea is, that you cannot rotate something with zero diameter.
Not sure what you mean by zero diameter, I'd put it differently:

Imagine an object that is perfectly symmetrical around some axis. Then, classically, it will be impossible for you to make it spin (nor stop it) around that axis, because there is no microscopic handle for friction to push against. You can't even tell whether it is already spinning. So whatever ang. momentum it might have about that axis is irrelevant to the problem (or absorbed into a mass term), since the other degrees of freedom can never exchange any energy with this rotational mode.

Molecule:

in standard kinetic gas theory
1 atom in molecule 0 degrees
2 atoms in molecule 2 degrees
3 - infinity atoms in molecule 3 degrees
Thanks, but can you explain why that is? Can't you rotate a diatomic molecule about all three coordinate axes?

Suppose we have our nuclei at (0,1) and (0,-1) (i.e., we position them along the y-axis). Rotating the molecule about the y-axis doesn't really change it at all if we idealize the nuclei as uniform spheres. Is that why we don't count that as a legitimate rotation?

I think you have to use quantum mechanics to get a rigorous reason for not getting a contribution for rotations about the axis of a diatomic molecule. Classically, even a diatomic molecule with a very small (non-zero) moment of inertia about its axis would contribute the same amount to the hamiltonian from that axis as from the other two. My stat. mech. textbook went into detail with this, but it's buried away and I can't find an alternate explanation on the web.

My interpretation always was: the kinetic theory of gases doesn't say that a diatomic molecule can't rotate around its long axis. It's just that, if it does, the mass will be very close to the axis of rotation, so the rotational KE would be negligibly small. Rotating around the other two axes, the mass is far from the axis, and rot. KE becomes significant. So, total of three translational degrees of freedom plus two rotational degrees of freedom. If there's some small error there, the model already was only an approximation anyway, as soon the initial assumptions said that temperature isn't too low, and pressure or density isn't too high, so that we can neglect all intermolecular forces.

Thanks, but can you explain why that is? Can't you rotate a diatomic molecule about all three coordinate axes?

Suppose we have our nuclei at (0,1) and (0,-1) (i.e., we position them along the y-axis). Rotating the molecule about the y-axis doesn't really change it at all if we idealize the nuclei as uniform spheres. Is that why we don't count that as a legitimate rotation?
In Quantum Mechanics, it is not possible for a rotational mode to have any amount of energy, it must be in one of the energy levels of the system. If you simply model a diatomic molecule as a rigid rotor, the rotational energy levels are proportional to 1/I where I is the moment of inertia. About the y axis the moment of inertia is very small hence energy of the excited states is large and generally not readily excited at low temperatures.

My interpretation always was: the kinetic theory of gases doesn't say that a diatomic molecule can't rotate around its long axis. It's just that, if it does, the mass will be very close to the axis of rotation, so the rotational KE would be negligibly small. Rotating around the other two axes, the mass is far from the axis, and rot. KE becomes significant. So, total of three translational degrees of freedom plus two rotational degrees of freedom. If there's some small error there, the model already was only an approximation anyway, as soon the initial assumptions said that temperature isn't too low, and pressure or density isn't too high, so that we can neglect all intermolecular forces.
I agree with this.Any component of rotation about the long axis can be considered as negligible since the moment of inertia about this axis can be considered as negligible.

I agree with this.Any component of rotation about the long axis can be considered as negligible since the moment of inertia about this axis can be considered as negligible.
This argument is misleading and very incomplete, because if you reduce the moment of inertia classically this degree of freedom would not be negligible, and would even dominate for small changes in energy. Classically the dimensional argument is the best and quantum mechanically it goes like this. Energy for the quantum mechanical rotor:

$$E_l = \frac{\hbar^2 }{ 2I} l \left(l+1\right)$$

Where E is the energy of the states with quantum number l and I is the moment of inertia. So the Energy you can save in this degree of freedom is much higher than in the other degrees of freedom. But this state is "frozen" because the gas doesn't have enough energy to get even to the first energetic level.

This argument is misleading and very incomplete, because if you reduce the moment of inertia classically this degree of freedom would not be negligible, and would even dominate for small changes in energy. Classically the dimensional argument is the best and quantum mechanically it goes like this. Energy for the quantum mechanical rotor:

$$E_l = \frac{\hbar^2 }{ 2I} l \left(l+1\right)$$

Where E is the energy of the states with quantum number l and I is the moment of inertia. So the Energy you can save in this degree of freedom is much higher than in the other degrees of freedom. But this state is "frozen" because the gas doesn't have enough energy to get even to the first energetic level.
So let me explain it in greater detail.A diatomic molecule is two atoms joined by a bond,it can be imagined as having a dumbbell shape.Its rotational motion can be resolved into three perpendicular axes so therefore we might expect it to have three rotational degrees of freedom.The moment of inertia about the bond axis however is small enough to ignore due to the small size of the atoms.If this were not the case then monoatomic molecules would have three rotational degrees of freedom.Your quantum theory explanation may be more elegant but this one, also, works pretty well.

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The problem I have with the small size argument is the virial theorem, unless you completely forbid this degree of freedom, classically an equal part of energy will go into this degree of freedom as in all the others. No matter that it has to spin incredibly fast. So IMHO classically you have to say that it has zero width, otherwise it could and would spin.

The problem I have with the small size argument is the virial theorem, unless you completely forbid this degree of freedom, classically an equal part of energy will go into this degree of freedom as in all the others. No matter that it has to spin incredibly fast. So IMHO classically you have to say that it has zero width, otherwise it could and would spin.
I'm not familiar with the virial theorem but I agree that it does seem odd to assume a zero width.I just content myself with the fact that it is a simplifying assumption only and that any gas equations we derive are approximately correct and break down under those conditions that the assumptions break down

A note just in case this might help the discussion. Every textbook I have seen introduce the the kinetic theory of gases begins with a list of assumptions. One of those assumptions is always that quantum mechanical effects are being neglected and Newtonian behavior is assumed. But to what ends this beginning would lead, or whether the assumptions have been changed for newer texts, I'm not the best qualified person to comment.