Why do we say linear molecules only have 2 rotational degrees of freedom? Why does the third 'frozen' one not count?

Question

It is possible to excite rotations around the axes perpendicular to the bond of a linear molecule. However, rotation around the axis along the bond of the molecule would require huge energies, due to the much smaller moment of inertia which is inversely proportional to the energy. This is explained in more detail here: In counting degrees of freedom of a linear molecule, why is rotation about the axis not counted? 1

Therefore, out of three possible rotational degrees of freedom, only two are applicable to our linear molecule. The third rotation can't be accessed and is 'frozen out'. Thus a linear molecule is said to have two rotational degrees of freedom, as this third parameter would give us no extra information about the system.

• Non-linear molecules have 3N degrees of freedom in total. We know 3 are translational, 3 are rotational (all are allowed for non-linear molecules) so the remaining 3N-6 are vibrational.

• Linear molecules have 3 translational and only 2 rotational, and to keep a total of 3N degrees of freedom, they now need 3N-5 vibrational.

Why does the 'frozen' out rotational degree of freedom for linear molecules not count? As stated above, knowledge of it gives no extra information about the system, but at what point does a degree of freedom count as 'frozen out' and need to be compensated for by other (vibrational) degrees of freedom? And is there a deeper, physical argument beyond "everything needs to add up to 3N"? Why 3N, in this case?

In the general case, any molecule can be cooled down so that various degrees of freedom are gradually frozen out. Are these degrees of freedom replaced by others, and how does the system then know how to vibrate, say, in a different way?

I'm aware that I'm missing something here, possibly a fundamental flaw in how I currently understand the concept of degrees of freedom, and I greatly appreciate any help.

Answer 1

The number of degrees of freedom isn't $3N$, it's $3N$ minus the number of contraints.

The $N=2$ is a special case since with two particles molecules can't help being linear. However for $N>2$ if the molecule is linear you are constraining the motion of the third particle because it can only move along the line joining the other two. Likewise for higher values of $N$. This reduces the total number number of degrees of freedom, so the transaltional, rotational and vibrational degrees of freedom do not have to add up to $3N$.

Answer 2

Let us consider one of the rotational degrees of freedom of a molecule. Let us assume that the moment of inertia for the relevant principal axis is $I$. Then we can get the spacing of energy levels for this degree of freedom from the quantization rule $I\omega=n\hbar$ (I omit some constant factors and neglect some other things here and everywhere else), so for the level that is closest to the ground level $I\omega=\hbar$. The statistical probability of the system to be at a level with energy $E$ is proportional to $\exp(-E/\theta)$, where $\theta$ is temperature. For a rotational level, $E=I\omega^2=I(\hbar/I)^2=\hbar^2/I$. Unless the energy is comparable with or less than the temperature, the probability is close to zero, so the temperature where the probability is not negligible is of the order of $\hbar^2/I$. For the axis of a linear molecule, the moment of inertia is very small, so the temperature at which energy levels above the ground level are important is extremely high (e.g., it can be higher than the temperature of dissociation of the molecule). Therefore, at temperatures of practical interest, rotation around the axis of the linear molecule is not important for thermodynamic properties.

Answer 3

The OP is mixing up some terminology. When we talk of frozen modes, we don't usually mean the third rotational mode. We mean the first two modes, the classical modes that don't contribute to the specific heat of H2 at low temperature. It's a quantum mechanical effect. The reason there is no third rotational mode is strictly classical, and it's because when we analyze the molecule as point masses connected by a rod, there really is no third rotational mode. It's just not there. No one says it's "frozen".