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While counting the degrees of freedom of a diatomic molecule, We neglect the rotation about the axis of the molecule stating the reason that it's energy is negligible. I agree with this reasoning, and have no questions about it.

But if we look at the derivation of the equipartition theorem (Here), then we see that any independent energy which is a quadratic function of momentum or position should have an energy of $\frac{1}{2}kT$.

But the above derivation never assumes that the energy should not be neglegible, or anything else of that sort. So according to the derivation, even the rotation about the axis of the molecule should have a energy of $\frac{1}{2}kT$.

So, why don't we count the rotation about the axis as a degree of freedom?

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  • $\begingroup$ @Nick My question is about the contradiction between the derivation of the equipartition theorum and the accepted reasoning of why it is not counted $\endgroup$ Aug 23 '17 at 17:15
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The real reason that you don't count rotation around the molecule's narrow axis is that that degree of freedom is frozen out. The formula for rotational kinetic energy that is relevant here is given by: $$K = \frac{1}{2} \vec{L} \cdot I^{-1} \vec{L},$$ where $I$ is the moment of inertia tensor. If we choose our coordinates to diagonalize $I$ for a molecule we get (ignoring numerical factors fixed by the geometry): $$I \sim m\left[\begin{array}{ccc} a_0^2 & 0 & 0 \\ 0 & a_0^2 & 0 \\ 0 & 0 & r_{\mathrm{nuc}}^2 \end{array}\right],$$ where $a_0$ is the Bhor radius ($\approx 5\times10^{-11}\operatorname{m}$), $r_{\mathrm{nuc}}$ radius of the nucleus ($\approx 1\times 10^{-15}m$), and $m$ is the mass of the atoms (basically the nuclei).

Because angular momentum is quantized in packets of size $\hbar \approx 10^{-34} \operatorname{m}^2\operatorname{kg} \operatorname{s}^{-1}$ it takes about $K \sim \frac{\hbar^2}{m a_0^2} \approx 7\operatorname{meV}$ (that's milli-electron-Volts, $T\sim 90 \operatorname{K}$) to set an $\mathrm{H}_2$ molecule spinning about its long axes, but about $K \sim \frac{\hbar^2}{m r_{\mathrm{nuc}}^2} \approx 7\operatorname{MeV}$ of energy (mega-electron-Volts, $T\sim 8 \times 10^{10}\operatorname{K}$) to set it spinning about its narrow axis. Needless to say, that much thermal energy would quickly dissociate the molecule and ionize the atoms.

Bottom line, it takes about $10^{10}$ times as much energy to get a diatomic molecule spinning along its narrow axis than the long axes, making that degree of freedom very effectively frozen out.

It is correct that the equipartition theorem would, classically, put the lie to the claim that the energy stored in rotations around the narrow axis contain negligible energy - classically it would have its $kT/2$, too. This is one of those times that quantum mechanics is unavoidable for explaining a phenomenon.

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  • $\begingroup$ Thank you very much! this is one of the questions that bothered me for a long time, +1 $\endgroup$
    – Samà
    Aug 23 '17 at 18:49
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You can imagine the diatomic molecule as two punctual atomic nuclei surrounded by an electronic cloud. The whole configuration is entirely symmetric under rotation around the axis. In other words, saying that the molecule rotates around this axis has no real meaning, and should not be counted among its degrees of freedom.

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  • $\begingroup$ why the downvote? $\endgroup$ Aug 23 '17 at 17:39
  • $\begingroup$ Down voted because it's simply wrong. $\endgroup$ Aug 23 '17 at 17:45

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