Distribution of collisions in gas explained in Feynman lectures

Question

I am referring to this part of Feynman lectures "39–5 The ideal gas law" (https://www.feynmanlectures.caltech.edu/I_39.html#Ch39-F2) where he shows that the Ideal gas law also works for diatomic molecules by showing the mean kinetic energy of the center of a mass of the molecule is calculated the same way.

$$ \left\langle \frac{1}{2} M v_{\text{CM}}^2 \right\rangle = \frac{3}{2} k T \\ $$ With the internal velocities $v_A$ and $v_B$ of the molecule it can be shown that $$ \left\langle \frac{1}{2} M v_{\text{CM}}^2 \right\rangle = \frac{3}{2} k T + \frac{m_A m_B \left\langle \textbf{v}_A \cdot \textbf{v}_B \right\rangle}{M}. $$ The dotproduct of the two velocities needs to be zero on average, which implies that there is no preferred direction. Feynman proofs that by \begin{align} \left\langle (\textbf{v}_A-\textbf{v}_B)\cdot \textbf{v}_{CM} \right\rangle = 0. \end{align} $\textbf{v}_A-\textbf{v}_B$ is called the relative velocity. Why is it trivial that the relative velocity does not have a preferred direction, which with some calculations shows that the internal velocities $v_A$ and $v_B$ dont have either.

Answer 1

Imagine that the relative velocity between the two atoms in a diatomic molecule had a preferred direction. Then $v_A-v_B\neq 0$ on average. You would then expect the atoms to be moving at different speeds on average, so after a while one atom would be much farther away from the other atom than where they started. This would not be a diatomic molecule anymore, it would be two atoms drifting apart!

The point is that something must be binding the two atoms together for them to form a diatomic molecule. Their relative velocities have to be constrained so that they "stay together as a pair" on average. Sure, they can move relative to each other at any instant in time, but on average they should not be drifting apart or moving closer together or else they won't comprise a stable molecule.