Orthogonality of two randomly chosen velocity vectors in the kinetic theory of gas and the relative velocity

Question

I was looking for the average value of the relative velocity of ideal gas molecules in the kinetic theory. The following is from this article.

The vector notation is not mathematically rigorous, just take into account that.
But they say that $vec{v_1} cdot vec{v_2} = 0$ on average, meaning $vec{v_1}$ and $vec{v_2}$ are orthogonal on average. Is there any good physical reasoning for that?

Nihar Karve · Accepted Answer

As the commenters suggest, compute the average value. $theta$ takes on a uniform distribution from $0$ to $2pi$. So by the Law of the Unconscious Statistician, $$mathbb{E}[vec{x}cdotvec{y}] = mathbb{E}[|vec{x}||vec{y}|costheta]  = int_0^{2pi}|vec{x}||vec{y}|costhetafrac{1}{2pi-0} dtheta  = frac{1}{2pi}|vec{x}||vec{y}|int_0^{2pi}costheta dtheta = 0$$
Intuitively, it is because angles show a uniform spread between $0$ and $2pi$, and the cosine in the dot product is symmetric about $pi$.

JEB · Answer

Note that you are working in the center-of-mass of the system. If the gas is say, a ballon, and you find that:
$$ c_{12} equiv langle vec v_1 cdot vec v_2 rangle ne 0 $$
then the system is moving. If $c_{12}$ is small, then it is drifting slowly. Imagine if the ballon is moving much faster than the average speed from Maxwell's distribution (say, it's on the ISS, and your coordinates are fixed in Houston), then:
$$ c_{12} approx (7,{rm km/s})^2 $$
and that is going to bias the $langle v_i^2 rangle$ high, and the dot product expectation value needs to be subtracted off to get the result in the ballon frame.

Orthogonality of two randomly chosen velocity vectors in the kinetic theory of gas and the relative velocity

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