# Why is the $\langle v_{x}^{2} \rangle=\frac{1}{3} \langle v^2 \rangle$?

For a randomly moving particle. Or, I suppose that 1/3 could generalise to 1/n, where n is the non rotational degrees of freedom for that particle.

Related reference Kinetic Theory of Gasses.

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 I don't believe that this is the case, I can certainly set up states whose expectation value of velocity in a certain direction is zero: $e^{i p x}$ has $\langle v_y \rangle \sim \langle \partial_y \rangle = 0$. – DJBunk Sep 6 '12 at 16:41 He may have been using it in a context so that I didn't understand that it was a special case, but Feynman in Chapter 39 of Vol 1 disagrees with you. – Alyosha Sep 6 '12 at 18:04 @DJBunk I think this question concerns a statistical average in a large population of particles, not the probabilistic average of quantum mechanics. – David Zaslavsky♦ Sep 6 '12 at 18:15 Yes, that's correct. – Alyosha Sep 6 '12 at 18:16 +1 kinetic theory of gases blew my mind 20 years ago, and you brought it all back with this question. – ja72 Sep 6 '12 at 18:23

If the system is rotationally invariant, then we should have $\langle v_x^2 \rangle = \langle v_y^2 \rangle = \langle v_z^2 \rangle$. Thus $\langle v^2 \rangle = \langle (v_x^2 + v_y^2 + v_z^2 )\rangle$ which gives $\langle v^2 \rangle = 3 \langle v_x^2 \rangle$. Your comment about generalization to n dimensions is also correct.
Use dollar signs $around your code – Physics Monkey Sep 6 '12 at 18:31 It's not true. The averages of two quantities can be equal without the quantities being equal. In your example above, clearly the x velocity and y velocity are two different physical quantities. It just happens that their squares have the same average value when you look at many particles. – Physics Monkey Sep 6 '12 at 18:40 @Alyosha: the expectation value is linear, so in general you have$\langle a + b \rangle = \langle a \rangle + \langle b \rangle$. – Fabian Sep 6 '12 at 18:49 Generally$\langle A + B \rangle = \langle A \rangle + \langle B \rangle\$, this is more or less build into the setup. After all expectation values are computed by evaluating certain sums or integrals. – orbifold Sep 6 '12 at 18:53