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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 Z 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

1 Answer 1

up vote 8 down vote accepted

My understanding is that this question is being asked in the context of the kinetic theory of classical gases. In that context, here is the argument:

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.

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How do you write Latex here? –  Alyosha Sep 6 '12 at 18:23
    
Use dollar signs $ around your code –  Physics Monkey Sep 6 '12 at 18:31
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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
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@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
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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

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