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.

  • $\begingroup$ 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 $\langle v_y \rangle \sim \langle \partial_y \rangle = 0$. $\endgroup$ – DJBunk Sep 6 '12 at 16:41
  • $\begingroup$ 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. $\endgroup$ – Meow Sep 6 '12 at 18:04
  • $\begingroup$ @DJBunk I think this question concerns a statistical average in a large population of particles, not the probabilistic average of quantum mechanics. $\endgroup$ – David Z Sep 6 '12 at 18:15
  • $\begingroup$ Yes, that's correct. $\endgroup$ – Meow Sep 6 '12 at 18:16
  • $\begingroup$ +1 kinetic theory of gases blew my mind 20 years ago, and you brought it all back with this question. $\endgroup$ – ja72 Sep 6 '12 at 18:23

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.

  • $\begingroup$ How do you write Latex here? $\endgroup$ – Meow Sep 6 '12 at 18:23
  • $\begingroup$ Use dollar signs $ around your code $\endgroup$ – Physics Monkey Sep 6 '12 at 18:31
  • 3
    $\begingroup$ 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. $\endgroup$ – Physics Monkey Sep 6 '12 at 18:40
  • 4
    $\begingroup$ @Alyosha: the expectation value is linear, so in general you have $\langle a + b \rangle = \langle a \rangle + \langle b \rangle$. $\endgroup$ – Fabian Sep 6 '12 at 18:49
  • 1
    $\begingroup$ 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. $\endgroup$ – orbifold Sep 6 '12 at 18:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.