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Random walks and conservation laws in average

  1. May 28, 2009 #1
    Perpetual motion idea: Sit on a bench with a collimated laser, single slit and detector screen.

    This prepares particles with identical and symmetric spatial wave-functions, so each measurement transfers an independent, random quantity of transverse momentum (from a symmetric distribution) to the screen. So the total amount of transverse momentum transferred to the screen increases in proportion to the http://en.wikipedia.org/wiki/Random_walk" [Broken] of the total number of particles measured so far.

    Hence, the bench begins to move transversely. Also, if the photons do not lose or gain energy as they move from the laser to the screen, despite adopting transverse velocity when the collimated beam undergoes diffraction, then some of their longitudinal momentum has disappeared, so the bench also accelerates longitudinally. Accelerations without any external force and the energy accounting has broken.

    Or not, but how not? I'm arguing that it is insufficient for QM to only "on average" obey conservation laws, and I think the resolution has to do with entanglement prior to the measurement. Any ideas?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 29, 2009 #2

    DrChinese

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    Ok, so if you have a laser and you are imagining that there is recoil from the light being emitted. Of course we know that requires energy in, so there isn't much opportunity for perpertual motion from that.

    Second, you are saying that the traverse momentum imparted to the screen is at the expense of momentum in the direction of the laser beam (longitudinal direction). And that leads to there being more longitudinal recoil than there is longitudinal momentum transfer.

    But I would say that the "recoil" would match the transferred momentum. I guess that if the position of the absorption on the screen is known precisely, there would be a large uncertainty in the momentum itself. So it would seem that the accounting would work out fine.
     
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