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A quantum particle at rest viewed from a merry-go-round.

  1. Jul 27, 2012 #1
    Feynman's Lectures on Physics has an interesting graphic in volume 3, page 7_3, "Fig. 7-1. Relativistic transformation of the amplitude of a particle at rest in the x-t systems.", see scan below. Say ψ is the wavefunction of a particle at rest in 3D space, ψ = exp[-iEt], hbar = 1.

    If I now move with some velocity v the particle at rest now has some momentum so ψ --> ψ' = exp[-i(-mv.r' - Et')]?

    If on the other hand, instead of going some velocity v, I instead rotate say on a merry-go-round how will ψ transform? Will I say the particle now has angular momentum?

    Related question? What does a large field of simultaneous clocks look like from the reference frame of someone riding a merry-go round?

    From the three points of view,at rest, moving, and rotating, what do the hyperplanes ψ = constant look like? Feynman's graphic gives clue for the first two? Does rotation cause the hyperplanes to curve or are they "flat"?

    Does this problem have anything to do with the "magic" of boosts and rotations forming a group? Boosting around some point results in some rotation?

    Thanks for any hints or help!
     

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  3. Jul 28, 2012 #2

    Simon Bridge

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    You mean, if the observer is in a non-inertial reference frame?

    Hint: What normally happens in this case?
     
  4. Jul 28, 2012 #3
    If I'm on a spinning merry-go-round with no roof and it's raining vertically I would say from my frame of reference the rain has angular momentum? The further from the center the greater the momentum? Relativity adds an additional "twist" because of Thomas Precession?
     
  5. Jul 31, 2012 #4

    Bill_K

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    I give up -- what?
     
  6. Aug 1, 2012 #5

    Simon Bridge

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    Bill_K the Science Advisor ... is trying to trip me up :)
    It's a pedagogical gambit Bill - admittedly I've been neglecting this thread... but since you are here, perhaps you'd like to try answering the question?
     
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