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Boosting four-acceleration

  1. Oct 26, 2011 #1
    In an instantaneous co-moving inertial frame, the four-acceleration vector reduces to (0,a).

    Why then does applying a Lorentz transformation to the above vector not produce the general form?
     
  2. jcsd
  3. Oct 26, 2011 #2
    That's because the co-moving frame has a changing speed,
    and because a general result cannot be obtained from a special result.
    The simple form (0,a) is valid only for a small lapse of time.

    Using the full expression for the acceleration at any time in the once-co-moving frame,
    would allow you to derive the expression in any other frame.
    The general expression is Lorentz invariant.
     
  4. Oct 26, 2011 #3
    Yes, that makes sense to me. The problem I'm having is one of imagination. I have always pictured the Lorentz transformation as a kind of rotation. So I imagine a (0,a) vector being rotated from one position to another, and a one-to-one correspondence between a vector in the v=0 frame and the set of the same vector in all the other frames.

    I know it works for four-momentum. That is, if one starts with the specific v=0 case [itex] m_0 (c,0) [/itex] and Lorentz transforms it, it turns into the general case
    [tex] \gamma m_0 (c,\bf {v}) [/tex]

    I will give it more thought.
     
  5. Oct 26, 2011 #4

    Are you able to explain that? Most people seem to picture & refer to it as a rotation, but I can't envision it.
     
  6. Oct 26, 2011 #5
    Well, if you graph xi ([itex] i^2 = -1 [/itex]) on the horizontal axis versus ct on the vertical, then it's a rotation, and the angle is the rapidity. I suspect that some find that idea objectionable - it's misleading in some ways - but I still like it.

    Correction: the angle is rapidity times i. Pretty weird, yes.
     
    Last edited: Oct 27, 2011
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