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Analysis of accelerated clock

  1. Apr 12, 2013 #1

    JM

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    A clock that moves in a polygonal path is an accelerated clock, correct? It appears that such a clock is usually analyzed using equations derived for inertial clocks. 1. What is the justification for using inertial equations to analyze an accelerating clock? 2. Has anyone used general relativity to analyze the accelerating clock?
    JM
     
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  3. Apr 12, 2013 #2

    ghwellsjr

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    Yes.

    Einstein stated exactly this scenario in his 1905 paper introducing Special Relativity at the end of section 4. He merely assumed that what is "proved for a polygonal line is also valid for a continuously curved line".

    Why bother?
     
  4. Apr 12, 2013 #3

    Nugatory

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    To expand on this point, because OP may be misunderstanding something about the difference between SR and GR....

    General relativity is the general (well, well of course! why else would we call it "general" relativity) theory that works when there may be gravity and curvature effects. Special relativity works in the case of zero (or at least negligibly small) gravity and curvature; this is an interesting and important special case of the general problem.

    So:
    1) If a problem can be solved with SR, GR is guaranteed to produce the same results, as SR is just GR when gravity and curvature don't matter.
    2) SR works just fine for accelerated motion. You'll sometimes come across pop-sci books and crummy introductory texts that say that you need GR to handle acceleration; that's a common misconception, and it's wrong.

    The clock on a polygonal path is accelerated, but it's still in a flat spacetime with no gravitational effects, so SR works just fine.
     
  5. Apr 12, 2013 #4

    Dale

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    Yes, when necessary.
    Because sometimes you have both motion and gravitational effects in the same experiment. E.g. the famous Hafele-Keating experiments where planes with clocks were flown around the world.
     
  6. Apr 12, 2013 #5

    pervect

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    To move in a perfect polygon requires bursts of infinite acceleration. So it's a bit non-physical - an actual object would probably only move in an approximately polygonal path, the corners would be rounded rather than sharp.

    Certainly, though, an idealized object moving in an idealized polygon (with the required bursts of infinite acceleration) isn't an object in inertial motion, whatever else you want to say about it.
     
  7. Apr 12, 2013 #6
    Not really, the clock could stop when it gets to a vertex and starts in the new direction.

    Would you call it infinite acceleration when an object accelerates from zero velocity?
     
  8. Apr 12, 2013 #7

    pervect

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    I'm not sure I understand the question - or your apparent disagreement.

    An object can accelerate from zero velocity to a non-zero velocity over a period of time, in which case the acceleration is finite.

    For an object that starts out at rest to acquire non-zero velocity in zero time would require infinite acceleration.

    Truly infinite accelerations are not very physical. It is sometimes used as an approximation to a case where an object acquires a non-zero velocity in an unspecified time which is considered to be "short" - for example, impacts.
     
  9. Apr 12, 2013 #8
    So then what is your point?

    The clock comes to a complete halt at each vertex, and continues in a different direction? What is infinite about that?
     
  10. Apr 12, 2013 #9

    A.T.

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    The idea is that it doesn't make a difference, if it is one clock moving in a polygon, or multiple inertial clocks, moving in straight lines and meeting at the polygons corners to synchronize there (pass on the elapsed proper time).
     
  11. Apr 12, 2013 #10

    DrGreg

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    The original post never stated whether this was a polygonal path in space or a polygonal path in spacetime. You're assuming space but other posters have assumed spacetime.
     
  12. Apr 12, 2013 #11

    PAllen

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    Of course a polygonal path in spacetime is either a closed spacelike path or a CTC, neither of which presumably what the OP really meant ...
     
  13. Apr 12, 2013 #12

    DrGreg

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    Indeed. The term "polygonal path", in a spacetime context, usually refers to an "open ended polygon" rather than a true (closed) polygon.
     
  14. Apr 13, 2013 #13

    George Jones

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    Restating pervect's point:

    If the corners of the polygon are "rounded off" by a region of constant acceleration, and if this region is taken to be arbitrarily, then the difference between this situation involving acceleration and the result for the polygon becomes arbitrarily small.
     
  15. Apr 13, 2013 #14

    George Jones

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    :confused:
    The acceleration.

    In more detail, taking a distributional derivative gives a Heaviside step function, and taking a distribution derivative again gives a Dirac Delta function.
     
  16. Apr 13, 2013 #15

    A.T.

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    Passionflower means slowing down smoothly to stop at the corner, then turning, then accelerating smoothly. This way it is possible to move in a spatial polygon with finite acceleration.

    But I don't think this is the OPs issue. I suspect the OP thinks about moving on a polygon with constant speed, and why this can be modeled with inertial clocks, despite the infinite acceleration in this case. The reason is that the durrarion of the acceleration goes towards zero too. So it doesn't make a difference, if it is one clock moving in a polygon, or multiple inertial clocks, moving in straight lines and meeting at the polygons corners to synchronize there (pass on the elapsed proper time).
     
  17. Apr 14, 2013 #16

    Dale

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    That is a good way of stating it. Assuming that is what the OP is after, then it leads directly to this:

    The time on a clock is given by
    [itex]d\tau^2=dt^2-dx^2-dy^2-dz^2[/itex]
    [itex](d\tau/dt)^2=1-(dx/dt)^2-(dy/dt)^2-(dz/dt)^2[/itex]
    [itex]1/\gamma^2=1-v^2[/itex]

    So acceleration doesn't matter.
     
    Last edited: Apr 14, 2013
  18. Apr 14, 2013 #17

    JM

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    Thank you for the comment. I didn't see a proof for a polygonal path. Thats what I'm looking for. Can you point it out to me? How does the light postulate apply to an accelerated frame?
    I prefer to work in the context of the 1905 paper.
    JM
     
  19. Apr 14, 2013 #18

    JM

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    Thanks for your input. In the context of the 1905 paper the step from inertial to accelerated does not appear to be a trivial matter. But I don't recall ever seeing a specific analysis describing the process. If K is inertial and K' is accelerated how is the light postulate applied to relate the two frames?
    JM
     
  20. Apr 14, 2013 #19

    JM

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    Thanks, DaleSpam. Can you give me a reference with the derivation for this?
    JM
     
  21. Apr 14, 2013 #20

    PAllen

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    Einstein never worked with accelerated frames except perhaps in the context of GR. For SR, he worked with accelerated motion described in an inertial frame, which is all you need. Any problem, even what an arbitrary observer sees or measures, can be analyzed in any frame - there is no need or requirement to use a frame in which 'Bob' is at rest in order to analyze what Bob seed.

    Accelerated frames in SR are complicated (especially for non-uniform acceleration), and the light postulate does not apply to them. An accelerated observer can measure the speed of light to be different from what an inertial observer measures.

    Summary:
    - The Lorentz transform and axioms of SR are meant only for inertial frames
    - Accelerated motion, including what an accelerated observer sees or measures can be analyzed in any inertial frame.
    - It is possible to define and use accelerated frames in SR (though Einstein never did so) using mathematical techniques originally developed for GR. Use of these techniques doesn't mean an accelerated frame requires GR (a theory of gravity) .
     
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