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Is c invariant in Accelerating frames?

  1. Aug 4, 2009 #1
    Simple question.
    I would like to know if there is a definitive answer , consensus in the field, on the question of the measurement of light in an accelerating system.
    Whether one way measurements from the front to the back and vice versa would result in (c +v) = (c-v) = c as usual??

    Thanks,,, any answers appreciated.
     
  2. jcsd
  3. Aug 4, 2009 #2

    Ich

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    No .
     
  4. Aug 4, 2009 #3

    Dale

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    I agree with Ich as long as you are talking about the coordinate speed of light. If you make a local experimental measurement of the speed of light you will always obtain c, even in an accelerating frame.
     
  5. Aug 5, 2009 #4
    Hi DaleSpam Im confused
    In this context ,,what do you mean by the coordinate speed? I was asking about measuring the speed in a single coordinate system. Frame. Ship.

    From the front to the back [one way] and from the back to the front [one way]

    Do you mean that these measurements would not be isotropic and return c but that very local measurements taken at either the front or back would ,,or what??



    Thanks for your reply
     
  6. Aug 5, 2009 #5

    Dale

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    Most coordinate systems have three space coordinates and one time coordinate. The coordinate velocity is then the derivative of the three spatial coordinates wrt the time coordinate. This value is dependent on the coordinate system used, particularly the synchronization convention defined by the time coordinate.

    Contrast this with the four-velocity which is the unit tangent to the worldline and is thus defined in a coordinate-independent geometric fashion.
    AFAIK a one-way measurement of the speed of light is always coordinate dependent since it relies on the synchronization convention chosen.
    A two way measurement (eg a mirror at a known distance) will always return c as long as the curvature of spacetime is negligible over the measurement.
     
  7. Aug 5, 2009 #6

    Dale

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    Most coordinate systems have three space coordinates and one time coordinate. The coordinate velocity is then the derivative of the three spatial coordinates wrt the time coordinate. This value is dependent on the coordinate system used, particularly the synchronization convention defined by the time coordinate.

    Contrast this with the four-velocity which is the unit tangent to the worldline and is thus defined in a coordinate-independent geometric fashion.
    AFAIK a one-way measurement of the speed of light is always coordinate dependent since it relies on the synchronization convention chosen.
    A two way measurement (eg a mirror at a known distance) will always return c as long as the curvature of spacetime is negligible over the measurement.
     
  8. Aug 6, 2009 #7

    Ich

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    Not for accelerating observers, even in flat spacetime.
     
  9. Aug 6, 2009 #8

    sylas

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    In an accelerating space ship, any local measurement taken over a sufficiently small region will figure the speed of light as c.

    However, the distance from the front of the ship to the back of the ship is ambiguous; it can be defined in different ways. In brief; remote distances are a co-ordinate issue, and hence so too are speeds at a remote location within the ship.

    Here are two ways to define distances in the ship. One is to place a clock at some fixed location, and then put mirrors everywhere else in the ship, and see how long it takes light to get there and back, by your clock. Using this distance co-ordinate, the speed of light is everywhere constant. (By definition; you use this as an assumption to derive the distance!)

    The other way is to move your clock all over the ship, and use it as a "light ruler" at every point, and mark off distances that way. This will give you different distances from the "radar" method. It will also mean that the speed of light varies in different parts of the ship, from the perspective of an observer with a clock seeing how long it takes light to get from one point to another remote point.

    If you are the front of the ship, clocks at the back of the ship are running slow, and distances are also compressed. The speed of light is reduced, because by YOUR clock, it takes longer for light to get to a point at the rear and then back to your clock than distance/c.

    Conversely, if you are at the rear then the speed of light at the front of the ship is more than c, in this co-ordinate system.

    Cheers -- sylas
     
    Last edited: Aug 6, 2009
  10. Aug 6, 2009 #9

    Dale

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    Yes, even for accelerating observers. Of course, the clock and the mirror need to move in a Born-rigid fashion.
     
  11. Aug 6, 2009 #10

    atyy

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    For free falling guys, "locality" is determined by negligible curvature.

    Accelerating guys can also set up locally Minkowski coordinates, but for them "local" is determined by negligible curvature and their acceleration.

    Curvature produces second order deviations, acceleration produces first order deviations from Minkowski coordinates.

    http://www.csun.edu/~vcmth00m/fermi.pdf
     
  12. Aug 7, 2009 #11

    Ich

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    I think sylas explained why this would not work: You have to pick a definition of distance. The most sensible would be IMHO the "light ruler at every position", but it does not matter which one you choose. As long as we agree that forward distance is the same as backward distance, the measured two way speed of light will be different in both directions.
     
  13. Aug 7, 2009 #12

    Dale

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    I am not sure what you mean by this, especially how "different in both directions" applies to a two-way measurement of the speed of light.
     
  14. Aug 7, 2009 #13

    Ich

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    If you pick one specific line (from point A to point B), and assign to it one and only one length, by whatever procedure you like,
    then you will determine different two way light speeds: A-B-A differs from B-A-B.
    Both trips take the same amount of coordinate time, but in the first experiment you measure A's proper time (I think that's how a two-way measurement is defined), in the second B's. Both are different due to time dilation.
     
    Last edited: Aug 7, 2009
  15. Aug 7, 2009 #14

    Dale

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    The two measurements may not take the same amount of coordinate time, but now I have doubts about my assertion. I will have to work out the math for a couple of different coordinate systems.
     
  16. Aug 7, 2009 #15

    Ich

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    In static coordinates they have to; you can simply rearrange
    A-B-A -> A-B + B-A
    B-A-B -> B-A + A-B = A-B + B-A,
    because time translation changes nothing.
     
  17. Aug 7, 2009 #16

    Dale

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    That is not true in general for non-inertial coordinate systems.
     
  18. Aug 7, 2009 #17

    Ich

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    That's why I said "In static coordinates...", meaning A and B being at constant coordinate positions in a time-independent metric like Rindler or Schwarzschild coordinates.
     
  19. Aug 7, 2009 #18

    Dale

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    Sorry, I missed that. Anyway, I will do the math and see. I am not as confident as I initially was.
     
  20. Aug 8, 2009 #19
    Hi Sylas

    I am not sure how to interpret this. If there is a ruler running the length of the ship it would seem that the observed measurements must be the same at both ends. So wherein is the possibiity of ambiguity? Clocks at disparate locations can disagree on measurements but it is hard to picture what it might mean wrt to rulers and distance.
    Are you suggesting that even though the observed measurements might be the same at both ends ,somehow the "actual" distance might be different? Or are you assuming that the front and the back are two different frames with different metrics with the ship as a whole being some hybrid interpolation of the two ???

    Then I would ask , what is the real point of this procedure. It is only relevant to a single clock at a single location and, as you pointed out,,, "by definition" it can only return c under any conditions and therefore not only is that information without real meaning, so are any distance measurements derived from the process.

    Excuse me but isnt there an inherant problem with this method of moving a clock to determine distance or degree of desynchronization?
    Doesn't the basic foundation of SR include the recognition that the act of moving a clock must result in time dilation in the clock and consequently loss of synchronization.
    I had always assumed that the light method of synchronization for clocks that are spatially separated was, in part, a responce to this inherant problem.???

    I understand completely this description and concept. But isn't it simply a projection onto the accelerating frame of assumptions derived from calculations made in an inertial frame.
    Is it possible to arrive at these conclusions based on our knowledge of physics ,completely within the reference frame of the accelerating system on first priciples alone???
    Thanks
     
  21. Aug 8, 2009 #20

    Hurkyl

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    The ambiguity is that this solution is by no means the only reasonable one.

    What your ship-long ruler does for you is that it defines a system of spatial coordinates. You can then define lengths inside of your one-dimensional rocket in terms of some function defined upon those coordinates.

    But this notion of length isn't particularly closely related to the notion of distance defined by the geometry of space-time -- and a priori, we have no good reason to think it consistent with any other reasonable scheme for measuring distances.
     
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