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Speed of light paradox

  1. Oct 21, 2012 #1

    bgq

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    Hi, there is something I can't understand:

    Consider a stationary observer at A. Now consider an observer B in a train that moves with constant velocity v with respect to A. In the train, B tries to measure the speed of light using an empty tube of length L0 (proper length as measured by B). He sends a light signal at extremity E, the signal reaches extremity F (where a mirror exists) and return back to the extremity E. B measure this duration T0 (proper period as measured by B).



    Now B measure the speed of light as:

    CB = 2L0/T0

    Now according to A the length of the tube is contracted and the time is dilated, so he measure the speed of light as:

    CA = 2L/T = (2L0/γ)/(γT0) = (L0/T0)/(γ^2) = CB/(γ^2)

    which is different from the value measure by B (divided by Gamma squared)

    What did I miss in this analysis?

    thanks.
     

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  3. Oct 21, 2012 #2

    jtbell

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    Staff: Mentor

    Probably relativity of simultaneity.
     
  4. Oct 21, 2012 #3

    bgq

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    I didn't use anything related to simultaneity. The emitting and reception of light events are not simultaneous to both observers.
     
  5. Oct 21, 2012 #4

    Doc Al

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    From A's frame the tube is moving. The light going from E to F does not merely travel a distance L0/γ since F moves while the light is in transit.
     
  6. Oct 21, 2012 #5

    bgq

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    According to the theory of special relativity, the propagation of the light is not affected by the motion of the tube.
     
  7. Oct 21, 2012 #6

    Doc Al

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    Right! So?

    Each frame sees the light travel at speed c with respect to their own frame.
     
  8. Oct 21, 2012 #7
    The speed of light is the same (side note, there are things that are changed about the observation of light, but they have nothing to do with this question), but the position of the tube's extremities are changing with time. You have taken into account the contraction of the tube, but not the fact that the "stationary" observer also sees it moving. The normal derivation of SR uses light moving perpendicular to the motion of the reference frame for this reason.

    Essentially what you missed is that the light is moving at c for the "stationary" observer, but distance the light has to travel on the way back will be shorter than the way to the mirror since the "extremity E" has moved toward "extremity F" as time progresses.
     
  9. Oct 21, 2012 #8

    bgq

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    The light moves towards the extremity E with a speed c (not c+v), so the time will be the same (dilated of course). The issue here is that light does not behave in the same classical way. The motion of the tube has nothing to do with the propagation of the light (According to the postulates of SR).
     
  10. Oct 21, 2012 #9

    Doc Al

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    Yes, both frames will see the light travel at speed c with respect to themselves.
    The same as what?
    The motion of the tube certainly affects the time it takes for the light to go from E to F (and from F to E) as seen by A. It doesn't affect the speed, of course, but it does affect the distance and thus the travel time.
     
  11. Oct 21, 2012 #10

    Doc Al

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    The time it takes for the light to go from E to F and back is T0 in frame B and γT0 in frame A. That part is correct.

    But the distance traveled by the light (according to A) is not simply 2L0/γ.
     
  12. Oct 21, 2012 #11
    It has nothing to do with the propagation of the light, but it does close the distance that the light has to move before it reaches the extremity. You are correct that the tube motion has nothing to do with the light's motion, but the distance that the light travels before hitting "extremity E" will be less than the length contracted distance because the tube is moving toward the light. The light's velocity is not changing, but the point where you will stop measuring the time IS changing. The "stationary" observer will measure the light moving at c in one direction and the tube moving v in the other direction. That means that the distance between the light and the end of the tube is changing at -c-v as measured by the stationary observer. Nothing is actually moving at that speed so there is no paradox.
     
  13. Oct 21, 2012 #12

    bgq

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    Why not? If it is not the distance, then what is the distance as measured by A?
     
  14. Oct 21, 2012 #13

    Doc Al

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    Let the time it takes for the light to go from E to F according to A be called T1. Then the distance traveled by the light must be (again, according to A):
    cT1 = L0/γ + vT1

    That's the length of the tube (in A's frame) plus the distance the tube has moved (in A's frame).

    Do a similar analysis for the return trip.
     
  15. Oct 21, 2012 #14

    bgq

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    This leads to a contradiction:

    cT1 = L0/γ + vT1

    cT1 - vT1 = L0

    T1(c-v) = L0

    This shows that the speed of light becomes c-v (not c); I believe that the distance is L0/γ. It arises directly from lorentz transformations, so we don't have too much to deal with this.
     
  16. Oct 21, 2012 #15

    Doc Al

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    No contradiction at all.

    Clearly not. Note that I made use of the speed of light as c when I wrote that the distance = speed*time = cT1.

    You are confusing the speed of light measured by A, which is of course simply c, with the rate at which the light and the end of the tube move towards each other as seen by A. That is c-v, but as already noted that is not the speed of the light.
    That's the length of the tube as seen by A. But as already explained, that's not the distance that light travels according to A.
     
  17. Oct 21, 2012 #16

    bgq

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    The light travels from F to E, but covers a distance not equal to FE. How is this possible?
     
  18. Oct 21, 2012 #17

    Doc Al

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    Forget relativity for the moment. Imagine you are on a train moving at some speed. You walk from the rear of the car to the front of the car. As seen from the train tracks, would you not agree that you traveled a greater distance than just the length of the car? Same idea here.
     
  19. Oct 21, 2012 #18

    Mentz114

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    The light 4-vector in the rest frame of the emitter is a null vector (ω,k,0,0), with ω/k=c=1 (the last equality because I'm using gemetric units). If this vector is boosted by β , the result is another null vector ( because the proper length is invariant) (√(1+β)(1-β), √(1+β)(1-β), 0, 0). The velocity of propagation is obviously still 1 ( ratio of first and second components).

    So the Lorentz transformation ensures that the speed of propagation is unchanged, but the frequency and wave-number are identically Doppler shifted.
     
  20. Oct 21, 2012 #19

    bgq

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    This is OK according to Galilean relativity, where the speed is changed which ensures the change in the distance in the same invariant time. In SR there is a critical point is that the speed of light is not affected by either the observer or source or both. Anyway, even in Galilean relativity the distance is not changed between E and F, the speed does.
     
  21. Oct 21, 2012 #20

    Dale

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    You seem unable to grasp the fact that the distance traveled is not equal to the length of the tube. I suggest that you really think about the scenario that Doc Al posted and try to understand the difference between length and distance traveled. If you are unwilling to do that then it might help to write down the definitions of distance traveled and length.
     
  22. Oct 21, 2012 #21

    Erland

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    From the stationary observer's perspective:

    If T1 is the time of light travel in the same direction as the motion af the train, then

    c*T1 = L/g +v*T1, which gives T1 = (L/g)/(c-v). Likewise, the time of light travel in the opposite direction is T2 = (L/g)/(c+v). Adding this gives, after a little algebra:
    T1+T2 = 2gT, where T is the one way time measured by the moving observer, just as it should be. Thus, the assumption that the stationary observer measures the velocity to c leads to the right answer for the time.
     
    Last edited: Oct 21, 2012
  23. Oct 21, 2012 #22

    bgq

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    Hmm, I think I now understand the difference between length and the covered distance, but unfortunately this leads to the same result:

    d = (L0/γ + vT1) + (L0/γ - vT1) = 2L0
     
  24. Oct 21, 2012 #23

    Doc Al

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    t1 ≠ t2
     
  25. Oct 21, 2012 #24

    Dale

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    I cannot decipher that formula in this context. There should be two speeds, one for the speed of the object and one for the signal. Also, there should be two times, one for reaching the far end, and one for returning to the near end.
     
  26. Oct 22, 2012 #25
    The situation is as following:

    A sees B moving at vrel, with a ruler/tube in front of him. At some point, an event happens. B shoots some photons towards the direction of the ruler/tube.
    Both draw a coordinate system where they consider this event to happen at x=0/t=0.

    As you correctly calculated, the time for the lightbeam to travel the ruler and back seen from B's point of view is 2Lo/c


    A will see the same ruler length contracted L = Lo*1/γ

    So LOGICALLY, as light travels at c for both observers. If the ruler was not moving, it would be trivial 2L/c. Unfortunately the ruler end is moving away of the lightbeam, while the back is moving towards it on the turn around trip.
    So we have L/(c-vrel) for the time it takes to move towards the end, while we have L/(c+v) for the time it takes to move back.
    L/(c-vrel) + L/(c+vrel) is the time it takes seen from A's point of view.



    Time-interval expansion(dilation) is measured LOCALLY in B's rest frame. Meaning that two events which happen at the SAME space but at different times and are measured to have a time interval of t seconds, are measured in A's rest frame to have a higher time-interval (dilation/expansion).
    They do NOT happen at the same space-distance in A's frame.

    In your example, the two local events seen from B's point of view are:
    E1) B shoots lightbeam at x=0/t=0 towards the back of the rocket.
    E2) Lightbeam returns to b after being reflected at x=0/t= 2Lo/c



    This is NOT equivalent to length contraction, because while similarly here we measure the distance between two events at the SAME time seen from B's rest frame.
    We however do NOT measure the distance between those two events in A's frame which of course happen at different times, just like in the case of time dilation the events A measures happen at a different space-distance.
    INSTEAD we measure the distance between two events which happen at the SAME time on the back and front of an object which is moving at vrel.

    I did not think this out, but i find it quite remarkable.
    It seems to be a human centric view. Whatever seems to make more sense to us.
     
    Last edited: Oct 22, 2012
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