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Two different events?

  1. Feb 9, 2015 #1
    Quick question about a relativity calculation. Scenario: conveyor belt going at 0.5c. At t= 0 Observer A is on the floor at x = 0 next to, but not on the conveyor belt. Observer A sees Observer B on the conveyor belt at t = 0 x = 0 smashing through a opaque sugar glass pane wearing body clock suit which is displaying the time as t'=0 and stamping observer B's forehead with that time. I'll refer to that as Event X. Observer B agrees with Observer A that its clock was saying t'=0, and that t'=0 has been stamped on its forehead at Event X. Observer C according to Observer A was at x = c on the conveyor belt at t = 0. With my calculations, Observer C would be stating that Event X happened at t' = -0.57735. Presumably this would mean that the B's clock would have been displaying t'=-0.57735 as it smashed through the opaque sugar glass, and presumably that would suggest -0.57735 had been stamped on the forehead. Which would seem to describe a different event. Have I understood this correctly?
     
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  3. Feb 9, 2015 #2

    PeterDonis

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    Is Observer C using the same primed coordinates that Observer B is? If so, he must assign the same t' value to event X that B does, and you've already stipulated that B assigns t' = 0 to that event. The simplest way to do the Lorentz transformation from unprimed to primed coordinates is to put the spatial origin of the primed coordinates at B's location; then event X is at t' = 0, x' = 0, and Observer C will assign those coordinates to it.

    If you are getting another answer for the t' value that Observer C assigns, then he cannot be using the same primed coordinates that B is. That complicates the analysis, because it means the origins of the the coordinates (C's vs. A's unprimed coordinates) do not coincide, which means there will be extra terms in the Lorentz transformation.

    No. The time B's clock actually displays is an invariant; it's the same regardless of what coordinates you choose. The event is not defined by its coordinates; it's defined by what happens there--in this case, B smashing through the glass at the same instant as his clock displays the value 0.

    If Observer C's coordinates are such that he assigns some t' value other than 0 to event X, then that means, as above, that he cannot be using the same coordinates as B is, since B's coordinates assign t' = 0 to event X (and those coordinates are set up so that this t' value matches the actual observed display of B's clock at that event). So in Observer C's coordinates, the t' value he assigns to events on B's worldline does not match the observed reading of B's clock at those events.
     
  4. Feb 9, 2015 #3

    PeroK

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    Assuming C is at rest with respect to A, then C sees what A sees.

    If C defines time in the moving frame to be t' = 0 at point c when t = 0, then C has defined a different primed frame with a different origin.
     
  5. Feb 9, 2015 #4

    PeterDonis

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    The OP says that Observer C is on the conveyor belt; that indicates that he is at rest with respect to B, not A.
     
  6. Feb 9, 2015 #5

    PeroK

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    In that case, if B and C are in the same frame, why would they disagree about the time of an event?

    Question to the OP.
     
  7. Feb 9, 2015 #6
    I may have made a mistake in the calculation.

    gamma = 1/sqrt(1 - ((v * v) / (c * c)))
    = 1/sqrt(1 - ((0.5c * 0.5c) / (c * c))
    = 1/sqrt(1 - 0.25)
    = 1/sqrt(0.75)
    = 1.1547 (approx)

    t' = gamma * (t - ((v * x)/(c * c)))

    So x and t are from Observer A's perspective, and used to calculate the perspectives of Observer B and Observer C

    So for Observer B:
    t'= 1.1547 (0 - ((0.5c * 0) / (c * c)))
    t'= 0

    for Observer C:
    t'0 = 1.1547 (0 - ((0.5c * c) / (c * c)))
    t'0 = 1.1547 (-0.5)
    t'0 = -0.57735

    Have I made a mistake?
     
  8. Feb 9, 2015 #7

    PeterDonis

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    Why is there a -0.5 here? If Observer C is calculating the t' coordinate of event X, he should be using the t and x coordinates of event X, not his own.
     
  9. Feb 9, 2015 #8
    Sorry I have I can see it, the Event X is at x = 0 and so the calculation for Observer C is inappropriate.
     
  10. Feb 9, 2015 #9
    Yes thanks you're right :)
     
  11. Feb 9, 2015 #10
    What if the body clock suit also stamped on the Observer B's forehead the time it saw on Observer A's body clock? I'm assuming what I was using as the Observer C calculation
    t' = 1.1547 (0 - ((0.5c * c) / (c * c)))
    t' = 1.1547 (-0.5)
    t' = -0.57735

    was when Observer C is saying t = 0 at x = 0, yet wouldn't there be evidence on Observer B's head that from Observer C's rest frame the event of Observer A's clock equalling t = 0 was at t' = 0, or is it that they are both equally right concerning what time it was when t = 0?
     
  12. Feb 9, 2015 #11
    Or is it when t = 0 at x = c?
     
  13. Feb 9, 2015 #12
    Sorry I think I've been using the equations with one at rest at x = 0, t = 0 and the other at velocity v at x, and then using the equations for an account of how one sees the other. Which as you can see has lead to some confusion on my part with me flip flopping over what the x represents. But I can now see that t' and t are both considered the same for all events in their respective rest frames, and the equation I used for observer C just shows that if there had of been an observer D at rest with observer A at x = c then observer C would say its clock wasn't in synch with observer A's.
     
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