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Simultaneous events in different frames of reference

  1. May 10, 2013 #1
    1. The problem statement, all variables and given/known data

    In frame F there are two lights on the x-axis at D and -D (D=0.6x10^9 m ) which flash simultaneously when t=0. There is another frame F' which moves at v=0.8c in standard configuration with F.
    I need to work out when observers standing at the origin of both frame F and F' will see the light flashes, and also in frame F' if the speed is -0.8c instead.



    2. Relevant equations

    [tex] x' = γ(x-vt) [/tex]
    [tex] ct' = γ(ct - \frac{vx}{c}) [/tex]


    3. The attempt at a solution

    So I think for observers in frame F it will be 2 sec but I'm not sure how to work out when observers at the origin in frame F' will see the flashes. I've got the distance to both points in F' and the time the flashes take place in F' but I am not certain what to do with it.
     
    Last edited: May 10, 2013
  2. jcsd
  3. May 10, 2013 #2

    TSny

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    If you've worked out this much, then it should be easy to find the time interval that it takes the light to travel from the location of a flash to the observer. What is the speed of light in F'?
     
  4. May 10, 2013 #3
    It should be c, like in all reference frames. I'm just getting slightly confused by the fact that frame F' is moving. And by which flash the observer sees at what time.

    The flashes occur at 2.666 seconds in frame F'. Do I need to work out distance that it's moved in this time, add it to the original distance D and then divide by c?
    If so, I get this to be 6 seconds. But I'm not sure whether this is the same for both flashes of light or not.
     
  5. May 10, 2013 #4

    TSny

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    Only one of the flashes occurs at time t' = +2.667 s in F'.
    In F' one of the flashes occurs at time t' = 2.667 s. But the observer in F' won't see the flash until the light gets to her. So, you'll just need to find the additional time it takes the light to travel to the observer. Similarly for the other flash.
     
  6. May 10, 2013 #5
    The other one should be at t' = -2.667s.

    I'm doing 1x10^9 / 3x10^8 = 3.333 s
    3.333 + 2.666 = 6 s.

    And the other frame would be 3.333 - 2.666 = 0.666 s?
     
  7. May 10, 2013 #6

    TSny

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    That looks good to me. [Edit: "other frame" -> "other flash"]
     
    Last edited: May 10, 2013
  8. May 10, 2013 #7
    great, thanks for your help :)
     
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