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Two athletes time compensation

  1. Oct 4, 2016 #1
    1. The problem statement, all variables and given/known data
    In a reference system S two athletes are aligned at a distance d relative to each other on the Y axis to make a run parallel to the x axis. Two starters, one next to each athlete, shoot their guns out at slightly different times, offsetting the advantage of better athlete. The time difference in S is delta T. For which time difference there will be a reference system S 'in which there is not compensation, and which time difference there is a reference system S' in which there is a real compensation ?.
    Determine explicitly the Lorentz transformation that leads to S' suitable for each of the possibilities listed above, calculating the speed of S' to S and the space-time positions (events) of each athlete in the S system.

    2. Relevant equations
    Lorentz tranformation Equation

    3. The attempt at a solution

    I don´t really understand the exercise, but firs I think you must asume that the coordenates of one of the athletes are at the origin, and the other has the displacement d.
    r1=(ct1,0,0,0)
    r2=(ct2, d,0,0)

    When you have two events ussually you use the space time interval

    (delta S)^2 =(ct2-ct1)^2-(x2-x1)^2-(y2-y1)^2-(z2-z1)^2

    But i dont understant what it means no compensation real compensation?
     
  2. jcsd
  3. Oct 4, 2016 #2

    PeroK

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    I don't understand the question either. I think it's asking for a frame where the starting guns fire simultaneously, but the rest is confusing.
     
  4. Oct 4, 2016 #3
    Maybe there are some problem with translation, the problem is in spanish, but I agree, if there is no compensation from S' the two starting guns fire simultaneously, so delta t must be zero. So

    Δs^2 < 0
    But how I show this?
     
  5. Oct 4, 2016 #4

    PeroK

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    I'm going offline now but perhaps someone else can help.

    You need to think of your starting coordinates and what direction S' needs to move.
     
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