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Time difference of events when moving at relativistic speeds

  1. Apr 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Lets say that Planet A and Planet B are moving in the the same inertial reference frame. The distance between them is 8.3 light minutes. Event A occurs on Planet A at t=0, and Event B occurs on Planet B at t=2 minutes. If an observer is travelling from Planet A to B at 0.8c, what is the time difference between the two events?


    2. Relevant equations


    3. The attempt at a solution

    I want to say that we just need to use the Lorentz transformation, plugging in 2 minutes as the proper time, but it's probably not that simple.
     
  2. jcsd
  3. Apr 8, 2013 #2
    probably not.
    event A occurs at tA'=0, xA'=0 in planet frame (taking A as the origin).
    event B occurs at tB'=2 min, xB'=8.3 light minutes in planet frame.

    find out when these are in the moving frame tA, tB.

    you get the piece due to time dilation, but also a piece due to the fact
    that planet B is not sitting at the origin.
     
  4. Apr 8, 2013 #3
    So if I use Lorentz transformations and say tA = γ(t'A - v/c2 xA), similarly for tB, I get tA = 0 obviously, and tB = -464s, meaning that Event B occurs 464s before Event A in the reference frame of the observer? I hope I'm understanding this properly.
     
  5. Apr 17, 2013 #4
    More help to Clarify

    This should just be a straightforward application of time dilation.

    Assume your ship's origin lines up with A's origin. x=0, t=0, x'=0, t'=0.

    Now B is at rest with respect to A, so they are in the same frame, call this the ground frame.

    Your event is going to take place in the ground's frame at (x=8.3 light-mins, t = 2 min).

    Thus, use x'=γ(x-vt) where you plug in the ground's frame x and t from above. The x' that pops out is the coordinate where the ship observes the event to take place.

    Use the next Lorentz equation: t'=γ(t-vx/c^2) where you plug in the ground frame's x,t from above. This will give the t' that the ship observes. If you get an overall minus on the t', that's okay, that just means it happened before the orgiins lined up in the ship's frame.
     
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