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Spacelike separation. Finding a specific frame.

  1. Feb 1, 2016 #1
    1. The problem:
    If the Sun blows up at some instant and four minutes later we on Earth sit down to eat lunch, these two events are separated by a spacelike interval. The explosion of the sun cannot have influenced us at the time we sat down because it takes 8 minutes for light to reach us from the Sun. Since the interval is spacelike there must be a reference frame in which the two events are simultaneous. Find the speed of this frame relative to the Sun-Earth rest frame.

    My attempt:
    So I know that two events spacelike, there is a frame where they happen simultaneously. This frame will have a velocity, v>c. In the above situation, for a person on earth to experience sitting down and seeing the sun blow up at the same time, they either need to sit down 4 minutes later or for the sun to blow up 4 minutes later as Δt = 4 minutes in the frame. I want to find the frame where Δt = 0. Thus I need a time delation of 4 minutes....? The only thing I can think of in terms or equations is t' = γt and find V where t =1s and t' = 241s (4*60)But that gives me a V < C which I know isn't right. Can anyone suggest some equations or if my train of thought is wrong?
     
  2. jcsd
  3. Feb 1, 2016 #2

    PeterDonis

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    No, it won't. The events (Sun exploding and our eating lunch) happen at the same time in this frame, but they don't happen at the same place.

    Time dilation is a ratio, not an offset.

    What you need to take a look at is the equations for a Lorentz transformation (note that we are ignoring the gravity of the Sun and Earth and assuming SR applies, which isn't actually true but it is a good enough approximation for this problem). You have the coordinates of the two events (Sun exploding and our eating lunch) in one frame, the frame in which the Sun and Earth are at rest (note that we are also ignoring the Earth's motion around the Sun, which again isn't correct but is a good enough approximation for this problem); the time coordinates of the two events in this frame differ by 4 minutes. You want to find a Lorentz transformation to a frame in which the two events have the same time coordinate.
     
  4. Feb 1, 2016 #3

    Ibix

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    If the frame exists its relative velocity must be less than c because you can't travel faster than light.

    The time dilation and length contraction formulae are special cases that are not applicable here. You will need to work with the full Lorentz transforms, which is good advice in general for relativity problems.

    Although you do not need to for this problem, I would strongly recommend drawing space-time diagrams of the events in the two frames. It is very helpful in building up intuition.
     
  5. Feb 1, 2016 #4
    So from a Lorentz Transformation, t' = γ(t - vx/c2) where t'=t and solve for v from this equation? If I set this up, solving for v seems quite difficult. My other thought it to use the velocity transformation, Ux' = Ux - v / (1-(vUx/c2)) But what is my Ux and v in this equation?
     
  6. Feb 1, 2016 #5

    Orodruin

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    The times t and t' are not the time coordinates of the two different events. They are the time coordinates of the same event in different frames!
     
  7. Feb 1, 2016 #6
    So I want to take my frame S (t) to be the sun-earth frame where the events are separated by 4 minutes, and then my S' to be the sun-earth frame(t') in which the two events are simultaneous.
     
  8. Feb 1, 2016 #7

    Orodruin

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    Yes, I suggest using an original frame such that one of the events is in the space-time origin...
     
  9. Feb 1, 2016 #8
    So I put frame S at a point where the origin is where the sun explodes so that the separation between the two events is just described by the Earth event. Then, for frame S, x = distance from sun to earth. t = 4 minutes. In S' I want t' = 0 and x'=γ(x-vt)?
     
  10. Feb 1, 2016 #9

    Orodruin

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    Yes, although you really do not care what x' is, it is not part of the question.
     
  11. Feb 1, 2016 #10
    So from my lorentz equation, t' = γ(t-(vx/c2)) I set t' = 0 and solve for v from there...
     
  12. Feb 1, 2016 #11

    Orodruin

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    Yes.
     
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