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Special Relativity, simultaneity and spatial separation

  1. Apr 7, 2013 #1
    1. The problem statement, all variables and given/known data
    To an observer in frame [itex]S[/itex], two events are simultaneous and occur 600km apart. What is the time measured by an observer in frame [itex]S'[/itex] who measures their spatial separation to be 1200km? Explain the sign of your answer.


    2. Relevant equations

    Lorentz Transformation

    [itex] t' = \gamma(t-vx/c^2)[/itex]

    or

    [itex] x' = \gamma(x - vt) [/itex]

    3. The attempt at a solution

    The question doesn't give a value for v, and as such is making it a little difficult to wrap my head round. My attempt was to insert in the first equation t=0, x=600 and rearrange a little which gave me;

    [itex] t' = \dfrac{-600v^2}{c^2}[/itex]

    Not sure if this is what the question wants me to stop at or not.
     
  2. jcsd
  3. Apr 7, 2013 #2

    mfb

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    Staff: Mentor

    The given spatial separation in the second frame allows to calculate v.
    The units of your final formula look wrong.
     
  4. Apr 7, 2013 #3

    Andrew Mason

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    If you just apply the Lorentz transformation for length: x'1-x'2 you would get:

    [itex]L' = x'_1 - x'_2 = \gamma(x_1-vt_1) - \gamma(x_2-vt_2) = \gamma(x_1-x_2 + v(t_2-t_1))[/itex]

    Knowing the relationship between t1 and t2 (ie. the times of the two events in the S frame) this simplifies nicely. You know x1-x2 and x'1-x'2 so you can solve for γ and from γ you can find v. Then you can then apply the Lorentz transformation for time to find t1' - t2'.

    AM
     
  5. Apr 8, 2013 #4
    Have you learned yet that if you square the equation for x' and subtract the square of the equation for ct', you get:
    [tex](x')^2-(ct')^2=(x)^2-(ct)^2[/tex]
    This equation is independent of the relative velocity v, and is a fundamental characteristic of the geometry of flat space-time. This should help you solve your problem.
     
  6. Apr 8, 2013 #5
    Thank you everyone for your responses. They helped greatly.
    So, with the separation in frame S'(x'2-x'1) equalling 1200km and applying the Lorentz transformation to each event's position in that frame one would get;

    [itex] x'_2 - x'_1 = \gamma (x_2 - x_1 - v(t_2 - t_1))[/itex]

    [itex] 1200 = \gamma (600 - v(0))[/itex]

    [itex] \gamma = 200 [/itex]

    Now solve for v. Afterwards

    [itex] \Delta t' = \gamma (\Delta t + \dfrac{v \Delta x'}{c^2}) [/itex]
    [itex] \Delta t' = \gamma \dfrac{v \Delta x'}{c^2} [/itex]

    Now solve for Δt' using calculated v?
     
  7. Apr 8, 2013 #6

    Andrew Mason

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    γ=200?? Check that again! The rest is fine except you are using x' in the Lorentz transformation for t'! It should be x.

    AM
     
  8. Apr 8, 2013 #7
    2 sorry haha. Thank you.
     
  9. Apr 8, 2013 #8
    [tex](x')^2-(ct')^2=(x)^2-(ct)^2[/tex]
    [tex]1200^2-(ct')^2=600^2[/tex]
    [tex](ct')^2=1200^2-600^2[/tex]
    [tex]ct'=600\sqrt{3}[/tex]
     
  10. Apr 9, 2013 #9

    Andrew Mason

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    Ok. But it is not so easy to determine v from that. Using the Lorentz transformations you can determine both v and t'.

    AM
     
  11. Apr 9, 2013 #10
    The original question didn't ask for v.
     
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