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Time difference between 2 events in moving frame

  1. Oct 22, 2013 #1
    1. The problem statement, all variables and given/known data

    Two events occur at the same time in inertial frame S and are separated by a distance of 1km along the x-axis. What is the time difference between these two events as measured in frame S' moving with constant velocity along x and in which their spatial separation is measured to be 2km?


    2. Relevant equations

    [itex]x'=γ(x-vt)[/itex]
    [itex]t'=γ(t-\frac{vx}{c^2})[/itex]

    [itex]x=γ(x'+vt')[/itex]
    [itex]t=γ(t'+\frac{vx}{c^2})[/itex]

    The question should be able to be solved using just Lorentz transformation equations

    3. The attempt at a solution

    Frame S:
    Event 1: [itex](x_1, t_1)[/itex]
    Event 2: [itex](x_2, t_2)[/itex]
    [itex]t_2-t_1=0[/itex]
    [itex]x_2-x_1=1000[/itex]

    Frame S':
    Event 1: [itex](x'_1, t'_1)[/itex]
    Event 2: [itex](x'_2, t'_2)[/itex]
    [itex]t_2-t_1=?[/itex]
    [itex]x_2-x_1=2000[/itex]


    Now,

    [itex]x'_2 - x'_1 = γ(x_2 - vt_2) -γ(x_1 - vt_1)[/itex]
    [itex]2000 = γ[x_2 - x_1 -v(t_2-t_1)][/itex]
    [itex]2000 = γ[1000 -0][/itex]
    [itex]γ=2[/itex]

    Using [itex]γ = \frac{1}{√1-\frac{v^2}{c^2}}[/itex]

    I get [itex]v = \frac{√3}{2}c[/itex]


    Now to find the time difference in the S' frame:

    [itex]t'_2 - t'_1 = γ(t_2-\frac{vx_2}{c^2}) - γ(t_1-\frac{vx_1}{c^2}) [/itex]
    [itex]t'_2 - t'_1 = γ(t_2-t_1-\frac{v}{c^2}(x_2-x_1))[/itex]
    [itex]t'_2 - t'_1 = γ(-\frac{v}{c^2}(x_2-x_1))[/itex]

    Putting in the values I get a time difference of -5.77x10^-6
    It's negative so this can't be correct, so I like help on where I went wrong!
     
    Last edited: Oct 22, 2013
  2. jcsd
  3. Oct 22, 2013 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    Your answer is correct. The sign is immaterial since we only know the absolute value of (x2 - x1)

    AM
     
  4. Oct 22, 2013 #3
    I see, thanks!
     
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