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Homework Help: 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



    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]

    Frame S':
    Event 1: [itex](x'_1, t'_1)[/itex]
    Event 2: [itex](x'_2, t'_2)[/itex]


    [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]

    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

    User Avatar
    Science Advisor
    Homework Helper

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

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