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General Relativity Problem (frames of reference)

  1. Sep 8, 2016 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    A clock moving at v = (3/5)c reads 12:00 as it passes us in our frame of reference, how far away will it be (in light hours) when it reads 1:00.

    2. Relevant equations
    I denote a prime to mean the reference frame of the clock at rest. I use regular lettering to denote 'our' frame of reference.

    c = speed of light

    t' = t/sqrt(1-v^2/c^2)
    L' = L*sqrt(1-v^2/c^2)

    L = v*t
    3. The attempt at a solution

    Alright, I use Lorentz contraction to solve this problem.

    First I find the length L using v and t = 1 hr. L = (3/5)c*1 hr = (3/5)c light hours.
    Now, I use Lorentz contraction to find L'
    L' = (3/5)c*sqrt(1-9/25)
    L' = 12c/25
     
  2. jcsd
  3. Sep 8, 2016 #2

    SammyS

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    First of all, 12c/25 has units of speed, not distance.

    When one hour elapses in "our" frame of reference, the clock has traveled 3/5 light⋅hour (in our frame).

    From "our" point of view the moving clock runs slower than normal, so when the clock reads 1:00, it will have traveled farther than 3/5light⋅hours.
     
  4. Sep 10, 2016 #3

    RJLiberator

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    Thank you for your assistance SammyS.

    I see the difference now in light hours. When saying 12c/25, that is incorrect as it is a speed. I should have said 12/25 light hours. Noted.

    In our FOR it is clear that the clock traveled a distance of 3/5 light hours.
    So I can use time dilation here to see:

    [tex] t' = \frac{1hr} {\sqrt{1-\frac{\frac{9c^2} {25}} {c^2}}} [/tex]
    [tex]t' = 1.25 hr [/tex]

    Therefore, we say the distance will be 1.25 hr * 3c/5 = 3/4 light hours.

    3/4 light hours > 3/5 light hours

    Is that now a correct line of reasoning?
     
  5. Sep 10, 2016 #4

    SammyS

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    Yes. That's what I get.
     
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