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Problem in Lorentz transformation

  1. May 25, 2012 #1
    1. The problem statement, all variables and given/known data

    Please see the attached file and advise me my solution is right or not! Thanks!

    2. Relevant equations

    ~ d/t = v
    ~ Lorentz transformation : x'=(x-Vt)/(√(1-(V^2)/c^2), t' = [t-(V^2)/(c^2)]/√1-(V^2)/c^2),

    3. The attempt at a solution

    a) According to Cheryl, d/t =v, 15/(c/2) = 10^-7s

    From Tom on ground,
    i) when the light from the spark reaches mirror A
    t1 = L/(c+V) = 15m/(3x10^8 + c/2 ) = 3.33 x 10^-8s

    ii) when the light from spark reaches mirror B
    t2 = L/(c-V) = 15m/(3x10^8 - c/2 ) = 10^-7s

    iii) when the light from spark returns to Cheryl from mirror A
    t2 = L/(c-V) = 15m/(3x10^8 - c/2 ) = 10^-7s

    iv) when the light from spark returns to Cheryl from mirror B
    t1 = L/(c+V) = 15m/(3x10^8 + c/2 ) = 3.33 x 10^-8s

    b) How to apply the Lorentz transformation equation in the same four events take place?? please advise!

    My approach:

    t=10^-7s, d = 15m, v=c/2

    x'=(x-Vt)/(√(1-(V^2)/c^2),
    = 15-(c/2 x 10^-7)/(√(1-(V^2)/c^2) = 0

    t' = [t-(V^2)/(c^2)]/√1-(V^2)/c^2),

    = 8.66 x 10^-8s


    c) They disagree that ligt from the spark reachs mirror A at the same time as mirror B, because when the light from the spark reaches mirror A of clock slow down and length contract. They also disagree the light returning from mirror A reaches Chery l at the same time as the light returning from mirror B, because the light returning from mirror B of clock slow down and length contract.

    d) They agree it, because the clock is not slow down and length contract.

    e) Using t' = t√1-v^2/c^2)

    =10^-7s√1-c/v^2/c^2)

    =8.66 x 10^-8s

    "Moving clocks run slow"
    ΔT = ΔT0/√1-v^2/c^2
     
    Last edited: May 25, 2012
  2. jcsd
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