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Leading clocks lag

  1. Oct 8, 2006 #1
    This is rod and shed problem (rod is moving with v towards shed)
    Length of rod measured when the rod is stationary = L0
    Length of shed measured when the shed is stationary l0 = 0.5 L0

    Two clocks are fixedd to the front and back of the shed. Explain how these clocks may be synchronised.
    I did this. (Emitting two photons from the middle of the shed. Triggering the clocks at the back and front of the shed simultaneously)

    According to the Special Theory of Relativity, observers stationary relative to the rod judge that the two shed-based clocks are not synchronised, but that in fact the leading clock lags the trailing clock by the amount l0v/c^2. Explain qualitative why this should be so.

    I can explain this quantitatively.
    Here is my answer. The colcks fixed to the front and back of the shed are synchronised in the shed's frame of reference but from the rod's frame of reference shed is moving which result in observers from the rod's f.o.r (frame of reference) seeing the photon going in the direction of motion of the shed arrive later to one clock (say clock at the front of shed) then the photon going in the different direction of motion arrive to the other clock (say clock at the back of shed). So the clock at the back is triggered before the clock at the front which result in time difference between two clocks and therefore them being not synchronised anymore. The amount of time lag by the front clock is l0v/c^2
    let say point f is front of shed, point b is back of shed, and point m is the middle of the shed. Shed's velocity is v.

    clock f is at f, clock b is at b. f and b are moving along x axis with speed v, so fm = mb = l0/2Gamma, being shortened by lorentz contraction.
    tf = time at front clock
    tb = time at back clock
    c = speed of light

    ctf = l0/2Gamma + vtf
    ctb = l0/2Gamma - vtb

    time difference = tf - tb = l0 / (2Gamma*(c-v))
    but moving clocks run slow so actual time difference is (tf-tb) divided by Gamma which is l0v/c^2
    (I skipped some algebra here for simplicity)

    So the question from me is that. I don't know how to say it qualitatively. I can prove it by using Maths. So please anybody help me :(

    ps. Apologies for wrong spellings and bad grammar.
  2. jcsd
  3. Oct 10, 2006 #2
    The first paragraph in your explanation is the qualitative reasoning. You simply and conceptually explain why it is that the front clock should lag the trailing one. The fact that you then go on and quantitatively derive the [itex]l_0 v / c^2[/itex] relationship is extra credit as far as I'm concerned :)
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