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Lorentz transformation: time dilation

  1. Feb 3, 2010 #1
    1.) problem statement
    Relativistic protons that have a certain speed "v" are selected by measuring the time it takes the proton to travel between two detectors separated by a distance "L". Each detector produces an electronic pulse of very short duration (LaTeX Code: \\Delta t << L/v) when a proton passes through it. A coincidence circuit is made by delaying the pulse from the first detector by an amount L/v. The signals from the two detectors are fed into a logic circuit that produces an output pulse if the pulses arrive at the same time. For input pulses that arrive at the same time as measured in the laboratory frame, calculate the time difference between arrival of the input pulses as measured in the rest frame of the proton.


    2. Relevant equations
    LaTeX Code: \\Delta t = LaTeX Code: \\Delta t'/ (1-v^2/c^2)^1/2 where LaTeX Code: \\Delta t' is the time elapsed in reference frame of moving particle.


    3. The attempt at a solution

    I take the LaTeX Code: \\Delta t elapsed in the reference frame of the lab to = L/v

    So my question is, when I switch to the frame of the proton - the length of the path it travels is now moving towards it - do I have to do a length contraction ?? Or will my time dilation equation take this into account?

    I got:

    (delta) t (proton) = 1/(1-v^2/c^2)^1/2 L/v

    just solving the time dilation equation in the reference frame of the proton seeing the apparatus moving towards it. (Thus the time in the lab is ticking slower relative to the proton's frame.)
     
  2. jcsd
  3. Feb 4, 2010 #2
    In the rest frame of proton, the distance L would be contracted because L is measured by an observer in the lab. So in the frame of moving particle we have:

    In the frame of proton: [tex]L'=(1-v^2/c^2)^{1/2} L[/tex] and thus

    [tex]Delta t' = (1-v^2/c^2)^{1/2} \Delta t =(1-v^2/c^2)^{1/2}(1-v^2/c^2)^{1/2} L/v= (1-v^2/c^2) L/v[/tex].

    Thus in the frame of the moving proton time elapses faster than in the frame of lab relative to it or similarly time for an observer in the lab ticks slower relative to the proton's frame.

    AB
     
  4. Feb 4, 2010 #3

    vela

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    I think the wording of the problem is unclear, or it's a trick question.

    If the logic circuit produces a pulse in the rest frame, it will produce one in the proton's frame. Therefore...
     
  5. Feb 4, 2010 #4
    I took for granted that the pulses are of the same nature in any frame but affected by the frame-rigged system of SR!

    If my assumption is true, what do you say about my calculation vela!?

    AB
     
    Last edited: Feb 4, 2010
  6. Feb 4, 2010 #5

    vela

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    I think you calculated the lab frame [itex]\Delta t[/itex] incorrectly. It should just be L/v, right?
     
  7. Feb 4, 2010 #6
    But isn't it measured by an observer in the lab? Remember that we want to calculate quantities from proton's view point, right? The proton has no information about the [tex]\Delta t [/tex] that the observer in the lab measures. So in proton's frame, the lab's L/v must be contracted, right?

    AB
     
  8. Feb 4, 2010 #7

    vela

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    In the lab frame, the time [itex]\Delta t[/itex] is the uncontracted distance divided by the proton's speed, so [itex]\Delta t=L/v[/itex]. [itex]\Delta t[/itex] should only depend on what's measured in the lab frame. You used the length in the proton's frame and divided it by the time as measured in the proton's frame.

    In the proton's frame, the length is contracted to [itex]L' = L/\gamma[/itex], but time is dilated so [itex]\Delta t'=\Delta t/\gamma[/itex]. These two effects offset each other perfectly so that both observers agree on the proton's speed, [itex]L'/\Delta t' = L/\Delta t = v[/itex].
     
  9. Feb 4, 2010 #8
    Yup, I got you now!

    Thanks for teaching me something really profound!
     
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