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Relativity Q / Characteristic decay times

  1. Dec 13, 2012 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    1)The pion is an unstable subatomic particle whose characteristic decay time in its own frame is 26ns. A beam of such particles emerges from a source at point X in the laboratory travelling with speed ##v = \sqrt{3}/2\, c## in the laboratory frame. Determine i) the decay time of the particles in the laboratory frame and ii) the distance (as measured in the lab) from the point X to the point Y at which it is found that the concentration of mesons in the beam is reduced to 1/e its value at the source.

    2) Given a clock which is accurate to 1ns over a period of 1 second, estimate the speed that the clock has to attain for relativistic effects to become noticable.

    3. The attempt at a solution
    i) If we denote ##\tau_o## as the char decay time in the meson frame, then ##\tau## will be the char decay time in the lab frame. Just so I am clear on the meanings of these symbols: ##\tau_o## is the time interval of 2 events with the same spatial coordinates and so this interval is measured in the rest frame of some clock. ##\tau## is the time interval of 2 events without the same spatial coords so is measured in a frame moving relative to the stationary clock. Do these make sense? So for the question: in the lab frame, the char decay time is just ##\tau = \tau_o \cdot \gamma##.

    ii)It is a well known fact that 1/e of the sample remaining occurs at the char decay time. So the answer to this is, given we are concerned with the lab frame, ##d = \tau \cdot v##?

    2) When it says 'relativistic effects become noticable', I presume they just mean that the tick of the clock becomes longer and so the clock becomes less accurate? So in it's rest frame it is accurate to 1 ± 10-9s. I am not sure where to go next with this question. Any hints? Many thanks.
     
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  3. Dec 15, 2012 #2

    CAF123

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    Can anyone confirm what I did in the first question and help me with the second?
    Thanks!
     
  4. Dec 16, 2012 #3

    CAF123

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    For 2), should I assume the clock travels some distance , say 1 m, and do v = d/t?
     
  5. Dec 16, 2012 #4

    haruspex

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    It means that the shift in timing becomes large enough that it is detectable, despite the clock's only being accurate to 1 in 109. I.e., at what speed is the rate shifted by that much?
     
  6. Dec 17, 2012 #5

    CAF123

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    I wouldn't be sure how to compute that. Any hints?
     
  7. Dec 17, 2012 #6

    haruspex

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    Hey, you're the studying relativity. I just look it up as necessary on the net... where I see:
    t = t0/(1-v2/c2)1/2
    So you want the value of v at which (1-v2/c2)1/2 < 1 - 10-9.
     
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