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Urgent Question about the muon lifetime experiment

  1. Dec 10, 2009 #1
    1. The problem statement, all variables and given/known data

    In using the experiment where a scintillator is connected up to PMTs, and the data is then recorded on the computer.... The muon has already lived outside the scintillator for a large amount of time before it decays in the scintillator, so how is it that the experiment still works in providing the actual lifetime of the muon??
  2. jcsd
  3. Dec 10, 2009 #2
    just a quick guess but your measuring the energy or momentum using the scintillator yeh?

    If so ΔxΔp = hbar

    But cΔx = Δt
    and cΔp = ΔE

    So you get ΔEΔT = hbar
    So ΔT = hbar/ΔE

    Try using this concept to get the average lifetime of the muon, i dont think you can calculate the exact lifetime but the resolution using the energy it has instead of measuring the time...if that makes sense...
  4. Dec 10, 2009 #3
    My guess would be: a stopped muon starts your clock; whenever the corresponding decay positron gets registered, the clock is stopped. Neither energy nor impulse of the muon would be of interest in this case.

    Your analysis will provide an average lifetime for muons in a muon ensemble. Basically, you assume that this is a similar scenario to nuclear decay, which follows:

    [tex]N(t)=N(0)\cdot e^{(-\lambda t)}[/tex]

    If your data shows exponential dependence of this kind, you are able to determine the mean lifetime [itex]\tau[/itex], as [itex]\tau=1/ \lambda[/itex]. It doesn't matter how long the muon has been outside your experiment*. This information wouldn't change the time constant of your exponential function. Further investigation would require you to consider the finite measurement time/resolution etc. (e.g., http://en.wikipedia.org/wiki/Maximum_likelihood" [Broken]).

    *As long as you treat a stopped muon as if it was 'free'. This assumption is reasonable in this experiment for [itex]\mu^+[/itex]
    Last edited by a moderator: May 4, 2017
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