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Muon decay in flight - distribution of distance

  1. Nov 1, 2016 #1
    1. The problem statement, all variables and given/known data
    The average lifetime of muons at rest is τμ0 = 2.2 μs. A laboratory measurement on the decay in flight of muons in a beam emerging from a particle accelerator yields an average lifetime of τμ = 6.6 μs, as measured in the lab frame Σ.
    (g)
    [3 points] Given a large ensemble of these muons, some of the muons will travel farther than others in the laboratory frame Σ. The average distance traveled is the number you found in part (f). Graph the distribution of the distances traveled in the laboratory frame Σ by this ensemble of muons. Carefully label your axes, and show both horizontal and vertical scales.

    2. Relevant equations
    Δx = βcτμ = 1881 m for average distance part (f)

    3. The attempt at a solution
    I'm missing some point what is happening in such ensemble. Please give me some hint. Thank you.
     
  2. jcsd
  3. Nov 1, 2016 #2

    mfb

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    What is the time-distribution of decays in the muon rest frame? It is the same for all radioactive decays.

    Unrelated:
    The problem statement should specify that all muons have the same energy, or give some information about the energy distribution if you are supposed to take the distribution into account.
     
  4. Nov 3, 2016 #3
    Oh, yes, I looked at that by no-particle eyes. There occur standard particle decay. And it is governed by decay formula. Where probability of decay is $$ f(t)=\lambda e^{-\lambda t} $$ Where ## \lambda= 1/\tau_\mu ##. But I want to express it in lenghts. So I use my formula from (f) above. So than I have
    $$ f(x)=\frac{1}{\beta c \tau_\mu}e^{-\frac{x}{\tau_\mu}} $$.
    Do you agree?
    Because first of all I expected Gauss distribution with mean value at ## \Delta x ## and parameter ## \frac{1}{\beta c \tau_\mu} ## . But this is probably something other.
     
  5. Nov 3, 2016 #4

    mfb

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    It is an exponential distribution. You forgot time dilation, however.
     
  6. Nov 6, 2016 #5
    Heh, ok :) So, if I substitute ##t->\gamma \tau## and then result of (f) will be it alright?
    $$ f(x)=\frac{1}{\gamma \tau_\mu} e^{-\frac{\tau}{\tau_\mu}}=\frac{\beta c}{\gamma \Delta x} e^{-\frac{x}{\Delta x}}$$
     
  7. Nov 6, 2016 #6

    mfb

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    As it is a distribution, you need ##\int_0^\infty f(x) = 1##. Does that work out?

    The exponential term should take time dilation into account - it needs a gamma factor somewhere.
     
  8. Nov 6, 2016 #7
    I did it properly, and my results is
    $$f(x)= \frac{1}{\Delta x}e^{-\frac{x}{\Delta x}} $$
    but important is: ##\Delta x## and ##x## is average length in laboratory frame and length in laboratory frame. ##\gamma## doesn't appear in exponent because anything is expressed in same frame.
    The integral which you mentioned is 1 for this distribution.
    What do you mean now?
     
  9. Nov 6, 2016 #8

    mfb

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    Your ##\Delta x## should include a gamma-factor, as the average length in the laboratory frame depends on it.
     
  10. Nov 6, 2016 #9
    Hmmm ok I'm something missing.
    $$ f(t)=\frac{1}{\tau_0}e^{-\frac{t}{\tau_0}}$$ is general formula for decay (let's say in rest frame it means what see flying particale). Where ##\tau_0 ## is mean lifetime. Also $$ f(t)=\frac{1}{\tau'_0}e^{-\frac{t'}{\tau'_0}}$$ is same formula but expressed in moving frame with respect to particle (laboratory frame). Do you agree?
    If I want to express second forumla (in laboratory frame) in terms of rest frame I use ##t=t'/\gamma ## (x=0) and transform also ##\tau=\tau'/\gamma## and gamma-factors are deleted each other.
    Please let me now what I'm missing or what is wrong.
    Thank you.
     
  11. Nov 6, 2016 #10

    mfb

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    The left side didn't change, therefore the right side should be identical in both expressions. It is not.
     
  12. Nov 6, 2016 #11
    The second equation has wrong left side there should be ##f(t')##. Now is it right?
     
    Last edited: Nov 6, 2016
  13. Nov 6, 2016 #12

    mfb

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    Then you just changed the variable, but that is possible.

    I think it is way easier to just work in the lab frame. Calculate the mean lifetime in the lab, calculate the mean length, then write down the exponential decay with this mean length, done.
     
  14. Nov 7, 2016 #13
    Ok. One again. ##\tau_\mu=\gamma \tau_{\mu0} ##, where ##\tau_{\mu0} ## is lifetime at rest. ##\Delta x=\beta c \tau_\mu ## is path which travel in lab frame. ## x,t## is distance and time measured in lab frame. Then
    $$ f(t)=\frac{1}{\tau_{\mu} }e^{-\frac{t}{\tau_{\mu}}}dt=\frac{1}{\gamma\tau_{\mu0} }e^{-\frac{t}{\gamma\tau_{\mu0}}}dt$$
    $$ f(x)=\frac{\beta c}{\Delta x}e^{-\frac{x \beta c}{\Delta x}}dx$$ Is it now ok?
     
  15. Nov 7, 2016 #14

    mfb

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    Correct. You can simplify it and remove new variables by plugging in your equation for ##\Delta x## and then for ##\tau_\mu##
     
  16. Nov 7, 2016 #15
    Do you mean only this?
    $$f(x)=\frac{\beta c}{\Delta x}e^{-\frac{x \beta c}{\Delta x}} dx= \frac{1}{ \tau_{\mu}}e^{-\frac{x }{\tau_{\mu}}} dx$$
     
  17. Nov 7, 2016 #16

    mfb

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    Something is wrong with units now (edit: also in the post before). An additional factor ##\beta c## which should not be there.
    One more step and you have the muon lifetime (in its rest frame) in it instead of the effective lifetime in the lab.
     
  18. Nov 7, 2016 #17
    Ok I should use ##x=\beta c t ## for ##t##. So finally? :) And thank you for patience :)
    $$f(x)=\frac{1}{\beta c\gamma\tau_{\mu 0}}e^{-\frac{x}{\beta c \gamma\tau_{\mu 0}}} dx$$
     
  19. Nov 7, 2016 #18

    mfb

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    Looks right.
     
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