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Muon decay in particle accelerator

  1. Oct 23, 2007 #1
    1. The problem statement, all variables and given/known data
    The most precise measurement of the anomalous magnetic moment of the muon (g-2) is performed at Brookhaven (USA). In this experiment muons are held in a circular orbit in a 14 m diameter ring, with momentum 3.098 GeV. On average how many laps of the ring the muon will do before decaying? (hint: assume the the muons goes at the speed of light).


    2. Relevant equations
    None given


    3. The attempt at a solution
    I figure this is related to reltivity, so the muon's clock is running slow...
    T=[tex]\gamma T_{0}[/tex]
    (T=gamma T0)
    but if v is assumed to be c, the gamma factor is infinity?

    But I'm confused about how to work out a decay from energy when no decay time is given?

    Is there a list of formulas anywhere that would apply to this situation?

    Thanks!
     
    Last edited: Oct 23, 2007
  2. jcsd
  3. Oct 23, 2007 #2
    That question is rather silly, since it says that you must assume it goes at the speed of light, but if that was the case, then it would have to be massless, and since it wouldn't experience time, it would never decay. It may be a trick question, but as it stands, the answer would be an infinite number, since it doesn't experience time, that said, it couldn't go at c, since it has mass, but then if it was massless, it wouldn't be confined using traditional methods - you would need a gravitational field.
     
  4. Oct 23, 2007 #3

    Dick

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    The question isn't so much silly, as badly phrased. Get the gamma by comparing muon energy to rest mass. Use the gamma to get lifetime in the lab frame. Now use the hint. You don't have to use gamma to compute v because you know it's REALLY close to c. Just use v=c to find how many laps.
     
  5. Oct 23, 2007 #4
    So:
    muon energy = 3.098 c GeV
    speed of light in GeV = ??
    muon rest mass = 0.1066 GeV
    Gamma = (3.098c - 0.1066)/(0.1066*c^2) = ??
    T = gamma T0
    c/T = distance
    distance / 2*pi*7
    = number of laps

    is that correct? and how do I fill in the blanks?
    Thanks!
     
  6. Oct 23, 2007 #5

    Dick

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    Are you making these formulas up? Your units are all wrong. Relativistic energy is gamma*(rest mass)*c^2. Did you look up T0? What's the lifetime of the muon in the rest frame? Energy is measured in GeV. Mass is measured in GeV/c^2. Or energy is measured in joules and mass in kg. You might wish to convert to the usual units. But how can you say distance=c/T??? c is in m/sec and T is in sec. If you go back and put correct units on everything I think it will help you a lot.
     
  7. Oct 23, 2007 #6
    Ok, thanks...
    so gamma = relativistic energy/(restmass*c^2)
    muon rest mass = 105.6 MeV/c2 = 0.1056 GeV/c2
    gamma = 3.098GeV/(0.1056 GeV) = 29.337
    t(rest) = gamma*t(muon)
    but i don't know the lifetime of the muon in the rest frame as I wasn't given a decay rate or lifetime or anything... so how can this question be completed?

    Thanks for all your help!
     
  8. Oct 23, 2007 #7

    Dick

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    Much better. For the muon lifetime you just have to look it up. There is no formula.
     
  9. Oct 23, 2007 #8
    Ok, so I think I have it now... Thanks again for all your help!

    gamma = relativistic energy/(restmass*c^2)
    muon rest mass = 105.6 MeV/c2 = 0.1056 GeV/c2
    gamma = 3.098GeV/(0.1056 GeV) = 29.337
    t(rest) = gamma*t(muon)

    mean muon lifetime = 2.2ms

    t = 0.0022s*29.337 = 0.0645414s

    d=s*t
    d=3*10^8m/s * 0.0645414 = 19 362 420m
    number of laps = 19 362 420/(pi*7*2)
    so 440 232 laps?
    seems like a very large number...
     
  10. Oct 23, 2007 #9

    Dick

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    Well, it's not as big as you think. Because for the muon lifetime the 2.2 is in microseconds, not milliseconds. But other than that you've made a great improvement!
     
  11. Oct 23, 2007 #10
    thankyou :-)
    so more like 440 laps then? I guess that's more believable...
     
  12. Oct 24, 2007 #11

    Dick

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    I'll buy that. High energy muons can penetrate thousands of meters into the ground. Not millions.
     
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