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Muon Production

  1. I was reading about Fermilab moving their new storage ring to the Muon Campus for the Muon g-2 experiment. I was curious about how the produce the Muons. I understand that protons hit a graphite target producing pions that quickly decay into Muons. How much energy are is required? How much energy do the protons have in order to produce the pions? And what is the energy of the exiting Muons and final electrons?
     
  2. jcsd
  3. Vanadium 50

    Vanadium 50 17,636
    Staff Emeritus
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    The muons are selected to be at tghe magic momentum of 3.094 GeV. The protons are at 8 GeV. The pions are in between.
     
  4. e.bar.goum

    e.bar.goum 491
    Science Advisor

    These of slides have a nice summary of the experiment. For the record, the pions have a momentum of 3.1 GeV according to the slides. Which makes sense, since
    [itex]\pi^+ \rightarrow \mu^+ + \nu_\mu[/itex]
    https://indico.cern.ch/event/234546/session/9/contribution/20/material/slides/1.pdf

    ETA: The slides also explain the concept of "magic momentum", and will be interesting to anybody who is into accelerator physics.
     
  5. why are you calling the momentum magic? is there something extraordinary/interesting about that value? or did you want to sound poetic?
     
  6. e.bar.goum

    e.bar.goum 491
    Science Advisor

    The slides I linked to had the explanation for the concept of magic momentum. In a magnetic field, muons will move in horizontal circular motion, as you require in a storage ring, but you will also inevitably have some vertical component. The way around this would be to use electrostatic quadrupoles, but that adds more complications.

    [itex] \omega_a = \frac{e}{mc}(a_\mu B - (a_\mu - \frac{1}{\gamma^2 - 1}(B \times E )) [/itex]

    But then you need to measure E. But if you choose γ=29.3, the coefficient goes to 0, which corresponds to 3.09 GeV, and

    [itex] \omega_a = \frac{eB}{mc}a_\mu[/itex]

    Thus, magic momentum.
     
  7. mfb

    Staff: Mentor

    This looks a bit circular - you use aμ to determine the best energy to measure aμ. But I'm sure they took this small effect into account, and there is indeed just a single aμ value that fits to observations (so it is possible to solve this circular argument).
     
  8. Vanadium 50

    Vanadium 50 17,636
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    The muon magnetic moment is known to something like 11 decimal places, so the magic momentum is also known to something to a few parts per billion. The beam momentum has a spread of a few parts per thousand. So there's no problem with circularity. If you like, think of it as the momentum where the effect of the electric field is smallest, rather than identically zero.
     
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