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 18,488
    Staff Emeritus
    Science Advisor
    Education Advisor

    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. 801
    Science Advisor
    Education 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]

    ETA: The slides also explain the concept of "magic momentum", and will be interesting to anybody who is into accelerator physics.
  5. ChrisVer

    ChrisVer 2,403
    Gold Member

    why are you calling the momentum magic? is there something extraordinary/interesting about that value? or did you want to sound poetic?
  6. 801
    Science Advisor
    Education 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 18,488
    Staff Emeritus
    Science Advisor
    Education Advisor

    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.
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?

Draft saved Draft deleted