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Muon Production 
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#1
Jul3114, 08:34 PM

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I was reading about Fermilab moving their new storage ring to the Muon Campus for the Muon g2 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
Jul3114, 09:17 PM

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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.



#3
Jul3114, 09:28 PM

P: 234

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/...l/slides/1.pdf ETA: The slides also explain the concept of "magic momentum", and will be interesting to anybody who is into accelerator physics. 


#4
Aug114, 04:15 AM

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Muon Production



#5
Aug114, 04:22 AM

P: 234

[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. 


#6
Aug214, 03:59 PM

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P: 12,053

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



#7
Aug214, 05:28 PM

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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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