# Muon Production

1. Jul 31, 2014

### Bluecom

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. Jul 31, 2014

Staff Emeritus
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. Jul 31, 2014

### e.bar.goum

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
$\pi^+ \rightarrow \mu^+ + \nu_\mu$
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.

4. Aug 1, 2014

### ChrisVer

why are you calling the momentum magic? is there something extraordinary/interesting about that value? or did you want to sound poetic?

5. Aug 1, 2014

### e.bar.goum

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.

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

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

$\omega_a = \frac{eB}{mc}a_\mu$

Thus, magic momentum.

6. Aug 2, 2014

### 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).

7. Aug 2, 2014