# Homework Help: Relativistic momentum - muon decay

1. Mar 9, 2010

### tib

1. The problem statement, all variables and given/known data

Unstable, subatomic particles called muons have a rest energy of 105.7 MeV and a speed of 0.994c. If a muon were to decay and produce an electron and a photon, what would be the momentum of the electron as measured by an observer in the muon's frame? HINT: assume that the electro goes in the +x direction, the photon goes in the -x direction, and the muon is at rest in its own frame.

2. Relevant equations

E = mc^2 + K; p = gamma*m*v; numerous other forms of relativistic energy equations

3. The attempt at a solution

By the hint, (muon is at rest) we can assume that the initial momentum is zero. By conservation of momentum, initial momentum = final momentum thus, momentum of the electron = momentum of the photon.

This is as far as I can get. The posted answer is 52.8 MeV/c.

2. Mar 9, 2010

### vela

Staff Emeritus
You need to conserve total energy as well.

Do you know how to use four-vectors?

3. Mar 9, 2010

### tib

i do not know how to use four-vectors

4. Mar 9, 2010

### vela

Staff Emeritus
OK, no problem. So far you have

0 = pe-pγ

What equation do you get from conservation of energy?

EDIT: Some advice: write it in terms of Ee, Eγ, and any relevant masses.

Also, the subscript γ is meant to label the energy and momentum of the photon; it's not the gamma that shows up in the relativity equations.

Last edited: Mar 9, 2010
5. Mar 9, 2010

### tib

if we treat the muon at rest, Ei=Emuon=Eo=105.7 MeV

Ef=Ee+Eγ

I know Eγ=hf, but I feel like that's going in the wrong direction for the problem; also I know Ee=γmec2, but without the Eγ, I feel that's a deadend as well.

6. Mar 9, 2010

### vela

Staff Emeritus
Right. I'd just write it as Ei=mμc2. It's easier to plug the numbers in at the end.
Yeah, generally, you want to solve for energies and momenta directly. Throwing in things like gamma, velocity, and frequency just complicate the algebra.

One of the most useful relations you have for these types of problems is

$$(mc^2)^2 = E^2 - (pc)^2$$

So the idea is to rearrange the terms in the equations slightly, then square both equations, subtract one from the other, and use this relation to simplify the result.

7. Mar 9, 2010

### tib

I can see where I (think) I should go, by the equation you gave and by knowing the actual answer, I can see I find p=E/2c. If this is correct, we know by conservation of momentum that pe=pgamma; how do we know that the energy of the muon is split equally between the electron and photon?

Additionally, I still don't see the math to get to p=E/2c; using (mc2)2=E2-(pc)2 for the Ei; the result is 0=-(pc)2, because E=Eo=mc2 -- which does make sense, but isn't useful

And for Ef, photons are massless, but have momentum, so I don't see how to apply that equation to Ef.

8. Mar 9, 2010

### vela

Staff Emeritus
The energy isn't split evenly between the electron and photon.

The relation (mc2)2=E2-(pc)2 is good for each particle, so for the electron, you'd have (mec2)2=Ee2-(pec)2. The photon has no rest mass, so the relation reduces to 0=Eγ2-(pγc)2, or Eγ=|pγ|c.

9. Mar 10, 2010

### Count Iblis

It is easy to learn.

Also, note that the energy is split approximately evenly between the photon and electron, as a result of the electron being extremely relativistic.