Relativistic momentum - muon decay

[tib]

Homework Statement

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.

Homework Equations

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

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.

 

[vela]

You need to conserve total energy as well.

Do you know how to use four-vectors?

 

[tib]

i do not know how to use four-vectors

 

[vela]

OK, no problem. So far you have

0 = p_e-p_γ

What equation do you get from conservation of energy?

EDIT: Some advice: write it in terms of E_e, E_γ, 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.

 

[tib]

if we treat the muon at rest, E_i=E_muon=E_o=105.7 MeV

E_f=E_e+E_γ

I know E_γ=hf, but I feel like that's going in the wrong direction for the problem; also I know E_e=γm_ec², but without the E_γ, I feel that's a deadend as well.

 

[vela]

if we treat the muon at rest, E_i=E_muon=E_o=105.7 MeV

Right. I'd just write it as E_i=m_μc². It's easier to plug the numbers in at the end.

E_f=E_e+E_γ

I know E_γ=hf, but I feel like that's going in the wrong direction for the problem; also I know E_e=γm_ec², but without the E_γ, I feel that's a deadend as well.

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.

 

[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 p_e=p_gamma; 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 E_i; the result is 0=-(pc)², because E=E_o=mc² -- which does make sense, but isn't useful

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

 

[vela]

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 (m_ec²)²=E_e²-(p_ec)². The photon has no rest mass, so the relation reduces to 0=E_γ²-(p_γc)², or E_γ=|p_γ|c.

 

[Count Iblis]

i do not know how to use four-vectors

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.