Conservation of Four-Momentum Muon decay

[ssmfour]

Homework Statement

Muon with mass 80m is at rest. Muon decays into photon and particle Qm. The photon has an energy of E=60m in the +x direction. Find all of the other components.

Homework Equations

Vx= Px/E
m^2=E^2-P^2
Pt=E

Ignoring the Y and Z components of the Four-Momentum as they are always 0

The Attempt at a Solution

[80m,0]=[E,Px] + [E2,P2x]

Vx=Px/E
Px=(Vx)(E)
Px=(1)(E) or Px=(-1)(E)

Couldn't figure out if the photon itself moved in the +x direction or if just the energy was positive and the particle Qm would move in the +x direction. Can figure out the rest from there the sign just messes me up.

Thanks

 

[anathema]

for your help!
Thank you for your post. I am a scientist and I would be happy to help you solve this problem.

First, let's define some variables. Let m represent the mass of the muon, E represent the energy of the photon, and Qm represent the energy of the particle Qm.

From the given information, we know that the muon is at rest, which means that its initial Four-Momentum is [m,0]. After the decay, the muon has turned into a photon and a particle Qm, so we can write the final Four-Momentum as [E,Px] + [Qm,P2x].

Using the equations you provided, we can write:

Vx= Px/E
and m^2=E^2-P^2
and Pt=E

Since we know that the photon has an energy of E=60m in the +x direction, we can write:

E=60m
Vx=1 (since the photon is moving in the +x direction)
Px=E=60m

Now, we can use these values to solve for the energy and momentum of the particle Qm.

m^2=E^2-P^2
m^2=(60m)^2-Px^2
Px^2=(60m)^2-m^2
Px=sqrt[(60m)^2-m^2]

Pt=E
Pt=60m (since the energy of the particle Qm is equal to the energy of the photon)

Therefore, the final Four-Momentum for the particle Qm is [60m, sqrt[(60m)^2-m^2]].

I hope this helps you solve the problem. Let me know if you have any further questions.