- #1

Frostman

- 115

- 17

- Homework Statement
- An antiparticle of mass ##m## and energy ##E## hits a stationary particle in the laboratory and the system annihilates itself producing two photons, which are directed along the direction of the motion of the incident antiparticle.

1 - Calculate the energies of the two photons in the laboratory's reference system.

2 - Calculate the energies in the reference system integral with the incident particle, explicitly carrying out the Lorentz transformation. Discuss the result obtained.

- Relevant Equations
- Conservation of the quadri-momenta

I set up this problem this way:

##p_a^{\mu}=(E, \sqrt{E^2-m^2}, 0, 0)##

##p_b^{\mu}=(m, 0, 0, 0)##

##p_c^{\mu}=(2E_\gamma, 2E_\gamma, 0, 0)##

I have chosen to consider the two photons as a single particle of energy equal to ##2E_\gamma##. At this point I applied conservation of the quadri-momenta:

##E+m=2E_\gamma##

##\sqrt{E^2-m^2}=2E_\gamma##

From the first equation I find the first point: ##E_\gamma=\frac {E+m}2##. But if I put this in the second equation I don't have the equality. Why?

I'm supposing that the particle ##a## is a ##e^+## and particle ##b## is an ##e^-##, because the statement said that the system annihilates itself.

##p_a^{\mu}=(E, \sqrt{E^2-m^2}, 0, 0)##

##p_b^{\mu}=(m, 0, 0, 0)##

##p_c^{\mu}=(2E_\gamma, 2E_\gamma, 0, 0)##

I have chosen to consider the two photons as a single particle of energy equal to ##2E_\gamma##. At this point I applied conservation of the quadri-momenta:

##E+m=2E_\gamma##

##\sqrt{E^2-m^2}=2E_\gamma##

From the first equation I find the first point: ##E_\gamma=\frac {E+m}2##. But if I put this in the second equation I don't have the equality. Why?

I'm supposing that the particle ##a## is a ##e^+## and particle ##b## is an ##e^-##, because the statement said that the system annihilates itself.