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- Homework Statement
- In the annihilation process ##e^+e^- \rightarrow \gamma \gamma##, seen in the reference system of the laboratory, in which the electron is stationary and the positron has energy E, it may happen that one of the two photons is emitted at right angles to the direction of incidence of the positron. Calculate, in this case, the energy and the scattering angle of the other photon.

- Relevant Equations
- Conservation of the four-momentum

I consider the laboratory system. The four momentums in this reference system are respectively:

##p^\mu = \big(\sqrt{|p|^2+m^2}, 0, 0, |p| \big)##

##p'^\mu= \big(m, 0, 0, 0 \big)##

##k^\mu = E\big(1, 0, 1, 0\big)##

##k'^\mu = E'\big(1, 0, -\sin \varphi, \cos \varphi \big)##

I used conservation of four-momentum:

##(k^\mu)^2=(p^\mu+p'^\mu-k'^\mu)^2##

And I find out that:

##E'=\frac{m^2+m\sqrt{|p|^2+m^2}}{m+\sqrt{|p|^2+m^2}-|p|\cos\varphi}##

But I still have two unknown variables...

##p^\mu = \big(\sqrt{|p|^2+m^2}, 0, 0, |p| \big)##

##p'^\mu= \big(m, 0, 0, 0 \big)##

##k^\mu = E\big(1, 0, 1, 0\big)##

##k'^\mu = E'\big(1, 0, -\sin \varphi, \cos \varphi \big)##

I used conservation of four-momentum:

##(k^\mu)^2=(p^\mu+p'^\mu-k'^\mu)^2##

And I find out that:

##E'=\frac{m^2+m\sqrt{|p|^2+m^2}}{m+\sqrt{|p|^2+m^2}-|p|\cos\varphi}##

But I still have two unknown variables...