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Electron-Position Pair Kinetics

An electron-positron pair moves with velocity v=c(beta) in frame O. Annihilation of the pair produces two photons of energies E1 and E2 emerging at angles theta(1) and theta(2), with respect to the velocity (v). Find expressions for the energies E1 and E2 in terms of the electron mass, the speed (beta) and angle theta(1).

Solution Preview

We can solve this problem by performing a Lorentz transform to the rest frame of the positron. Working in c = 1 units, the four-momentum vector of the photons can be denoted as:

q1 = (E1, p1)


q2 = (E1,p2)

where |p1| = E1. Let's choose the x-component in the direction of the velocity, then p1 = (p1x,p1y,p1z), and
p1x = |p1|cos(theta1) = E1 cos(theta1). Similarly, we have |p2| = E2, and p2x = |p2|cos(theta2) = E2 cos(theta2). The components of q1 and q2 in the rest frame of the positron are obtained by performing a Lorentz boost with velocity v in the x-direction:

E1' = gamma(v) (E1 - v p1x) = gamma(v) [E1 - v E1 cos(theta1)] = gamma(v) ...

Solution Summary

We present a detail computation of the energies, the speed and the angle.