 Solution

At the threshold found in i, both products are stationary in the centre-of-momentum frame, so they have the same velocity in the laboratory frame. Take the energetic incident photon along $+x$. The total energy is $E+\epsilon$ and total momentum is $(E-\epsilon )/c$, so the centre-of-momentum frame moves atFor the numerical threshold,This is extremely close to, but strictly below, the speed of light. Each product has energy $(E+\epsilon )/2$ in the laboratory frame. Above threshold the same ratio of total momentum to total energy determines the motion of the pair's centre-of-momentum frame, but the individual electron and Positron need not have a common velocity.