Hint 8

By symmetry both photons must have the same energy $E_\gamma$. Taking the ${\bf e}_1$ axis to be the direction of the pion's motion, the 4-momenta of the photons are:

$${\sf p}_{\gamma 1}=(E_\gamma/c)(\cos\theta,\,\sin\theta,\,0,\... ... p}_{\gamma 2}=(E_\gamma/c) (\cos\theta,\,-\sin\theta,\,0,\,i).$$

Conservation of 4-momentum implies that the sum of these is equal to the 4-momentum of the original pion,

$${\sf p}_{\pi}=\gamma(v) m_\pi(v,0,0,ic).$$

Hence energy conservation requires that $2E_\gamma=\gamma(v) m_\pi c^2$and so $E_\gamma=339$ MeV.

To find the angle $\theta$, we can square the total 4-momentum to get

$$({\sf p}_{\gamma 1}+{\sf p}_{\gamma 2})^2 =2{\sf p}_{\gamma 1}\cdot{\sf p}_{\gamma 2}=-m_\pi^2 c^2.$$

This gives $\theta=0.2$ rad. (An alternative method to get $\theta$ is to use $p_1$ conservation.)