I am trying to find the photon energies of the decay $\pi_0 \rightarrow \gamma\gamma$ and their dependence on the pion energy $E_{\pi}$, its initial velocity $\beta$ and the scattering angle between the photon and initial pion trajectory $\theta$ in the lab frame.
Assuming ($\star$) one photon travels in the direction that the $\pi_0$ was travelling, I can get the photon energies with conservation of energy and momentum like this:
$$E_{\pi}=E_1+E_2$$
$$p_{\pi} = \frac{1}{c}(E_1-E_2)\quad \text{with} \quad p_{\gamma_{1,2}}=\frac{E_{\gamma_{1,2}}}{c}$$
to
$$E_{1,2}=\frac{1}{2}(E_{\pi}\pm cp_{\pi})$$
But ($\star$) can't be the general answer because in the laboratory frame, the photons might be emitted at an angle $\theta$ to the original $\pi_0$ direction. So I thought I'd say $$p_{\pi}=p_{1,2}\cos\theta$$
which would change my result to:
$$E_{1,2}=\frac{E_{\pi}}{2\pm\cos\theta}.$$
Can anyone confirm this result? I am missing an explicit dependency on the initial $\pi_0$ velocity $\beta$ here. Because the next step would be to confirm the photon energies are limited by $$E_{\pi}(1\pm\beta)/2.$$
Best Answer
These type of particle decay problems are almost always easiest to consider by starting in the center of momentum frame of the parent particle (the $\pi^0$ here).
Your basic analysis should start with,
Once you have those, the answer you are looking for is easily found by boosting to the lab frame.