# Photon energies of neutral pion decay

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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 is 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 $$E_{\pi}(1\pm\beta)/2.$$

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