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

Closed as per community consensus as the post is undergraduate level asked Jan 10, 2015
closed Jun 22, 2015

It might be easier if you solve it in rest frame and then transform back to lab frame. Anyway this is not graduate-level-upwards, vote to close. @Upvoters please do the voting in the linked post.

What is it then?