# Photon energies after neutral pion decay

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

• Without explicitly answering, you should first consider the rest-frame of the pion (where $E_1 = E_2 = m_\pi/2$ and then make the Lorentz transformation in the direction of the $\pi^0$ direction to get the answer in the lab frame. – Paganini Jan 10 '15 at 18:23

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).
Consider a pion with initial quadrimomentum (in units such that $c=1=\hbar$) $$p=(E,\mathbf{p})$$ and two final photons with quadrimomenta $$k_1=(\omega_1, \mathbf{k}_1) \qquad k_2=(\omega_2, \mathbf{k}_2).$$ Conservation of energy and momentum: $p=k_1+k_2.$ Thus: $$k_2=p-k_1$$ squaring and using the mass-shell relation $p^2=m_\pi^2$, $k_1^2=0=k_2^2$ $$0=m_\pi^2-2p\cdot k_1$$ where $\cdot$ denotes the 4-dimensional vector product defined by $\eta=diag(+1,-1,-1,-1).$ Expanding we have $$m_\pi^2=2E\omega_1-2|\mathbf{p}|\omega_1\cos\theta$$ where we have used $\omega_1=|\mathbf{k}_1|$, and defined $\theta$ as scattering angle. Therefore: $$\omega_1(E, \theta)=\frac{m_\pi^2}{2\left(E-\sqrt{E^2-m_\pi^2}\cos\theta\right)}=|\mathbf{k_1}|.$$ Going back to $c\not=1\not=\hbar$ units we have: $$\hbar\omega_1(E, \theta)=\frac{m_\pi^2c^4}{2\left(E-\sqrt{E^2-m_\pi^2c^4}\cos\theta\right)}=|\mathbf{k_1}|.$$