Photon energies after neutral pion decay

Question

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

Answer 1

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,

1. What is the rest energy of the pion?
2. Two photons are generated here, what do the conservation laws tell you about the direction and energy of the daughter products?

Once you have those, the answer you are looking for is easily found by boosting to the lab frame.

Answer 2

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: $$ \hbar\omega_1(E, \theta)=\frac{m_\pi^2c^4}{2\left(E-\sqrt{E^2-m_\pi^2c^4}\cos\theta\right)}=|\mathbf{k_1}|. $$