My problem relates to the decay of a neutral pion, more precisely the decay into two photons.
If the pion the rests the momenta of the two photons are contrarily. If the pion still has some momentum at its decay it will be transferred to the photons. (EDIT: The momenta given by the rest enrgy of the pion will always be contrarily. In case the pion is still moving there will be some extra momentum and the photons will most likely not move in opposite directins)
If the photons appear under a certain angle to the pion's direction I can calculate the photons momenta with simple geometry. But what happens if one photon appears in the same direction as the pion and the other one in the opposite direction. If the pion has a mass of 135MeV/c^2 and a momentum of 135MeV/c the photons share them in equal parts (!?). In case of the back photon the momenta would cancel out thus having no momentum.
How do I handle this scenario?
Postscript: (The superscripted numbers are indizes not exponents) Given a pion with mass $m_\pi$ and momentum $\vec{p_\pi}$. The decay will create two photons $\gamma_1$,$\gamma_2$ that share equal parts of $m_\pi$ (follows from conservation of momentum CoM)
$$E_{\pi_{rest}}=m_\pi c^2 \qquad \Rightarrow \qquad E_\gamma=\frac{E_{\pi_{rest}}}{2}$$ $$\left|\vec{p^1_\gamma}\right|=\frac{E_\gamma}{c}=\frac{m_\pi c}{2} \quad (1)$$
CoM leads to
$$\vec{p^1_{\gamma_1}}=-\vec{p^1_{\gamma_2}} \quad (2)$$
$p_\pi$ will be devided onto both photons (Assume the parts are equal)
$$\vec{p^2_{\gamma_1}}=\vec{p^2_{\gamma_2}}=\frac{\vec{p_{\pi}}}{2} \quad (3)$$
Thus:
$$\vec{p_{\gamma_1}}=\vec{p^1_{\gamma_1}}+\vec{p^2_{\gamma_1}} \\ \vec{p_{\gamma_2}}=\vec{p^1_{\gamma_2}}+\vec{p^2_{\gamma_2}}$$
If now $m_\pi c=\left|\vec{p_\pi}\right| \quad $ and $ \quad \vec{p^1_\gamma} \parallel \vec{p_\pi}$ using (1),(2) and (3)
$$\vec{p_{\gamma_1}}=\frac{\vec{p_{\pi}}}{2}+\frac{\vec{p_{\pi}}}{2}=\vec{p_{\pi}} \\ \vec{p_{\gamma_2}}=\frac{\vec{p_{\pi}}}{2}-\frac{\vec{p_{\pi}}}{2}=0$$
The total momentum is still $p_\pi$ but the second photon has no momentum. How do I interpret this? Is there only one photon or is $p_\pi$ devided unequal?
