Consider the pion photoproduction reaction

$$
\gamma+p \rightarrow p+\pi^{\circ}
$$

where the rest energy is $938 \mathrm{MeV}$ for the proton and $135 \mathrm{MeV}$ for the neutral pion.

If the initial proton is at rest in the laboratory, find the laboratory threshold gamma-ray energy for this reaction to "go".
The quantity $E^{2}-P^{2} c^{2}$ is invariant under Lorentz transformation and for an isolated system is the same before and after a reaction. The threshold $\gamma$-ray energy is that for which the final state particles are all at rest in the center of mass frame. Thus

$$
\left(E_{\gamma}+m_{p} c^{2}\right)^{2}-\left(\frac{E_{\gamma}}{c}\right)^{2} c^{2}=\left(m_{p}+m_{\pi}\right)^{2} c^{4}
$$

where $E_{\gamma}$ is the energy of the photon and $\frac{E_{\gamma}}{c}$ its momentum, giving

$$
E_{\gamma}=\frac{\left(m_{\pi}^{2}+2 m_{\pi} m_{p}\right) c^{4}}{2 m_{p} c^{2}}=144.7 \mathrm{MeV}
$$

as the threshold $\gamma$-ray energy.