I am a little unsure how to answer the following question, Find the energies of two photons emitted in opposite directions along the pion's original line of motion if the pion has a r.m.e of 500MEV and is moving with a Kinetic energy of 0.8 Gev.
I was planing to use the equation $$ 4E1E2 = (mc^2)^2$$ where $m$ is the rest mass of the pion. My issue is, how do I account for the kinetic energy of the pion?
All you need to do is conserve energy and momentum in the lab frame.
Firstly you conserve energy in lab frame:
\begin{equation} E_{\gamma 1} + E_{\gamma2} = E_{\pi} = 1.3GeV \end{equation}
Then you work out what the pion's momentum was (still in the lab frame) using the mass-energy-momentum relation where the $E_\pi$ is the total kinetic and mass energy:
\begin{equation} E_{\pi}^2 = m_{\pi}^2c^4 + p_{\pi}^2c^2 \end{equation}
Then you apply the relation connecting the photon energy and momentum: \begin{equation} p_{\gamma} = \frac{E_{\gamma}}{c} \end{equation}
And conserve momentum (still in the lab frame): \begin{equation} \vec{p}_{\gamma1} + \vec{p}_{\gamma2} = \vec{p}_\pi \end{equation}
Note that $\vec{p}_{\gamma1}$ and $\vec{p}_{\gamma2}$ are going to be in opposite directions and one of them will be in the same direction as $\vec{p}_\pi$ so you can relate the moduli of the vectors like so (if you choose $\gamma_1$ to be the one going "forward":
\begin{equation} p_{\gamma1} - p_{\gamma2} = p_\pi \end{equation}
Between these four equations you can solve for $E_{\gamma1}$ and $E_{\gamma2}$.
