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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 $E$ 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 $E_{\gamma2}$.

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 $E_{\gamma2}$.

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} = 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 $E$ is the total kinetic and mass energy:

\begin{equation} E_{\pi}^2 = m_{\pi}^2c^4 + p_{\pi}^2c^2 \end{equation}

Then you conserve momentum (still in the lab frame): \begin{equation} \vec{p}_{\gamma1} + \vec{p}_{\gamma2} = \vec{p}_\pi \end{equation}

And apply the relation connecting the photon energy and momentum: \begin{equation} p_{\gamma} = \frac{E_{\gamma}}{c} \end{equation}

Between these four equations you can solve for $E_{\gamma1}$ and $E_{\gamma2}$. 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 $\vec{p}_\pi$.

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 $E$ 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 $E_{\gamma2}$.

You can use the conservation of energy in the rest frame of the decaying pion to find the energies (and momenta) of the photons then transform back into the lab frame to see what the energies look like there.

Alternatively you can apply the conservation of four momentum in the lab frame to start with which will be a bit simpler if you've learned about four momentum.

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} = 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 $E$ is the total kinetic and mass energy:

\begin{equation} E_{\pi}^2 = m_{\pi}^2c^4 + p_{\pi}^2c^2 \end{equation}

Then you conserve momentum (still in the lab frame): \begin{equation} \vec{p}_{\gamma1} + \vec{p}_{\gamma2} = \vec{p}_\pi \end{equation}

And apply the relation connecting the photon energy and momentum: \begin{equation} p_{\gamma} = \frac{E_{\gamma}}{c} \end{equation}

Between these four equations you can solve for $E_{\gamma1}$ and $E_{\gamma2}$. 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 $\vec{p}_\pi$.