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Final Lab Frame Scattered Electron 4-momentum components
We can perform a Lorentz transformation from the Center of Mass frame, with zero total momentum, to the Lab frame.
[math]\left( \begin{matrix}E'_{1}+E'_{2}\\ p'_{1(x)}+p'_{2(x)} \\p'_{1(y)}+ p'_{2(y)} \\ p'_{1(z)}+p'_{2(z)}\end{matrix} \right)=\left(\begin{matrix}\gamma^* & 0 & 0 & \beta^* \gamma^*\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ \beta^* \gamma^* & 0 & 0 & \gamma^* \end{matrix} \right) . \left( \begin{matrix}E^*_{1}+E^*_{2}\\ p^*_{1(x)}+p^*_{2(x)} \\ p^*_{(1(y)}+p^*_{2(y)} \\ p^*_{1(z)}+p^*_{2(z)}\end{matrix} \right)[/math]
| [math]p^*_{1(x)}= p'_{1(x)}[/math]
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| [math]p^*_{1(y)}= p'_{1(y)}[/math]
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[math]\Longrightarrow\begin{cases}
E'=\gamma E^*+\beta \gamma p^*_{z} \\
p'_{z}=\beta \gamma E^*+ \gamma p^*_{z}
\end{cases}[/math]
[math]\Longrightarrow\begin{cases}
\gamma = \frac{E'}{E^*} \\
\beta =\frac{p'_{z}}{E'}
\end{cases}[/math]
Without knowing the values for gamma or beta, we can use the fact that lengths of the two 4-momenta are invariant
[math]s={\mathbf P^*} ^2=(E^*_{1}+E^*_{2})^2-(\vec p\ ^*_{1}+\vec p\ ^*_{2})^2[/math]
[math]s={\mathbf P}^2=(E'_{1}+E'_{2})^2-(\vec p\ '_{1}+\vec p\ '_{2})^2[/math]
Setting these equal to each other, we can use this for the collision of two particles of mass m1 and m2. Since the total momentum is zero in the Center of Mass frame, we can express total energy in the center of mass frame as
[math](E^*_{1}+E^*_{2})^2-(\vec p\ ^*_{1}+\vec p\ ^*_{2})^2=s=(E'_{1}+E'_{2})^2-(\vec p\ '_{1}+\vec p\ '_{2})^2[/math]
[math](E^*)^2-(\vec p\ ^*)^2=(E'_{1}+E'_{2})^2-(\vec p\ ')^2[/math]
[math](E^*)^2=(E')^2-(\vec p\ ')^2[/math]
[math](E^*)^2+(\vec{p'})^2=(E')^2[/math]
[math]E'=\sqrt{(E^*)^2+(\vec p\ ')^2}[/math]
Since momentum is conserved, the initial momentum from the incident electron and stationary electron is still the same in the Lab frame, therefore [math]p\ '_{Lab}=11000 MeV[/math]
[math]E'=\sqrt{(106.031 MeV)^2+(11000 MeV)^2}\approx 11000.511 MeV[/math]
This is the same as the initial total energy in the Lab frame, which should be expected since scattering is considered to be an elastic collision.
Since,
[math]E'\equiv E'_{1}+E'_{2}[/math]
| [math]\Longrightarrow E'_{1}=E'-E'_{2}[/math]
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Using the relation
[math]E'^2\equiv p^2+m^2[/math]
| [math]\Longrightarrow p_{1}^{'\ 2}=E_{1}^{'\ 2}-m_{1}^2=E_{1}^{'\ 2}-(.511 MeV)^2[/math]
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Using
[math]p^2 \equiv p_x^2+p_y^2+p_z^2[/math]
| [math]\Longrightarrow p_{1(z)}'=\sqrt {p_{1}^{'\ 2}-p_{1(x)}^{'\ 2}-p_{1(y)}^{'\ 2}}[/math]
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