Final Lab Frame Scattered Electron 4-momentum components

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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]

[math]p^*_{1(y)}= p'_{1(y)}[/math]

[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]

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]

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]