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.

$\left( \begin{matrix}E'_{1}+E'_{2}\\ p'_{1(x)}+p'_{2(x)} \\p'_{1(y)}+ p'_{2(y)} \\ p'_{1(z)}+p'_{2(z)}\end{matrix} \right)=\left(\begin{matrix}\gamma^* & 0 & 0 & \beta^* \gamma^*\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ \beta^* \gamma^* & 0 & 0 & \gamma^* \end{matrix} \right) . \left( \begin{matrix}E^*_{1}+E^*_{2}\\ p^*_{1(x)}+p^*_{2(x)} \\ p^*_{(1(y)}+p^*_{2(y)} \\ p^*_{1(z)}+p^*_{2(z)}\end{matrix} \right)$

$p^*_{1(x)}= p'_{1(x)}$

$p^*_{1(y)}= p'_{1(y)}$

$\Longrightarrow\begin{cases} E'=\gamma E^*+\beta \gamma p^*_{z} \\ p'_{z}=\beta \gamma E^*+ \gamma p^*_{z} \end{cases}$

$\Longrightarrow\begin{cases} \gamma = \frac{E'}{E^*} \\ \beta =\frac{p'_{z}}{E'} \end{cases}$

Without knowing the values for gamma or beta, we can use the fact that lengths of the two 4-momenta are invariant

$s={\mathbf P^*} ^2=(E^*_{1}+E^*_{2})^2-(\vec p\ ^*_{1}+\vec p\ ^*_{2})^2$

$s={\mathbf P}^2=(E'_{1}+E'_{2})^2-(\vec p\ '_{1}+\vec p\ '_{2})^2$

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

$(E^*_{1}+E^*_{2})^2-(\vec p\ ^*_{1}+\vec p\ ^*_{2})^2=s=(E'_{1}+E'_{2})^2-(\vec p\ '_{1}+\vec p\ '_{2})^2$

$(E^*)^2-(\vec p\ ^*)^2=(E'_{1}+E'_{2})^2-(\vec p\ ')^2$

$(E^*)^2=(E')^2-(\vec p\ ')^2$

$(E^*)^2+(\vec{p'})^2=(E')^2$

$E'=\sqrt{(E^*)^2+(\vec p\ ')^2}$

Since momentum is conserved, the initial momentum from the incident electron and stationary electron is still the same in the Lab frame, therefore $p\ '_{Lab}=11000 MeV$

$E'=\sqrt{(106.031 MeV)^2+(11000 MeV)^2}\approx 11000.511 MeV$

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,

$E'\equiv E'_{1}+E'_{2}$

$\Longrightarrow E'_{1}=E'-E'_{2}$

Using the relation

$E'^2\equiv p^2+m^2$

$\Longrightarrow p_{1}^{'\ 2}=E_{1}^{'\ 2}-m_{1}^2=E_{1}^{'\ 2}-(.511 MeV)^2$

Using

$p^2 \equiv p_x^2+p_y^2+p_z^2$

$\Longrightarrow p_{1(z)}'=\sqrt {p_{1}^{'\ 2}-p_{1(x)}^{'\ 2}-p_{1(y)}^{'\ 2}}$