Calculations of 4-momentum components
Initial Lab Frame
Assume the incoming electron has momentum of 11000 MeV in the positive z direction.
[math]p_{(e^-)z Lab}\equiv p_{(e^-)Lab}=11000 MeV[/math]
The moller electron is initially at rest
[math]p_{(m)Lab}\equiv 0[/math]
The total energy in this frame
[math]E_{Lab}\equiv \sqrt{p_{Lab}^2+m_{Lab}^2}[/math]
[math]E_{Lab}\equiv \sqrt{(p_{(e^-)Lab}+p_{(m)Lab})^2+(m_{(e^-)Lab}+m_{(m)Lab})^2}[/math]
[math]E_{Lab}\equiv \sqrt{(11000 MeV)^2+(.511 MeV+.511 MeV)^2}\approx 11000 MeV[/math]
Center of Mass Frame
Using the definition
[math]E^2\equiv p^2c^2+m^2c^4[/math]
[math]\Longrightarrow p^2=\frac {E^2}{c^2}-m^2 c^2[/math]
We can use 4-momenta vectors, i.e. [math]{\mathbf P}\equiv \left(\begin{matrix} E/c\\ p_x \\ p_y \\ p_z \end{matrix} \right)[/math]
we can use the fact that the scalar product of a 4-momenta with itself,
[math]{\mathbf P_1}\cdot {\mathbf P_1}=E_1E_1-\vec p_1\cdot \vec p_1 c^2=m^2c^4=s[/math]
is invariant
Using this, the sum of two 4-momenta forms a 4-vector as well
[math]{\mathbf P_1}+ {\mathbf P_2}= \left( \begin{matrix}E_1+E_2\\\vec p_1 c+\vec p_2 c\end{matrix} \right)= {\mathbf P}[/math]
The length of this four-vector is an invariant as well
[math]{\mathbf P^2}=({\mathbf P_1}+{\mathbf P_2})^2=(E_1+E_2)^2-(\vec p_1 c+\vec p_2 c)^2=s[/math]
[math]{\mathbf P^2}=({\mathbf P_1}+{\mathbf P_2})^2=(E_1+E_2)^2-(\vec p_1 +\vec p_2 )^2=s[/math]
(with c=1)
For incoming electrons moving only in the z-direction, we can write
[math]{\mathbf P_1}+ {\mathbf P_2}= \left( \begin{matrix}E_1+E_2\\ 0 \\ 0 \\ p_{1z}+p_{2z}\end{matrix} \right)={\mathbf P}[/math]
We can perform a Lorentz transformation to the Center of Mass frame, with zero total momentum
[math]\left( \begin{matrix}E_{1 CM}+E_{2 CM}\\ 0 \\ 0 \\ 0\end{matrix} \right)=\left(\begin{matrix}\gamma_{CM} & 0 & 0 & -\beta_{CM} \gamma_{CM}\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ -\beta_{CM} \gamma_{CM} & 0 & 0 & \gamma_{CM} \end{matrix} \right) . \left( \begin{matrix}E_{1 Lab}+E_{2 Lab}\\ 0 \\ 0 \\ p_{1z Lab}+p_{2z Lab}\end{matrix} \right)[/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}_{CM}^2=(E_{1 CM}+E_{2 CM})^2-(\vec p_{1 CM}+\vec p_{2 CM})^2[/math]
[math]s={\mathbf P}_{Lab}^2=(E_{1 Lab}+E_{2 Lab})^2-(\vec p_{1 Lab}+\vec p_{2 Lab})^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 CM}+E_{2 CM})^2-(\vec p_{1 CM}+\vec p_{2 CM})^2=s=(E_{1 Lab}+E_{2 Lab})^2-(\vec{p_{1 Lab}}+\vec p_{2 Lab})^2[/math]
[math](E_{CM})^2-(\vec p_{CM})^2=(E_{1 Lab}+E_{2 Lab})^2-(\vec{p_{1 Lab}}+\vec p_{2 Lab})^2[/math]
[math](E_{CM})^2=(E_{1 Lab}+E_{2 Lab})^2-(\vec{p_{1 Lab}}+\vec p_{2 Lab})^2[/math]
[math]E_{CM}=[(E_{1 Lab}+E_{2 Lab})^2-(\vec{p_{1 Lab}}+\vec p_{2 Lab})^2]^{1/2}[/math]
[math]E_{CM}=[E_{1 Lab}^2+2E_{1 Lab}E_{2 Lab}+E_{2 Lab}^2-\vec p_{1 Lab} . \vec p_{2 Lab} -\vec p_{1 Lab} . \vec p_{1 Lab} -\vec p_{2 Lab} . \vec p_{1 Lab} -\vec p_{2 Lab} . \vec p_{2 Lab} ]^{1/2}[/math]
[math]E_{CM}=[(E_{1 Lab}^2- p_{1 Lab}^2 )+(E_{2 Lab}^2-p_{2 Lab}^2 )+2E_{1 Lab}E_{2 Lab}-\vec p_{1 Lab} . \vec p_{2 Lab} -\vec p_{2 Lab} . \vec p_{1 Lab} ]^{1/2}[/math]
[math]E_{CM}=[(E_{1 Lab}^2- p_{1 Lab}^2 )+(E_{2 Lab}^2-p_{2 Lab}^2 )+2E_{1 Lab}E_{2 Lab}-p_{1 Lab} p_{2 Lab}\cos(\theta_{Lab}) - p_{2 Lab} p_{1 Lab}\cos(\theta_{Lab}) ]^{1/2}[/math]
[math]E_{CM}=[m_{1 Lab}^2+m_{2 Lab}^2+2E_{1 Lab}E_{2 Lab}-p_{1 Lab} p_{2 Lab}\cos(\theta_{Lab}) - p_{2 Lab} p_{1 Lab}\cos(\theta_{Lab}) ]^{1/2}[/math]
[math]E_{CM}=[m_{1 Lab}^2+m_{2 Lab}^2+2E_{1 Lab}E_{2 Lab}-2p_{1 Lab} p_{2 Lab}\cos(\theta_{Lab}) ]^{1/2}[/math]
[math]E_{CM}=[m_{1 Lab}^2+m_{2 Lab}^2+2E_{1 Lab}E_{2 Lab}-2p_{1 Lab} p_{2 Lab}\cos(\theta_{Lab}) ]^{1/2}[/math]
Using the relations [math]\beta\equiv \vec p/E\Longrightarrow \vec p_{Lab}=\beta_{Lab} E_{Lab}[/math]
[math]E_{CM}=[m_{1 Lab}^2+m_{2 Lab}^2+2E_{1 Lab}E_{Lab}2(1-\beta_{1 Lab}\beta_{2 Lab}\cos(\theta_{Lab}))]^{1/2}[/math]
where [math] θ_{Lab} [/math] is the angle between the particles in the Lab frame.
[math]E_{CM}=[m_1^2+m_2^2+2E_{1 Lab}E_{2 Lab}(1-\beta_{1 Lab}\beta_{2 Lab}\cos(\theta_{Lab}))]^{1/2}[/math]
In the frame where one particle (m2 Lab) is at rest
[math]\Longrightarrow \beta_{2 Lab}=0[/math]
[math]\Longrightarrow p_{2 Lab}=0[/math]
which implies,
[math] E_{2 Lab}=[p_{2 Lab}^2+m_{2 Lab}^2]^{1/2}=m_{2 Lab}[/math]
| [math]E_{CM}=(m_{1 Lab}^2+m_{2 Lab}^2+2E_{1 Lab} m_{2 Lab})^{1/2}=(.511MeV^2+.511MeV^2+2(\sqrt{(11000 MeV)^2+(.511 MeV)^2})(.511 MeV))^{1/2}\approx 106.030760886 MeV[/math]
|
where [math]E_{1 Lab}=\sqrt{p_{1 Lab}^2+m_{1 Lab}^2}\approx 11000 MeV[/math]
Inspecting the Lorentz transformation to the Center of Mass frame:
[math]\left( \begin{matrix}E_{1 CM}+E_{2 CM}\\ 0 \\ 0 \\ 0\end{matrix} \right)=\left(\begin{matrix}\gamma_{CM} & 0 & 0 & -\beta_{CM} \gamma_{CM}\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ -\beta_{CM} \gamma_{CM} & 0 & 0 & \gamma_{CM} \end{matrix} \right) . \left( \begin{matrix}E_{1 Lab}+E_{2 Lab}\\ 0 \\ 0 \\ p_{1z Lab}+p_{2z Lab}\end{matrix} \right)[/math]
For the case of a stationary electron, this simplifies to:
[math]\left( \begin{matrix} E_{CM} \\ p_{xCM} \\ p_{y CM} \\ p_{z CM}\end{matrix} \right)=\left(\begin{matrix}\gamma_{CM} & 0 & 0 & -\beta_{CM} \gamma\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ -\beta_{CM} \gamma_{CM} & 0 & 0 & \gamma_{CM} \end{matrix} \right) . \left( \begin{matrix}E_{1 Lab}+m_{2Lab}\\ 0 \\ 0 \\ p_{1z Lab}+0\end{matrix} \right)[/math]
which gives,
[math]\Longrightarrow\begin{cases}
E_{CM}=\gamma_{CM} (E_{1 Lab}+m_{2 Lab})-\beta_{CM} \gamma_{CM} p_{1z Lab} \\
p_{zCM}=-\beta_{CM} \gamma_{CM}(E_{1 Lab}+m_{2 Lab})+\gamma_{CM} p_{1z Lab}
\end{cases}[/math]
Solving for [math]\beta_{CM}[/math], with [math]p_{zCM}=0[/math]
[math]\Longrightarrow \beta_{CM} \gamma_{CM}(E_{1 Lab}+m_{2 Lab})=\gamma_{CM} p_{1z Lab}[/math]
| [math]\Longrightarrow \beta_{CM}=\frac{p_{1 Lab}}{(E_{1 Lab}+m_{2 Lab})}[/math]
|
Similarly, solving for [math]\gamma_{CM}[/math] by substituting in [math]\beta_{CM}[/math]
[math]E_{CM}=\gamma_{CM} (E_{1 Lab}+m_{2 Lab})-\frac{p_{1 Lab}}{(E_{1 Lab}+m_{2 Lab})} \gamma_{CM} p_{1z Lab}[/math]
[math]E_{CM}=\gamma_{CM} \frac{(E_{1 Lab}+m_2)^2}{(E_{1 Lab}+m_{2 Lab})}-\gamma_{CM}\frac{(p_{1z Lab})^2}{(E_{1 Lab}+m_{2 Lab})}[/math]
Using the fact that [math]E_{CM}=[(E_{1 Lab}+E_{2 Lab})^2-(\vec p_{1 Lab}+\vec p_{2 Lab})^2]^{1/2}[/math]
[math]E_{CM}=\gamma_{CM} \frac{E_{CM}^2}{(E_{1 Lab}+m_{2 Lab})}[/math]
| [math]\Longrightarrow \gamma_{CM}=\frac{(E_{1 Lab}+m_{2 Lab})}{E_{CM}}[/math]
|
This gives the momenta of the particles in the center of mass to have equal magnitude, but opposite directions
Using the relation
[math]\left( \begin{matrix} E_{1 CM}+E_{2 CM} \\ p_{1xCM}+p_{2xCM} \\ p_{1y CM}+p_{2yCM} \\ p_{1z CM}+p_{2zCM}\end{matrix} \right)=\left(\begin{matrix}\gamma_{CM} & 0 & 0 & -\beta_{CM} \gamma_{CM}\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ -\beta_{CM} \gamma_{CM} & 0 & 0 & \gamma_{CM} \end{matrix} \right) . \left( \begin{matrix}E_{1 Lab}+m_{2 Lab}\\ 0 \\ 0 \\ p_{1z Lab}+0\end{matrix} \right)[/math]
[math]\Longrightarrow \left( \begin{matrix} E_{2 CM} \\ p_{2xCM} \\ p_{2yCM} \\ p_{2zCM}\end{matrix} \right)=\left(\begin{matrix}\gamma_{CM} & 0 & 0 & -\beta_{CM} \gamma_{CM}\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ -\beta_{CM} \gamma_{CM} & 0 & 0 & \gamma_{CM} \end{matrix} \right) . \left( \begin{matrix}m\\ 0 \\ 0 \\ 0\end{matrix} \right)[/math]
[math]\Longrightarrow\begin{cases}
E_{2CM}=\gamma_{CM} (m_{2 lab}) \\
p_{2zCM}=-\beta_{CM} \gamma_{CM} (m_{2 Lab})
\end{cases}[/math]
[math]\Longrightarrow\begin{cases}
E_{2CM}=\frac{(E_{1 Lab}+m_{2 Lab})}{E_{CM}} (m_2) \\
p_{2zCM}=-\frac{p_{1 Lab}}{(E_{1 Lab}+m_2)} \frac{(E_{1 Lab}+m_{2 Lab})}{E_{CM}} (m_{2 Lab})
\end{cases}[/math]
[math]\Longrightarrow\begin{cases}
E_{2CM}=\frac{(11000 MeV+.511 MeV)}{106.031 MeV} (.511 MeV) \approx 53.0129177 MeV\\
p_{2zCM}=-\frac{11000 MeV}{106.031 MeV} (.511 MeV) \approx -53.013 MeV
\end{cases}[/math]
[math]\Longrightarrow \left( \begin{matrix} E_{1 CM}\\ p_{1xCM} \\ p_{1y CM} \\ p_{1z CM}\end{matrix} \right)=\left(\begin{matrix}\gamma_{CM} & 0 & 0 & -\beta_{CM} \gamma_{CM}\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ -\beta_{CM} \gamma_{CM} & 0 & 0 & \gamma_{CM} \end{matrix} \right) . \left( \begin{matrix}E_{1 Lab}\\ 0 \\ 0 \\ p_{1z Lab}\end{matrix} \right)[/math]
[math]\Longrightarrow\begin{cases}
E_{1CM}=\gamma_{CM} (E_{1 Lab})-\beta_{CM} \gamma_{CM} p_{1z Lab} \\
p_{1zCM}=-\beta_{CM} \gamma(E_{1 Lab})+\gamma_{CM} p_{1z Lab}
\end{cases}[/math]
[math]\Longrightarrow\begin{cases}
E_{1CM}=\frac{(E_{1 Lab}+m_{2 Lab})}{E_{CM}} (E_{1 Lab})-\frac{p_{1zLab}}{(E_{1 Lab}+m_2)} \frac{(E_{1 Lab}+m_{2 Lab})}{E_{CM}} p_{1z Lab} \\
p_{1zCM}=-\frac{p_{1 Lab}}{(E_{1 Lab}+m_2)} \frac{(E_{1 Lab}+m_{2 Lab})}{E_{CM}}(E_{1 Lab})+\frac{(E_{1 Lab}+m_{2 Lab})}{E_{CM}} p_{1z Lab}
\end{cases}[/math]
[math]\Longrightarrow\begin{cases}
E_{1CM}=\frac{(11000 MeV+.511 MeV)}{106.031 MeV} (11000 MeV)-\frac{11000 MeV}{106.031 MeV} 11000 MeV \approx 53.013 MeV\\
p_{1zCM}=-\frac{11000 MeV}{106.031 MeV}(11000 MeV)+\frac{(11000 MeV+.511 MeV)}{106.031 MeV} 11000 MeV \approx 53.013 MeV
\end{cases}[/math]
[math]p_{1 CM} =\sqrt{p_{1xCM}^2+p_{1yCM}^2+p_{1zCM}^2}\Longrightarrow p_{1 CM}=p_{1zCM}[/math]
[math]p_{2 CM} =\sqrt{p_{2xCM}^2+p_{2yCM}^2+p_{2zCM}^2}\Longrightarrow p_{2 CM}=p_{2zCM}[/math]
| [math] p_{1 CM}=p_{2 CM}\approx 53.013 MeV[/math]
|
| [math]E_{1 CM}= E_{2 CM}\approx 53.013 MeV[/math]
|
Moller electron Lab Frame
Finding the correct kinematic values starting from knowing the momentum of the Moller electron, [math]p_{(m)Lab}[/math] , in the Lab frame,
Using [math]\theta=\arccos(\frac{p_{z(m) Lab}}{p_{(m)Lab}})[/math]
| [math]\Longrightarrow {p_{z(m) Lab}=p_{(m)Lab}\cos(\theta)}[/math]
|
Checking on the sign resulting from the cosine function, we are limited to:
| [math]0^\circ \le \theta \le 60^\circ \equiv 0 \le \theta \le 1.046\ Radians[/math]
|
Since, [math]\frac{p_{(m)zLab}}{p_{(m)Lab}}=cos(\theta)\Longrightarrow p_{(m)zLab}\ should\ always\ be\ positive[/math]
Similarly, [math]\phi=\arccos(\frac{p_{x(m) Lab}}{p_{xy(m) Lab}})[/math]
where [math]p_{xy(m) Lab}=\sqrt{p_{x(m) Lab}^2+p_{y(m)Lab}^2}[/math]
[math]p_{xy(m) Lab}=(p_{x(m) Lab}^2+p_{y(m) Lab}^2)^2[/math]
and using [math]p_{(m)Lab}^2=p_{x(m) Lab}^2+p_{y(m) Lab}^2+p_{z(m) Lab}^2[/math]
this gives [math]p_{(m)Lab}^2=p_{xy(m) Lab}^2+p_{z(m) Lab}^2[/math]
[math]\Longrightarrow p_{(m)Lab}^2-p_{z(m) Lab}^2=p_{xy(m) Lab}^2[/math]
[math]\Longrightarrow p_{xy(m) Lab}=\sqrt{p_{(m)Lab}^2-p_{z(m) Lab}^2}[/math]
which gives[math]\phi = \arccos(\frac{p_{x(m) Lab}}{\sqrt{p_{(m)Lab}^2-p_{z(m) Lab}^2}})[/math]
| [math]\Longrightarrow p_{x(m) Lab}=\sqrt{p_{(m)Lab}^2-p_{z(m) Lab}^2} \cos(\phi)[/math]
|
Similarly, using [math]p_{(m)Lab}^2=p_{x(m) Lab}^2+p_{y(m) Lab}^2+p_{z(m) Lab}^2[/math]
[math]\Longrightarrow p_{(m)Lab}^2-p_{x(m) Lab}^2-p_{z(m) Lab}^2=p_{y(m) Lab}^2[/math]
| [math]p_{y(m) Lab}=\sqrt{p_{(m)Lab}^2-p_{x(m) Lab}^2-p_{z(m) Lab}^2}[/math]
|
Checking on the sign from the cosine results for [math]\phi[/math]
| [math]-\pi \le \phi \le \pi\ Radians[/math]
|
| [math]For\ 0 \le \phi \le \frac{-\pi}{2}\ Radians[/math]
|
| x=POSITIVE
|
| y=NEGATIVE
|
| [math]For\ 0 \le \phi \le \frac{\pi}{2}\ Radians[/math]
|
| x=POSITIVE
|
| y=POSITIVE
|
| [math]For\ \frac{-\pi}{2} \le \phi \le -\pi\ Radians[/math]
|
| x=NEGATIVE
|
| y=NEGATIVE
|
| [math]For\ \frac{\pi}{2} \le \phi \le \pi\ Radians[/math]
|
| x=NEGATIVE
|
| y=POSITIVE
|
Moller electron Center of Mass Frame
Relativistically, the x and y components remain the same in the conversion from the Lab frame to the Center of Mass frame, since the direction of motion is only in the z direction.
| [math]p_{x(m) CM}\Leftrightarrow p_{x(m) Lab}[/math]
|
| [math]p_{y(m) CM}\Leftrightarrow p_{y(m) Lab}[/math]
|
| [math]p_{z(m) CM}=\sqrt{p_{(m)CM}^2-p_{x(m) CM}^2-p_{y(m) CM}^2}[/math]
|
where previously it was shown
| [math] p_{(m) CM}\approx 53.013 MeV[/math]
|
| [math]E_{(m) CM}\approx 53.013 MeV[/math]
|
Electron Center of Mass Frame
Relativistically, the x, y, and z components have the same magnitude, but opposite direction, in the conversion from the Moller electron's Center of Mass frame to the electron's Center of Mass frame.
| [math]p_{x(m) CM}= -p_{x(e^-) CM}[/math]
|
| [math]p_{y(m) CM}= -p_{y((e^-) CM}[/math]
|
| [math]p_{z(m) CM}=-p_{z((e^-) CM}[/math]
|
where previously it was shown
| [math]p_{(e^-) CM}\approx 53.013 MeV[/math]
|
| [math]E_{(e^-) CM}\approx 53.013 MeV[/math]
|
Electron Lab Frame
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_{(e^-) Lab}+E_{(m) Lab}\\ p_{(e^-)z Lab}+p_{(m)z Lab} \\p_{(e^-)z Lab}+ p_{(m)z Lab} \\ p_{(e^-)z Lab}+p_{(m)z Lab}\end{matrix} \right)=\left(\begin{matrix}\gamma_{CM} & 0 & 0 & \beta_{CM} \gamma_{CM}\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ \beta_{CM} \gamma_{CM} & 0 & 0 & \gamma_{CM} \end{matrix} \right) . \left( \begin{matrix}E_{(e^-) CM}+E_{(m) CM}\\ p_{(e^-)x CM}+p_{(m)x CM} \\ p_{(e^-)y CM}+p_{(m)y CM} \\ p_{(e^-)z CM}+p_{(m)z CM}\end{matrix} \right)[/math]
| [math]p_{x(e^-) CM}= p_{x(e^-) Lab}[/math]
|
| [math]p_{y(e^-) CM}= p_{y((e^-) Lab}[/math]
|
[math]\Longrightarrow\begin{cases}
E_{Lab}=\gamma_{Lab} (E_{ CM})+\beta_{Lab} \gamma_{Lab} p_{z CM} \\
p_{z Lab}=\beta_{Lab} \gamma_{Lab}(E_{ CM})+ \gamma_{Lab}p_{z CM}
\end{cases}[/math]
[math]\Longrightarrow\begin{cases}
\gamma_{Lab} = \frac{E_{ Lab}}{E_{CM}} \\
\beta_{Lab} =\frac{p_{zLab}}{E_{LAB}}
\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}_{CM}^2=(E_{(e^-) CM}+E_{(m) CM})^2-(\vec p_{(e^-) CM}+\vec p_{(m) CM})^2[/math]
[math]s={\mathbf P}_{Lab}^2=(E_{(e^-) Lab}+E_{(m) Lab})^2-(\vec p_{(e^-) Lab}+\vec p_{(m) Lab})^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 CM}+E_{2 CM})^2-(\vec p_{1 CM}+\vec p_{2 CM})^2=s=(E_{1 Lab}+E_{2 Lab})^2-(\vec{p_{1 Lab}}+\vec p_{2 Lab})^2[/math]
[math](E_{CM})^2-(\vec p_{CM})^2=(E_{1 Lab}+E_{2 Lab})^2-(\vec{p_{Lab}})^2[/math]
[math](E_{CM})^2=(E_{Lab})^2-(\vec{p_{Lab}})^2[/math]
[math](E_{CM})^2+(\vec{p_{Lab}})^2=(E_{Lab})^2[/math]
[math]E_{Lab}=\sqrt{(E_{CM})^2+(\vec{p_{Lab}})^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_{Lab}=\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_{Lab}\equiv E_{(e^-)Lab}+E_{(m)Lab}[/math]
| [math]\Longrightarrow E_{(e^-)Lab}=E_{Lab}-E_{(m)Lab}[/math]
|
Using the relation
[math]E^2\equiv p^2+m^2[/math]
| [math]\Longrightarrow p_{(e^-)Lab}^2=E_{(e^-)Lab}^2-m_{(e^-)Lab}^2=E_{(e^-)Lab}^2-(.511 MeV)^2[/math]
|
Using
[math]p^2 \equiv p_x^2+p_y^2+p_z^2[/math]
| [math]\Longrightarrow p_{(e^-)z Lab}=\sqrt{p_{(e^-)Lab}^2-p_{(e^-)xLab}^2-p_{(e^-)Lab}^2}[/math]
|
DV_MollerTrackRecon#Calculations_of_4-momentum_components
