Center of Mass for Stationary Target

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4.1.1 Center of Mass for Stationary Target

For an incoming electron of 11GeV striking a stationary electron we would expect:

Inspecting the Lorentz transformation to the Center of Mass frame:

$\left( \begin{matrix}E^*_{1}+E^*_{2}\\ 0 \\ 0 \\ 0\end{matrix} \right)=\left(\begin{matrix}\gamma^* & 0 & 0 & -\beta^* \gamma^*\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ -\beta^* \gamma^* & 0 & 0 & \gamma^* \end{matrix} \right) . \left( \begin{matrix}E_{1}+E_{2}\\ 0 \\ 0 \\ p_{1(z)}+p_{2(z)}\end{matrix} \right)$

For the case of a stationary electron, this simplifies to:

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

which gives,

$\Longrightarrow\begin{cases} E^*=\gamma^* (E_{1}+m)-\beta^* \gamma^* p_{1(z)} \\ p^*_{z}=-\beta^* \gamma^*(E_{1}+m)+\gamma^* p_{1(z)} \end{cases}$

Solving for $\beta^*$, with $p^*_{z}=0$

$\Longrightarrow \beta^* \gamma^*(E_{1}+m)=\gamma^* p_{1(z)}$

$\Longrightarrow \beta^*=\frac{p_{1}}{(E_{1}+m)}$

Similarly, solving for $\gamma^*$ by substituting in $\beta^*$

$E^*=\gamma^* (E_{1}+m)-\frac{p_{1}}{(E_{1}+m)} \gamma^* p_{1(z)}$

$E^*=\gamma^* \frac{(E_{1}+m)^2}{(E_{1}+m)}-\gamma^*\frac{(p_{1(z)})^2}{(E_{1}+m)}$

Using the fact that $E^*=[(E_{1}+E_{2})^2-(\vec p_{1}+\vec p_{2})^2]^{1/2}$

$E^*=\gamma^* \frac{E^*\ ^2}{(E_{1}+m)}$

$\Longrightarrow \gamma^*=\frac{(E_1+m)} {E^*}$

Using the relation

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

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

$\Longrightarrow\begin{cases} E^*_{2}=\gamma^* (m) \\ p^*_{2(z)}=-\beta^* \gamma^* (m) \end{cases}$

$\Longrightarrow\begin{cases} E^*_{2}=\frac{(E_{1}+m)}{E^*} (m) \\ p^*_{2(z)}=-\frac{p_{1}}{(E_{1}+m)} \frac{(E_{1}+m)}{E^*} (m_{2}) \end{cases}$

$\Longrightarrow\begin{cases} E^*_{2}=\frac{(11000 MeV+.511 MeV)}{106.031 MeV} (.511 MeV) \approx 53.015 MeV\\ p^*_{2(z)}=-\frac{11000 MeV}{106.031 MeV} (.511 MeV) \approx -53.013 MeV \end{cases}$

$\Longrightarrow \left( \begin{matrix} E^*_{1}\\ p^*_{1(x)} \\ p^*_{1(y)} \\ p^*_{1(z)}\end{matrix} \right)=\left(\begin{matrix}\gamma^* & 0 & 0 & -\beta^* \gamma^*\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ -\beta^* \gamma^* & 0 & 0 & \gamma^* \end{matrix} \right) . \left( \begin{matrix}E_{1}\\ 0 \\ 0 \\ p_{1(z)}\end{matrix} \right)$

$\Longrightarrow\begin{cases} E^*_{1}=\gamma^* (E_{1})-\beta^* \gamma^* p_{1(z)} \\ p^*_{1(z)}=-\beta^* \gamma^*(E_{1})+\gamma^* p_{1(z)} \end{cases}$

$\Longrightarrow\begin{cases} E^*_{1}=\frac{(E_{1}+m)}{E^*} (E_{1})-\frac{p_{1(z)}}{(E_{1}+m)} \frac{(E_{1}+m)}{E^*} p_{1(z)} \\ p^*_{1(z)}=-\frac{p_{1}}{(E_{1}+m)} \frac{(E_{1}+m)}{E^*}(E_{1})+\frac{(E_{1}+m)}{E^*} p_{1(z)} \end{cases}$

$\Longrightarrow\begin{cases} E^*_{1}=\frac{(11000 MeV+.511 MeV)}{106.031 MeV} (11000 MeV)-\frac{11000 MeV}{106.031 MeV} 11000 MeV \approx 53.015 MeV\\ p^*_{1(z)}=-\frac{11000 MeV}{106.031 MeV}(11000 MeV)+\frac{(11000 MeV+.511 MeV)}{106.031 MeV} 11000 MeV \approx 53.013 MeV \end{cases}$

$p^*_{1} =\sqrt {(p^*_{1(x)})^2+(p^*_{1(y)})^2+(p^*_{1(z)})^2} \Longrightarrow p^*_{1}=p^*_{1(z)}$

$p^*_{2} =\sqrt {(p^*_{2(x)})^2+(p^*_{2(y)})^2+(p^*_{2(z)})^2} \Longrightarrow p^*_{2}=p^*_{2(z)}$

This gives the momenta of the particles in the center of mass to have equal magnitude, but opposite directions.

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  Figure 5.1.3: Initial momentum for lab frame electron incident at 11GeV as seen in the center of mass frame.

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  Figure 5.1.4: Initial momentum for a lab frame stationary particle as seen in the center of mass frame.

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  Figure 5.1.5: Initial incoming angle theta for a lab frame electron incident at 11GeV as seen in the center of mass frame.

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  Figure 5.1.6: Initial incoming angle theta for a lab frame stationary electron as seen in the center of mass frame.

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