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:

[math]\displaystyle{ \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) }[/math]

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

[math]\displaystyle{ \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) }[/math]

which gives,

[math]\displaystyle{ \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} }[/math]

Solving for [math]\displaystyle{ \beta^* }[/math], with [math]\displaystyle{ p^*_{z}=0 }[/math]

[math]\displaystyle{ \Longrightarrow \beta^* \gamma^*(E_{1}+m)=\gamma^* p_{1(z)} }[/math]

[math]\displaystyle{ \Longrightarrow \beta^*=\frac{p_{1}}{(E_{1}+m)} }[/math]

Similarly, solving for [math]\displaystyle{ \gamma^* }[/math] by substituting in [math]\displaystyle{ \beta^* }[/math]

[math]\displaystyle{ E^*=\gamma^* (E_{1}+m)-\frac{p_{1}}{(E_{1}+m)} \gamma^* p_{1(z)} }[/math]

[math]\displaystyle{ E^*=\gamma^* \frac{(E_{1}+m)^2}{(E_{1}+m)}-\gamma^*\frac{(p_{1(z)})^2}{(E_{1}+m)} }[/math]

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

[math]\displaystyle{ E^*=\gamma^* \frac{E^*\ ^2}{(E_{1}+m)} }[/math]

[math]\displaystyle{ \Longrightarrow \gamma^*=\frac{(E_1+m)} {E^*} }[/math]

Using the relation

[math]\displaystyle{ \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) }[/math]

[math]\displaystyle{ \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) }[/math]

[math]\displaystyle{ \Longrightarrow\begin{cases} E^*_{2}=\gamma^* (m) \\ p^*_{2(z)}=-\beta^* \gamma^* (m) \end{cases} }[/math]

[math]\displaystyle{ \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} }[/math]

[math]\displaystyle{ \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} }[/math]

[math]\displaystyle{ \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) }[/math]

[math]\displaystyle{ \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} }[/math]

[math]\displaystyle{ \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} }[/math]

[math]\displaystyle{ \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} }[/math]

[math]\displaystyle{ p^*_{1} =\sqrt {(p^*_{1(x)})^2+(p^*_{1(y)})^2+(p^*_{1(z)})^2} \Longrightarrow p^*_{1}=p^*_{1(z)} }[/math]

[math]\displaystyle{ p^*_{2} =\sqrt {(p^*_{2(x)})^2+(p^*_{2(y)})^2+(p^*_{2(z)})^2} \Longrightarrow p^*_{2}=p^*_{2(z)} }[/math]

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

• 
  Figure 5.1.3: Initial momentum for lab frame electron incident at 11GeV as seen in the center of mass frame.
• 
  Figure 5.1.4: Initial momentum for a lab frame stationary particle as seen in the center of mass frame.

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