Difference between revisions of "Center of Mass for Stationary Target"

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<center><math>\textbf{\underline{Navigation}}</math>
 
<center><math>\textbf{\underline{Navigation}}</math>
  
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[[VanWasshenova_Thesis#Weighted_Isotropic_Distribution_in_Lab_Frame|<math>\triangle </math>]]
 
[[VanWasshenova_Thesis#Weighted_Isotropic_Distribution_in_Lab_Frame|<math>\triangle </math>]]
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[[CED_Verification_of_DC_Angle_Theta_and_Wire_Correspondance|<math>\vartriangleright </math>]]
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For an [[DV_Calculations_of_4-momentum_components#Center_of_Mass_Frame|incoming electron of 11GeV striking a stationary electron]] we would expect:
 
For an [[DV_Calculations_of_4-momentum_components#Center_of_Mass_Frame|incoming electron of 11GeV striking a stationary electron]] we would expect:
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Inspecting the Lorentz transformation to the Center of Mass frame:
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<center><math>\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></center>
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For the case of a stationary electron, this simplifies to:
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<center><math>\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></center>
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which gives,
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<center><math>\Longrightarrow\begin{cases}
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E^*=\gamma^* (E_{1}+m)-\beta^* \gamma^* p_{1(z)} \\
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p^*_{z}=-\beta^* \gamma^*(E_{1}+m)+\gamma^* p_{1(z)}
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\end{cases}</math></center>
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Solving for <math>\beta^*</math>, with <math>p^*_{z}=0</math>
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<center><math>\Longrightarrow \beta^* \gamma^*(E_{1}+m)=\gamma^* p_{1(z)}</math></center>
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{| class="wikitable" align="center"
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| style="background: gray"      | <math>\Longrightarrow \beta^*=\frac{p_{1}}{(E_{1}+m)}</math>
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|}
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Similarly, solving for <math>\gamma^*</math> by substituting in <math>\beta^*</math>
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<center><math>E^*=\gamma^* (E_{1}+m)-\frac{p_{1}}{(E_{1}+m)} \gamma^* p_{1(z)}</math></center>
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<center><math>E^*=\gamma^* \frac{(E_{1}+m)^2}{(E_{1}+m)}-\gamma^*\frac{(p_{1(z)})^2}{(E_{1}+m)}</math></center>
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Using the fact that <math>E^*=[(E_{1}+E_{2})^2-(\vec p_{1}+\vec p_{2})^2]^{1/2}</math>
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<center><math>E^*=\gamma^* \frac{E^*\ ^2}{(E_{1}+m)}</math></center>
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{| class="wikitable" align="center"
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| style="background: gray"      | <math>\Longrightarrow \gamma^*=\frac{(E_1+m)} {E^*}</math>
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|}
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Using the relation
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<center><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}+m\\ 0 \\ 0 \\ p_{1(z)}+0\end{matrix} \right)</math></center>
  
  
 
<center><math>\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></center>
 
<center><math>\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></center>
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<center><math>\Longrightarrow\begin{cases}
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E^*_{2}=\gamma^* (m) \\
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p^*_{2(z)}=-\beta^* \gamma^* (m)
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\end{cases}</math></center>
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<center><math>\Longrightarrow\begin{cases}
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E^*_{2}=\frac{(E_{1}+m)}{E^*} (m) \\
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p^*_{2(z)}=-\frac{p_{1}}{(E_{1}+m)} \frac{(E_{1}+m)}{E^*} (m_{2})
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\end{cases}</math></center>
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<center><math>\Longrightarrow\begin{cases}
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E^*_{2}=\frac{(11000 MeV+.511 MeV)}{106.031 MeV} (.511 MeV) \approx 53.015 MeV\\
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p^*_{2(z)}=-\frac{11000 MeV}{106.031 MeV} (.511 MeV) \approx -53.013 MeV
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\end{cases}</math></center>
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<center><math>\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></center>
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<center><math>\Longrightarrow\begin{cases}
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E^*_{1}=\gamma^* (E_{1})-\beta^* \gamma^* p_{1(z)} \\
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p^*_{1(z)}=-\beta^* \gamma^*(E_{1})+\gamma^* p_{1(z)}
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\end{cases}</math></center>
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<center><math>\Longrightarrow\begin{cases}
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E^*_{1}=\frac{(E_{1}+m)}{E^*} (E_{1})-\frac{p_{1(z)}}{(E_{1}+m)} \frac{(E_{1}+m)}{E^*} p_{1(z)} \\
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p^*_{1(z)}=-\frac{p_{1}}{(E_{1}+m)} \frac{(E_{1}+m)}{E^*}(E_{1})+\frac{(E_{1}+m)}{E^*} p_{1(z)}
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\end{cases}</math></center>
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<center><math>\Longrightarrow\begin{cases}
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E^*_{1}=\frac{(11000 MeV+.511 MeV)}{106.031 MeV} (11000 MeV)-\frac{11000 MeV}{106.031 MeV} 11000 MeV \approx 53.015 MeV\\
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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
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\end{cases}</math></center>
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<center><math>p^*_{1} =\sqrt {(p^*_{1(x)})^2+(p^*_{1(y)})^2+(p^*_{1(z)})^2} \Longrightarrow p^*_{1}=p^*_{1(z)}</math></center>
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<center><math>p^*_{2} =\sqrt {(p^*_{2(x)})^2+(p^*_{2(y)})^2+(p^*_{2(z)})^2} \Longrightarrow p^*_{2}=p^*_{2(z)}</math></center>
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This gives the momenta of the particles in the center of mass to have equal magnitude, but opposite directions.
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<center><gallery widths=500px heights=400px>
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File:Init_e_Mom_CM.png|'''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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File:Init_Mol_Mom_CM.png|'''Figure 5.1.4:''' Initial momentum for a lab frame stationary particle as seen in the center of mass frame.
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</gallery></center>
  
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<center>[[File:Init_e_Mom_CM.png|600 px]][[File:Init_Mol_Mom_CM.png|600 px]]</center>
 
  
  
  
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<center>[[File:Init_e_Theta_CM.png|600 px]][[File:Init_Mol_Theta_CM.png|600 px]]</center>
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<center><gallery widths=500px heights=400px>
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File:Init_e_Theta_CM.png|'''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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File:Init_Mol_Theta_CM.png|'''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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</gallery></center>
  
  
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<center><math>\textbf{\underline{Navigation}}</math>
 
<center><math>\textbf{\underline{Navigation}}</math>
  
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[[VanWasshenova_Thesis#Determining_wire-theta_correspondence|<math>\vartriangleleft </math>]]
+
[[Uniform_distribution_in_Energy_and_Theta_LUND_files|<math>\vartriangleleft </math>]]
 
[[VanWasshenova_Thesis#Weighted_Isotropic_Distribution_in_Lab_Frame|<math>\triangle </math>]]
 
[[VanWasshenova_Thesis#Weighted_Isotropic_Distribution_in_Lab_Frame|<math>\triangle </math>]]
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[[CED_Verification_of_DC_Angle_Theta_and_Wire_Correspondance|<math>\vartriangleright </math>]]
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[[Phase_space_Limiting_Particles|<math>\vartriangleright </math>]]
  
 
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Latest revision as of 18:36, 30 May 2017

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

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


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


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


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


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


[math]E^*=\gamma^* \frac{E^*\ ^2}{(E_{1}+m)}[/math]
[math]\Longrightarrow \gamma^*=\frac{(E_1+m)} {E^*}[/math]



Using the relation

[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}+m\\ 0 \\ 0 \\ p_{1(z)}+0\end{matrix} \right)[/math]


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


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


[math]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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