Difference between revisions of "Scattered and Moller Electron Energies in CM Frame"

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=Scattered and Moller Electron energies in CM=
 
=Scattered and Moller Electron energies in CM=
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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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We can use the Mandelstam variable s, the square of the center of mass energy, to find <math>E^*</math>
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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>
 
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<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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{| class="wikitable" align="center"
 
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| style="background: gray"      | <center><math>E^*=E^*_1+E^*_2=\sqrt{2m(m+E_1)}</math></center>
 
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|}
 

Revision as of 01:43, 16 June 2017

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Scattered and Moller Electron energies in CM

We can use the Mandelstam variable s, the square of the center of mass energy, to find [math]E^*[/math]

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