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| <center><math>\textbf{\underline{Navigation}}</math> | | <center><math>\underline{\textbf{Navigation}}</math> |
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| [[Limits_based_on_Mandelstam_Variables|<math>\vartriangleleft </math>]] | | [[Total_Energy_in_CM_Frame|<math>\vartriangleleft </math>]] |
| [[VanWasshenova_Thesis#Initial_Lab_Frame_4-momentum_components|<math>\triangle </math>]] | | [[VanWasshenova_Thesis#Initial_4-momentum_Components|<math>\triangle </math>]] |
| [[Special_Case_of_Equal_Mass_Particles|<math>\vartriangleright </math>]] | | [[Final_Lab_Frame_Moller_Electron_4-momentum_components_in_XZ_Plane|<math>\vartriangleright </math>]] |
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| </center> | | </center> |
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| =Scattered and Moller Electron energies in CM= | | =Scattered and Moller Electron energies in CM= |
| 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>
| | We can use the Mandelstam variable s, the square of the center of mass energy, to find <math>E^*</math> |
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| | <center><math>s \equiv \left({\mathbf P_1^*}+ {\mathbf P_2^{*}}\right)^2</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> | | <center><math>s \equiv \mathbf P_1^{*2}+2 \mathbf P_1^* \mathbf P_2^*+ \mathbf P_2^{*2}</math></center> |
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| which gives,
| | As shown earlier, the square of a 4-momentum is |
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| <center><math>\Longrightarrow\begin{cases} | | <center><math>\mathbf P^{2} \equiv m^2</math></center> |
| 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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| | This gives, |
| | <center><math>s \equiv m_1^{2}+2 \mathbf P_1^* \mathbf P_2^*+ m_2^{2}</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>
| | For the case <math>m_1=m_2=m</math> |
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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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| | <center><math>s \equiv 2m^{2}+2 \mathbf P_1^* \mathbf P_2^*</math></center> |
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| | Using the relationship |
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| Similarly, solving for <math>\gamma^*</math> by substituting in <math>\beta^*</math>
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| | <center><math>\mathbf P_1 \cdot \mathbf P_2 = E_{1}E_{2}-(\vec p_1 \vec p_2)</math></center> |
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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> | | <center><math>s \equiv 2m^2+2(E_1^*E_2^*-\vec p \ _1^* \vec p \ _2^*)</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>
| | In the center of mass frame of reference, |
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| | <center><math>E_1^*=E_2^* \quad and \quad \vec p \ _1^*=-\vec p \ _2^*</math></center> |
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| <center><math>E^*=\gamma^* \frac{E^*\ ^2}{(E_{1}+m)}</math></center>
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| {| class="wikitable" align="center"
| | <center><math>s \equiv 2m^2+2E_1^{*2}+2\vec p_1 \ ^{*2} </math></center> |
| | style="background: gray" | <math>\Longrightarrow \gamma^*=\frac{(E_1+m)} {E^*}</math>
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| |}
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| | Using the relativistic energy equation |
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| | <center><math>E^2 \equiv \vec p_1 \ ^2+m^2</math></center> |
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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>s \equiv 2m^2+2m^2+2\vec p_1 \ ^{*2}+\vec p_1 \ ^{*2})</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> | | <center><math>s=4(m^2+\vec p_1 \ ^{*2})=(2E_1^*)^{2}=E^{*2}</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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| | {| class="wikitable" align="center" |
| | | style="background: gray" | <math>\Rightarrow E_1^*=\frac{106.030760886 MeV}{2}=53.015380443 MeV=E_2^*</math> |
| | |} |
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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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| | ---- |
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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>\underline{\textbf{Navigation}}</math> |
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| <center><math>\Longrightarrow\begin{cases} | | [[Total_Energy_in_CM_Frame|<math>\vartriangleleft </math>]] |
| E^*_{1}=\gamma^* (E_{1})-\beta^* \gamma^* p_{1(z)} \\
| | [[VanWasshenova_Thesis#Initial_4-momentum_Components|<math>\triangle </math>]] |
| p^*_{1(z)}=-\beta^* \gamma^*(E_{1})+\gamma^* p_{1(z)}
| | [[Final_Lab_Frame_Moller_Electron_4-momentum_components_in_XZ_Plane|<math>\vartriangleright </math>]] |
| \end{cases}</math></center>
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| | | </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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| |}
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[math]\displaystyle{ \underline{\textbf{Navigation}} }[/math]
[math]\displaystyle{ \vartriangleleft }[/math]
[math]\displaystyle{ \triangle }[/math]
[math]\displaystyle{ \vartriangleright }[/math]
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]\displaystyle{ E^* }[/math]
[math]\displaystyle{ s \equiv \left({\mathbf P_1^*}+ {\mathbf P_2^{*}}\right)^2 }[/math]
[math]\displaystyle{ s \equiv \mathbf P_1^{*2}+2 \mathbf P_1^* \mathbf P_2^*+ \mathbf P_2^{*2} }[/math]
As shown earlier, the square of a 4-momentum is
[math]\displaystyle{ \mathbf P^{2} \equiv m^2 }[/math]
This gives,
[math]\displaystyle{ s \equiv m_1^{2}+2 \mathbf P_1^* \mathbf P_2^*+ m_2^{2} }[/math]
For the case [math]\displaystyle{ m_1=m_2=m }[/math]
[math]\displaystyle{ s \equiv 2m^{2}+2 \mathbf P_1^* \mathbf P_2^* }[/math]
Using the relationship
[math]\displaystyle{ \mathbf P_1 \cdot \mathbf P_2 = E_{1}E_{2}-(\vec p_1 \vec p_2) }[/math]
[math]\displaystyle{ s \equiv 2m^2+2(E_1^*E_2^*-\vec p \ _1^* \vec p \ _2^*) }[/math]
In the center of mass frame of reference,
[math]\displaystyle{ E_1^*=E_2^* \quad and \quad \vec p \ _1^*=-\vec p \ _2^* }[/math]
[math]\displaystyle{ s \equiv 2m^2+2E_1^{*2}+2\vec p_1 \ ^{*2} }[/math]
Using the relativistic energy equation
[math]\displaystyle{ E^2 \equiv \vec p_1 \ ^2+m^2 }[/math]
[math]\displaystyle{ s \equiv 2m^2+2m^2+2\vec p_1 \ ^{*2}+\vec p_1 \ ^{*2}) }[/math]
[math]\displaystyle{ s=4(m^2+\vec p_1 \ ^{*2})=(2E_1^*)^{2}=E^{*2} }[/math]
| [math]\displaystyle{ \Rightarrow E_1^*=\frac{106.030760886 MeV}{2}=53.015380443 MeV=E_2^* }[/math]
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[math]\displaystyle{ \underline{\textbf{Navigation}} }[/math]
[math]\displaystyle{ \vartriangleleft }[/math]
[math]\displaystyle{ \triangle }[/math]
[math]\displaystyle{ \vartriangleright }[/math]