Difference between revisions of "DV Calculations of 4-momentum components"

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==Calculations of 4-momentum components==
 
  
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===Verification===
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[[Verification of Relativistic Components]]
  
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===Conclusion===
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[[Verification_of_Relativistic_Components#Conclusion | Conclusion]]
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----
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===Initial Lab Frame===
 
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Assume the incoming electron has momentum of 11000 MeV in the positive z direction.
 
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<center><math>p_{1\ z}\equiv p_{1}=11000 MeV</math></center>
 
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The moller electron is initially at rest
 
−
 
 
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<center><math>p_{2}\equiv 0</math></center>
 
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The total energy in this frame
 
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<center><math>E\equiv \sqrt{p^2+m^2}</math></center>
 
−
 
 
−
 
 
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<center><math>E\equiv \sqrt{(p_{1}+p_{2})^2+(m_{1}+m_{2})^2}</math></center>
 
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{| class="wikitable" align="center"
 
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| style="background: gray"      | <math>\Longrightarrow E\equiv \sqrt{(11000 MeV)^2+(.511 MeV+.511 MeV)^2}\approx 11000 MeV</math>
 
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|}
 
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===Center of Mass Frame===
 
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Using the definition
 
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<center><math>E^2\equiv p^2c^2+m^2c^4</math></center>
 
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−
 
 
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<center><math>\Longrightarrow p^2=\frac {E^2}{c^2}-m^2 c^2</math></center>
 
−
 
 
−
 
 
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We can use 4-momenta vectors, i.e. <math>{\mathbf P}\equiv \left(\begin{matrix} E/c\\ p_x \\ p_y \\ p_z \end{matrix} \right)</math>
 
−
 
 
−
 
 
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we can use the fact that the scalar product of a 4-momenta with itself,
 
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<center><math>{\mathbf P_1}\cdot {\mathbf P_1}=E_1E_1-\vec p_1\cdot \vec p_1 c^2=m_1^2c^4=s</math></center>
 
−
 
 
−
 
 
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is invariant
 
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Using this, the sum of two 4-momenta forms a 4-vector as well
 
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<center><math>{\mathbf P_1}+ {\mathbf P_2}= \left( \begin{matrix}E_1+E_2\\\vec p_1 c+\vec p_2 c\end{matrix} \right)= {\mathbf P}</math></center>
 
−
 
 
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The length of this four-vector is an invariant as well
 
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<center><math>{\mathbf P^2}=({\mathbf P_1}+{\mathbf P_2})^2=(E_1+E_2)^2-(\vec p_1 c+\vec p_2 c)^2=(m_1c^4+m_2c^4)^2=s</math></center>
 
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−
 
 
−
 
 
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−
 
 
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<center><math>{\mathbf P^2}=({\mathbf P_1}+{\mathbf P_2})^2=(E_1+E_2)^2-(\vec p_1 +\vec p_2 )^2=s</math></center>
 
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<center> ''(with c=1)'' </center>
 
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−
 
 
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For incoming electrons moving only in the z-direction, we can write
 
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−
 
 
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<center><math>{\mathbf P_1}+ {\mathbf P_2}= \left( \begin{matrix}E_1+E_2\\ 0 \\ 0 \\ p_{1z}+p_{2z}\end{matrix} \right)={\mathbf P}</math></center>
 
−
 
 
−
 
 
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We can perform a Lorentz transformation to the Center of Mass frame, with zero total momentum
 
−
 
 
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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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Without knowing the values for gamma or beta, we can use the fact that lengths of the two 4-momenta are invariant
 
−
 
 
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<center><math>s={\mathbf P^*}^2=(E^*_{1}+E^*_{2})^2-(\vec p_{1}+\vec p_{2})^2</math></center>
 
−
 
 
−
 
 
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<center><math>s={\mathbf P}^2=(E_{1}+E_{2})^2-(\vec p_{1}\ ^*+\vec p_{2}\ ^*)^2</math></center>
 
−
 
 
−
 
 
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<center><math>\Longrightarrow m_{1}=m^*_{1}\ ; m_{2}=m^*_{2}</math></center>
 
−
 
 
−
 
 
−
Setting these equal to each other, we can use this for the collision of two particles of mass m<sub>1</sub> and m<sub>2</sub>.  Since the total momentum is zero in the Center of Mass frame, we can express total energy in the center of mass frame as
 
−
 
 
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<center><math>(E^*_{1}+E^*_{2})^2-(\vec p^*_{1}+\vec p^*_{2})^2=s=(E_{1}+E_{2})^2-(\vec{p_{1}}+\vec p_{2})^2</math></center>
 
−
 
 
−
 
 
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<center><math>(E^*)^2-(\vec p^*)^2=(E_{1}+E_{2})^2-(\vec{p_{1}}+\vec p_{2})^2</math></center>
 
−
 
 
−
 
 
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<center><math>(E^*)^2=(E_{1}+E_{2})^2-(\vec{p_{1}}+\vec p_{2})^2</math></center>
 
−
 
 
−
 
 
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<center><math>E^*=[(E_{1}+E_{2})^2-(\vec{p_{1}}+\vec p_{2})^2]^{1/2}</math></center>
 
−
 
 
−
 
 
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<center><math>E^*=[E_{1}^2+2E_{1}E_{2}+E_{2}^2-\vec p_{1}  . \vec p_{2} -\vec p_{1}  . \vec p_{1} -\vec p_{2}  . \vec p_{1} -\vec p_{2}  . \vec p_{2} ]^{1/2}</math></center>
 
−
 
 
−
 
 
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<center><math>E^*=[(E_{1}^2- p_{1}^2 )+(E_{2}^2-p_{2}^2 )+2E_{1}E_{2}-\vec p_{1}  . \vec p_{2} -\vec p_{2}  . \vec p_{1} ]^{1/2}</math></center>
 
−
 
 
−
 
 
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<center><math>E^*=[(E_{1}^2- p_{1}^2 )+(E_{2}^2-p_{2}^2 )+2E_{1}E_{2}-p_{1} p_{2}\cos(\theta) - p_{2} p_{1}\cos(\theta) ]^{1/2}</math></center>
 
−
 
 
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−
 
 
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<center><math>E^*=[m_{1}^2+m_{2}^2+2E_{1}E_{2}-p_{1} p_{2}\cos(\theta) - p_{2} p_{1}\cos(\theta) ]^{1/2}</math></center>
 
−
 
 
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<center><math>E^*=[m_{1}^2+m_{2}^2+2E_{1}E_{2}-2p_{1} p_{2}\cos(\theta) ]^{1/2}</math></center>
 
−
 
 
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<center><math>E^*=[m_{1}^2+m_{2}^2+2E_{1}E_{2}-2p_{1} p_{2}\cos(\theta) ]^{1/2}</math></center>
 
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Using the relations <math>\beta\equiv \vec p/E\Longrightarrow \vec p=\beta E</math>
 
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<center><math>E^*=[m_{1}^2+m_{2}^2+2E_{1}E_{Lab}2(1-\beta_{1 Lab}\beta_{2 Lab}\cos(\theta_{Lab}))]^{1/2}</math></center>
 
−
 
 
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<center>where <math> θ_{Lab} </math> is the angle between the particles in the Lab frame.</center>
 
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<center><math>E_{CM}=[m_1^2+m_2^2+2E_{1 Lab}E_{2 Lab}(1-\beta_{1 Lab}\beta_{2 Lab}\cos(\theta_{Lab}))]^{1/2}</math></center>
 
−
 
 
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In the frame where one particle ''(m<sub>2 Lab</sub>)'' is at rest
 
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<center><math>\Longrightarrow \beta_{2 Lab}=0</math></center>
 
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<center><math>\Longrightarrow p_{2 Lab}=0</math></center>
 
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which implies,
 
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<center><math> E_{2 Lab}=[p_{2 Lab}^2+m_{2 Lab}^2]^{1/2}=m_{2 Lab}</math></center>
 
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{| class="wikitable" align="center"
 
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| style="background: gray"      | <math>E_{CM}=(m_{1 Lab}^2+m_{2 Lab}^2+2E_{1 Lab} m_{2 Lab})^{1/2}=(.511MeV^2+.511MeV^2+2(\sqrt{(11000 MeV)^2+(.511 MeV)^2})(.511 MeV))^{1/2}\approx 106.030760886 MeV</math>
 
−
|}
 
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where <math>E_{1 Lab}=\sqrt{p_{1 Lab}^2+m_{1 Lab}^2}\approx 11000 MeV</math>
 
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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 CM}+E_{2 CM}\\ 0 \\ 0 \\ 0\end{matrix} \right)=\left(\begin{matrix}\gamma_{CM} & 0 & 0 & -\beta_{CM} \gamma_{CM}\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ -\beta_{CM} \gamma_{CM} & 0 & 0 & \gamma_{CM} \end{matrix} \right) . \left( \begin{matrix}E_{1 Lab}+E_{2 Lab}\\ 0 \\ 0 \\ p_{1z Lab}+p_{2z Lab}\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_{CM} \\ p_{xCM} \\ p_{y CM} \\ p_{z CM}\end{matrix} \right)=\left(\begin{matrix}\gamma_{CM} & 0 & 0 & -\beta_{CM} \gamma\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ -\beta_{CM} \gamma_{CM} & 0 & 0 & \gamma_{CM} \end{matrix} \right) . \left( \begin{matrix}E_{1 Lab}+m_{2Lab}\\ 0 \\ 0 \\ p_{1z Lab}+0\end{matrix} \right)</math></center>
 
−
 
 
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which gives,
 
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<center><math>\Longrightarrow\begin{cases}
 
−
E_{CM}=\gamma_{CM} (E_{1 Lab}+m_{2 Lab})-\beta_{CM} \gamma_{CM} p_{1z Lab} \\
 
−
p_{zCM}=-\beta_{CM} \gamma_{CM}(E_{1 Lab}+m_{2 Lab})+\gamma_{CM} p_{1z Lab}
 
−
\end{cases}</math></center>
 
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Solving for <math>\beta_{CM}</math>, with <math>p_{zCM}=0</math>
 
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<center><math>\Longrightarrow \beta_{CM} \gamma_{CM}(E_{1 Lab}+m_{2 Lab})=\gamma_{CM} p_{1z Lab}</math></center>
 
−
 
 
−
 
 
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{| class="wikitable" align="center"
 
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| style="background: gray"      | <math>\Longrightarrow \beta_{CM}=\frac{p_{1 Lab}}{(E_{1 Lab}+m_{2 Lab})}</math>
 
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|}
 
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Similarly, solving for <math>\gamma_{CM}</math> by substituting in <math>\beta_{CM}</math>
 
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−
 
 
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<center><math>E_{CM}=\gamma_{CM} (E_{1 Lab}+m_{2 Lab})-\frac{p_{1 Lab}}{(E_{1 Lab}+m_{2 Lab})} \gamma_{CM} p_{1z Lab}</math></center>
 
−
 
 
−
 
 
−
<center><math>E_{CM}=\gamma_{CM} \frac{(E_{1 Lab}+m_2)^2}{(E_{1 Lab}+m_{2 Lab})}-\gamma_{CM}\frac{(p_{1z Lab})^2}{(E_{1 Lab}+m_{2 Lab})}</math></center>
 
−
 
 
−
 
 
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Using the fact that <math>E_{CM}=[(E_{1 Lab}+E_{2 Lab})^2-(\vec p_{1 Lab}+\vec p_{2 Lab})^2]^{1/2}</math>
 
−
 
 
−
 
 
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<center><math>E_{CM}=\gamma_{CM} \frac{E_{CM}^2}{(E_{1 Lab}+m_{2 Lab})}</math></center>
 
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{| class="wikitable" align="center"
 
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| style="background: gray"      | <math>\Longrightarrow \gamma_{CM}=\frac{(E_{1 Lab}+m_{2 Lab})}{E_{CM}}</math>
 
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|}
 
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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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Using the relation
 
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<center><math>\left( \begin{matrix} E_{1 CM}+E_{2 CM} \\ p_{1xCM}+p_{2xCM} \\ p_{1y CM}+p_{2yCM} \\ p_{1z CM}+p_{2zCM}\end{matrix} \right)=\left(\begin{matrix}\gamma_{CM} & 0 & 0 & -\beta_{CM} \gamma_{CM}\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ -\beta_{CM} \gamma_{CM} & 0 & 0 & \gamma_{CM} \end{matrix} \right) . \left( \begin{matrix}E_{1 Lab}+m_{2 Lab}\\ 0 \\ 0 \\ p_{1z Lab}+0\end{matrix} \right)</math></center>
 
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<center><math>\Longrightarrow \left( \begin{matrix} E_{2 CM} \\ p_{2xCM} \\ p_{2yCM} \\ p_{2zCM}\end{matrix} \right)=\left(\begin{matrix}\gamma_{CM} & 0 & 0 & -\beta_{CM} \gamma_{CM}\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ -\beta_{CM} \gamma_{CM} & 0 & 0 & \gamma_{CM} \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_{2CM}=\gamma_{CM} (m_{2 lab}) \\
 
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p_{2zCM}=-\beta_{CM} \gamma_{CM} (m_{2 Lab})
 
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\end{cases}</math></center>
 
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<center><math>\Longrightarrow\begin{cases}
 
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E_{2CM}=\frac{(E_{1 Lab}+m_{2 Lab})}{E_{CM}} (m_2) \\
 
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p_{2zCM}=-\frac{p_{1 Lab}}{(E_{1 Lab}+m_2)} \frac{(E_{1 Lab}+m_{2 Lab})}{E_{CM}} (m_{2 Lab})
 
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\end{cases}</math></center>
 
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<center><math>\Longrightarrow\begin{cases}
 
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E_{2CM}=\frac{(11000 MeV+.511 MeV)}{106.031 MeV} (.511 MeV) \approx 53.0129177 MeV\\
 
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p_{2zCM}=-\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 CM}\\ p_{1xCM} \\ p_{1y CM} \\ p_{1z CM}\end{matrix} \right)=\left(\begin{matrix}\gamma_{CM} & 0 & 0 & -\beta_{CM} \gamma_{CM}\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ -\beta_{CM} \gamma_{CM} & 0 & 0 & \gamma_{CM} \end{matrix} \right) . \left( \begin{matrix}E_{1 Lab}\\ 0 \\ 0 \\ p_{1z Lab}\end{matrix} \right)</math></center>
 
−
 
 
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<center><math>\Longrightarrow\begin{cases}
 
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E_{1CM}=\gamma_{CM} (E_{1 Lab})-\beta_{CM} \gamma_{CM} p_{1z Lab} \\
 
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p_{1zCM}=-\beta_{CM} \gamma(E_{1 Lab})+\gamma_{CM} p_{1z Lab}
 
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\end{cases}</math></center>
 
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<center><math>\Longrightarrow\begin{cases}
 
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E_{1CM}=\frac{(E_{1 Lab}+m_{2 Lab})}{E_{CM}} (E_{1 Lab})-\frac{p_{1zLab}}{(E_{1 Lab}+m_2)} \frac{(E_{1 Lab}+m_{2 Lab})}{E_{CM}} p_{1z Lab} \\
 
−
p_{1zCM}=-\frac{p_{1 Lab}}{(E_{1 Lab}+m_2)} \frac{(E_{1 Lab}+m_{2 Lab})}{E_{CM}}(E_{1 Lab})+\frac{(E_{1 Lab}+m_{2 Lab})}{E_{CM}} p_{1z Lab}
 
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\end{cases}</math></center>
 
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<center><math>\Longrightarrow\begin{cases}
 
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E_{1CM}=\frac{(11000 MeV+.511 MeV)}{106.031 MeV} (11000 MeV)-\frac{11000 MeV}{106.031 MeV} 11000 MeV \approx 53.013 MeV\\
 
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p_{1zCM}=-\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 CM} =\sqrt{p_{1xCM}^2+p_{1yCM}^2+p_{1zCM}^2}\Longrightarrow p_{1 CM}=p_{1zCM}</math></center>
 
−
 
 
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<center><math>p_{2 CM} =\sqrt{p_{2xCM}^2+p_{2yCM}^2+p_{2zCM}^2}\Longrightarrow p_{2 CM}=p_{2zCM}</math></center>
 
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−
 
 
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{| class="wikitable" align="center"
 
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| style="background: gray"      | <math> p_{1 CM}=p_{2 CM}\approx 53.013 MeV</math>
 
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|}
 
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{| class="wikitable" align="center"
 
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| style="background: gray"      | <center><math>E_{1 CM}= E_{2 CM}\approx 53.013 MeV</math></center>
 
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|}
 
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===Moller electron Lab Frame===
 
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Finding the correct kinematic values starting from knowing the momentum of the Moller electron, <math>'''p_{(m)'''Lab}</math> ,  in the Lab frame,
 
−
 
 
−
<center>[[File:Xz_(m)LAB.png]]</center>
 
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<center>Using <math>\theta=\arccos \left(\frac{p_{z(m) Lab}}{p_{(m)Lab}}\right)</math></center>
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"      | <math>\Longrightarrow {p_{z(m) Lab}=p_{(m)Lab}\cos(\theta)}</math>
 
−
|}
 
−
 
 
−
 
 
−
Checking on the sign resulting from the cosine function, we are limited to:
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: red"      | <math>0^\circ \le \theta \le 60^\circ \equiv 0 \le \theta \le 1.046\ Radians</math>
 
−
|}
 
−
 
 
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Since,
 
−
<center><math>\frac{p_{(m)zLab}}{p_{(m)Lab}}=cos(\theta)</math></center>
 
−
 
 
−
 
 
−
<center><math>\Longrightarrow p_{(m)zLab}\ should\ always\ be\ positive</math></center>
 
−
<center>[[File:Xy_Lab.png]]</center>
 
−
 
 
−
<center>Similarly, <math>\phi=\arccos \left( \frac{p_{x(m) Lab}}{p_{xy(m) Lab}} \right)</math></center>
 
−
 
 
−
 
 
−
<center>where <math>p_{xy(m) Lab}=\sqrt{p_{x(m) Lab}^2+p_{y(m)Lab}^2}</math></center>
 
−
 
 
−
 
 
−
<center><math>p_{xy(m) Lab}=(p_{x(m) Lab}^2+p_{y(m) Lab}^2)^2</math></center>
 
−
 
 
−
 
 
−
<center>and using <math>p_{(m)Lab}^2=p_{x(m) Lab}^2+p_{y(m) Lab}^2+p_{z(m) Lab}^2</math></center>
 
−
 
 
−
 
 
−
<center>this gives <math>p_{(m)Lab}^2=p_{xy(m) Lab}^2+p_{z(m) Lab}^2</math></center>
 
−
 
 
−
 
 
−
<center><math>\Longrightarrow p_{(m)Lab}^2-p_{z(m) Lab}^2=p_{xy(m) Lab}^2</math></center>
 
−
 
 
−
 
 
−
<center><math>\Longrightarrow p_{xy(m) Lab}=\sqrt{p_{(m)Lab}^2-p_{z(m) Lab}^2}</math></center>
 
−
 
 
−
 
 
−
<center>which gives<math>\phi = \arccos \left( \frac{p_{x(m) Lab}}{\sqrt{p_{(m)Lab}^2-p_{z(m) Lab}^2}}\right)</math></center>
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"      | <math>\Longrightarrow p_{x(m) Lab}=\sqrt{p_{(m)Lab}^2-p_{z(m) Lab}^2} \cos(\phi)</math>
 
−
|}
 
−
 
 
−
 
 
−
<center>Similarly, using <math>p_{(m)Lab}^2=p_{x(m) Lab}^2+p_{y(m) Lab}^2+p_{z(m) Lab}^2</math></center>
 
−
 
 
−
 
 
−
<center><math>\Longrightarrow p_{(m)Lab}^2-p_{x(m) Lab}^2-p_{z(m) Lab}^2=p_{y(m) Lab}^2</math></center>
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"      | <math>p_{y(m) Lab}=\sqrt{p_{(m)Lab}^2-p_{x(m) Lab}^2-p_{z(m) Lab}^2}</math>
 
−
|}
 
−
 
 
−
 
 
−
Checking on the sign from the cosine results for <math>\phi</math>
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: red"      | <math>-\pi \le \phi \le \pi\ Radians</math>
 
−
|}
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: red"      | <math>For\ 0 \le \phi \le \frac{-\pi}{2}\ Radians</math>
 
−
|-
 
−
| <center>x=POSITIVE</center>
 
−
|-
 
−
| <center>y=NEGATIVE</center>
 
−
|}
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: red"      | <math>For\ 0 \le \phi \le \frac{\pi}{2}\ Radians</math>
 
−
|-
 
−
| <center>x=POSITIVE</center>
 
−
|-
 
−
| <center>y=POSITIVE</center>
 
−
|}
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: red"      | <math>For\ \frac{-\pi}{2} \le \phi \le -\pi\ Radians</math>
 
−
|-
 
−
| <center>x=NEGATIVE</center>
 
−
|-
 
−
| <center>y=NEGATIVE</center>
 
−
|}
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: red"      | <math>For\ \frac{\pi}{2} \le \phi \le \pi\ Radians</math>
 
−
|-
 
−
|<center> x=NEGATIVE</center>
 
−
|-
 
−
| <center>y=POSITIVE</center>
 
−
|}
 
−
 
 
−
===Moller electron Center of Mass Frame===
 
−
 
 
−
Relativistically, the x and y components remain the same in the conversion from the Lab frame to the Center of Mass frame, since the direction of motion is only in the z direction.
 
−
 
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"      | <math>p_{x(m) CM}\Leftrightarrow p_{x(m) Lab}</math>
 
−
|}
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"  | <math>p_{y(m) CM}\Leftrightarrow p_{y(m) Lab}</math>
 
−
|}
 
−
 
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"  | <math>p_{z(m) CM}=\sqrt{p_{(m)CM}^2-p_{x(m) CM}^2-p_{y(m) CM}^2}</math>
 
−
|}
 
−
 
 
−
 
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"      | <math>\Longrightarrow E_{(m) CM}=\sqrt{\frac{m(m+E_{(e^-) Lab})}{2}}</math>
 
−
|}
 
−
 
 
−
===Electron Center of Mass Frame===
 
−
Relativistically, the x, y, and z components have the same magnitude, but opposite direction, in the conversion from the Moller electron's Center of Mass frame to the electron's Center of Mass frame.
 
−
 
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"      | <math>p_{x(m) CM}= -p_{x(e^-) CM}</math>
 
−
|}
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"  | <math>p_{y(m) CM}= -p_{y((e^-) CM}</math>
 
−
|}
 
−
 
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"  | <math>p_{z(m) CM}=-p_{z((e^-) CM}</math>
 
−
|}
 
−
 
 
−
where previously it was shown
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"      | <math>p_{(e^-) CM}\approx 53.013 MeV</math>
 
−
|}
 
−
 
 
−
 
 
−
 
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"      | <math>E_{(e^-) CM}\approx 53.013 MeV</math>
 
−
|}
 
−
 
 
−
===Electron Lab Frame===
 
−
 
 
−
We can perform a Lorentz transformation from the Center of Mass frame, with zero total momentum, to the Lab frame.
 
−
 
 
−
 
 
−
<center><math>\left( \begin{matrix}E_{(e^-) Lab}+E_{(m) Lab}\\ p_{(e^-)z Lab}+p_{(m)z Lab} \\p_{(e^-)z Lab}+ p_{(m)z Lab} \\ p_{(e^-)z Lab}+p_{(m)z Lab}\end{matrix} \right)=\left(\begin{matrix}\gamma_{CM} & 0 & 0 & \beta_{CM} \gamma_{CM}\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ \beta_{CM} \gamma_{CM} & 0 & 0 & \gamma_{CM} \end{matrix} \right) . \left( \begin{matrix}E_{(e^-) CM}+E_{(m) CM}\\ p_{(e^-)x CM}+p_{(m)x CM} \\ p_{(e^-)y CM}+p_{(m)y CM} \\ p_{(e^-)z CM}+p_{(m)z CM}\end{matrix} \right)</math></center>
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"      | <math>p_{x(e^-) CM}= p_{x(e^-) Lab}</math>
 
−
|}
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"  | <math>p_{y(e^-) CM}= p_{y((e^-) Lab}</math>
 
−
|}
 
−
 
 
−
 
 
−
<center><math>\Longrightarrow\begin{cases}
 
−
E_{Lab}=\gamma_{Lab} (E_{ CM})+\beta_{Lab} \gamma_{Lab} p_{z CM} \\
 
−
p_{z Lab}=\beta_{Lab} \gamma_{Lab}(E_{ CM})+ \gamma_{Lab}p_{z CM}
 
−
\end{cases}</math></center>
 
−
 
 
−
 
 
−
<center><math>\Longrightarrow\begin{cases}
 
−
\gamma_{Lab} = \frac{E_{ Lab}}{E_{CM}} \\
 
−
\beta_{Lab} =\frac{p_{zLab}}{E_{LAB}}
 
−
\end{cases}</math></center>
 
−
 
 
−
 
 
−
 
 
−
 
 
−
Without knowing the values for gamma or beta, we can use the fact that lengths of the two 4-momenta are invariant
 
−
 
 
−
<center><math>s={\mathbf P}_{CM}^2=(E_{(e^-) CM}+E_{(m) CM})^2-(\vec p_{(e^-) CM}+\vec p_{(m) CM})^2</math></center>
 
−
 
 
−
 
 
−
<center><math>s={\mathbf P}_{Lab}^2=(E_{(e^-) Lab}+E_{(m) Lab})^2-(\vec p_{(e^-) Lab}+\vec p_{(m) Lab})^2</math></center>
 
−
 
 
−
 
 
−
Setting these equal to each other, we can use this for the collision of two particles of mass m<sub>1</sub> and m<sub>2</sub>.  Since the total momentum is zero in the Center of Mass frame, we can express total energy in the center of mass frame as
 
−
 
 
−
<center><math>(E_{1 CM}+E_{2 CM})^2-(\vec p_{1 CM}+\vec p_{2 CM})^2=s=(E_{1 Lab}+E_{2 Lab})^2-(\vec{p_{1 Lab}}+\vec p_{2 Lab})^2</math></center>
 
−
 
 
−
 
 
−
<center><math>(E_{CM})^2-(\vec p_{CM})^2=(E_{1 Lab}+E_{2 Lab})^2-(\vec{p_{Lab}})^2</math></center>
 
−
 
 
−
 
 
−
<center><math>(E_{CM})^2=(E_{Lab})^2-(\vec{p_{Lab}})^2</math></center>
 
−
 
 
−
 
 
−
<center><math>(E_{CM})^2+(\vec{p_{Lab}})^2=(E_{Lab})^2</math></center>
 
−
 
 
−
 
 
−
<center><math>E_{Lab}=\sqrt{(E_{CM})^2+(\vec{p_{Lab}})^2}</math></center>
 
−
 
 
−
Since momentum is conserved, the initial momentum from the incident electron and stationary electron is still the same in the Lab frame, therefore <math>p_{Lab}=11000 MeV</math>
 
−
 
 
−
 
 
−
<center><math>E_{Lab}=\sqrt{(106.031 MeV)^2+(11000 MeV)^2}\approx 11000.511 MeV</math></center>
 
−
 
 
−
 
 
−
This is the same as the initial total energy in the Lab frame, which should be expected since scattering is considered to be an elastic collision.
 
−
 
 
−
 
 
−
Since,
 
−
 
 
−
<center><math>E_{Lab}\equiv E_{(e^-)Lab}+E_{(m)Lab}</math></center>
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"      | <math>\Longrightarrow E_{(e^-)Lab}=E_{Lab}-E_{(m)Lab}</math>
 
−
|}
 
−
 
 
−
Using the relation
 
−
 
 
−
<center><math>E^2\equiv p^2+m^2</math></center>
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"      | <math>\Longrightarrow p_{(e^-)Lab}^2=E_{(e^-)Lab}^2-m_{(e^-)Lab}^2=E_{(e^-)Lab}^2-(.511 MeV)^2</math>
 
−
|}
 
−
 
 
−
 
 
−
Using
 
−
 
 
−
<center><math>p^2 \equiv p_x^2+p_y^2+p_z^2</math></center>
 
−
 
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"      | <math>\Longrightarrow p_{(e^-)z Lab}=\sqrt{p_{(e^-)Lab}^2-p_{(e^-)xLab}^2-p_{(e^-)Lab}^2}</math>
 
−
|}
 
−
 
 
  
 
[[DV_RunGroupC_Moller#Momentum distributions in the Center of Mass Frame]]
 
[[DV_RunGroupC_Moller#Momentum distributions in the Center of Mass Frame]]
  
 
[[DV_MollerTrackRecon#Calculations_of_4-momentum_components]]
 
[[DV_MollerTrackRecon#Calculations_of_4-momentum_components]]

Latest revision as of 20:25, 31 May 2017

Verification

Verification of Relativistic Components

Conclusion

Conclusion

DV_RunGroupC_Moller#Momentum distributions in the Center of Mass Frame

DV_MollerTrackRecon#Calculations_of_4-momentum_components

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