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

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=Calculations of 4-momentum components=
 
  
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===Verification===
 +
[[Verification of Relativistic Components]]
  
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==Initial Conditions==
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===Conclusion===
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[[Verification_of_Relativistic_Components#Conclusion | Conclusion]]
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----
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===Lab Frame===
 
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<center>[[File:lab.png]]</center>
 
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<center>'''Figure 1: Definition of variables in the Lab Frame '''</center>
 
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Begining with the assumption that the incoming electron, p<sub>1</sub>,  has momentum of 11000 MeV in the positive z direction.
 
−
 
 
−
 
 
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<center><math>\vec p_{1(z)}\equiv \vec p_{1}=11000 MeV \widehat {z}</math></center>
 
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−
 
 
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We can also assume the Moller electron, p<sub>2</sub>, is initially at rest
 
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−
 
 
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<center><math>\vec p_{2}\equiv 0</math></center>
 
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−
 
 
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This gives the total energy in this frame as
 
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−
 
 
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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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<center>[[File:CM.png]]</center>
 
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<center>'''Figure 2: Definition of variables in the Center of Mass Frame'''</center>
 
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Starting with the definition for the total relativistic energy:
 
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<center><math>E^2\equiv p^2c^2+m^2c^4</math></center>
 
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<center><math>\Longrightarrow \frac {E^2}{c^2}-p^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 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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and using 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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<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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<center><math>{\mathbf P_1}+ {\mathbf P_2}= \left( \begin{matrix}E_1+E_2\\ 0 \\ 0 \\ p_{1(z)}+p_{2(z)}\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=(m_{1}^*+m_{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=(m_{1}+m_{2})^2</math></center>
 
−
 
 
−
 
 
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<center><math>\Longrightarrow (m_{1}^*+m_{2}^*)^2=(m_{1}+m_{2})^2</math></center>
 
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−
 
 
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Using the fact that
 
−
 
 
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<center><math>\begin{cases}
 
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m_{1}=m_{2} \\
 
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m_{1}^*=m_{2}^*
 
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\end{cases}</math></center>
 
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<center><math>(m_{1}^*+m_{1}^*)^2=(m_{1}+m_{1})^2</math></center>
 
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<center><math>2m_{1}^*=2m_{1}</math></center>
 
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<center><math>m_{1}^*=m_{1}</math></center>
 
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<center><math>\Longrightarrow m_{1}=m^*_{1}\ ; m_{2}=m^*_{2}</math></center>
 
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−
 
 
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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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−
 
 
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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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<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_{2}(1-\beta_{1}\beta_{2}\cos(\theta))]^{1/2}</math></center>
 
−
 
 
−
 
 
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<center>where <math> \theta </math> is the angle between the particles in the Lab frame.</center>
 
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In the frame where one particle ''(m<sub>2</sub>)'' is at rest
 
−
 
 
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<center><math>\Longrightarrow \beta_{2}=0</math></center>
 
−
 
 
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<center><math>\Longrightarrow p_{2}=0</math></center>
 
−
 
 
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which implies,
 
−
 
 
−
 
 
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<center><math> E_{2}=[p_{2}^2+m_{2}^2]^{1/2}=m_{2}</math></center>
 
−
 
 
−
 
 
−
 
 
−
 
 
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{| class="wikitable" align="center"
 
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| style="background: gray"      | <math>E^*=(m_{1}^2+m_{2}^2+2E_{1} m_{2})^{1/2}=(.511MeV^2+.511MeV^2+2(\sqrt{(11000 MeV)^2+(.511 MeV)^2})(.511 MeV))^{1/2}\approx 106.030760886 MeV</math>
 
−
|}
 
−
 
 
−
where <math>E_{1}=\sqrt{p_{1}^2+m_{1}^2}\approx 11000 MeV</math>
 
−
 
 
−
 
 
−
 
 
−
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_{2}\\ 0 \\ 0 \\ p_{1(z)}+0\end{matrix} \right)</math></center>
 
−
 
 
−
 
 
−
which gives,
 
−
 
 
−
 
 
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<center><math>\Longrightarrow\begin{cases}
 
−
E^*=\gamma^* (E_{1}+m_{2})-\beta^* \gamma^* p_{1(z)} \\
 
−
p^*_{z}=-\beta^* \gamma^*(E_{1}+m_{2})+\gamma^* p^*_{1(z)}
 
−
\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_{2})=\gamma^* p_{1(z)}</math></center>
 
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−
 
 
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{| class="wikitable" align="center"
 
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| style="background: gray"      | <math>\Longrightarrow \beta^*=\frac{p_{1}}{(E_{1}+m_{2})}</math>
 
−
|}
 
−
 
 
−
 
 
−
 
 
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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_{2})-\frac{p_{1}}{(E_{1}+m_{2})} \gamma^* p_{1(z)}</math></center>
 
−
 
 
−
 
 
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<center><math>E^*=\gamma^* \frac{(E_{1}+m_2)^2}{(E_{1}+m_{2})}-\gamma^*\frac{(p_{1(z)})^2}{(E_{1}+m_{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>
 
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<center><math>E^*=\gamma^* \frac{E^*\ ^2}{(E_{1}+m_{2})}</math></center>
 
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{| class="wikitable" align="center"
 
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| style="background: gray"      | <math>\Longrightarrow \gamma^*=\frac{(E_1+m_2)} {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_{2}\\ 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}
 
−
E^*_{2}=\gamma^* (m_{2}) \\
 
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p^*_{2(z)}=-\beta^* \gamma^* (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{(E_{1}+m_{2})}{E^*} (m_2) \\
 
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p^*_{2(z)}=-\frac{p_{1}}{(E_{1}+m_2)} \frac{(E_{1}+m_{2})}{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 53.0129177 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_{2})}{E^*} (E_{1})-\frac{p_{1(z)}}{(E_{1}+m_2)} \frac{(E_{1}+m_{2})}{E^*} p_{1(z)} \\
 
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p^*_{1(z)}=-\frac{p_{1}}{(E_{1}+m_2)} \frac{(E_{1}+m_{2})}{E^*}(E_{1})+\frac{(E_{1}+m_{2})}{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 53.013 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.
 
−
 
 
−
{| class="wikitable" align="center"
 
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| style="background: gray"      | <math> p^*_{1}=p^*_{2}\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}= E^*_{2}\approx 53.013 MeV</math></center>
 
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|}
 
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−
==Final Conditions==
 
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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^'_{2}</math> ,  in the Lab frame,
 
−
 
 
−
<center>[[File:xz_lab.png]]</center>
 
−
<center>'''Figure 3: Definition of Moller electron variables in the Lab Frame in the x-z plane.'''</center>
 
−
 
 
−
<center>Using <math>\theta '_2=\arccos \left(\frac{p^'_{2(z)}}{p^'_{2}}\right)</math></center>
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"      | <math>\Longrightarrow {p^'_{2(z)}=p^'_{2}\cos(\theta '_2)}</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 '_2 \le 60^\circ \equiv 0 \le \theta '_2 \le 1.046\ Radians</math>
 
−
|}
 
−
 
 
−
Since,
 
−
<center><math>\frac{p^'_{2(z)}}{p^'_{2}}=cos(\theta '_2)</math></center>
 
−
 
 
−
 
 
−
<center><math>\Longrightarrow p^'_{2(z)}\ should\ always\ be\ positive</math></center>
 
−
<center>[[File:xy_lab.png]]</center>
 
−
<center>'''Figure 4: Definition of Moller electron variables in the Lab Frame in the x-y plane.'''</center>
 
−
 
 
−
<center>Similarly, <math>\phi '_2=\arccos \left( \frac{p^'_{2(x) Lab}}{p^'_{2(xy)}} \right)</math></center>
 
−
 
 
−
 
 
−
<center>where <math>p_{2(xy)}^'=\sqrt{(p_{2(x)}^')^2+(p^'_{2(y)})^2}</math></center>
 
−
 
 
−
 
 
−
<center><math>(p^'_{2(xy)})^2=(p^'_{2(x)})^2+(p^'_{2(y)})^2</math></center>
 
−
 
 
−
 
 
−
<center>and using <math>p^2=p_{(x)}^2+p_{(y)}^2+p_{(z)}^2</math></center>
 
−
 
 
−
 
 
−
<center>this gives <math>(p^'_{2})^2=(p^'_{2(xy)})^2+(p^'_{2(z)})^2</math></center>
 
−
 
 
−
 
 
−
<center><math>\Longrightarrow (p'_{2})^2-(p'_{2(z)})^2 = (p'_{2(xy)})^2</math></center>
 
−
 
 
−
 
 
−
<center><math>\Longrightarrow p_{2(xy)}^'=\sqrt{(p^'_{2})^2-(p^'_{2(z)})^2}</math></center>
 
−
 
 
−
 
 
−
<center>which gives<math>\phi '_2 = \arccos \left( \frac{p_{2(x)}'}{\sqrt{p_{2}^{'\ 2}-p_{2(z)}^{'\ 2}}}\right)</math></center>
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"      | <math>\Longrightarrow p_{2(x)}'=\sqrt{p_{2}^{'\ 2}-p_{2(z)}^{'\ 2}} \cos(\phi)</math>
 
−
|}
 
−
 
 
−
 
 
−
<center>Similarly, using <math>p_{2}^2=p_{2(x)}^2+p_{2(y)}^2+p_{2(z)}^2</math></center>
 
−
 
 
−
 
 
−
<center><math>\Longrightarrow p_{2}^{'\ 2}-p_{2(x)}^{'\ 2}-p_{2(z)}^{'\ 2}=p_{2(y)}^{'\ 2}</math></center>
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"      | <math>p_{2(y)}'=\sqrt{p_{2}^{'\ 2}-p_{2(x)}^{'\ 2}-p_{2(z)}^{'\ 2}}</math>
 
−
|}
 
−
 
 
−
 
 
−
Checking on the sign from the cosine results for <math>\phi '_2</math>
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: red"      | <math>-\pi \le \phi '_2 \le \pi\ Radians</math>
 
−
|}
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: red"      | <math>For\ 0 \le \phi '_2 \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 '_2 \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 '_2 \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 '_2 \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^*_{2(x)}\Leftrightarrow p_{2(x)}'</math>
 
−
|}
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"  | <math>p^*_{2(y)}\Leftrightarrow p_{2(y)}'</math>
 
−
|}
 
−
 
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"  | <math>p^*_{2(z)}=\sqrt {(p^*_2)^2-(p^*_{2(x)})^2-(p^*_{2(y)})^2}</math>
 
−
|}
 
−
 
 
−
 
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"      | <math>\Longrightarrow E^*_{2}=\approx 53.013 MeV</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^*_{2(x)}= -p^*_{1(x)}</math>
 
−
|}
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"  | <math>p^*_{2(y)}= -p^*_{1(y)}</math>
 
−
|}
 
−
 
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"  | <math>p^*_{2(z)}=-p^*_{1(z)}</math>
 
−
|}
 
−
 
 
−
where previously it was shown
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"      | <math>p^*_{1}\approx 53.013 MeV</math>
 
−
|}
 
−
 
 
−
 
 
−
 
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"      | <math>E^*_{1}\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'_{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}+E^*_{2}\\ p^*_{1(x)}+p^*_{2(x)} \\ p^*_{(1(y)}+p^*_{2(y)} \\ p^*_{1(z)}+p^*_{2(z)}\end{matrix} \right)</math></center>
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"      | <math>p^*_{1(x)}= p'_{1(x)}</math>
 
−
|}
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"  | <math>p^*_{1(y)}= p'_{1(y)}</math>
 
−
|}
 
−
 
 
−
 
 
−
<center><math>\Longrightarrow\begin{cases}
 
−
E'=\gamma E^*+\beta \gamma p^*_{z} \\
 
−
p'_{z}=\beta \gamma E^*+ \gamma p^*_{z}
 
−
\end{cases}</math></center>
 
−
 
 
−
 
 
−
<center><math>\Longrightarrow\begin{cases}
 
−
\gamma = \frac{E'}{E^*} \\
 
−
\beta =\frac{p'_{z}}{E'}
 
−
\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^*} ^2=(E^*_{1}+E^*_{2})^2-(\vec p\ ^*_{1}+\vec p\ ^*_{2})^2</math></center>
 
−
 
 
−
 
 
−
<center><math>s={\mathbf P}^2=(E'_{1}+E'_{2})^2-(\vec p\ '_{1}+\vec p\ '_{2})^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}+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>
 
−
 
 
−
 
 
−
<center><math>(E^*)^2-(\vec p\ ^*)^2=(E'_{1}+E'_{2})^2-(\vec p\ ')^2</math></center>
 
−
 
 
−
 
 
−
<center><math>(E^*)^2=(E')^2-(\vec p\ ')^2</math></center>
 
−
 
 
−
 
 
−
<center><math>(E^*)^2+(\vec{p'})^2=(E')^2</math></center>
 
−
 
 
−
 
 
−
<center><math>E'=\sqrt{(E^*)^2+(\vec p\ ')^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'=\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'\equiv E'_{1}+E'_{2}</math></center>
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"      | <math>\Longrightarrow E'_{1}=E'-E'_{2}</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_{1}^{'\ 2}=E_{1}^{'\ 2}-m_{1}^2=E_{1}^{'\ 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_{1(z)}'=\sqrt {p_{1}^{'\ 2}-p_{1(x)}^{'\ 2}-p_{1(y)}^{'\ 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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