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

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==Calculations of 4-momentum components==
 
  
 +
===Verification===
 +
[[Verification of Relativistic Components]]
  
−
 
+
===Conclusion===
−
 
+
[[Verification_of_Relativistic_Components#Conclusion | Conclusion]]
−
 
+
----
−
 
 
−
===Initial Lab Frame===
 
−
 
 
−
Assume the incoming electron has momentum of 11000 MeV in the positive z direction.
 
−
 
 
−
<center><math>\vec p_{1(z)}\equiv \vec p_{1}=11000 MeV </math>ẑ</center>
 
−
 
 
−
The moller electron is initially at rest
 
−
 
 
−
<center><math>\vec p_{2}\equiv 0</math></center>
 
−
 
 
−
The total energy in this frame
 
−
 
 
−
<center><math>E\equiv \sqrt{p^2+m^2}</math></center>
 
−
 
 
−
 
 
−
<center><math>E\equiv \sqrt{(p_{1}+p_{2})^2+(m_{1}+m_{2})^2}</math></center>
 
−
 
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"      | <math>\Longrightarrow E\equiv \sqrt{(11000 MeV)^2+(.511 MeV+.511 MeV)^2}\approx 11000 MeV</math>
 
−
|}
 
−
 
 
−
===Center of Mass Frame===
 
−
Using the definition
 
−
 
 
−
 
 
−
<center><math>E^2\equiv p^2c^2+m^2c^4</math></center>
 
−
 
 
−
 
 
−
<center><math>\Longrightarrow p^2=\frac {E^2}{c^2}-m^2 c^2</math></center>
 
−
 
 
−
 
 
−
We can use 4-momenta vectors, i.e. <math>{\mathbf P}\equiv \left(E/cpxpypz
\right)</math>
 
−
 
 
−
 
 
−
we can use the fact that the scalar product of a 4-momenta with itself,
 
−
 
 
−
 
 
−
<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>
 
−
 
 
−
 
 
−
is invariant
 
−
 
 
−
Using this, the sum of two 4-momenta forms a 4-vector as well
 
−
 
 
−
<center><math>{\mathbf P_1}+ {\mathbf P_2}= \left( E1+E2→p1c+→p2c
\right)= {\mathbf P}</math></center>
 
−
 
 
−
 
 
−
The length of this four-vector is an invariant as well
 
−
 
 
−
<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>
 
−
 
 
−
 
 
−
 
 
−
 
 
−
 
 
−
<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>
 
−
<center> ''(with c=1)'' </center>
 
−
 
 
−
 
 
−
For incoming electrons moving only in the z-direction, we can write
 
−
 
 
−
 
 
−
<center><math>{\mathbf P_1}+ {\mathbf P_2}= \left( E1+E200p1(z)+p2(z)
\right)={\mathbf P}</math></center>
 
−
 
 
−
 
 
−
We can perform a Lorentz transformation to the Center of Mass frame, with zero total momentum
 
−
 
 
−
 
 
−
<center><math>\left( E∗1+E∗2000
\right)=\left(γ∗00−β∗γ∗01000010−β∗γ∗00γ∗
\right) . \left( E1+E200p1(z)+p2(z)
\right)</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>
 
−
 
 
−
 
 
−
<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
 
−
 
 
−
<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_{1}}+\vec p_{2})^2</math></center>
 
−
 
 
−
 
 
−
<center><math>(E^*)^2=(E_{1}+E_{2})^2-(\vec{p_{1}}+\vec p_{2})^2</math></center>
 
−
 
 
−
 
 
−
<center><math>E^*=[(E_{1}+E_{2})^2-(\vec{p_{1}}+\vec p_{2})^2]^{1/2}</math></center>
 
−
 
 
−
 
 
−
<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>
 
−
 
 
−
 
 
−
<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>
 
−
 
 
−
 
 
−
<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>
 
−
 
 
−
 
 
−
 
 
−
<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>
 
−
 
 
−
 
 
−
<center><math>E^*=[m_{1}^2+m_{2}^2+2E_{1}E_{2}-2p_{1} p_{2}\cos(\theta) ]^{1/2}</math></center>
 
−
 
 
−
 
 
−
 
 
−
<center><math>E^*=[m_{1}^2+m_{2}^2+2E_{1}E_{2}-2p_{1} p_{2}\cos(\theta) ]^{1/2}</math></center>
 
−
 
 
−
 
 
−
Using the relations <math>\beta\equiv \vec p/E\Longrightarrow \vec p=\beta E</math>
 
−
 
 
−
 
 
−
<center><math>E^*=[m_{1}^2+m_{2}^2+2E_{1}E_{2}(1-\beta_{1}\beta_{2}\cos(\theta))]^{1/2}</math></center>
 
−
 
 
−
 
 
−
<center>where <math> \theta </math> is the angle between the particles in the Lab frame.</center>
 
−
 
 
−
 
 
−
 
 
−
 
 
−
In the frame where one particle ''(m<sub>2</sub>)'' is at rest
 
−
 
 
−
 
 
−
<center><math>\Longrightarrow \beta_{2}=0</math></center>
 
−
 
 
−
 
 
−
<center><math>\Longrightarrow p_{2}=0</math></center>
 
−
 
 
−
 
 
−
which implies,
 
−
 
 
−
 
 
−
<center><math> E_{2}=[p_{2}^2+m_{2}^2]^{1/2}=m_{2}</math></center>
 
−
 
 
−
 
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| 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:
 
−
 
 
−
 
 
−
<center><math>\left( E∗1+E∗2000
\right)=\left(γ∗00−β∗γ∗01000010−β∗γ∗00γ∗
\right) . \left( E1+E200p1(z)+p2(z)
\right)</math></center>
 
−
 
 
−
 
 
−
For the case of a stationary electron, this simplifies to:
 
−
 
 
−
<center><math>\left( E∗p∗xp∗yp∗z
\right)=\left(γ∗00−β∗γ∗01000010−β∗γ∗00γ∗
\right) . \left( E1+m200p1(z)+0
\right)</math></center>
 
−
 
 
−
 
 
−
which gives,
 
−
 
 
−
 
 
−
<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>
 
−
 
 
−
 
 
−
Solving for <math>\beta^*</math>, with <math>p^*_{z}=0</math>
 
−
 
 
−
<center><math>\Longrightarrow \beta^* \gamma^*(E_{1}+m_{2})=\gamma^* p_{1(z)}</math></center>
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"      | <math>\Longrightarrow \beta^*=\frac{p_{1}}{(E_{1}+m_{2})}</math>
 
−
|}
 
−
 
 
−
 
 
−
 
 
−
Similarly, solving for <math>\gamma^*</math> by substituting in <math>\beta^*</math>
 
−
 
 
−
 
 
−
<center><math>E^*=\gamma^* (E_{1}+m_{2})-\frac{p_{1}}{(E_{1}+m_{2})} \gamma^* p_{1(z)}</math></center>
 
−
 
 
−
 
 
−
<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>
 
−
 
 
−
 
 
−
Using the fact that <math>E^*=[(E_{1}+E_{2})^2-(\vec p_{1}+\vec p_{2})^2]^{1/2}</math>
 
−
 
 
−
 
 
−
<center><math>E^*=\gamma^* \frac{E^*\ ^2}{(E_{1}+m_{2})}</math></center>
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"      | <math>\Longrightarrow \gamma^*=\frac{(E_1+m_2)} {E^*}</math>
 
−
|}
 
−
 
 
−
 
 
−
 
 
−
This gives the momenta of the particles in the center of mass to have equal magnitude, but opposite directions
 
−
 
 
−
 
 
−
Using the relation
 
−
 
 
−
<center><math>\left( E∗1+E∗2p∗1(x)+p∗2(x)p∗1(y)+p∗2(y)p∗1(z)+p∗2(z)
\right)=\left(γ∗00−β∗γ∗01000010−β∗γ∗00γ∗
\right) . \left( E1+m200p1(z)+0
\right)</math></center>
 
−
 
 
−
 
 
−
<center><math>\Longrightarrow \left( E∗2p∗2(x)p∗2(y)p∗2(z)
\right)=\left(γ∗00−β∗γ∗01000010−β∗γ∗00γ∗
\right) . \left( m000
\right)</math></center>
 
−
 
 
−
 
 
−
<center><math>\Longrightarrow\begin{cases}
 
−
E^*_{2}=\gamma^* (m_{2}) \\
 
−
p^*_{2(z)}=-\beta^* \gamma^* (m_{2})
 
−
\end{cases}</math></center>
 
−
 
 
−
 
 
−
<center><math>\Longrightarrow\begin{cases}
 
−
E^*_{2}=\frac{(E_{1}+m_{2})}{E^*} (m_2) \\
 
−
p^*_{2(z)}=-\frac{p_{1}}{(E_{1}+m_2)} \frac{(E_{1}+m_{2})}{E^*} (m_{2})
 
−
\end{cases}</math></center>
 
−
 
 
−
 
 
−
<center><math>\Longrightarrow\begin{cases}
 
−
E^*_{2}=\frac{(11000 MeV+.511 MeV)}{106.031 MeV} (.511 MeV) \approx 53.0129177 MeV\\
 
−
p^*_{2(z)}=-\frac{11000 MeV}{106.031 MeV} (.511 MeV) \approx -53.013 MeV
 
−
\end{cases}</math></center>
 
−
 
 
−
 
 
−
<center><math>\Longrightarrow \left( E∗1p∗1(x)p∗1(y)p∗1(z)
\right)=\left(γ∗00−β∗γ∗01000010−β∗γ∗00γ∗
\right) . \left( E100p1(z)
\right)</math></center>
 
−
 
 
−
 
 
−
<center><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></center>
 
−
 
 
−
 
 
−
 
 
−
<center><math>\Longrightarrow\begin{cases}
 
−
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)} \\
 
−
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)}
 
−
\end{cases}</math></center>
 
−
 
 
−
 
 
−
 
 
−
<center><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 53.013 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></center>
 
−
 
 
−
 
 
−
 
 
−
<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>
 
−
 
 
−
 
 
−
<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>
 
−
 
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"      | <math> p^*_{1}=p^*_{2}\approx 53.013 MeV</math>
 
−
|}
 
−
 
 
−
 
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"      | <center><math>E^*_{1}= E^*_{2}\approx 53.013 MeV</math></center>
 
−
|}
 
−
 
 
−
===Moller electron Lab Frame===
 
−
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_(m)LAB.png]]</center>
 
−
 
 
−
<center>Using <math>\theta=\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)}</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>
 
−
|}
 
−
 
 
−
Since,
 
−
<center><math>\frac{p^'_{2(z)}}{p^'_{2}}=cos(\theta)</math></center>
 
−
 
 
−
 
 
−
<center><math>\Longrightarrow p^'_{2(z)}\ should\ always\ be\ positive</math></center>
 
−
<center>[[File:Xy_Lab.png]]</center>
 
−
 
 
−
<center>Similarly, <math>\phi=\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 = \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</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^*_{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-(p^*_{2(x)})^2-(p^*_{2(y)})^2}</math>
 
−
|}
 
−
 
 
−
 
 
−
 
 
−
 
 
−
{| class="wikitable" align="center"
 
−
| style="background: gray"      | <math>\Longrightarrow E^*_{2}=\sqrt{\frac{m(m+E_{1})}{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^*_{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( E1+E2p1(x)+p2(x)p1(y)+p2(y)p1(z)+p2(z)
\right)=\left(γ∗00β∗γ∗01000010β∗γ∗00γ∗
\right) . \left( E∗1+E∗2p∗1(x)+p∗2(x)p∗(1(y)+p∗2(y)p∗1(z)+p∗2(z)
\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>
 
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<center><math>\Longrightarrow\begin{cases}
 
−
\gamma = \frac{E}{E^*} \\
 
−
\beta =\frac{p_{z}}{E}
 
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\end{cases}</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
 
−
 
 
−
<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>
 
−
 
 
−
 
 
−
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>
 
−
 
 
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<center><math>(E^*)^2-(\vec p^*)^2=(E_{1}+E_{2})^2-(\vec{p})^2</math></center>
 
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<center><math>(E^*)^2=(E)^2-(\vec{p})^2</math></center>
 
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<center><math>(E^*)^2+(\vec{p})^2=(E)^2</math></center>
 
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<center><math>E=\sqrt{(E^*)^2+(\vec{p})^2}</math></center>
 
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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>
 
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<center><math>E=\sqrt{(106.031 MeV)^2+(11000 MeV)^2}\approx 11000.511 MeV</math></center>
 
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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.
 
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Since,
 
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<center><math>E\equiv E_{1}+E_{2}</math></center>
 
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{| class="wikitable" align="center"
 
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| style="background: gray"      | <math>\Longrightarrow E_{1}=E-E_{2}</math>
 
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|}
 
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Using the relation
 
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<center><math>E^2\equiv p^2+m^2</math></center>
 
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{| class="wikitable" align="center"
 
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| style="background: gray"      | <math>\Longrightarrow p_{1}^2=E_{1}^2-m_{1}^2=E_{1}^2-(.511 MeV)^2</math>
 
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|}
 
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Using
 
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<center><math>p^2 \equiv p_x^2+p_y^2+p_z^2</math></center>
 
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{| class="wikitable" align="center"
 
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| style="background: gray"      | <math>\Longrightarrow p_{1(z)}=\sqrt{p_{1}^2-p_{1(x)}^2-p_{1(y)}^2}</math>
 
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|}
 
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[[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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