Scattered and Moller Electron Energies in CM Frame - Revision history

Vanwdani at 18:54, 15 May 2018

2018-05-15T18:54:33Z

Vanwdani: /* Scattered and Moller Electron energies in CM */

2017-06-16T01:53:14Z

Scattered and Moller Electron energies in CM

Vanwdani at 01:52, 16 June 2017

2017-06-16T01:52:40Z

Vanwdani at 01:50, 16 June 2017

2017-06-16T01:50:47Z

Vanwdani: /* Scattered and Moller Electron energies in CM */

2017-06-16T01:49:58Z

Scattered and Moller Electron energies in CM

Vanwdani at 01:45, 16 June 2017

2017-06-16T01:45:30Z

Vanwdani: /* Scattered and Moller Electron energies in CM */

2017-06-16T01:43:33Z

Scattered and Moller Electron energies in CM

Vanwdani at 22:30, 15 June 2017

2017-06-15T22:30:02Z

Vanwdani at 22:11, 15 June 2017

2017-06-15T22:11:40Z

Vanwdani: Created page with "=Scattered and Moller Electron energies in CM= Inspecting the Lorentz transformation to the Center of Mass frame: $\left( \begin{matrix}E^_{1}+E^_{2}\\ 0 \\ 0 \\…"$

2017-05-31T17:59:17Z

Created page with "=Scattered and Moller Electron energies in CM= Inspecting the Lorentz transformation to the Center of Mass frame: <center><math>\left( \begin{matrix}E^*_{1}+E^*_{2}\\ 0 \\ 0 \\…"

New page

=Scattered and Moller Electron energies in CM=
Inspecting the Lorentz transformation to the Center of Mass frame:

<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>

For the case of a stationary electron, this simplifies to:

<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>

which gives,

<center><math>\Longrightarrow\begin{cases}
E^*=\gamma^* (E_{1}+m)-\beta^* \gamma^* p_{1(z)} \\
p^*_{z}=-\beta^* \gamma^*(E_{1}+m)+\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)=\gamma^* p_{1(z)}</math></center>

{| class="wikitable" align="center"
| style="background: gray" | <math>\Longrightarrow \beta^*=\frac{p_{1}}{(E_{1}+m)}</math>
|}

Similarly, solving for <math>\gamma^*</math> by substituting in <math>\beta^*</math>

<center><math>E^*=\gamma^* (E_{1}+m)-\frac{p_{1}}{(E_{1}+m)} \gamma^* p_{1(z)}</math></center>

<center><math>E^*=\gamma^* \frac{(E_{1}+m)^2}{(E_{1}+m)}-\gamma^*\frac{(p_{1(z)})^2}{(E_{1}+m)}</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)}</math></center>

{| class="wikitable" align="center"
| style="background: gray" | <math>\Longrightarrow \gamma^*=\frac{(E_1+m)} {E^*}</math>
|}

Using the relation

<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>\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>\Longrightarrow\begin{cases}
E^*_{2}=\gamma^* (m) \\
p^*_{2(z)}=-\beta^* \gamma^* (m)
\end{cases}</math></center>

<center><math>\Longrightarrow\begin{cases}
E^*_{2}=\frac{(E_{1}+m)}{E^*} (m) \\
p^*_{2(z)}=-\frac{p_{1}}{(E_{1}+m)} \frac{(E_{1}+m)}{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 53.015 MeV\\
p^*_{2(z)}=-\frac{11000 MeV}{106.031 MeV} (.511 MeV) \approx -53.013 MeV
\end{cases}</math></center>

<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>

<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)}{E^*} (E_{1})-\frac{p_{1(z)}}{(E_{1}+m)} \frac{(E_{1}+m)}{E^*} p_{1(z)} \\
p^*_{1(z)}=-\frac{p_{1}}{(E_{1}+m)} \frac{(E_{1}+m)}{E^*}(E_{1})+\frac{(E_{1}+m)}{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 53.015 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>

This gives the momenta of the particles in the center of mass to have equal magnitude, but opposite directions.

{| class="wikitable" align="center"
| style="background: gray" | <center><math>E^*=E^*_1+E^*_2=\sqrt{2m(m+E_1)}</math></center>
|}

@@ Line 1: / Line 1: @@
-<center><math>\textbf{\underline{Navigation}}</math>
+<center><math>\underline{\textbf{Navigation}}</math>
 [[Total_Energy_in_CM_Frame|<math>\vartriangleleft </math>]]
@@ Line 74: / Line 74: @@
-<center><math>\textbf{\underline{Navigation}}</math>
+<center><math>\underline{\textbf{Navigation}}</math>
 [[Total_Energy_in_CM_Frame|<math>\vartriangleleft </math>]]

@@ Line 65: / Line 65: @@
 {| class="wikitable" align="center"
-| style="background: gray"      | <math>\Rightarrow E_1=\frac{106.030760886 MeV}{2}=53.015380443 MeV=E_2</math>
+| style="background: gray"      | <math>\Rightarrow E_1^*=\frac{106.030760886 MeV}{2}=53.015380443 MeV=E_2^*</math>
 |}

@@ Line 1: / Line 1: @@
 <center><math>\textbf{\underline{Navigation}}</math>
-[[Limits_based_on_Mandelstam_Variables|<math>\vartriangleleft </math>]]
+[[Total_Energy_in_CM_Frame|<math>\vartriangleleft </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>]]
 </center>
@@ Line 76: / Line 76: @@
 <center><math>\textbf{\underline{Navigation}}</math>
-[[Limits_based_on_Mandelstam_Variables|<math>\vartriangleleft </math>]]
+[[Total_Energy_in_CM_Frame|<math>\vartriangleleft </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>]]
 </center>

← Older revision		Revision as of 01:50, 16 June 2017
Line 67:		Line 67:
	\| style="background: gray" \| <math>\Rightarrow E_1=\frac{106.030760886 MeV}{2}=53.015380443 MeV=E_2</math>		\| style="background: gray" \| <math>\Rightarrow E_1=\frac{106.030760886 MeV}{2}=53.015380443 MeV=E_2</math>
	\|}		\|}
		+
		+
		+
		+
		+	----
		+
		+
		+	<center><math>\textbf{\underline{Navigation}}</math>
		+
		+	[[Limits_based_on_Mandelstam_Variables\|<math>\vartriangleleft </math>]]
		+	[[VanWasshenova_Thesis#Initial_4-momentum_Components\|<math>\triangle </math>]]
		+	[[Special_Case_of_Equal_Mass_Particles\|<math>\vartriangleright </math>]]
		+
		+	</center>

@@ Line 49: / Line 49: @@
-<center><math>s_{CM} \equiv 2m^2+2E_1^{*2}+2\vec p_1 \ ^{*2} </math></center>
+<center><math>s \equiv 2m^2+2E_1^{*2}+2\vec p_1 \ ^{*2} </math></center>
@@ Line 57: / Line 57: @@
-<center><math>s_{CM} \equiv 2m^2+2m^2+2\vec p_1 \ ^{*2}+\vec p_1 \ ^{*2})</math></center>
+<center><math>s \equiv 2m^2+2m^2+2\vec p_1 \ ^{*2}+\vec p_1 \ ^{*2})</math></center>
-<center><math>s_{CM}=4(m^2+\vec p_1 \ ^{*2})=(2E_1^*)^{2}=E^{*2}</math></center>
+<center><math>s=4(m^2+\vec p_1 \ ^{*2})=(2E_1^*)^{2}=E^{*2}</math></center>

← Older revision		Revision as of 01:45, 16 June 2017
Line 12:		Line 12:

	We can use the Mandelstam variable s, the square of the center of mass energy, to find <math>E^*</math>		We can use the Mandelstam variable s, the square of the center of mass energy, to find <math>E^*</math>
		+
		+	<center><math>s \equiv \left({\mathbf P_1^}+ {\mathbf P_2^{}}\right)^2</math></center>
		+
		+
		+	<center><math>s \equiv \mathbf P_1^{2}+2 \mathbf P_1^ \mathbf P_2^+ \mathbf P_2^{2}</math></center>
		+
		+
		+	As shown earlier, the square of a 4-momentum is
		+
		+
		+	<center><math>\mathbf P^{2} \equiv m^2</math></center>
		+
		+	This gives,
		+	<center><math>s \equiv m_1^{2}+2 \mathbf P_1^* \mathbf P_2^*+ m_2^{2}</math></center>
		+
		+
		+	For the case <math>m_1=m_2=m</math>
		+
		+
		+
		+	<center><math>s \equiv 2m^{2}+2 \mathbf P_1^* \mathbf P_2^*</math></center>
		+
		+	Using the relationship
		+
		+
		+	<center><math>\mathbf P_1 \cdot \mathbf P_2 = E_{1}E_{2}-(\vec p_1 \vec p_2)</math></center>
		+
		+
		+
		+	<center><math>s \equiv 2m^2+2(E_1^E_2^-\vec p \ _1^* \vec p \ _2^*)</math></center>
		+
		+
		+	In the center of mass frame of reference,
		+
		+	<center><math>E_1^=E_2^ \quad and \quad \vec p \ _1^=-\vec p \ _2^</math></center>
		+
		+
		+	<center><math>s_{CM} \equiv 2m^2+2E_1^{2}+2\vec p_1 \ ^{2} </math></center>
		+
		+
		+	Using the relativistic energy equation
		+
		+	<center><math>E^2 \equiv \vec p_1 \ ^2+m^2</math></center>
		+
		+
		+	<center><math>s_{CM} \equiv 2m^2+2m^2+2\vec p_1 \ ^{2}+\vec p_1 \ ^{2})</math></center>
		+
		+
		+	<center><math>s_{CM}=4(m^2+\vec p_1 \ ^{2})=(2E_1^)^{2}=E^{*2}</math></center>

← Older revision		Revision as of 01:43, 16 June 2017
Line 9:		Line 9:

	=Scattered and Moller Electron energies in CM=		=Scattered and Moller Electron energies in CM=
−	~~Inspecting the Lorentz transformation to the Center of Mass frame:~~


−	<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>
−
−
−	~~For~~ the ~~case of a stationary electron, this simplifies to:~~
−
−	<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>
−
−
−	~~which gives,~~
−
−
−	~~<center><math>\Longrightarrow\begin{cases}~~
−	~~E^=\gamma^ (E_{1}+m)-\beta^* \gamma^* p_{1(z)} \\~~
−	~~p^_{z}=-\beta^ \gamma^(E_{1}+m)+\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)=\gamma^ p_{1(z)}</math></center>~~
−
−
−	~~{\| class="wikitable" align="center"~~
−	~~\| style="background: gray" \| <math>\Longrightarrow \beta^*=\frac{p_{1}}{(E_{1}+m)}</math>~~
−	\|}
−
−
−
−	~~Similarly~~, ~~solving for <math>\gamma^</math> by substituting in <math>\beta^</math>~~
−
−
−	~~<center><math>E^=\gamma^ (E_{1}+m)-\frac{p_{1}}{(E_{1}+m)} \gamma^* p_{1(z)}</math></center>~~
−
−
−	~~<center><math>E^=\gamma^ \frac{(E_{1}+m)^2}{(E_{1}+m)}-\gamma^*\frac{(p_{1(z)})^2}{(E_{1}+m)}</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)}</math></center>~~
−
−	~~{\| class="wikitable" align="center"~~
−	~~\| style="background: gray" \| <math>\Longrightarrow \gamma^=\frac{(E_1+m)} {E^}</math>~~
−	\|}
−
−
−
−
−	~~Using the relation~~
−
−	<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>\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>\Longrightarrow\begin{cases}~~
−	~~E^_{2}=\gamma^ (m) \\~~
−	~~p^_{2(z)}=-\beta^ \gamma^* (m)~~
−	~~\end{cases}</math></center>~~
−
−
−	~~<center><math>\Longrightarrow\begin{cases}~~
−	~~E^_{2}=\frac{(E_{1}+m)}{E^} (m) \\~~
−	~~p^_{2(z)}=-\frac{p_{1}}{(E_{1}+m)} \frac{(E_{1}+m)}{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 53.015 MeV\\~~
−	~~p^*_{2(z)}=-\frac{11000 MeV}{106.031 MeV} (.511 MeV) \approx -53.013 MeV~~
−	~~\end{cases}</math></center>~~
−
−
−	<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>
−
−
−	~~<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)}{E^} (E_{1})-\frac{p_{1(z)}}{(E_{1}+m)} \frac{(E_{1}+m)}{E^*} p_{1(z)} \\~~
−	~~p^_{1(z)}=-\frac{p_{1}}{(E_{1}+m)} \frac{(E_{1}+m)}{E^}(E_{1})+\frac{(E_{1}+m)}{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 53.015 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>~~
−
−
−	~~This gives~~ the ~~momenta~~ of ~~the particles in~~ the center of mass to ~~have equal magnitude, but opposite directions.~~
−
−
−
−
−	~~{\| class="wikitable" align="center"~~
−	~~\| style="background: gray" \| <center>~~<math>E^~~=E^_1+E^*_2=\sqrt{2m(m+E_1)}~~</math>~~</center>~~
−	\|}

← Older revision		Revision as of 22:11, 15 June 2017
Line 1:		Line 1:
		+	<center><math>\textbf{\underline{Navigation}}</math>
		+
		+	[[Limits_based_on_Mandelstam_Variables\|<math>\vartriangleleft </math>]]
		+	[[VanWasshenova_Thesis#Initial_Lab_Frame_4-momentum_components\|<math>\triangle </math>]]
		+	[[Special_Case_of_Equal_Mass_Particles\|<math>\vartriangleright </math>]]
		+
		+	</center>
		+
		+
	=Scattered and Moller Electron energies in CM=		=Scattered and Moller Electron energies in CM=
	Inspecting the Lorentz transformation to the Center of Mass frame:		Inspecting the Lorentz transformation to the Center of Mass frame:

Scattered and Moller Electron Energies in CM Frame - Revision history

Vanwdani at 18:54, 15 May 2018

Vanwdani: /* Scattered and Moller Electron energies in CM */

Vanwdani at 01:52, 16 June 2017

Vanwdani at 01:50, 16 June 2017

Vanwdani: /* Scattered and Moller Electron energies in CM */

Vanwdani at 01:45, 16 June 2017

Vanwdani: /* Scattered and Moller Electron energies in CM */

Vanwdani at 22:30, 15 June 2017

Vanwdani at 22:11, 15 June 2017

Vanwdani: Created page with "=Scattered and Moller Electron energies in CM= Inspecting the Lorentz transformation to the Center of Mass frame: \left( \begin{matrix}E^*_{1}+E^*_{2}\\ 0 \\ 0 \\…"

Vanwdani: Created page with "=Scattered and Moller Electron energies in CM= Inspecting the Lorentz transformation to the Center of Mass frame: $\left( \begin{matrix}E^_{1}+E^_{2}\\ 0 \\ 0 \\…"$