[math]\displaystyle{ \textbf{\underline{Navigation}} }[/math]
[math]\displaystyle{ \vartriangleleft }[/math]
[math]\displaystyle{ \triangle }[/math]
[math]\displaystyle{ \vartriangleright }[/math]
Special Case of Equal Mass Particles
For incoming electrons moving only in the z-direction, we can write
[math]\displaystyle{ {\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]
We can perform a Lorentz transformation to the Center of Mass frame, with zero total momentum
[math]\displaystyle{ \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]
Without knowing the values for gamma or beta, we can utalize the fact that lengths of the two 4-momenta are invariant
[math]\displaystyle{ {\mathbf P^*}^2=(E^*_{1}+E^*_{2})^2-(\vec p\ ^*_{1}+\vec p\ ^*_{2})^2=(m_{1}^*+m_{2}^*)^2 }[/math]
[math]\displaystyle{ {\mathbf P}^2=(E_{1}+E_{2})^2-(\vec p_{1}+\vec p_{2})^2=(m_{1}+m_{2})^2 }[/math]
This gives,
[math]\displaystyle{ \Longrightarrow (m_{1}^*+m_{2}^*)^2=(m_{1}+m_{2})^2 }[/math]
Using the fact that
[math]\displaystyle{ \begin{cases}
m_{1}=m_{2} \\
m_{1}^*=m_{2}^*
\end{cases} }[/math]
since the rest mass energy of the electrons remains the same in inertial frames.
Substituting, we find
[math]\displaystyle{ (m_{1}^*+m_{1}^*)^2=(m_{1}+m_{1})^2 }[/math]
[math]\displaystyle{ 2m_{1}^*=2m_{1} }[/math]
[math]\displaystyle{ m_{1}^*=m_{1} }[/math]
[math]\displaystyle{ \Longrightarrow m_{1}=m^*_{1}\ ; m_{2}=m^*_{2} }[/math]
This confirms that the mass remains constant between the frames of reference.