Special Case of Equal Mass Particles

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Special Case of Equal Mass Particles

For incoming electrons moving only in the z-direction, we can write

[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]

We can perform a Lorentz transformation to the Center of Mass frame, with zero total momentum

[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]

Without knowing the values for gamma or beta, we can utalize the fact that lengths of the two 4-momenta are invariant

[math]{\mathbf P^*}^2=(E^*_{1}+E^*_{2})^2-(\vec p\ ^*_{1}+\vec p\ ^*_{2})^2=(m_{1}^*+m_{2}^*)^2[/math]

[math]{\mathbf P}^2=(E_{1}+E_{2})^2-(\vec p_{1}+\vec p_{2})^2=(m_{1}+m_{2})^2[/math]

This gives,

[math]\Longrightarrow (m_{1}^*+m_{2}^*)^2=(m_{1}+m_{2})^2[/math]

Using the fact that

[math]\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](m_{1}^*+m_{1}^*)^2=(m_{1}+m_{1})^2[/math]

[math]2m_{1}^*=2m_{1}[/math]

[math]m_{1}^*=m_{1}[/math]

[math]\Longrightarrow m_{1}=m^*_{1}\ ; m_{2}=m^*_{2}[/math]

This confirms that the mass remains constant between the frames of reference.

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