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Scattered and Moller Electron energies in CM
We can use the Mandelstam variable s, the square of the center of mass energy, to find [math]E^*[/math]
[math]s \equiv \left({\mathbf P_1^*}+ {\mathbf P_2^{*}}\right)^2[/math]
[math]s \equiv \mathbf P_1^{*2}+2 \mathbf P_1^* \mathbf P_2^*+ \mathbf P_2^{*2}[/math]
As shown earlier, the square of a 4-momentum is
[math]\mathbf P^{2} \equiv m^2[/math]
This gives,
[math]s \equiv m_1^{2}+2 \mathbf P_1^* \mathbf P_2^*+ m_2^{2}[/math]
For the case [math]m_1=m_2=m[/math]
[math]s \equiv 2m^{2}+2 \mathbf P_1^* \mathbf P_2^*[/math]
Using the relationship
[math]\mathbf P_1 \cdot \mathbf P_2 = E_{1}E_{2}-(\vec p_1 \vec p_2)[/math]
[math]s \equiv 2m^2+2(E_1^*E_2^*-\vec p \ _1^* \vec p \ _2^*)[/math]
In the center of mass frame of reference,
[math]E_1^*=E_2^* \quad and \quad \vec p \ _1^*=-\vec p \ _2^*[/math]
[math]s \equiv 2m^2+2E_1^{*2}+2\vec p_1 \ ^{*2} [/math]
Using the relativistic energy equation
[math]E^2 \equiv \vec p_1 \ ^2+m^2[/math]
[math]s \equiv 2m^2+2m^2+2\vec p_1 \ ^{*2}+\vec p_1 \ ^{*2})[/math]
[math]s=4(m^2+\vec p_1 \ ^{*2})=(2E_1^*)^{2}=E^{*2}[/math]
| [math]\Rightarrow E_1=\frac{106.030760886 MeV}{2}=53.015380443 MeV=E_2[/math]
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