Scattered and Moller Electron Energies in CM Frame

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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]\displaystyle{ E^* }[/math]

[math]\displaystyle{ s \equiv \left({\mathbf P_1^*}+ {\mathbf P_2^{*}}\right)^2 }[/math]

[math]\displaystyle{ 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]\displaystyle{ \mathbf P^{2} \equiv m^2 }[/math]

This gives,

[math]\displaystyle{ s \equiv m_1^{2}+2 \mathbf P_1^* \mathbf P_2^*+ m_2^{2} }[/math]

For the case [math]\displaystyle{ m_1=m_2=m }[/math]

[math]\displaystyle{ s \equiv 2m^{2}+2 \mathbf P_1^* \mathbf P_2^* }[/math]

Using the relationship

[math]\displaystyle{ \mathbf P_1 \cdot \mathbf P_2 = E_{1}E_{2}-(\vec p_1 \vec p_2) }[/math]

[math]\displaystyle{ 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]\displaystyle{ E_1^*=E_2^* \quad and \quad \vec p \ _1^*=-\vec p \ _2^* }[/math]

[math]\displaystyle{ s \equiv 2m^2+2E_1^{*2}+2\vec p_1 \ ^{*2} }[/math]

Using the relativistic energy equation

[math]\displaystyle{ E^2 \equiv \vec p_1 \ ^2+m^2 }[/math]

[math]\displaystyle{ s \equiv 2m^2+2m^2+2\vec p_1 \ ^{*2}+\vec p_1 \ ^{*2}) }[/math]

[math]\displaystyle{ s=4(m^2+\vec p_1 \ ^{*2})=(2E_1^*)^{2}=E^{*2} }[/math]

[math]\displaystyle{ \Rightarrow E_1=\frac{106.030760886 MeV}{2}=53.015380443 MeV=E_2 }[/math]