The quantity is known as the
[math]\displaystyle{ u \equiv \left({\mathbf P_1^*}- {\mathbf P_2^{'*}}\right)^2=\left({\mathbf P_2^{*}}- {\mathbf P_1^{'*}}\right)^2 }[/math]
In the CM Frame
[math]\displaystyle{ {\mathbf P_1^{*}}=-{\mathbf P_2^{*}} }[/math]
[math]\displaystyle{ {\mathbf P_1^{'*}}=-{\mathbf P_2^{'*}} }[/math]
[math]\displaystyle{ E_1^{*}=E_1^{'*}=E_2^{*}=E_2^{'*} }[/math]
[math]\displaystyle{ \left | \vec p_1^* \right |=\left | \vec p_1^{'*} \right |=\left | \vec p_2^* \right |=\left | \vec p_2^{'*} \right | }[/math]
[math]\displaystyle{ u =\left({\mathbf P_1^*}- {\mathbf P_2^{'*}}\right)^2=\left({\mathbf P_2^{*}}- {\mathbf P_1^{'*}}\right)^2 }[/math]
[math]\displaystyle{ u={\mathbf P_1^{*2}}+ {\mathbf P_2^{'*2}}-2 {\mathbf P_1^*} {\mathbf P_2^{'*}}={\mathbf P_2^{*2}}+ {\mathbf P_1^{'*2}}-2 {\mathbf P_2^*} {\mathbf P_1^{'*}} }[/math]
[math]\displaystyle{ u=2m^2-2E_1^*E_2^{'*}+2 \vec p_1^* \vec p_2^{'*}=2m^2-2E_2^*E_1^{'*}+2 p_2^* p_1^{'*} }[/math]
[math]\displaystyle{ u=2m^2-2E_1^{*2}+2 \left | p_1^{*2}\right | \cos \theta_{1^*\ 2^{'*}}=2m^2-2E_2^{*2}+2 \left | p_2^{*2}\right | \cos \theta_{2^*\ 1^{'*}} }[/math]
where [math]\displaystyle{ \theta_{1^*\ 2^{'*}} }[/math] and [math]\displaystyle{ \theta_{2^*\ 1^{'*}} }[/math]is the angle between the before and after momentum in the CM frame
Using the relativistic relation [math]\displaystyle{ E^2=m^2+p^2 }[/math] this reduces to
[math]\displaystyle{ u=-2p_1^{*2}+2 \left | p_1^{*2}\right | \cos \theta_{1^*\ 2^{'*}}=-2p_2^{*2}+2 \left | p_2^{*2}\right | \cos \theta_{2^*\ 1^{'*}} }[/math]
[math]\displaystyle{ u=-2p_1^{*2}(1- \cos \theta_{1^*\ 2^{'*}})=-2p_2^{*2}(1-\cos \theta_{2^*\ 1^{'*}}) }[/math]
For [math]\displaystyle{ \theta_{1^*1^{'*}}=90^{\circ} }[/math], by symmetry this implies [math]\displaystyle{ \theta_{1^*2^{'*}}=270^{\circ} }[/math]
[math]\displaystyle{ u=-2p_1^{*2} }[/math]
This can be rewritten again using the relativistic energy relation [math]\displaystyle{ E^2=m^2+p^2 }[/math]
[math]\displaystyle{ u=2(m^{2}-E_1^{*2})=2(m^{2}-E_2^{*2}) }[/math]
In the Lab Frame
[math]\displaystyle{ u={\mathbf P_1^{2}}+ {\mathbf P_2^{'2}}-2 {\mathbf P_1} {\mathbf P_2^{'}}={\mathbf P_2^{2}}+ {\mathbf P_1^{'2}}-2 {\mathbf P_2} {\mathbf P_1^{'}} }[/math]
[math]\displaystyle{ u=2m^2-2E_1E_2^{'}+2 \vec p_1 \vec p_2^{'}=2m^2-2E_2E_1^{'}+2 p_2 p_1^{'} }[/math]
with [math]\displaystyle{ p_2=0 }[/math]
and [math]\displaystyle{ E_2=m }[/math]
[math]\displaystyle{ u=2m^2-2E_1E_2^{'}+2 \vec p_1 \vec p_2^{'}=2m^2-2mE_1^{'} }[/math]
[math]\displaystyle{ u=2m^2-2E_1E_2^{'}+2 \left | p_1 \right | \left | p_2^{'} \right | \cos \theta_{1\ 2^{'}}=2m^2-2mE_1^{'} }[/math]
Maximum Moller Scattering Angle Theta in Lab Frame
Since u is invariant between frames
[math]\displaystyle{ u=2m^2-2E_1E_2^{'}+2 \left | p_1 \right | \left | p_2^{'} \right | \cos \theta_{1\ 2^{'}}=2(m^2-E_2^{*2}) }[/math]
with[math]\displaystyle{ E_2^{*} \approx 53\ MeV }[/math] for [math]\displaystyle{ E_1=11000\ MeV }[/math]
As found earlier, the Moller electron has a maximum energy possible of:
[math]\displaystyle{ E_2^{'}=5500\ MeV }[/math]
Using the relativistic energy relation, [math]\displaystyle{ E^2=m^2+p^2 }[/math]
[math]\displaystyle{ p=\sqrt {E^2-m^2} \rightarrow p \approx E for E \gg m }[/math]