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]
[math]\displaystyle{ \theta \approx 0^{\circ} }[/math]
There is no scattering, or no momentum transfer at 90 degrees since the incident momentum direction is the same as the scattered momentum direction. However, at a certain angle enough momentum must be transferred to provide the ionization energy to create a Moller electron.
[math]\displaystyle{ \theta=90^{\circ} }[/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]
Minimum 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 }[/math] for [math]\displaystyle{ E \gg m }[/math]
[math]\displaystyle{ \therefore p_2^{'} \approx E_2^{'} }[/math]
Rewriting the expression relating the terms
[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]
[math]\displaystyle{ u=2m^2-2(11000\ MeV)(5500\ MeV) +2(11000\ MeV)(5500\ MeV) \cos \theta_{1\ 2^{'}}=2(m^2-(53\ MeV)^{2}) }[/math]
Solving for the angle theta
[math]\displaystyle{ 2m^2-2(11000\ MeV)(5500\ MeV) +2(11000\ MeV)(5500\ MeV) \cos \theta_{1\ 2^{'}}=2(m^2-(53\ MeV)^{2}) }[/math]
[math]\displaystyle{ (11000\ MeV)(5500\ MeV)(1 - \cos \theta_{1\ 2^{'}})=(53\ MeV)^{2} }[/math]
[math]\displaystyle{ (1 - \cos \theta_{1\ 2^{'}})=\frac{(53\ MeV)^{*2}}{(11000\ MeV)(5500\ MeV)} }[/math]
[math]\displaystyle{ 1 - \cos \theta_{1\ 2^{'}}=4.64297520661\times 10^{-5} }[/math]
[math]\displaystyle{ \cos \theta_{1\ 2^{'}}=1-4.64297520661\times 10^{-5} }[/math]
[math]\displaystyle{ \theta_{1\ 2^{'}}=\arccos .999953570248 }[/math]
[math]\displaystyle{ \theta_{1\ 2^{'}}=.009636400914\ radians=.55^{\circ} }[/math]
The Moller electron is traveling near, but not on the beamline.
Maximum Moller Scattering Angle Theta in Lab Frame
Since u is invariant between frames
[math]\displaystyle{ u=2m^2-2E_1^*E_2^{'*}+2 \left | p_1^* \right | \left | p_2^{'*} \right | \cos \theta_{1\ 2^{'}}=2m^2-2E_1E_2^{'}+2 \left | p_1 \right | \left | p_2^{'} \right | \cos \theta_{1\ 2^{'}} }[/math]
[math]\displaystyle{ E_1^*=53\ MeV=p_1^* }[/math]
[math]\displaystyle{ E_1=11000\ MeV=p_1 }[/math]
[math]\displaystyle{ -E_1^*E_2^{'*}+ \left | p_1^* \right | \left | p_2^{'*} \right | \cos \theta_{1\ 2^{'}}\approx -E_1E_2^{'}+ \left | p_1 \right | \left | p_2^{'} \right | \cos \theta_{1\ 2^{'}} }[/math]
[math]\displaystyle{ E_1^*E_2^{'*}- \left | p_1^* \right | \left | p_2^{'*} \right | \cos \theta_{1\ 2^{'}}\approx E_1E_2^{'}- \left | p_1 \right | \left | p_2^{'} \right | \cos \theta_{1\ 2^{'}} }[/math]
[math]\displaystyle{ E_1^{*2}(1- \cos \theta_{1\ 2^{'}})\approx E_1(E_2^{'}- \left | p_2^{'} \right | \cos \theta_{1\ 2^{'}}) }[/math]
[math]\displaystyle{ (53\ MeV)^{*2}(1- \cos \theta_{1\ 2^{'}})\approx (11000\ MeV)(E_2^{'}- \left | p_2^{'} \right | \cos \theta_{1\ 2^{'}}) }[/math]
[math]\displaystyle{ (255363636364\ MeV)(1- \cos \theta_{1\ 2^{'}})\approx (E_2^{'}- \left | p_2^{'} \right | \cos \theta_{1\ 2^{'}}) }[/math]