Limit of Scattering Angle Theta in Lab Frame

The quantity is known as the

[math]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]{\mathbf P_1^{*}}=-{\mathbf P_2^{*}}[/math]

[math]{\mathbf P_1^{'*}}=-{\mathbf P_2^{'*}}[/math]

[math]E_1^{*}=E_1^{'*}=E_2^{*}=E_2^{'*}[/math]

[math]\left | \vec p_1^* \right |=\left | \vec p_1^{'*} \right |=\left | \vec p_2^* \right |=\left | \vec p_2^{'*} \right |[/math]

[math]u =\left({\mathbf P_1^*}- {\mathbf P_2^{'*}}\right)^2=\left({\mathbf P_2^{*}}- {\mathbf P_1^{'*}}\right)^2[/math]

[math]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]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]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]\theta_{1^*\ 2^{'*}}[/math] and [math]\theta_{2^*\ 1^{'*}}[/math]is the angle between the before and after momentum in the CM frame

Using the relativistic relation [math]E^2=m^2+p^2[/math] this reduces to

[math]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]u=-2p_1^{*2}(1- \cos \theta_{1^*\ 2^{'*}})=-2p_2^{*2}(1-\cos \theta_{2^*\ 1^{'*}})[/math]

For [math]\theta_{1^*1^{'*}}=90^{\circ}[/math], by symmetry this implies [math]\theta_{1^*2^{'*}}=270^{\circ}[/math]

[math]u=-2p_1^{*2}[/math]

This can be rewritten again using the relativistic energy relation [math]E^2=m^2+p^2[/math]

[math]u=2(m^{2}-E_1^{*2})=2(m^{2}-E_2^{*2})[/math]

In the Lab Frame

[math]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]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]p_2=0[/math]

and [math]E_2=m[/math]

[math]u=2m^2-2E_1E_2^{'}+2 \vec p_1 \vec p_2^{'}=2m^2-2mE_1^{'}[/math]

[math]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]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] E_2^{*} \approx 53\ MeV[/math] for [math]E_1=11000\ MeV[/math]

As found earlier, the Moller electron has a maximum energy possible of:

[math]E_2^{'}=5500\ MeV[/math]

Using the relativistic energy relation, [math]E^2=m^2+p^2[/math]

[math]p=\sqrt {E^2-m^2} \rightarrow p \approx E[/math] for [math]E \gg m[/math]

[math]\therefore p_2^{'} \approx E_2^{'}[/math]

Rewriting the expression relating the terms

[math]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]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]2m^2-2(11000\ MeV)(5500\ MeV) +2(11000\ MeV)(5500\ MeV) \cos \theta_{1\ 2^{'}}=2(m^2-(53\ MeV)^{2})[/math]

[math](11000\ MeV)(5500\ MeV)(1 - \cos \theta_{1\ 2^{'}})=(53\ MeV)^{2}[/math]

[math](1 - \cos \theta_{1\ 2^{'}})=\frac{(53\ MeV)^{*2}}{(11000\ MeV)(5500\ MeV)}[/math]

[math]1 - \cos \theta_{1\ 2^{'}}=4.64297520661\times 10^{-5}[/math]

[math]\cos \theta_{1\ 2^{'}}=1-4.64297520661\times 10^{-5}[/math]

[math]\theta_{1\ 2^{'}}=\arccos .999953570248[/math]

[math]\theta_{1\ 2^{'}}=.009636400914\ radians=.55^{\circ}[/math]

The Moller electron is traveling near, but not on the beamline.