Limit of Scattering Angle Theta in Lab Frame

The quantity is known as the

$u \equiv \left({\mathbf P_1^*}- {\mathbf P_2^{'*}}\right)^2=\left({\mathbf P_2^{*}}- {\mathbf P_1^{'*}}\right)^2$

In the CM Frame

${\mathbf P_1^{*}}=-{\mathbf P_2^{*}}$

${\mathbf P_1^{'*}}=-{\mathbf P_2^{'*}}$

$E_1^{*}=E_1^{'*}=E_2^{*}=E_2^{'*}$

$\left | \vec p_1^* \right |=\left | \vec p_1^{'*} \right |=\left | \vec p_2^* \right |=\left | \vec p_2^{'*} \right |$

$u =\left({\mathbf P_1^*}- {\mathbf P_2^{'*}}\right)^2=\left({\mathbf P_2^{*}}- {\mathbf P_1^{'*}}\right)^2$

$u={\mathbf P_1^{*2}}+ {\mathbf P_2^{'*2}}-2 {\mathbf P_1^*} {\mathbf P_2^{'*}}={\mathbf P_2^{*2}}+ {\mathbf P_1^{'*2}}-2 {\mathbf P_2^*} {\mathbf P_1^{'*}}$

$u=2m^2-2E_1^*E_2^{'*}+2 \vec p_1^* \vec p_2^{'*}=2m^2-2E_2^*E_1^{'*}+2 p_2^* p_1^{'*}$

$u=2m^2-2E_1^{*2}+2 \left | p_1^{*2}\right | \cos \theta_{1^*\ 2^{'*}}=2m^2-2E_2^{*2}+2 \left | p_2^{*2}\right | \cos \theta_{2^*\ 1^{'*}}$

where $\theta_{1^*\ 2^{'*}}$ and $\theta_{2^*\ 1^{'*}}$is the angle between the before and after momentum in the CM frame

Using the relativistic relation $E^2=m^2+p^2$ this reduces to

$u=-2p_1^{*2}+2 \left | p_1^{*2}\right | \cos \theta_{1^*\ 2^{'*}}=-2p_2^{*2}+2 \left | p_2^{*2}\right | \cos \theta_{2^*\ 1^{'*}}$

$u=-2p_1^{*2}(1- \cos \theta_{1^*\ 2^{'*}})=-2p_2^{*2}(1-\cos \theta_{2^*\ 1^{'*}})$

$\theta \approx 0^{\circ}$

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.

$\theta=90^{\circ}$

For $\theta_{1^*1^{'*}}=90^{\circ}$, by symmetry this implies $\theta_{1^*2^{'*}}=270^{\circ}$

$u=-2p_1^{*2}$

This can be rewritten again using the relativistic energy relation $E^2=m^2+p^2$

$u=2(m^{2}-E_1^{*2})=2(m^{2}-E_2^{*2})$

In the Lab Frame

$u={\mathbf P_1^{2}}+ {\mathbf P_2^{'2}}-2 {\mathbf P_1} {\mathbf P_2^{'}}={\mathbf P_2^{2}}+ {\mathbf P_1^{'2}}-2 {\mathbf P_2} {\mathbf P_1^{'}}$

$u=2m^2-2E_1E_2^{'}+2 \vec p_1 \vec p_2^{'}=2m^2-2E_2E_1^{'}+2 p_2 p_1^{'}$

with $p_2=0$

and $E_2=m$

$u=2m^2-2E_1E_2^{'}+2 \vec p_1 \vec p_2^{'}=2m^2-2mE_1^{'}$

$u=2m^2-2E_1E_2^{'}+2 \left | p_1 \right | \left | p_2^{'} \right | \cos \theta_{1\ 2^{'}}=2m^2-2mE_1^{'}$

Minimum Moller Scattering Angle Theta in Lab Frame

Since u is invariant between frames

$u=2m^2-2E_1E_2^{'}+2 \left | p_1 \right | \left | p_2^{'} \right | \cos \theta_{1\ 2^{'}}=2(m^2-E_2^{*2})$

with$E_2^{*} \approx 53\ MeV$ for $E_1=11000\ MeV$

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

$E_2^{'}=5500\ MeV$

Using the relativistic energy relation, $E^2=m^2+p^2$

$p=\sqrt {E^2-m^2} \rightarrow p \approx E$ for $E \gg m$

$\therefore p_2^{'} \approx E_2^{'}$

Rewriting the expression relating the terms

$u=2m^2-2E_1E_2^{'}+2 \left | p_1 \right | \left | p_2^{'} \right | \cos \theta_{1\ 2^{'}}=2(m^2-E_2^{*2})$

$u=2m^2-2(11000\ MeV)(5500\ MeV) +2(11000\ MeV)(5500\ MeV) \cos \theta_{1\ 2^{'}}=2(m^2-(53\ MeV)^{2})$

Solving for the angle theta

$2m^2-2(11000\ MeV)(5500\ MeV) +2(11000\ MeV)(5500\ MeV) \cos \theta_{1\ 2^{'}}=2(m^2-(53\ MeV)^{2})$

$(11000\ MeV)(5500\ MeV)(1 - \cos \theta_{1\ 2^{'}})=(53\ MeV)^{2}$

$(1 - \cos \theta_{1\ 2^{'}})=\frac{(53\ MeV)^{*2}}{(11000\ MeV)(5500\ MeV)}$

$1 - \cos \theta_{1\ 2^{'}}=4.64297520661\times 10^{-5}$

$\cos \theta_{1\ 2^{'}}=1-4.64297520661\times 10^{-5}$

$\theta_{1\ 2^{'}}=\arccos .999953570248$

$\theta_{1\ 2^{'}}=.009636400914\ radians=.55^{\circ}$

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

$u=2m^2-2E_1^*E_2^{'*}+2 \left | p_1^* \right | \left | p_2^{'*} \right | \cos \theta_{1\ 2^{'}}=2(m^2-2E_1E_2^{2})$