Limit of Energy in Lab Frame

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The t quantity is known as the square of the 4-momentum transfer

$t \equiv \left({\mathbf P_1^*}- {\mathbf P_1^{'*}}\right)^2=\left({\mathbf P_2^{*}}- {\mathbf P_2^{'*}}\right)^2$

In the CM Frame

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

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

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

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

$t =\left({\mathbf P_1^*}- {\mathbf P_1^{'*}}\right)^2=\left({\mathbf P_2^{*}}- {\mathbf P_2^{'*}}\right)^2$

$t={\mathbf P_1^{*2}}+ {\mathbf P_1^{'*2}}-2 {\mathbf P_1^*} {\mathbf P_1^{'*}}={\mathbf P_2^{*2}}+ {\mathbf P_2^{'*2}}-2 {\mathbf P_2^*} {\mathbf P_2^{'*}}$

$t=2m^2-2E_1^*E_1^{'*}+2 \vec p_1^* \vec p_1^{'*}=2m^2-2E_2^*E_2^{'*}+2 p_2^* p_2^{'*}$

$t=2m^2-2E_1^{*2}+2 \left | p_1^{*2}\right | \cos \theta_{1^*\ 1^{'*}}=2m^2-2E_2^{*2}+2 \left | p_2^{*2}\right | \cos \theta_{2^*\ 2^{'*}}$

where $\theta_{1^*\ 1^{'*}}$ and $\theta_{2^*\ 2^{'*}}$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

$t=-2p_1^{*2}+2 \left | p_1^{*2}\right | \cos \theta_{1^*\ 1^{'*}}=-2p_2^{*2}+2 \left | p_2^{*2}\right | \cos \theta_{2^*\ 2^{'*}}$

$t=-2p_1^{*2}(1- \cos \theta_{1^*\ 1^{'*}})=-2p_2^{*2}(1-\cos \theta_{2^*\ 2^{'*}})$

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

There is no scattering, or no momentum transfer at 0 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.

$\theta=90^{\circ}$

The maximum momentum is transfered at 90 degrees, i.e. $\cos 90^{\circ}=0$

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

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

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

$\theta=180^{\circ}$

The maximum momentum is transfered at 180 degrees, i.e. $\cos 180^{\circ}=-1$

$t=-2p_1^{*2}(1- \cos \theta_{1^*\ 1^{'*}})=-2p_2^{*2}(1-\cos \theta_{2^*\ 2^{'*}})$

$t=-4p_1^{*2}$

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

$t=4(m^{2}-E_1^{*2})=4(m^{2}-E_2^{*2})$

In the Lab Frame

$t={\mathbf P_1^{2}}+ {\mathbf P_1^{'2}}-2 {\mathbf P_1} {\mathbf P_1^{'}}={\mathbf P_2^{2}}+ {\mathbf P_2^{'2}}-2 {\mathbf P_2} {\mathbf P_2^{'}}$

$t=2m^2-2E_1E_1^{'}+2 \vec p_1 \vec p_1^{'}=2m^2-2E_2E_2^{'}+2 p_2 p_2^{'}$

with $p_2=0$

and $E_2=m$

$t=2m^2-2mE_2^{'}=2(m^2-E_2^{'}m)$

Maximum Moller Energy in Lab Frame

Since t is invariant between frames

$t=2(m^2-E_2^{'}m)=2(m^2-E_2^{*2})$

$\rightarrow E_2^{'}=\frac{E_1^{*2}}{m}$

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

The Moller electron has a maximum energy possible of:

$E_2^{'}=5500\ MeV$

Minimum Moller Energy in Lab Frame

Since t is invariant between frames

$t=2(m^2-E_2^{'}m)=0$

$m^2=E_2^{'}m$

$m\gt E_2^{'}$

This implies that the Moller electron has a non-zero momentum, hence it's total energy is more than it's rest mass energy. The momentum that the Moller electron would have would have to be transfered from the incident electron to the "stationary" electron bound to the detector. The binding energy of an electron bound to a hydrogen atom is 13.6 eV

$t=-2p_1^{*2}(1- \cos \theta_{1^*\ 1^{'*}})=-2p_2^{*2}(1-\cos \theta_{2^*\ 2^{'*}})=2(m^2-E_2^{'}m)$

$-2p_2^{*2}(1-\cos \theta_{2^*\ 2^{'*}})=2(m^2-E_2^{'}m)$

$(m^2-E_2^{*2})(1-\cos \theta_{2^*\ 2^{'*}})=-m(E_2^{'}-m)$

$((.511\ MeV)^2-(53\ MeV)^{*2})(1-\cos \theta_{2^*\ 2^{'*}})=-(.511\ MeV)(E_2^{'}-(.511\ MeV))$

$5496553579.26\ eV(1-\cos \theta_{2^*\ 2^{'*}})=(E_2^{'}-(.511\ MeV))$

At $\theta_{2^*\ 2^{'*}}=0^{\circ}$

$5496553579.26\ eV+(.511\ MeV)=E_2^{'}$

$.511\ MeV=E_2^{'}$

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