The t quantity is known as the square of the 4-momentum transfer
[math]t \equiv \left({\mathbf P_1^*}- {\mathbf P_1^{'*}}\right)^2=\left({\mathbf P_2^{*}}- {\mathbf P_2^{'*}}\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]t =\left({\mathbf P_1^*}- {\mathbf P_1^{'*}}\right)^2=\left({\mathbf P_2^{*}}- {\mathbf P_2^{'*}}\right)^2[/math]
[math]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^{'*}}[/math]
[math]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^{'*}[/math]
[math]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'}[/math]
where [math]\theta_{1\ 1'}[/math] and [math]\theta_{2\ 2'}[/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]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'}[/math]
[math]t=-2p_1^{*2}(1- \cos \theta_{1\ 1'})=-2p_2^{*2}(1-\cos \theta_{2\ 2'})[/math]
The maximum momentum is transfered at 90 degrees, i.e. [math]\cos 90^{circ}=0[/math]
[math]t=-2p_1^{*2}[/math]
This can be rewritten again using the relativistic energy relation [math]E^2=m^2+p^2[/math]
[math]t=-2(m^{2}-E_1^{*2})=-2(m^{2}-E_2^{*2})[/math]
In the Lab Frame
[math]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^{'}}[/math]
[math]t=2m^2-2E_1E_1^{'}+2 \vec p_1 \vec p_1^{'}=2m^2-2E_2E_2^{'}+2 p_2 p_2^{'}[/math]
with [math]p_2=0[/math]
and [math]E_2=m[/math]
[math]t=2m^2-2mE_2^{'}=2(m^2-E_2^{'}m)[/math]
Between Frames
[math]t=2(m^2-E_2^{'}m)=2(m^2-E_2^{*2})[/math]
[math]\rightarrow E_2^{'}=\frac{E_1^{*2}}{m}[/math]
with[math] E_2^{*} \equiv 53\ MeV[/math] for [math]E_1=11000\ MeV[/math]
The Moller electron has a maximum energy possible of:
[math]E_2^{'}=5500\ MeV[/math]
[math]\textbf{\underline{Navigation}}[/math]
[math]\vartriangleleft [/math]
[math]\triangle [/math]
[math]\vartriangleright [/math]