Limit of Energy in Lab Frame

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]

[math]\theta=0^{\circ}[/math]

There is no scattering, or no momentum transfer since the incident momentum direction is the same as the scattered momentum direction

[math]\theta=90^{\circ}[/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]

[math]\theta=180^{\circ}[/math]

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

[math]t=-2p_1^{*2}(1- \cos \theta_{1^*\ 1^{'*}})=-2p_2^{*2}(1-\cos \theta_{2^*\ 2^{'*}})[/math]

[math]t=-4p_1^{*2}[/math]

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

[math]t=4(m^{2}-E_1^{*2})=4(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]

Maximum Moller Energy in Lab Frame

Since t is invariant 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^{*} \approx 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]

Minimum Moller Energy in Lab Frame

Since t is invariant between frames

[math]t=2(m^2-E_2^{'}m)=0[/math]

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[math]\textbf{\underline{Navigation}}[/math]

[math]\vartriangleleft [/math] [math]\triangle [/math] [math]\vartriangleright [/math]