Limit of Energy in Lab Frame

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

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

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

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

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

[math]\displaystyle{ t =\left({\mathbf P_1^*}- {\mathbf P_1^{'*}}\right)^2=\left({\mathbf P_2^{*}}- {\mathbf P_2^{'*}}\right)^2 }[/math]

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

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

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

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

The maximum momentum is transfered at 90 degrees, i.e. [math]\displaystyle{ \cos 90^{\circ}=0 }[/math]

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

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

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

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

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

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

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

[math]\displaystyle{ t=4(m^{2}-E_1^{*2})=4(m^{2}-E_2^{*2}) }[/math]

In the Lab Frame

[math]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ p_2=0 }[/math]

and [math]\displaystyle{ E_2=m }[/math]

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

[math]\displaystyle{ \rightarrow E_2^{'}=\frac{E_1^{*2}}{m} }[/math]

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

The Moller electron has a maximum energy possible of:

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

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

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