Limit of Energy in Lab Frame: Difference between revisions
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<center><math>t \equiv \left({\mathbf P_1^*}- {\mathbf P_1^{'*}}\right)^2=\left({\mathbf P_2^{*}}- {\mathbf P_2^{'*}}\right)^2</math></center> | <center><math>t \equiv \left({\mathbf P_1^*}- {\mathbf P_1^{'*}}\right)^2=\left({\mathbf P_2^{*}}- {\mathbf P_2^{'*}}\right)^2</math></center> | ||
=In the CM Frame= | |||
<center><math>{\mathbf P_1^{*}}=-{\mathbf P_2^{*}}</math></center> | <center><math>{\mathbf P_1^{*}}=-{\mathbf P_2^{*}}</math></center> | ||
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=In the Lab Frame= | |||
<center><math>t={\mathbf P_1^{2}}+ {\mathbf P_1^{'2}}-2 {\mathbf P_1} {\mathbf P_1^{'}}</math></center> | <center><math>t={\mathbf P_1^{2}}+ {\mathbf P_1^{'2}}-2 {\mathbf P_1} {\mathbf P_1^{'}}</math></center> | ||
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<center><math>=2m^2-2E_1E_1^'+2 p_1 p_1^'=2m^2-2E_2E_2^'+2 p_2 p_2^'</math></center> | <center><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></center> | ||
Revision as of 15:42, 15 March 2018
The t quantity is known as the square of the 4-momentum transfer
In the CM Frame
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
The maximum momentum is transfered at 90 degrees, i.e. [math]\displaystyle{ \cos 90^{circ}=0 }[/math]
This can be rewritten again using the relativistic energy relation [math]\displaystyle{ E^2=m^2+p^2 }[/math]
In the Lab Frame
with [math]\displaystyle{ p_2=0 }[/math]
and [math]\displaystyle{ E_2=m }[/math]
[math]\displaystyle{ \vartriangleleft }[/math] [math]\displaystyle{ \triangle }[/math] [math]\displaystyle{ \vartriangleright }[/math]