[math]\textbf{\underline{Navigation}}[/math]
[math]\vartriangleleft [/math]
[math]\triangle [/math]
[math]\vartriangleright [/math]
t Channel
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]
[math]t \equiv \left({\mathbf P_1^*}- {\mathbf P_1^{'*}}\right)^2[/math]
[math]t \equiv \mathbf P_1^{*2}-2 \mathbf P_1^* \mathbf P_1^{'*}+ \mathbf P_1^{'*2}[/math]
[math]t \equiv 2m_1^2-2E_1^*E_1^{'*}+2 \vec p \ _1^* \vec p \ _1^{'*}[/math]
In the center of mass frame of reference,
[math] E^* \equiv E_1^*=E_1^{'*} = E_2^*=E_2^{'*} = E_1^*=E_2^*[/math]
and
[math]|p^*| \equiv | \vec p \ _1^*|=| \vec p \ _1^{'*}| =| \vec p \ _2^*|=| \vec p \ _2^{'*}|[/math]
and [math]\theta_1[/math] is the angle between [math]\vec p \ _1^* [/math] and [math] \vec p \ _1^{'*}[/math]
[math]t \equiv 2m_1^*-2E_1^{*2}+2 |p |^{*2}cos\ \theta[/math]
Using the relativistic term for Energy
[math]E^2=\vec p \ ^2+m^2[/math]
[math]t \equiv -2 p \ _1^{*2}(1-cos\ \theta)[/math]
[math]\textbf{\underline{Navigation}}[/math]
[math]\vartriangleleft [/math]
[math]\triangle [/math]
[math]\vartriangleright [/math]
