Difference between revisions of "S-Channel"

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Revision as of 22:23, 8 June 2017

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

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

s Channel

The s quantity is known as the square of the center of mass energy (invariant mass)

[math]s \equiv \left({\mathbf P_1^*}+ {\mathbf P_2^{*}}\right)^2=\left({\mathbf P_1^{'*}}+ {\mathbf P_2^{'*}}\right)^2[/math]



[math]s \equiv \left({\mathbf P_1^*}+ {\mathbf P_2^{*}}\right)^2[/math]


[math]s \equiv \mathbf P_1^{*2}+2 \mathbf P_1^* \mathbf P_2^*+ \mathbf P_2^{*2}[/math]


As shown earlier, the square of a 4-momentum is


[math]\mathbf P^{2} \equiv m^2[/math]


[math]s \equiv m^{2}+2 \mathbf P_1^* \mathbf P_2^*+ m_2^{2}[/math]


This gives

[math]s \equiv 2m^{2}+2 \mathbf P_1^* \mathbf P_2^*[/math]


Similarly, the scalar product of two 4-momentums

[math]s \equiv 2m^2+2(E_1^*E_2^*-\vec p_1^* \vec p_2^*)[/math]


In the center of mass frame of reference,

[math]E_1^*=E_2^* \quad and \quad \vec p_1^*=-\vec p_2^*[/math]


[math]s \equiv 2m^2+2(E_1^{*2}+\vec p_1^{*2} )[/math]


Using the relativistic energy equation

[math]E^2 \equiv p^2+m^2[/math]


[math]s \equiv 2m^2+2((m^2+p_1^{*2})+p_1^{*2})[/math]


[math]s=4(m_{CM}^2+p_{CM}^2)[/math]




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

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