Difference between revisions of "Limits based on Mandelstam Variables"

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Similarly, by the relativistic definition of energy
 
Similarly, by the relativistic definition of energy
  
<center><math>E^2=p^2+m^2</math></center>
+
<center><math>E^2 \equiv p^2+m^2</math></center>
  
 
where both particles have the same mass, this implies  
 
where both particles have the same mass, this implies  
  
<center><math> \left({\mathbf P_1^*}+ {\mathbf P_2^{*}}\right)^2= \left(\begin{matrix}2E\\0 \\ 0 \\ 0 \end{matrix} \right)^2</math></center>
+
<center><math> \left({\mathbf P_1^*}+ {\mathbf P_2^{*}}\right)^2=2E^2</math></center>

Revision as of 09:10, 2 June 2017

Limits based on Mandelstam Variables

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


In the center of mass frame, the momentum of the particles interacting are equal and opposite, i.e. [math]p_1=-p_2[/math]. However, the 4-momentum still retains an energy component, which as a scalar quantity, can not be countered by another particle's direction of motion.


[math]{\mathbf P_1^*}\equiv \left(\begin{matrix} E_1\\ p_{x_1} \\ p_{y_1} \\ p_{z_1} \end{matrix} \right) \ \ \ \ {\mathbf P_2^*}\equiv \left(\begin{matrix} E_2\\ p_{x_2} \\ p_{y_2} \\ p_{z_2} \end{matrix} \right)[/math]


[math] \left({\mathbf P_1^*}+ {\mathbf P_2^{*}}\right)^2=\left( \left(\begin{matrix} E_1\\ p_{x_1} \\ p_{y_1} \\ p_{z_1} \end{matrix} \right)+\left(\begin{matrix} E_2\\ p_{x_2} \\ p_{y_2} \\ p_{z_2} \end{matrix} \right) \right)^2[/math]


[math] \left({\mathbf P_1^*}+ {\mathbf P_2^{*}}\right)^2=\left( \left(\begin{matrix} E_1\\ p_{x_1} \\ p_{y_1} \\ p_{z_1} \end{matrix} \right)+\left(\begin{matrix} E_2\\ -p_{x_1} \\ -p_{y_1} \\ -p_{z_1} \end{matrix} \right) \right)^2[/math]


[math] \left({\mathbf P_1^*}+ {\mathbf P_2^{*}}\right)^2= \left(\begin{matrix} E_1+E_2\\0 \\ 0 \\ 0 \end{matrix} \right)^2[/math]


Similarly, by the relativistic definition of energy

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

where both particles have the same mass, this implies

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