Limits based on Mandelstam Variables

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

Limits based on Mandelstam Variables

Since the Mandelstam variables are the scalar product of 4-momenta, which are invariants, they are invariants as well. The sum of these invariant variables must also be invariant as well. Find the sum of the 3 Mandelstam variables when the two particles have equal mass in the center of mass frame gives:

[math]s+t+u=(4(m^2+ p \ ^{*2}))+(-2 p \ ^{*2}(1-cos\ \theta))+(-2 p \ ^{*2}(1+cos\ \theta))[/math]

[math]s+t+u \equiv 4m^2[/math]


[math]s \equiv 4(m^2+\vec p \ ^{*2})[/math]

This implies

[math]s \ge 4m^2[/math]

In turn, this implies

[math] t \le 0 \qquad u \le 0[/math]

At the condition both t and u are equal to zero, we find

[math] t = 0 \qquad u = 0[/math]

[math]-2 p \ ^{*2}(1-cos\ \theta) = 0 \qquad -2 p \ ^{*2}(1+cos\ \theta) = 0[/math]

[math](-2 p \ ^{*2}+2 p \ ^{*2}cos\ \theta) = 0 \qquad (-2 p \ ^{*2}-2 p \ ^{*2}cos\ \theta) = 0[/math]

[math]2 p \ ^{*2}cos\ \theta = 2 p \ ^{*2} \qquad -2 p \ ^{*2}cos\ \theta = 2 p \ ^{*2}[/math]

[math]\cos\ \theta = 1 \qquad \cos\ \theta = -1[/math]

[math]\Rightarrow \theta_{t=0} = \arccos \ 1=0^{\circ} \qquad \theta_{u=0} = \arccos \ -1=180^{\circ}[/math]

Holding u constant at zero we can find the maximum of t



[math]t=4m^2-4m^2+ 4p \ ^{*2})[/math]

[math]t=4p \ ^{*2}[/math]

[math]-2 p \ ^{*2}(1-cos\ \theta)=4p \ ^{*2}[/math]

[math](1-cos\ \theta)=-2[/math]

[math]-cos\ \theta=-3[/math]

[math] \theta=\arccos 3[/math]