Difference between revisions of "Limits based on Mandelstam Variables"

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<center><math> t \le 0  \qquad u \le 0</math></center>
 
<center><math> t \le 0  \qquad u \le 0</math></center>
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At the condition both t and u are equal to zero, we find
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<center><math> t = 0  \qquad u = 0</math></center>
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<center><math>-2 p \ ^{*2}(1-cos\ \theta) = 0  \qquad -2 p \ ^{*2}(1+cos\ \theta) = 0</math></center>
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<center><math>(-2 p \ ^{*2}+2 p \ ^{*2}cos\ \theta) = 0  \qquad (-2 p \ ^{*2}-2 p \ ^{*2}cos\ \theta) = 0</math></center>
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<center><math>2 p \ ^{*2}cos\ \theta = 2 p \ ^{*2}  \qquad -2 p \ ^{*2}cos\ \theta = 2 p \ ^{*2}</math></center>
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<center><math>cos\ \theta = 1  \qquad cos\ \theta = -1</math></center>

Revision as of 23:52, 9 June 2017

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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]


Since

[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]