Limits based on Mandelstam Variables: Difference between revisions

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<center><math>s+t+u=(4(m^2+\vec p \ ^{*2}))+(-2 p \ ^{*2}(1-cos\ \theta))+(-2 p \ ^{*2}(1+cos\ \theta))</math></center>
<center><math>s+t+u=(4(m^2+ p \ ^{*2}))+(-2 p \ ^{*2}(1-cos\ \theta))+(-2 p \ ^{*2}(1+cos\ \theta))</math></center>




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<center><math>s \ge 4m^2</math></center>
<center><math>s \ge 4m^2</math></center>
In turn, this implies
<center><math> t \le 0  \qquad u \le 0</math></center>

Revision as of 23:44, 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]\displaystyle{ s+t+u=(4(m^2+ p \ ^{*2}))+(-2 p \ ^{*2}(1-cos\ \theta))+(-2 p \ ^{*2}(1+cos\ \theta)) }[/math]


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


Since

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


This implies

[math]\displaystyle{ s \ge 4m^2 }[/math]


In turn, this implies


[math]\displaystyle{ t \le 0 \qquad u \le 0 }[/math]