Limits based on Mandelstam Variables: Difference between revisions

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<center><math>\Rightarrow  \theta_{t=0} = \arccos \ 1=0^{\circ}  \qquad \theta_{u=0} = \arccos \ -1=180^{\circ}</math></center>
<center><math>\Rightarrow  \theta_{t=0} = \arccos \ 1=0^{\circ}  \qquad \theta_{u=0} = \arccos \ -1=180^{\circ}</math></center>


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




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<center><math>t_{max}=4m^2-4m^2- 4p \ ^{*2}</math></center>
<center><math>t_{max}=4m^2-4m^2- 4p \ ^{*2}</math></center>


The maximum transfer of momentum would be




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<center><math>-2 p \ ^{*2}(1-cos\ \theta)=-4p \ ^{*2}</math></center>
 
<center><math>-2 p \ ^{*2}(1-cos\ \theta_{t=max})=-4p \ ^{*2}</math></center>






<center><math>(1-cos\ \theta)=2</math></center>
<center><math>(1-cos\ \theta_{t=max})=2</math></center>






<center><math>-cos\ \theta=1</math></center>
<center><math>-cos\ \theta_{t=max}=1</math></center>





Revision as of 21:48, 12 June 2017

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

[math]\displaystyle{ \vartriangleleft }[/math] [math]\displaystyle{ \triangle }[/math] [math]\displaystyle{ \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]\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]


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


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


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


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


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


[math]\displaystyle{ \cos\ \theta = 1 \qquad \cos\ \theta = -1 }[/math]


[math]\displaystyle{ \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 minimum of t


[math]\displaystyle{ s+t_{max} \equiv 4m^2 }[/math]


[math]\displaystyle{ \Rightarrow t_{max}=4m^2-s }[/math]


[math]\displaystyle{ t_{max}=4m^2-4m^2- 4p \ ^{*2} }[/math]



The maximum transfer of momentum would be


[math]\displaystyle{ t_{max}=-4p \ ^{*2} }[/math]



[math]\displaystyle{ -2 p \ ^{*2}(1-cos\ \theta_{t=max})=-4p \ ^{*2} }[/math]


[math]\displaystyle{ (1-cos\ \theta_{t=max})=2 }[/math]


[math]\displaystyle{ -cos\ \theta_{t=max}=1 }[/math]


[math]\displaystyle{ \theta_{t=max} \equiv \arccos -1 }[/math]


The domain of the arccos function is from −1 to +1 inclusive and the range is from 0 to π radians inclusive (or from 0° to 180°). We find as expected for u=0 at [math]\displaystyle{ \theta=180^{\circ} }[/math]

[math]\displaystyle{ \theta_{t=max}=180^{\circ} }[/math]