Difference between revisions of "Limits based on Mandelstam Variables"

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However, from the definition of u
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However, from the definition of u being invariant between frames of reference
  
<center><math>u \equiv \left({\mathbf P_1^*}- {\mathbf P_2^{'*}}\right)^2=\left({\mathbf P_2^*}- {\mathbf P_1^{'*}}\right)^2</math></center>
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<center><math>u \equiv \left({\mathbf P_1^*}- {\mathbf P_2^{'*}}\right)^2=\left({\mathbf P_2^*}- {\mathbf P_1^{'*}}\right)^2=\left({\mathbf P_1}- {\mathbf P_2^{'}}\right)^2=\left({\mathbf P_2^}- {\mathbf P_1^{'}}\right)^2</math></center>
  
  

Revision as of 17:04, 13 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]


[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 minimum of t


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


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


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



The maximum transfer of momentum would be


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



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


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


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


[math] \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]\theta=180^{\circ}[/math]

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


However, from the definition of u being invariant between frames of reference

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


When u=0, this implies 4 different scenarios

[math]\mathbf P_1^*= \mathbf P_2^{'*}=\mathbf P_2^*= \mathbf P_1^{'*}=0[/math]


This is the simple solution which would imply no collision.


[math]\mathbf P_1^*= \mathbf P_2^{'*}; \mathbf P_2^*= \mathbf P_1^{'*}=0 [/math]


[math]\mathbf P_1^*= \mathbf P_2^{'*}=0;\mathbf P_2^*= \mathbf P_1^{'*}[/math]


These two cases show a stationary particle receiving all the momentum of an incident particle. This is not possible for equal mass particles.


[math]\mathbf P_1^*= \mathbf P_2^{'*}=\mathbf P_2^*= \mathbf P_1^{'*}[/math]



[math]t \equiv \left({\mathbf P_1^*}- {\mathbf P_1^{'*}}\right)^2=\left({\mathbf P_2^{*}}- {\mathbf P_2^{'*}}\right)^2[/math]