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| − | However, from the definition of u being invariant between frames of reference | + | However, from the definition of t being invariant between frames of reference |
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| − | <center><math>u \equiv \overbrace{\left({\mathbf P_1^*}- {\mathbf P_2^{'*}}\right)^2=\left({\mathbf P_2^*}- {\mathbf P_1^{'*}}\right)^2}^{CM\ FRAME}=\overbrace{\left({\mathbf P_1}- {\mathbf P_2^{'}}\right)^2 = \left({\mathbf P_2}- {\mathbf P_1^{'}}\right)^2}^{LAB\ FRAME}</math></center> | + | <center><math>u \equiv \overbrace{\left({\mathbf P_1^*}- {\mathbf P_1^{'*}}\right)^2=\left({\mathbf P_2^*}- {\mathbf P_2^{'*}}\right)^2}^{CM\ FRAME}=\overbrace{\left({\mathbf P_1}- {\mathbf P_1^{'}}\right)^2 = \left({\mathbf P_2}- {\mathbf P_2^{'}}\right)^2}^{LAB\ FRAME}</math></center> |
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| − | When u=0, this implies 4 different scenarios
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| − | <center> <math>\mathbf P_1^*= \mathbf P_2^{'*}=\mathbf P_2^*= \mathbf P_1^{'*}=0</math></center>
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| − | This is the simple solution which would imply no collision.
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| − | <center><math>\mathbf P_1^*= \mathbf P_2^{'*}; \mathbf P_2^*= \mathbf P_1^{'*}=0 </math></center>
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| − | <center> <math>\mathbf P_1^*= \mathbf P_2^{'*}=0;\mathbf P_2^*= \mathbf P_1^{'*}</math></center>
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| − | These two cases show a stationary particle receiving all the momentum of an incident particle. This is not possible for equal mass particles.
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| − | <center> <math>\mathbf P_1^*= \mathbf P_2^{'*}=\mathbf P_2^*= \mathbf P_1^{'*}</math></center>
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| − | <center><math>t \equiv \left({\mathbf P_1^*}- {\mathbf P_1^{'*}}\right)^2=\left({\mathbf P_2^{*}}- {\mathbf P_2^{'*}}\right)^2</math></center>
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Revision as of 17:10, 13 June 2017
[math]\textbf{\underline{Navigation}}[/math]
[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]
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 t being invariant between frames of reference
[math]u \equiv \overbrace{\left({\mathbf P_1^*}- {\mathbf P_1^{'*}}\right)^2=\left({\mathbf P_2^*}- {\mathbf P_2^{'*}}\right)^2}^{CM\ FRAME}=\overbrace{\left({\mathbf P_1}- {\mathbf P_1^{'}}\right)^2 = \left({\mathbf P_2}- {\mathbf P_2^{'}}\right)^2}^{LAB\ FRAME}[/math]