Difference between revisions of "Limits based on Mandelstam Variables"

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

Revision as of 01:56, 14 June 2017


[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]


[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]


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

[math]s \equiv \overbrace{\left({\mathbf P_1^*}+ {\mathbf P_2^{*}}\right)^2=\left({\mathbf P_1^{'*}}+ {\mathbf P_2^{'*}}\right)^2}^{CM\ FRAME}=\overbrace{\left({\mathbf P_1}+ {\mathbf P_2}\right)^2 = \left({\mathbf P_1^{'}}+ {\mathbf P_2^{'}}\right)^2}^{LAB\ FRAME}[/math]

In the center of mass frame of reference,

[math]E_1^*=E_2^*=E^* \quad and \quad \vec p \ _1^*=-\vec p \ _2^*= \vec p \ ^*[/math]

[math]s \equiv 2m^2+2(E_1^{*2}+\vec p \ ^{*2} )[/math]

Using the relativistic energy equation

[math]E^2 \equiv \vec p \ ^2+m^2[/math]

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

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

[math]\frac{s-4m^2}{4}= \vec p \ ^{*2}[/math]

[math]t=-2 p \ ^{*2}(1-cos\ \theta)=\frac{-2(s-4m^2)}{4}(1-cos\ \theta)[/math]

[math]\frac{-2t}{s-4m^2}=(1-cos\ \theta)[/math]

[math]cos\ \theta=1-\frac{-2t}{s-4m^2}[/math]