Limits based on Mandelstam Variables

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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]

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 maximum of t

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

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

[math]\displaystyle{ t=4m^2-4m^2+ 4p \ ^{*2}) }[/math]

[math]\displaystyle{ t=4p \ ^{*2} }[/math]

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

[math]\displaystyle{ (1-cos\ \theta)=-2 }[/math]

[math]\displaystyle{ -cos\ \theta=-3 }[/math]

[math]\displaystyle{ \theta_{max} \equiv \arccos 3 }[/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°). This implies for arccos 3, the range will include imaginary numbers. Knowing that the range of the cosine function is -1 to +1 inclusive and the domain to be any angle

[math]\displaystyle{ x = \arccos{3} }[/math]

[math]\displaystyle{ cos(x) = \cos{\arccos{3}} }[/math]

[math]\displaystyle{ \cos{x} = 3 }[/math]

From Euler's formula

[math]\displaystyle{ \cos{x} = \frac{e^{ix} + e^{-ix}}{2} }[/math]

[math]\displaystyle{ \frac{e^{ix} + e^{-ix}}{2} = 3 }[/math]

[math]\displaystyle{ e^{ix} + e^{-ix} = 6 }[/math]

Multiply with [math]\displaystyle{ e^{ix} }[/math]

[math]\displaystyle{ e^{2ix} + 1 = 6e^{ix} }[/math]

Letting [math]\displaystyle{ y = e^{ix} }[/math]

We get an quadratic equation:

[math]\displaystyle{ y^2 - 6y + 1 = 0 }[/math]

[math]\displaystyle{ y = (6 ± √32)/2 }[/math]

[math]\displaystyle{ y_1 = 5.828427 = e^{ix} }[/math]

[math]\displaystyle{ y_2 = 0.171573 = e^{ix} }[/math]

Apply ln on both sides gives the solution for arccos 3:

[math]\displaystyle{ x_1 = \frac{ln(5.828427) }{ i}=-1.762747i }[/math]

[math]\displaystyle{ x_2 = \frac{ln(0.171573) }{ i}=-1.76275i }[/math]