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:

$s+t+u=(4(m^2+ p \ ^{*2}))+(-2 p \ ^{*2}(1-cos\ \theta))+(-2 p \ ^{*2}(1+cos\ \theta))$

$s+t+u \equiv 4m^2$

Since

$s \equiv 4(m^2+\vec p \ ^{*2})$

This implies

$s \ge 4m^2$

In turn, this implies

$t \le 0 \qquad u \le 0$

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

$t = 0 \qquad u = 0$

$-2 p \ ^{*2}(1-cos\ \theta) = 0 \qquad -2 p \ ^{*2}(1+cos\ \theta) = 0$

$(-2 p \ ^{*2}+2 p \ ^{*2}cos\ \theta) = 0 \qquad (-2 p \ ^{*2}-2 p \ ^{*2}cos\ \theta) = 0$

$2 p \ ^{*2}cos\ \theta = 2 p \ ^{*2} \qquad -2 p \ ^{*2}cos\ \theta = 2 p \ ^{*2}$

$\cos\ \theta = 1 \qquad \cos\ \theta = -1$

$\Rightarrow \theta_{t=0} = \arccos \ 1=0^{\circ} \qquad \theta_{u=0} = \arccos \ -1=180^{\circ}$

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

$s+t \equiv 4m^2$

$\Rightarrow t=4m^2-s$

$t=4m^2-4m^2+ 4p \ ^{*2})$

$t=4p \ ^{*2}$

$-2 p \ ^{*2}(1-cos\ \theta)=4p \ ^{*2}$

$(1-cos\ \theta)=-2$

$-cos\ \theta=-3$

$\theta_{max} \equiv \arccos 3$

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

$\theta = \arccos{3}$

$\cos{\theta} = \cos{\arccos{3}}$

$\cos{\theta} = 3$

From Euler's formula

$\cos{x} = \frac{e^{i \theta} + e^{-i \theta}}{2}$

$\frac{e^{i \theta} + e^{-i \theta}}{2} = 3$

$e^{i \theta} + e^{-i \theta} = 6$

Multiply with $e^{i \theta}$

$e^{2i \theta} + 1 = 6e^{i \theta}$

Letting $y = e^{i \theta}$

We get an quadratic equation:

$y^2 - 6y + 1 = 0$

$y = (6 ± √32)/2$

$y_1 = 5.828427 = e^{i \theta}$

$y_2 = 0.171573 = e^{i \theta}$

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

$\theta_1 = \frac{ln(5.828427) }{ i}=-1.76275i$

$\theta_2 = \frac{ln(0.171573) }{ i}=1.76275i$