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 minimum of t

$s+t_{max} \equiv 4m^2$

$\Rightarrow t_{max}=4m^2-s$

$t_{max}=4m^2-4m^2- 4p \ ^{*2}$

The maximum transfer of momentum would be

$t_{max}=-4p \ ^{*2}$

$-2 p \ ^{*2}(1-cos\ \theta_{t=max})=-4p \ ^{*2}$

$(1-cos\ \theta_{t=max})=2$

$-cos\ \theta_{t=max}=1$

$\theta_{t=max} \equiv \arccos -1$

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 $\theta=180^{\circ}$

$\theta_{t=max}=180^{\circ}$

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

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

In the center of mass frame of reference,

$E_1^*=E_2^*=E^* \quad and \quad \vec p \ _1^*=-\vec p \ _2^*= \vec p \ ^*$

$s \equiv 2m^2+2(E_1^{*2}+\vec p \ ^{*2} )$

Using the relativistic energy equation

$E^2 \equiv \vec p \ ^2+m^2$

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

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

$\frac{s-4m^2}{4}= \vec p \ ^{*2})$