Two-Dimensional Four-Point Kinematics (2)

You have two slow momenta ($P_{2}$ and $P_{4}$) and two fast momenta ($P_{1}$ and $P_{3}$).

Energy-Momentum Vectors

The four external momenta can be written as

$$ P_{1} = \begin{pmatrix} w_{1} \sinh{(\rho_{1})} & w_{1} \cosh{(\rho_{1})} \end{pmatrix}; $$

$$ P_{2} = \begin{pmatrix} m_{2} \cosh{(\xi_{2})} & m_{2} \sinh{(\xi_{2})} \end{pmatrix}; $$

$$ P_{3} = \begin{pmatrix} w_{3} \sinh{(\rho_{3})} & w_{3} \cosh{(\rho_{3})} \end{pmatrix}; $$

$$ P_{4} = \begin{pmatrix} m_{4} \cosh{(\xi_{4})} & m_{4} \sinh{(\xi_{4})} \end{pmatrix}. $$

The conservation of energy leads to

$$ w_{1} \sinh{(\rho_{1})} + m_{2} \cosh{(\xi_{2})} = w_{3} \sinh{(\rho_{3})} + m_{4} \cosh{(\xi_{4})}. $$

Similarly, the conservation of momentum leads to

$$ w_{1} \cosh{(\rho_{1})} + m_{2} \sinh{(\xi_{2})} = w_{3} \cosh{(\rho_{3})} + m_{4} \sinh{(\xi_{4})}. $$

These two expressions can be combined to yield two equivalent expressions:

$$ w_{1} \exp{(\rho_{1})} + m_{2} \exp{(\xi_{2})} = w_{3} \exp{(\rho_{3})} + m_{4} \exp{(\xi_{4})}; $$

$$ \frac{w_{1}}{\exp{(\rho_{1})}} - \frac{m_{2}}{\exp{(\xi_{2})}} = \frac{w_{3}}{\exp{(\rho_{3})}} - \frac{m_{4}}{\exp{(\xi_{4})}}. $$

You can introduce three 2-Mandelstam invariants:

$$ s = -(P_{1} + P_{2})^{2} = -(P_{3} + P_{4})^{2}; $$

$$ t = -(P_{1} - P_{3})^{2} = -(P_{2} - P_{4})^{2}; $$

$$ u = -(P_{1} - P_{4})^{2} = -(P_{2} - P_{3})^{2}. $$

Due to the conservation laws, you have

$$ s + t + u = - w_{1}^{2} + m_{2}^{2} - w_{3}^{2} + m_{4}^{2}. $$

You can introduce six Regge-Mandelstam invariants:

$$ r_{12} \equiv -\frac{P_{1} \cdot P_{2}}{w_{1} m_{2}} = \frac{s + w_{1}^{2} - m_{2}^{2}}{2 w_{1} m_{2}} = \sinh{(\rho_{1} - \xi_{2})}; $$

$$ r_{34} \equiv -\frac{P_{3} \cdot P_{4}}{w_{3} m_{4}} = \frac{s + w_{3}^{2} - m_{4}^{2}}{2 w_{3} m_{4}} = \sinh{(\rho_{3} - \xi_{4})}; $$

$$ r_{13} \equiv \frac{P_{1} \cdot P_{3}}{w_{1} w_{3}} = \frac{w_{1}^{2} + w_{3}^{2} + t}{2 w_{1} w_{3}} = \cosh{(\rho_{1} - \rho_{3})}; $$

$$ r_{24} \equiv -\frac{P_{2} \cdot P_{4}}{m_{2} m_{4}} = \frac{m_{2}^{2} + m_{4}^{2} - t}{2 m_{2} m_{4}} = \cosh{(\xi_{2} - \xi_{4})}; $$

$$ r_{14} \equiv \frac{P_{1} \cdot P_{4}}{w_{1} m_{4}} = \frac{u + w_{1}^{2} - m_{4}^{2}}{2 w_{1} m_{4}} = \sinh{(\xi_{4} - \rho_{1})}; $$

$$ r_{23} \equiv \frac{P_{2} \cdot P_{3}}{m_{2} w_{3}} = \frac{u - m_{2}^{2} + w_{3}^{2}}{2 m_{2} w_{3}} = \sinh{(\xi_{2} - \rho_{3})}. $$

Since the hyperbolic cosine satisfies

$$ \cosh{(x)} \geq 1; $$

you have the condition:

$$ -(w_{1} - w_{3})^{2} \leq t \leq (m_{2} - m_{4})^{2} . $$

This condition defines the range of values allowed for $t$. Since the hyperbolic sine satisfies

$$ -\infty < \sinh{(x)} < \infty; $$

you have the conditions

$$ -\infty < s < \infty, \qquad -\infty < u < \infty. $$

That is, $s$ and $u$ appear to be physically unbounded.