You have two incoming slow momenta (\(P_{1}\) and \(P_{2}\)) and two outgoing slow momenta (\(P_{3}\) and \(P_{4}\)).
Energy-Momentum Vectors
Each of the four external slow momenta can be written as
$$ P_{i} = \begin{pmatrix} m_{i} \cosh{(\xi_{i})} & m_{i} \sinh{(\xi_{i})} \end{pmatrix}. $$
Here \(m_{i} > 0\) and \(-\infty < \xi_{i} < \infty\).
Conservation
The conservation of energy leads to
$$ \sum_{i = 1}^{2} m_{i} \cosh{(\xi_{i})} = \sum_{j = 3}^{4} m_{j} \cosh{(\xi_{j})}. $$
Similarly, the conservation of momentum leads to
$$ \sum_{i = 1}^{2} m_{i} \sinh{(\xi_{i})} = \sum_{j = 3}^{4} m_{j} \sinh{(\xi_{j})}. $$
These two expressions can be combined to yield two equivalent expressions:
$$ \sum_{i = 1}^{2} m_{i} \exp{(\xi_{i})} = \sum_{j = 3}^{4} m_{j} \exp{(\xi_{j})}; $$
$$ \sum_{i = 1}^{2} \frac{m_{i}}{\exp{(\xi_{i})}} = \sum_{j = 3}^{4} \frac{m_{j}}{\exp{(\xi_{j})}}. $$
2-Mandelstam Invariants
The most relevant 2-Mandelstam invariant is
$$ s = - \Vert P_{1} + P_{2} \Vert^{2} = - \Vert P_{3} + P_{4} \Vert^{2}. $$
Regge-Mandelstam Invariants
The most relevant Regge-Mandelstam invariants are:
$$ r_{12} \equiv -\frac{P_{1} \cdot P_{2}}{m_{1} m_{2}} = \frac{s - m_{1}^{2} - m_{2}^{2}}{2 m_{1} m_{2}} = \cosh{(\xi_{1} - \xi_{2})}; $$
$$ r_{34} \equiv -\frac{P_{3} \cdot P_{4}}{m_{3} m_{4}} = \frac{s - m_{3}^{2} - m_{4}^{2}}{2 m_{3} m_{4}} = \cosh{(\xi_{3} - \xi_{4})}. $$
Since the hyperbolic cosine satisfies
$$ \cosh{(x)} \geq 1; $$
you have the conditions:
$$ s \geq (m_{1} + m_{2})^{2}, \qquad s \geq (m_{3} + m_{4})^{2}. $$
Note that you can write the 2-Mandelstam invariant in terms of the masses and the difference in rapidities:
$$ s = m_{1}^{2} + m_{2}^{2} + 2 m_{1} m_{2} \cosh{(\xi_{1} - \xi_{2})} = m_{3}^{2} + m_{4}^{2} + 2 m_{3} m_{4} \cosh{(\xi_{3} - \xi_{4})}. $$
Solving for the Outgoing Data
If you introduce variables \(f_{i}\) via
$$ f_{i} = \exp{(\xi_{i})}; $$
then the conservation laws can be rewritten as
$$ A = m_{3} f_{3} + m_{4} f_{4}, \qquad B = \frac{m_{3}}{f_{3}} + \frac{m_{4}}{f_{4}}. $$
Here
$$ A = m_{1} f_{1} + m_{2} f_{2}; $$
and
$$ B = \frac{m_{1}}{f_{1}} + \frac{m_{2}}{f_{2}}. $$
Note that
$$ A B = - \Vert P_{1} + P_{2} \Vert^{2} = s. $$
You can write
$$ A = \sqrt{s} C, \qquad B = \frac{\sqrt{s}}{C}. $$
Then it follows that
$$ C = \sqrt{\frac{A}{B}} = \sqrt{f_{1} f_{2}} \sqrt{\frac{m_{1} f_{1} + m_{2} f_{2}}{m_{1} f_{2} + m_{2} f_{1}}}. $$
Solving for \(f_{3}\) and \(f_{4}\) gives
$$ f_{3} = \frac{AB + m_{3}^{2} - m_{4}^{2} + \sqrt{\left[ AB - (m_{3} + m_{4})^{2} \right] \left[ AB - (m_{3} - m_{4})^{2} \right]}}{2 m_{3} B}; $$
$$ f_{4} = \frac{AB - m_{3}^{2} + m_{4}^{2} - \sqrt{\left[ AB - (m_{3} + m_{4})^{2} \right] \left[ AB - (m_{3} - m_{4})^{2} \right]}}{2 m_{4} B}. $$
In terms of \(s\) and the variables \(f_{1}\) and \(f_{2}\) you have
$$ f_{3} = \left(\frac{s + m_{3}^{2} - m_{4}^{2} + \sqrt{\left[ s - (m_{3} + m_{4})^{2} \right] \left[ s - (m_{3} - m_{4})^{2} \right]}}{2 \sqrt{s} m_{3}}\right) \sqrt{f_{1} f_{2}} \sqrt{ \frac{m_{1} f_{1} + m_{2} f_{2}}{m_{1} f_{2} + m_{2} f_{1}} }; $$
$$ f_{4} = \left(\frac{s - m_{3}^{2} + m_{4}^{2} - \sqrt{\left[ s - (m_{3} + m_{4})^{2} \right] \left[ s - (m_{3} - m_{4})^{2} \right]}}{2 \sqrt{s} m_{4}}\right) \sqrt{f_{1} f_{2}} \sqrt{ \frac{m_{1} f_{1} + m_{2} f_{2}}{m_{1} f_{2} + m_{2} f_{1}} }. $$
Solving for the rapidities gives
$$ \xi_{3} = \frac{\xi_{1} + \xi_{2}}{2} + \frac{1}{2} \log{\left( \frac{m_{1} \exp{(\xi_{1} - \xi_{2})} + m_{2}}{m_{1} + m_{2} \exp{(\xi_{1} - \xi_{2})}} \right)} + \log{\left(\frac{s + m_{3}^{2} - m_{4}^{2} + \sqrt{\left[ s - (m_{3} + m_{4})^{2} \right] \left[ s - (m_{3} - m_{4})^{2} \right]}}{2 \sqrt{s} m_{3}}\right)}; $$
$$ \xi_{4} = \frac{\xi_{1} + \xi_{2}}{2} + \frac{1}{2} \log{\left( \frac{m_{1} \exp{(\xi_{1} - \xi_{2})} + m_{2}}{m_{1} + m_{2} \exp{(\xi_{1} - \xi_{2})}} \right)} + \log{\left(\frac{s - m_{3}^{2} + m_{4}^{2} - \sqrt{\left[ s - (m_{3} + m_{4})^{2} \right] \left[ s - (m_{3} - m_{4})^{2} \right]}}{2 \sqrt{s} m_{4}}\right)}. $$
A simpler relation between \(\xi_{3}\) and \(\xi_{4}\) follows from \(r_{34}\):
$$ \exp{(\xi_{3} - \xi_{4})} = \frac{s - m_{3}^{2} - m_{4}^{2} + \sqrt{\left[ s - (m_{3} + m_{4})^{2} \right] \left[ s - (m_{3} - m_{4})^{2} \right]}}{2 m_{3} m_{4}}. $$
