You have four incoming slow momenta (\(P_{1}\), \(P_{2}\), \(P_{3}\), and \(P_{4}\)) and two outgoing slow momenta (\(P_{5}\) and \(P_{6}\)).
Energy-Momentum Vectors
Each of the six 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
$$ \sum_{i = 1}^{4} m_{i} \cosh{(\xi_{i})} = \sum_{j = 5}^{6} m_{j} \cosh{(\xi_{j})}. $$
Similarly, the conservation of momentum leads to
$$ \sum_{i = 1}^{4} m_{i} \sinh{(\xi_{i})} = \sum_{j = 5}^{6} m_{j} \sinh{(\xi_{j})}. $$
These two expressions can be combined to yield two equivalent expressions:
$$ \sum_{i = 1}^{4} m_{i} \exp{(\xi_{i})} = \sum_{j = 5}^{6} m_{j} \exp{(\xi_{j})}; $$
$$ \sum_{i = 1}^{4} \frac{m_{i}}{\exp{(\xi_{i})}} = \sum_{j = 5}^{6} \frac{m_{j}}{\exp{(\xi_{j})}}. $$
2-Mandelstam Invariants
You can introduce fifteen 2-Mandelstam invariants, but only four are relevant:
$$ s_{12} = -\Vert P_{1} + P_{2}\Vert^{2}, \qquad s_{13} = -\Vert P_{1} + P_{3}\Vert^{2}, \qquad s_{14} = -\Vert P_{1} + P_{4}\Vert^{2}; $$
$$ s_{56} = -\Vert P_{5} + P_{6} \Vert^{2}. $$
3-Mandelstam Invariants
You can introduce ten 3-Mandelstam invariants, but only one is relevant:
$$ s_{123} = -\Vert P_{1} + P_{2} + P_{3} \Vert^{2} = -\Vert P_{4} - P_{5} - P_{6} \Vert^{2}. $$
Regge-Mandelstam Invariants
You can introduce fifteen Regge-Mandelstam invariants, but only four are relevant:
$$ r_{12} \equiv -\frac{P_{1} \cdot P_{2}}{m_{1} m_{2}} = \frac{s_{12} - m_{1}^{2} - m_{2}^{2}}{2 m_{1} m_{2}} = \cosh{(\xi_{1} - \xi_{2})}; $$
$$ r_{13} \equiv -\frac{P_{1} \cdot P_{3}}{m_{1} m_{3}} = \frac{s_{13} - m_{1}^{2} - m_{3}^{2}}{2 m_{1} m_{3}} = \cosh{(\xi_{1} - \xi_{3})}; $$
$$ r_{14} \equiv -\frac{P_{1} \cdot P_{4}}{m_{1} m_{4}} = \frac{s_{14} - m_{1}^{2} - m_{4}^{2}}{2 m_{1} m_{4}} = \cosh{(\xi_{1} - \xi_{4})}; $$
$$ r_{56} \equiv -\frac{P_{5} \cdot P_{6}}{m_{5} m_{6}} = \frac{s_{56} - m_{5}^{2} - m_{6}^{2}}{2 m_{5} m_{6}} = \cosh{(\xi_{5} - \xi_{6})}. $$
Since the hyperbolic cosine satisfies
$$ \cosh{(x)} \geq 1; $$
you have the conditions:
$$ s_{ij} \geq (m_{i} + m_{j})^{2}. $$
These conditions define the range of values allowed for the 2-Mandelstam invariants. Since the 3-Mandelstam invariants can be written in terms of the 2-Mandelstam invariants, you also find conditions on the 3-Mandelstam invariants:
$$ s_{123} \geq (m_{1} + m_{2})^{2} + (m_{2} + m_{3})^{2} + (m_{3} + m_{1})^{2} - m_{1}^{2} - m_{2}^{2} - m_{3}^{2}. $$
Hence
$$ s_{123} \geq (m_{1} + m_{2} + m_{3})^{2}. $$
Regge-Sudakov Invariants
You can introduce four Regge-Sudakov invariants via
$$ 2 r_{ij} = R_{ij} + \frac{1}{R_{ij}}. $$
Since the Regge-Mandelstam invariants are hyperbolic cosines, the Regge-Sudakov invariants are exponentials:
$$ r_{ij} = \cosh{(\xi_{i} - \xi_{j})} \quad \Longrightarrow \quad R_{ij} = \exp{(\xi_{i} - \xi_{j})}. $$
In terms of the masses and the 2-Mandelstam invariants you have
$$ R_{ij} = \frac{s_{ij} - m_{i}^{2} - m_{j}^{2} + \sqrt{ \left( s_{ij} - m_{i}^{2} - m_{j}^{2} \right)^{2} - 4 m_{i}^{2} m_{j}^{2}}}{2 m_{i} m_{j}}. $$
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
$$ A = m_{5} f_{5} + m_{6} f_{6}, \qquad B = \frac{m_{5}}{f_{5}} + \frac{m_{6}}{f_{6}}. $$
Here
$$ A = \sum_{i = 1}^{4} m_{i} f_{i}; $$
and
$$ B = \sum_{i = 1}^{4} \frac{m_{i}}{f_{i}}. $$
Note that
$$ A B = - \Vert P_{1} + P_{2} + P_{3} + P_{4} \Vert^{2} = s_{56}. $$
Let
$$ A = \sqrt{s_{56}} C, \qquad B = \frac{\sqrt{s_{56}}}{C}. $$
Then
$$ C = \sqrt{\frac{A}{B}}. $$
In terms of the \(m_{i}\) and the \(f_{i}\) you have
$$ C = \sqrt{f_{1} f_{2} f_{3} f_{4}} \sqrt{ \frac{m_{1} f_{1} + m_{2} f_{2} + m_{3} f_{3} + m_{4} f_{4}}{m_{1} f_{2} f_{3} f_{4} + m_{2} f_{3} f_{4} f_{1} + m_{3} f_{4} f_{1} f_{2} + m_{4} f_{1} f_{2} f_{3}} }. $$
In this form, \(C\) is symmetric in all variables. Another way to write \(C\) is
$$ C = f_{1} \sqrt{ \frac{m_{1} + \dfrac{m_{2}}{R_{12}} + \dfrac{m_{3}}{R_{13}} + \dfrac{m_{4}}{R_{14}}}{m_{1} + m_{2} R_{12} + m_{3} R_{13} + m_{4} R_{14}} }. $$
Solving for \(f_{5}\) and \(f_{6}\) gives
$$ f_{5} = \frac{AB + m_{5}^{2} - m_{6}^{2} + \sqrt{\left[ AB - (m_{5} + m_{6})^{2} \right] \left[ AB - (m_{5} - m_{6})^{2} \right]}}{2 m_{5} B}; $$
$$ f_{6} = \frac{AB - m_{5}^{2} + m_{6}^{2} - \sqrt{\left[ AB - (m_{5} + m_{6})^{2} \right] \left[ AB - (m_{5} - m_{6})^{2} \right]}}{2 m_{6} B}. $$
