You have two slow momenta (\(P_{2}\) and \(P_{4}\)) and two null momenta (\(P_{1}\) and \(P_{3}\)).
Energy-Momentum Vectors
The four external momenta can be written as
$$ P_{1} = \begin{pmatrix} k_{1} & k_{1} \end{pmatrix}; $$
$$ P_{2} = \begin{pmatrix} m_{2} \cosh{(\xi_{2})} & m_{2} \sinh{(\xi_{2})} \end{pmatrix}; $$
$$ P_{3} = \begin{pmatrix} k_{3} & k_{3} \end{pmatrix}; $$
$$ P_{4} = \begin{pmatrix} m_{4} \cosh{(\xi_{4})} & m_{4} \sinh{(\xi_{4})} \end{pmatrix}. $$
Here \(k_{1} > 0\), \(m_{2} > 0\), \(k_{3} > 0\), and \(m_{4} > 0\).
Conservation
The conservation of energy leads to
$$ k_{1} + m_{2} \cosh{(\xi_{2})} = k_{3} + m_{4} \cosh{(\xi_{4})}. $$
Similarly, the conservation of momentum leads to
$$ k_{1} + m_{2} \sinh{(\xi_{2})} = k_{3} + m_{4} \sinh{(\xi_{4})}. $$
These two expressions can be combined to yield two equivalent expressions:
$$ 2k_{1} + m_{2} \exp{(\xi_{2})} = 2k_{3} + m_{4} \exp{(\xi_{4})}; $$
$$ \frac{m_{2}}{\exp{(\xi_{2})}} = \frac{m_{4}}{\exp{(\xi_{4})}}. $$
The second expression leads to
$$ \xi_{4} = \xi_{2} + \log{\left( \frac{m_{4}}{m_{2}} \right)}. $$
Solving for \(k_{3}\) gives
$$ k_{3} = k_{1} + \frac{m_{2}}{2} \exp{(\xi_{2})} - \frac{m_{4}}{2} \exp{(\xi_{4})} = k_{1} + \left( \frac{m_{2}^{2} - m_{4}^{2}}{2 m_{2}} \right) \exp{(\xi_{2})}. $$
2-Mandelstam Invariants
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
$$ s + t + u = m_{2}^{2} + m_{4}^{2}. $$
Regge-Mandelstam Invariants
You can introduce six Regge-Mandelstam invariants:
$$ r_{12} \equiv -\frac{P_{1} \cdot P_{2}}{k_{1} m_{2}} = \frac{s - m_{2}^{2}}{2 k_{1} m_{2}} = \exp{(-\xi_{2})}; $$
$$ r_{34} \equiv -\frac{P_{3} \cdot P_{4}}{k_{3} m_{4}} = \frac{s - m_{4}^{2}}{2 k_{3} m_{4}} = \exp{(-\xi_{4})}; $$
$$ r_{13} \equiv -\frac{P_{1} \cdot P_{3}}{k_{1} k_{3}} = 0; $$
$$ r_{24} \equiv -\frac{P_{2} \cdot P_{4}}{m_{2} m_{4}} = \frac{m_{2}^{2} + m_{4}^{2}}{2 m_{2} m_{4}} = \cosh{(\xi_{2} - \xi_{4})}; $$
$$ r_{14} \equiv -\frac{P_{1} \cdot P_{4}}{k_{1} m_{4}} = \frac{m_{4}^{2} - u}{2 k_{1} m_{4}} = \exp{(-\xi_{4})}; $$
$$ r_{23} \equiv -\frac{P_{2} \cdot P_{3}}{m_{2} k_{3}} = \frac{m_{2}^{2} - u}{2 m_{2} k_{3}} = \exp{(-\xi_{2})}. $$
Since the hyperbolic cosine satisfies
$$ \cosh{(x)} \geq 1; $$
you have the condition:
$$ m_{2}^{2} + m_{4}^{2} \geq 2 m_{2} m_{4}. $$
Indeed, \(r_{13} = 0\) implies that \(t = 0\) and that the vector \(P_{t} = P_{1} - P_{3} = P_{4} - P_{2}\) is null. Since \(t = 0\), you also have
$$ s + u = m_{2}^{2} + m_{4}^{2}. $$
Since the exponential function is always positive, you also have
$$ r_{12} = r_{23} \geq 0, \qquad r_{14} = r_{34} \geq 0. $$
The equality constraint implies that
$$ \frac{s - m_{2}^{2}}{m_{2}^{2} - u} = \frac{k_{1}}{k_{3}} = \frac{m_{4}^{2} - u}{s - m_{4}^{2}}. $$
In terms of \(s\) only, you find that
$$ \frac{k_{1}}{k_{3}} = \frac{s - m_{2}^{2}}{s - m_{4}^{2}}. $$
You can use this to solve for \(k_{3}\):
$$ k_{3} = \left( \frac{s - m_{4}^{2}}{s - m_{2}^{2}} \right) k_{1}. $$
Combining this with the previous result gives \(k_{1}\) in terms of \(\xi_{2}\):
$$ k_{1} = \left( \frac{s - m_{2}^{2}}{2 m_{2}} \right) \exp{(\xi_{2})}; $$
and thus
$$ k_{3} = \left( \frac{s - m_{4}^{2}}{2 m_{2}} \right) \exp{(\xi_{2})} = \left( \frac{s - m_{4}^{2}}{2 m_{4}} \right) \exp{(\xi_{4})}. $$
The positivity constraint gives
$$ \frac{s - m_{2}^{2}}{2 k_{1} m_{2}} \geq 0 \quad \Longrightarrow \quad s \geq m_{2}^{2}. $$
Similarly, you also need \(s \geq m_{4}^{2}\). Both of these conditions mean that the vector \(P_{s} = P_{1} + P_{2} = P_{3} + P_{4}\) is slow and can be written as
$$ P_{s} = \begin{pmatrix} \sqrt{s} \cosh{(\xi_{s})} & \sqrt{s}\sinh{(\xi_{s})} \end{pmatrix}. $$
The composite slow rapidity \(\xi_{s}\) is given by
$$ \xi_{s} = \xi_{2} + \log{\left( \frac{\sqrt{s}}{m_{2}} \right)}. $$
Note that
$$ s - m_{2}^{2} = m_{4}^{2} - u \geq 0, \qquad s - m_{4}^{2} = m_{2}^{2} - u \geq 0. $$
This means that the vector \(P_{u} = P_{1} - P_{4} = P_{3} - P_{2}\) does not have a fixed rapidity and there are three regimes.
