You have two incoming slow momenta (P1 and P2) and two outgoing slow momenta (P3 and P4). In order to be consistent with the non-relativistic result, I will assume elasticity:
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Energy-Momentum Vectors
The four external slow momenta can be written as
P_{1} = \begin{pmatrix} m_{1} \cosh{(\xi_{1})} & m_{1} \sinh{(\xi_{1})} \end{pmatrix};
P_{2} = \begin{pmatrix} m_{2} \cosh{(\xi_{2})} & m_{2} \sinh{(\xi_{2})} \end{pmatrix};
P_{3} = \begin{pmatrix} m_{1} \cosh{(\xi_{3})} & m_{1} \sinh{(\xi_{3})} \end{pmatrix};
P_{4} = \begin{pmatrix} m_{2} \cosh{(\xi_{4})} & m_{2} \sinh{(\xi_{4})} \end{pmatrix}.
Conservation
The conservation of energy leads to
m_{1} \cosh{(\xi_{1})} + m_{2} \cosh{(\xi_{2})} = m_{1} \cosh{(\xi_{3})} + m_{2} \cosh{(\xi_{4})}.
Similarly, the conservation of momentum leads to
m_{1} \sinh{(\xi_{1})} + m_{2} \sinh{(\xi_{2})} = m_{1} \sinh{(\xi_{3})} + m_{2} \sinh{(\xi_{4})}.
These two expressions can be combined to yield two equivalent expressions:
m_{1} \exp{(\xi_{1})} + m_{2} \exp{(\xi_{2})} = m_{1} \exp{(\xi_{3})} + m_{2} \exp{(\xi_{4})};
\frac{m_{1}}{\exp{(\xi_{1})}} + \frac{m_{2}}{\exp{(\xi_{2})}} = \frac{m_{1}}{\exp{(\xi_{3})}} + \frac{m_{2}}{\exp{(\xi_{4})}}.
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 = 2 m_{1}^{2} + 2 m_{2}^{2}.
Regge-Mandelstam Invariants
You can introduce six Regge-Mandelstam invariants:
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_{1} m_{2}} = \frac{s - m_{1}^{2} - m_{2}^{2}}{2 m_{1} m_{2}} = \cosh{(\xi_{3} - \xi_{4})};
r_{13} \equiv -\frac{P_{1} \cdot P_{3}}{m_{1}^{2}} = \frac{2m_{1}^{2} - t}{2 m_{1}^{2}} = \cosh{(\xi_{1} - \xi_{3})};
r_{24} \equiv -\frac{P_{2} \cdot P_{4}}{m_{2}^{2}} = \frac{2m_{2}^{2} - t}{2 m_{2}^{2}} = \cosh{(\xi_{2} - \xi_{4})};
r_{14} \equiv -\frac{P_{1} \cdot P_{4}}{m_{1} m_{2}} = \frac{m_{1}^{2} + m_{2}^{2} - u}{2 m_{1} m_{2}} = \cosh{(\xi_{1} - \xi_{4})};
r_{23} \equiv -\frac{P_{2} \cdot P_{3}}{m_{1} m_{2}} = \frac{m_{1}^{2} + m_{2}^{2} - u}{2 m_{1} m_{2}} = \cosh{(\xi_{2} - \xi_{3})}.
Since the hyperbolic cosine satisfies
\cosh{(x)} \geq 1;
you have the conditions:
s \geq (m_{1} + m_{2})^{2}, \qquad t \leq 0, \qquad u \leq (m_{1} - m_{2})^{2}.
These conditions define the range of values allowed for the 2-Mandelstam invariants. The first one says that s only takes positive values above the threshold value (m_{1} + m_{2})^{2}. Thus, the vector P_{s} = P_{1} + P_{2} = P_{3} + P_{4} is slow and you can write
P_{s} = \begin{pmatrix} \sqrt{s} \cosh{(\xi_{s})} & \sqrt{s} \sinh{(\xi_{s})} \end{pmatrix}.
The second condition says that t only takes negative values. This means that the vector P_{t} = P_{1} - P_{3} = P_{4} - P_{2} is fast and you can write:
P_{t} = \begin{pmatrix} \sqrt{-t} \sinh{(\rho_{t})} & \sqrt{-t} \cosh{(\rho_{t})} \end{pmatrix}.
Note that
P_{s} \cdot P_{t} = \sqrt{s} \sqrt{-t} \sinh{(\xi_{s} - \rho_{t})}.
When \xi_{s} = 0 and \rho_{t} = 0, you have the center-of-momentum frame.
Since r_{12} = r_{34}, you have
\cosh{(\xi_{1} - \xi_{2})} = \cosh{(\xi_{3} - \xi_{4})}.
This has two solutions:
\xi_{1} - \xi_{2} = \xi_{3} - \xi_{4};
or
\xi_{1} - \xi_{2} = \xi_{4} - \xi_{3}.
The first solution implies that
\xi_{3} = \xi_{1} - \xi_{2} + \xi_{4}.
Combining this with the conservation law gives
\xi_{3} = \xi_{1}, \qquad \xi_{4} = \xi_{2}.
This is the forward solution. The second solution is compatible with r_{14} = r_{23}. Indeed, since you can write the slow rapidity in terms of the slow speed as
\xi = \operatorname{atanh}{(v)},
it follows that
\operatorname{atanh}{(v_{1})} - \operatorname{atanh}{(v_{2})} = \operatorname{atanh}{(v_{4})} - \operatorname{atanh}{(v_{3})}.
This is the relativistic analog of the relation
u_{1} - u_{2} = v_{2} - v_{1};
found for the non-relativistic problem.
Although r_{13} \neq r_{24}, you have
t = 2 m_{1}^{2} (1 - r_{13}) = 2 m_{2}^{2} (1 - r_{24}).
This means that
m_{1}^{2} (1 - \cosh{(\xi_{1} - \xi_{3})}) = m_{2}^{2} (1 - \cosh{(\xi_{2} - \xi_{4})}).
But recall that
\sinh^{2}{\left( \frac{x}{2} \right)} = \frac{\cosh{(x)} - 1}{2}.
Thus,
m_{1}^{2} \sinh^{2}{\left( \frac{\xi_{1} - \xi_{3}}{2} \right)} = m_{2}^{2} \sinh^{2}{\left( \frac{\xi_{2} - \xi_{4}}{2} \right)}.
Solving for the Outgoing Data
You can first write
\xi_{4} = \xi_{1} - \xi_{2} + \xi_{3}.
Then it follows that
\exp{(\xi_{4})} = \frac{\exp{(\xi_{1})} \exp{(\xi_{3})}}{\exp{(\xi_{2})}}.
From the conservation law you find
m_{1} \exp{(\xi_{1})} + m_{2} \exp{(\xi_{2})} = m_{1} \exp{(\xi_{3})} + m_{2} \left( \frac{\exp{(\xi_{1})} \exp{(\xi_{3})}}{\exp{(\xi_{2})}} \right).
This gives
\exp{(\xi_{3})} = \left[\frac{ m_{1} \exp{(\xi_{1})} + m_{2} \exp{(\xi_{2})} }{m_{1} \exp{(\xi_{2})} + m_{2} \exp{(\xi_{1})} }\right] \exp{(\xi_{2})}.
Plugin this back into the first expression gives
\exp{(\xi_{4})} = \left[\frac{ m_{1} \exp{(\xi_{1})} + m_{2} \exp{(\xi_{2})} }{m_{1} \exp{(\xi_{2})} + m_{2} \exp{(\xi_{1})} }\right] \exp{(\xi_{1})}.
The speeds can be introduced back via the relation
v = \tanh{(\xi)} \quad \Longrightarrow \quad \exp{(2 \xi)} = \frac{1 + v}{1 - v}.
You find
\frac{\sqrt{1 + v_{3}}}{\sqrt{1 - v_{3}}} = \frac{\sqrt{1 + v_{2}}}{\sqrt{1 - v_{2}}} \frac{m_{1} \sqrt{(1 + v_{1})(1 - v_{2})} + m_{2} \sqrt{(1 - v_{1})(1 + v_{2})}}{m_{1} \sqrt{(1 - v_{1})(1 + v_{2})} + m_{2} \sqrt{(1 + v_{1})(1 - v_{2})}};
\frac{\sqrt{1 + v_{4}}}{\sqrt{1 - v_{4}}} = \frac{\sqrt{1 + v_{1}}}{\sqrt{1 - v_{1}}} \frac{m_{1} \sqrt{(1 + v_{1})(1 - v_{2})} + m_{2} \sqrt{(1 - v_{1})(1 + v_{2})}}{m_{1} \sqrt{(1 - v_{1})(1 + v_{2})} + m_{2} \sqrt{(1 + v_{1})(1 - v_{2})}}.
This leads to the identity
\frac{(1 + v_{3})(1 - v_{4})}{(1 - v_{3})(1 + v_{4})} = \frac{(1 + v_{2})(1 - v_{1})}{(1 - v_{2})(1 + v_{1})}.
When m_{2} = m_{1}, this solution reduces to
\xi_{3} = \xi_{2}, \qquad \xi_{4} = \xi_{1}.
This is the exchange solution.
The article in Wikipedia on elastic collisions was helpful and it was fun to verify its results.
Composite Rapidities
The slow rapidity \xi_{s} associated to P_{s} can be writen as
\xi_{s} = \log{\left( \frac{m_{1} \exp{(\xi_{1})} + m_{2} \exp{(\xi_{2})}}{\sqrt{s}} \right)} = \log{\left( \frac{m_{1} \exp{(\xi_{3})} + m_{2} \exp{(\xi_{4})}}{\sqrt{s}} \right)}.
Similarly, the fast rapidity \rho_{t} associated to P_{t} can be written as
\rho_{t} = \log{\left( \frac{m_{1} \exp{(\xi_{1})} - m_{1} \exp{(\xi_{3})}}{\sqrt{-t}} \right)} = \log{\left( \frac{m_{2} \exp{(\xi_{4})} - m_{2} \exp{(\xi_{2})}}{\sqrt{-t}} \right)}.
Regimes
The condition on u does not allow for a specific kind of rapidity for the vector P_{u} = P_{1} - P_{4} = P_{3} - P_{2}. Depending on the values of s and t, you can have all three cases.
Slow u-channel
In the region
0 < u \leq (m_{1} - m_{2})^{2}
the vector P_{u} is slow and given by
P_{u} = \begin{pmatrix} \sqrt{u} \cosh{(\xi_{u})} & \sqrt{u} \sinh{(\xi_{u})} \end{pmatrix}.
Then
P_{s} \cdot P_{u} = - \sqrt{s} \sqrt{u} \cosh{(\xi_{s} - \xi_{u})};
and
P_{t} \cdot P_{u} = - \sqrt{-t} \sqrt{u} \sinh{(\rho_{t} - \xi_{u})}.
The composite slow rapidity \xi_{u} is given by
\xi_{u} = \log{\left( \frac{m_{1} \exp{(\xi_{1})} - m_{2} \exp{(\xi_{4})}}{\sqrt{u}} \right)} = \log{\left( \frac{m_{1} \exp{(\xi_{3})} - m_{2} \exp{(\xi_{2})}}{\sqrt{u}} \right)}.
Null u-channel
If u = 0, then the vector P_{u} is null and can be written as
P_{u} = \begin{pmatrix} k_{u} & k_{u} \end{pmatrix}.
Then
P_{s} \cdot P_{u} = \sqrt{s} k_{u} \exp{(-\xi_{s})};
and
P_{t} \cdot P_{u} = \sqrt{-t} k_{u} \exp{(\rho_{t})}.
You also have
k_{u} = m_{1} \cosh{(\xi_{1})} - m_{2} \cosh{(\xi_{4})} = m_{1} \sinh{(\xi_{1})} - m_{2} \sinh{(\xi_{4})}.
From here it follows that
\xi_{4} = \xi_{1} - \log{\left( \frac{m_{1}}{m_{2}} \right)}.
Using the conservation law, you can also write
k_{u} = m_{1} \cosh{(\xi_{3})} - m_{2} \cosh{(\xi_{2})} = m_{1} \sinh{(\xi_{3})} - m_{2} \sinh{(\xi_{2})}.
From here it follows that
\xi_{3} = \xi_{2} + \log{\left( \frac{m_{1}}{m_{2}} \right)}.
Note that
\xi_{1} + \xi_{2} = \xi_{3} + \xi_{4}.
Fast u-channel
If u < 0, then the vector P_{u} is fast and given by
P_{u} = \begin{pmatrix} \sqrt{-u} \sinh{(\rho_{u})} & \sqrt{-u} \cosh{(\rho_{u})} \end{pmatrix}.
Then
P_{s} \cdot P_{u} = \sqrt{s} \sqrt{-u} \sinh{(\xi_{s} - \rho_{u})};
and
P_{t} \cdot P_{u} = \sqrt{-t} \sqrt{-u} \cosh{(\rho_{t} - \rho_{u})}.
The composite fast rapidity \rho_{u} can be written as
\rho_{u} = \log{\left( \frac{m_{1} \exp{(\xi_{1})} - m_{2} \exp{(\xi_{4})}}{\sqrt{-u}} \right)} = \log{\left( \frac{m_{1} \exp{(\xi_{3})} - m_{2} \exp{(\xi_{2})}}{\sqrt{-u}} \right)}.
