A slow energy-momentum vector in two-dimensional spacetime has the form
$$ P = \begin{bmatrix} \pm m \cosh{(\xi)} & m \sinh{(\xi)} \end{bmatrix}. $$
Here \(m > 0\) and \(\xi\) is the slow rapidity. The slow speed is
$$ v = \tanh{(\xi)}. $$
Since the hyperbolic tangent is bounded,
$$ -1 < \tanh{(\xi)} < 1; $$
the slow speed is bounded:
$$ -1 < v < 1. $$
However, the square slow speed is only bounded from above:
$$ v^{2} < 1. $$
The rapidity in terms of the speed is
$$ \exp{(\xi)} = \sqrt{\frac{1 + v}{1 - v}}. $$
Then it follows that
$$ \cosh{(\xi)} = \frac{1}{\sqrt{1 - v^{2}}}, \qquad \sinh{(\xi)} = \frac{v}{\sqrt{1 - v^{2}}}. $$
These are familiar expressions.
A fast energy-momentum vector in two-dimensional spacetime has the form
$$ P = \begin{bmatrix} w \sinh{(\rho)} & \pm w \cosh{(\rho)} \end{bmatrix}. $$
Here \(w > 0\) and \(\rho\) is the fast rapidity. The fast speed is
$$ u = \coth{(\rho)}. $$
Since the hyperbolic cotangent is bounded,
$$ \coth{(\rho)} < -1, \qquad \coth{(\rho)} > 1; $$
the fast speed is bounded:
$$ u < -1, \qquad u > 1. $$
However, the square fast speed is only bounded from below:
$$ u^{2} > 1. $$
The fast rapidity in terms of the fast speed is
$$ \exp{(\rho)} = \sqrt{\frac{u + 1}{u - 1}}. $$
Then it follows that
$$ \cosh{(\rho)} = \frac{u}{\sqrt{u^{2} - 1}}, \qquad \sinh{(\rho)} = \frac{1}{\sqrt{u^{2} - 1}}. $$
Slow and Slow
Consider working with two slow momenta (with rapidities \(\xi_{1}\) and \(\xi_{2}\), and speeds \(v_{1}\) and \(v_{2}\)). It follows that
$$ \exp{(\xi_{1} - \xi_{2})} = \sqrt{\frac{(1 + v_{1})(1 - v_{2})}{(1 - v_{1})(1 + v_{2})}}. $$
Thus
$$ \cosh{(\xi_{1} - \xi_{2})} = \frac{1 - v_{1} v_{2}}{\sqrt{(1 - v_{1}^{2})(1 - v_{2}^{2})}}, \qquad \sinh{(\xi_{1} - \xi_{2})} = \frac{v_{1} - v_{2}}{\sqrt{(1 - v_{1}^{2})(1 - v_{2}^{2})}}; $$
and hence
$$ \tanh{(\xi_{1} - \xi_{2})} = \frac{v_{1} - v_{2}}{1 - v_{1} v_{2}}. $$
This relation is related to the velocity addition formula (here it is subtraction).
Fast and Fast
Now consider working with two fast momenta (with rapidities \(\rho_{1}\) and \(\rho_{2}\), and speeds \(u_{1}\) and \(u_{2}\)). It follows that
$$ \exp{(\rho_{1} - \rho_{2})} = \sqrt{\frac{(u_{1} + 1)(u_{2} - 1)}{(u_{1} - 1)(u_{2} + 1)}}. $$
Thus
$$ \cosh{(\rho_{1} - \rho_{2})} = \frac{u_{1} u_{2} - 1}{\sqrt{(u_{1}^{2} - 1)(u_{2}^{2} - 1)}}, \qquad \sinh{(\rho_{1} - \rho_{2})} = \frac{u_{2} - u_{1}}{\sqrt{(u_{1}^{2} - 1)(u_{2}^{2} - 1)}}; $$
and hence
$$ \coth{(\rho_{1} - \rho_{2})} = \frac{u_{1} u_{2} - 1}{u_{2} - u_{1}}. $$
This relation is related to the velocity addition formula (here it is subtraction).
Slow and Fast
Now consider working with one slow momentum (with rapidity \(\xi\) and speed \(v\)) and one fast momentum (with rapidity \(\rho\) and speed \(u\)). It follows that
$$ \exp{(\xi - \rho)} = \sqrt{ \frac{(1 + v)(u - 1)}{(1 - v)(u + 1)} }. $$
Thus
$$ \cosh{(\xi - \rho)} = \frac{u - v}{\sqrt{(1 - v^{2})(u^{2} - 1)}}, \qquad \sinh{(\xi - \rho)} = \frac{uv - 1}{\sqrt{(1 - v^{2})(u^{2} - 1)}}; $$
and hence
$$ \tanh{(\xi - \rho)} = \frac{uv - 1}{u - v}. $$
