Quantities invariant by Lorentz transform

Question

If I rotate a triangle lying on the two-dimensional plane both the lengths of the sides and the angles formed are invariant (ie, they are the same before and after the rotation). In a 2D Lorentz transformation on the (t,x) plane, we know that the spacetime interval between points is invariant. What is in this case the invariant quantity corresponding to angles in the Euclidean case?

Answer 1

The Special relativity analogue of the angles in a triangle is the “rapidity” $\theta$ between timelike segments. The relative velocity $v=c\tanh\theta$ and the time dilation factor is $\gamma=\cosh\theta$ and the Doppler factor is $k=\exp\theta$.

ADDED IN EDIT:
Analogous to an "angle" as the intercepted arc length of the "unit circle", the "rapidity" can be interpreted as the intercepted arc length [using the Minkowski metric] of the "unit hyperbola".

ADDED IN RESPONSE TO THE COMMENT:
In the Clock Effect, we have a triangle with timelike legs.
We can define the "relative rapidity between future timelike legs" as the Minkowski-arclength of the intercepted arc on the future unit-hyperbola.

In this diagram, we have rapidities $\theta$, $\phi$, and $\eta$.
Given the rapidity $\theta$,
we have relative-velocity $v_{OP,OP'}=\tanh\theta$
and [relative] time-dilation factor $\gamma_{OP,OP'}=\cosh\theta$
... and similarly for the other rapidities: $\phi$ and $\eta$.

[Note: There is no hyperbolic arc intercepted by $P'O$ and $P'Z$. So, it's not clear how to define the "rapidity at P' inside the triangle".]

With the chosen values,
$v_{OP,OP'}=\tanh\theta=6/10$ (so $\gamma_{OP,OP'}=\cosh\theta=10/8)$
and $v_{PZ,P'Z}=\tanh\phi=-6/10$ (so $\gamma_{PZ,P'Z}=\cosh\phi=10/8$).
This is in analogy to the Euclidean case.

(I've drawn the spacetime diagram using rotated graph paper so that the tickmarks along the various segments can be counted.)