Figure 1: The geometric relationship between inertial reference frames. Hyperbolic projections first in the small (red) triangles, then in the large (blue) triangles yields the standard formulas for the Lorentz transformation, expressed as a hyperbolic rotation.

The relationship between two inertial frames is shown in Figure 1. As usual, the vertical (black) axis represents the worldline of an observer at rest, and the horizontal (black) axis represents a line of simultaneity for such an observer. This inertial observer describes events using coordinates ($x$,$t$). Similarly, the slanted timelike (blue) axis represents the worldline of an observer moving to the right with speed $v=\tanh\beta$, and the slanted spacelike (blue) axis represents a line of simultaneity for such an observer. This observer describes events using coordinates ($x'$,$t'$). Several instances of the hyperbolic angle $\beta$ are indicated (green); (some) right angles are indicated by (red) dots.

The geometric interpretation of the coordinates ($x$, $t$) and ($x'$, $t'$) of an arbitrary point (upper right) as “distances” to the appropriate axis are shown in the figure. Starting with the small (red) right triangle at the bottom, the adjacent leg has length $x'$, so the opposite leg (shown in red) has length $x'\tanh\beta$. Looking now at the large (blue) right triangle on the right, the hypotenuse has length $t'+x'\tanh\beta$, whereas the adjacent leg has length $t$. Thus, \begin{equation} t = (t'+x'\tanh\beta)\cosh\beta = t'\cosh\beta + x'\sinh\beta . \label{lor1} \end{equation} Exactly the same construction applied to the other two triangles yields the same result with $t$ and $x$ switched, namely \begin{equation} x = (x'+t'\tanh\beta)\cosh\beta = x'\cosh\beta + t'\sinh\beta . \label{lor2} \end{equation} Equations (\ref{lor1}) and (\ref{lor2}) are the Lorentz transformation taking the coordinates ($x'$,$t'$) of the moving observer into the coordinates ($x$,$t$) of the observer at rest, expressed as a hyperbolic rotation.

There are several ways to derive the inverse transformation, from the observer at rest to the moving observer. By symmetry, the only thing that changes is the sign of the relative velocity, so that $\beta$ is replaced by $-\beta$. Thus, \begin{align} t'&= t\cosh\beta - x\sinh\beta ,\nonumber\\ x'&= x\cosh\beta - t\sinh\beta , \label{lor3} \end{align} as can also be easily verified algebraically, using the fundamental identity of hyperbolic geometry, namely \begin{equation} \cosh^2\beta - \sinh^2\beta = 1 . \end{equation} It is also straightforward to adapt Figure 1 to the case where $\beta<0$, then swap the primed and unprimed variables. Finally, and perhaps most interestingly, it is possible to use the figure as drawn, with a different choice of triangles, to derive (\ref{lor3}) directly.