# Rotational systems in General Relativity

Assuming one rotates with a constant acceleration. Does he see curved space? what about constant velocity? I'd like to get some mathematical or intuitive explanation.

$$ds^2~=~\Big(1~-~\frac{r^2\Omega^2}{c^2}\Big)dt^2~-~r^2d\theta^2~-~2r^2\frac{\Omega}{c} d\theta dt~-~dr^2$$
This has an event horizon at $r^2\Omega^2/c^2 = 1$, but note that this is a coordinate horizon not a true horizon.
• How do you define a "true horizon" ? One that cannot be eliminated by any coordinates transformation ? If I remember, the Schwarzschild black hole metric in the $u \, v$ Kruskal-Szekeres coordinates doesn't show the horizon. – Cham Feb 15 '18 at 16:11