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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.

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Rotating the observer doesn't change the curvature of the spacetime, so obviously the spacetime is still flat. However the coordinate system used in the rest frame of the observer is now curved i.e. the spacetime isn't curved but the coordinates are.

The transformation of the metric into a rotating frame is described in Lawrence Crowell's answer to Special Relativistic Time Dilation -- A computer in a very fast centrifuge. The result is:

$$ ds^2~=~\Big(1~-~\frac{r^2\Omega^2}{c^2}\Big)dt^2~-~r^2d\theta^2~-~2r^2\frac{\Omega}{c} d\theta dt~-~dr^2 $$

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.

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  • $\begingroup$ 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. $\endgroup$ – Cham Feb 15 '18 at 16:11

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