Is Schwarzschild spacetime in Eddington-Finklestein coordinates flat at Schwarzschild radius?

Question

Schwarzschild metric in Eddington-Finklestein coordinates reads

$$ds^2=-\left(1-\frac{2M}{r}\right)dv^2+2dvdr+r^2d\Omega^2$$

My professor claimed that at $r=2M$, this metric becomes Minkowski metric. I was suspicious about his claim, so I asked him about it after class and he hastily told me the below coordinate transform

$$t=v-r\qquad x=v+r$$

Using his coordinate transform, I obtained this metric

$$ds^2=-\left(1-\frac{2M}{r}\right)(dx+dt)^2+2(dx^2-dt^2)+r^2d\Omega^2$$

By setting $r=2M$, the first term vanishes and the metric becomes $ds^2=-2dt^2+2dx^2+r^2d\Omega^2$. I am not sure if this is the metric for the flat spacetime. Is the spacetime really flat to an observer hovering around Schwarzschild radius? If so, this observer is in an inertial reference frame?

Answer 1

Choice of coordinates does not change the nature of the spacetime around a black hole.

Spacetime always looks flat locally. How locally such an approximation will work depends on the second derivatives of the metric that determine the spacetime curvature. The curvature is non-zero at the event horizon and depends on the black hole mass - smaller for more massive black holes.

Answer 2

weeab00 wrote: "Is the spacetime really flat to an observer hovering around Schwarzschild radius?"

Locally the spacetime is flat at every r except r=0, but the local patch becomes smaller the nearer you get to the singularity, and you can't hover at the Schwarzschildradius since that requires infinite force.

If your local rulers fall with the escape velocity like the local reference observers in Gullstrand-Painlevé raindrop coordinates the whole covariant spatial metric becomes flat though, see https://arxiv.org/abs/gr-qc/0411060, but that is not the case with Eddington-Finkelstein.

In the latter the ingoing lightrays have a flat 45° angle on the spacetime diagram though, so maybe you or your professor mixed something up.

weeab00 wrote: "If so, this observer is in an inertial reference frame?"

Free falling, i.e. force free observers are always in an inertial frame of reference, but your hovering observer is not since he needs force in order to hover.