# If the spacetime singularity at a black hole's event horizon is an artificial one then how can it be real?

If the metric belonging to the spacetime of a black hole is expressed as the Schwarzschild metric, it turns out that there is a coordinate singularity at the event horizon. If we use an alternative metric (for example one of these) then the expression of spacetime is regular. How can this be if time stops at the event horizon only in Schwarzschild coordinates?
One thing I don't fully understand either is why the volume of a black hole, when calculated in the Schwarzschild metric, is zero, while it's non-zero when calculated in another coordinate system. How can this be?

As a fairly close analogy for what happens with Schwarzschild coordinates, suppose you replace Cartesian $$(x,y)$$ coordinates for the Euclidean plane with $$(x,z)$$, where $$z=y/x$$.

There is a bijection between these coordinate systems for all points with $$x\ne 0$$, but the points with $$x=0, y\ne 0$$ are not covered by $$(x,z)$$ coordinates at all, while the origin is covered redundantly (all $$(0,z)$$ with $$z\in\mathbb R$$ map to the origin).

If you plot in $$(x,z)$$ coordinates a line (geodesic) that passes above or below the origin, it will appear to go off to infinity at $$x=0$$. It has to, because there's a "barrier" at $$x=0$$ in $$(x,z)$$ coordinates. The barrier represents the origin, the line doesn't go through the origin, so it has to go around the barrier, but the barrier is of infinite length, so the line has to go to infinity.

In Schwarzschild coordinates, ignoring the angular coordinates, the line $$r=r_s$$ maps redundantly to the origin $$X=T=0$$ of Kruskal-Szekeres coordinates. When you take $$r\to r_s$$ while keeping $$t$$ finite, you do not approach the event horizon (which is the ray $$X=T>0$$), but rather the origin.* Geodesics that approach $$r=r_s$$ go off to infinity because they are heading toward the real event horizon, which is at infinity in these coordinates. It's not due to time stopping at the horizon. It's entirely due to the $$t$$ coordinate exhibiting pathological behavior near the horizon.

I've never heard the claim that the interior volume in Schwarzschild coordinates is zero. I found this paper which says it's zero, but their argument involves, no joke, integrating from $$r_s$$ to $$r_s$$. I would rather say that the interior volume is just undefined, because the constant-$$t$$ interior surface isn't spacelike.

Not long ago I read a paper arguing that the interior volume should rather be taken to be the derivative with respect to $$Δt$$ of an interior four-volume of thickness $$Δt$$. This has the nice property of giving the same result (namely $$\frac43\pi r_s^3$$) in all static coordinate systems, including Schwarzschild. Unfortunately, I now can't find the paper.

* $$X=T=0$$ is part of the event horizon of the maximally entended Schwarzschild vacuum solution, but it's not a part of the physically relevant part of it, which is the half "plane" $$X+T>0$$. The origin is more or less the sphere at $$t=-\infty,r=r_s$$ in Gullstrand–Painlevé and similar coordinate systems, which realistically is long before the black hole formed.

• The article you linked is precisely the article from where I got the information about the volumes! Very clear answer! So the horizon is in fact the origin, so to speak? – Methadont Jun 5 at 21:46