Singularity of a black hole: point or solid sphere?

Question

A black hole is defined by its event horizon. The event horizon has a Schwarzschild radius of, $$r_s=\dfrac{2GM}{c^2}$$ Technically, this means that any body of mass, $M$, with a radius smaller than its Schwarzschild radius, $r_s$, can be a black hole. In other words, a collapsed star which forms some unknown spherical body so dense that it warps space time so much that light cannot escape its gravitational field at some radius would form a black hole.

So, my question is:

Why do most people think of a black hole as having an infinitely dense point (the singularity) when there is a clear possibility that the black hole could instead be caused by a highly dense sphere which has an event horizon?

Answer 1

It's all about time. The singularity (of a Schwarzschild black hole) is not a point in space, it is a spacelike surface in the future. Within the event horizon, spacetime has been warped in such a way that the future direction is now towards the singularity. Arriving there is as certain as arriving at next week. This is why you can't imagine a rigid body sitting inside the event horizon and avoiding being further squashed. If that rigid body lasts in time, then it too arrives at the singularity (having been first spaghettified) because all timelike lines go there.

The above, as I said, is for the simplest kind of black hole, named after Schwarzschild and Droste. That is a black hole with no angular momentum, which is very rare in practice. Any astronomical black hole is likely to have considerable amounts of angular momentum. In this case it is called a Kerr black hole and now the singularity has a different character, in the vacuum solution at least. It is like a ring in space, and one can imagine in principle a spaceship, or perhaps a solid structure, which manages to avoid arriving at the singularity. However, any matter in the region interior to the horizon will disturb spacetime away from the strict Kerr form so it is quite a difficult problem to say exactly what will happen.

Answer 2

Hritik RC asked: "Why do most people think of a black hole as having an infinitely dense point (the singularity) when there is a clear possibility that the black hole could instead be caused by a highly dense sphere which has an event horizon?"

Since the question is in the category GR I'll focus on the relativistic and ignore the quantum perspective, but we don't have a theory of quantum gravity yet anyways, so it is what it is:

The collapsing star is a sphere converging to its horizon radius in the frame of an outside observer due to the gravitational time dilation that makes the collapse slow down asymptotically in his frame of reference.

In the local frame on the other hand not even an infinite counterforce could counteract the gravitational pull, mathematically you would need an imaginary force $\rm F \approx G M m/r^2/\sqrt{1-r_s/r}$ to stop the collapse, which doesn't exist, so the collapse must proceed.

When therefore finally the radial coordinate $\rm r \to 0$, the cirumference of the collapsing matter also approaches $0$ since the $g_{\theta \theta}=\rm r^2$, so the cirumference of the former star is simply $\rm C=2\pi r$, which becomes a point with no surface when $\rm r \to 0$.

In a way its still an infinite surface though, but the base vectors of the surface are no longer in the $\rm \{\theta, \ \phi \}$ but in the $\rm \{t, \ r \}$ direction (while the $g_{\theta \theta}$ and $g_{\phi \phi}$ shrink to $0$, the $g_{\rm tt}$ and $g_{\rm rr}$ blow up to $\pm \infty$) since you hit the singularity with infinite relative velocity $\rm v \approx c \sqrt{r_s/r}$ which shifts the hypersurface of simultaneity by $90°$ on the spacetime diagram and is similar to zero velocity with flipped spacetime axes, see here.

Answer 3

If the hypothetical compact object you're imagining isn't a point, but instead a sphere of radius $R < R_S$, then there must be some force counteracting gravity exactly when $r=R$, where $r$ is the circumference of of a circle wrapping the equator of the sphere divided by $2\pi$ (we have to use this definition of length because it's the one that is easiest to work with when GR makes the geometry very curved).

But forces don't act instantly, and it takes time for one part of the sphere to act on another part. Because $R < R_S$, any force coming from the interior of the sphere would need to (locally) travel faster than light to affect another part of the sphere at a larger radius. Assuming the equivalence principle, which says that special relativity holds locally within an arbitrary spacetime, that's not possible. That's kind of the definition of $R_S$: even light must move "inwards" within this radius. So there's no classical force that could keep the sphere from collapsing, even in principle, and you have a total gravitational collapse.

Answer 4

You are mixing up the singularity with the black hole horizon. For simplicity, let's focus only on the nonrotating uncharged Black Hole. The standard coordinates (Scharzschild) break down at the event horizon. But this is just an artefact of the coordinates, There are different coordinate systems that can describe the black hole inside the event horizon.

For example you can look at the Kruskal–Szekeres coordinates:

https://en.wikipedia.org/wiki/Kruskal%E2%80%93Szekeres_coordinates

You can also look for Penrose diagrams.

Inside the black hole, there is a limit to how far the coordinate system can be extended. The reason is, that geometrical quantities diverge in a coordinate independent way. This limit, which can also be described as a point (for the simplest black holes), is called the singularity.

Interestingly though, the singularity does not lie in the space like centre of the black hole but in the time like future.

Answer 5

Why do most people think of a black hole as having an infinitely dense point (the singularity) when there is a clear possibility that the black hole could instead be caused by a highly dense sphere which has an event horizon?

Both option are misconceptions.

There is no infinitely dense point, because the singularity (which would be this point) is not a point in spacetime. The mass is nowhere. It is not localized. All of the mass is due to the presence of curvature on the spacetime. It is weird, but that's relativity for you.

Furthermore, the singularity is not a place, it is an instant. $r=0$ means a time instant, not a position in space. This is because $r$ measures time within the event horizon.

Finally, there are very strong results in general relativity that tell things will eventually collapse to a singularity. In Schwarzschild spacetime, it is a common exercise to prove that anything that enters the black hole will hit the singularity in finite time. Furthermore, in much more generality, there are the singularity theorems, which essentially say that if a black hole forms, so does a singularity. Roger Penrose was awarded the Nobel Prize in Physics for this discovery.

Answer 6

Why do most people think of a black hole as having an infinitely dense point (the singularity)

Einstein field equations describes gravitation of every body in the universe. Canonical solution to these general relativity field equations is Schwarzschild metric :

$$ {ds}^{2}=\left(1-{\frac {r_{\mathrm {s} }}{r}}\right)c^{2}\,dt^{2}-\left(1-{\frac {r_{\mathrm {s} }}{r}}\right)^{-1}\,dr^{2}-r^{2}{d\Omega }^{2}, \tag 1$$

This metric has one singularity which does not go away on coordinate transformation, namely when $r=0,$ limit of (1) equation becomes :

$$ \lim_{r\to 0} \left[ \left(1-{\frac {r_{\mathrm {s} }}{r}}\right)c^{2}\,dt^{2}-\left(1-{\frac {r_{\mathrm {s} }}{r}}\right)^{-1}\,dr^{2}-r^{2}{d\Omega }^{2} \right] = - \infty, \tag 2$$

So, roughly speaking space-time element at this point becomes infinite. Any two events which is seemingly very close to this singularity are separated in time by "Googolplex" time interval (joke). At exact singularity point as (2) limit shows space-time breaks at all and it's not possible to say "when" or "where" about event (if any) happening at singularity in the center.

That's why.

Caveat: Your question makes sense in such view, that what exactly happens at the center is believed to be known when we invent fully complete quantum gravity theory. In the past quantum mechanics has eliminated some black body radiation singularities ("ultraviolet catastrophe"). So it's natural to expect that quantum gravity could eliminate gravitational singularities once again. If this will happen, maybe really at the center is not a "naked Einstein singularity", but some "quantum object" instead.

Last most intriguing scenario is that neither general relativity, nor quantum mechanics models are applicable to the singularity object. Maybe for resolution of this conundrum we need not mix relativity with quantum world laws, but to invent totally new Physics, whatever it could be. Sounds strange, but given that we are trying to merge GR and QM for almost $100~\text{years}$ with no success, this may indicate that we are going not in that direction.