# Singularity of a black hole: point or solid sphere? [duplicate]

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?

• What force prevents your highly dense sphere from collapsing? Commented May 14 at 19:31
• I would think that most questions beginning "Why do most people think..." are off topic here. Commented May 14 at 20:28
• What got you to think it? Commented May 14 at 20:46
• Possible duplicates: physics.stackexchange.com/q/24934/2451 and links therein. Commented May 15 at 7:40
• The answers below are all somewhat misleading: they are correct descriptions of the Schwarzschild metric (classical GR, nonrotating BH). It's worth noting that: (a) All astrophysical BHs likely have nonzero angular momentum. (b) It is likely that classical GR simply does not describe physics near the center of an actual BH, because there will be quantum gravity effects in the region of super-planckian curvature/density. Both these things drastically change the answer. Of course it's also important to understand the textbook GR answer as long as its context is clear. Commented May 15 at 7:53

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.

• Andrew, assuming that nothing can overcome the flow of time in this instance (and so if I fell in feet first, blood could no longer reach my head) how is the gravity of this astrophysical object affecting other objects and not just itself? Commented May 15 at 9:42
• Good question, answer is here: physics.stackexchange.com/questions/648767/… Commented May 15 at 10:47
• P.S. the gist of the answer referred to in previous comment is: to find the mass influencing spacetime at any given event, examine the past light cone of that event. (And for Schwarzschild-Droste case, the singularity is not in the past light cone of any event, which you can deduce from a Penrose diagram for example.) But if you allow for black hole evaporation then things get more complicated. Commented May 15 at 10:50
• @Wookie You cannot fall feet forward to the singularity just like you cannot move feet forward to midnight. The singularity is a moment of time, there is no direction in space pointing to it. So your blood flow is unaffected. Secondly, as Andrew explains, the singularity is not gravitationally attractive. It is simply a moment of time in the future. In other words, the singularity is not the cause, but a result of gravity. Commented May 15 at 16:13
• @AndrewSteane Your statement that the singularity is a spacelike surface is misleading, especially without clarifying its geometry. Strictly mathematically speaking it is an infinitely long coordinate line removed from the manifold, a non-existing region of spacetime. $r=0$ is the equation of a line, but not a point of surface. However, since this line is removed, the description of spacetime near it becomes asymptotic and is the 3D hypersurface of the 4D hypercylinder of the spherical type (spherinder) spacelike along and timelike across: math.stackexchange.com/questions/2929400 Commented May 15 at 16:24

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.

• This answer is correct +1 but a couple point may be confusing to readers: “the circumference of the former star is simply C=2πr, which becomes a point with no surface when r→0.” - What becomes a point is the circumference of space, not the singularity. Plus the circumference has no surface, so this statement can be improved. Also, “In a way its still an infinite surface though” - Mathematically it is a coordinate Euclidean line (see my comment under the answer by Andrew for details). Commented May 15 at 16:46
• @safesphere - in that case I will clarify: 1) in all coordinates that use r as one of their axes (like Schwarzschild Droste, Gullstrand Painlevé and Eddington Finkelstein) it shrinks to a point, the Kruskal Szekeres coordinates where it does not don't use r as an axis. 2) The circumference of the star isn't its surface, but the conversion factor from one to the other is simply 4r since the gθθ and gφφ are euclidean. Commented May 15 at 17:33
• it shrinks to a point” - This still is unclear. What do you mean by “it”? The circumference (correct) or the entire space (wrong)? On your second point, once you are at the horizon, it’s not a thin surface anymore, but an infinitely deep space (due to the length contraction). Upon crossing, this depth is represented on the inside by the infinite spacelike coordinate $t$ and there is no evident “surface” anywhere. Formally you still can define a spherical slice as $r=r_o,t=t_o$ but due to the nature of these coordinates flipping you don’t see any spherical surface annywhere inside. Commented May 16 at 3:57

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.

• Surely, a gravity-counteracting force would be emitted from the spinning progenitor that allows the event horizon to form. Commented May 15 at 9:37
• @safesphere I agree that the Schwarzschild singularity is a Euclidean line and not a sphere, as is clear from its Penrose diagram. But I don't think the physics of my argument is wrong: the interior has a trapped surface, so both future-directed null expansions are negative. In that way, the argument I made applies, say, to a spherical collapsing star. That's the essence of the point I was trying to make, not about the topology of the singularity (other comments covered that already), but otherwise, I agree with you! Commented May 15 at 16:55
• The future evolution of the black hole is the sphere of the horizon shrinking down to the singularity. The $t$ coordinate switches from being timelike outside to spacelike inside. This means that everything that fell to the horizon at different times $t$ emerges on the inside at different spatial locations $t$. The inner space becomes a 3D hypersurface of the 4D hypercylinder of the spherical type. This implies that the sphere of the shrinking horizon remains a sphere, as you stated, but the third spatial dimension becomes infinite instead of pointing to the center. Commented May 16 at 2:34

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.

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.

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.