Can black holes form in a finite amount of time?

One thing I know about black holes is that an object gets closer to the event horizon, gravitation time dilation make it move more slower from an outside perspective, so that it looks like it take an infinite amount of time for the object to reach the event horizon. It seems like a similar process should slow the formation of the black hole itself: As the star collapses, its gravitational time dilation make itself collapse more slowly. This make me wonder, are what astronomers claim to be black holes really black holes, or are they stars that progressively make themselves more similar to one without actually reaching the stage of having an event horizon?

EDIT: Contemplating one answer, I realize the question is ambiguous. What does finite time mean in general relativity. Here is a less ambiguous question: Is there a connected solution of 3+1 dimensional general relativity with one space-like slice not have a singularity, and another space-like slice having one.

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Spacelike singularities occur in uncharged nonrotating black holes. You must then distinguish between physical singularities and coordinate singularities. For example, the Schwarzchild metric has a coordinate singularity at the Schwarzchild radius that may be eliminated by a change in coordinates, but I doubt this is what you meant. – Gordon Feb 12 '11 at 6:48
I suggest to replace "singularity" by "an intersection with an event horizon" in your rephrased question, since you want to know about black hole formation and not singularity formation. The answer is then "yes", with the Vaidya solution being the simplest example. See e.g. Fig. 4 in arxiv.org/abs/0809.2213 for its Penrose diagram. – Daniel Grumiller Feb 12 '11 at 10:15
duplicated by physics.stackexchange.com/q/21319 – Ben Crowell Nov 16 '14 at 22:50
Also note that for the Schwarzschild spacetime, the singularity is spacelike, so this answer to your question is "yes" for the case of the Kruskal extension of the Schwarzschild spacetime. – Jerry Schirmer Nov 17 '14 at 0:10

As a consequence of the singularity theorems, it is not only possible but (arguably) inevitable for singularities to form in a finite amount of "time" in a physically reasonable spacetime. The word "time" in this context means "proper time along a specific timelike geodesic". For example, if there is a trapped surface* in spacetime, then a singularity will appear within a finite amount of proper time (along a timelike geodesic) in the future of that surface; so, an observer sitting in a collapsing star will reach the singularity in finite time. Thus, the collapse of matter is one possible way to create a singularity "out of nothing". If your spacetime is globally hyperbolic and you foliate it by Cauchy surfaces you can say in a much more "universal" way that the singularity didn't exist at time [;t_0;] and came to exist at time [;t_1;].

I should point out that the singularities are a generic feature of physically reasonable spacetimes; take a look at the Hawking-Penrose theorem - it applies in very general situations.

Also, as the original question was about black holes and not singularities, I should advise you to make a clear distinction between the two concepts. Trapped surfaces form due to the condensation of matter (this is the famous Schoen-Yau theorem), and under a certain extra hypothesis, these surfaces will be hidden inside black holes. This extra hypothesis is the well-known (weak) Cosmic Censorship Conjecture (CCC). If it does not hold, gravitational collapse can create naked singularities, that is, singularities not "causally hidden" by the event horizon of a black hole. Much of what is known in general about black holes depend crucially on the CCC.

*A trapped surface is a two-dimensional spacelike compact surface such that the null geodesics departing from it are accelerating towards each other - mathematically, we say that the expansion of the congruence of future-directed null geodesics orthogonal to the surface is negative.

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I think the question actually was about the possibility of formation of trapped surfaces in finite time. You say "if there are trapped surfaces then there is singularity" but can the trapped surfaces formate in finite time themselves? – Anixx Apr 10 '11 at 10:26
@Anixx Yes. See the Schoen-Yau paper on formation of apparent horizons (the relevant result is theorem 2). – Rodrigo Barbosa Jun 24 '11 at 18:20
I don't think Schoen and Yau actually answers @Anixx 's question. It states that when sufficient mass is already concentrated, then there must exist a trapped surface. For actual dynamical formation of apparent horizons you need something like Pin Yu's recent paper (or, going historically, you have the pioneering work of Oppenheimer-Snyder on dust, Christodoulou on scalar field in spherical symmetry in the 90s, Christodoulou on vacuum gravitational collapse in 2009 etc.) – Willie Wong Jun 25 '11 at 12:41
@Willie Wong thank for the remark. Indeed the main question here is whether sufficient mass can be concentrated in finite time to form a trapped surface since the closer mass particles approach each other, the slower they move due to gravitational time slowing effect. – Anixx Jun 26 '11 at 16:40

The conceptual key here is that time dilation is not something that happens to the infalling matter. Gravitational time dilation, like special-relativistic time dilation, is not a physical process but a difference between observers. When we say that there is infinite time dilation at the event horizon we don't mean that something dramatic happens there. Instead we mean that something dramatic appears to happen according to an observer infinitely far away. An observer in a spacesuit who falls through the event horizon doesn't experience anything special there, sees her own wristwatch continue to run normally, and does not take infinite time on her own clock to get to the horizon and pass on through. Once she passes through the horizon, she only takes a finite amount of clock time to reach the singularity and be annihilated. (In fact, this ending of observers' world-lines after a finite amount of their own clock time, called geodesic incompleteness, is a common way of defining the concept of a singularity.)

When we say that a distant observer never sees matter hit the event horizon, the word "sees" implies receiving an optical signal. It's then obvious as a matter of definition that the observer never "sees" this happen, because the definition of a horizon is that it's the boundary of a region from which we can never see a signal.

People who are bothered by these issues often acknowledge the external unobservability of matter passing through the horizon, and then want to pass from this to questions like, "Does that mean the black hole never really forms?" This presupposes that a distant observer has a uniquely defined notion of simultaneity that applies to a region of space stretching from their own position to the interior of the black hole, so that they can say what's going on inside the black hole "now." But the notion of simultaneity in GR is even more limited than its counterpart in SR. Not only is simultaneity in GR observer-dependent, as in SR, but it is also local rather than global.

Is there a connected solution of 3+1 dimensional general relativity with one space-like slice not have a singularity, and another space-like slice having one.

This is a sophisticated formulation, but I don't think it succeeds in getting around the fundamental limitations of GR's notion of "now." Figure 1 is a Penrose diagram for a spacetime that contains a black hole formed by gravitational collapse of a cloud of dust.[Seahra 2006]

On this type of diagram, light cones look just like they would on a normal spacetime diagram of Minkowski space, but distance scales are highly distorted. The upright line on the left represents an axis of spherical symmetry, so that the 1+1-dimensional diagram represents 3+1 dimensions. The quadrilateral on the bottom right represents the entire spacetime outside the horizon, with the distortion fitting this entire infinite region into that finite area on the page. Despite the distortion, the diagram shows lightlike surfaces as 45-degree diagonals, so that's what the event horizon looks like. The triangle is the spacetime inside the event horizon. The dashed line is the singularity, which is spacelike. The green shape is the collapsing cloud of dust, and the only reason it looks smaller at early times is the distortion of the scales; it's really collapsing the whole time, not expanding and then recontracting.

In figure 2, E is an event on the world-line of an observer. The red spacelike slice is one possible "now" for this observer. According to this slice, no dust particle has ever fallen in and reached the singularity; every such particle has a world-line that intersects the red slice, and therefore it's still on its way in.

The blue spacelike slice is another possible "now" for the same observer at the same time. According to this definition of "now," none of the dust particles exists anymore. (None of them intersect the blue slice.) Therefore they have all already hit the singularity.

If this was SR, then we could decide whether red or blue was the correct notion of simultaneity for the observer, based on the observer's state of motion. But in GR, this only works locally (which is why I made the red and blue slices coincide near E). There is no well-defined way of deciding whether red or blue is the correct way of extending this notion of simultaneity globally.

So the literal answer to the quoted part of the question is yes, but I think it should be clear that this doesn't establish whether infalling matter has "already" hit the singularity at some "now" for a distant observer.

Although it may seem strange that we can't say whether the singularity has "already" formed according to a distant observer, this is really just an inevitable result of the fact that the singularity is spacelike. The same thing happens in the case of a Schwarzschild spacetime, which we think of as a description of an eternal black hole, i.e., one that has always existed and always will. On the similar Penrose diagram for an eternal black hole, we can still draw a spacelike surface like the red one, representing a definition of "now" such that the singularity doesn't exist yet.

Figure 3 shows the situation if we take into account black hole evaporation. For the observer at event E$_1$, we still have spacelike surfaces like the blue one according to which the matter has "already" hit the singularity, and others like the red according to which it hasn't. However, suppose the observer lives long enough to be at event E$_2$. There is no spacelike surface through E$_2$ that intersects the cloud of infalling dust. Therefore the observer can infer at this time that all the infalling matter has hit the singularity. This makes sense, of course, because the observer has seen the Hawking radiation begin and eventually cease, meaning that the black hole no longer exists and its history is over.

Seahra, "An introduction to black holes," http://www.math.unb.ca/~seahra/resources/notes/black_holes.pdf

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What a great answer! – Emil Mar 22 '15 at 11:17
One question (as I'm still struggling a bit with Penrose diagrams): In figure 2, would there also always be a sensible "now slice" for which nothing has crossed the event horizon yet? – Emil Mar 22 '15 at 11:57
According to the last picture E2 sees Hawking radiation all at one moment – Anixx Sep 10 '15 at 13:57
@Anixx No. It is a conformal diagram, that corner at the right end of the dotted line is simply the last burst of Hawking radiation. The black hole got smaller ever since the horizon formed (or since its temperature dropped below background) just like that infalling matter was always falling inwards even though the picture makes it look like it is going out. The diagram makes causal structure easy to see, other structure not so easy. So regardless of how big the event horizon gets or shrinks we draw it as a 45 degree line to indicate that the inside stays in, it doesn't mean the size is fixed. – Timaeus Sep 17 '15 at 1:55
@Timaeus light travels along the 45 degree line so according this diagram all Hawking radiation travels along the horizon tll the final burst. – Anixx Sep 17 '15 at 7:01

You are simply looking at it from an observer's viewpoint. Yes, looking from outside, matter tends to asymptotically approach but never reach the event horizon. If you were part of that matter spiraling into a black hole, there would be no problem reaching the horizon, crossing it, and going right down to the singularity. The event horizon is not a physical barrier. You could be free falling, and your time would not be infinitely dilated. So the answer is yes they can form easily in a finite time.

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"If you were part of that matter spiraling into a black hole, there would be no problem reaching the horizon, crossing it, and going right down to the singularity." - why if the time of the black hole existence is finite? The BH will simply evaporate befgore you reach the horizon. Probably you got this quote from a very old book that does not accout for BH evaporation. – Anixx Jun 25 '11 at 11:01
@Anixx: Exactly the opposite is the case. Black hole evaporation is the only thing that ever makes it possible for a distant observer to say yes, the infalling matter has definitely hit the singularity. See figure 3 in my answer and the explanation below. – Ben Crowell Nov 16 '14 at 22:34
Assuming most things in the unverse are not spiralling into a black hole, including us, doesn't that means that in our frame of reference, nothing have ever crossed an event horizon? – lvella Aug 27 '15 at 6:21

To begin with, there is a connected solution of 3+1 GR in which particles fall to the singularity in finite time. In particular, Gullstrand-Painleve coordinates do this. The big difference with Schwarzschild coordinates is that the speed of light depends on direction: light moves into a black hole faster than it moves out. See:
http://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates

For the formation of a black hole in these coordinates, see:

Phys.Rev.D79:101503,2009, J. Ziprick, G. Kunstatter, Spherically Symmetric Black Hole Formation in Painlevé-Gullstrand Coordinates
http://arxiv.org/abs/0812.0993

For the generalization of Gullstrand-Painleve coordinates to the rotating black hole, see the very readable paper that gives an intuitive explanation for what is going on, see:
Am.J.Phys.76:519-532,2008, Andrew J. S. Hamilton, Jason P. Lisle, The river model of black holes http://arxiv.org/abs/gr-qc/0411060

Note: that the above paper is peer reviewed and shows that yes indeed, particles falling past the event horizon travel with speeds greater than the 1 (in GP coordinates). In GR, the speeds of objects depend on the choice of coordinates. Consequently, this exceeding of the speed 1 is not equivalent to exceeding the speed of light. In GP coordinates, a light beam moving towards the singularity inside the event horizon also moves at speed greater than 1. Consequently, there is no violation of special relativity.

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Thanks for giving refrences. – Itai Bar-Natan Feb 12 '11 at 19:05
Another reference I like (g) is my paper on Gullstrand-Painleve coordinates, which shows how to write them as F=ma: Int.J.Mod.Phys.D18:2289-2294,2009, "The Force of Gravity in Schwarzschild and Gullstrand-Painleve Coordinates", arxiv.org/abs/0907.0660 – Carl Brannen Feb 12 '11 at 20:17
Gullstrand-Painleve coordinates use local time of the free-falling object. On the event horizon this time corresponds to the infinite time of the external observer. Thus the infalling object still cannot reach horizon in finite time. Also in Gullstrand-Painleve coordinates the infalling object reaches the local speed of light at the horizon and even higher speed inside which is impossible for any object that has internal structure. For an object that has mass its kinetic energy will also exceed that of its mass which contradicts the conservation of energy. – Anixx Apr 10 '11 at 10:43
So yes, such coordinates are possible. No, no massive object can follow this path. – Anixx Apr 10 '11 at 10:46
@Anixx; Re: "no massive object can follow this path." Rather than arguing the point, I'm going to note that I'm the only amateur who's ever won an honorable mention at the annual gravitation essay contest, and that the topic of my paper was GP coordinates: arxiv.org/abs/0907.0660 For more information on GP coordinates, see "The River Model of Black Holes" cited in the answer. – Carl Brannen Apr 10 '11 at 21:08

What astronomers claim to be black holes are objects that "progressively make themselves more similar to [a black hole] without actually reaching the stage of having an event horizon", as they reckon. That's assuming that GR is valid, since all such claims depend on GR's equations. Plenty of books on GR note that black holes are perhaps better named "frozen stars" from a distant observer's perspective.

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Who ever is +1-ing this is a joker. – Killercam May 31 '13 at 14:30

I wouldn't know that any of the answers above has shown that, from an outsider's perspective, anything can ever reach the horizon, which was essentially the question of the OP. From an in-falling observer's point of view, there is no problem because kinematic time dilation and gravitational time contraction of the rest of the universe ("looking" in the rear view mirror of an in-falling observer's space ship, let's say) cancel each other out, exactly. But from an outsider's point of view, this is never the case.

Btw, that does not contradict any of the phenomena of black holes like jets and accretion disk-dependent events.

P.S. I suggest to not first think about e.g. light signals when we want to understand the nature of relativistic phenomena. The reality can always be "retro-engineered" from light signals, their arrival time and location and their redshifts, but that's very complicated and confusing to begin with. In order to understand things, we should first assume an instant information transfer.

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It seems to me that Ben's answer does discuss both the infalling & external observer. – Kyle Kanos Nov 16 '15 at 22:53