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Since I am more confused by the answers given in this site to the many variants and duplicates of this question, with some arguing that from the point of view of the falling observer, it happens in finite time, and the issue is a matter of GR frame of reference (in Can black holes form in a finite amount of time?) and others saying that everything falling into a black hole will always asymptotically falls towards the event horizon, but never actually crossing it (in How can anything ever fall into a black hole as seen from an outside observer?), I am going to pose this question as a thought experiment, hoping that I will be able to make sense out of the answer, and get to a conclusion myself:

Imagine I am standing on altitude $h$ above a non-rotating black hole of mass $M$. I am not in orbit, but I am not falling because I am in a rocket that perfectly counters the gravity, keeping me stationary. I have with me a magic ball. It is magic because it can fly like Superman, thrusting with any finite amount of force. So, no matter how close it gets to the event horizon, as long as it doesn't crosses it, it can escape flying radially outwards. Now I drop my ball from the rocket, and it free falls radially into the black hole. It can decide at any moment to use its powers to try to climb up back to the rocket, but I don't know when or if that will happen.

So, how much time I must wait to be completely sure that my ball crossed the event horizon and will never return?

Qmechanic
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lvella
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    Maybe you might consider including the thought experiment tag if you want to bring magic into it. It worked for Einstein:) –  Aug 27 '15 at 19:45
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    In case your ball is infinitely lighter than the black hole, the answer is infinity. You can never be sure it is not coming back. But in reality your ball has a finite mass which can not be neglected. Its mass is to be added to the black hole's mass $M$, therefore increasing its size. An outside observer will see that his ball got sucked into the black hole approximately when the ball reaches this (increased) gravitational radius. – Prof. Legolasov Aug 27 '15 at 19:50
  • @Jim ok, just did – Prof. Legolasov Aug 27 '15 at 20:31
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    I would love to see this answered in terms of a nice Penrose diagram by one of the GR pros. (I never quite got the hang of them myself, but I find them hugely illustrative with a bit of expert guidance.) – Emil Aug 28 '15 at 07:55
  • "rocket that free falls radially into the black hole" Does the rocket free fall into the black hole? If not, then you should edit your post. – stuffu Aug 28 '15 at 10:24

3 Answers3

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Since another answer claims that a massive magic device would form in finite time I have to disagree. You have to wait forever, but only because your device is magic.

The simplest problems are the spherically symmetric ones. And if you can get things close to an event horizon and magically bring them away as long as they stay outside then it is possible to not even know if the black hole forms.

It is widely known that it takes finite time for two black holes to merge into a single black hole; this has been proved in the corresponding numerical computations.

This question wasn't about the real world, it was about the real world where there are magic devices that can move on timelike curves whenever they feel like it. Which is a useful thought experiment for understanding the geometry of a black hole.

Step one. Draw a Kruskal-Szekeres diagram for a star of mass M+m and pick an event of Schwarzschild $r=r_0$ and Schwarzschild $t=t_0.$

Step two. Draw a time like curve heading to the event horizon. Consider the region that has Schwarzschild t bigger than $t_0$ and has $r$ bigger than that curve at that Schwarzschild $t.$

This is a region of spacetime that sees a spherical shell of mass $m$ starting at $r=r_0$ and $t=t_0$ and heading down into an event horizon of a mass $M$ black hole.

Step three. Now pick any event in this region of spacetime. Which is any point outside the black hole event horizon provided it is farther out than the thing lower down. So it is ancient, waiting for the new bigger black hole to form. Say it has an $r=r_{old}$ and a $t=t_{old}.$

Step four. Trace its past lightcone. Now pick any $\epsilon>0$ and trace that cone back until it reaches the surface of Schwarzschild $r=(M+m)(2+\epsilon).$ And find that Schwarzschild $t_{young}$ where that event (past lone intersecting the surface Schwarzschild $r=(M+m)(2+\epsilon)$) occurs. As long as the magic spherically symmetric shell of mass $m$ stays at Schwarzschild r smaller than $r=(M+m)(2+\epsilon)$ until after Schwarzschild $t=t_{young}$ then it can engage its magic engines and come back up and say hi to the person a $r=r_{old}.$

And the person won't see it until after the event $r=r_{old},$ $t=t_{old}.$

Which means. No matter how long you wait outside, the magic spherical shell of mass $m$ could still return to you so it most definitely has not crossed the event horizon of the original mass $M$ black hole and not even the larger event horizon for the mass $M+m$ black hole of it plus the original black hole.

We do use the magic ability to come up. If you are willing to leave some of the substance behind it could shoot off a large fraction of itself and use that to have rest of it escape.

But real everyday substances can't get thin enough to fit into that small region just outside the horizon so you can't make a device that does this out of ordinary materials.

But as far as your logic goes, this process would take infinite time and therefore is impossible.

We want to know if you can tell whether the magic device joined the black hole. The answer is no exactly because it takes an infinite amount of Schwarzschild time.

Earlier answer follows ...

For instance imagine a bunch of thin shells of matter. You can have flat space on the inside and then have a little bit of curvature between the two inner most shells. And have it get more and more curved on the outside of each sucessive shell until outside all of them it looks like a star of mass $M.$

Each shell is like two funnels sewn together with a deeper funnel always on the outside and all sewn together where they have the same circumference at the place they are sewn together.

So now how do I know we can never know if anything crosses an event horizon. If they crossed an event horizon then the last bit to cross has a final view, what they see with their eyes or cameras as they cross. And if there is something they see that hasn't crossed yet when they cross that thing can run away and wait as many millions or billions of years as long as you want. And where ever and whenever they are they, the people outside, will still see the collapsing shells from before they crossed the event horizon.

So now imagine a different universe. One where they didn't form a black hole or cross an event horizon. But all the shells got really close, so close that everything up to the point looks the same to the person in the future. Then they turn around and come back.

So we never saw a single thing cross the event horizon. And if there are magic ways to get away as long as you haven't crossed the event horizon then there is no amount of time to wait before you know they cross.

Because no matter how long you wait they still might not cross the horizon or they might cross it and you don't know yet.

With the spherical symmetry it is easy to see that what I say works because there are really nice pictures for the spherical symmetry case where you can see what is and isn't possible. So you can pick a radius and a time and I can draw a point on a graph and trace back to find out how close the magic device has to get before it turns around. As long as things can wait until they are really really close then you can't tell if they have crossed an event horizon.

The other answer is just plain wrong. If you take a collapsing star of mass $M+m$ then you can find where an arbitrarily distant time sees the infalling body. And as long as you waited until that point then the magic device can escape.

Timaeus
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  • Im sorry, but I couldn't understand anything you wrote. What sphere is you talking about in the second paragraph? What shell? Then they suddenly became plural: what shells? Is there a black hole inside the shells? Please, assume the ball I was talking about has the mass of a tennis ball... – lvella Aug 28 '15 at 06:08
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    @lvella You can make it as little or as big as you want. You can't tell if or whether it crosses as long as it can squeeze as close to the horizon as you want and still escape then you don't know if it crossed. – Timaeus Aug 28 '15 at 07:24
  • @Timaeus no offense, but what you wrote can hardly be true. It is widely known that it takes finite time for two black holes to merge into a single black hole; this has been proved in the corresponding numerical computations. But as far as your logic goes, this process would take infinite time and therefore is impossible. P.S. I am keeping an open mind and recognize that my answer might turn out to be incorrect, but right now I am not convinced enough. Maybe you can convince me? – Prof. Legolasov Aug 28 '15 at 19:58
  • @Hindsight Edited. You can't know when a shell has joined because whatever you see could just be the portion of the journey before they came back. If they can get small then they could turn back at any nonzero distance from the horizon and getting closer makes them come back later so if you haven't seen them it just might be that you haven't waited long enough. And that holds no matter how long you wait. So wildly known results are wrong or else you misunderstand them. – Timaeus Aug 28 '15 at 21:19
  • @Timaeus my proposition was that it takes finitely long for two black holes to merge. After that time (which could be calculated) an outside observer can not expect the magical black-hole ball to come back. – Prof. Legolasov Aug 28 '15 at 21:31
  • @Hindsight But you are wrong. Pick any time for an outside observer and if you claim he black hole is formed by then put down some money and place a bet against someone that can read a Kruskal-Szekeres diagram. They will take your time and compute an earlier time and a radius just a little bit bigger than the 2(M+m) event horizon and program the device to descend farther than that and wit longer than that and then come out and collect money from you. You'd lose money making a bet. That how wrong you are. You make me doubt the merger result too if you think they are similar. – Timaeus Aug 28 '15 at 21:38
  • @Timaeus why wouldn't they be similar? Suppose your ball is a black hole, too. – Prof. Legolasov Aug 28 '15 at 21:39
  • @Timaeus isn't your argumentation based on the approximation of the background metric generated by the black hole? The whole point I am trying to make is that the ball also has a gravitational field, and the metric of the two-body problem has no analytical expression. Therefore your analysis has to be incorrect? – Prof. Legolasov Aug 28 '15 at 21:41
  • @Hindsight Are you even reading what I write? If your object is a spherical shell of mass m outside a black hole of mass M then there is an analytic solution. Just sew a mass M solution to a mass M+m solution along the surface that is the location of the spherical shell of mass m that we are free to move along any time like curve by hypothesis. We could make it fall in from infinity and then break into two layers one going in and one going out when it gets to a critical surface area if we want a specific example. I'm not approximating anything and you seem to not read what I write. – Timaeus Aug 28 '15 at 21:45
  • @Hindsight It isnt the mass of the ball that is the problem. It is the size of the ball. You have to be able to get really really thin to hide from observers in the very distant future. If you can get thin enough you can outlive any finite lifetime object on the outside and have it die thinking you might go in. Since you thought it was mass not thickness that is the problem I gave an example with a perfectly thin object with a finite mass and it never forms a black hole. – Timaeus Aug 28 '15 at 21:49
  • @Timaeus Wikipedia is not a very reliable source, but it is what I just dug up: https://en.wikipedia.org/wiki/Two-body_problem_in_general_relativity In this article, it is said that "If both masses are considered to contribute to the gravitational field, as in binary stars, the Kepler problem can be solved only approximately." If you are implying that this is wrong and you could somehow solve this problem by doing something with this $(M + m)$ solution of yours, then please be so kind to give an appropriate reference. – Prof. Legolasov Aug 28 '15 at 21:55
  • @Timaeus I am sorry, you have made an edit to your answer and now it contains something completely different. I haven't had read it when I was writing to you the last time, but now I have. – Prof. Legolasov Aug 28 '15 at 22:01
  • @Timaeus having said that, I still think you are wrong. It is not OK to just draw a space-time diagram for the $(M + m)$ black hole and go from there. Because in the described experiment an $M$ black hole dynamically becomes an $(M + m)$ black hole. It wasn't there from the start. Please stay patient, I have admitted that I might be wrong after all. – Prof. Legolasov Aug 28 '15 at 22:03
  • @Hindsight I can't tell if you read what I write. You said it was the mass that matters. That's wrong. Pure wrong. Its the size of the object that matters. So to show you are wrong it being the mass I make a different object with the same mass but one that is thinner (a spherical shell) and show that you can have the finite mass but zero thickness and then it never crosses. This proves that it was the thickness of your example not the finite mass that makes things different. If you ignore that I have a thin shell then you will be wrong thinking it's $m\neq 0$ rather than size that matters. – Timaeus Aug 28 '15 at 22:04
  • @Timaeus you can't have an object of zero thickness because it will become a black hole of the gravitational radius which is a function of its mass. In this case the answer still would be finite. – Prof. Legolasov Aug 28 '15 at 22:05
  • @Hindsight It's not different in the slightest in content. I merely wrote it to be harder to be misunderstood. – Timaeus Aug 28 '15 at 22:05
  • @Timaeus I am going to ask you a very specific question. We have two objects: a huge astrophysical black hole and a small one (the "ball"). Would it or would it not take finite time (as seen by an outside observer) for them to merge? – Prof. Legolasov Aug 28 '15 at 22:08
  • @Hindsight As for zero thinness, you are not allowed to argue about that because it is part of the OP question that the whole thing can get as close as it wants so it has to be thin. The question isn't about whether the object itself is a black hole it is about whether it can escape the real black hole or whether it merges with it. It doesn't merge. As for it being a black hole there is no such thing as a critical amount of energy per unit area. You can put a finite amount of energy into a perfectly thin shell and if the surface area is large enough it won't form an event horizon. Just draw it – Timaeus Aug 28 '15 at 22:11
  • @Hindsight As for dynamically becoming larger that is the exact scenario I described if they are programmed to fall you get a black hole with all the object inside. If they are programmed to blast off at a critical radius then they can trick you, with a smaller critical radius for larger times of you waiting. But always works for any time. – Timaeus Aug 28 '15 at 22:13
  • @Timaeus yes I am allowed to argue with that since it is GR we are talking about and there is a very concrete, physical prediction of GR: if an object is thinner than its gravitational radius (which is determined by its mass), than it becomes a black hole. And when two black holes merge (when their horizons become one) there is no way an object inside this single horizon can escape it. That was my point the whole time, and your answer is unphysical since you are forgetting that the ball always has a gravitational radius. – Prof. Legolasov Aug 28 '15 at 22:13
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    @Timaeus now I get what you are saying. So basically, you think that despite the fact that they would take a finite time to merge, we could never be sure that during the last second our ball hasn't launched its magical powers. And if it has, it could take much longer to fly away, so we have to keep waiting. True? – Prof. Legolasov Aug 28 '15 at 22:18
  • Let us continue this discussion in chat. – Prof. Legolasov Aug 28 '15 at 22:21
  • @Timaeus but in that case the problem only arises when we treat the ball as being a point. But we know it has a mass. Therefore it has a gravitational radius. Therefore it is not a point. Hence the time is finite after all. Or is my last conclusion wrong? – Prof. Legolasov Aug 28 '15 at 22:51
  • @Hindsight As mentioned in chat, just because you have a finite mass m doesn't mean it has to have a minimum thinness, as long as you distribute it over a large surface area it can avoid being a black hole. So that's exactly what I did. Because I was trying to show you that it was the thinness not the massness that matters. – Timaeus Aug 28 '15 at 23:01
  • @Hindsight Yes you are wrong. As I said in chat and as I said here you are wrong to go straight from "there is a mass m > 0" to "there is a gravitational radius r > 0 and the object is a black hole of surface area $4\pi r^2$ because as you must be perfectly aware you can spread a mass m around a black hole of mass M in a concentric spherical shell or surface area $4\pi(2(M+m)(1+\epsilon))^2$ and the mass m will not be inside an event horizon if $\epsilon >0$ and since this is the example I talk about all the time please don't blame me if I fail to know why you ignore it. – Timaeus Aug 30 '15 at 23:49
  • @Hindsight And if you are asking whether your conclusion holds of it is a "point mass" with its own event horizon then I'll point out like I always have that that violates the idea of getting as close as you want. But if you want to ask that as a separate question then it can be addressed. – Timaeus Aug 30 '15 at 23:52
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    I finally understood this answer! (I think!) If it was not magic, there would be limit on the amount of kinetic energy the ball could produce in order to escape (based on the ball's mass, since $E=mc^2$, so much of its mass could be fuel), so there is a critical distance visible from outside. Beyond that, for a non-magic device, there is no return. But since it is magic and can escape as long as it hasn't touched the event horizon, the observer can never know if it will return, since closing the gap to the event horizon would take an infinite amount of the outside observer's proper time. – lvella Apr 08 '17 at 05:31
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In case your ball is infinitely lighter than the black hole, the answer is infinity. You can never be sure it is not coming back. But in reality your ball has a finite mass which can not be neglected. Its mass is to be added to the black hole's mass M, therefore increasing its size. An outside observer will see that his ball got sucked into the black hole approximately when the ball reaches this (increased) gravitational radius.

(Based on my comment above)

UPDATE (CAUTION): This answer actually might turn out to be wrong, as suggested by two (independent) people in the comments below. I don't understand any of their argumentation so far, but there is a real possibility that I am wrong (and in this case I want to understand why). Please do not consider my answer to be an absolute truth until this issue is resolved.

Guys, please try to back up your opinion with some references. Maybe somebody who understands this subject better than me should comment.

P.S. The "increased" gravitational radius should be understood only as an analogy which gives a reasonable approximation to the time $t$ when a ball can be considered to be sucked into the black hole. An actual GR-based calculation here requires a solution of the two-body problem, which, as far as I know, is not solved in GR and can only be calculated numerically.

To @Timaeus and @Nathaniel: I will try to reformulate my answer a little. If our ball is massive, then it must have a gravitational radius. We can think of it as of a small black hole.

As I mentioned above, we do not know how two black holes would interact when their horizons intersect. But when the distance between two black holes is large enough and the mass of the ball is much less than the mass of our astrophysical black hole, we could approximate this behaviour with a ball's geodesic worldline in the gravitational field of the black hole.

BUT: this only works when the distance between the black hole and the ball is large.

So I have proposed a heuristic method of calculating the maximal time that we need to wait in order to determine that our ball does not come back. I take the $r_i$ which is the (unphysical) increased gravitational radius of the total black hole:

$$ r _i > r. $$

And I see this radius as some kind of a threshold on distance. When our ball reaches $r _i$ it already passes the point of no return, since the horizons begin to merge and this process (I believe) is irreversible.

Now Mr. @Timaeus can draw his beloved spacetime diagram and see that it takes finite time for the ball to reach $r _i$, turn on its drive and thrust its way back to the space station.

This is how I arrive at my initial conclustion: it takes finite time to become sure that the ball is not coming back.

As a special case I would consider a massless probe ball which by definition does not increase the gravitational radius: $r _i = r$. In this case the answer becomes infinite.

Prof. Legolasov
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  • Comments are not for extended discussion; this conversation has been moved to chat. – David Z Aug 31 '15 at 02:42
  • The answer is infinite for any mass. When a massive object approaches a black hole, the EH gradually increases as the object gets closer. The key here is that the radius of the EH increases before the massive object crosses it. Never does the EH abruptly "jump" to suddenly consume the approaching object. So the object never crosses the EH. Furthermore, all other falling objects "frozen" at the EH move to a larger radius as the radius of the EH increases. So nothing ever crosses the EH. This is pretty obvious, because the objects follow their timelike geodesics, which never cross the EH. – safesphere May 14 '18 at 05:48
  • @safesphere but numeric simulations suggest that black holes merge in a finite time (as measured from the infinite distance). Is it a contradiction? – Prof. Legolasov May 21 '18 at 14:02
  • It doesn't seem to be a contradiction. The key here is energy conservation. Everything hovering over both event horizons must keep hovering and there doesn't seem to be a reason why this wouldn't be the case. When two black holes approach, their event horizons expand and eventually meet. All other falling objects frozen in time near either event horizon keep their distance to the event horizon, but not to the center of the black hole and therefore move out with the event horizon as it expands. Timelike geodesics still never cross the event horizons as they merge. What would a contradiction be? – safesphere May 21 '18 at 15:33
  • @safesphere how is two black holes merging different from a massive object falling into the black hole? If that massive object is a point mass, then it is another black hole – and we are looking at a merger. If the mass is negligible, than the test particle approximation is valid – and your reasoning is well motivated in this case. But I'm talking about the first case. – Prof. Legolasov May 21 '18 at 16:04
  • A change in size or shape of an EH, including a merger of two of them, doesn't violate energy conservation. An EH is just a set of lightlike geodesics, not a material object. A massive object crossing an EH does violate energy conservation and for this reason never happens (time stops). For a remote observer there is nothing inside a black hole. When a star collapses, an EH expands from the center by pushing all matter out. All matter remains at the EH with nothing inside. A merger of two black holes is like a merger of two soap bubbles into a bigger one. Same total mass/energy, bigger size. – safesphere May 21 '18 at 20:04
  • @safesphere but a massive object is surrounded by an event horizon! – Prof. Legolasov May 21 '18 at 20:06
  • This view is very popular and almost consensus, but it is incorrect. The massive object is spread thin like a "layer of paint" on the EH. There is nothing inside the EH, not because it's "an empty space" there, but because there is no such place in our universe as inside the EH. An EH is not a spacelike surface that something could "cross". An EH is not a "place", but is lightlike. Crossing it has the same physical meaning as exceeding the speed of light. Impossible for two reasons, energy conservation and the hyperbolic geometry - the "inside" is not a part of our spacetime. – safesphere May 21 '18 at 20:18
  • @safesphere in the canonical hole, there’s singularity mass inside. I believe astrophysical holes are presumed to be canonical? – Prof. Legolasov May 21 '18 at 20:21
  • I think we've beaten this horse to death :) The concept of singularity is misguided. It violates energy conservation among other things. "Canonical" is a label. Don't be confused by labels. Gravity getting out from the EH is nonsense. The total energy of a BH gravitational field is exactly $E=Mc^2$, so all mass/energy of the BH is outside. A rest mass of a falling object decreases during the fall and becomes zero at the EH. Even if it were possible to "cross" a lightlike geidesic, there would be nothing left there to cross with zero mass. There's no "inside", the radius is no longer spacelike. – safesphere May 21 '18 at 20:38
  • @safesphere I see your point, and you are making a lot of sense. Just to clarify – all black holes are collapsed stars? Even the mergers observed by LIGO? Has this been experimentally confirmed, or is it just a possibility? – Prof. Legolasov May 21 '18 at 21:13
  • The idea of a star collapsing into a "singularity" has been around for a long time due to the inertia of thinking and applying the every day common sense logic to uncommon objects. It is not how the collapse happens. An EH is an object of a finite (and growing) size, but no "volume" inside. When a star collapses, the EH forms as a point in the center and grows in size by pushing the star matter out from the center. In the end, all star matter is spread thin around the EH with no singularity "inside" (there is no even such a place as "inside" where the singularity could've been "located"). – safesphere May 21 '18 at 21:25
  • @safesphere again, I totally get your point. I'm interested in experimental confirmation of this interpretation. – Prof. Legolasov May 21 '18 at 21:31
  • Agreed. It would be very interesting indeed. One possibility is the EH Telescope project. They should publish the images of Sagittarius A* (the black hole in the center of of the Milky Way) any month now. They claim they can resolve the details like dimples on a golf ball in LA while looking from NY. So hopefully nice images. – safesphere May 21 '18 at 21:38
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Timaeus already provided you with the correct answer but I would like to add another (higher-level) way to think about this issue:

The event horizon, by its very definition, is the set of events whose (interior) future light cones don't intersect future infinity, let alone your future wordline as an outside observer. Put differently, the information of anything crossing the event horizon can – by definition! – never escape the black hole. But your question

So, how much time I must wait to be completely sure that my ball crossed the event horizon and will never return?

is basically asking "When will I obtain the aforementioned information as an outside observer?" But again, by the very definition of the event horizon, this is impossible, so the answer is: Never. You must wait an infinite amount of time.

balu
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  • What is impossible is to receive the information from inside the black hole, but I am perfectly fine calculating it, so I see no inherent contradiction between my request and the definition of event horizon. – lvella Oct 07 '20 at 09:05
  • Not quite – it is also impossible to receive information from the event horizon, i.e. the boundary of the inside of the black hole. In any case, I didn't mean to imply that there is a contradiction between your request and the definition of the event horizon. I just wanted to present a different way of arriving at Timaeus's answer. – balu Oct 07 '20 at 10:43