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?

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    $\begingroup$ Maybe you might consider including the thought experiment tag if you want to bring magic into it. It worked for Einstein:) $\endgroup$ – user81619 Aug 27 '15 at 19:45
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    $\begingroup$ 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. $\endgroup$ – Prof. Legolasov Aug 27 '15 at 19:50
  • $\begingroup$ @Jim ok, just did $\endgroup$ – Prof. Legolasov Aug 27 '15 at 20:31
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    $\begingroup$ 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.) $\endgroup$ – Emil Aug 28 '15 at 7:55
  • $\begingroup$ "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. $\endgroup$ – stuffu Aug 28 '15 at 10:24

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.

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  • $\begingroup$ 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... $\endgroup$ – lvella Aug 28 '15 at 6:08
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    $\begingroup$ @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. $\endgroup$ – Timaeus Aug 28 '15 at 7:24
  • $\begingroup$ @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? $\endgroup$ – Prof. Legolasov Aug 28 '15 at 19:58
  • $\begingroup$ @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. $\endgroup$ – Timaeus Aug 28 '15 at 21:19
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    $\begingroup$ @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? $\endgroup$ – Prof. Legolasov Aug 28 '15 at 22:18

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.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – David Z Aug 31 '15 at 2:42
  • $\begingroup$ 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. $\endgroup$ – safesphere May 14 '18 at 5:48
  • $\begingroup$ @safesphere but numeric simulations suggest that black holes merge in a finite time (as measured from the infinite distance). Is it a contradiction? $\endgroup$ – Prof. Legolasov May 21 '18 at 14:02
  • $\begingroup$ 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? $\endgroup$ – safesphere May 21 '18 at 15:33
  • $\begingroup$ @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. $\endgroup$ – Prof. Legolasov May 21 '18 at 16:04

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