# Can something (again) ever fall through the event horizon?

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.