# B Falling into a black hole (seen from outside)

1. Feb 23, 2016

### gerald V

I refer to the time coordinate of an outside observer (ideally hoovering over the BH at large constant distance). Can (s)he actually see something falling into the hole within finite time, that means actually disappearing and making the Event horizon larger? If no, all the merging of black holes, the area increasing law, the Information paradox etc. would make no sense, it seems to me.
If one regards the Schwarzschild metric, an infall cannot happen, since the infalling object never reaches the horizon. However, one MUST NOT use the Schwarzschild metric, since this describes an isolated BH. The correct solution of the Einstein field equations for a BH plus something falling in is crucially different, the more as the equations are nonlinear. These correct equations describe the infall as taking place within finite time (in the said time coordinate).

Would any distinguished expert be so friendly to say whether the above is right or wrong. In the latter case, what else is correct? Thank you in advance.

2. Feb 23, 2016

### Staff: Mentor

She will see the object falling towards the hole; the light the object emits as it gets closer and closer to the horizon will take longer and longer to get back out to her, so it will appear to her that the object is falling more and more slowly. But this is an optical illusion; it does not mean the object actually is slowing down.

She will never see any light emitted by the object at or below the event horizon; that light is trapped and will never escape. She will, however, be able to tell indirectly that the horizon got larger, from the change in, for example, the hole's gravitational lensing effect. Those observations involve light that always stays outside the horizon, so it can reach her; but the amount of bending of the light will increase, and she can detect that.

No, this is not correct. You are confusing the Schwarzschild geometry with Schwarzschild coordinates. Unfortunately, people use the word "metric" to refer to both things, which is a recipe for misunderstanding.

The Schwarzschild geometry is a 4-dimensional geometric object that contains an event horizon and a black hole region inside it, and objects moving in this spacetime geometry can certainly fall inside the horizon and enter the black hole region. That is true independently of any coordinates that any observer might choose. The fact that light emitted at or below the horizon can never get out is also a geometric fact about this spacetime, independent of any coordinates; it's just the way the geometry works.

Schwarzschild coordinates are what are often referred to as "the coordinates of the outside observer". But they're just coordinates; they have no physical meaning. So the fact that the "time" coordinate in these coordinates goes to infinity as you approach the horizon doesn't mean anything physically; it's just an artifact of those particular coordinates. One way to see this is to note that there are other coordinates, such as Painleve, Eddington-Finkelstein, or Kruskal coordinates, which do not have that problem; they cover the entire region both outside and inside the horizon, all the way down to $r = 0$. But the really correct way to see it is to compute invariants--things like the time that would elapse on an infalling observer's clock as he falls. Those invariants are finite at and inside the horizon, showing that physically, there is a region of spacetime there and objects can fall into it. Other invariants can be computed that tell you why light emitted at or inside the horizon can't get out.

It also describes one that is exactly spherically symmetric, i.e., that has exactly zero angular momentum. (And exactly zero electric charge.) Obviously no real black hole will exactly match this idealized model. But that's true of all models in physics; there is always an element of approximation involved. For many real black holes, the Schwarzschild metric is a perfectly good approximation. Indeed, it's a good approximation in many other situations as well, such as for modeling the spacetime around the Earth--even though the Earth is rotating and is not precisely spherical.

The Schwarzschild geometry is a correct solution of the Einstein field equations plus something falling in--in the approximation where the mass of the something falling in is small compared to the mass of the hole. We know it's a good approximation in this case because we have numerical solutions for the same case, which look the same as the Schwarzschild solution, to a very good approximation.

Please give a reference for this claim.

3. Feb 23, 2016

### mgkii

Somebody watching from outside the black hole (assume it's not eating anything else, nothing to obscure the observer's view, etc) would see the person or object falling towards the black hole, but getting slower all the time and appearing to stop as they get closer and closer to the event horizon, and also fading in brightness, eventually fading to nothing. On the inside of the event horizon, that person or object has gone right down into... whatever it is that's in there. But the light from that person will not escape past the event horizon

4. Feb 23, 2016

### mgkii

Looks like PeterDonis beat me to the punch. And much more comprehensively as well!! In the words of the bard... "Wot 'e said".

5. Feb 28, 2016

### gerald V

Dear PeterDonis, thanks a lot, this really helps me.

* I hopefully have correctly understood the following:

- It is kind of dumb to speak about "seeing" something in the context of a fall into a BH

- If an object falls into a BH, this occurs within finite proper time of any observer hoovering or orbiting outside the horizon

- Despite of the lack of a visual signal, the observer can verify that the infall actually took place by estimating the spacetime geometry as those of a black hole more massive (and possibly with different angular momentum and charge) as the original one

* What remains a bit unclear to me (although it is not a big point of concern):

I am aware that the Schwarzschild geometry describes a nonrotating hole. But I have thought that for those case the usual Schwarzschild coordinates are physical in the sense that the time coordinate is identic with the proper time of an observer hoovering at those position. Close to the horizon, hoovering means accelerating like hell, what explains the strong discrepancy with the proper time of a freely falling object.

* Having clarified that the infall takes place within finite proper time of an outside observer, I now would like to raise a further question:

Does the outside observer lose information as a consequence of the infall? Don't the gravitational waves produced by the infall - as recently detected by Ligo - carry with them all the information on the anisotropies of the former configuration? Together with the new parameters of the hole, doesn't this allow to completely reconstruct the object fallen in (in principle)? I am aware of Hawking radiation, firewalls etc. - but these appear as quite a different story to me.

6. Feb 28, 2016

### Staff: Mentor

Not dumb, just limited. If you're outside the horizon, you will never see (as in, receive light from) anything that happens at or inside the horizon. But you can still see things that happen outside the horizon, including infalling objects that haven't yet crossed the horizon.
zio
That isn't what I said; I said that the infalling object reaches the horizon in a finite proper time according to its own clock. (Actually, I only implied it; but it's correct.)

Whether or not the object reaches the horizon in a finite time according to an observer outside the horizon, is a question that does not have a unique, well-defined answer; it depends on how you choose your coordinates. There is no absolute meaning to the "time", according to a given observer, at which events that are spatially separated from that observer take place.

In most cases, he doesn't even have to estimate the geometry, because the mass, charge, and angular momentum of the infalling object are so small, compared to those of the hole, that they can be ignored. So the observer can just use the geometry of the original hole, and compute that the object will fall in in a finite proper time by its own clock.

No, it isn't. The Schwarzschild time coordinate only matches the proper time of an observer who is at rest at infinity. And it is only well-defined at all outside the horizon.

We don't know for sure. See below.

The waves observed by LIGO were produced by two black holes merging, not by an ordinary object falling into a black hole. The latter process will not produce any observable GWs.

Classically, you don't even need the new parameters of the hole or any GWs. Classical physics is deterministic; the infalling object's initial conditions already tell you everything about what is going to happen to it in the future, so no new information is added when it falls into the hole.

The whole "information loss" problem is a quantum problem, not a classical one; it arises because we believe that classical GR breaks down close to the singularity inside the black hole, and some sort of quantum gravity effect takes over. Quantum mechanics, as we understand it, does not allow for information loss (this property is called "unitarity"); but we do not have a satisfactory picture of how the quantum information contained in the infalling object is preserved when it reaches the area at the center of the hole that, classically, is where the singularity is. Hawking radiation, firewalls, etc. are all part of our attempts to construct such a satisfactory picture, which, as I've said, are still not complete.

7. Aug 18, 2016

### gerald V

Thank you.

May I come back to the classical collapse and your reference to spacelike separated events. I know the instructive expample where at Alpha Centauri it is decided to send a spacefleet to Earth. In coordinates where I am at rest, this event can yet have taken place, while in coordinates where a person passing by in a car is at rest, this still has not yet taken place. Of course, we do not even know right now that there is anything of such kind going on. But at some point in (our) time we will get the information, at latest when the spacefleet arrives. And then we know that thence such a decision definitely was taken, as a coordinate-independent truth. There may be some observers in the universe unable of even having received a light-fast signal generated at those decision, but this is not what I am referring to. I am referring to us who have the fleet before our very eyes.

- The LIGO event which occured before our very eyes (or ears), does it say that the two black holes definitely fell into each other, and now are one single hole, as a coordinate-independent truth?

- Or is any black hole only an "as if" black hole, for us outside. The book by Sexl-Urbantke says that for an outside observer a star never collapses to a black hole in the strict sense, rather the observable parameters change in such way that one can speak about a black hole in practical terms.

- Is there any principle difference whether two black holes fall into each other or some other object falls into a black hole (I am aware that the gravitational waves generated will be very weak in the latter case, but I am asking for the principle)?

8. Aug 18, 2016

### Staff: Mentor

Yes.

This is misstating it. The correct statement is the outside observer's coordinates, at least the ones being used in this example, do not cover the region at and inside the hole's horizon, so they cannot describe the portion of the collapse that occurs in that region. But that portion of the collapse can still be a coordinate-independent truth. There is no guarantee that any particular set of coordinates can cover all of spacetime; and if it doesn't, then you can't draw conclusions about what does or does not happen, in a coordinate-independent truth sense, based on what the coordinates can or cannot describe.

I'm not sure what you mean by a "principle difference". Both can be coordinate-independent truths, if that's what you mean.