# B Crashing into an event horizon

1. Jun 8, 2016

### .Scott

I was just reading a review http://physics.aps.org/articles/v9/62 of a recent paper authored by Physics luminaries (http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.116.231301). The paper itself is behind a paywall - and I suspect the math is more than I have the energy for.

The gist of the article is that a QM mechanism is shown for keeping at least some of the EM information on the surface of the black hole - and that a corresponding QM mechanism may be found for gravity.

But I am wondering how much can be said about the mass and gravity information of material falling into a black hole without getting into QM. So I'll start by saying as much as I think I can, and let the more knowledgeable Physicists here correct and continue the story.

My thoughts can be described by considering these scenarios:

In our practice round, Alice will fall through an opening in a Dyson sphere and ultimately collide with the star at the center. Meanwhile, Bob will be on the outside of the sphere keeping close track of gravity. He will notice continuous change as Alice accelerates towards the star. Upon reaching the star, Bob will note that the mass of the star and Alice have combined to form a shared trajectory. Moments later, Bob will notice rockets firing on the Dyson sphere keeping the sphere centered around the new stellar trajectory.

In our second round, Alice will dust herself off and, as is often the case, she will dive into a black hole while Bob looks on. Now, of course, from Alice's point of view she cruises past that horizon on the way to oblivion. Or perhaps you prefer a wall of fire - it doesn't matter because her voyage beyond the horizon is not part of our universe - it is pragmatically a fiction.

On the other hand, the event horizon has a significant affect on what Bob sees. The entire trajectory of the black hole will change, just as the star did. But there will be no evidence of a collision with anything at the center of the black hole because Alice's real-world (Bob's world) trajectory never crosses the event horizon.

So, standing next to Bob and his array of gravity-measuring devices, what do we see? As Alice makes her final approach to the event horizon, her affects are limited by the speed of light to a small region of the event horizon. And it is further limited by the fact that information about her cannot travel through the non-existent interior of the event horizon. The new mass and trajectory of the black hole must be communicated across or above its surface at the speed of light. And since it is not possible to communicate this mass (or anything) across the event horizon surface, it must be communicated well above the surface - at or in the vicinity of the photon sphere.

So the mass is not absorbed by the black hole in an instant, or anything close to an instant and much of the affect will not be seen as emanating directly from the event horizon crossing point. Instead, information about the mass profile of Alice will be briefly sustained in a region surrounding the black hole and well above the event horizon.

Wouldn't Bob see Alice's gravitational trajectory start to change as she approached the photon sphere?

I would think that from that point on her mass would move across that sphere forming a hemisphere, then a full sphere, then a more and more evenly distributed sphere until her mass contribution was indistinguishable from the overall mass of the black hole.

And if that is the path for adding mass to the black hole, I would think that black hole evaporation would follow the reverse path.

2. Jun 8, 2016

### Staff: Mentor

No, he won't. He will see no change in gravity at all. He sees a change when Alice falls past him; but he doesn't see any further change in gravity when Alice is falling inside the sphere towards the star.

No; he will observe the mass of the star-sphere-Alice system to have already increased once Alice passes him on the way in. He will be able to tell that the sphere adjusts its motion slightly; but that will look to him like an internal change in the star-sphere-Alice system that doesn't change its external gravity; its external gravity will already have changed, when Alice fell past him, as above.

Only in the sense that he can't see a "collision" of Alice with an object. But everything else works just the same--in particular, Bob still sees the gravity of the hole-sphere-Alice system change when Alice falls past him, so it doesn't matter that he can't see a collision of Alice with something at the center, because that's not what triggers the change in gravity that he sees.

You need to rethink how you are imagining these scenarios in the light of the above.

Also, none of this is relevant to the paper you linked to.

3. Jun 8, 2016

### Staff: Mentor

Are you assuming a spherically symmetric Alice shell collapsing into the hole? If not then I am not sure about this. A point Alice should emit GWs as she falls in.

4. Jun 8, 2016

### Staff: Mentor

Technically, yes. For an Alice with a mass much smaller than that of the system she's falling into, I don't think it makes too much difference, as long as we assume that she falls right past Bob (i.e., along a radial line that passes through Bob's position). See below.

If she is falling in radially, I'm not sure if she should, at least not in the usual sense. If she is falling in radially, then, if we idealize her as a point mass, there should be a discontinuity in the curvature of spacetime on the future light cone of her worldline. This discontinuity could be viewed as a "gravitational wave front", but it would not be like a normal GW (for one thing, I'm not sure it would have the same quadrupole character as normal GWs). I have not worked through detailed math on this, though.

If Alice falls right past Bob, the propagation of the discontinuity doesn't really matter, because Bob will be "inside" the discontinuity as soon as Alice passes him. If Alice falls in along a different radial line, then yes, Bob will not see the change in gravity until the discontinuity passes him.

5. Jun 8, 2016

### BoGGoG

I haven't read it, but the paper can be found here: https://arxiv.org/abs/1601.00921
arxiv is always a good first place to search for papers =)

6. Jun 8, 2016

### Staff: Mentor

Well reasoned, as always!

7. Jun 8, 2016

### .Scott

Peter, you have never been so wrong.
If Alice was the shell of a sphere collapsing into a smaller shall, then you would be right. While inside the Alice shell, Bob would see no gravitational effect. Then as he passed to the outside there would be an effect that would be constant from that point on.

But that is not what is happening. The gravitation pull from Alice will always be proportional to the inverse square of the distance to her and since that is changing continuously until she hits the star, the gravitational pull from Alice will be changing continuously. Also, the pull of gravity from Alice will always be towards Alice - so as the direction towards her changes, so will that vector.

Crossing past through an opening in the Dyson sphere will not in any way change this.

The connection is not to the main thrust of the article, which as I understand it deals with EM information. But from what I gather, the article also mentions that there is a further pursuit to account for the gravitational information. That's an interesting notion. Since gravity carries information, it must follow the same limits as information carried by light.

I think the general assumption has been that a collision with the event horizon would be of the same nature as crossing through an opening in a sphere. But I don't think that is possible.

8. Jun 8, 2016

### .Scott

It is a thought experiment. If you need to make Alice more massive - go ahead. Let's make her $10^{22}$Kg.
Also, I am not limiting her to passing near Bob. In fact, Bob is allowed to have an array of gravity sensors allowing him to track Alice to whatever precision he would like (within theoretical physical limits). Although, to keep the situation potentially analogous to the black hole situation, there should be no sensors in or near the Dyson sphere.

9. Jun 8, 2016

### Staff: Mentor

Nope. We are not using Newtonian gravity here; we are in a regime where it gives wrong answers. We are using GR, and in GR, gravity is not an inverse square force. You are welcome to use math to analyze this scenario, but it has to be the right math, and Newtonian gravity is not the right math here.

10. Jun 8, 2016

### Staff: Mentor

Yes, and that's exactly what I described in my previous posts. I said that there will be an abrupt change in spacetime curvature on the future light cone of Alice's worldline.

Can you give any references to back up this statement?

11. Jun 8, 2016

### .Scott

OK, I thought you were talking about the Dyson sphere situation. If you are talking about the black holes, then I partially agree with you. Her gravitational pull will eventually look like a shell around the black hole - because of the limits within the photon sphere. Are those the GR effects you are talking about?

12. Jun 8, 2016

### Staff: Mentor

What the article is talking about has nothing to do with what you are imagining. Your scenario can be analyzed using just classical GR.

13. Jun 8, 2016

### Staff: Mentor

No. Even in the Dyson sphere situation, if you are trying to make any kind of useful comparison with a black hole scenario, you have to allow the sphere to be compact enough for gravity to be strong near its surface. In that regime you can't use Newtonian gravity, and you can't view gravity as an inverse square force.

Can you give a reference to back up this statement? As far as I can tell, you are waving your hands and making intuitive guesses without any math behind them.

14. Jun 8, 2016

### .Scott

(this is regarding the assumption that approaching an event horizon is of the same nature as approaching a Dyson sphere)
There are innumerable descriptions of what Bob sees when Alice drops into a black hole. I have never seen anything mention of Alice's mass appearing to do anything other than follow Alice's apparent track onto the event horizon.
Lack of any mention suggests that nothing special was suspected by the authors.

15. Jun 8, 2016

### Staff: Mentor

The "innumerable descriptions" you speak of don't talk about Alice's mass at all; Alice is assumed to be a "test object" with negligible mass. The descriptions only talk about light emitted by Alice outward towards Bob, and how it appears when it reaches Bob. So all of those descriptions are irrelevant to this discussion.

"Nothing special" in the sense of Alice being a test object with negligible mass, not an object whose mass affects anything about the scenario.

16. Jun 8, 2016

### .Scott

What's special about the gravity at the event horizon is not simply its strength, but the escape velocity - and the resulting information limiting effects.
(re: Her gravitational pull will eventually look like a shell around the black hole - because of the limits within the photon sphere.)
That's my central argument. If we put the black hole at (x,y,z)=(0,0,0), Bob at (0,0,1000), and Alice at the event horizon at (1,0,0), then what does Bob see of Alice's trajectory? Information about her trajectory cannot travel directly to Bob. It has to first climb up to the neighborhood of photon sphere before it can even begin traveling towards Bob. So Bob will see this mass as spreading across that region of the photon sphere - or something roughly like that. Of course, in the end, it will all seem to be nothing more than a bigger black hole. But how do you transition from two trajectories to a combined trajectory - as viewed from outside the black hole?

In the case of a Dyson sphere, it's easy. There's a collision with a mass in the middle of the sphere. But that can't happen with a black hole - not from Bob's point of view.

17. Jun 8, 2016

### Staff: Mentor

True but irrelevant. It's still the case that gravity cannot be described in Newtonian terms anywhere close to the horizon. In fact, it can't be described in Newtonian terms well outside the photon sphere.

And that's the statement I'm asking you to back up with a reference, because I don't know where it's coming from and it doesn't look like anything I'm familiar with.

What coordinates are you using here? I'm not aware of any in which this would be a valid description.

I don't see how this follows. Once again, it seems to me like you are waving your hands and making intuitive guesses without any actual math to back them up.

You don't. The black hole is not an "object" in the ordinary sense, and it doesn't have a "trajectory" in the ordinary sense. If you view things from very, very far away, you can treat the hole like any other isolated massive object and assign it a trajectory as an approximation, but that approximation breaks down when you get close enough to the hole (where "close enough" is still well outside the photon sphere, as above). If you remain in the regime where the approximation is valid, you can't even tell that the central object is a black hole; it's basically a point mass with no discernible internal structure.

18. Jun 8, 2016

### .Scott

OK. You seem to know the math. If I surround a black hole with tidal and gravitational sensors at a distance of 1000 Scharzschild radii and drop a moon into it, I will certainly "see" a black hole as a point source or spherical mass with a trajectory. I will also "see" the moon also as a point source or spherical mass. If I continue to run the measurements through the collision process, what would the apparent result be to a creature such as myself with Newtonian habits? Do I see an abrupt collision? If so, is it at the event horizon and is it like a collision? Or does it correspond more closely to a collision within the event horizon? Or is it not so much a collision as a slow blending of the trajectories?

My feeble intuition seems to suggest that "slow blending". But it sounds to me as though you can determine better.

19. Jun 8, 2016

### Staff: Mentor

This is probably far enough for the approximation I mentioned to be valid; I'll assume it is in what follows.

As I said, in this approximation you can't even tell what the two objects are--whether they are both black holes, both ordinary objects, or one is a black hole and the other is an ordinary object. The collision just looks like any other collision of two point masses to you in this approximation. But that's because in this approximation you can't tell anything about either object's internal structure anyway, whether it collides with something else or not.

20. Jun 8, 2016

### .Scott

OK, but you haven't said whether this is because of a theoretical limit on the ability to observe. If I was observing a small planet striking a very large liquid one, I could, in theory, use gravity and tidal measurements to determine something about the collision. I might even be able to tell whether there were large hunks of planetary mass moving around within the large ocean after the collision.

But what happens as my instruments approach as collisions are occurring on a black hole? Does the measuring process completely fall apart before I can discern the transition from 2 trajectories to 1?

21. Jun 9, 2016

### Staff: Mentor

If you are using the Newtonian approximation, you are already imposing limits on your model that are far stricter than our theoretical ability to observe, or even on our current technological ability to observe. We currently have the technological ability to observe many phenomena that cannot be properly modeled by Newtonian gravity. So if you want to model things at the level we can actually observe, you can't use Newtonian gravity.

Basically you are trying to have it both ways. You want a simple model like Newtonian gravity that you can intuitively grasp, but you also want the greatest possible accuracy in observation. Those two things are not compatible; you have to pick one.

Yes, you could; but you might not be able to use Newtonian gravity to do it.

You are assuming that at that level of accuracy of measurement, what you observe must include "the transition from 2 trajectories to 1". My whole point is that this assumption is false. At that level of accuracy of measurement, what you observe does not look like a simple collision of two masses. It looks much more complicated than that. In order to make it look like a simple collision of masses, you have to back way out to a much coarser measurement scale, where you can use the Newtonian approximation and ignore all the internal details of what happens on smaller scales.

Once again, you can't have it both ways. Either you accept that your accurate measurements won't show you a simple collision, or you accept that your coarse measurements that show a simple collision leave out a lot of finer details that could be observed with more accurate measurements. Pick one.