# Free fall into BH and horizons

1. Nov 10, 2014

### tzimie

Apparent horizon recedes in front of the free falling observer, this is quite obvious.

My question is, what is AROUND the observer. Spagettification stretches the observer vertically, but compresses horizontally. Do horizontal tidal forces form (apparent) white horizons around the observer?

2. Nov 10, 2014

### Staff: Mentor

If you mean that light from the hole's horizon continues to appear to come from in front of the observer, even after he has fallen inside the horizon, yes, this is true (because light emitted at various angles from the horizon will curve around and come at the observer from the front), but note that it assumes that the horizon itself is a visible object. Actually, it isn't. So this "apparent horizon" (which, btw, can have other meanings, so it's important to be clear about what you're using it to refer to) is really an idealization, not something that would actually be seen by an infalling observer.

I'm not sure I understand exactly what you mean by this, but I think the answer is no.

3. Nov 11, 2014

### tzimie

Ok, let me put it this way.
Imagine a spherical shell of dust particles, surrounding the black hole. All these particles don't have any orbiting momentum and they are at the same distance. If the shell initially is far enough from BH, "observers" sitting on every dust particle see each other. Well, almost all (some can be behind the BH). However, let's place the shell the way so even particles on opposite sides of BH are visible (because of GR lensing effect)

Now they begin their free fall into BH. Do these observers lose sight of each other at some moment of time? Without math I doubt 2 falling particles from opposite sites would be visible to each other until the end. Do you agree?

4. Nov 11, 2014

### Grinkle

I don't understand - once an observer passes through the event horizon, photons previously inaccessible to the observer can now reach the observer. Won't the observer now see things that were not visible before crossing the horizon? I get the sense that this is not just semantics of a horizon per se being visible vs things behind a conceptual horizon suddenly becoming visible. I think my mental picture of 'suddenly one can see everything inside the black hole' is probably not consistent with what your are saying.

5. Nov 11, 2014

### Staff: Mentor

Yes.

Yes.

Yes. But the horizon itself won't be one of them, because the horizon itself is not a visible object. It's just a boundary between two regions of spacetime.

6. Nov 11, 2014

### Grinkle

Now I think we are into a semantic discussion.

I argue that it is valid to claim having observed an event horizon by noting the transition point where what is behind the horizon suddenly starts to interact with you. Of course no one outside the horizon can receive your claims.

To give you an idea of how I am thinking by analogy I submit that one observes the human "blind spot" (link below if you are not familiar with the reference) in a similar indirect fashion.

http://en.wikipedia.org/wiki/Blind_spot_(vision)

If your point is that the event horizon has no physical manifestation that one can directly interact with, I am still not convinced, thinking that the event horizon can be called that sphere in space surrounding a singularity that emits Hawking radiation. Am I wrong - maybe that is different from the event horizon. It wouldn't be the first time.

7. Nov 11, 2014

### Staff: Mentor

But you have no way of knowing when that point is. The event horizon is a global phenomenon, not a local one; there is nothing local to tell you when you've passed it, and none of the information you are receiving has markers on it to tell you that it came from an object inside the horizon.

You appear to be making an unwarranted analogy with an ordinary horizon on Earth, where indeed you can tell when something is above or below it, because you can see the horizon itself. This is, as I just said, an unwarranted analogy; a black hole's event horizon does not work the same way. Perhaps that makes the term "horizon" unfortunate, but we're stuck with that terminology.

8. Nov 11, 2014

### Staff: Mentor

This analogy does not seem at all useful to me.

I think the best answer at this point is that we're not sure. The reason is that there is another type of horizon, called an "apparent horizon" (no connection with the way that term was used earlier in the thread, which is why I mentioned before that the term had other meanings), which is a 2-sphere at which, roughly speaking, radially outgoing light no longer moves outward (it stays at the same location, at that same 2-sphere). For an idealized black hole that never gains or loses any mass, the apparent horizon and the event horizon coincide; for a hole that gains or loses mass very slowly, they almost coincide.

This is important for two reasons. First, it is possible to detect an apparent horizon using measurements that, while not strictly "local" (you have to look at a family of radially outgoing light rays that cover at least a significant portion of a 2-sphere), does not require knowing the entire future of the spacetime, as locating the event horizon does. Second, if you look at the details of how the prediction of Hawking radiation is derived, many physicists think that derivation really predicts Hawking radiation from an apparent horizon, not an event horizon.

This still doesn't necessarily make an apparent horizon "visible" (as a sphere of Hawking radiation), at least not to an observer falling through it. According to Hawking's original derivation, an observer falling into a black hole would not observe any Hawking radiation at all as he passed the horizon (this would apply to either kind of horizon); only an observer who accelerates in order to "hover" outside the horizon would observe Hawking radiation. There are various speculations about quantum effects (referred to by the general term "firewalls") that would produce radiation at a horizon (most of these, IIRC, predict it for an apparent horizon) that would be observable by observers freely falling through the horizon. However, I don't think any of these speculations have gained wide acceptance, and certainly none of them have led to any testable predictions.

9. Nov 12, 2014

### tzimie

I am talking about an apparent horizon only.
Obviously, Hawking radiation is observer-dependent as it comes from an apparent horizon (an example - Unruh effect where there is no absolute horizon at all). Obviously, no observer can cross his own apparent horizon - otherwise it would be completely destroyed. Based on "no drama" approach, nothing happens to free falling observers when they cross an absolute horizon of black hole. However, as an apparent horizon recedes in front of them as they fall, they should see Hawking radiation - more and more intense as they approach singularity. This is quite obvious.

What is more interesting, do they see white apparent horizons around them, as they get infinitely contracted horizontally? Hawking radiation is extremely intensive per se, but is also extremely red-shifted in case of black hole. Hawking radiation from white horizons should be extremely intensive PLUS blue shifted - and it should fill the space inside the black hole (relative to free falling observer!!!) with radiation.

10. Nov 12, 2014

### Grinkle

Of course - thanks. That makes sense to me.

I need to read more / digest the apparent horizon vs event horizon concepts.

11. Nov 12, 2014

### Staff: Mentor

You're not using this term correctly here; what you are calling an "apparent horizon" here is usually called a "Rindler horizon". See below.

This describes a Rindler horizon, associated with a family of accelerated observers. There are similarities between the Unruh effect, associated with a Rindler horizon, and Hawking radiation, associated with a black hole horizon (apparent or absolute--see my comments in a previous post), but they are not the same thing, and Hawking radiation is not associated with a Rindler horizon, it's associated with a black hole horizon.

Now you're using the term "apparent horizon" in yet another sense (the sense you used it in your OP), to denote the fact that light emitted from a black hole horizon can be curved by the spacetime geometry inside the horizon so that it comes at an infalling observer from the front. This is a purely optical effect that has nothing at all to do with Hawking radiation. This is why I cautioned you early in this thread to be careful about terminology.

12. Nov 12, 2014

### tzimie

So, what Hawking radiation is coming from, from what kind of horizon? Hawking radiation is derived for the static BH, when all horizons are the same. What's about the dynamic case?

As you remember, there was a thread about nested black holes. A galaxy of small black holes, all of a sudden becoming a supermassive black hole? So, an observer inside the galaxy (but outside the smaller black holes) observes hawking radiation from individual black holes. It is crazy to imagine that hawking radiation from small BHs suddenly stops when the galaxy becomes one big black hole and big absolute horizon consumes all the galaxy. For an observer locally nothing should happen

I believed that Hawking radiation is coming from Rindler horizon, which I was mistakenly calling an 'apparent'. For the static black hole there is no difference, for Unruh effect it works well, then an observer inside BH should also see Hawking radiation.

Last edited: Nov 12, 2014
13. Nov 12, 2014

### Staff: Mentor

See post #8; at this point we're not sure. (Personally, I think we'll eventually settle on a model where it comes from an apparent horizon--i.e., from a surface where outgoing light stays at the same place. But we're not at that level of knowledge yet.)

This is a good point, and it's one reason why I personally believe we'll eventually find that Hawking radiation comes from apparent horizons. (Note that, for this to work in the scenario you describe in the above quote, you also need to take into account that apparent horizons are observer-dependent, as I said in a post in the nested black holes thread.)

Not necessarily. If Hawking radiation is associated with an apparent horizon, we still need to determine whether the radiation is only emitted outward (meaning it would only be visible outside the apparent horizon), or whether it is emitted inward as well. A straightforward analysis based on Hawking's original derivation would say that it is only emitted outward. I think this is another area in which we're simply not sure at this point.

14. Nov 12, 2014

### tzimie

Thank you, nice to know that some things I suspected are, at least, not false.

I would be nice to have Hawking radiation inside the BH, because on millimeter scale radiation should be so intense, that the high energy matter density would affect the space time curvature. And free falling observers, observing radiation on the same level as they are and even above them, could deduce (here I am not sure, as mass conservation in GR is tricky) that some of the mass of BH exists in a form of radiation and not in the singularity. In a limit, near singularity an observer would see most of the mass in a form of radiation, so singularity is effectively "flattened". Please don't interpret it as a personal theory, I am just thinking about the wonderful option to eliminate singularity even far away from Planck scale...

15. Nov 12, 2014

### Staff: Mentor

Some of the "firewall" type proposals are along these lines. But note that if this is true, the state of the quantum field can't be anything like what we currently consider to be a "vacuum" state (even taking into account the fact, which I think I mentioned before, that exactly which field state is the "vacuum" state is observer-dependent). In other words, all that energy density has to be there in the quantum field in the first place, for it to appear as radiation.

Only if the radiation were static, i.e., not falling into the singularity along with them. If the radiation is falling into the singularity, the proper deduction for a free-falling observer seeing it above them would be that they are somewhere inside the collapsing object that will form the black hole.

However, if the radiation is static, then there can't actually be a horizon present in the first place, because spacetime inside a horizon can't be static. At least, we know that's always true for an event horizon. I'm not sure if it's been shown to be always true for an apparent horizon, but it seems like it ought to be true, since inside an apparent horizon, even radially outgoing light moves inward, so since nothing can move outward faster than radially outgoing light, it ought to be impossible for anything to stay at the same radius (since that would require not moving inward, which would be moving outward faster than radially outgoing light).

16. Nov 12, 2014

### tzimie

But why can't the inner structure be similar to Kerr black hole - with a second horizon inside around a 'flattened' and 'smeared' singularity, which, in such case, becomes just a time-like cloud?

17. Nov 12, 2014

### Staff: Mentor

The problem with the region inside the inner horizon in Kerr spacetime is, first, that it's not physically reasonable, because of the presence of closed timelike curves, and second, it couldn't form in the first place, because there is no possible model of collapse of a rotating object that leads to the presence of that region in the spacetime. Basically, in any such collapse process, the energy density of the collapsing matter would become infinite at the instant the inner horizon formed. (Actually, I don't know if an analysis has been done of this for ordinary matter, but it has been done for ingoing radiation in Kerr spacetime, and the blueshift of the radiation becomes infinite at the inner horizon, which corresponds to infinite energy density.)

18. Nov 16, 2014

### tzimie

Yes, I know that Kerr solution is not realistic. I was just thinking about the second horizon, but without rotation, it was just an analogy, apparently, a bad one.

As most of the realistic Black Holes are rotating, it means that almost 100 after the discovery of GR, we still have no idea what is inside :(

19. Nov 16, 2014

### Staff: Mentor

It's not quite that bad. Although in one respect it's actually worse than your statement implies. The Schwarzschild solution for the interior of a non-rotating black hole has a problem too: it's unstable against small perturbations. (So is the portion of the Kerr solution between the outer and inner horizons.) Since no real black hole, even a non-rotating one, is perfectly symmetric, there will always be small perturbations in a real hole; and their presence invalidates all of the simple exact solutions we know for black holes (the Kerr-Newman family of solutions).

For a non-rotating black hole, there is a (very complicated) exact solution, called the BKL solution, that describes a non-rotating black hole interior which is stable against small perturbations. In this solution, the closer you get to the singularity, the spacetime curvature fluctuates more and more chaotically. (This solution, with the direction of time reversed, has also been proposed as a description of the universe close to the Big Bang--that's what the Wikipedia link above describes.) I don't know if a rotating version of this type of solution exists.

However, even in the absence of exact solutions, we can still do numerical simulations of the spacetime curvature inside a black hole, whether it's non-rotating or rotating. Such simulations can tell us a lot about the possibilities even if we don't have any simple exact solutions. So we do have ways of investigating what's inside.

20. Nov 18, 2014

### tzimie

Hm, but WHY is it unstable?
BH is perfectly stable from the outside ("ho hair")
Inside everything (including perturbations) have a limited proper time to exist, so if BH is, say, asymmetric, it becomes symmetric very soon.

Or are you talking about "mass inflation" I was reading about in wiki few years ago (and it did not make any sense to me for multiple reasons) - but now this subject (mass inflation) has disappeared from wiki?