# Diving into a Black Hole

1. Nov 13, 2009

### LukeD

So I'm trying to work out what an observer sees as he dives into a black hole (either free fall, or under acceleration).

According to Wikipedia, as you accelerate toward an event horizon, it should shrink away from you.

Does this happen under free fall or do I need to actively accelerate toward the black hole?

Starting from the Schwartzchild Metric: $$ds^2 = -(1-\frac{2M}{r}) dt^2 + (1-\frac{2M}{r})^{-1} dr^2 + r^2 (d\theta^2 + sin^2 \theta d\phi^2)$$
(which shows that the observer at infinity has a horizon at r = 2M)

and the path of a radially free falling observer:
$$\frac{dt}{dr} = -(\frac{2M}{r})^{-1/2}(1-\frac{2M}{r})^{-1}$$

How do I do a transformation that will let me see where the free falling observer (or some other observer I construct) sees a horizon?

2. Nov 13, 2009

### thehangedman

From what I can remember, nobody ever makes it "across" the event horizon. Time slows and space shrinks as you get closer, reaching zero / infinite levels at the barrier. The result is you actually can never get in. The infinities in the equation really make solving that puzzle mathematically problematic. You end up having to divide everything into two "parts", solutions on the inside vs solutions on the outside. I think it highly likely that a more proper set of equations, namely those that bring in quantum effects, will resolve the issues, but I could not claim to know that solution.

3. Nov 13, 2009

### JesseM

No, an observer falling in reaches the horizon in finite proper time (time as measured by his own clock). It's true that in Schwarzschild coordinates it takes an infinite amount of coordinate time to reach the horizon, but that's just a weakness of that particular coordinate system, you can pick other coordinate systems like Kruskal-Szekeres coordinates (see my post #4 here for a quick description) where there are no coordinate infinities at the horizon.

4. Nov 13, 2009

### thehangedman

We are both right. Observers watching the person "fall in" would see that he never makes it, which was what I was saying. More to the posters question though, the person falling sees things as if he is just falling in (proper time). Finite time to one = infinite time to another. The problem isn't that, so much as what "infinite time" really means physically.

Take this now and apply it to the model used by Hawking to describe radiation from the black hole. The particle falling into the black hole, in it's own proper time, would see a finite time to enter the hole (thus reducing the BH energy), while it's anti-particle pair, the one that escapes, leaves in a finite time, on a scale that those of us sitting outside the BH would consider finite. But we, sitting outside, would never observe the first particle entering the BH. Making it impossible in our frame of reference to ever see a BH radiate away it's energy.

In the end, all these questions are problematic for two reasons: 1. Infinities pose problems in several different ways. Pure math may have no issue, but they certainly make experiments difficult (I can't live that long...), and 2. The model is actually incomplete. It's fun to try and guess what happens to particles and the void near a black hole, but nobody really knows because nobody has a valid model for GR + QFT.

5. Nov 13, 2009

### JesseM

But that's just a visual phenomenon. It is similarly true that an observer undergoing constant proper acceleration in flat SR spacetime will have a Rindler horizon and will see any non-accelerating object take an infinite time to cross this horizon, but no physicist would claim there is any question about the fact that the non-accelerating observer does cross it in finite proper time (and finite coordinate time in any inertial coordinate system).
Is this your own original argument or have you seen a physicist make it? If the former, please note the IMPORTANT! Read before posting thread. I believe it's perfectly possible in classical GR for observers to see the size of the horizon expand even though they never see infalling objects actually reach it, and the page here seems to say that the "virtual particle pair at the horizon" explanation for Hawking radiation is just a heuristic picture, that it doesn't really correspond in any exact way to the actual quantum field theoretic calculations that predict Hawking radiation.
Well, in Rindler coordinates there would be coordinate infinities on the Rindler horizon seen by an accelerating observer in flat spacetime too, do you think there is anything "problematic" about the claim that a non-accelerating observer can easily cross this horizon (which is really nothing more than the edge of a future light cone which the accelerating observer never crosses)? If you allow arbitrary non-inertial coordinate systems as in GR, you can come up with coordinate systems which have infinities anywhere you like, including one that says it would take an infinite amount of coordinate time for you to reach the nearest door in the room you're currently in...despite that I don't think there's any question that you can leave the room!
I believe it's only at the Planck scale (and you'd only reach that kind of energy within a microscopic distance of the location that GR predicts a singularity) that there are significant problems integrating GR with quantum theory, at lower energies I think most physicists would bet that our existing theory of quantum field theory in curved spacetime can be trusted in its predictions.

Last edited: Nov 13, 2009
6. Nov 14, 2009

### yuiop

The trouble with using Rindler coordinates to prove that the event horizon is just a coordinate singularity and not a real singularity is that Rindler coordinates fail to predict what happens at the real singularity. An observer falling through the Rindler Horizon continues to fall forever and never arrives at the equivalent of the central singularity. Therefore Rindler coordinates are not a perfect analogy of a black hole and because they fail to show what a real singularity looks like, its makes the case that Rindler coordinates "prove" that the event horizon is not a real singularity, suspect. In other words, we can equally use Rindler coordinates to "prove" that the centre of a black hole is not a real singularity either and that a free falling observer requires infinite proper time to arrive at the central singularity (if it exists).

7. Nov 14, 2009

### Dmitry67

Could you provide references/examples?

8. Nov 14, 2009

### pervect

Staff Emeritus
I don't think anyone said such a thing? The Rindler coordinates don't actually prove that the event horzion IS a coordinate singularity - the actual proof of that is more technical. However, it does turn out to be the case that both the Rindler coordinates and the Schwarzschild coordinates for a black hole are both examples of singular coordinate systems which exhibit event horizons. Therefore they are useful to check out some statements which are supposed to be true about event horizons.

Note carefully that what happens AFTER we pass through the event horizon is not the same in both cases. The usefulness of the Rindler example is to provide some insight in what happens when we pass the event horizon, not to predict the ultimate fate of the particle.

You may be misunderstanding the point. The point isn't that they never arive at the central singularity - the point is that the "time" at which they cross the event horizon is infinite in the Rindler coordinate system.

Some people, including several in this and other threads, make the mistake of confusing bad behavior of the coordinates with bad behavior of the physics. The point is that even though the time at which the object or signal crosses the Rindler horizon is "infinte" in that badly behaved coordinate system, it doesn't mean that it "doesn't happen or that it never happens. Just as in the case of someone falling into a black hole, the person passing through the Rindler horizon won' notice anything special, and they will cross it in a finite amount of proper time.

A finer point is that one can use the same arguments advanced about black holes not existing because of their event horizons to prove that the Earth doesn't exist if someone gets in a spaceship and accelerates away hard and fast enough.

Now since philosophy isn't science, one might acutally be able to get away with this philosophical position if one is really determined enough, but I think the counterexample shows that this sort of thinking is not particularly useful. "Existence" isn't a very well defined concept, but I think it's commonly accepted for the most part that the Earth doesn't stop existing if someone else travels in such a manner that they can't observe it.

Last edited: Nov 14, 2009
9. Nov 14, 2009

### JesseM

Rindler coordinates deal with flat spacetime, not black holes, so there is no "real singularity" in this case. I never made the claim that the case of an observer falling through the Rindler horizon is analogous to the case of an observer falling through the BH event horizon in every respect (aside from the issue of the central singularity, there is also the issue of tidal forces), the analogy is just intended to show that certain types of coordinate singularities are not real physical singularities. Physicists agree that the event horizon of a black hole is just a coordinate singularity that can be made to disappear by choosing a different coordinate system, whereas they agree that the central singularity of a black hole is a real physical singularity which will have infinite curvature no matter what coordinate system you choose.
Rindler coordinates don't "prove" this, an analysis of the actual Schwarzschild black hole spacetime using a different choice of coordinate system than Schwarzshcild coordinates proves it. Rindler coordinates are just an analogy to show that there are examples of coordinate singularities in one coordinate system (the Rindler horizon in Rindler coordinates) which disappear when you choose a different coordinate system (an inertial coordinate system), and physicists have shown that the coordinate singularity at the event horizon of a black hole in Schwarzschild coordinates is of this type.
I doubt very much you can do that! Again, the discussion of Rindler coordinates was not meant to "prove" anything, the proof that the infinities in Schwarzschild coordinates at the black hole's event horizon are just a coordinate singularity, analogous to the Rindler horizon in Rindler coordinates, would have nothing to do with flat spacetime or Rindler coordinates, you'd have to look it up in a GR textbook. If you don't trust me that physicists have in fact proved this, I can look up a reference.

I brought up the analogy of the Rindler horizon just to disprove the argument that there must be some genuine mystery about the BH event horizon since outside observers will never see anyone cross it and the time to reach it is infinite in Schwarzschild coordinates; the point is that you can say exactly the same thing about the Rindler horizon/Rindler coordinates, and no one thinks there is a genuine mystery about whether you can cross a Rindler horizon, so this shows that the argument about the BH event horizon is insufficient to demonstrate that we should find anything mysterious about it either. In other words, I'm just trying to make the negative point that thehangedman's argument fails, not to prove the positive case that the BH event horizon is just a coordinate singularity like the Rindler horizon in Rindler coordinates (though again, it's not hard to verify that this is the case by consulting a GR textbook).

Last edited: Nov 14, 2009
10. Nov 16, 2009

### yuiop

Upon further reflection, I will retract my statement that Rindler coordinates can equally be used to "prove" that there is no central singularity, because Rindler coordinates do not tell us anything about events below the event horizon. I will also concede that Rindler coordinates provide a nice demonstration that a free falling observer measuring a finite time to reach the event horizon and an accelerating observer that is stationary in Rindler coordinates measuring an infinite time to reach the event horizon is not a paradox in itself. Also, I would point out that Rindler coordinates (actually Rindler coordinates transformed to Minkowski coordinates) tell us that a free falling observer arrives at the event horizon in finite proper time, but does not demonstrate that a free falling observer actually passes through the event horizon, although no one in this thread has made that claim.

Last edited: Nov 16, 2009
11. Nov 16, 2009

### ExecNight

Well obviously the observer reaches the horizon in finite proper time by my clock too. Otherwise i wouldn't be able to observe any blackholes, but simply a phenomenon, a ray of light caused by the objects that are still falling towards the event horizon.

Considering that the event horizon is simply a threshold where light can't find a least resistant path in space-time other than...

Well i should stop there, because we have no idea what is beyond the event horizon. Even talking about gravitational forces or mass or a singularity in the middle is just speculation and philosophy then. The common concept of a singularity comes from the idea that matter can not just dissappear. Well obviously it didn't dissappear, simply formed a blackhole.

While we are into that, in my humble opinion( :) ) a blackhole's behaviour will dramatically change as the observer changes its relativistic speed around the black hole. So way alot more speculations and a dead end right there.

So maybe we should start making some speculations in search for the truth who knows.

12. Nov 16, 2009

### JesseM

In fact this is exactly what you would observe if you could see arbitrarily large frequencies of light (and if light was emitted continuously rather than in a quantum manner)--a "frozen star" with all the objects that have fallen in, as well as the outer edge of the star that formed the original black hole, poised on the edge of the event horizon but never quite reaching it, with their light appearing more and more redshifted as they get closer and closer. See this section of the Usenet Physics FAQ:

13. Nov 16, 2009

### ExecNight

So you mean if i accelerate towards that redshifting light on the blackhole and somewhat be able to cause the exact required effect on the wavelenght.

Wow that would be really fun!

14. Nov 18, 2009

### thehangedman

Can someone just give a definitive answer to the posters question? Would an outside observer, obviously in their own reference frame, ever see (is the time infinite) an object cross the even horizon? It's obvious from the discussion here, that a coordinate system can be created where the time it takes for the falling object to reach the event horizon is finite, but I think that misses the point of the question.

It's quite relevant, since we sitting on Earth fall into that category. Would we, sitting here with our telescopes, ever see something cross the event horizon? Would it take infinite time for us to see that?

15. Nov 18, 2009

### George Jones

Staff Emeritus
If we choose to stay here on Earth, we can never see anything cross the event horizon.

16. Nov 18, 2009

### JesseM

Visually they would never see it reach the horizon, I already explained that in my post #12 to ExecNight (this has nothing to do with what reference frames you use, all frames agree on the answer to local questions like whether a beam of light emitted by a falling object at the horizon will ever reach an observer hovering outside the horizon, regardless of whether the frame says the object crosses at finite coordinate time). See the long quote from the FAQ which I included in that post.

However, I don't think this is what the original poster was actually asking...LukeD was asking whether the event horizon would seem to visually shrink as you were approaching it, if you were freefalling rather than accelerating towards the horizon (it might help if LukeD provided a link to the wikipedia page which says the accelerating observer would see the horizon shrink).

17. Nov 18, 2009

### thehangedman

Thank you once again George for your help.