# Falling into BH question

1. Nov 23, 2008

### nicksauce

I was watching a NOVA program, and they claimed that if you were falling into a black hole, when you crossed the event horizon you would be vaporized by the intense light that's built up at the surface that hasn't been able to escape the BH. This seems plausible, I suppose, but I have never heard such a thing before; in fact everything I've read said that you wouldn't notice anything while crossing the horizon. So is there any truth to this claim?

2. Nov 23, 2008

### atyy

3. Nov 24, 2008

### George Jones

Staff Emeritus
(Non-extremal) Rotating black holes have two horizons. The outer horizon, the boundary of the black hole, is an event horizon, and nothing too special happens as an observer crosses this horizon. The inner horizon is a Cauchy horizon, and some calculations indicate that measured energy density diverges at the Cauchy horizon. The Burki Ori paper to which atyy linked is about the Cauchy horizon (of Reissner-Nordstrom black holoes).

4. Nov 24, 2008

### wolram

Please forgive this intrusion, but why is it that we know so much about black holes, i do not want to be antagonistic but what do we realy know about them?

5. Nov 24, 2008

### nicksauce

Ok thanks, that makes sense. I'm a little confused about the two horizons for rotating black holes. Is the outer horizon the same thing as the ergosphere, or is that something different all together?

6. Nov 25, 2008

### George Jones

Staff Emeritus
No, the boundary for the ergosphere is outside the event horizon, except at the poles, where it touches the event horizon. The event horizon is a boundary of no return, but, using rockets, observers can cross and re-cross the boundary of the ergosphere as many times as they like.

7. Dec 3, 2008

### MuggsMcGinnis

Do you know if anyone has published a solution to General Relativity that defines a frame of reference from which it would be possible to observe something to pass beyond an event horizon? That is, see it disappear?

Perhaps naively, I presume that, if it can happen, there must be some possibility of observing it. Or, at least some context from which the observation could be defined.

I'm referring, specifically, to objects with non-zero rest mass. Can an observer, in any reference frame, observe an object with non-zero rest mass, fall (disappear) through a Schwarzschild event horizon?

Thanks

Last edited: Dec 3, 2008
8. Dec 3, 2008

### Chronos

Reaching the event horizon would not be pleasant. Photons from the outside universe would be severely blue shifted.

9. Dec 3, 2008

### MuggsMcGinnis

I can't disagree about how pleasant the trip would be... I'm just trying to understand if it is (observably) finite in duration.

By my reasoning, as an object falls closer to the event horizon, time (for it) slows.
Since the event horizon has a non-zero temperature, it's radiating a certain wattage of power per unit of surface area.
As time slows, the quantity of energy per second will increase.
Therefore, as the object falls closer to the event horizon, it will cook.

10. Dec 4, 2008

### Chronos

It will not cook from the inside out, rather from the outside in. The horizon the observer infalls toward is redshifted relative to the observer. It is the external universe that is time accelerated and blue shifted relative to the observer.

11. Dec 5, 2008

### George Jones

Staff Emeritus
Since you refer to a black hole's temperature, I assume you're talking about Hawking radiation. Here what Birrell and Davies, pages 268-269, has to say.
For a freely falling observer, this cooking occurs not at the event horizon, but as the observer approaches the singularity. For a hoverig observer, cooking occurs "at" the event horizon.

12. Dec 7, 2008

### MuggsMcGinnis

Regarding my previous posting... anybody have an answer?

I have yet to find equations for calculating how long it will take for an object with non-zero rest mass to be seen to fall through an event horizon. I thought that the Kruskal extensions might provide a means of calculating this, but I don't seem to be clever enough to figure it out.

Anybody?

Last edited: Dec 7, 2008
13. Dec 7, 2008

### Chronos

An external observer will never see the infalling party cross the event horizon. A time dilation thing.

14. Dec 7, 2008

### MuggsMcGinnis

Chronos

That's what I thought.

What bothers me is that the presumption that something can fall to (much less, through) an event horizon in finite time depends upon there being a true reality' that contradicts the observable reality.

Imagine an observer external to the event horizon throws an object toward the center of mass of the black hole'. No matter how long the observer waits, the in-falling object will always appear to not have reached the event horizon.

In fact, the observer can interact, causally, with the object, by bouncing light off of it. The object will approach the event horizon at an ever-slowing rate... a rate asymptotically approaching zero as the distance above the Schwarzschild radius approaches zero and as time approaches infinity.

If it's possible to interact with the in-falling object, no matter how long one waits, how can we presume that the object actually falls through the event horizon in finite time?

It seems that such a presumption is contrary to direct observation.

15. Dec 7, 2008

### xantox

It will take exactly the same time as the black hole takes to disappear completely by Hawking evaporation.

16. Dec 7, 2008

### MuggsMcGinnis

Yes, if one includes Hawking radiation. However, I was just considering the idealized case of a static Schwarzschild event horizon.

Regarding some previous posts in this (somewhat divergent) thread: It's easy to see how Hawking radiation would cook any in-falling object. As the black hole shrinks, its temperature increases and it radiates more energy per second. The in-falling object is getting closer to this surface that's getting hotter. Also, time is slowing for the in-falling object so the quantity of energy it sees radiating from the event horizon is even greater, per unit of time. All of these factors combine so that (I'm guessing) the apparent temperature that the in-falling object is subjected to approaches infinite as the remaining lifetime of the event horizon approaches zero. It would end with one infinitesimal instant of infinite temperature.

17. Dec 7, 2008

### xantox

By building this spacetime in general relativity and then measuring the proper time on the infalling body worldline.

No, the infalling (free-falling) body will cross the event horizon right away, when the black hole is big, and the temperature of the horizon is low. Distant observers will see in the same distant future the last photons emitted in the distant past by the infalling body mixed with the photons from the black hole final explosion by evaporation.

18. Dec 7, 2008

### MuggsMcGinnis

And the external observer will see the in-falling object disappear through the event horizon?

19. Dec 8, 2008

### xantox

Yes, in the sense that he will receive at that moment the very last theoretical photons which may be possibly emitted by the object before entering the black hole.

20. Jan 4, 2009

### skeptic2

Xantox, I wonder if you could explain your answer from post #17 a little. It is clear that to a distant observer an infalling object will appear to slow down, turn red and dim as it approaches the event horizon. However, apart from this optical effect, from the inertial frame of the distant observer, isn't time increasingly dilated and space radially contracted close to the horizon? It would seem that the object would still have to cross an infinite expanse of contracted space during increasingly dilated time. The infalling object wouldn't just appear to slow down and stop, to the distant observer it in fact, would. Could you please explain how the free falling object is able to cross the event horizon right away? I recognize of course that in its own inertial frame the infalling object would fall to the center of the BH rapidly as the event horizon evaporates beneath it.