# Two Black Hole Questions

1. Sep 4, 2009

### mjacobsca

1) If you could hover over a black hole's event horizon (let's say it was a BIG hole with lower tidal forces and you could hover 1 mile out), and stuck a 1mi long pole over the event horizon such that 1mm of the tip of the 1mi pole were over the horizon, would the entire pole be ripped out of your hands? Could you pull the 1mm tip back out? I assume you couldn't.

2) I'm still slightly confused by Hawking Radiation. Almost all examples I've read give an example of two virtual particles developing OUTSIDE of the event horizon, and one particle being pulled into the event horizon while the other is escaping. I'm unclear as to how the black hole loses mass when it is adding particles from outside. One example I read said that the particle that gets pulled into the horizon gains gravitational energy from the black hole and turns into a real particle, hence the hole loses mass. However, even if the black hole lost gravitational energy to the particle that got sucked in, isn't the energy sucked back into the black hole? How is energy lost. I'm hoping one of you can give a hobbyist physics enthusiast a non-mathematical explanation of the whole phenomena.

Mike

2. Sep 4, 2009

### Haelfix

1) From a free falling observers point of view, he/she does not have any information available to actually know when they pass the event horizon or even where it is. There is nothing unusual about the particular point locally. The observer will still feel whatever tidal forces he felt 1 m before and 1 m after.

The rough analogy is to think of fish swimming down a river where the current is steadily getting faster. The fictitious point where the fish can no longer turn around and go back to his starting point is the 'event horizon', but from its vantage point there is no way off knowing where or when exactly that is, everything is still perfectly uniform and smooth before and after. The fish would need to know the global properties of the river to make that assessment, information that in the case of blackholes it does not and cannot have.

2) Technically, the blackhole doesn't lose mass like you would think (as in chunks of it coming off). Rather it absorbs 'negative energy'. The explanation you were given is morally correct, the virtual particle 'borrows' energy from the blackholes gravitational field (somewhat reminiscent of the Dirac sea analogy). Its a little hard to give the precise details here, b/c the process is still not understood perfectly well and it involves some rather heavy duty mathematical machinery so the word analogy has to stay loose. For instance, exactly what does one mean by 'particle', and what does one mean by 'energy' or the 'gravitational field of a bh' . Its a bit involved to give precise meanings to these rather fuzzy concepts which are very far removed from the classical counterparts.

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3. Sep 4, 2009

### xantox

The proper acceleration felt by a hovering observer at radius r tends to infinity when r tends to 2M (the event horizon), for black holes of any size. This means that lower portions of a hovering pole would be subject to increasing and diverging stresses, ie it would break or vaporize as soon as you start to lower it sufficiently near the horizon.

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4. Sep 4, 2009

### nutgeb

I agree with this but with a qualification -- I think Haelfix's description is correct, that from the local perspective of the pole, the fact that part of it has crossed the event horizon creates only a (relatively tiny) incremental increase in acceleration, not a stark contrast from one end of the pole to the other. At the radial distance of the event horizon, the tidal discrepency in acceleration as between one end of the pole to the other may be fairly small, particularly if the mass of the BH is large. But if the pole is very, very long, tidal forces could come into play, causing the pole to break.

I think the specific answer to the OP is that the hovering person holding the outer end of the pole would never be able to pull hard enough to bring the inward end of the pole back out over the event horizon -- or even to prevent it from accelerating ever faster toward the BH. If the pole-holder is capable of arbitrarily increasing her pulling force progressively, eventually the material of the pole will stretch to its breaking point, and any broken piece(s) of the pole that extend at all into the event horizon at the time of the break will continue accelerating into the BH. The remaining piece held by the pole-holder would of course stay entirely outside the event horizon.

5. Sep 4, 2009

### xantox

If a pole is hovering, it shall break before any part of it crosses the horizon.

6. Sep 4, 2009

### nutgeb

I don't know where you got the idea that gravitational acceleration becomes infinite at the event horizon. That's not correct. Gravitational acceleration increases smoothly at 1/r2 as the horizon is approached and crossed. Nothing unusual happens to the acceleration gradient at the horizon.

Gravitational acceleration approaches infinity at the singularity at the center of the BH, not at the horizon.

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7. Sep 4, 2009

### JesseM

There is no "gravitational force" in GR as there is in Newtonian gravity. A freefalling observer experiences no G-forces anywhere, but non-inertial observers do just like they do in SR, and the G-force experienced by an observer hovering non-inertially at constant Schwarzschild radius does go to infinity as you approach the horizon. This is discussed on this page, in the paragraph that begins with However, this acceleration is expressed in terms of the Schwarzschild radial parameter r, whereas the hovering observer’s radial distance r' must be scaled by the “gravitational boost” factor. The equation they give for the "proper local acceleration" felt by a hovering observer is $$\frac{-m}{r^2\sqrt{1 - 2m/r}}$$, which does go to infinity as r approaches the Schwarzschild radius 2m.
It's tidal forces (and curvature) that approach infinity at the singularity, not "gravitational acceleration" (of course it is impossible to calculate the proper acceleration of a hovering observer at any radius less than the Schwarzschild radius, because such an observer would have to be moving faster than light).

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8. Sep 5, 2009

### nutgeb

Jesse, I am convinced your statements are wrong, but I will leave it to others to contradict or confirm them.

9. Sep 5, 2009

### JesseM

Are you convinced that the page I linked to is wrong? It says clearly in that paragraph and multiple subsequent ones that the proper acceleration experienced by a hovering observer (the G-force) goes to infinity as you approach the horizon. You can find the same equation for proper acceleration experienced by a hovering observer in textbooks too--for example, see pages 3-31 to 3-32 of https://www.amazon.com/Exploring-Black-Holes-Introduction-Relativity/dp/020138423X, see pervect's post #4.

I have observed that you tend to make a lot of incorrect statements about GR based on non-quantitative arguments that wrongly blend GR concepts with classical ones (for example, the concept of a gravitational force which drops off according to an inverse-square law, an idea which is not fundamental in GR though it can be recovered in the Newtonian limit). If your response to being corrected is just to say "I disagree" but then not actually explain why you disagree, then you will never learn anything new.

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10. Sep 5, 2009

### DrGreg

JesseM is correct. The equation he quotes can also be found in the book

Woodhouse, N M J (2007), General Relativity, Springer, ISBN 978-1-84628-486-1, p.99​

(which applies under the convention that units are chosen where c = 1 and G = 1). You can find the lecture notes on which this book was closely based http://people.maths.ox.ac.uk/~nwoodh/gr/index.html [Broken], p.54 (Lecture 12).

You are perhaps being confused by the fact that in Schwartzschild coordinates the radial spatial component of four-acceleration is indeed given m/r2. But after applying to local metric, the proper acceleration is the formula given by JesseM (what you actually feel and can be measured by an accelerometer, and which dictates the force required to provide the acceleration).

If the "acceleration due to gravity" (i.e. the proper acceleration of a hovering particle) were finite at the event horizon, it would be possible for massive particles to hover at the event horizon by applying a finite force. Such hovering is impossible; only massless particles can hover at the horizon.

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11. Sep 5, 2009

### nutgeb

OK, I stand corrected. Dr Greg, I appreciate the way you give explanations without superfluous strutting of ego.

Question: If the locally measured acceleration becomes infinite as the horizon is reached, then why doesn't the locally measured velocity also immediately become infinite? Presumably the locally measured rate of acceleration (1st derivative of velocity) must return to a finite level once the observer attains a non-zero inward velocity at or inside the horizon.

12. Sep 5, 2009

### JesseM

Strutting of ego? I didn't talk about how great I was, I just criticized you for repeatedly refusing to respond to critiques of your claims (apparently you only do this when the critiques come from me, out of some personal grudge I imagine)
There's never a physically possible situation where the acceleration measured by an accelerometer is infinite, because it's impossible for any massive observer to hover on the horizon. The locally measured acceleration can become arbitrarily large at smaller and smaller finite distances from the horizon of course, but "locally" means you're talking about an infinitesimally small region of both space and time, so the change in velocity (as measured in the locally inertial coordinate system of a freefalling observer passing the 'hovering' object) should be infinitesimal in this region too regardless of the magnitude of the acceleration.

A non-massive object like a photon can hover on the horizon, though of course it can't carry its own accelerometer. But there is a sense in which the photon's acceleration can be seen as infinite from the perspective of a freefalling observer passing it, as discussed on the first page I linked to:
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13. Sep 5, 2009

### skeptic2

If I'm not mistaken, if a person hovering 1 mile above an event horizon (from his perspective) dangles a 1 mile pole towards the event horizon, due to the contraction of spacetime, the pole will not even reach the event horizon.

14. Sep 5, 2009

### nutgeb

The strutting that occurs on forums such as this from time to time can include (a) putting others down, (b) nitpicking casual uses of terminology as a tactic to discredit the other party's knowledge, (c) digressing into off-topic discussions that are of little relevance as a tactic to bolster the credibility of one's on-topic assertions, and (d) needing to have the last word on every subject.
Repeatedly? I do try to avoid expanding on my first attempt at an explanation unless and until I'm ready to say something coherent.
Are you specifically saying that the "small region of both space and time" is always exactly the right size such that the "arbitrarily large" acceleration always results in the observer crossing the horizon at a locally measured velocity of exactly c, regardless of the observer's initial velocity?

Also, if an observer hovering infintessimally close outside the horizon gets accelerated across the horizon, they are going to locally measure that their velocity continues to accelerate in a finite way as they approach the center. Since they had locally measured the acceleration to be arbitrarily large and approaching infinite while they were still hovering, then once they attain a non-zero inward velocity they must thereafter locally measure their rate of acceleration (change in velocity) to be finite and no longer arbitrarily large.

15. Sep 5, 2009

### Haelfix

Its probably worth putting numbers in here. 1 mile is not very long, and is pretty localized. Further whats the magnitude of the tidal forces we are talking about (or whats the mass of the black hole in question) which impacts on how much burning an observer on a rocket ship has to accelerate to stay hovering.

16. Sep 5, 2009

### JesseM

I'm a stickler for correctness and I don't like to see incorrect claims bandied about. This may result in putting down arguments that are incorrect (I haven't put you down as a person because I don't know anything about you), or making an issue of correcting wrong claims others make even if they aren't central to what was being argued (the charge of wanting to have the last word is unfair, since what I really want is for the other person to stick to the topic being discussed and either explain in detail what points of mine they disagree with and why, or else come to an agreement). This isn't ego-strutting, it's just a minor obsession with accuracy in every detail.
But if you make some claim that you can't justify in a rigorous way, then you shouldn't be making it on this forum in the first place--the rules state that this forum is not meant as a place for speculative discussion of claims that are not justifiable in mainstream SR and GR (of course the rules allow people to ask questions about whether a particular nonrigorous conceptual argument can be made rigorous in a GR context, but quite often you make definite assertions using ill-defined conceptual arguments). And yes, this has happened repeatedly in discussions I've had with you, as in this thread where you were making various arguments which were ill-defined or outright impossible in a GR context (like talking about the length of an object dragged from one region of Schwarzschild spacetime to another, even though extended rigid bodies are impossible in GR), or this thread where you were claiming the relativistic Doppler effect causes light to become more blueshifted than would be predicted by the classical Doppler formula when as I argued the opposite seems to be true, or this recent thread where you asserted that redshift could be derived from energy considerations and never provided a derivation or acknowledged that you weren't sure that such a derivation would actually be possible. I've also noticed a number of other threads that I didn't participate in where you also seemed to be using ill-defined arguments to try to come to definite conclusions which are not an accepted part of mainstream GR, like this one and this one and this one and this one.
Again, the size is understood to be infinitesimal when we are talking about "local" measurements. As measured in the locally inertial frame of a freefalling observer crossing the horizon, the horizon is always measured to be moving outward at exactly c.
A massive observer can't hover infinitesimally close the horizon, they can only hover at some finite distance from it (an infinitesimal is smaller than any possible finite real number greater than zero).
You can't locally measure your velocity relative to the center since the center is not in the same infinitesimal region of spacetime as you. Also note that the singularity has a timelike separation from an observer in the horizon, not a spacelike one, so talking about your velocity relative to the center is analogous to talking about your velocity relative to the Big Crunch in a closed universe. The Schwarzschild coordinate system confuses things by having a radial coordinate which is physically timelike inside the horizon (and a time coordinate which is physically spacelike inside the horizon), it's less confusing if you use a coordinate system where the time coordinate is always timelike and the radial coordinate is always spacelike, such as Kruskal-Szekeres coordinates (see the bottom section of this page). Kruskal coordinates have the nice property that light rays always look like straight diagonal lines at 45 degrees in this coordinate system, while timelike worldlines always have an angle that's closer to vertical than 45 degrees. In Kruskal diagrams you can see that the singularity is a timelike future one that everyone inside the horizon will inevitably hit as their worldline takes them forward in time, much like the Big Crunch in a closed universe.
The thing about acceleration approaching infinity as you approach the horizon is specifically for hovering observers--a freefalling observer's accelerometer will show a reading of zero at every point up to and including the horizon, and accelerating observers who are not hovering will measure different G-forces from hovering observers even when they cross paths at the same point in spacetime. It's simply impossible for a massive observer to hover at any radius inside the horizon, so the question of how the formula for acceleration experienced by hovering observers continues past the horizon isn't a meaningful one.

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17. Sep 5, 2009

### DrGreg

That's not my interpretation of JesseM's comments. His style differs from mine, but I think he was just trying to get you to think before you make definitive statements that might actually turn out to be false.
I think JesseM has pretty much answered this, but to clarify:

The acceleration here is the acceleration of a hovering object relative to a local freefalling object, rather than the other way round. To be honest, I haven't done the calculation to see whether these two accelerations are the same or not. I suspect it probably is provided the object is falling from rest as you measure it.

Objects with mass simply cannot hover at or below the event horizon, so there is no answer to what their proper acceleration would be there. Also note that in relativity, the speed of an observer affects the value of any accelerations they measure.

I don't think it's directly relevant, but it's perhaps worth noting that a hovering observer observing a falling object will say the object decelerates and stops at the event horizon, taking an infinite time to get there. This is just making the point that we need to stick with local measurements of acceleration; remote measurements just get distorted by spacetime curvature!

The falling observer measures his own acceleration to be zero at all times, and, once through the event horizon, there cannot be any local hovering landmarks to compare against. Making a remote measurement of the singularity isn't feasible because it's not local.

Although it's pretty meaningless to talk of the proper acceleration of a photon, we can think of a massive particle accelerating towards the speed of light. To reach that limit is a finite time would require a proper acceleration tending to infinity. Therefore, loosely speaking, we could say "the proper acceleration of something at the speed of light is infinite", recognising that this is a bit of an abuse of the terminology.

As a concluding remark, in general relativity, when we mention velocity, acceleration, etc, we need to be clear what is being measured relative to what, and whether it is a "proper" invariant measurement, or a local or remote coordinate measurement.

18. Sep 5, 2009

### nutgeb

Agreed. I did not mean to suggest otherwise.
I understand that with respect to an observer at infinity. But will a hovering observer who is fairly near the horizon see it that way also? Certainly that hovering observer will see a plunging object pass by him at a high rate of velocity.
Fair enough. I understand the difficulty of the falling observer measuring his own change in proper velocity as a function of time inside the horizon, given the lack of stationary landmarks and the lack of any feeling of acceleration in the seat of his pants. And any calculation the falling observer does of his proper velocity needs to take into account the "distance boost", as compared to the Schwarzschild r coordinate. Nevertheless, the falling observer measures a finite proper time ("wristwatch time"), to fall from the horizon to the center, which, (disregarding the fact of his own eventual tidal demise), enables a calculation that (a) his proper velocity must have increased progressively after crossing the horizon, and (b) his derived proper acceleration must have been finite and not "arbitrarily large." Such a calculation will show that his proper velocity inside the horizon was "timelike" in the sense that the figure will be faster than c, but I don't see how that precludes the notion that a finite proper distance also has been traversed between the horizon and the center. Or perhaps after accounting for the "distance boost", the falling observer will calculate that the distance he traveled from the horizon to the center was infinite (?)
Agreed. I've tried to do that, and I appreciate you keeping me honest if I get it wrong.

19. Sep 5, 2009

### atyy

The formula JesseM and DrGreg gave is infinite at the event horizon, and I'm not sure it's valid there, but it also doesn't matter, since only hovering above the horizon needs to be considered, and for any finite distance above the horizon, no matter how small, the acceleration is finite.

20. Sep 5, 2009

### xantox

As already thoroughly answered by the previous posters, this is a general relativistic result – it seems you were applying the newtonian law. Another non-intuitive fact is that in the hovering frame, radial distances are stretched, so that a hovering pole would need to be very long to actually reach the horizon.

Static tidal forces diverge too in the limit of the horizon for hovering observers.

I obtain the following for a supermassive black hole of 10000 solar masses. Its horizon radius is 29530 km. Hovering 1 km away from the horizon would imply for an astronaut weighing 70kg on earth to weigh 10^14kg. Interestingly, the predicted Unruh temperature should still be fairly cold out there at 10^-6K (but this will also diverge in the limit of the horizon), though the radiation flux from the rest of the universe should be quite intense. It would be interesting to calculate it as well as the Hawking luminosity on that frame.

The black hole should probably appear from that hovering frame as a luminous region (per Hawking radiation) encompassing almost all sky excepted for a small tunnel of intense light on the other side.

21. Sep 5, 2009

### nutgeb

Thanks, I understand that.

Going back to the OP, it seems that the infinite acceleration does matter for the long pole dipped into the horizon from outside. That indicates that, if the outward end of the pole is held with arbitrarily great strength by the outside hoverer, the pole must break as soon as the inward end approaches the horizon.

One thing that confuses me is that if the pole-holder is infinitely far from the horizon, he will feel the same BH's "surface gravity" pulling on the inward tip of the pole at the horizon at a finite force of only 1/4m (in geometric unit figures). In which case the pole won't necessarily break (?) The Raine & Thomas textbook seems to explain this as a redshifting phenomenon.

22. Sep 5, 2009

### DrGreg

Yes, and yes. The point is that the hovering observer is accelerating, and if they are close to the horizon, accelerating at an extremely high rate. You know, even in flat spacetime, with no gravity, if you drop an apple from an accelerating rocket, the apple never "falls" more than a distance of c2/a below the rocket, before it slows down and almost stops, relative to the rocket's accelerating frame. (Look up "Rindler coordinates".) When the acceleration becomes enormous, that "stopping distance" can become very small!

(To digress from this conversation and address the original question of this thread, under such huge proper acceleration, apart from the fact that a human would be crushed to death by the G-force and his rocket will eventually disintegrate under the stress, as you lower the pole, the pole's weight will be enormous, approaching infinity as you lower it, and at some point either you'll let go or it will break under the huge tension force within it.)

I'll have to pull you up over the term "proper velocity". There's no such thing in the sense you are using it here. (There is something called "proper velocity" or "celerity" in relativity, but it's not relevant to what's being discussed here.)

In view of this, the rest of your post doesn't really make much sense. I'll have to go back to my last post: when you say "velocity", velocity relative to what?

The proper acceleration of the falling object is zero. That's a coordinate-independent fact. It's the proper acceleration of a succession of floating observers that increases towards infinity as you drop to the horizon, and there aren't any beyond it. The speed of these floaters relative to you increases to c as you drop to the horizon, but there aren't any beyond it.

There are others on this forum who have more expertise in this area than me, but I'm not sure if it really makes any sense to talk of "distance travelled" inside the horizon, again, relative to what exactly?

23. Sep 5, 2009

### DrGreg

If any formula gives you an answer of infinity, that's the mathematics' way of saying the formula isn't valid in those circumstances. In this case it's not valid because there are no hovering massive particles to be measured at or below the horizon.

24. Sep 5, 2009

### Haelfix

"The black hole should probably appear from that hovering frame as a luminous region (per Hawking radiation) encompassing almost all sky excepted for a small tunnel of intense light on the other side. "

I don't think I buy that, b/c it breaks the equivalence principle. A local observer should not be able to detect hawking radiation. Instead he/she will detect Unruh radiation (b/c the hovering observer is in an accelerated frame) but he/she cannot know that some of that radiation escapes away to infinity as Hawking radiation.

25. Sep 5, 2009

### atyy

Does the pole break because of the increasing proper acceleration, or because it is not "born rigidly accelerated". Is there any analogue of born rigid acceleration" that would enable the pole to hover stresslessly?