Some "Easy" stuff about black holes

[folmerveeman]

Hi there xkcd, after doing some basic research on black holes, I've come up with a couple of concepts that baffle me, and a couple of questions regarding those concepts
So could you guys maybe enlighten me?

1. Ok, so 'normal' black holes have one point of singularity, yet rotating black holes have a 'singularity ring'. What would conditions be like within this singularity?

2. Given that when you are sufficiently close to the black hole's event horizon all the light gets sucked into it, could you use this to achieve practical invisibility?

3. Given that black holes have three properties, Current, Mass and Angular Momentum. If the mass increases, does the black hole expand? or the singularity? or both?

Cheers!:)

[Scyrus]

Hello, I am not well versed in the matter, in fact this is info I dug up as well, but I believe I can answer one or two questions to some extent.


1. You are describing a Kerr black hole, which is essentialy a rotating black hole which, instead of being a single point singularity, is a rotating singularity: a 2-dimensional ring with non-zero radius. This black hole is not spherical, it is slightly elliptical (the equatorial bulge being on the plane of the ring itself, and the poles perpendicular to the center of said ring). Incoming objects will gain some angular momentum in the direction of the rings rotation before crossing the horizon. Most likely, a great ammount of dust would form an accretion disc.
EDIT: I have read that perhaps it is possible for an object to cross the even horizon and avoid the singularity by making use of the rotating properties of the black hole, and it could theoretically act as a wormhole. This has no physical meaning as of yet but could make an interesting plot point in a space sci-fi. I am curious, are these questions born of your own curiosity or do you intend to use them in a story? It'd be cool

2. Perhaps it would be possible to use gravitational lensing to conceal yourself depending on where you and external observants are.

3. When matter falls into a black hole, it increases its mass (much as it would happen with any massive object). We understand black holes as being singularities, so they do not have volume, however, an increase in its mass would also mean an increase in its Schwarzschild radius, meaning that although the singularity itself would not "expand", the radius of the black hole's even horizon would increase. In a pratical sense, although the body itself does not expand, its even horizon does grow.

[TestTubeGames]

I try not to think too much about the singularity, personally. It is, after all, unobservable for those outside the black hole... and a region where strange quantum gravity effects will take hold.

To your second point, the event horizon marks the boundary between where light can escape and can't escape. Of course, if close to the horizon, to escape the light would have to be directed at the right angle (away from the black hole). And by the time the light got away from the hole it would be tremendously redshifted (visible light would shift into the non-visible realm). But there would still be detectable light (microwaves... or radio waves, that kind of thing). I'm not sure what would qualify for practical invisibility. I imagine any human that close to a black hole would be so whacked out and confused by the spacetime warping, that a single human hiding near the event horizon wouldn't stand out much. Even if they were giving off light! But I wonder if there aren't easier ways to hide...

To your final point, if the mass increases, the black hole does expand. Since I again shy away from thinking too hard about the singularity, I'll say we know the event horizon does expand in that instance. The more massive the hole, the stronger the pull of gravity, the larger the influence on its surroundings... thus the bigger the event horizon.

[folmerveeman]

Hello, I am not well versed in the matter, in fact this is info I dug up as well, but I believe I can answer one or two questions to some extent.


1. You are describing a Kerr black hole, which is essentialy a rotating black hole which, instead of being a single point singularity, is a rotating singularity: a 2-dimensional ring with non-zero radius. This black hole is not spherical, it is slightly elliptical (the equatorial bulge being on the plane of the ring itself, and the poles perpendicular to the center of said ring). Incoming objects will gain some angular momentum in the direction of the rings rotation before crossing the horizon. Most likely, a great ammount of dust would form an accretion disc.

2. Perhaps it would be possible to use gravitational lensing to conceal yourself depending on where you and external observants are.

3. When matter falls into a black hole, it increases its mass (much as it would happen with any massive object). We understand black holes as being singularities, so they do not have volume, however, an increase in its mass would also mean an increase in its Schwarzschild radius, meaning that although the singularity itself would not "expand", the radius of the black hole's even horizon would increase. In a pratical sense, although the body itself does not expand, its even horizon does grow.

1. So the singularity would be a two dimensional disk? The way I read it it was described as some sort of a ring or donut, with a hole in the middle, which was the origin of my question

2. Please do elaborate!

3. So what the layman calls the black hole is actually the black hole + event horizon? Interesting.

[folmerveeman]

To your second point, the event horizon marks the boundary between where light can escape and can't escape. Of course, if close to the horizon, to escape the light would have to be directed at the right angle (away from the black hole). And by the time the light got away from the hole it would be tremendously redshifted (visible light would shift into the non-visible realm). But there would still be detectable light (microwaves... or radio waves, that kind of thing). I'm not sure what would qualify for practical invisibility. I imagine any human that close to a black hole would be so whacked out and confused by the spacetime warping, that a single human hiding near the event horizon wouldn't stand out much. Even if they were giving off light! But I wonder if there aren't easier ways to hide...

But what if it was some sort of probe or space station, perhaps in a rapid orbit so it doesn't get sucked in?

[Scyrus]

1. So the singularity would be a two dimensional disk? The way I read it it was described as some sort of a ring or donut, with a hole in the middle, which was the origin of my question

2. Please do elaborate!

3. So what the layman calls the black hole is actually the black hole + event horizon? Interesting.

1. Yeah it's a ring but it has no thickness, that's why I said 2-dimensional. I am unsure what characteristic of the collapsing star would define the radius of the ring, but the bigger the radius, the more elliptical it would be.

2. This would fall in the category of near-complete sci-fi, and even though Wikipedia articles aren't very good for this, according to this, a ring black hole may act as a wormhole and can be traversed to travel in spacetime, but I would say it would have to be an extremely accurate and calculated path or you'd be ripped apart by gravity. It could act as a plot twist however, a character determining a black hole to be rotating thanks to observational clues and diving in in the hopes of coming out unscathed somewhere/somewhen. But again, this part here is only speculative and theoretical. It does bring a smile to imagine this hypothetically being possible in our universe but, alas, we know but a drop on an ocean!

3. Yes, some people define black hole as being the singularity itself and the event horizon, since whatever crosses the horizon will soon be part of the singularity and not even light can escape, but the core itself has 0 volume and does not grow. And FYI, I am a layman. I am hoping some other, more learned xkcdians will come and shed more light and reason on the answers you seek .

[eSOANEM]

1. We can't really say anything about the singularity itself because we're pretty sure that what we do know is wrong (or at best, not entirely accurate) because there will be very significant quantum gravity effects going on and, without a consistent theory of quantum gravity, we're going to have trouble working out what they are.

Because of this, talking about the singularity is, for the most part, unhelpful and so by "black-hole" I will be referring to the event horizon.

2. Sort of. As an object falls past the event horizon, to a distant observer it looks like the person never quite reaches the horizon, they just seem to slow down and slow down in such a way that they will never be observed to cross the horizon. Of course, because all the other infalling matter will be incredibly hot (because it gains a lot of energy falling into the black hole, it will heat up) you'd probably be invisible by shear virtue of how bright everything around you would be.

3. The black hole expands. If the angular momentum increases, it becomes more elliptical, if the charge increases there won't be much obvious change in the shape/size. Position and momentum (the other two variables describing a non-quantum black hole) will, fairly obviously not affect the black hole's size or shape.

[thoughtfully]

A few clarifications.

Real "normal" black holes rotate. Maybe "normal" applies to the Schwartzchild solution as the model for a black hole, but in nature a nonrotating hole would be pretty damn strange.

As has been pointed out, the singularity is terra incognita, but pretty much nobody believes that they're really points (or rings) with zero volume and infinite density.

If light is getting sucked into the black hole, you've been sucked for a good while already. The effect you'll get is a delay, not invisibility. Not that simultaneity means a whole lot in this context

In fact, to an outside observer, you never actually cross the horizon, you just sit there. Until the black hole evaporates. Oops, those silly quantum effects again!

The radius of a nonrotating black hole is directly proportional to its mass. For a rotating one, it's still pretty nearly so.

[SlyReaper]

Speaking of black hole evaporation and time dilation: does that mean even an infalling observer would never observe himself crossing the event horizon? What I mean is, time would appear to speed up as he approaches the event horizon, so the black hole's rate of evaporation would appear to him to be increasing. Eventually, the event horizon below him appears to be giving off so much energy, it just incinerates the observer who never crossed it.

Is this wrong? Because it's always said that while an observer at infinity would never see an object cross the event horizon, an infalling observer would cross it in finite relative time. But I don't know if that takes black hole evaporation into account.

[Scyrus]

Speaking of black hole evaporation and time dilation: does that mean even an infalling observer would never observe himself crossing the event horizon? What I mean is, time would appear to speed up as he approaches the event horizon, so the black hole's rate of evaporation would appear to him to be increasing. Eventually, the event horizon below him appears to be giving off so much energy, it just incinerates the observer who never crossed it.

Is this wrong? Because it's always said that while an observer at infinity would never see an object cross the event horizon, an infalling observer would cross it in finite relative time. But I don't know if that takes black hole evaporation into account.

Theorizing what happens to a falling observer in a black hole is most likely the hardest part on which to make assumptions, because there are many factors involved, the forces, time dilation, redshifting, spaghettification of the observer, etc. (etc. means I dont know any more )
So from here on, its mere speculation on my part, I'd like to see other people's opinions:

But if you are headed towards the horizon, relativity becomes important, what appears to "speed up" is in fact the rest of the Universe, while in reality it is the observer approaching the black hole that slows down, because the black hole's escape velocity is much higher than c, an observant crossing the horizon will be accelerated to relativistic speeds (perhaps 0,999c?). However, it's not like the falling observer would see his arms in slow-mo, he would see himself just fine because in his point of view it is the rest of the universe that has gone haywire. But if he is falling into the horizon, whatever hawking radiation from the black hole's evaporation would experience the same time dilation as the observer, and for energy to incinerate him it must be in the same space as him. And since the rest of the universe is speeding up relatively to the observant, and he himself is traveling at increasing speeds, the last thing a falling observer would see would be, assuming he is falling slowly enough for the signals to reach the brain and process them:
a. If he were facing the black hole, the last thing he would see would be complete darkness, as the moment he crosses the horizon, even the incoming light from the outside will never reflect off the black hole or any other object and will never bounce off the observer's eyes.
b. If he were facing the rest of the universe (back turned to the black hole) he would see a blur of white. Why? Because relativity would say that, relative to you, everything in the Universe is now travelling at the speed of light, and everything would be seen as traveling at c (light).
c. If you fell feet-first into the horizon you would see what an external observant would probably see of you: redshifting. Your lower body would simply become dark red until it crossed the horizon, at which point nothing would be seen because your feet are no longer there and no light can reflect off of them. What would you feel? Not sure, but no doubt the sphaggettification is faster than your axons, so, you would just stop feeling your body.

/end speculation

[starslayer]

@SlyReaper: No, an infalling observer sees nothing strange as he falls in. The black hole does appear to be forever below him, but that's because you can still see the outside universe from inside the hole, whereas nothing closer to the singularity is coming up. Time doesn't speed up for the infalling observer, but he does see the universe increasingly concentrated into a blueshifted band at eye level. This website has a number of good videos/simulations of what happens when something falls into a black hole.

So the singularity would be a two dimensional disk?

No, it's a ring, not a disk; an observer could pass through the center of it without hitting anything*. As eSOANEM mentioned, though, we don't expect the singularities of real black holes to be like that, since quantum gravity is going to take over at some point, and true singularities (that is, things having no dimension in some way) almost certainly don't exist.

*He will, however, inevitably hit the singularity at some point.

[eSOANEM]

Based on my understanding of a black hole, an infalling observer (who, for the sake of argument, we'll assume is in a space ship with a bow and stern and which allows the observer to see in all directions simultaneously) would see the following:

A long way from the black hole, they will notice the gravitational lensing and, assuming the black hole is feeding, the accretion disk.

As the ship falls into the accretion disk, assuming there is one and that the ship can survive the enormous temperature, the observer will see the light from the accretion disk shifted. The direction of this shift will depend on the velocity of the ship and the distance of the ship from the horizon. The reason for this is that the gravitational potential will try to redshift the light below the ship whilst, because he is approaching the black hole, the ordinary Doppler effect will try to blueshift it. It would not surprise me if, for a free falling observer, these two effects cancel each other out exactly.

So the ship falls through the accretion disk and possibly sees the light shifted red or blue in front or behind him depending on his position and speed relative to the black hole.

As the ship approaches the horizon, tidal forces will stretch the ship, these forces will, eventually lead to spaghettification of the ship and astronaut, but for now let's ignore these and pretend they somehow survive.

Counteracting the tidal forces, the observer will also see gravitational time dilation affecting his ship such that the bow appears to slow down and the stern appears to speed up. This will make the ship appear to be squished lengthways and will, to a certain extent, cancel out the effect of the tidal forces.

Again, I haven't done any calculations, but looking in very handwavey terms at the tidal forces and gravitational time dilation, I think that, for most of the fall the tidal forces will win until very close to the horizon itself where the time dilation will kick in.

What this means is that as he approaches the horizon, first the ship will stretch before the bow appears to suddenly slow down and rush towards the astronaut.



Beyond that, I have no idea. From what I know, the astronaut should, assuming he survives, pass through the horizon without much noticeable happening however he should not see the bow of the ship cross. I think he will see the bow fly past him as he falls though followed by the stern (which by now well have stretched to a long way behind him.

Throughout this, last section, I have assumed that the time dilation observed by the infalling astronaut is the same as for a distant observer moving at the same speed, this is very possibly not the case.

@SlyReaper: No, an infalling observer sees nothing strange as he falls in. The black hole does appear to be forever below him, but that's because you can still see the outside universe from inside the hole, whereas nothing closer to the singularity is coming up. Time doesn't speed up for the infalling observer, but he does see the universe increasingly concentrated into a blueshifted band at eye level. This website has a number of good videos/simulations of what happens when something falls into a black hole.

An infalling observer who is a point mass certainly will not see anything peculiar in the region of the horizon (assuming there is no other infalling matter). I'm less certain as to what will be observed is there is other matter falling in too be it the spaceship in my example or matter from an accretion disk or both.

[Diadem]

3. So what the layman calls the black hole is actually the black hole + event horizon? Interesting.

When physicists talk about black holes, they always talk about "the area enclosed by an event horizon", never the singularity itself, unless they explicitly say otherwise.

The black hole (the area enclosed by the event horizon) is an actual physical thing. It's the relevant part. We can't see what happens inside, and we will never be able to see. So there's not much practical sense in talking about it. It falls outside physics. Though for the record, almost no physicist believes there's an actual singularity there. Singularities happen all over the place in physics, and they always mean the same thing: "Your theory is not applicable here". Try calculating the gravitational attraction between two astronomical bodies. You generally start by assuming they are point masses. But then what happens if they get close enough together? The attraction becomes infinite. Is that an actual singularity? No of course not, it just means your theory is not applicable there, because they aren't really point masses.

Black Holes are predicted by General Relativity. But general relativity is (like all theories!) an incomplete theory. It doesn't and can't properly describe quantum effects. And a singularity, that is, lots of mass compressed to a very small region of space, sounds very quantumy. But we can't use quantum mechanics to describe what happens either, because quantum mechanics is also incomplete, it doesn't and can't describe gravity. To accurately describe what happens in the centre of a black hole, we need a theory of quantum gravity. We've been trying to construct that theory for a few decades now, but so far without success.

We still know black holes exist though. That's definitely General Relativity territory. Though we aren't sure if they evaporate. That's a quantum effect. So our understanding is incomplete there.

Speaking of black hole evaporation and time dilation: does that mean even an infalling observer would never observe himself crossing the event horizon? What I mean is, time would appear to speed up as he approaches the event horizon, so the black hole's rate of evaporation would appear to him to be increasing. Eventually, the event horizon below him appears to be giving off so much energy, it just incinerates the observer who never crossed it.

Well this is already true even without black hole evaporation. Black holes are not exactly friendly to organic life. A typical black hole has enough radiation circling it to make a brave man weep. And melt. And explode.

But yes, if you fall into a classical (classical here means: classic gr. No evaporation) static-size black hole, that would take a finite time for you, but an infinite time for an outside observer. That means that if the black hole disappears sometime in the future for an outside observer, you'll never reach it. However, this is ignoring one important fact.

Is this wrong? Because it's always said that while an observer at infinity would never see an object cross the event horizon, an infalling observer would cross it in finite relative time. But I don't know if that takes black hole evaporation into account.

This is already not entirely true even for classical black holes.

The reason is: If you fall into a black hole, the black hole grows. First there was a black hole with mass M and radius R. Now there's a black hole with mass M + you with radius R + some very small amount (an 80 kg person would grow a black hole by 1.2 * 10^-25 m). In other words, if you get close enough to a black hole, the black hole will grow to encompass you. This happens in a finite time for an outside observer (and obviously also in a finite time for you).

Though now we're doing black hole dynamics and that is very very hard. For example in the above scenario the angular momentum and centre of mass of the black hole would also change. To say exactly what happens is impossible analytically, you'd have to resort to very heavy duty computational methods.



*edit* Also one final note. There's some talk here about tidal forces and spaghettification. I'd like to quickly point out that this is only a significant problem with small (around 1 solar mass) black holes. Bigger black holes have much smaller tidal effects. Fall into one of those supermassive black holes that exist in the centre of galaxies, and you won't even notice the tidal forces.

[SlyReaper]

Speaking of black hole evaporation and time dilation: does that mean even an infalling observer would never observe himself crossing the event horizon? What I mean is, time would appear to speed up as he approaches the event horizon, so the black hole's rate of evaporation would appear to him to be increasing. Eventually, the event horizon below him appears to be giving off so much energy, it just incinerates the observer who never crossed it.

Is this wrong? Because it's always said that while an observer at infinity would never see an object cross the event horizon, an infalling observer would cross it in finite relative time. But I don't know if that takes black hole evaporation into account.

Theorizing what happens to a falling observer in a black hole is most likely the hardest part on which to make assumptions, because there are many factors involved, the forces, time dilation, redshifting, spaghettification of the observer, etc. (etc. means I dont know any more )
So from here on, its mere speculation on my part, I'd like to see other people's opinions:

But if you are headed towards the horizon, relativity becomes important, what appears to "speed up" is in fact the rest of the Universe, while in reality it is the observer approaching the black hole that slows down, because the black hole's escape velocity is much higher than c, an observant crossing the horizon will be accelerated to relativistic speeds (perhaps 0,999c?). However, it's not like the falling observer would see his arms in slow-mo, he would see himself just fine because in his point of view it is the rest of the universe that has gone haywire. But if he is falling into the horizon, whatever hawking radiation from the black hole's evaporation would experience the same time dilation as the observer, and for energy to incinerate him it must be in the same space as him. And since the rest of the universe is speeding up relatively to the observant, and he himself is traveling at increasing speeds, the last thing a falling observer would see would be, assuming he is falling slowly enough for the signals to reach the brain and process them:
a. If he were facing the black hole, the last thing he would see would be complete darkness, as the moment he crosses the horizon, even the incoming light from the outside will never reflect off the black hole or any other object and will never bounce off the observer's eyes.
b. If he were facing the rest of the universe (back turned to the black hole) he would see a blur of white. Why? Because relativity would say that, relative to you, everything in the Universe is now travelling at the speed of light, and everything would be seen as traveling at c (light).
c. If you fell feet-first into the horizon you would see what an external observant would probably see of you: redshifting. Your lower body would simply become dark red until it crossed the horizon, at which point nothing would be seen because your feet are no longer there and no light can reflect off of them. What would you feel? Not sure, but no doubt the sphaggettification is faster than your axons, so, you would just stop feeling your body.

/end speculation

I already knew this, but how does this tie in with hawking radiation? If the rest of the universe appears to be speeding up, then the black hole's rate of evaporation would also appear to be speeding up. Everything I've ever read or seen about what an observer would see as he falls into a black hole seems to assume a non-feeding, non-evaporating black hole. But by our current understanding, there's no such thing as a non-evaporating black hole.

To clarify, I'm strictly talking about the moments before the observer crosses the event horizon, so the energy released through hawking radiation would always be coming from below.

[eSOANEM]

*edit* Also one final note. There's some talk here about tidal forces and spaghettification. I'd like to quickly point out that this is only a significant problem with small (around 1 solar mass) black holes. Bigger black holes have much smaller tidal effects. Fall into one of those supermassive black holes that exist in the centre of galaxies, and you won't even notice the tidal forces.

This is a good point, and one I forgot to consider.

This would have the effect (in my example), I think, of making the ship stay about the same length until suddenly the bow seems to start slowing down enormously due to the gravitational time dilation as the bow approaches the horizon.

I already knew this, but how does this tie in with hawking radiation? If the rest of the universe appears to be speeding up, then the black hole's rate of evaporation would also appear to be speeding up. Everything I've ever read or seen about what an observer would see as he falls into a black hole seems to assume a non-feeding, non-evaporating black hole. But by our current understanding, there's no such thing as a non-evaporating black hole.

To clarify, I'm strictly talking about the moments before the observer crosses the event horizon, so the energy released through hawking radiation would always be coming from below.

The rest of the universe doesn't speed up. An outside observer sees the infalling one to slow down, but that doesn't mean the infalling observer sees the outside one speed up.

I don't know if the infaller would actually see any time dilation on the distant observer other than the standard time dilation from SR.

If the black hole is evaporating, I think (assuming I've understood the heuristics correctly) he should see nothing special at all. The heuristics say that hawking radiation is just what quantum fluctuations look like on the horizon to a distant observer. I don't know how accurate the heuristics are, but seeing as the infalling observer will be moving very quickly at this point, even if there is a deviation from this heuristic, they will probably pass the horizon far too quickly to see anything unusual in the quantum foam.

[Robert'); DROP TABLE *;]

The rest of the universe doesn't speed up. An outside observer sees the infalling one to slow down, but that doesn't mean the infalling observer sees the outside one speed up.

...Doesn't it? I thought gravitational time dilation was symmetrical, i.e. Alice thinks she's faster than Bob, therefore Bob thinks he's slower than Alice.

[SlyReaper]

That has always been my understanding. Surely if it wasn't symmetrical, you could violate causality? Fly close to an event horizon, the rest of the universe sees you slow down, but you see the rest of the universe pass at the same speed -> fly away again, and see yourself back at the event horizon. Perhaps fly back to the event horizon again to interact with your past self.

[eSOANEM]

The rest of the universe doesn't speed up. An outside observer sees the infalling one to slow down, but that doesn't mean the infalling observer sees the outside one speed up.

...Doesn't it? I thought gravitational time dilation was symmetrical, i.e. Alice thinks she's faster than Bob, therefore Bob thinks he's slower than Alice.

That has always been my understanding. Surely if it wasn't symmetrical, you could violate causality? Fly close to an event horizon, the rest of the universe sees you slow down, but you see the rest of the universe pass at the same speed -> fly away again, and see yourself back at the event horizon. Perhaps fly back to the event horizon again to interact with your past self.

Yeah, you're right. I was getting it completely mixed up with SR time dilation. Apologies. I will edit my previous post accordingly.

[SlyReaper]

Okay, so what you're saying is hawking radiation is only evident to a distant observer, but not to an infalling one? But that doesn't really make sense either, since the infalling observer sees the event horizon as being below him (and "distant", given how stretched spacetime is this close to a black hole) at all times until he's dead. If you have two observers, one distant, one infalling, the distant observer will see the infalling one asymptotically approaching the event horizon, and be in the same space as where this hawking radiation is emanating from. So why does the infalling observer not see the same? If the distant observer detects n particles emitted by a given region around the infalling person in m ticks of a distant clock, why would the infalling observer not also see n particles emitted in m ticks of the same distant clock?

I've probably not explained that very well, but I can't think of how better to say it.

[eSOANEM]

It's more that, in a brief period of time (assuming I've understood the heuristics correctly), hawking radiation looks like the normal quantum foam of virtual particles popping in and out of existance. It's only over a long (at some level) time that you notice that they're not recombining.

As the infalling observer shoots past the horizon at a considerable fraction of c, it seems to me that, unless he already knows exactly where to look, he's not going to see anything out of the ordinary and that even if he does know, he still might not.

Now, this is all predicated on some pretty shakey ground (the heuristics are definitely not accurate and I may be pushing them too far). Really this is a QFT in curved space problem and, from previous questions, I get the impression that doogly seems to know more about it than most.

[legend]

What is usually considered to be the radius of a black hole R_s=2M\frac{G}{c^2} is actually just the size of the event horizon as seen from a stationary, infinite far away observer. But if you are falling into the black hole, there is absolutely nothing special going on at this point and, unless you are using some advanced tracking device, there is no way you can actually tell when you cross this point. This means the size and the shape of the event horizon are observer dependent concepts. Because the Hawking radiation is produced by event horizons this means that the temperature and intensity of the radiation also depends on the observer and a free falling doesn't actually see it*.

*He might actually see some radiation, I've never really calculated it.

[eSOANEM]

The term "event horizon" is usually used to mean the absolute horizon in the context of a black hole.

The absolute horizon is the surface from which escaping light rays do not reach infinity even after an infinite amount of time. Because of this, light rays from within the absolute horizon will never reach any point outside it in a finite amount of time. As its name suggests, the location, size and shape of the horizon is not frame dependent.

The apparent horizon, is the surface for which the escaping light rays of any surface inside it converge. For escaping rays to converge, the field must be stronger than for them merely to not escape (because, for them to converge they must be bent back on themselves, to not escape, they do not need to be bent so much); because of this, the apparent horizon is necessarily inside the absolute horizon. As its name suggests, the apparent horizon is observer dependent.

In order for Hawking radiation to actually lead to evaporation in any realistic black hole (a black hole which has always existed will have the two horizons coincide in its frame) the radiation must be produced at the absolute horizon the location of which will be agreed upon by both the infalling and distant observers.

[thoughtfully]

An observer in free fall sees no Hawking Radiation. Read Leo Susskind's Black Hole Wars for some discussion on the topic. One way to see this is to look at Hawking Radiation as the Equivalence Principle equivalent to Unruh Radiation. No acceleration, no radiation.

Another thing to note is is that spacetime need not be all that curved at the horizon. For BHs at the cores of large galaxies, it's pretty flat.

[Yakk]

That seems strange.

A light black hole doesn't generate a significant amount of gravitational force at beyond a very short distance, yet the Hawking radiation it produces is extremely hot and high energy.

If you are in freefall towards that black hole, you don't see any Hawking radiation, but if you are stationary and accelerating away from it at a ridiculously tiny rate, you see this huge explosion?

t_ev =~ 1.33E-17 M^3 s / kg^3

So a 1 tonne black hole takes 0.0000000133 seconds to release 9E19 J of energy. And you only see this Hawking radiation if you are not in free fall towards the 1 tonne object (which generates negligible acceleration).

There is something wrong here?

[SlyReaper]

^^ What Yakk said.

[legend]

We're talking about surface gravity here. Because the size of the black hole goes with M and the gravitational acceleration with MR^-2 the surface gravity is actually inversely proportional to the mass, if we use the analogy from the unruh radiation the temperature is proportional to the acceleration thus inversely proportional to the mass and everything works out in the end.

An event horizon is a surface from beyond which no information can reach an observer. Yes when you look at a black hole you see an event horizon, but there really is nothing special about black holes and there are many other (hypothetical) scenarios where they can arose. E.g. because the universe is expanding there is a surface beyond which the expansion is "faster than c" and thus no information car reach us. This also creates an event horizon. Another example would be a uniformly accelerating observer in a flat space time. In this case there is also a region of space from which information can not reach the accelerating observer and thus the observer sees an event horizon. The second example also illustrates that you actually can have horizons in SR.
Now whenever you have (any type of) an event horizon you see it radiating particles at a certain temperature. Because where one sees a horizon is observer dependent the radiation is as well. Choosing a special observer (the infinitely far away one), using his point of view to construct something and than trying to assign some global physical meaning to it makes no sense. There are no preferred reference frames in the universe.

@thoughtfully:
Are you sure you actually see no radiation? Originally that's what I though too. But "normal" black holes should obey the censorship hypothesis, so no matter where you are and how you are falling you should see a horizon and thus some nonzero temperature. The unruh argument doesn't really work here, because it has no singularities.

[gmalivuk]

We're talking about surface gravity here. Because the size of the black hole goes with M and the gravitational acceleration with MR^-2 the surface gravity is actually inversely proportional to the mass, if we use the analogy from the unruh radiation the temperature is proportional to the acceleration thus inversely proportional to the mass and everything works out in the end.

Except at the event horizon of *any* black hole, the surface gravity is "infinite", in the sense that no finite force outward will prevent a massive object from falling in.

[SU3SU2U1]

But "normal" black holes should obey the censorship hypothesis, so no matter where you are and how you are falling you should see a horizon and thus some nonzero temperature. The unruh argument doesn't really work here, because it has no singularities.

These statements aren't quite right. An observer falling in to a black hole won't see the horizon, everything will look locally flat. Try changing from the Schwarzchild coordinates to the coordinates of an infalling observer. Also, the unruh observer will see a Rindler horizon- which is very much related to the radiation he sees.

The reason this happens is because of a strange facet of quantum field theory. The interesting thing about the theory is that it doesn't come with particles at the start- you have to make some decisions to define what a particle is (in particular, you have to choose a Feynman propagator by separating positive and negative energies). Lorentz observers can all agree what a "particle" means. However- one VERY interesting thing is that arbitrary observers can't. Thats the basis for the Unruh effect and the basis for the Hawking radiation. An infalling observer and an observer stationary at infinity can't agree on what a particle is.

[eSOANEM]

Choosing a special observer (the infinitely far away one), using his point of view to construct something and than trying to assign some global physical meaning to it makes no sense. There are no preferred reference frames in the universe.

I presume this addressed at me and my discussion about absolute and apparent horizons.

The absolute horizon does use an observer at infinity in its derivation however it is never working in his frame.

The only assumption it makes in claiming to be absolute, is that if an observer at infinity (alternatively an arbitrarily distant observer) is not reached by light rays leaving a surface in an infinite period of time, then no observer a finite distance from the surface will have been reached in a finite period of time. This is true for all asymptotically flat universes and should provide a good approximation in approximately asymptotically flat universes (like ours appears to be).

So, as spacetime as we observe is pretty close to asymptotically flat, it seems reasonable to assume the absolute horizon is observer independent.

[Yakk]

Now, a freefalling observer as it passes the event horizon won't see anything special right where they are -- but I suspect they'll see the event horizon warp around themselves and close off behind them?

In front of them ... won't there be something weird? From an external observers perspective, they will take forever to cross the event horizon, but they will move slower and slower (and cross it after "ticking" an asymptotically finite number of ticks, assuming they are broadcasting a clock of some kind).

I'm trying to talk about the time where you are "crossing the event horizon" -- putting a clock at your front and at your rear -- but I suspect what happens is, even if there is nearly zero tidal forces, when your front clock crosses the event horizon and your back clock hasn't, by the time your back clock gets the front clock's tick it has crossed the event horizon. Hence "you are moving at c" -- in a sense, the light is actually going forward, and you actually "catch up" to the light from the front clock rather than it coming towards you. And to the observer at infinity, the front clock slows down more rapidly than the back clock, and they never get the front clock's "final tick" from the other side of the event horizon.

I really should look at some of those "freefalling into black hole" videos again.

[gmalivuk]

Now, a freefalling observer as it passes the event horizon won't see anything special right where they are -- but I suspect they'll see the event horizon warp around themselves and close off behind them?

No, even when you cross the horizon in freefall you'll still only see the black hole as taking up some finite amount of the visual field in front of you.

[Yakk]

No event horizon behind you? I think I got where my intuition went wrong -- you are always going to be followed by some light falling in behind you. So there can't be an event horizon back there, eh?

[starslayer]

There is, but it's a coordinate effect; the hole itself still appears to be below you. The link I posted shows this in at least one of its videos. Where you can see something like what you described happen is if you slowly lower yourself into the black hole; if you do that, relativistic beaming concentrates your field of vision into a small cone in front of you, and you see an ever smaller disk of light in front of you surrounded by blackness. What's actually happening is that since you are moving very close to c, the beaming is allowing you to see what's behind you, namely, the black hole.

[Diadem]

You can still see the rest of the universe from inside a black hole. Nothing forbids light from falling into a black hole, after all. And you still can't see anything deeper into the black hole, because you can't just not escape from a black hole, you must fall down. I'm honestly not quite sure what the view from inside a black hole looks like. I assume one region of space is black, while another looks normal, but I don't know the relative sizes of those regions.

Interestingly, the above is only true for a non-rotating black hole. A rotating black hole has two horizons. The outer one is the point beyond which you can't return, and beyond which everything must fall down towards the centre of the black hole. But rotating black holes have an inner horizon too, beyond which space goes back to normal. In this region of space you can move around freely in all directions, except that you can't pass the inner horizon again to go out.

In this region of space you'd be able to observe the singularity. With a good enough starship and enough energy reverses, you might even live there for a while, observing what is going on. Though if I recall correctly gravity in this region is strong enough to overcome any known other force, so you'd need some pretty impressive propulsion system.

[eSOANEM]

I'm trying to talk about the time where you are "crossing the event horizon" -- putting a clock at your front and at your rear -- but I suspect what happens is, even if there is nearly zero tidal forces, when your front clock crosses the event horizon and your back clock hasn't, by the time your back clock gets the front clock's tick it has crossed the event horizon. Hence "you are moving at c" -- in a sense, the light is actually going forward, and you actually "catch up" to the light from the front clock rather than it coming towards you. And to the observer at infinity, the front clock slows down more rapidly than the back clock, and they never get the front clock's "final tick" from the other side of the event horizon.

My intuition tells me that you cannot see the clock in front of you cross the horizon for the same reason a distant observer won't. I am very confused about what will happen because of this. In my head I have an image of the front clock appearing to fly back towards you but I have no idea how accurate this is.

I think I need to sketch the 1-dimensional case to get my intuition a bit better.

[Yakk]

You can see it -- but when you see the light on the front of your ship from the far side of the event horizon, it is because you have already crossed the event horizon and "caught up" to the light.

The space-time you are "sitting" on as you cross the event horizon is moving at c towards the singularity. So the light moving backwards doesn't have to leave the event horizon to move towards you at speed c -- instead, it moves towards the singularity, while moving relative to local space towards you at speed c.

[eSOANEM]

That sounds a lot more reasonable than the situation I was imagining.

That said, shouldn't it be possible to approach the horizon at arbitrarily low speeds (I'm talking possibility here rather than whether it would ever be even remotely practical to build the drive necessary to do so)? In such a case, shouldn't it be possible to notice the front failing to slip over the horizon because you're still outside it.

As I say, I need to sketch the diagram but it's late here, I will do so in the morning.

[Yakk]

No. One way of thinking about the event horizon is the spot where space time is moving at c towards the singularity. You can only move at c relative to local space time... So you cannot enter at an arbitrarily slow speed. Of course this is just a just so story -- the real story is do the math.

[starslayer]

No. One way of thinking about the event horizon is the spot where space time is moving at c towards the singularity. You can only move at c relative to local space time... So you cannot enter at an arbitrarily slow speed. Of course this is just a just so story -- the real story is do the math.

That still means you can enter at an arbitrarily slow speed, though. At the event horizon, if you're firing your rockets at 0.99999c or whatever, you will only move towards the singularity at 0.00001c. The key point is that you have no choice but to move towards the singularity, although you can still try to fight the flow of spacetime.

[eSOANEM]

Ok, I've sketched the diagrams now, a freefalling observer will definitely not see the horizon so there won't be any weird illusions going on with the bow. The only thing I'm seeing from my diagram is a slight stretch but that's probably an error in my sketch, if it's not, it'll just be the tidal effects. Now, for a non-freefalling observer entering a black hole (i.e. one approaching the horizon at a constant rate (or trying to), this is also the case.

I guess I need to train my intuition better.