Clarifying Black Hole Horizons: An Examination of Observer Perspectives

[wabbit]

Your problem, as others have told you is that you are thinking of the horizon as some sort of local surface, like a barrier or something of that nature.

Not really, I was actually saying the Schwarzschild horizon doesn't exist for an infalling observer. But yes I was referring to the specific surface defined by a value of the Schwarzschild radial coordinate (calling this Schwarzschild horizon now to avoid ambiguity) and trying to figure out how different observers experience or describe this surface, especially as they fall across it.

Well, anyway, I should probably just calculate some trajectories to lay to rest those head and feet misgivings.

 

[bcrowell]

This may indeed be the source of my confusion, thanks for pointing this out. I was using the "light doesn't escape" definition (e.g. in my feet to head example) but also probably others without making a clear distinction. (I also referred to the Schwarzschild horizon as "the black hole horizon", distinct from the no-light-escape horizon defined by a given observer. )

Let's stick with light doesn't escape then. I will try to be more specific and to always say "Schwarzschild horizon" for the usual one, and "observer X horizon" for the light-doesn't-escape horizon as defined by observer X.

You need to spell out in more detail what you mean by "light doesn't escape." The standard definition is that light doesn't escape to null infinity. That's not an observer-dependent definition.

 

[bcrowell]

No, I am imagining an observer jumping off a hovering spacecraft and free-falling in across the horizon.

But that shredding, I do find hard to reconcile with "nearly Minkowski", it means that if I fall across infalling, and start a 1g rocket just as my head is above but my feet inside(*), I will be torn apart ?

(*)above/inside the Schwarzschild horizon as usually defined, corresponding to a fixed radial Schwarzschild coordinate.

In order to hover, you need some external force, such as the force from the deck of a spaceship pressing upward on your feet. That force is the force that destroys your body.

 

[wabbit]

You need to spell out in more detail what you mean by "light doesn't escape." The standard definition is that light doesn't escape to null infinity. That's not an observer-dependent definition.

Right. Actually as I realized after that I was using something slightly different, and this may simply be incorrect : that light doesn't travel from inside to outside. This is what the feet to head contradicts from the viewpoint of the infalling observer, but if this property doesn't hold (locally) than there is no issue, and this looks like the source of my confusion.

 

[wabbit]

In order to hover, you need some external force, such as the force from the deck of a spaceship pressing upward on your feet. That force is the force that destroys your body.

I understand. It's just hard to imagine a 1g force ripping my body apart in near-flat spacetime (well, not quite flat in that example, but something I am accustomed to here on earth) - like jumping off into free fall from 50cm above the Schwarzschild horizon, moving apart at 1g from said spacecraft , and then quickly (after just 60cm fall) , grabbing hold of the spacecraft with my hand, my shoulder will be torn apart. Maybe I just need to get used to it.

Actually 1g is rather strong, we might as well say 0.1g

Unless the g-field is highly inhomogenous there, so that my feet are pulled in much more strongly - but I don't think that's the case near the Schwarzschild horizon of a large black hole (I am relying on the EP/"nothing special happens as a free falling observer crosses the horizon" here. This is valid I presume if the spacetime curvature can be approximated by a constant g field in a small (a few meters) region that crosses the horizon.)

 

[wabbit]

One last question, hopefully specific enough to be either confirmed or infirmed unambiguously: I thought that the metric in a small region crossing the Schwarzschild horizon of a very large black hole is very close to corresponding to a constant g field. So a hovering observer at constant Schwarzschild radial and angular coordinates in this region is equivalent to a Rindler observer accelerating at that same g in Minkowski spacetime. I am assumimg that (locally) his Rindler horizon coincides with the black hole Schwarzschild horizon. Is this wrong?

 

[Khashishi]

I understand. It's just hard to imagine a 1g force ripping my body apart in near-flat spacetime (well, not quite flat in that example, but something I am accustomed to here on earth) - like jumping off into free fall from 50cm above the Schwarzschild horizon, moving apart at 1g from said spacecraft , and then quickly (after just 60cm fall) , grabbing hold of the spacecraft with my hand, my shoulder will be torn apart. Maybe I just need to get used to it.

If you are stationary near the Schwarzschild horizon, then you are extremely height-contracted. Or, in your perspective, the event horizon will be very far away. So you need to be more clear on what you mean by 50cm.

 

[wabbit]

If you are stationary near the Schwarzschild horizon, then you are extremely height-contracted. Or, in your perspective, the event horizon will be very far away. So you need to be more clear on what you mean by 50cm.

Yes ! this was ambiguous in my formulation, perhaps this is the key.
As I see it, from the viewpoint of the hovering observer, the Schwarzschild horizon is at an infinite proper distance - an object free falling away from him will take an infinite time (proper time of the hoverer) to reach the horizon. This is correct?
From the free falling observer after he jumps off, that Schwarzschild horizon is however a finite proper distance away. This is the distance (measured by the falling observer in his own frame, by timing light signals or even using a ruler I guess), that I am using to define those 50 cm. Those 50 cm are a very long (even infinite if it crosses the horizon ) distance from the viewpoint of the hovering observer.

So I should not have said "hovering 50cm above the horizon". This is meaningless, no one can hover at a finite distance (as measured by him) over the horizon. This should be instead "hovering at a point where the proper distance of a free falling observer to the horizon would be 50cm".

Maybe this changes things, I need to think about this point. Thanks for mentionning that.

 

[Haelfix]

Again, in this business it's really important to only ask questions that make sense based on what an observer has access too. A free falling observer does not have access to the location of the horizon. Only the observer at infinity, using charts that cover the whole spacetime can make that analysis and decide what did or did not happen. Of course that observer sees a large amount of red shift as the infalling observer is squashed against the horizon.

Nevertheless he can deduce based on his knowledge of GR that the observer must have crossed the horizon, where it met him at the speed of light and then passed him and that from this observers point of view, nothing special happened.

Now, a slightly more refined problem often given to students as homework, is the following experiment. Take the point of view of a hovering 'static' observer, on his rocket ship. Now give this person a long fishing pole (lets make it a light year across) and let him dip the pole down through the horizon (that observer won't know exactly where the horizon is, but let's say we arrange it so that he gets lucky). What happens?

 

[wabbit]

OK I would say it must take an infinite (static observer's proper) time for the pole to reach the horizon, so the static hoverer will just see the end of the pole moving farther and farther away as he dips it assuming the "dipping" is just letting the pole slip in his hand in free fall, or that he exerts a force to slow down the dipping.

But regarding the previous case I don't agree that the observer "at infinity" can know that the free falling observer has crossed the horizon since this only happens at infinite proper time for him, which is to say never. Is this incorrect?

 

[Nugatory]

But regarding the previous case I don't agree that the observer "at infinity" can know that the free falling observer has crossed the horizon since this only happens at infinite proper time for him, which is to say never. Is this incorrect?

The event which is the infaller crossing the horizon is not on the worldline of the observer at infinity, so there is no meaningful way to associate it with any proper time on that worldline. It's no more correct to say that it happens "at infinite proper time" than any other time.

However, there are two interesting events on the worldline of the observer at infinity:

One is the last point at which the observer at infinity will be able to send a radio message to the infaller and expect a reply. If sent before that point the message will reach the infaller before he has passed the horizon, so the infaller's reply will eventually (possibly many millennia later) make it back to the observer at infinity. Messages sent after that point will still reach the infaller, but the infaller will already be inside the horizon and the infaller's reply will not make it back out.

The second interesting point is the last point at which a message sent by the observer at infinity will reach the infaller at all; messages sent after that point will not reach the infaller before the infaller reaches the singularity. Imagine that the infaller is your beloved spouse, thrown into the black hole by some appalling mechanical failure on your spaceship hovering outside the black hole. There is no saving your spouse once the horizon is crossed, but you would wish for him/her to hear your promise of undying love before their inevitable death in the singularity... You do not have infinite time to get that message transmitted, in fact you don't have very long at all before it is too late and your spouse will die alone and uncomforted by any parting words from you.

You could reasonably choose any point on your worldline after the first to be the time when the horizon crossing "happens" and any point after the second to be when the infaller's meeting with the central singularity "happens". There's no need to wait for infinite time to pass on your worldline.

 

[PeterDonis]

Because of their definition, absolute horizons won't exist for evaporating black holes.

That's not correct--at least, it's not correct for the simplest model of an evaporating black hole, the one Hawking originally constructed. A Penrose diagram for this model is here:

http://en.wikipedia.org/wiki/Black_hole_information_paradox#Hawking_radiation

The triangular region at the upper left, with the singularity (sawtooth line) at the top, is all a black hole region in the absolute sense; no event in this region can send light signals to future null infinity. The boundary of this region (the 45 degree line going up and to the right from r = 0 to the right end of the singularity) is the absolute horizon.

There are other models of "black hole" evaporation where there is no absolute horizon--but putting the term "black hole" in quotes emphasizes the fact that in these models, there is no black hole--no singularity as well as no absolute horizon. In these models, there are only apparent horizons that exist temporarily but later disappear. However, we don't know yet which kind of model is the correct one.

 

[PeterDonis]

If you are stationary near the Schwarzschild horizon, then you are extremely height-contracted. Or, in your perspective, the event horizon will be very far away.

This is not correct. The proper distance to the horizon for a stationary observer is finite, and goes to zero as the $r$ coordinate of the stationary observer goes to $2M$. Don't misinterpret the meaning of the $(1 - 2M / r)^{-1}$ in the line element; it doesn't mean that distances get "infinitely stretched" as the horizon is approached, even though it looks that way on the surface. You need to actually compute the integral from some finite $r > 2M$ to $r = 2M$; you will see that it has the properties I described above.

 

[PeterDonis]

from the viewpoint of the hovering observer, the Schwarzschild horizon is at an infinite proper distance - an object free falling away from him will take an infinite time (proper time of the hoverer) to reach the horizon. This is correct?

As Nugatory said, there is no invariant meaning to "how much time" it takes for the infalling object to reach the horizon (though he also pointed out a very reasonable criterion for assigning a finite time to that event--what proper time the hovering observer would have to send a light signal in order for it to reach the infalling object before it crosses the horizon). There is, however, an invariant meaning to the proper distance from the hovering observer to the horizon; it's the integral I described in my previous post, in response to Khashishi. As I noted in that post, the proper distance is not infinite.

 

[wabbit]

As Nugatory said, there is no invariant meaning to "how much time" it takes for the infalling object to reach the horizon (though he also pointed out a very reasonable criterion for assigning a finite time to that event--what proper time the hovering observer would have to send a light signal in order for it to reach the infalling object before it crosses the horizon). There is, however, an invariant meaning to the proper distance from the hovering observer to the horizon; it's the integral I described in my previous post, in response to Khashishi. As I noted in that post, the proper distance is not infinite.

For the time, I was not assuming an invariant meaning, but a coordinate chart defined by the observer using as time his proper time. If I understand what you and Nugatory are saying, there is no matural choice of such a chart. I was implicitly thinking of time and distance as measured by the observer, extending his local chart as far as he can, and presuming that for the hovering observer this chart would extend to the Schwarzschild horizon and no further, and would (unambiguously) assign an infinite time (which I was calling "proper time of the hovering observer") to the event of the infalling observer crossing the Schwarzschild horizon.

I prefer to stay with a hovering observer than using the observer at infinity, for concreteness, though my understanding is that Nugatory's statements about the latter would also apply to the former.

Annway, at least now I understand that I don't understand anything here. Oh well.

 

[PeterDonis]

If I understand what you and Nugatory are saying, there is no natural choice of such a chart.

Correct.

I was implicitly thinking of time and distance as measured by the observer, extending his local chart as far as he can

This is one possible way of doing it, yes. But it has limitations; see below.

presuming that for the hovering observer this chart would extend to the Schwarzschild horizon and no further

To be precise, it is singular at the horizon, so it only covers the region outside the horizon.

would (unambiguously) assign an infinite time (which I was calling "proper time of the hovering observer") to the event of the infalling observer crossing the Schwarzschild horizon.

No, that's not what this chart does. As above, this chart is singular at the horizon, which means it cannot assign a "time" to events on the horizon at all. The chart is simply not well-defined there. To assign "time" to events on the horizon at all, you must switch to a chart that is not singular there.

Pop science presentations often are sloppy about this, saying that "infinite time" is assigned to events on the horizon. This just illustrates why you can't learn science from pop science presentations.

(Note, btw, that even though this chart is singular at the horizon, we can still take limits as the horizon is approached, and so we can still verify some calculations. For example, we can compute the proper time for a free-falling object to go from some radius $r_1$ to some smaller radius $r_2$; then we can take the limit of this as $r_2 \rightarrow 2M$ and find that the limit is finite, indicating that the free-falling object will reach the horizon in a finite proper time by its own clock. Or we can compute the proper distance between two hovering observers at radius $r_1$ and $r_2$, and take the limit of that as $r_2 \rightarrow 2M$, and find that the limit is finite, indicating that any hovering observer has only a finite proper distance to the horizon. So we can justify the statements made earlier in this thread about such things even without switching charts, strictly speaking. But of course it's a lot easier to see how things work at the horizon if you switch to a chart that's not singular there.)

 

[jartsa]

I am confused about black hole horizons and such common statements as "light cannot escape from inside the horizon".

The way I currently understand it is as follows :

1. Horizons are always relative to an observer, and what is called "the black hole horizon" is just a shorthand for "the black hole horizon relative to a hovering observer", i.e. one with fixed spatial Schwarzschild coordinates. For a large back hole, this is essentially his Rindler horizon.

2. For other observers, the black hole horizon is different and may not exist.

3. For a free falling observer headed directly towards the singularity (constant angular Schwarzschild coordinates), I am guessing there is a naked singularity - and that light can in fact escape from the Schwarzschild interior region (as defined by the hovering observer) from his viewpoint, as the usual descrption of freefall across the horizon suggests.

4. For an oberver in free fall on a circular orbit, I wonder if there is a horizon, perhaps the same as for the hovering observer ?

I am looking for clarification as to which of the above statements are correct, and if not in which way they are wrong.

Thanks for your help.

When observer hovers far from event horizon, far away from him there seems to be a place where things seem to freeze. When the observer starts free falling down, he sees those frozen things stay frozen, except for things that are hovering, observer sees those things start to slowly unfreeze when he starts to free fall.

When observer hovers near an event horizon, he sees the place where everything seems to be frozen to be close to him. When said observer starts to free fall down, he sees the frozen things start to unfreeze quickly. Nothing seems to stay frozen in this case, for some odd reason ... probably has something to do with frozen light unfreezing.

 

[wabbit]

The event which is the infaller crossing the horizon is not on the worldline of the observer at infinity, so there is no meaningful way to associate it with any proper time on that worldline. It's no more correct to say that it happens "at infinite proper time" than any other time.

However, there are two interesting events on the worldline of the observer at infinity:

One is the last point at which the observer at infinity will be able to send a radio message to the infaller and expect a reply. If sent before that point the message will reach the infaller before he has passed the horizon, so the infaller's reply will eventually (possibly many millennia later) make it back to the observer at infinity. Messages sent after that point will still reach the infaller, but the infaller will already be inside the horizon and the infaller's reply will not make it back out.
.

I did a calculation for the round trip of a photon from the hovering observer to some point which can be the inflalling observer reflecting that photon, and I find that time (proper time of the hoverer) is infinite. So if this is correct, any time in his future, the hoverer will still be receiving messages saying "not crossed yet, getting closer".

There may be errors here of course - here it goes :

Taking $c=1$ and considering only trajectories with constant angular Schwarzschild coordinates, r and t being the remaining Schwarzschild coordinates,
$$d\tau^2=(1-\frac{r_s}{r})dt^2-(1-\frac{r_s}{r})^{-1}dr^2$$
For a photon moving radially at $r>r_s$, this gives $$\frac{dt}{dr}=\epsilon\frac{r}{r-r_s}$$where the $\epsilon=+1$ is for a photon moving toward infinity, $\epsilon=-1$ for one moving toward the hole.
Integrating this yields
$$t-t_0=\epsilon\left(r-r_0+r_s\ln\frac{r-r_s}{r_0-r_s}\right)$$
For a photon emitted at $(r_0,t_0)$ toward the hole, reflected at $(r_1,t_1)$ and received back at $(r_0,t_2)$ this gives
$$\Delta t=_{def}t_1-t_0=t_2-t_1=\frac{t_2-t_0}{2}=r_0-r_1+r_s\ln\frac{r_0-r_s}{r_1-r_s}$$
And $lim_{r_1\rightarrow r_s^+}\Delta t=+\infty$

On the other hand, for a hovering observer at $r_0$, $$\frac{d\tau_{hov}}{dt}=\sqrt{1-\frac{r_s}{r_0}}$$ so $\Delta \tau_{hov}=\sqrt{1-\frac{r_s}{r_0}}\Delta t$ which also tend to infinity as $r_1\rightarrow r_s$

 

[PeterDonis]

I did a calculation for the round trip of a photon from the hovering observer to some point which can be the inflalling observer reflecting that photon, and I find that time (proper time of the hoverer) is infinite.

Yes, but think about what this calculation is telling you. What it is telling you is that, if the hovering observer sends light signals down to the infalling observer, who reflects them back, then as the infalling observer approaches the horizon, the time the hovering observer has to wait to receive the return signal increases without bound.

But now, suppose that the hovering observer sends light signals which are "time stamped" with the time of emission; and suppose that, when the infalling observer reflects the signals, he adds a second "time stamp" giving the time on his clock at the instant of reflection. Then, as the return signals come back, the hovering observer will see the two time stamps approach finite values, even though the time at which he receives the return signals increases without bound. The finite value approached by the first time stamp (the time of emission) is what Nugatory called "the last point at which the observer at infinity will be able to send a radio message to the infaller and expect a reply". The finite value approached by the second time stamp (the time of reflection) is the proper time on the infaller's clock at which he crosses the horizon.

 

[Nugatory]

did a calculation for the round trip of a photon from the hovering observer to some point which can be the infalling observer reflecting that photon, and I find that time (proper time of the hoverer) is infinite. So if this is correct, any time in his future, the hoverer will still be receiving messages saying "not crossed yet, getting closer".

The round trip time is always finite, although it becomes arbitrarily large as the infaller gets arbitrarily close to the horizon. When the infaller reaches the horizon, the round trip time isn't infinite - it's undefined, because the light signal doesn't make the round trip at all. But it is a great stretch, and one that I doubt you would accept in any other context, to say that because light from an event never reaches your eyes that event never happens (or doesn't happen until an infinite amount of time has passed).

Suppose that every five seconds you send a time stamped radio message to the infaller, and when he receives it he replies "I saw your message dated <whatever>". You send a sequence of messages reading "12:00:05", "12:00:10", "12:00:15", ... and you receive (perhaps millennia later) messages reading "I saw your 12:00:05 message", and "I saw your 12:00:10 message", but no matter how long you wait you will never ever receive a reply to your "12:00:15" message. I can reasonably describe this as the infaller passing through the horizon sometime between receiving my 12:00:10 message and my 12:00:15 message.

As for when these three events (infaller received 12:00:10, infaller passed horizon, infaller received 12:00:15) "happened"? None of them are on my worldline, so I have to map them to events that are on my wordline and to which I can assign proper times. So I choose a coordinate system that covers both my worldline and the infaller's worldline; note the time coordinate of the event on the infaller's worldline; then identify the point on my worldline that has the same time coordinate so happened "at the same time". Whatever my wristwatch reads at that point... that's "when" the event happened.
1) This is equivalent to the procedure that we follow in ordinary flat Minkowski spacetime for assigning times to events off our worldline. The only difference is that in the flat spacetime case, there is an obvious and natural choice of coordinate system, so we don't notice the extent to which "at the same time" is just a convention based on our choice of coordinate system. In curved spacetime, we don't have this luxury.
2) Schwarzschild coordinates are unusable for this purpose because they don't cover both worldlines.
3) The Kruskal time coordinate works just fine however, and there is no difficulty converting the Kruskal time coordinate of a point on the observer's worldine to the time displayed by the observer's wristwatch at that point. Of course, a different choice of coordinates would produce different results... which is why we say that simultaneity is a convention.

There is, however, a very great difference between saying that there is no single universally accepted definition of when the infaller's crossing the horizon "happens", and saying that it never "happens" or that it "happens" after infinite time.

 

[PeterDonis]

I can reasonably describe this as the infaller passing through the horizon sometime between receiving my 12:00:10 message and my 12:00:15 message.

And if you have the infaller add his own timestamp to his replies, e.g., "I saw your 12:00:05 message at 1:00:15", "I saw your 12:00:10 message at 1:00:30" (I'm just making those numbers up, but it's fairly simple to calculate realistic ones), then you can also reasonably say that the infaller crossed the horizon some time between 1:00:30, by his clock, and the time when he would have received the 12:00:15 message (which you can calculate by observing the time stamps he gives on previous replies--for the above data, it would be something like 1:00:45 by his clock).

 

[yoron]

"One way of stating the equivalence principle is that spacetime is always locally Minkowski. Therefore a local experiment like the one you're describing would not give an unusual result. As a free-falling observer, you see nothing special happening. You can see your feet at all times. An observer using Schwarzschild coordinates would explain this by saying that although the light from your feet didn't escape the event horizon, your eyes crossed the horizon in time to see that light."

Beautifully expressed Ben, and the way I like to think of it too. Remember wondering what speed one locally would reach at the instant crossing that event horizon once, finally deciding that it should hinge on ones definition of where the 'singularities center' is, just behind that event horizon? Or? Seems to exist different versions there.

although speeds may be a inconsistent definition as it is a geodesic (free fall) :) relativity is tricky.

 

[PeterDonis]

Remember wondering what speed one locally would reach at the instant crossing that event horizon once

Speed relative to what? Speed is always relative. The horizon is moving at $c$ relative to you when you cross it, but that's because the horizon is an outgoing lightlike surface; it's no different than a light ray moving at $c$ relative to you as it passes you. Relative to other objects, you can be moving at any speed up to $c$, depending on the details of the individual case, just as in any other local inertial frame.

finally deciding that it should hinge on ones definition of where the 'singularities center' is, just behind that event horizon?

If you are crossing the event horizon, the singularity is in your future. It is a moment of time, not a place in space.

 

[yoron]

Well "Speed relative to what? Speed is always relative. The horizon is moving at c relative to you when you cross it, but that's because the horizon is an outgoing lightlike surface; it's no different than a light ray moving at c relative to you as it passes you. Relative to other objects, you can be moving at any speed up to c, depending on the details of the individual case, just as in any other local inertial frame."

Yes, that's what I first though too when starting to wonder about it. But then I started to look up definitions of what a center should be :). The one you point out should then be equivalent to the one in where a singularities 'center' is 'everywhere', passing that event horizon, which in fact seem the one to be preferred as it seems the most straightforward. But I seem to remember other definitions too. And when it comes to uniform motions they all are observer dependent, unless we prove a 'gold standard' globally, I know that :) That's why I used 'local' writing about passing the event horizon.

The last one though? Space becomes time and time becomes space? Not sure I will agree on that one, think I've seen other definitions too?

One simple reason to why I think it is equivalent to a center being 'everywhere' is me assuming a initial acceleration for the in-falling 'observer', making him the one initiating a 'speed' relative the Black hole. In the case of there being no initiator, just geodesics converging to a center, speeds becomes a very confusing definition. In that case the only time a 'speed' exist it must be a result of a initial acceleration, locally defined. Relativity is filled with traps. then again, how was it we set light to a constant? Uniformly moving, no accelerations involved.

 

[wabbit]

"One way of stating the equivalence principle is that spacetime is always locally Minkowski. Therefore a local experiment like the one you're describing would not give an unusual result. As a free-falling observer, you see nothing special happening. You can see your feet at all times. An observer using Schwarzschild coordinates would explain this by saying that although the light from your feet didn't escape the event horizon, your eyes crossed the horizon in time to see that light."

Beautifully expressed Ben, and the way I like to think of it too. Remember wondering what speed one locally would reach at the instant crossing that event horizon once, finally deciding that it should hinge on ones definition of where the 'singularities center' is, just behind that event horizon? Or? Seems to exist different versions there.

although speeds may be a inconsistent definition as it is a geodesic (free fall) :) relativity is tricky.

Thanks, the explanation from the Schwarzschild observer is very natural.

It must be true however also from the viewpoint of the infalling observer since light doesn't cross the Schwarzschild horizon from inside to outside (I had a doubt about this but checking their path in Kruskall coordinates removed that doubt).

Looking at a Kruskal diagram I can indeed see light rays from my feet to my head, and they reach my eyes after the head has passed the Schwarzschild horizon, however "slowly" I am crossing doesn't matter. While I am not sure I would know how to translate it into equations, I find PeterDonis' mention that "the horizon is moving towards you at light speed" very evocative here at least.

The apparent paradox I was having trouble with seems to arise from that "slowly" which I was previously interpreting as something like / proportional to dr/dtau - which makes no sense since r isn't a valid coordinate across the horizon, and isn't even spatial inside.

 

[PeterDonis]

I started to look up definitions of what a center should be :).

That's not a good approach, because the singularity does not satisfy many such definitions. A much better approach is to look at the actual geometry of Schwarzschild spacetime and see where the singularity fits in it.

Space becomes time and time becomes space?

That's not what I said. I said the singularity is a moment of time, not a place in space. That does not mean space becomes time and time becomes space. It means your understanding of the actual geometry of this spacetime is incorrect.

Wabbit's advice to look at a Kruskal diagram is good; see, for example, here:

http://en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates

Note that the singularity (the blue hyperbola at the top of the region marked "II") is a spacelike line, not a timelike line--in other words, it's always more horizontal than vertical. That is a characteristic of a moment of time, not a place in space; a place in space will be described by a curve that is always more vertical than horizontal. Note also that there is no "switching of space and time"; timelike lines are always more vertical than horizontal, and spacelike lines are always more horizontal than vertical. (Light rays are always 45 degree lines, just as in standard spacetime diagrams in special relativity. That is one reason why the Kruskal diagram is so useful.)

It's true that pop science presentations often say things like "space and time switch places inside the horizon". That just illustrates why it's not a good idea to learn science from pop science presentations.

 

[wabbit]

Regarding the infinite time, I understand that there isn't a unique time coordinate for an observer, however in this context I find the midpoint of a photon round trip quite natural, and since however proper time is proportional to Schwarzschild t this fits all together rather nicely. Using this particular hoverer's time, the infalling observer never reaches the horizon (or reaches it at infinite time). This "never" is of course relative to the hoverer, but it does correspond to the fact that the horizon is not observable by him, ever. So it is more than a matter of coordinate choice, it reflects a property of the horizon relative to the hoverer.
So my perhaps idiosyncratic use of "an event that never happen for an observer" is "an event outside of any of that observer's future past lightcones". Obviously this is not the accepted meaning of the term, if any, so I'll try to keep it private now : )

 

[Asher Weinerman]

the infalling observer never reaches the horizon (or reaches it at infinite time). This "never" is of course relative to the hoverer,

The event which is the infaller crossing the horizon is not on the worldline of the observer at infinity, so there is no meaningful way to associate it with any proper time on that worldline. It's no more correct to say that it happens "at infinite proper time" than any other time.

I notice there is a lot of confusion regarding how long it takes to cross the event horizon as I tried to explore in a different thread.
Measuring time by bouncing "time-stamped light signals off an infalling observer may help with measuring proper time which of course will be finite, but it has little to do with the coordinate time t. Don't be scared of Schwartzchild coordinates - they are valid right up to but not including the event horizon. You can set up stationary clocks right up to but not including the event horizon, and these clocks will tell you that coordinate time increases without bound as the observer approaches the horizon. This is a valid result. You can transform into any coordinates you want such as Kruzkal Sekerez or whatever and say that now the "time" coordinate is finite, but you are just obscuring the physical meaning by using coordinates whose physical meaning is obscure.

I'm not an expert, but as indicated in the other thread, it is possible that horizon crossing may occur in finite coordinate time by realizing that the infalling object has non-zero mass, and therefore the problem loses its spherical symmetry and the horizon expands to engulf the infalling object as the mass of the black hole increases. I've been told that numerical simulations of this two-body problem indicate that coordinate time is finite for this to occur.

There are real problems with coordinate time t increasing without bound. Wabbit is correct in stating that these phenomena never occur, regardless of coordinate transformations which are just different ways of describing the same physical situation. In fact, no black holes could ever form because the matter could never cross the horizon to make the hole. Also they would evaporate before forming. Never is a long time. Don't be fooled by coordinate transformations.

 

[PAllen]

I notice there is a lot of confusion regarding how long it takes to cross the event horizon as I tried to explore in a different thread.
Measuring time by bouncing "time-stamped light signals off an infalling observer may help with measuring proper time which of course will be finite, but it has little to do with the coordinate time t. Don't be scared of Schwartzchild coordinates - they are valid right up to but not including the event horizon. You can set up stationary clocks right up to but not including the event horizon, and these clocks will tell you that coordinate time increases without bound as the observer approaches the horizon. This is a valid result. You can transform into any coordinates you want such as Kruzkal Sekerez or whatever and say that now the "time" coordinate is finite, but you are just obscuring the physical meaning by using coordinates whose physical meaning is obscure.

I'm not an expert, but as indicated in the other thread, it is possible that horizon crossing may occur in finite coordinate time by realizing that the infalling object has non-zero mass, and therefore the problem loses its spherical symmetry and the horizon expands to engulf the infalling object as the mass of the black hole increases. I've been told that numerical simulations of this two-body problem indicate that coordinate time is finite for this to occur.

There are real problems with coordinate time t increasing without bound. Wabbit is correct in stating that these phenomena never occur, regardless of coordinate transformations which are just different ways of describing the same physical situation. In fact, no black holes could ever form because the matter could never cross the horizon to make the hole. Also they would evaporate before forming. Never is a long time. Don't be fooled by coordinate transformations.

And what is your criteria for physically meaningful coordinates? Who chooses and how? The reality is that GR and SR are defined by the feature that physically meaningful quantities are invariants. One of the the 3 or so founding axioms of GR per Einstein (e.g. right from the beginning) was general covariance, which means (among other things) that any statement that is true only in one (or some) coordinates is not a physical statement at all.

By your (il)logic, a ball dropped from a uniformly accelerating rocket 'never' gets more than 1/a (with 'a' being the acceleration in natural coordinates) in distance from the rocket, because that is what is true in Rindler coordinates, which are the coordinates analogous to SC coordinates for a uniformly accelerating rocket.

 

[wabbit]

In fact, no black holes could ever form because the matter could never cross the horizon to make the hole.

Yes, I find this is a fun fact (using that "relative never" meaning) - though a (apparent, relative) horizon does form, and after a (short) time in the distant observer's view the situation is practically indistinguishable from that associated with a black hole. It does however also seem to mean that, relative to that distant hoverer, classically no information is ever lost in a Black hole (I am not claiming this relative to all observers, only to hoverers).