Communication Below The Event Horizon

azoundria

New member
The event horizon of a black hole is the point at which radio communication or other forms of radiation such as light cannot escape. However, this point changes as you approach the black hole.

According to Wikipedia "all event horizons appear to be some distance away from any observer, and objects sent towards an event horizon never appear to cross it from the sending observer's point of view".

In addition, "other objects that had entered the horizon along the same radial path but at an earlier time would appear below the observer but still above the visual position of the horizon, and if they had fallen in recently enough the observer could exchange messages with them before either one was destroyed by the gravitational singularity".

If you are at a position outside of the event horizon, you can communicate with a position just outside the event horizon. However, if I understand correctly, from that position the event horizon will be different (closer to the black hole). Therefore, this position could communicate with something even closer to the black hole, just outside of the closer event horizon.

Couldn't that chaining process then be continued to get communication closer and closer into the black hole?

jangofett287

Heat shield 'tester'
No.
By "appear to be some distance away from any observer" they mean that the event horizon is a non-zero distance. This distance can be infinitely small. You just can't observe something cross an event horizon in finite your-time.
Get ready, this is where relativity gets fun.

Let's consider 2 spaceships and a black hole.
One spaceship sets off towards the black hole. Let's say this spaceship has a big clock on the outside. From it's point of view, the clock ticks normally, the ship flies normally, the black hole's apparent size gets bigger as you'd expect and he crosses the event horizon never to be seen again.
From the other spaceship, which stayed still, the first spaceship's clock will appear to slow down asymptotically, which is to say that it will start slowing down and keep slowing down, but never quite stop. If the second spaceship measures the first spaceship's velocity relative to the black hole they will find the same thing. If they try to communicate with the first ship they'll find the communication travel time takes longer and longer exponentially, even though the first ship appears to be slowing down asymptotically.
What if the second spaceship had a clock, and the first spaceship (the one going into the hole) looked back at it?
Well they would see that clock speed up exponentially, even as their own clock kept running normally.

qbit

New member
I've often wondered (knowing very little about astrophysics and relativity) if you crossed the event horizon of a black hole, if you would experience falling into it for an amount of time approaching infinity. Would people watching you fall in see it happen in "normal" time and the people falling in would feel like they are falling forever?

Excuse my ignorance on this, but what is the theoretical basis that if a human fell into a black hole that they would be crushed? I know it seems like the obvious answer due to what we understand about a singularity, but is it possible that spacetime is curved such that we retain our "quantum configuration" but are compressed way beyond the planck constant? Thus actually continuing to live and function as normal in the singularity? This would be consistent with the endless falling (if that's even what would be experienced).

Paul

Urwumpe

Not funny anymore
Donator
https://en.wikipedia.org/wiki/Tidal_force And you're pulled apart, not crushed. Basically if your going down feet first eventually your feet are experiencing noticeably more gravity than your head, because the gravitational gradient of a black hole is so steep. Eventually the difference gets so high that you're pulled apart. For more info see https://en.wikipedia.org/wiki/Spaghettification

The fun really starts when you start to consider the spacecraft to approach a super-massive black hole at a significant portion of the speed of light...

azoundria

New member
I'm not sure if this answers my question. Or at least, I'm still confused.

What I understood from Wikipedia was that, even if the event horizon is crossed, aka we go beyond that event horizon from the standpoint of a stationary outside observer, we still never reach it. In other words, it's a little like chasing a rainbow where you can never get to the bottom of it. From the standpoint of a ship entering a black hole, they never reach the event horizon, because if I understand, the event horizon moves as you approach the black hole, because your perspective changes.

I also understood that two ships which had both crossed the event horizon could communicate with one another.

Therefore, based on this, you could postulate that if a massive ring of ships were all to enter the event horizon, we would have circular communication between all ships around the entire loop. For that moment of time, which I agree would be heavily distorted, there would be no restriction to communication between those ships. From the standpoint of each ship, it hasn't yet reached the event horizon, and it can communicate with the others at the same time pace.

Now, supposing that one ship entered a small fraction of a second later. Would it then be excluded from this ring of communication entirely? Obviously, time would flow faster, however, this could be adjusted for.

Also according to Wikipedia, "other objects that had entered the horizon along the same radial path but at an earlier time would appear below the observer but still above the visual position of the horizon, and if they had fallen in recently enough the observer could exchange messages with them before either one was destroyed by the gravitational singularity". This makes me believe that it wouldn't.

Therefore, if each ship in our ring is instead offset only slightly from the other (a spiral of sorts) and all falling together into the black hole, why wouldn't this allow us to get communication past the event horizon? At what particular point would it break down?

perseus

The space agencies proliferate, soon there will be trips, to black holes. I recommend paying only the outward journey, and pay the return trip, once at home, your heirs will thank you, hehe

azoundria

New member
The space agencies proliferate, soon there will be trips, to black holes. I recommend paying only the outward journey, and pay the return trip, once at home, your heirs will thank you, hehe
That's bound to be the trip of a lifetime for sure.

Ravenous

Donator
Donator
I'm not sure if this answers my question. Or at least, I'm still confused.

What I understood from Wikipedia was that, even if the event horizon is crossed, aka we go beyond that event horizon from the standpoint of a stationary outside observer, we still never reach it.

What I believe this means, is when you get close to the horizon its apparent position (apparent to you) depends on where you are (and/or on your velocity)

Even when you're "under" the horizon, you can still look up and see what's above you. Your view is distorted, because space is highly curved in this region, but you can still see incoming light.

As you continue to fall you'll be able to see a (narrow?) region of sky above you, but you will never feel like you've crossed the horizon. (Those of us safely in orbit won't see this effect though.)

Of course slightly after all this, you'll hit the centre and that will be messy, though luckily it will be an infinitely small mess. If the tidal forces don't kill you on the way in of course.

Urwumpe

Not funny anymore
Donator
Of course slightly after all this, you'll hit the centre and that will be messy, though luckily it will be an infinitely small mess. If the tidal forces don't kill you on the way in of course.

Well, the more fascinating question is: Can a naked singularity actually exist in reality or can we only see a forming black hole, frozen 99.999% along the way because of time dilation? If we fall towards it, do we see a naked singularity or are we seeing the collapse?

And if we see it form, could we maybe get inside the singularity BEFORE it actually forms?

Linguofreak

Well-known member
I'm not sure if this answers my question. Or at least, I'm still confused.

What I understood from Wikipedia was that, even if the event horizon is crossed, aka we go beyond that event horizon from the standpoint of a stationary outside observer, we still never reach it. In other words, it's a little like chasing a rainbow where you can never get to the bottom of it. From the standpoint of a ship entering a black hole, they never reach the event horizon, because if I understand, the event horizon moves as you approach the black hole, because your perspective changes.

I also understood that two ships which had both crossed the event horizon could communicate with one another.

Therefore, based on this, you could postulate that if a massive ring of ships were all to enter the event horizon, we would have circular communication between all ships around the entire loop. For that moment of time, which I agree would be heavily distorted, there would be no restriction to communication between those ships. From the standpoint of each ship, it hasn't yet reached the event horizon, and it can communicate with the others at the same time pace.

Now, supposing that one ship entered a small fraction of a second later. Would it then be excluded from this ring of communication entirely? Obviously, time would flow faster, however, this could be adjusted for.

Also according to Wikipedia, "other objects that had entered the horizon along the same radial path but at an earlier time would appear below the observer but still above the visual position of the horizon, and if they had fallen in recently enough the observer could exchange messages with them before either one was destroyed by the gravitational singularity". This makes me believe that it wouldn't.

Therefore, if each ship in our ring is instead offset only slightly from the other (a spiral of sorts) and all falling together into the black hole, why wouldn't this allow us to get communication past the event horizon? At what particular point would it break down?

If A and B are falling into a black hole, and A is just outside and B is just inside the event horizon, they can exchange messages for a while, but B's first message to A, sent while A is still outside the horizon, won't reach A until A crosses the horizon and is thus out of contact with the outside world. If A manages to stop short of the horizon and escape, none of B's messages will reach him, since they can't cross the event horizon moving outwards.

n72.75

Tutorial Publisher
Donator
I think part of the problem with these discussions of falling into black holes is that "observer" refers to a point in space, not a human that occupies multiple points in space.

Conceivably you could make a black hole that was large enough tidal forces would not kill you at the event horizon, however the temporal gradient would pose a massive problem.

Andy44

owner: Oil Creek Astronautix
I think part of the problem with these discussions of falling into black holes is that "observer" refers to a point in space, not a human that occupies multiple points in space.

Conceivably you could make a black hole that was large enough tidal forces would not kill you at the event horizon, however the temporal gradient would pose a massive problem.

IIRC from Hawking's book such black holes do exist.

I also seem to remember him saying that if a man were magically transported into a black hole, he would instantly transport infinitely into the future, while an outside observer with the magic ability to see the man would see the man appear to freeze in time.

So if you were to go in head first and look back at your feet, I am guessing you'd see your lower body age and whither and decay to dust at a rate which increases with the distance from your face, which sounds pretty painful.

Linguofreak

Well-known member
So if you were to go in head first and look back at your feet, I am guessing you'd see your lower body age and whither and decay to dust at a rate which increases with the distance from your face, which sounds pretty painful.

Nope. You'd only get such a stark difference in rate of time between head and feet if there were extreme tidal forces between your head and feet, such that you'd be dead in a millisecond anyways.

The structure of General Relativity is such that our entire galaxy could have crossed the event horizon of a trillion-light-year wide black hole yesterday and we'd be totally unable to tell that we weren't in flat space from any experiment we could perform within the confines of our galaxy (of course, the outside universe would look very different, so we know that isn't the case). In fact, for a sufficiently massive black hole (or, equivalently, for an experimental setup on a sufficiently small scale in space and time) the spacetime arbitrarily deep into a black hole (relative to its radius) will look completely flat and normal, except at the singularity itself, so we could be 999,999,999,999/1,000,000,000,000ths of the way from the event horizon to the singularity of a truly enormous black hole (say with a radius of Graham's Number light years) and not notice anything out of the ordinary for trillions of years on end.

Of course, that's assuming General Relativity. Depending what gravity ends up looking like when we take quantum mechanics into account (which is still an area of active research as General Relativity and quantum mechanics don't play well together mathematically within the limits of our current mathematical knowledge), things could end up looking quite different. Assuming that the idea of Hawking Radiation is correct, I think it is quite likely that we will find that black holes proper never actually form, but that objects that look arbitrarily close to one from the outside do, and that when falling towards such an object, an infalling observer will be incinerated by Hawking Radiation, whatever the mass of the object. (Black holes of much size at all don't give of much radiation as seen from the outside, but an infalling observer would observe the radiation to become blueshifted, both because of their inward velocity and because there would be less and less gravitational resist between them and the area just above the horizon where the radiation originates).

Fizyk

Member
I think the main misunderstanding here comes from the fact that things look quite different for free-falling observers and for observers that are at rest with respect to the black hole.

The free-falling observers won't really notice anything when crossing the event horizon (as long as they are small enough compared to the size of the black hole). This is a direct consequence of the fundamental assumptions behind relativity - that all inertial observers experience the same laws of physics and that gravitational forces are (locally) indistinguishable from inertial forces. Free-falling observer is an inertial one and if it's small enough, then everything is local.

If the observer becomes too big, we can start having problems, as different parts of the observer will accelerate at different rates. This is what would cause the effect called "spaghettification" for a human falling into a stellar-mass black hole. We are, however, small enough not to experience any unpleasantries when falling into supermassive black holes.

Things start looking quite differently if the observer tries to stay at rest outside of the black hole. The acceleration necessary to do that becomes infinite at the horizon - which means that there is a region near the horizon in which the tidal forces will be lethal for anyone trying to hang out there. It's this huge acceleration that you have to experience that makes all the difference.

While you are firing your engines at full thrust in order to stay above the horizon, you will indeed be unable to see anything fall inside the black hole. Everything that falls in will become increasingly redshifted to you and eventually "frozen" on the horizon, with no communication with it being possible (it will receive your signals, but its responses will never reach you).

If you turn off the engines, however, you will begin free falling - which means that you will fall below the horizon, eventually, and you will be able to receive all the responses that weren't able to reach you before. The horizon "disappears" from your perspective, as I described above. You and everything that fell before you are just objects in space that looks like empty, flat space (again: locally - you will notice tidal forces at large enough distances). Of course there is this slight issue of a singularity in your near future, but if you disregard that, it's really like floating in empty space.

By the way - you don't need a black hole to have an event horizon. Surprisingly, a constantly accelerating observer in a space with no gravity will also "see" an event horizon at a distance of $\frac{c^2}{a}$ behind them. Anything that is sent from the other side of that horizon will never reach the observer - unless they stop accelerating, that is. Then there is no horizon again. It's a bit weird, but this is how it works.

Urwumpe

Not funny anymore
Donator
I think the main misunderstanding here comes from the fact that things look quite different for free-falling observers and for observers that are at rest with respect to the black hole.

Do you know any kind of scientific article about the formation of black holes? I only know approximations that pretend the black holes to be existing from big bang on, but none that deals with the space-time consequences of a collapsing star.

Fizyk

Member
Do you know any kind of scientific article about the formation of black holes? I only know approximations that pretend the black holes to be existing from big bang on, but none that deals with the space-time consequences of a collapsing star.
No, unfortunately. I'm a bit out of touch with the scientific literature.

One thing I remember on that topic, though, was that you could replace a spherically symmetric region of the Schwarzschild solution with a solution for a region filled with matter, and it was possible to do that in a way that satisfied the Einstein's equations in the whole space-time (on the boundary between the solutions as well). You can draw a few conclusions for that:
1. Spherically symmetric changes to the distribution of mass in the central body don't matter to the solution for the outside.
2. Drawing further from that, if a star were to collapse in a spherically symmetric manner, we'd just have more and more spacetime "taken" by the Schwarzschild solution, and less and less "taken" by the one with matter.
3. It turns out that you can make some calculations about the pressure inside the star, and if the star is compressed in a region with a radius smaller than 9/8 of the Schwarzschild radius, the pressure in the center turns out to be infinite - which means that you can't have stable objects smaller than that.

I can try googling for some sources about that in the afternoon :tiphat:

Linguofreak

Well-known member
I think the main misunderstanding here comes from the fact that things look quite different for free-falling observers and for observers that are at rest with respect to the black hole.

One thing I've been musing over is the different ways the distance to the horizon can be treated. If we have a chain of yardsticks from infinity to the horizon that is, for a brief instant, both at zero velocity and in freefall everywhere along its length, the length of that chain, if I'm not mistaken, will actually be infinite. Of course, after any time at all passes the chain will cease to be stationary, and, due to tidal forces it will either break up or cease to be in freefall everywhere (or both), but we can use the positions of the yardsticks during the instant that they were stationary to define a spatial coordinate system.

If mark space with that coordinate system and drop an observer in, they will see the spacing of the radial coordinates length contracted as they accelerate inwards, until, when they hit the horizon, the coordinate spacing will be zero (as they are moving at c relative to the frame the coordinates are drawn in).

If we mark space with the same coordinates and have the observer hover just over the horizon, they will see the coordinates get more and more closely spaced as they look closer to the horizon, and will see the spacing drop to zero at a point (the event horizon) as far below them as the acceleration horizon they would observe if accelerating in free space with the same acceleration as they are using to hover.

---------- Post added at 19:47 ---------- Previous post was at 19:46 ----------

No, unfortunately. I'm a bit out of touch with the scientific literature.

One thing I remember on that topic, though, was that you could replace a spherically symmetric region of the Schwarzschild solution with a solution for a region filled with matter, and it was possible to do that in a way that satisfied the Einstein's equations in the whole space-time (on the boundary between the solutions as well). You can draw a few conclusions for that:
1. Spherically symmetric changes to the distribution of mass in the central body don't matter to the solution for the outside.
2. Drawing further from that, if a star were to collapse in a spherically symmetric manner, we'd just have more and more spacetime "taken" by the Schwarzschild solution, and less and less "taken" by the one with matter.
3. It turns out that you can make some calculations about the pressure inside the star, and if the star is compressed in a region with a radius smaller than 9/8 of the Schwarzschild radius, the pressure in the center turns out to be infinite - which means that you can't have stable objects smaller than that.

I can try googling for some sources about that in the afternoon :tiphat:

Of course, the big thing missing here is that a real black hole would almost certainly more resemble a Kerr rather than a Schwarzschild hole.

Fizyk

Member
One thing I've been musing over is the different ways the distance to the horizon can be treated. If we have a chain of yardsticks from infinity to the horizon that is, for a brief instant, both at zero velocity and in freefall everywhere along its length, the length of that chain, if I'm not mistaken, will actually be infinite. Of course, after any time at all passes the chain will cease to be stationary, and, due to tidal forces it will either break up or cease to be in freefall everywhere (or both), but we can use the positions of the yardsticks during the instant that they were stationary to define a spatial coordinate system.
I don't think being in freefall matters here really - we should just be able to say that we have a chain of stationary yardsticks (the last yardstick above the horizon will actually have to end some distance from the horizon, because being exactly stationary on the horizon is equivalent to moving at c, which no massive object can do). This does define a sensible radial coordinate.

If mark space with that coordinate system and drop an observer in, they will see the spacing of the radial coordinates length contracted as they accelerate inwards, until, when they hit the horizon, the coordinate spacing will be zero (as they are moving at c relative to the frame the coordinates are drawn in).
Sounds reasonable, but I have to admit I'm not really sure about it. The big problem here is that we have to define how the length of the sticks is being measured, as it's not really obvious in a curved space-time.

If we mark space with the same coordinates and have the observer hover just over the horizon, they will see the coordinates get more and more closely spaced as they look closer to the horizon, and will see the spacing drop to zero at a point (the event horizon) as far below them as the acceleration horizon they would observe if accelerating in free space with the same acceleration as they are using to hover.
Now this I'm even less sure about. Why would the coordinates (or ends of the sticks) be spaced closer and closer near the horizon to a hovering observer?
I can't see any reason for that happening (but again - how do we measure distances?).

Of course, the big thing missing here is that a real black hole would almost certainly more resemble a Kerr rather than a Schwarzschild hole.
Yes, that's true. Schwarzschild black holes probably don't even exist at all (nothing has angular momentum exactly equal to 0).

EDIT:
I can try googling for some sources about that in the afternoon :tiphat:
As it turns out, there is a pretty nice article on Wikipedia: https://en.wikipedia.org/wiki/Interior_Schwarzschild_metric

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