Black holes

[sol invictus]

Another seemingly consistent possibility is that the probe will never fall into the singularity, but will be blasted to pieces by the Hawking radiation of all the matter that had fallen in earlier, which gets converted to energy and hurled back at the probe before it has a chance to fall in itself.

So long as you never jump into the hole yourself, that description is consistent with every experiment you'll ever be able to do.

This would have been made possible by the fact that the horizon is not a real event horizon (as evidenced by black hole evaporation) and so it's not really the case that all paths must inevitably lead into the singularity.

What? No, that's not the reason. The blasting to pieces happens outside the horizon, and there's no rule that "all paths must lead to the singularity" outside the horizon.

This would imply that the probe, from its viewpoint as it falls, would see the black hole under it accelerate in time and its Hawking radiation increase, until the probe sees the black hole for the explosion that it really is and becomes part of the explosion, instead of reaching the singularity.

Nope. The probe sees little or no Hawking radiation. From its point of view, it falls through the horizon with nothing special happening and gets crushed at the singularity. Surprisingly, that statement is consistent with the one above.

 

[Ziggurat]

But coordinates exist in which it never crosses the horizon. In those coordinates, it gets frozen at the larger (final) radius of the event horizon.

I can't really say that this answer is wrong, but it doesn't make sense to me. One can watch a probe get very close to the event horizon. Whatever coordinates one wishes to use, it LOOKS really close. If lots of mass is then added so that the event horizon expands significantly, what do we observe? Because we can expand the event horizon past where we saw the probe. So unless the probe appears to get pushed outwards when we do so (and maybe that's what happens, I don't know), then I don't understand how we can't observe it crossing the event horizon during such an expansion. And if we do observe it crossing the event horizon, then it would seem that any coordinate system where it never does would be a bad coordinate system. Does the probe really get pushed outwards from out perspective? Or am I missing something here?

 

[sol invictus]

I can't really say that this answer is wrong, but it doesn't make sense to me. One can watch a probe get very close to the event horizon. Whatever coordinates one wishes to use, it LOOKS really close. If lots of mass is then added so that the event horizon expands significantly, what do we observe? Because we can expand the event horizon past where we saw the probe. So unless the probe appears to get pushed outwards when we do so (and maybe that's what happens, I don't know), then I don't understand how we can't observe it crossing the event horizon during such an expansion. And if we do observe it crossing the event horizon, then it would seem that any coordinate system where it never does would be a bad coordinate system. Does the probe really get pushed outwards from out perspective? Or am I missing something here?

From outside we can never observe anything crossing the event horizon, by definition. Remember that event horizons are not local - their location and size cannot be determined by any local measurement, because they depends on the future.

We cannot really "expand the event horizon past where we saw the probe", because if we saw it inside the event horizon, we too would be inside the event horizon (and the premise of the question was that we remain outside). The event horizon is a null (speed of light) surface, not a fixed radius. A signal inside it never reaches infinity.

The areal radius (the radius defined in terms of the area of a sphere surrounding the hole) is something that at least conceivably could be measured by the probe as it falls and sent out as a radio report that is received by a distant observer. We could ask whether the sequence of reported areal radii will be monotonically decreasing. If not, you could say that the probe had been "pushed out" as the horizon grew. But I'm almost sure it would be monotonic (I could check) assuming the probe is freely falling into the hole.

You ask instead whether the last areal radius report received must always be greater than the final horizon radius. The answer to that is clearly "no" - the probe could just sit at the origin sending signals before the hole forms at all, in which case all reports received will be "radius=zero" - but such results can be interpreted as meaning the signal was sent at an early stage in the hole's evolution (like before it formed) and passed through the collapsing matter before it accumulated and made a black hole.

 

[Ziggurat]

So what would we observe if we put an object between two black holes, and the two black holes collided? What would we see happen to the object?

 

[sol invictus]

So what would we observe if we put an object between two black holes, and the two black holes collided? What would we see happen to the object?

Its signals would get increasingly redshifted, and eventually disappear entirely.

 

[Ziggurat]

Its signals would get increasingly redshifted, and eventually disappear entirely.

I'm not sure how that squares with the idea that you can't observe anything cross the event horizon. Does the horizon change discontinuously during collision? Because otherwise, it seems that it shouldn't actually disappear.

 

[Caper]

I clicked on this topic expecting it to be about Stephen Hawking visiting strip clubs.

 

[sol invictus]

I'm not sure how that squares with the idea that you can't observe anything cross the event horizon. Does the horizon change discontinuously during collision? Because otherwise, it seems that it shouldn't actually disappear.

Nothing is discontinuous. The signals you'd see were emitted before the object crossed the event horizon.

A simpler situation is a black hole that formed at some time, for example from a spherically symmetric, thin shell of collapsing matter or radiation. At some time, the event horizon nucleates at the origin and expands outward at the speed of light, just in time to meet the collapsing shell at the moment it crosses its own Schwarzschild radius.

Suppose a probe is sitting at the origin, emitting beeps. Beeps emitted before the horizon nucleated reach infinity. Beeps emitted after do not. Beeps emitted very near the time the horizon appeared are very red shifted and take a very long time to get to infinity.

 

[Ziggurat]

Nothing is discontinuous. The signals you'd see were emitted before the object crossed the event horizon.

But if you stopped seeing a signal, how is that different from seeing the object cross the event horizon? Something's not adding up for me.

A simpler situation is a black hole that formed at some time, for example from a spherically symmetric, thin shell of collapsing matter or radiation. At some time, the event horizon nucleates at the origin and expands outward at the speed of light, just in time to meet the collapsing shell at the moment it crosses its own Schwarzschild radius.

Suppose a probe is sitting at the origin, emitting beeps. Beeps emitted before the horizon nucleated reach infinity. Beeps emitted after do not. Beeps emitted very near the time the horizon appeared are very red shifted and take a very long time to get to infinity.

But you never stop seeing the probe, it just gets more and more red shifted. But in the scenario I gave, you're claiming that we stop seeing the probe. I can't reconcile these things yet. I get the simple example, but I want to get the more complex example, and so far I don't.

I think I might now get my earlier question about the uniformly expanding black hole. It seemed strange that the probe would appear to get pushed outwards, but on reflection, that's consistent: if we drop a probe at the front end of a moving black hole, the black hole will appear to carry the probe along with it, in front of the event horizon. Getting pushed outwards doesn't seem any different, even if it isn't intuitive.

So I'll tell you what I suspect happens in the scenario I described earlier. I believe the event horizons of the two approaching black holes will stretch out towards one another and eventually connect in the middle. The probe will appear to be pushed outwards in a ring around where these two black holes connect and merge.

 

[sol invictus]

But if you stopped seeing a signal, how is that different from seeing the object cross the event horizon? Something's not adding up for me.

You don't see it as it crosses, or after it crosses - you see it before it crosses. Maybe it's just a semantic issue?

But you never stop seeing the probe, it just gets more and more red shifted. But in the scenario I gave, you're claiming that we stop seeing the probe. I can't reconcile these things yet.

You really stop seeing it. At some point, it emits the last photon/beep before it crosses. That photon/beep arrives at some finite time, and after that, there are never any more.

Moreover, there's Hawking radiation. I (and others) maintain that the Hawking photons you detect would, if you could somehow measure them accurately enough, indicate that the object burned up near the event horizon, thermalized with it, and then was re-radiated. But that's a considerably more subtle issue than what we've been discussing.

I think I might now get my earlier question about the uniformly expanding black hole. It seemed strange that the probe would appear to get pushed outwards, but on reflection, that's consistent: if we drop a probe at the front end of a moving black hole, the black hole will appear to carry the probe along with it, in front of the event horizon. Getting pushed outwards doesn't seem any different, even if it isn't intuitive.

Except that's not what happens. You can see that quite easily from the conformal diagram.

So I'll tell you what I suspect happens in the scenario I described earlier. I believe the event horizons of the two approaching black holes will stretch out towards one another and eventually connect in the middle. The probe will appear to be pushed outwards in a ring around where these two black holes connect and merge.

It clearly doesn't do that in any real sense. A solid object cannot turn into a ring without something pretty drastic happening to it, but in your scenario all the tidal forces can be very weak. And if the probe is reporting its position or acceleration, it will not report anything consistent with expanding into a ring.

I suppose it might temporarily appear that way as some kind of optical illusion/lensing effect before it gets sucked in, but I don't see how.

 

[Ziggurat]

You really stop seeing it. At some point, it emits the last photon/beep before it crosses. That photon/beep arrives at some finite time, and after that, there are never any more.

In that sense, a probe falling into an ordinary Schwarzchild black hole will disappear too. But I'm trying to figure out what happens in the classical limit, where we don't care about quantization of light. And in that limit, the probe will never truly disappear, but only get dimmer and dimmer.

Except that's not what happens. You can see that quite easily from the conformal diagram.

I get that, but I'm specifically interested in what it looks like from the outside.

It clearly doesn't do that in any real sense.

But that's still what it will appear to do, yes? I'm not looking at this as anything more than an optical effect, but that's still what I've been trying to figure out.

I suppose it might temporarily appear that way as some kind of optical illusion/lensing effect before it gets sucked in, but I don't see how.

Well, how is that any different than a probe falling straight in appearing to get "stuck" at the horizon? You said it yourself: "From outside we can never observe anything crossing the event horizon". If it disappears (classically), then we observed it cross. The only way I can see to not observe it cross is for it to appear to split into a ring, even if that never actually happens.

Again, ignore quantum effects for the moment, it's OK if quantum stuff throws a monkey wrench into everything, I'm still interested in the classical result.

 

[sol invictus]

But that's still what it will appear to do, yes? I'm not looking at this as anything more than an optical effect, but that's still what I've been trying to figure out.

I assume you have in mind that it appears to expand into a ring, rather than simply having its middle disappear? If so, I don't see why or how that would happen. If instead you simply mean that the middle will get dimmer faster than the edges, then sure.

Well, how is that any different than a probe falling straight in appearing to get "stuck" at the horizon? You said it yourself: "From outside we can never observe anything crossing the event horizon". If it disappears (classically), then we observed it cross. The only way I can see to not observe it cross is for it to appear to split into a ring, even if that never actually happens.

I really can't see why you think that. It can simply fade out - there's no reason for it to split into a ring.

 

[Brian-M]

But you never stop seeing the probe, it just gets more and more red shifted. But in the scenario I gave, you're claiming that we stop seeing the probe. I can't reconcile these things yet. I get the simple example, but I want to get the more complex example, and so far I don't.

This was cleared up earlier on in the thread. The idea that you never stop seeing something that falls into a black-hole is an outdated one from before they started applying quantum theory to the problem. When something falls into a black hole, it gets severely redshifted and quickly disappears.

But even without the quantum theory, it'd quickly become redshifted to the point that none of the photons are in the visible spectrum. And after a while it'd be redshifted to the point that none of the photons are even in the radio frequency range. Eventually it'd be redshifted to the point you'd have no practical way of building a receiver to observe it.

For all practical purposes, it'd disappear. But theoretically it'd still be detectable.

When you add quantum theory back into the mix, by the time it's disappeared for all practical purposes, it's also theoretically non-detectable.

So the only difference between purely classical theory and including quantum theory on the question of whether or not you'd be able to see an object long after it's fallen into a black-hole is the hypothetical point of whether or not you'd still be able to detect it's presence with some kind of device that'd probably need an antenna the size of a galaxy to pick up the ultra-ultra-ultra-ultra-ultra low frequency redshifted photons.

(Or at least, that's how I understand it. As always, I'm no expert and could be completely wrong.)

 

[Ziggurat]

I assume you have in mind that it appears to expand into a ring, rather than simply having its middle disappear?

Yes.

If so, I don't see why or how that would happen.

Because at a certain point, light from the near side of the probe can still reach you, but light from the far side will get sucked in because the black holes will have gotten closer together during the extra transit time.

And if it doesn't appear to expand, then we will have seen it cross the event horizon. Which I'm accepting that we cannot do. So I see no way out of a contradiction except to have the probe appear to expand into a ring. If we collide the black holes along the z axis, then our solution must be azimuthally symmetric, and I see no other way to do it.

I really can't see why you think that. It can simply fade out - there's no reason for it to split into a ring.

Classically you can't fade out, because there's no classical lower limit to the amount of radiation emitted or detected. You can get arbitrarily dim, but you can't actually disappear unless you've actually crossed the event horizon.

But even if you don't choose to go that route, even if you want to stick with quantum mechanics, you can still crank up the brightness of the probe. I see no reason to think that the probe must disappear from fading before we are confronted with the probe either appearing to split into a ring or crossing the event horizon.

 

[Ziggurat]

This was cleared up earlier on in the thread. The idea that you never stop seeing something that falls into a black-hole is an outdated one from before they started applying quantum theory to the problem.

That doesn't matter to me. I'm still interested in the classical result. Even if it's wrong, there still should be a classical result, and that's what I want to know.

 

[sol invictus]

Because at a certain point, light from the near side of the probe can still reach you, but light from the far side will get sucked in because the black holes will have gotten closer together during the extra transit time.

No, light from the far side will simply reach you later (and be more redshifted). That doesn't make the object look like a ring.

And if it doesn't appear to expand, then we will have seen it cross the event horizon.

Why? It sounds like you think there's a moment when the horizon has moved past the object or part of the object, and therefore we will stop receiving light from it. That's not how it works. Instead, you keep receiving light from it indefinitely (ignoring QM), it just keeps getting fainter and more red.

So I see no way out of a contradiction except to have the probe appear to expand into a ring. If we collide the black holes along the z axis, then our solution must be azimuthally symmetric, and I see no other way to do it.

I still have no idea what you think the "contradiction" is. Can you please be very specific? What exactly do you think is contradictory about always seeing the object as (say) a sphere, that simply gets dimmer and dimmer?

Classically you can't fade out, because there's no classical lower limit to the amount of radiation emitted or detected. You can get arbitrarily dim, but you can't actually disappear unless you've actually crossed the event horizon.

We're not communicating. By "fade out", I mean it gets dimmer and dimmer with time. That's all. If you want to ignore QM and pretend it emits continuous radiation, that's fine.

 

[Ziggurat]

No, light from the far side will simply reach you later (and be more redshifted).

No, sol, it won't. There will be a point in time when light emitted simultaneously from the far side and the near side cannot both reach you. The only way that could not happen is if the event horizons didn't connect in the middle, but rather repulsed each other. But that doesn't make any sense.

Why? It sounds like you think there's a moment when the horizon has moved past the object or part of the object

In the right frame of reference, that's exactly what happens, isn't it? When we're in, say Kruskal coordinates, the front end of the object passes the event horizon first, the back end later. So if in the falling frame we simultaneously emit light from the front and the back when the event horizon bisects out probe, then only the light from the back will leave.

So too here: the middle will get swallowed first, in the right frame. And in that frame, light from the sides can still escape. But that doesn't mean that it can escape in any direction. It cannot escape through the middle, even though it can still escape. So some signals will only be visible from one side.

and therefore we will stop receiving light from it. That's not how it works. Instead, you keep receiving light from it indefinitely (ignoring QM), it just keeps getting fainter and more red.

That does not suffice. I'm not saying that light isn't received indefinitely, but that doesn't mean it must be received indefinitely in any direction. That's plainly true even in the simple infalling case.

I still have no idea what you think the "contradiction" is. Can you please be very specific?

Classically, an object cannot disappear from view unless it's inside the event horizon. So either we see the object forever, or we see it pass the event horizon. If we cannot see it pass the event horizon, then we will see it forever. But I can find no possible answer for what it will look like when the two black holes merge except for it appearing to expand into a ring. So what am I wrong about? Am I wrong that classically we can see it forever? Or is there some solution other than appearing to turn into a ring which will let us see it forever?

What exactly do you think is contradictory about always seeing the object as (say) a sphere, that simply gets dimmer and dimmer?

How can we see it as a sphere? The solution must be azimuthally symmetric, because the setup of the problem is azimuthally symmetric. Unless this sphere appears to be wrapped around the entire merged black hole (which really makes no sense), I don't see how you can get a spherical probe to be azimuthally symmetric after the black holes merge.

We're not communicating. By "fade out", I mean it gets dimmer and dimmer with time. That's all. If you want to ignore QM and pretend it emits continuous radiation, that's fine.

Yes, that's exactly the setup I want to consider.

 

[sol invictus]

No, sol, it won't. There will be a point in time when light emitted simultaneously from the far side and the near side cannot both reach you.

"Simultaneously" in what frame? And why does it matter if the light was emitted simultaneously in any case? I thought you wanted the object to emit classical radiation continuously.

So too here: the middle will get swallowed first, in the right frame. And in that frame, light from the sides can still escape. But that doesn't mean that it can escape in any direction. It cannot escape through the middle, even though it can still escape. So some signals will only be visible from one side.

Nope. In your classical approximation, light emitted from the middle will continuously arrive at a distant observation point forever. You're missing something here, but I can't be sure what it is. Maybe you're forgetting that light takes longer to get out the nearer it comes to being swallowed by the horizon?

Classically, an object cannot disappear from view unless it's inside the event horizon. So either we see the object forever, or we see it pass the event horizon. If we cannot see it pass the event horizon, then we will see it forever. But I can find no possible answer for what it will look like when the two black holes merge except for it appearing to expand into a ring. So what am I wrong about? Am I wrong that classically we can see it forever? Or is there some solution other than appearing to turn into a ring which will let us see it forever?

The latter. It looks like a sphere (or perhaps an ellipsoid, depending on its actual shape) forever.

How can we see it as a sphere? The solution must be azimuthally symmetric, because the setup of the problem is azimuthally symmetric. Unless this sphere appears to be wrapped around the entire merged black hole (which really makes no sense), I don't see how you can get a spherical probe to be azimuthally symmetric after the black holes merge.

What?? I'm really at a loss here. How is a sphere not azimuthally symmetric? (In reality, if the probe is really spherical it might look ellipsoidal in cross section - I'm not sure.)

Look - do you agree that you will always receive light from the exact center? If not, there's your problem. You always will, because (if you like) light emitted right at the moment the horizon reaches the center will reach a distant observer after infinite time, and light emitted right before that moment will reach a distant observer after a very long time.

It occurs to me that perhaps you're thinking of viewing this thing from a different angle than I am. I've been assuming you're looking from a direction perpendicular to the axis connecting the two holes, so you can see what's between them. It's true that if instead the probe was directly behind one of the holes, it could get lensed into a ring. But that has nothing to do with the two black holes merging, nor does it require the object be anywhere near the horizon.

 

[Farsight]

Sorry to have ben away guys, I've been tied up with work on the house. I'll try to catch up.

 

[Farsight]

This is where you reamin wrong - no one is ignoring Schwarzschild coordinates ("mathematics "). What is happening is that you are ignoring all of the other coordinate systems.

I'm not. I've explained the issue with Kruskal–Szekeres coordinates repeatedly. Surely you understand it by now? The infalling observer is measured a reducing clock rate whilst the rate of his own processes are similarly reducing. He doesn't notice that the clock rate / process rate is getting slower and slower.

When you ignore all coordinate systems as GR does then you get the real situation where external observers measure time dilation as a clock approaches the event horizon while an observer with the clock measures no time dilation.

Yes, he doesn't measure any time dilation, because he's time dilated too. And in the end the time dilation goes infinite, like it does for the hypotherical observer travelling at c. And then he doesn't measure anything. If he fell into the black hole a billion years ago, he hasn't passed the event horizon yet. And there is no future date at which that you can claim that he has. His finite proper time is infinite with respect to time measured by observers in the universe at large, so the events that you would describe him observing never happen.

If you use a specific coordinate system then you get the results specific to that coordinate system where there may or may not be a singularity at the event horizon. That is why that is called a coordinate singularity. It is not real - it is an artifect of the coordinate system.

No coordinate system is real, they're all artefacts of measurement. And if you end up not measuring anything, it's a nonsense to claim that it's OK to use some coordinate system that pretends that those measurments are happening when they aren't.

One more time:
Alice decides to use Schwarzschild coordinates and she measures that there is a singularity at the event horizon.
Bob decides to use Kruskal–Szekeres coordinates and he measures that nothing special happens at the event horizon.

One more time: Bob is a popsicle, and has been that way for a billion years. He isn't measuring anything.