Black holes

[Farsight]

It's only infinite coordinate time in certain coordinate systems (like Schwarzchild coordinates). But in other coordinate systems (such as Kruskal–Szekeres coordinates), it happens in finite coordinate time as well. There isn't any reason to think that Schwarzchild coordinates are really more fundamental than other coordinates. In fact, given that the event horizon is a coordinate singularity in the Schwarzchild metric even though it's not a geometric singularity, there is in fact good reason to prefer other coordinates which do not have a coordinate singularity at the event horizon.

I beg to differ, Zig. See above plus previous posts of mine such as this. Kruskal–Szekeres coordinates ignore the end of time denoted by the (truncated) vertical peak in the Schwarzschild chart. The light cones get past it like it isn't there.

 

[Farsight]

Farsight, Hawking's word is not gospel and I have never treated it as gospel. I know that he has been wrong at least once (black holes & information) I stated that black holes are black (any light emitted within the event horizon cannot escape from the event horizon).

See post #163, I take it back, apologies.

There is little to discuss anout Hawking radiation.

It's one for another day, but think on this: if I'm right, Hawking radiation cannot exist.

Farsight, Don't pick a coordinate system!
Do not obsess on Schwarzschild coordinate system...

You've said this already in post 184, and now you're repeating it in 195. See above for my response.

 

[Farsight]

For simplicity, let's consider a simple Schwarzschild black hole, a supermassive one, so that tidal forces don't give us much trouble. We throw in a probe, and observe everything from a safe distance. (Unless specified otherwise, let all further observations be from the viewpoint of a distant observer.) The probe transmits as it approaches the horizon, reporting nothing out of the ordinary. Then its transmissions slow down and disappear. The probe is stuck there right above the event horizon, invisible, flattened and frozen in time. So far, so good.

Yes, so far so good. Some of the other guys will insist on telling the tale from the probe's viewpoint, forgetting that the probe's proper time takes infinite "universe time" to elapse.

Now, we grab a lot of matter, as much as the mass M of the black hole itself (it's a thought experiment, so let's not worry about the logistics) and throw it in as well. Yay! The black hole now has mass 2M and its horizon is considerably farther out.

Sounds as if you're asking how a black hole ever actually forms. That's a question that's been kicked around quite a lot actually, see for example about the formation and growth of black holes by Kevin Brown. He mentions the "frozen star" idea which I'm in favour of, though he doesn't favour it.

My first question is: from our viewpoint, where's the probe now? I see two major possibilities: either the probe is still where it was, i.e. at a point which is now below the horizon, or it actually moved out as the black hole's horizon expanded. Which is it?

My answer is below the horizon, wherein the black hole is a gravastar-like lump of frozen stress-energy. But note that the word "below" presents a difficulty, in that one cannot measure space and time in this region. In the wiki gravastar article it says This region is called a "gravitational vacuum", because it is a void in the fabric of space and time..

(I used to assume the former, but now I think the latter looks more likely.) - If the former is true, the logical consequence would be that the singularity is at the center of the black hole, surrounded by all the matter that fell in - our probe being a sample of it - that now fills the horizon. If the latter is true, the logical consequence would be that there is nothing inside the horizon, and everything that ever fell in, including the core of the original collapsing star, is still in a thin superdense bubble just outside the horizon, and the central singularity is not below the horizon, but exactly at it (from our distant viewpoint). Is the reasoning correct?

I don't think so. One could conceive of a Hokey Cokey scenario wherein the dance circle gets bigger as more people join it, and if you can't measure space and time within the circle you might well argue that a bubble is all you've got. But your reasoning doesn't cover the possibility of a frozen lump that gets bigger like a hailstone growing in a thunderstorm by accreting layers.

Now, let's put Hawking radiation in the picture. We get rid of CMBR (again, it's a thought experiment, so let's not worry about the logistics) and wait a really, really long time. The black hole should slowly radiate away. Let's wait until the black hole that remains is half the original mass (M/2). The event horizon has shrunk considerably. Regardless of where the probe was in question one, now it must be closer to the center - it has passed through where the event horizon had been at the time we threw the probe in.

Let's not put Hawking radiation in the picture. The infinite time dilation at the black hole event horizon means it takes forever for anything to happen down there. IMHO Hawking radiation ignores this. But going with the flow...

If during our superlong wait, we recorded everything, and then played it superfast, presumably there should be at least a little more transmission from the probe (albeit muffled by all the mass we threw in after the probe). My second question is: how much more would we hear from the probe? Would we just hear it get a little closer to the original location of the horizon at the time we threw it in, or would we actually hear it transmit about passing the original horizon location and falling further in, to where it is now? (The latter sounds more likely to me, but I'm not sure.)

I think it would be something like sending the probe down to a "hailstone" black hole, all the while seeing its transmission rate slow down until it appears to have stopped. It sticks to the surface, gets covered up and entombed within the hailstone. Then after a while the hailstone evaporates, and the probe is revealed. We will assume that the probe doesn't evaporate but the hailstone continues to do so. As it gets smaller our probe stays on the surface frozen, and then with a final flourish the hailstone evaporates completely, leaving our probe beeping normally. But having said all that, I don't think it's going to happen. Once that probe is in there, that's it, forever.

By now, a lot of the mass of the black hole has radiated away, and it is no longer there. My third question: can we tell which mass has disappeared and was converted to radiation? Was it the original collapsing star and the first of the infalling matter that, from our distant viewpoint, finally reached the singularity by this time? Was it the last of the infalling matter?

I would say that none of it is converted to radiation, but if you twisted my arm I'd say it was last in first out.

A related point to consider: if we replay in fast motion our recorded observation of the black hole radiating away, we realize that the black hole is essentially a huge ongoing explosion, slowed down in time from our distant viewpoint. My fourth question: what implications does this have for our probe? We threw our probe in, and since then, we've seen a huge amount of energy hurled back at us from the black hole. All this energy has already reached us by now, so does this mean the probe must have already encountered all this energy on its way down? Wouldn't it be incinerated by this enormous amount of energy? Was it already destroyed by now and itself returned to us in the form of Hawking radiation - or is it still hovering above the horizon, frozen in time, and all the energy radiated from the black hole shrinking underneath it somehow... went around it?

In truth the probe was destroyed at the event horizon, hammered flat by the total radial length contraction. Then it was destroyed again by being crushed under the subsequent infalling matter. Then if black holes did emit Hawking radiation, the probe would have been destroyed again by being radiated away. Things don't look too good for our probe!

Of course, it needn't be either-or; some of these questions may have answers I haven't thought of. I'll be glad to hear your thoughts.

A pleasure, big buy. Nice to see an intelligent thoughtful post by the way. You're a guy who thinks for himself.

 

[Farsight]

How can you tell? What mathematical, physical, or geometric test did you apply before making this determination?

What test do Schwarzschild coordinates pass, that Lemaitre coordinates fail?

(Is the test "Lemaitre coordinate systems fail because they don't prove me right, and I know I'm right"?)

They pass the does it ever actually happen? test. Tell the story of big guy's probe from the Lemaitre viewpoint, and you fail the fairytales in neverneverland test.

 

[Farsight]

Possibly, but for an outside observer, the probe hasn't crossed the event horizon, and will never have crossed it, no matter how long we wait. For the classical case, this would be the end of the story - the portion of the probe's flight below the horizon would simply lie beyond infinitely distant future for us, the remote observers. We couldn't hear transmissions from inside as we couldn't wait longer than infinitely long. For us, nothing could get out because nothing could get in, in the first place.

Good stuff, big guy. But think of the event horizon as the end of events. Then it's real enough.

But Hawking radiation screws everything up.

It does. We've got infinite time dilation such that events take forever to occur, and yet Hawking-radiation events occur. It's a contradiction in terms. And these Hawking radiation events demand this: "In order to preserve total energy, the particle that fell into the black hole must have had a negative energy". Pull the other one. There are no negative-energy particles.

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...

There is something else I haven't mentioned: that the probe loses its integrity and stops being matter on the way into the black hole.

Since both can't be true, I think theory should be able to rule out at least one of these possibilities. But I can't really tell which. Perhaps some of the experts here know more. And maybe this isn't even theoretically resolved yet and nobody knows. I wonder.

The one thing you can be sure of is that nobody knows for sure. Hawking radiation is often presented as a given, but we have no actual evidence for it. However we do have evidence that there are some very small very massive things out there, like at the centre of the galaxy, and they aren't shining.

 

[Farsight]

All: IMHO this is the thing to focus in on when trying to understand black holes:

...The event horizon is a null (speed of light) surface...

The coordinate speed of light varies in a non-inertial reference frame, like the room you're in. At the event horizon, judged by observers on Earth, the coordinate speed of light is zero. So if you could look though a gedanken telecope at an optical clock at the event horizon, you would see that it isn't ticking. An idealised optical clock features light moving between parallel mirrors, so it's the same for a light beam at the event horizon. You would see that it doesn't move. That's why it can't get out, not because it's fighting up through some "waterfall of infalling space". It's also the same for an observer at the event horizon. He isn't moving either, or thinking, and if light doesn't move, he isn't seeing. And of course since light from him doesn't move, you can't actually see him. Hence the gedanken telescope.

An important thing to note is that light can't get any slower than this. The coordinate speed of light can't get any lower, and gravitational time dilation can't get any more total. So there's no gradient in gravitational potential any more. Hence there's no local gravitational field, and things don't fall down towards the mooted point-singularity at the centre.

 

[edd]

Farsight - it's really not obvious that seeing a clock moving slowly (even to the point of stopping) is the same as that clock moving slowly or stopping locally.
That sort of time dilation in what you see would occur from cosmological redshift when I look at a distant galaxy, but you presumably wouldn't say that the clocks in a distant galaxy are running slowly. At least I'd hope not, as the situation there is symmetric and they would see our clocks running slowly.

 

[ALIENS]

iam afraid for black holes.

 

[Ziggurat]

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?

I think I know what I was missing now. I am absolutely correct that some light emitted from the far side cannot make it to the near side because the event horizon intervenes in the middle, but what I was missing is that light from earlier which passes through the middle just as the event horizon approaches will take an infinite amount of time to escape. So that's how we can continue to see the far side even though the event horizon intervenes.

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.)

A small sphere sitting in the x-y plane next to a big sphere at the origin is not azimuthally symmetric, because it's off to one side, and that side is different from all the others.

But I think I've figured out what I was missing: It looks like it's off to the side you view it at, but it will look that way from all sides. So if we orbit around the black hole in the x-y plane, the probe will appear to orbit with us. So in the sense that all viewing angles see something equivalent, we can recover azimuthal symmetry, even though it won't seem as such from any one viewpoint.

 

[sol invictus]

I think I know what I was missing now. I am absolutely correct that some light emitted from the far side cannot make it to the near side because the event horizon intervenes in the middle, but what I was missing is that light from earlier which passes through the middle just as the event horizon approaches will take an infinite amount of time to escape. So that's how we can continue to see the far side even though the event horizon intervenes.



A small sphere sitting in the x-y plane next to a big sphere at the origin is not azimuthally symmetric, because it's off to one side, and that side is different from all the others.

But I think I've figured out what I was missing: It looks like it's off to the side you view it at, but it will look that way from all sides. So if we orbit around the black hole in the x-y plane, the probe will appear to orbit with us. So in the sense that all viewing angles see something equivalent, we can recover azimuthal symmetry, even though it won't seem as such from any one viewpoint.

OK, although I still can't figure out exactly what situation you have in mind. Two black holes merging will form a bigger, spherical black hole centered on the point between them (assuming zero net momentum), so whatever probe was in between will just get swallowed up entirely. Viewed straight from the side, you'll see a fainter and fainter cross-section of the probe, probably distorted by lensing, but it won't appear to be a ring because you should always be able to see light propagating straight to you from the center.

Viewed from an angle a ring or partial ring is more possible, since (after all) lensing can turn spheres into rings when they are situated directly behind a massive lensing object. But again, that has little or nothing to do with things freezing on the horizon, it can happen even when the lens isn't a black hole.

We also haven't considered the fact that light can orbit the hole(s) once or several times before escaping in some direction, which might give rise to some pretty complicated optical effects as the merge takes place.

 

[ben m]

They pass the does it ever actually happen? test. Tell the story of big guy's probe from the Lemaitre viewpoint, and you fail the fairytales in neverneverland test.

I see my mistake now! I had forgotten to use circular reasoning. Thanks for setting me straight.

 

[Ziggurat]

OK, although I still can't figure out exactly what situation you have in mind. Two black holes merging will form a bigger, spherical black hole centered on the point between them (assuming zero net momentum), so whatever probe was in between will just get swallowed up entirely.

But it can't appear to get swallowed up, because that would mean that we would see it pass the event horizon, which is prohibited. So I've been trying to find out what it looks like, and the answer is bound to be strange.

Viewed from an angle a ring or partial ring is more possible, since (after all) lensing can turn spheres into rings when they are situated directly behind a massive lensing object. But again, that has little or nothing to do with things freezing on the horizon, it can happen even when the lens isn't a black hole.

The issue here is that the lens in question is weird. It's not a standard lensing effect, because the saddle configuration during a merger is very different from the spherical lensing of a static black hole.

We also haven't considered the fact that light can orbit the hole(s) once or several times before escaping in some direction, which might give rise to some pretty complicated optical effects as the merge takes place.

That's precisely the sort of thing I had in mind when I posited that it might appear to turn into a ring, though I don't think that's quite how it would play out anymore.

 

[Ziggurat]

I beg to differ, Zig. See above plus previous posts of mine such as this. Kruskal–Szekeres coordinates ignore the end of time denoted by the (truncated) vertical peak in the Schwarzschild chart. The light cones get past it like it isn't there.

What makes you so sure that it is there? Because the Schwarzchild coordinates say so? But there's no reason to favor them over any other coordinates. The event horizon is a coordinate singularity in the Schwarzchild metric, but that alone tells you something important: they're bad coordinates to use near the event horizon. You're taking a coordinate singularity and treating it like a genuine singularity, and that's a mistake. The event horizon is not a genuine singularity.

 

[Reality Check]

Farsight repeats his obsession with Schwarzschild coordinates

It's one for another day, but think on this: if I'm right, Hawking radiation cannot exist.

That is interestiung - you assert that you have the knowledge of GR and QM needed to show that Hawking radiation is impossible but you have displayed a total ignorance of one of the basic parts of GR (see below).

You've said this already in post 184, and now you're repeating it in 195. See above for my response.

You response just exposes your ignorance of GR yet again.
GR does not impose any specific coordinate system.
It is ignorant in the extreme to say that one coordinate system is somehow special.
Your obsession with Schwarzschild coordinates is such ignorance.

You are even ignorant about Schwarzschild coordinates ! There is no "end of time" in them.

 

[Thabiguy]

Thanks; the responses and the debate clarified some things, helped clear some misconceptions of mine and gave me a lot more to think about.

One thing I'd like to confirm: we all seem to easily agree that classically, when we throw in a probe, we won't receive any transmissions from it reporting about the probe flying through and past where the horizon was at the time we threw in the probe. Do I understand it correctly that this remains true even if we consider Hawking radiation - i.e. even if we wait long enough for the horizon to shrink and the entire black hole to evaporate, we still won't receive any transmissions reporting about the probe flying through and past where the horizon was at the time we threw it in? (I mean actual transmissions, no matter how red-shifted, not any quantum-forensic reconstructions from the Hawking radiation.)

 

[sol invictus]

But it can't appear to get swallowed up, because that would mean that we would see it pass the event horizon, which is prohibited.

Not at all - in reality, that's exactly how it will appear. It's only in a thought experiment where we ignore quantization of light and pretend we can detect arbitrarily faint and long-wavelength radiation that it doesn't.

The issue here is that the lens in question is weird. It's not a standard lensing effect, because the saddle configuration during a merger is very different from the spherical lensing of a static black hole.

Assuming the object is opaque and viewed straight from the side, you might see an ellipse in the center (the cross section of the front, lensed) AND some kind of hourglass shaped outline (the back, wrapped around).

 

[shadron]

I don't understand the spaghettification reference. I think I've heard it somewhere before as well, but not sure exactly what's meant by it.

Being ripped to small chunks I get. Vast tidal forces over very small distances will do that to you.

But presumably spaghettification is something different?

It is called spaghettification because tidal forces pull equally in opposite directions; that is why there is a tidal bulge on either side of the Earth, not just the one side close to the moon. If you were falling in and your head (or feet) was pointed towards the gravity source, your head and feet would be pulled apart by an inexorably increasing force until you were dismembered. But it doesn't stop there; each piece is likewise pulled apart, and then all those pieces are also pulled apart and so on, until you are a cloud of plasma.

No, no horizontal squeezing. There is no such component of the tidal force.

 

[Ziggurat]

Not at all - in reality, that's exactly how it will appear. It's only in a thought experiment where we ignore quantization of light and pretend we can detect arbitrarily faint and long-wavelength radiation that it doesn't.

Well, yes, but that classical scenario is explicitly what I'm considering.

 

[Ziggurat]

No, no horizontal squeezing. There is no such component of the tidal force.

Yes there is. For small objects, the horizontal forces scale as the square of the object size, whereas the longitudinal tidal forces scale linearly with the object size. And if our object is small compared to the length scales of our gravitational source, then the horizontal components becomes negligible, and we can generally ignore it. But it does exist. And when you get close enough to the singularity, the horizontal forces won't be negligible anymore, and there is significant squeezing.

 

[shadron]

Yes there is. For small objects, the horizontal forces scale as the square of the object size, whereas the longitudinal tidal forces scale linearly with the object size. And if our object is small compared to the length scales of our gravitational source, then the horizontal components becomes negligible, and we can generally ignore it. But it does exist. And when you get close enough to the singularity, the horizontal forces won't be negligible anymore, and there is significant squeezing.

OK, I stand corrected.