Black holes

[Farsight]

I didn't mean that the heat and light would be coming from inside the black hole, but would come from the surface of the collapsing star that hasn't yet reached the event horizon. From it's point of view, hardly any time has passed, so it should still be in the form of fiery hot plasma. Since it's still outside the event, horizon you should be able to experience the effects of it's radiant energy once you get close enough that it's not redshifted to the point of being unobservable. There'd be a huge amount of highly energetic motion in the plasma... relative to anything experiencing high levels of time dilation due to being in close proximity to the event horizon.

I'm sorry Brian, I can't empathise with this. It still leaves the has it happened yet? issue, and if it was the case, black holes wouldn't exist, they'd be "radio stars". The sort of black hole I'm describing is even more of a hole than the point-singularity black hole described in popular literature and elsewhere.

 

[Farsight]

I've already given you an example. Instead of "knocking it down", you ignored it. Here it is again:

[latex]$ds^2 = -(r-r_0) dt^2 + dr^2/(r-r_0) + dy^2 + dz^2$[/latex]

I didn't ignore it at all. You asked me a question in post #17 in Susskind-Hawking thread, and I replied in post #19 saying No. It isn't time that's stopped, it's light. So it takes forever to make a measurement, so you can't make any measurements, and as a result there is no actual metric.. You then said It's not just light or EM phenomena, Farsight, it's everything. All processes slow and stop on the horizon as viewed from outside, not just EM ones. You also said the metric I posted describes absolutely flat spacetime.

The coordinate speed of light is zero at r=r0. Clocks (including light clocks) run slower and slower if they are held at fixed, smaller and smaller r.

But does the speed of light vary. In a flat spacetime, it isn't varying.

In fact, this metric has exactly the properties you keep referring to in the black hole. Do you know what this metric describes, Farsight? It's not a black hole.

The Rindler horizon? Sorry, I have to go, I can't check it out.

 

[Ziggurat]

Continued from above.

No, not because we're "in different frames". A reference frame is an artefact of measurement. You measure things different to me because you're moving with respect to me.

You say that like it's two different things, but it isn't.

Again it's an artefact. There is no actual frame falling into the black hole.

Yes there is.

If it's you falling into the black hole, while I remain at a safe distance, your measurements are constantly changing with respect to mine.

And if you sit on earth, and I orbit in a space station, my measurements are also constantly changing with respect to you. So what?

Like I said above, first we deal with the motion of light at the event horizon

Kruskal coordinates handle that perfectly well. There's no problem with light at the event horizon.

I understand this, you don't, because you view coordinate systems and reference frames as real things instead of the artefacts that they are.

Ypu're wrong. You DO view a coordinate system as more than an artifact (not artefact, BTW). Except you view one particular coordinate system as being special.

How many times do I have to say it? I don't think any coordinate system is real.

Yet your entire argument is based on the Schwarzchild coordinate system. You keep referring to that one picture in MTW of the Schwarzchild coordinates, and insisting that it represents reality while the others do not.

What's real is light moving in this universe. Or not moving, as the case may be.

Light at the event horizon is moving. It's moving along the event horizon.

They represent a non-real solution.

Not so. They represent the exact same solution.

The undefined result at the Schwarzschild R=2M is where light isn't moving.

It's only undefined in the Schwarzchild metric because the Schwarzchild metric has a coordinate singularity at the event horizon. But it's not a real singularity. Do you even get the difference between a coordinate singularity and a genuine singularity?

It can't go any slower than not moving. Kruskal-Szekeres coordinates ignore that.

Not at all. Light most definitely moves at the event horizon in Kruskal coordinates. Or in any coordinate system which doesn't have a coordinate singularity.

No you cannot. Here's some pictures of the night sky. Take your pick, find one you like, one that matches what you'd see if you stepped outside. Now point out a reference frame. Do you still want to insist that you can point up to the night sky and say "look, there's a reference frame"? Seriously? Think carefully Zig.

Seriously. And what's more, obviously. There's a practically limitless number of reference frames to choose from. Pick a star, watch its motion, and you've got yourself a reference frame, the reference frame of that star. Hell, that's exactly what Gravity Probe B does to keep itself steady: it watches a guide star. Each star in the sky provides a reference frame.

Or better yet, we can even use the sky to find the unique reference frame of the universe, the co-moving reference frame. Just watch the CMB and look for a dipole moment in the temperature distribution, and that tells you how the co-moving frame is moving with respect to you.

Do not presume to lecture me. You know far less than you think you do.

 

[Ziggurat]

No I haven't, not a bit, you haven't addressed the issue. I must press you on this, because when you find you can't answer it you'll hopefully realise there's a gap in your understanding: you've got a light beam emitted vertically at the event horizon. It doesn't slow down, it doesn't fall back, it doesn't curve round. Why doesn't the light get out?

Because the null surface IS the event horizon. It travels along the event horizon. You have made a fundamental mistake in thinking that you can keep space and time completely separated, but you cannot. The Schwarzchild metric flatters this misconception, but even there it's simply not true.

It is the nub of it. We're discussing the nature of black holes. Explain why that light beam doesn't get out.

Because the null surface lies along the event horizon. Just like it does if we adopt an accelerating reference frame in special relativity. Were you aware of this? Did you know that event horizons exist in special relativity? Light is always going faster than our accelerating observer, but light at the event horizon will never reach our observer. Does that mean that light is motionless? No, it does not. Obviously. We just need to adopt a non-accelerating observer to see it.

Well, guess what: in GR, falling is the non-accelerating frame.

Seriously, the event horizon of a black hole is no different from the event horizon of an accelerating observer.

 

[W.D.Clinger]

Ypu're wrong. You DO view a coordinate system as more than an artifact (not artefact, BTW). Except you view one particular coordinate system as being special.

Artefact is the British spelling of artifact.

Yet your entire argument is based on the Schwarzchild coordinate system. You keep referring to that one picture in MTW of the Schwarzchild coordinates, and insisting that it represents reality while the others do not.

Word.

What I find curious is that he's shown parts (a) and (b) of MTW Figure 32.1 on many occasions, but has never said a word about part (c) of that figure or about the text in section 32.4 that refers to Figure 32.1. It's as though he doesn't own a copy of MTW himself, and knows nothing about it apart from what he's read on the World-Wide Web.

No I haven't, not a bit, you haven't addressed the issue. I must press you on this, because when you find you can't answer it you'll hopefully realise there's a gap in your understanding: you've got a light beam emitted vertically at the event horizon. It doesn't slow down, it doesn't fall back, it doesn't curve round. Why doesn't the light get out?

Because the null surface IS the event horizon. It travels along the event horizon. You have made a fundamental mistake in thinking that you can keep space and time completely separated, but you cannot. The Schwarzchild metric flatters this misconception, but even there it's simply not true.

I know what you mean, but I suspect your language will only confuse Farsight further. I think Farsight's "it doesn't curve round" means he's thinking in terms of spatial curvature, and I'm afraid your "travels along the event horizon" will conjure up an image of light travelling in some spatial direction that's perpendicular to the radial coordinate.

Farsight has shown no understanding of curvature along the time dimension. It might help to point out that any spatial point p on the event horizon that remains at p over time is following a null geodesic, and is travelling radially away from the central singularity at the speed of light.

Unfortunately, Farsight interprets that to mean the speed of light is zero at the event horizon. He doesn't realize that a photon directed radially inward from p also follows a null geodesic, travelling at the speed of light toward the central singularity and arriving there pretty quickly.

He seems quite determined not to understand things like that.

 

[sol invictus]

But does the speed of light vary.

Can't you check that for yourself? You have the metric.

But yes, the coordinate speed of varies. In particular it goes to zero at r=r0.

 

[Ziggurat]

Artefact is the British spelling of artifact.

Yeah, well, the British are wrong. After all, they also pronounce schedule as "shedule". Must be something wrong with their shools.

 

[dlorde]

Beeps emitted very near the time the horizon appeared are very red shifted and take a very long time to get to infinity.

Well, they would, wouldn't they?

There must be a better word than infinity here...

 

[dlorde]

.. People will tell you that if you're at the event horizon you see your own clock ticking normally, but it's a fallacy because it takes you forever in terms of "universe time" to see anything.

What is "universe time" ?

 

[Complexity]

What is "universe time" ?

Sigh...

That's an entirely new dimension of confusion.

So it goes.

 

[sol invictus]

Well, they would, wouldn't they?

There must be a better word than infinity here...

Despite how it sounds, that actually does mean something well-defined. But what I meant was "a distant observer".

 

[dlorde]

Despite how it sounds, that actually does mean something well-defined. But what I meant was "a distant observer".

Yes; I've seen infinity used in this context before - but it can look a bit odd taken out of context.

 

[DeiRenDopa]

Yeah, well, the British are wrong. After all, they also pronounce schedule as "shedule". Must be something wrong with their shools.

Yeah, funny they are.

Their shists have skisms; they skeme, but they shlep; they drink shnaps (in skooners) as they shmooze; their shmucks are real shlocks; their skolars go to skool; ...

 

[ctamblyn]

Yeah, funny they are.

Their shists have skisms; they skeme, but they shlep; they drink shnaps (in skooners) as they shmooze; their shmucks are real shlocks; their skolars go to skool; ...

I'm schocked and appalled by this schameful stereotyping.

 

[Zeuzzz]

The above posts are annoyingly amuzing for me.

Still, black holes remain more a mathematical extrapolation of current (extreme) experimental physics than a vigourously tested area of scientific study.

 

[Farsight]

No, not because we're "in different frames". A reference frame is an artefact of measurement. You measure things different to me because you're moving with respect to me.

You say that like it's two different things, but it isn't.

I say that because that's how it is. If you and I are together in space, in spacesuits, we will measure things, and exchange information over the radio, or merely touch helmets. Our measurements tally. However when you accelerate to some high speed relative to me, we no longer agree on our measurements. We are however fully aware of Lorentz invariance, and agree that these differences are a result of the difference in our motion.

Again it's an artefact. There is no actual frame falling into the black hole.

Yes there is.

No there isn't. Look everybody, there's Zig in his spacesuit, falling to Earth. We can see he's carrying rods and clocks, he's shining his light beams and making his measurements. But can we see him surrounded by some dotted-line grid that is his coordinate system? No. Can we see some rectangular box around him that is his reference frame? No. It's just a guy in a spacesuit falling through space.

And if you sit on earth, and I orbit in a space station, my measurements are also constantly changing with respect to you. So what?

Your measurements are different to mine because you're in a different environment. You are immersed in this environment, and you cannot detect that your local measurements are different until you exchange notes with me.

You're wrong. You DO view a coordinate system as more than an artifact (not artefact, BTW). Except you view one particular coordinate system as being special.

No I don't. My view is of this universe, and light moving through it. Along with the wave nature of matter. I'll tell you more about it sometime.

Yet your entire argument is based on the Schwarzchild coordinate system. You keep referring to that one picture in MTW of the Schwarzchild coordinates, and insisting that it represents reality while the others do not.

No it isn't. It's based on the motion of light in the universe. For example, if we set gravity aside, then the invariant spacetime interval of SR is founded upon this: the light-path lengths are the same.

Light at the event horizon is moving. It's moving along the event horizon.

Where gravitational time dilation is infinite according to observers at a great distance? Like me? I know what, I'll look at that light through my gedanken telescope. Has it moved yet? No. Let's give it half an hour. Has it moved yet? No. How about a year? Has it moved yet? No. And so it goes.

Not so. They represent the exact same solution.

Yes so. Because in my gedanken telescope I can also see you. Have you moved yet? No. You don't move, the light doesn't move, and you don't see things happening normally. You don't see anything. Not ever. Your finite proper time takes forever in the real world. The things you thing you'd be able to see are in neverneverland.

It's only undefined in the Schwarzchild metric because the Schwarzchild metric has a coordinate singularity at the event horizon. But it's not a real singularity.

Like I said, it isn't in a singularity in the usual sense. It's a c=0, not some infinity.

Do you even get the difference between a coordinate singularity and a genuine singularity?

Yes. You don't understand that you cannot eliminate this c=0 by choosing neverneverland coordinates where a stopped Zig in front of stopped light sees everything happening normally. Don't you get it yet? Nothing happens. Ever.

Not at all. Light most definitely moves at the event horizon in Kruskal coordinates. Or in any coordinate system which doesn't have a coordinate singularity.

Light doesn't move in a coordinate system. It moves through space. And if there is a location in space where that light is stopped, you can't make it move by flicking to a coordinate system that pretends that a stopped light-clock isn't stopped just because a stopped observer is sitting in front of it.

No you cannot. Here's some pictures of the night sky. Take your pick, find one you like, one that matches what you'd see if you stepped outside. Now point out a reference frame. Do you still want to insist that you can point up to the night sky and say "look, there's a reference frame"? Seriously? Think carefully Zig.

Seriously. And what's more, obviously. There's a practically limitless number of reference frames to choose from. Pick a star, watch its motion, and you've got yourself a reference frame, the reference frame of that star. Hell, that's exactly what Gravity Probe B does to keep itself steady: it watches a guide star. Each star in the sky provides a reference frame.

LOL. So the sky is full of a limitless number of reference frames is it? Why, there's so many of them, and they're so obvious that I can't see the stars any more. Zig, look at me:

SLAP!

Now wake up. The sky is full of stars, and galaxies, and gas and dust and all the other real things we can detect with our telescopes. It is not full of reference frames. A reference frame is not something which actually exists out there. Do not place mathematical abstractions such as this on a pedestal above the scientific evidence you can see.

Or better yet, we can even use the sky to find the unique reference frame of the universe, the co-moving reference frame. Just watch the CMB and look for a dipole moment in the temperature distribution, and that tells you how the co-moving frame is moving with respect to you.

I have no issue with the CMBR dipole anisotropy, which appears to offer a way to gauge our motion through the universe. But again the reference frame it provides is not some actual object within the universe.

Do not presume to lecture me. You know far less than you think you do.

Ditto.

As for the artefact, I'm John Duffield, I live in Poole in the UK. Pleased to meet you. And you are?

 

[Zeuzzz]

As for the artefact, I'm John Duffield, I live in Poole in the UK. Pleased to meet you. And you are?

Good luck with that, I've tried before to no avail.

 

[Farsight]

Because the null surface IS the event horizon. It travels along the event horizon.

So a beam of light directed straight up doesn't get out because it travels sideways? And no mention of the coordinate speed of light varying with gravitational potential? Or the speed of light? Or the infinite gravitational time dilation according to distant observers? Let's tease this one out. Let's say our vertical emitter is a small distance above the event horizon. It employs the hyperfine transition, an electromagnetic spin-flip "process" that emits light, which travels straight up. We see this light as redshifted. Now move the emitter to a point closer to the event horizon. We still see the light, but now it's even more redshifted, and we know that the emission process is running at a slower rate. We repeat this, and with some care, we now detect shortwave radio waves. Then longwave, and so on. At what point do those electromagnetic waves suddenly turn sideways, and at what speed do they then move with respect to distant observers? And with respect to the emission process?

You have made a fundamental mistake in thinking that you can keep space and time completely separated, but you cannot. The Schwarzchild metric flatters this misconception, but even there it's simply not true.

I have made no fundamental mistake, I've plumbed the fundamentals. There is no fundamental time, Zig. What's out there is space and light moving through it. And other things too, but we start with light. We use the motion of light to define our distance and time.

Because the null surface lies along the event horizon. Just like it does if we adopt an accelerating reference frame in special relativity. Were you aware of this? Did you know that event horizons exist in special relativity?

Of course I do.

Light is always going faster than our accelerating observer, but light at the event horizon will never reach our observer. Does that mean that light is motionless? No, it does not. Obviously. We just need to adopt a non-accelerating observer to see it.

In flat spacetime the speed of light through the universe is uniform. An observer moving through the universe doesn't change that. Look, there he is out in space. Can you see him? Yes. Now look just behind him? See that black cone? Can you see his Rindler horizon? No. You see plain old stars. There is no gravitational lensing surrounding a black region. This isn't in the same league as a black hole.

Well, guess what: in GR, falling is the non-accelerating frame.

See post #75 in a previous thread to appreciate that I'm fully aware of this.

Seriously, the event horizon of a black hole is no different from the event horizon of an accelerating observer.

See above. Seriously, it is.

 

[Farsight]

Can't you check that for yourself? You have the metric.

The metric doesn't tell me it varies. Experiment tells me that. It's as plain as the nose in front of my face. Experiments like the Shapiro delay. Experiments using super-accurate optical clocks which lose synchronisation when separated by only a foot of vertical elevation. They're optical clocks. They employ electromagnetic hyperfine transistions and electromagnetic waves. And we can simplify them to parallel-mirror light clocks, which we know will also lose synchronisation when at different elevations:

|--------------|
|--------------|

Can you see the time flowing in these experiments, sol? No. You see light moving through space. That's all that's there. Learn to see what's there instead of refusing to see it because your head is fog full of mathematical abstraction and artefact. Coordinate systems and reference frames and the metric are all derived from the motion of light through space, and just as Einstein said, it isn't constant, and you can see it isn't constant.

But yes, the coordinate speed of varies. In particular it goes to zero at r=r0.

So what happens when you take the lower of the two parallel-mirror light clocks down to the event horizon? Draw us all a picture. In your own time.

 

[Farsight]

Good luck with that, I've tried before to no avail.

It's OK either way. If everybody gives their real names you get a more civilised discussion. If it's only me, I've got the moral high ground.