Black holes

[Roboramma]

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.

In one pop science relativity book I have the author describes a scenario wherein a spaceship falls toward the earth. At the center of the ship are two apples, say five feet from each other (and their center is the center of the ship). As the ship falls toward the earth, assume there is a giant tunnel that passes through the center of the earth. We look at the apples: at first nothing seems to happen, but since they are being accelerated toward the center of the earth, in our spaceship we should see them move together (which I suppose becomes even more obvious if we take the ship out of the picture entirely).

Is that the same force that you're describing?

 

[Ziggurat]

In one pop science relativity book I have the author describes a scenario wherein a spaceship falls toward the earth. At the center of the ship are two apples, say five feet from each other (and their center is the center of the ship). As the ship falls toward the earth, assume there is a giant tunnel that passes through the center of the earth. We look at the apples: at first nothing seems to happen, but since they are being accelerated toward the center of the earth, in our spaceship we should see them move together (which I suppose becomes even more obvious if we take the ship out of the picture entirely).

Is that the same force that you're describing?

Yes.

 

[Farsight]

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.

It isn't the same, edd. If you and I have relative motion, I see your clock going slower than mine, and you see my clock going slower than yours. But when we're at different gravitational potentials, such that you're lower than me, I see your clock going slower than mine and you see mine going faster than yours. 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. If you fell in a billion years ago, you haven't seen anything yet.

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

You aren't using any reasoning, ben. You're just accepting what some book tells you. This is a skeptics forum. Don't be a patsy.

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.

There is a reason: a coordinate system is an artefact of measurement, not something real. And when the coordinate speed of light goes to zero as measured by distant observers like you and me, light doesn't move, and all electromagnetic processes grind to a halt. So sol (say) at the event horizon can't make any measurements. Coordinate systems like Kruskal-Szekeres employ a stopped observer along with a stopped light-clock, and make out that he sees his clock ticking normally when it isn't ticking at all. Ask yourself what a hypothetical observer travelling through the universe at the speed of light would see. We all know that we can't actually travel at c, but set that aside and ask the question.

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 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.

I'm not ignorant of GR, or Schwarzschild coordinates, and nor am I saying that one coordinate system is special. I'm saying all coordinate systems are artefacts of measurement, and it's a mistake to dismiss the undefined result at R=2M. And might I add that your "you are ignorant!" is no argument at all. We might expect that from a medieval bishop, but we don't expect that from a physicist.

 

[Brian-M]

A question...

If time slows down to a halt at the event horizon, wouldn't the surface of the star that collapsed to form the black hole still be just outside the event horizon forever falling in? We just won't be able to see it because it's redshifted to the point of being undetectable.

So even if you deliberately attempted to fall into a black hole by heading directly into it with your spaceship's impulse engines at full power, what you'd observe as you get closer to the event horizon would be that the space ahead of you would gradually get brighter and brighter (while at the same time you get dimmer and dimmer to an outside observer) until you find yourself plunging head-first into the fiery hot surface of a star before you even reached the horizon?

 

[edd]

Sheesh I wasn't trying to say its the same, Farsight. I was trying to point out a problem in your logic by using a different example, but never mind.

 

[Farsight]

Sorry Ed. Please do try to point out a flaw in my logic.

If time slows down to a halt at the event horizon, wouldn't the surface of the star that collapsed to form the black hole still be just outside the event horizon forever falling in? We just won't be able to see it because it's redshifted to the point of being undetectable.

Yes. That's why Oppenheimer and Volkoff called them frozen stars. See the history on wiki. Note though that that it isn't time that halts. It's motion. A clock "clocks up" some regular cyclic motion rather than the literal flow of time.

So even if you deliberately attempted to fall into a black hole by heading directly into it with your spaceship's impulse engines at full power, what you'd observe as you get closer to the event horizon would be that the space ahead of you would gradually get brighter and brighter (while at the same time you get dimmer and dimmer to an outside observer) until you find yourself plunging head-first into the fiery hot surface of a star before you even reached the horizon?

I don't think so. It's still a black hole, light can't get out. There's no motion and heat is an emergent property of motion, so there is no heat. You'd look straight ahead and all you'd see is black.

 

[sol invictus]

Please do try to point out a flaw in my logic.

Here:

There is a reason: a coordinate system is an artefact of measurement, not something real.

Fine.

And when the coordinate speed of light goes to zero as measured by distant observers like you and me, light doesn't move, and all electromagnetic processes grind to a halt.

Not fine, and contradicts the first statement.

Coordinate systems are artefacts, so you precisely cannot conclude anything about physics from the fact that a coordinate speed goes to zero.

That's obvious, but if you need more proof, there are perfectly good coordinate systems in which the coordinate speed of light goes to zero at any point in space you would like it to.

 

[Ziggurat]

There is a reason: a coordinate system is an artefact of measurement, not something real.

Then why are you treating one coordinate system like it's real?

And when the coordinate speed of light goes to zero as measured by distant observers like you and me, light doesn't move, and all electromagnetic processes grind to a halt. So sol (say) at the event horizon can't make any measurements. Coordinate systems like Kruskal-Szekeres employ a stopped observer along with a stopped light-clock, and make out that he sees his clock ticking normally when it isn't ticking at all.

You think it's stopped because there's an infinity in the Schwarzchild metric. But as I've pointed out repeatedly but you have failed to understand, this is a coordinate singularity. It is not a genuine singularity. And it's actually easy to tell the difference.

Ask yourself what a hypothetical observer travelling through the universe at the speed of light would see. We all know that we can't actually travel at c, but set that aside and ask the question.

Ask yourself why you can't travel at c. The answer is that using c as your velocity in the Lorentz transform produces... (wait for it) a genuine singularity in the transform. Not a coordinate singularity, but a genuine singularity. So there simply is no comparison between transforming to an invalid coordinate system like a light-speed observer and transforming to

I'm not ignorant of GR, or Schwarzschild coordinates, and nor am I saying that one coordinate system is special.

Yes, you really are treating it as special. That's exactly what you're doing when you make this claim:

I'm saying all coordinate systems are artefacts of measurement, and it's a mistake to dismiss the undefined result at R=2M.

No, it isn't. I already told you why. It's a coordinate singularity. And it's actually easy to tell the difference between a coordinate singularity and a genuine singularity in GR.

 

[Farsight]

Here:

Farsight: There is a reason: a coordinate system is an artefact of measurement, not something real.

Fine.

Farsight: And when the coordinate speed of light goes to zero as measured by distant observers like you and me, light doesn't move, and all electromagnetic processes grind to a halt.

Not fine, and contradicts the first statement.

No it doesn't. Remember the optical clocks losing synchronisation at different elevations, simplified to parallel-mirror light clocks:

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

The lower clock goes slower because the light goes slower. That's real. But coordinate systems aren't. So drop the word coordinate:

And when the speed of light goes to zero as measured by distant observers like you and me, light doesn't move, and all electromagnetic processes grind to a halt.

Coordinate systems are artefacts, so you precisely cannot conclude anything about physics from the fact that a coordinate speed goes to zero.

See above. The coordinate system is an artefact so you drop it and take the evidence at face value. Look at those optical clocks. There's no time flowing in them. The light goes slower in the lower clock than in the upper clock. It isn't merely the coordinate speed of light varying in a non-inertial reference frame, it's the speed of light. Only you can't measure it locally as being slower because the rate of all electromagnetic processes is reduced. You measure the slower light with a slower clock, and get the same answer.

That's obvious, but if you need more proof, there are perfectly good coordinate systems in which the coordinate speed of light goes to zero at any point in space you would like it to.

There are no examples other than a black hole event horizon where the speed of light goes to zero. If you beg to differ, put it up and I'll knock it down.

 

[Ziggurat]

No it doesn't. Remember the optical clocks losing synchronisation at different elevations, simplified to parallel-mirror light clocks:

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

The lower clock goes slower because the light goes slower. That's real. But coordinate systems aren't. So drop the word coordinate:

And when the speed of light goes to zero as measured by distant observers like you and me, light doesn't move, and all electromagnetic processes grind to a halt.

Small problem with your comparison: the parallel mirrors in your arrangement are stationary in the gravitational field. An infalling observer is not.

They aren't in the same reference frame, so the basis for your comparison is invalid to begin with.

 

[Farsight]

Then why are you treating one coordinate system like it's real?

I'm not. I'm treating the motion of light in the universe as real.

And when the coordinate speed of light goes to zero as measured by distant observers like you and me, light doesn't move, and all electromagnetic processes grind to a halt. So sol (say) at the event horizon can't make any measurements. Coordinate systems like Kruskal-Szekeres employ a stopped observer along with a stopped light-clock, and make out that he sees his clock ticking normally when it isn't ticking at all.

You think it's stopped because there's an infinity in the Schwarzchild metric. But as I've pointed out repeatedly but you have failed to understand, this is a coordinate singularity. It is not a genuine singularity. And it's actually easy to tell the difference.

See above. Strike out the word coordinate. I don't think it's stopped because there's an infinity in the Schwarzschild metric. I think it's stopped because the speed of light is zero. It isn't an infinity, it's a zero, so it isn't a singularity in the usual sense of the word.

Ask yourself why you can't travel at c. The answer is that using c as your velocity in the Lorentz transform produces... (wait for it) a genuine singularity in the transform. Not a coordinate singularity, but a genuine singularity. So there simply is no comparison between transforming to an invalid coordinate system like a light-speed observer and transforming to

I know why I can't travel at c. It's because of what pair production tells us. We can quite literally make an electron out of light. That electron has spin angular momentum and magnetic dipole moment, and you can diffract it. It has a wave nature wherein 4π / c^1½ applies. It's a spherical wave. You can't make it move linearly at c too. That's what underlies the Lorentz transform. But nevertheless there is a comparison, because we can say that you're travelling at very close to c through the universe. You might think everything is normal in your reference frame, but your thinking is slowed down so much that it hardly registers. A second to you is a thousand years to me, and I've got plenty of time to put as asteroid in your way. BLAM. You didn't even see it. Take it all the way to c and you see nothing at all. That's what "infinite time dilation" does to you.

Yes, you really are treating it as special. That's exactly what you're doing when you make this claim:

I'm saying all coordinate systems are artefacts of measurement, and it's a mistake to dismiss the undefined result at R=2M.

See above. I'm not treating one coordinate system as special. I'm looking past the coordinate systems at how the motion of light defines our distance and time, and thence our coordinate system.

No, it isn't. I already told you why. It's a coordinate singularity. And it's actually easy to tell the difference between a coordinate singularity and a genuine singularity in GR.

No, it isn't. And I repeat: take a look at this page from Misner/Thorne/Wheeler's Gravitation posted by a guy called Jesse. On the diagram on the left, the curve peaks to infinity at the event horizon. That's the gravitational time dilation tending to infinity, and coordinate time tending to forever. At the top of the peak, is the end of time, so there is no top to it. But it's "transformed away" using Kruskal-Szekeres coordinates. A mathematical conjuring trick is employed to do a hop skip and a jump over the end of time, by pretending that a stopped observer sitting in front of a stopped clock sees it ticking merrily away.

 

[Farsight]

Small problem with your comparison: the parallel mirrors in your arrangement are stationary in the gravitational field. An infalling observer is not.

Let's deal with the motion of light at the event horizon first. Then maybe we can talk about an infalling photon, then an infalling electron, and work up from there.

They aren't in the same reference frame, so the basis for your comparison is invalid to begin with.

Reference frames are artefacts too! You can't point up to the night sky and say "look, there's a reference frame". Those two light beams are moving through the space in the room you're in. One moves slower than the other, and that's it. There's nothing invalid about it at all.

It's a question of seeing the empirical evidence without letting artefacts cloud your view. Draw it on a piece of paper, only instead of light beams, draw racehorses. One gets to the end before the other. Now try to claim they're moving at the same speed.

 

[Ziggurat]

See above. Strike out the word coordinate. I don't think it's stopped because there's an infinity in the Schwarzschild metric. I think it's stopped because the speed of light is zero.

How do you know it's zero? Because of the infinity in the Schwarzchild metric.

In an infalling coordinate system, it never goes to zero.

It isn't an infinity, it's a zero, so it isn't a singularity in the usual sense of the word.

The speed of light is the reciprocal of that value.

I know why I can't travel at c. It's because of what pair production tells us. We can quite literally make an electron out of light.

Yeah, no. That's not why. Even classically, you're prohibited from reaching c. You don't need to resort to quantum mechanics here.

That electron has spin angular momentum and magnetic dipole moment, and you can diffract it. It has a wave nature wherein 4π / c^1½ applies. It's a spherical wave. You can't make it move linearly at c too. That's what underlies the Lorentz transform.

Yeah, sorry, but this is unadulterated nonsense. An electron is, in general, NOT a spherical wave. And historically, the Lorentz transforms precede quantum mechanics, so clearly the development of the Lorentz transforms cannot have had anything to do with spherical waves. And lastly, if you apply a Lorentz transform to a spherical wave, what do you get? Something that's not spherical. So nothing in this claim makes any sense whatsoever. You've really gone off the deep end here.

But nevertheless there is a comparison, because we can say that you're travelling at very close to c through the universe. You might think everything is normal in your reference frame, but your thinking is slowed down so much that it hardly registers. A second to you is a thousand years to me

Because we're in different frames.

The infalling frame is different than the stationary frame. You're drawing your conclusions about the infalling frame from the stationary frame. Yet you don't even see how your own example invalidates that.

See above. I'm not treating one coordinate system as special.

You keep saying that, but it's still not true. You have treated the stationary frame as special, and ignored the difference between the stationary frame and the infalling frame.

No, it isn't. And I repeat: take a look at this page from Misner/Thorne/Wheeler's Gravitation posted by a guy called Jesse. On the diagram on the left, the curve peaks to infinity at the event horizon.

Yeah, that's the coordinate singularity of the Schwarzchild coordinates. And you want to treat that coordinate singularity as special, because you want to treat the Schwarzchild metric as special. And you don't even understand that this is what you're doing.

That's the gravitational time dilation tending to infinity, and coordinate time tending to forever. At the top of the peak, is the end of time, so there is no top to it. But it's "transformed away" using Kruskal-Szekeres coordinates. A mathematical conjuring trick is employed to do a hop skip and a jump over the end of time, by pretending that a stopped observer sitting in front of a stopped clock sees it ticking merrily away.

And you've done it again: you have treated one coordinate system as special. You think that the Schwarzchild coordinates are the real ones, and we just get to the Kruskal coordinates by using tricks. The problem, though, is that Kruskal coordinates can stand on their own as a direct solution to the GR field equations. And from that perspective, the change of coordinates introduces a coordinate singularity that never previously existed when we transform to Schwarzchild coordinates from Kruskal coordinates.

 

[Farsight]

How do you know it's zero? Because of the infinity in the Schwarzchild metric.

No. Because the light can't get out. We have good evidence of something very small and very massive at the centre of the galaxy, so massive that it just has to be a black hole. Whatever our differences as to the interpretation, I'm sure we all agree that there are things out there that light cannot escape from. As for why, imagine you've got a light beam shining straight up from the event horizon. It doesn't rise up then fall back down like some lofted cannonball. It doesn't curve round and back towards the black hole because it's going straight up. But it doesn't get out. Ask yourself why, and please don't resort to the waterfall analogy that uses an "inflowing aether" to duck the issue.

Sorry, I have to go. I'll look at your other points later.

ETA: I haven't gone off at the deep end. It's just not in the textbooks yet. We really can make electrons via pair production, and we really can diffract them. Dig up the old Yukawa pion paper and look for spherical waves.

 

[Ziggurat]

You can't point up to the night sky and say "look, there's a reference frame".

Actually, that's exactly what you can do. In fact, it's a very frequently employed technique. I'm a little surprised you didn't know that.

 

[Ziggurat]

No. Because the light can't get out.

It can't get out in Kruskal coordinates. It can't get out in Eddington–Finkelstein coordinates. Yet in neither of these coordinates does time or light stop at the event horizon. So the fact that light can't get out doesn't tell us what you apparently think it tells us. You have, without even knowing it, based your entire conclusion about what happens on the Schwarzchild coordinates.

We have good evidence of something very small and very massive at the centre of the galaxy, so massive that it just has to be a black hole. Whatever our differences as to the interpretation, I'm sure we all agree that there are things out there that light cannot escape from.

Yes, but that has absolutely nothing to do with our disagreement.

As for why, imagine you've got a light beam shining straight up from the event horizon. It doesn't rise up then fall back down like some lofted cannonball. It doesn't curve round and back towards the black hole because it's going straight up. But it doesn't get out. Ask yourself why, and please don't resort to the waterfall analogy that uses an "inflowing aether" to duck the issue.

I don't have to do any of that. I can just look at the null surfaces in any coordinate system which doesn't have a coordinate singularity at the event horizon. And they will tell me what happens to the light: it won't escape. But time doesn't stop either, not in an infalling reference frame.

ETA: I haven't gone off at the deep end. It's just not in the textbooks yet. We really can make electrons via pair production, and we really can diffract them. Dig up the old Yukawa pion paper and look for spherical waves.

You misunderstand. I'm not saying you've gone off the deep end because you believe in pair production or electron diffraction or spherical waves. I'm saying you've gone off the deep end in thinking that these things underlie the Lorentz transformations. They don't. You do not need any quantum mechanics to explain the Lorentz transformations.

 

[sol invictus]

There are no examples other than a black hole event horizon where the speed of light goes to zero. If you beg to differ, put it up and I'll knock it down.

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]

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. 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.

 

[Brian-M]

I don't think so. It's still a black hole, light can't get out. There's no motion and heat is an emergent property of motion, so there is no heat. You'd look straight ahead and all you'd see is black.

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.

 

[Farsight]

Continued from above.

Because we're in different frames.

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.

The infalling frame is different than the stationary frame. You're drawing your conclusions about the infalling frame from the stationary frame. Yet you don't even see how your own example invalidates that.

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

You keep saying that, but it's still not true. You have treated the stationary frame as special, and ignored the difference between the stationary frame and the infalling frame.

Like I said above, first we deal with the motion of light at the event horizon, then we move on to the infalling photon, then an infalling electron, and then the infalling observer.

Yeah, that's the coordinate singularity of the Schwarzchild coordinates. And you want to treat that coordinate singularity as special, because you want to treat the Schwarzchild metric as special. And you don't even understand that this is what you're doing.

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

And you've done it again: you have treated one coordinate system as special. You think that the Schwarzchild coordinates are the real ones, and we just get to the Kruskal coordinates by using tricks.

How many times do I have to say it? I don't think any coordinate system is real. What's real is light moving in this universe. Or not moving, as the case may be.

The problem, though, is that Kruskal coordinates can stand on their own as a direct solution to the GR field equations. And from that perspective, the change of coordinates introduces a coordinate singularity that never previously existed when we transform to Schwarzchild coordinates from Kruskal coordinates.

They represent a non-real solution. The undefined result at the Schwarzschild R=2M is where light isn't moving. It can't go any slower than not moving. Kruskal-Szekeres coordinates ignore that.

You can't point up to the night sky and say "look, there's a reference frame".

Actually, that's exactly what you can do. In fact, it's a very frequently employed technique. I'm a little surprised you didn't know that.

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.

 

[Farsight]

It can't get out in Kruskal coordinates. It can't get out in Eddington–Finkelstein coordinates. Yet in neither of these coordinates does time or light stop at the event horizon. So the fact that light can't get out doesn't tell us what you apparently think it tells us. You have, without even knowing it, based your entire conclusion about what happens on the Schwarzchild coordinates.

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?

I'm sure we all agree that there are things out there that light cannot escape from.

Yes, but that has absolutely nothing to do with our disagreement.

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

I don't have to do any of that. I can just look at the null surfaces in any coordinate system which doesn't have a coordinate singularity at the event horizon. And they will tell me what happens to the light: it won't escape. But time doesn't stop either, not in an infalling reference frame.

I must be the only one here who has read A World Without Time: The Forgotten Legacy of Godel and Einstein? Time isn't flowing, Zig. That's just a figure of speech. Clocks clock up some kind of regular cyclical motion. Whether it's a pendulum clock, a quartz wristwatch, or an atomic clock, that's what they do. You can't open up a mechanical clock and point to the movement and say "look, there's time flowing". When a clock runs slow it's because the underlying motion is going slower. And when I give you two identical optical clocks, and you place one a foot lower than the other, take a good hard look at what's happening. Don't be distracted by thinking you can look at the null surfaces in any coordinate system. You can't actually do that. Do not elevate abstraction above scientific evidence to such an extent that you fail to examine the latter.

You misunderstand. I'm not saying you've gone off the deep end because you believe in pair production or electron diffraction or spherical waves. I'm saying you've gone off the deep end in thinking that these things underlie the Lorentz transformations. They don't. You do not need any quantum mechanics to explain the Lorentz transformations.

OK noted. Let's come back to that once we've dealt with the motion of light.