Black holes

[Farsight]

OK, you simply don't understand the nature of the curvature we're talking about then. The relevant curvature to this discussion is intrinsic rather than extrinsic.

We're discussing intrinsic curvature. You can read about it on wiki. See the bit that says:

"There is a key distinction between extrinsic curvature, which is defined for objects embedded in another space (usually a Euclidean space) in a way that relates to the radius of curvature of circles that touch the object, and intrinsic curvature, which is defined at each point in a Riemannian manifold".

We're talking about Riemann curvature, remember? The rubber sheet depicts it. The geodesic dome and the paper strip are analogies for it. And there was I thinking you had some sincerity edd. I am disappointed. RC is crushingly wrong, and people would rather be dishonest than admit it. It leaves a bad taste in the mouth.

ETA: all the more reason for me to think that this thread is dead.

 

[sol invictus]

We're discussing intrinsic curvature. You can read about it on wiki. See the bit that says:

"There is a key distinction between extrinsic curvature, which is defined for objects embedded in another space (usually a Euclidean space) in a way that relates to the radius of curvature of circles that touch the object, and intrinsic curvature, which is defined at each point in a Riemannian manifold".

We're talking about Riemann curvature, remember? The rubber sheet depicts it. The geodesic dome and the paper strip are analogies for it. And there was I thinking you had some sincerity edd. I am disappointed. RC is crushingly wrong, and people would rather be dishonest than admit it. It leaves a bad taste in the mouth.

ETA: all the more reason for me to think that this thread is dead.

Since you're such an expert, Farsight - what's the intrinsic curvature of a cylinder?

 

[edd]

So if you understand it why are you talking about flat surfaces as if they're curved?

 

[Farsight]

Since you're such an expert, Farsight - what's the intrinsic curvature of a cylinder?

Flat. Take a vertical slice through it and you're left with this: ||. Remember I said take a vertical slice through RC's cone and you're left with this: V. That's like a side view of Edd's bit of paper.

So if you understand it why are you talking about flat surfaces as if they're curved?

I'm not. RC is talking about curved regions as if they're flat. Have a read of this to learn about intrinsic and extrinisc curvature. It's all easy stuff.

 

[edd]

You are - you say it here for example

Yes. Say it's a 180-degree fold. Get a microscope and look closely at the folded region. The paper there is sharply curved like this: U. If you had a very long strip of paper and you made a large number of very small very slight "folds" or creases of say 0.00001 degrees apiece, then between them the paper is flat, but because they're there the strip isn't flat any more.

If this geometry is so easy for you why are you getting it wrong?

 

[D'rok]

Have a read of this to learn about intrinsic and extrinisc curvature. It's all easy stuff.

Wow. I read your link. It really is a good, clear explanation. Even I can see that you really put your foot in it with your answer to edd's question.

 

[sol invictus]

Flat.

So you think a cylinder is flat, but a single fold in a piece of paper has curvature? Doesn't it bother you that the fold is locally like a piece of a small cylinder (remember your "U")?

Probably not, since logical consistency has never been your strong point...

 

[edd]

OK, this might be illuminating. It's something that occurred to me while reading back on some recent posts.

Here's an embedding diagram - think rubber sheet analogy. It's not the Newtonian potential, lets be clear about that. I've taken a 2D piece of a Schwarzschild solution and embedded it in a 3D Euclidean volume so as to preserve its intrinsic curvature. It's a pretty standard thing to do - you've seen lots of diagrams like it before, I'm sure.



Below is another embedding diagram - again I've taken something with intrinsic curvature and embedded it in a 3D Euclidean volume in a way that preserves that curvature. Farsight - can you explain how you think the gravitational field for this solution differs? Don't worry if you think it's not a physically realisable solution.

 

[Farsight]

You are - you say it here for example

Yes. Say it's a 180-degree fold. Get a microscope and look closely at the folded region. The paper there is sharply curved like this: U. If you had a very long strip of paper and you made a large number of very small very slight "folds" or creases of say 0.00001 degrees apiece, then between them the paper is flat, but because they're there the strip isn't flat any more".

If this geometry is so easy for you why are you getting it wrong?

I'm not getting it wrong. RC got it wrong. Badly wrong.

So you think a cylinder is flat, but a single fold in a piece of paper has curvature? Doesn't it bother you that the fold is locally like a piece of a small cylinder (remember your "U")? Probably not, since logical consistency has never been your strong point...

There's mathematical flat and there's common-usage flat. We use analogies featuring the latter to explain the former. Like geodesic domes and rubber sheets. And hills. Hills are like mountains, but they're smaller, and they aren't jagged. They're smoother things, they're curved. They have curvature. And when they don't, they're flat hills. Which means they aren't hills. It's the same for a gravitational field.

 

[D'rok]

I'm not getting it wrong.

Sure you are. It went like this:

So if I take a sheet of paper and introduce a fold I've introduced some intrinsic curvature into the paper, is that right Farsight?

Yes...

Then you linked to this:

The notion of geodesic deviation enables us to distinguish two types of curvature in geometry.

The first is most familiar to us, extrinsic curvature. It arises whenever we have a surface that curves into a higher dimension. We have seen many examples. One of the simplest arises when a flat sheet of paper is bent or rolled up into a cylinder.

...

Take a flat Euclidean surface. The geometry on its surface will be Euclidean, obviously. That means, if we draw a triangle on the surface, its angles will sum to 180 degrees. Now roll that surface up into a cylinder. That means the surface has now acquired extrinsic curvature. However its intrinsic curvature has not changed; it is still intrinsically flat.

http://www.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/non_Euclid_variable/index.html#Intrinsic

You got it wrong. Badly wrong.

 

[ctamblyn]

And I've lost count of the number of times I've given you empirical evidence for the varying speed of light that can't vary to less than zero.

All the empirical evidence you have put forward has been entirely consistent with MTW GR.

You still haven't answered my request for empirical evidence that proves the MTW picture of GR is wrong while supporting FGR

 

[ctamblyn]

There's mathematical flat and there's common-usage flat.

Edd was very clear in his question. He was asking about intrinsic curvature.

Just admit your mistake and move on; it's not a biggie (and perhaps you'll win back some respect).

 

[Farsight]

OK, this might be illuminating. It's something that occurred to me while reading back on some recent posts. Here's an embedding diagram - think rubber sheet analogy. It's not the Newtonian potential, lets be clear about that.

No it isn't Newtonian potential. The curvature on this diagram isn't diminishing with distance. It isn't the same as this. Of course there's also a missing section in the centre, but that's not a problem. It's a good representation of a black hole, better than the representations where the centre goes down to a point singularity. See google images for some examples.

I've taken a 2D piece of a Schwarzschild solution and embedded it in a 3D Euclidean volume so as to preserve its intrinsic curvature. It's a pretty standard thing to do - you've seen lots of diagrams like it before, I'm sure.

Not lots. There's the odd one, like the "wormhole" here. The usual depictions of a black hole resembles Newtonian potential away from the centre for good reason. The intrinsic curvature you preserve resembles a portion of a closed universe, as per the spherical depiction in the wikipedia shape of the universe article. Only WMAP indicates that the universe is flat.

Below is another embedding diagram - again I've taken something with intrinsic curvature and embedded it in a 3D Euclidean volume in a way that preserves that curvature. Farsight - can you explain how you think the gravitational field for this solution differs? Don't worry if you think it's not a physically realisable solution.

Light veers down the gradient. So I'd say one's a black hole in a non-flat universe whilst the other's a "white hole" in a non-flat universe. There is a space-time parallel between climbing out of a black hole and the expanding universe, but since you mentioned gravitation field for this solution and there is no gravitational field in the overall universe (which after all didn't collapse when it was small and dense) I'd say these are not depictions of a contracting and expanding universe.

When it comes to embedding diagrams, don't get too excited about "higher dimensional spaces". The curvature we're talking about isn't the curvature of light, it's a curvature in the speed of light plotted with distance from the centre of a circular gravitating body. Since we use light to define out second and our metre, it's also a plot in the metrical properties of space. People call it curved spacetime, but space isn't curved, it's just inhomogeneous in a non-linear way.

 

[edd]

Ok. Again, who wants to buzz in?

 

[edd]

Just to add - there are reasons I disapprove of the rubber sheet analogy, but this is a particularly striking example of the misunderstanding it induces, I think.

 

[Farsight]

All the empirical evidence you have put forward has been entirely consistent with MTW GR. You still haven't answered my request for empirical evidence that proves the MTW picture of GR is wrong while supporting FGR

Have you not read the thread? Black holes are black. Light can't get out. There's no chicken-little "waterfall" of infalling space. A vertical light beam emitted at the event horizon doesn't slow down, it doesn't fall back, and it doesn't curve back round. Just as a light beam passing between two stars doesn't curve. Just as a light beam passing between two black holes doesn't curve. So why doesn't it get out? Because the thing you call the coordinate speed of light is zero. And the empirical evidence isn't that the coordinate speed of light varies in a non-inertial reference frame. The empirical evidence is that the speed of light varies with gravitational potential, just like Einstein said. And with gravitational time dilation going infinite at the event horizon as measured by you and me, that means that at that location the speed of light is zero. That's why the light doesn't get out. And that's why the point singularity and the neverland KS infalling observer is wrong.

Just admit your mistake and move on; it's not a biggie (and perhaps you'll win back some respect).

I haven't made a mistake. Now go look in the mirror. Just admit your mistake and move on. Only I'm afraid it is a biggie.

 

[ctamblyn]

Ok. Again, who wants to buzz in?

Buzz.

They're exactly the same thing. They both describe a t = constant, theta = pi, r > 2m slice of the Schwarzschild solution.

ETA: As ben m pointed out further down, I intended "theta=pi/2" (the equatorial plane) not "theta=pi" (the negative z-axis).

 

[Farsight]

Just to add - there are reasons I disapprove of the rubber sheet analogy, but this is a particularly striking example of the misunderstanding it induces, I think.

In your own time, Edd.

 

[edd]

Ctamblyn doesn't surprise me by buzzing in correctly - he's more competent than I. Both are identical - the choice of extrinsic curvature direction is irrelevant.
My intention was to either bring yet more of Farsight's misunderstanding to light or, in the event he correctly identified the two as identical that the two embedding diagrams could be stitched along a fold of identical z in order to demonstrate that an extrinsic fold has no bearing on an intrinsic curvature.

For a layman - the intrinsic curvature should be measurable by instruments constrained to live within the surface. By definition they can't feel if it's going 'up' or 'down' - and indeed that's a meaningless thing to ask. For anyone living solely on those surfaces they cannot distinguish them by laying down lines and angles and measuring them.

 

[ctamblyn]

Have you not read the thread?
(...snip...)

Like I've said several times now: if I were wrong, you could just link to the post where you provided some empirical evidence which shows GR is wrong while FGR is right. That post doesn't exist, though.

All the evidence you have presented so far is consistent with ordinary GR, while FGR is so vague that even you seem unsure what counts as potential evidence for it. Two of the things suggested - loss of synch between nuclear and e/m clocks, and "electron stripping" (your term) - turned out not to be potential evidence at all (you stated that failure to observe those phenomena would not affect your confidence in what I called the "core" of FGR/relativity+).

I haven't made a mistake.
(...snip...)

Edd asked, very clearly, whether the folded paper had intrinsic curvature and you said "yes". You were wrong. The folded paper is intrinsically flat for the same reasons that the cylindrical surface is.

Obviously you're free to keep digging if that's what you choose, but I suggest that it would look better for you if you'd just admit your mistake.