Black holes

[Ziggurat]

Don't be specious. A cube has six flat surfaces. A sphere has one surface, and it isn't flat. It's curved. And no finite region of that sphere is locally flat. Only an infinitesimal region is locally flat. A region of zero extent.

But we aren't limited to the equivalent of spheres. Not in geometry, not in reality. Yes, a sphere has no finite region of zero curvature. But that means nothing for something other than a sphere.

I didn't say that. Let me reiterate. Here's a picture of gravitational potential. If you take a small region, like at the bottom in the middle and say it's absolutely flat, and then shuffle over a little and say the next small region is absolutely flat, and then repeat ad infinitum, what you do is throw away the gravitational field. You end up with this instead of this. How much simpler do I have to make it before it sinks in?

I've been wondering the same thing. You've presented ONE potential, and you're using a claim about this one potential to draw categorical conclusions about all potentials. That is completely unjustified. And it is wrong.

Here's a really simple example: the gravitational potential of a spherical shell. What's the potential inside the shell? It's a constant. The potential is flat over the entire interior region. But it's curved outside that region. So that almost trivially simple solution contradicts your claim.

 

[ctamblyn]

Not that he's obliged to reply, but I'll assume that Farsight's lack of response to my posts #1063 and #1064 is an implicit consession of sorts. Either that or I'm on ignore. As for the lack of reply to #1065, I'd still appreciate an answer:

A spacetime metric that represents what, exactly? Here it is, it's like Pythagoras's theorem, we've got a spacetime interval and the usual y and z terms, but we've now got r terms. What do they represent exactly, and what real-world scenario does the overall expression describe?

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

Just out of interest... if you are unable to figure out what the terms above mean, how did you do it for the Schwarzschild metric? And how did you arrive at the conclusion that certain ranges of the Kruskal coordinates were unphysical, without first figuring out what the various terms meant? Did you just take someone's word for it?

ETA: Actually, this question is worth an answer too:

IMHO it would be more interesting if we had a black hole handy and could throw something straight in and see if we can detect electron stripping and 511KeV gamma photons etc.

OK, and if we could somehow do that and didn't detect any 511keV photons, would your confidence in Relativity+ be weakened (and by how much)? If you did detect those photons, would it be strengthened (and by how much)?

 

[ctamblyn]

It's space inside the spherical shell, not spacetime.

What?? I'm not talking about a shell that exists for an instant and then vanishes, Farsight. I'm talking about a rigid shell that can exist for an arbitrarily long time, so there's a spacetime region inside it.

Even in Farsight's view of the word in which spacetime is some mathematical fiction (ETA: and in which only space is real), he's still wrong - every spacelike slice of the shell's interior is a flat 3D space.

 

[ben m]

Farsight's flatness claim is really, really bizarre. Since he doesn't do math, I doubt he's receptive to the GR calculation showing the flat spacetime inside a spherical shell. But ... there's no such thing as a flat surface in ordinary geometry? Farsight can't form a mental image of a sphere with a dent in it? As in, the shape of a watermelon grown on a hard surface, which has finite curvature in most places and a big patch which can be perfectly flat?

I thought crackpots were supposed to be good at vague mental geometry problems, at the expense of everything else.

 

[dafydd]

Farsight's flatness claim is really, really bizarre. Since he doesn't do math, I doubt he's receptive to the GR calculation showing the flat spacetime inside a spherical shell. But ... there's no such thing as a flat surface in ordinary geometry? Farsight can't form a mental image of a sphere with a dent in it? As in, the shape of a watermelon grown on a hard surface, which has finite curvature in most places and a big patch which can be perfectly flat?

I thought crackpots were supposed to be good at vague mental geometry problems, at the expense of everything else.

His constant failure at physics is both amusing and educational.

 

[Reality Check]

You introduced the geodesic dome analogy. I was just going with the flow.

I introduce the geodesic dome analogy in the hope that it would educate you. You then went with a flow of ignorance about GR applied to that analogy.

I need the pretty pictures to explain it to you, because you don't understand it. The word "intrinsic" is a label that covers up a lack of understanding.

Wow - you really are determined to demonstarte you ignorance !
The word intrinsic is a label is used in the mathematics of spacetime. Curvature in spacetime is intrinsic. It iiddefined without reference to any enclosing space. GR has 4 dimensions. It is not a 4-space embedded in a 5-space where there can be a curvature defined with respect to that 5-space.

If your manifold is exactly flat locally in a region that is other than infinitesimal, it's exactly flat globally. It's like saying this part of the surface of the earth is exactly flat. Absolutely utterly flat. And the next part, and the next, and all other parts. Then the surface of the earth isn't a sphere any more, it's a flat plane. It isn't curved at all.

That is rather stupid. That is not what is stated. It is

• this patch of the surface of the earth is an exactly Minkowskian space. Absolutely utterly Minkowskian.
• And the next patch is also a different exactly Minkowskian space.
• And the next patch is also a different exactly Minkowskian space.
• and all other patches.

Then the surface of the earth isn't a smooth sphere any more - it is a geodesic sphere. It has a global curvature. It is locally Minkowskian ("flat") where every "flat" patch points in a different direction, i.e. it is not globally "flat".

The rest of your post is just repeating your ignorance of GR mathematics. There is no "tilt" becuse there is no magic plane to define that tilt.

 

[Farsight]

Farsight, why can you never admit that you were wrong about something? A person can't be right all the time.

I do. I was wrong the other day about neutrinos and photons arriving at the same time from a supernova. I said I was wrong and I said I was sorry. I've done it before and I'll do it again. It's because I admit to getting things wrong that I end up getting things right. Not everything, but most things.

 

[Farsight]

I introduce the geodesic dome analogy in the hope that it would educate you. You then went with a flow of ignorance about GR applied to that analogy.

Wow - you really are determined to demonstarte you ignorance !
The word intrinsic is a label is used in the mathematics of spacetime. Curvature in spacetime is intrinsic. It iiddefined without reference to any enclosing space. GR has 4 dimensions. It is not a 4-space embedded in a 5-space where there can be a curvature defined with respect to that 5-space.

That is rather stupid. That is not what is stated. It is

• this patch of the surface of the earth is an exactly Minkowskian space. Absolutely utterly Minkowskian.
• And the next patch is also a different exactly Minkowskian space.
• And the next patch is also a different exactly Minkowskian space.
• and all other patches.

Then the surface of the earth isn't a smooth sphere any more - it is a geodesic sphere. It has a global curvature. It is locally Minkowskian ("flat") where every "flat" patch points in a different direction, i.e. it is not globally "flat".

The rest of your post is just repeating your ignorance of GR mathematics. There is no "tilt" becuse there is no magic plane to define that tilt.

Oh boy. OK RC, shuffle sideways a foot or two so that you've got one foot on one patch, and the other foot on another. Are you in a region that's locally flat?

 

[edd]

Oh boy. OK RC, shuffle sideways a foot or two so that you've got one foot on one patch, and the other foot on another. Are you in a region that's locally flat?

Do you actually know what's meant by local, Farsight? Because you seem to be suggesting a situation where RC's feet are sufficiently widely spread that the term 'local' no longer applies.

 

[Farsight]

Do you actually know what's meant by local, Farsight? Because you seem to be suggesting a situation where RC's feet are sufficiently widely spread that the term 'local' no longer applies.

Sure I do edd. RC can keep his feet together and shuffle sideways to where one patch meets the other. He can have one foot on one patch and the other foot on another even when his feet are still together. Don't look for reasons to side with him because he's wrong.

Here's a really simple example: the gravitational potential of a spherical shell. What's the potential inside the shell? It's a constant. The potential is flat over the entire interior region. But it's curved outside that region. So that almost trivially simple solution contradicts your claim.

Shame there aren't any hollow planets. My "claim" is correct. The curvature you can see in the pictures of curved spacetime is Riemann curvature. No Riemann curvature, and your potential never gets off the flat and level. Then you can kiss your gravitational field goodbye.

 

[edd]

Sure I do edd. RC can keep his feet together and shuffle sideways to where one patch meets the other.

So now you're asking him to move his feet outside the region he's specified as flat, just so you can win the argument about whether the region he's in is flat.... great arguing tactic.

 

[Farsight]

Not that he's obliged to reply, but I'll assume that Farsight's lack of response to my posts #1063 and #1064 is an implicit consession of sorts. Either that or I'm on ignore.

No concession, and you're not on ignore. I've already answered your wifebeater questions about what you call FGR, and I'm not answering them again.

As for the lack of reply to #1065, I'd still appreciate an answer: Just out of interest... if you are unable to figure out what the terms above mean, how did you do it for the Schwarzschild metric?

All the terms are listed and defined, the scenario is adequately described, and even a child can spot the undefined result at R=2M.

And how did you arrive at the conclusion that certain ranges of the Kruskal coordinates were unphysical, without first figuring out what the various terms meant? Did you just take someone's word for it?[/i]

No. I arrived at the conclusion by understanding that t is a measure of local motion, and by appreciating that the speed of light reduces with gravitational potential, goes to zero at the event horizon, and can't get any lower than that. Light clocks don't tick because they're stopped, and putting a stopped observer in front of a stopped clock ain't going to make it start ticking again. Simple.

ETA: Actually, this question is worth an answer too: OK, and if we could somehow do that and didn't detect any 511keV photons, would your confidence in Relativity+ be weakened (and by how much)? If you did detect those photons, would it be strengthened (and by how much)?

It's hard to say. I said it would be more interesting if we had a black hole handy and could throw something straight in and see if we can detect electron stripping and 511KeV gamma photons etc. The electron stripping is the same sort of thing as do nuclear clocks stay synchronised with electromagnetic clocks as you change elevations. The underlying question there is is there any strength-of-space gradient in a proton? This isn't something I've examined, but I have to say I'm not fond of the idea of any kind of gravitational field at the nuclear level. I am however confident that if an electron travelling fast encounters a region where the local speed of light is less than its linear speed, it cannot survive as an electron. However note what I said about neutrinos on the relativity+ thread. The rotational aspect of electron spin can't just disappear, an electron on its own isn't going to turn into a 511keV photon. If it stayed with the proton then maybe yes we can force an annihilation, but otherwise we're going to see neutrinos. You can't use them to drive a power plant. If we don't see any photons or neutrinos I'd be thinking where did I go wrong? But I wouldn't then be losing confidence in the big picture. Instead I'd be looking for some error in some aspect of it.

 

[Farsight]

So now you're asking him to move his feet outside the region he's specified as flat, just so you can win the argument about whether the region he's in is flat.... great arguing tactic.

It's no tactic. He can claim that region x is flat and the adjacent region y is flat, but he can't get away with shoving the slight curvature that's there in x and y into some nil-extent region between them. What, you step sideways a smidge and now you're suddendly straddling an ultra-curved region of nil extent? It's nonsense edd. Total garbage.

Right, I've got to go. Catch you later.

 

[W.D.Clinger]

Farsight versus Einstein, part 4

Farsight continues to disagree with Einstein's definition of the gravitational field:

No Riemann curvature, and your potential never gets off the flat and level. Then you can kiss your gravitational field goodbye.

Nope. According to Einstein, the gravitational field is described by the connection.

It's a mathematical fact that the connection can be nonzero even when the Riemann curvature is zero. That has been pointed out several times. Einstein's words have been quoted, and a concrete counterexample to Farsight's claim has been provided:

http://www.internationalskeptics.com/forums/showpost.php?p=8188847&postcount=1034
http://www.internationalskeptics.com/forums/showpost.php?p=8194020&postcount=1066
http://www.internationalskeptics.com/forums/showpost.php?p=8201601&postcount=1083
​

 

[DeiRenDopa]

Farsight's flatness claim is really, really bizarre. Since he doesn't do math, I doubt he's receptive to the GR calculation showing the flat spacetime inside a spherical shell. But ... there's no such thing as a flat surface in ordinary geometry? Farsight can't form a mental image of a sphere with a dent in it? As in, the shape of a watermelon grown on a hard surface, which has finite curvature in most places and a big patch which can be perfectly flat?

I thought crackpots were supposed to be good at vague mental geometry problems, at the expense of everything else.

It's actually more bizarre than you may have imagined, if that's possible.

Consider these two Farsight posts:

It is true. When you measure the difference between the speed of light at the ceiling as opposed to the floor in a very very tall building, you find that on floor 100,000 the difference is less than it is in the basement. If you could take an equatorial slice through the planet and measure the speed of light at various locations, when you plot them out what you end up with is the upturned hat. When you look at one small region of this there's still a gradient in gravitational potential and a gradient in c. There's no discernible curvature of the gradient, you can't detect any tidal force, but things still fall down, and light still curves when it moves through space, so we say spacetime is curved. Way out a zillion miles to the left or right where there's no discernible gradient, your measurements can't detect any curvature of light, or things falling down, and then we say spacetime is flat.

It's a phrase we use to in conjunction with a gravitational field, which definitely exists. If it didn't, you wouldn't fall down. Note however that a gravitational field is a region where there is a gradient in gravitational potential. You could place two clocks within voids at the centre of two large but unequal masses. In these voids the clocks don't fall down, but they do tick at different rates because the conditions we label as gravitational potential are not the same. The clocks ticking at different rates tells us that gravitational potential does label something very real.

In other words, Farsight is well aware of the fact that gravitational gradients and potentials (his words, not mine) depend, in part, upon the distribution of mass (i.e. they can be something other than an "upturned hat").

However, when it comes to flatness, he clearly has not yet learned calculus (well, we already knew that; his physics is, after all, founded on pre-Newtonian concepts).

But more strange is that he has not yet - apparently - learned the need for internal consistency (with respect to his own ideas).

 

[Ziggurat]

Shame there aren't any hollow planets.

What, planets are the only source of gravity? No, any mass will produce gravity. And there sure as hell are spherical shells of mass. Furthermore, we're talking about what the theory of general relativity says. And the theory makes predictions even for things that might not currently exist. There is certainly nothing that prohibits spherical shells of mass.

My "claim" is correct. The curvature you can see in the pictures of curved spacetime is Riemann curvature. No Riemann curvature, and your potential never gets off the flat and level. Then you can kiss your gravitational field goodbye.

Argument by picture? You really are becoming the new Michael Mozina. But you're still wrong. You're taking a characteristic of particular mass distributions and drawing conclusions about the potentials for all possible mass distributions. That is both obviously illogical to do, and furthermore it's already been proven explicitly wrong by counterexample. So no, your "claim" (why would you put quotes around it?) is not correct at all.

 

[ctamblyn]

No concession, and you're not on ignore. I've already answered your wifebeater questions about what you call FGR, and I'm not answering them again.

What do you mean "again"? Not once did you produce experimental evidence that showed GR is wrong while FGR is right. If I'm wrong, then link to the post where you did.

All the terms are listed and defined, the scenario is adequately described, and even a child can spot the undefined result at R=2M.

But if you were given just the Schwarzshild line element and the coordinate ranges, how would you deduce the "meanings" of those coordinates?

No. I arrived at the conclusion by understanding that t is a measure of local motion, and by appreciating that the speed of light reduces with gravitational potential, goes to zero at the event horizon, and can't get any lower than that. Light clocks don't tick because they're stopped, and putting a stopped observer in front of a stopped clock ain't going to make it start ticking again. Simple.

Actually, forget my previous question as this explains the interesting bit anyway. Your assumption that light cannot get through the horizon and reach the central singularity guided you in your interpretation of the Schwarzshild metric. You didn't start with the metric - the solution to the GR field equations - and then try to figure out what the physical situation is. You have basically added an extra assumption into GR, amounting to the very assertion that you claimed to show that GR itself supported (that light cannot reach the central singularity). In other words, you have begged the question.
http://en.wikipedia.org/wiki/Begging_the_question

(...snip...)
If we don't see any photons or neutrinos I'd be thinking where did I go wrong? But I wouldn't then be losing confidence in the big picture. Instead I'd be looking for some error in some aspect of it.

So the lack of the 511 keV photons would not be evidence against the core of Relativity+, if you will. Therefore the presence of those photons could not logically be evidence in support of the core of Relativity+.

----------

Here's a proof, since I have a few minutes to spare: Let R be the event that "core" of Relativity+ is true, and P be the event that photons are observed. Then, by the normal properties of conditional probability,

Pr(R) = Pr(R | P)Pr(P) + Pr(R | ~P)Pr(~P),

where the tilde ~ denotes "not". Here, Pr(R) means "the probability that the core of Relativity+ is true before the experimental results are seen". Pr(R | P) means "the probability that the core of Relativity+ is true if the experiment detects photons", and Pr(R | ~P) means "the probability that the core of Relativity+ is true if the experiment does not detect photons".

However, you have said that not observing the photons has no effect on Pr(R). Thus,

Pr(R | ~P) = Pr(R).

Using that in the above equation together with Pr(~P) = 1 - Pr(P), some simple algebra yields the result

Pr(R | P) = Pr(R).

In other words, the probability that the core of Relativity+ is true if the experiment observes photons is just the same as the original probability that the core of Relativity+ is true. In other words, the observation of photons is not evidence for the core of Relativity+.

 

[Reality Check]

Oh boy. OK RC, shuffle sideways a foot or two so that you've got one foot on one patch, and the other foot on another. Are you in a region that's locally flat?

No. Duh !


One foot in a region that is flat and pointing in one direction. The other foot in a region that is flat and pointing in another direction! So you are doubly wrong

1. There is no one region that is flat.
2. There is curvature.

This does not change what you cannot understand: A geodesic dome is not a flat plane !

I suspect that you are just unable to understand what flat means in this context. You seem to think that it means that every flat patch on the geodesic dome

1. Has magically fallen onto a plane that has equally magically appeared in your head.
2. Has changed their orientation to match that magical plane.

What flat means for the geodesic dome is that each patch can be measured as flat without referring to any external structure. This is really good because there are no external structures. There is just the geodesic dome. So the definition of flat would be something: A region in which an triangle's internal angles add up to 180 degrees.

General Relativity does not use extrinsic curvature. Spacetime is not embedded in the higher dimention space that is required to define extrinsic curvature.

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.

 

[Farsight]

It's still garbage RC.

Anyway, I think this thread has run its course guys. I hope you've found it useful, and that some of your have learned some physics. And I sincerely hope that if you hear somebody waxing lyrical about a black hole being surrounded by infalling space, you'll take it with a pinch of salt next time and recognise it for the woo that it is.

 

[theprestige]

It's still garbage RC.

Anyway, I think this thread has run its course guys. I hope you've found it useful, and that some of your have learned some physics. And I sincerely hope that if you hear somebody waxing lyrical about a black hole being surrounded by infalling space, you'll take it with a pinch of salt next time and recognise it for the woo that it is.

After all this, your argument is defeated by a geodesic dome?