QM + GR = black holes cannot exist

My pleasure, Paddoboy

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Another reply for Billy T from Prof. Carroll:

 

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You stand corrected on your correction of brucep that the force of gravity is not down to spacetime curvature.

http://curious.astro.cornell.edu/question.php?number=649

 

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Here is also a more refined detailed description of gravity, spacetime curvature, and tidal affects.

http://www.einstein-online.info/spotlights/geometry_force

where in part it says......

"The same is true for any curved surface: a tiny region of such a surface looks almost exactly the same as part of a plane. This indistinguishability is exactly analogous to the elusiveness of gravity that has been described above: For a very small spacetime region, say, the elevator of a free-falling observer, gravity is absent. Over a brief observation period, the interior of the elevator looks as if it were part of the spacetime of special relativity, where there is no gravity at all. Only in a larger spacetime region, the differences become measurable. Residual gravity, tidal forces come into play. This is completely analogous to geometric curvature: The larger a region of our curved surface, the larger the deviations from flat geometry, for instance from the law stating that angles of a triangle always sum up to 180 degrees.

Einstein took this analogy seriously, and he found that he could make it much more precise.

mechanics take on an especially simple form: As long as no external force is acting on an object, it will move on a straight line through space-time: at a constant velocity along a straight path.

Now we add gravity to the situation, for instance by placing a massive sphere somewhere in space. InNewton's theory of gravity, this sphere will exert a force on all other masses around it. If we place a test particle in the vicinity, we see that its movement is deflected away from the usual straight space-time line - its path would be curved towards the sphere and it would become accelerated as it feels the sphere's attraction.

In Einstein's geometric theory of gravity, the situation is described in a completely different way: A mass that we place in an region of space will lead to a distortion of space-time. Empty spacetime is flat - it looks exactly like the spacetime of special relativity. Spacetime in the presence of masses is curved. In curved space-time, there are no straight lines - just as there are no straight lines on the surface of a sphere. The closest we can get to the notion of a straight line is a geodesic, a spacetime curve that is as straight as possible. Test particles in the vicinity of the massive sphere follow these geodesics. Gravity does not reflect them from their straight lines - it re-defines what it means to move on a straightest possible line.

Einstein's universe performs an ever-ongoing cosmic dance, with matter and space-time interacting: A given configuration of matter distorts spacetime geometry. This distorted geometry makes matter move in certain ways. The movement changes the matter configuration as the sources of gravity change their locations. With the matter configuration changed, spacetime geometry changes as well. Now that spacetime geometry is a bit different, it also acts on matter in a different way, matter moves, geometry changes, and so on in an endless dance.

So what is gravity, in Einstein's universe? Generally speaking, any distortion of spacetime geometry. More precisely, there are two sides to gravity: In part, gravity is an observer artefact: it can be made to vanish by going into free fall. Most of the gravity that we experience here on earth when we see objects falling to the ground is of this type, which we might call "relative gravity". The remainder of gravity, "intrinsic gravity", if you will, manifests itself in tidal forces, and is associated with a specific property of geometry: The curvature of spacetime.

 

Prof. Berg addresses Billy T's questions:

 

Billy T, Prof. Jensen's reply to your comment:

 

Then I guess you haven't read Brian Greene's new book yet. They have been looking for very rough estimates indeed. I wonder if it has to do with something from them using Planck Units to describe it. At distances so small, it seems like numbers could become overly inflated.

 

Honestly, I can't imagine me ever reading anything by Greene.

 

Sitting here alone I read this and chuckled, out loud. Then I had to get up and go into my library and check to see if I had any of his books... Yup there is one and to my suprize it was full of highlights and notes.

I had no idea that I had any of his books but I do have quite a few both texts and popular books on physics, so I did have to check. After reading several of the highlights and notes all I can say, is I really miss the LIKE button.., not intended as use for Greene, but for your comment above. I have one of his books and would not recommend it.., and never picked up another.

Some of the physicists in the popular press have something of value to say in a way most anyone can understand it. Some are just fun to read, which may just mean they hired better ghost writers. Some don't fit in either of those categories.

 

I think BillyT's. question was excellent, inasmuch as a) it had a simple Yes/No answer, and that b) nobody seems the agree what that simple answer might be

Although it is well known I am not a physicist, let me ask a VERY closely related, but slightly more general,question

Assume a light-emitting celestial body - a star. Assume further that light is emitted from this body in all directions.

Disregarding atmospheric scattering, I observe my star to be discrete - i.e. not "fuzzy" at the perimeter (is this true? If so why? I believe that R. Feynman has an answer, related to why light appears to always travels in straight line)

If it is, this seems to suggest to me that the only light from this star that I see consists on "more-or-less" parallel photon beams.

Question: If these parallel photon beams interact via self-gravity, is my image of the diameter of a star at, say, 1 million light years distance, and assuming I can calculate the "image shrinking" over this sort of distance, different from the "true" diameter of the star?

Granted that this "true diameter" may not be possible to determine, do astrophysicists take this into account when making such measurements? If Yes, then the answer to BillyT's question is Yes. If No, then No likewise (assuming these guys know what they are talking about)

 

I should give Greene credit, or at least the producers of his last PBS show, for putting voices on the air that dissent from his view. My beef with him is that he promotes his speculative physics as being far more certain than it is and his particular interpretations of quantum mechanics as far more certain and intellectually dominating an interpretation than I believe they are.

And, of course, he actually does the work! So good for him for pursuing difficult physics and promoting it. I just can't get behind his work. It is really hard to promote science along with the tentative nature of much of its frontiers--maybe that's impossible to do.

 

Photons self interact when they contribute to the local spacetime curvature. They don't pull on eachother locally or over distance. If they did then Newtons lensing equation would be correct. You won't find any Feynman diagram that will suggest otherwise. Light travels in straight lines because the local spacetime is flat to some infinitesimal limit over distance and mathematically flat tangent to every point on the spacetime manifold [the one Einstein used to model GR]. You won't find any empirical lensing observations that would suggest any image shrinking associated with the source of the light being lensed. What's being observed is the source isn't where it would be if there was no spacetime curvature over the global geodesic. The fly in this discussions ointment is assuming Newtons action at a distance gravitational model is remotely correct for doing this kind of an analysis. It it was the Newton prediction of dphi = 2M/r would be correct and the Einstein prediction of dphi = 4M/r would be empirically wrong.

 

So they don't always "contribute to local spacetime curvature?"

I never mentioned Feynman diagrams - they are a different kettle of fish altogether.

I am sorry, I cannot parse this comment.. Maybe you could re-phrase it?

Just so you know, my differential geometry is reasonably strong, and I find your comments here confusing (or dare I say confused) to say the least

Which might have some meaning if it were explained what "dphi" and "r" actually refer to (assuming I take M to be Mass)

Are you completely sure you speak from a sound knowledge base?

 

They interact with spacetime curvature over the entire path. That's the local gravitational interaction. The curvature is gravity. The Feynman diagrams predict how the particles of light interact with matter. Of course it doesn't predict photons pull on eachother. As far as I can determine neither does quantum gravity. You should use your differential geometry to get dphi = 4M/r. The prediction Eddington confirmed with his famous observation confirming the GR prediction. You don't understand what it means? It's the deflection predicted by GR expressed in geometricized units. Most commonly used in gravitational physics. For light passing the Sun it would be

M_sun = 1477 meter
R_sun ~ 7E8 meter

dphi = 4(1477m)/7E8m = .00000211

I'm pretty sure but you can't be until you learn something about the physics.

BTW I don't know differential geometry but I know how to use the metric equations and subsequently understand the physics of GR and the predictions it makes with respect to natural phenomena in the GR domain of applicability. It's confusing for me that you don't other than you're mostly interested in the mathematics but not as much the physics.

 

No I don't. Bruce said the local spacetime curvature relates to the local force of gravity, it doesn't, it relates to the tidal force at that location. The local "spacetime tilt" relates to the local force of gravity. And like I said, you need spacetime curvature to have this "spacetime tilt". Without it your spacetime stays flat and level like it is in the centre of the Earth. That's why Riemann curvature is the defining feature of a real gravitational field. And note this in the Wiki article: "The curvature tensor represents the tidal force experienced by a rigid body".

But see above, bruce is still saying "the curvature is gravity". It isn't.

 

Farsight, you got the idea of "tilt" from a document that claims the tilt is due to spacetime curvature. Please read your sources.

 

I didn't get the idea from that document. I've known about this for years. Curved spacetime is associated with Riemann curvature which is associated with the tidal force. You need it to get your gravitational potential off the flat and level. The force of gravity depends on the gradient in the potential, and the tidal force depends on the change in gradient, which is the spacetime curvature. Basic stuff.

 

And yet the first time you use the word "tilt" in this context you introduce a citation that disagrees with your "basic" claim.

Why use a source that says that your "basic" understanding of GR is incorrect?

 

Since brucep has decided to get personal, I shan't bother to respond

Instead I will talk to Farsight who at least is usually civil in disputes.

First a nit-pick: Einstein's field equation use the Ricci curvature tensor, not that of Riemann. A small point as they are closely related.

Now the field equations of GR state, in absurdly simplified form, that in the presence of a (poorly defined) non-zero mass-energy source, the Ricci curvature is likewise no-zero. Conversely, if the source is zero, then curvature is zero. Surely the can be no disagreement about that?

So spacetime IS either flat or it is not. But GR is a relativity theory, so each observer is entitled to consider herself in flat spacetime, whatever the "true"answer to the 1st sentence in this para, may be.

Flatness or otherwise can only be resolved by comparing 2 (or more) observer's concept of flatness. This is what the tidal force is.

In other words, it is not quite correct to say that "curvature relates to tidal force", rather it is REVEALED by tidal force. (by edit) - and since you cannot have a tidal force without "gravity", it seems fair to assume that the curvature that is revealed by this force IS gravity

And before anyone says that the "reality" of spacetime curvature (in the case of a single observer) requires a god-like entity to decide it, take another look at the field equations and how they protect the conservation laws.