QM + GR = black holes cannot exist

[paddoboy]

I see you've given up the Galileo fight. I'll consider that a moral victory.

Not at all, as is evidenced in my red hering comment in association with strawmen.

And no, I don't personally believe that black holes exist. The overwhelming evidence can be otherwise explained. You might have noticed that Tashja's expert opinions literally contradicted each other in certain aspects; that's a round-about way of saying that "we simply don't know yet".

That's your perogative to deny evidence for BH's as much as you like, but from an outside FoR, it does not appear to be very smart.
And no, I don't believe the expert opinions did contradict each other....In fact they down right declared that BH's and EH's certainly do exist, but admitted the unknown factor re quantum effects.

 

[brucep]

Geometric units. The solution is dimensionless. The slight acceleration is nonsense. It's a local inertial free fall frame. The same frame for Newton First Law Of Motion. If the object in free fall is predicted to experience accelerations then the derivation would reflect that. It doesn't.

For Newton coordinates

dr/dt = [2M/r]^1/2

For local Schwarzschild coordinates

dr_shell/dt_shell = [2M/r]^1/2

For local rain coordinates

dr/dt_rain = [2M/r]^1/2

None of these derivations predict tidal accelerations effect the rate of free fall. With tidal accelerations you get the nonsense prediction that I exampled. That the rate at r=2M exceeds c. Newton doesn't predict it. Einstein doesn't predict it. What seems to confuse folks (sometimes) is thinking gravity is a force. That the mass of the object in free fall interacts with the mass M via a force. Over the natural path local spacetime curvature is an infinitesimal. Over a very long path the tidal accelerations may change the path slightly. For example the thought experiment where the dude is in the falling elevator with two particles and eventually can measure a slight deviation of the particle path with respect to his path. The tidal accelerations don't change the rate of free fall.

For some reason the brackets for the equations won't post. You know where the brackets go. Wrote the equations out again to see if I can get the brackets. Didn't work. ? ? What happens is the brackets appear for an instant then change to what you see now. When I call up edit the brackets are there. When I post they're gone. The brackets are gone from all the equations of motion I wrote down. At least on this IPAD. Could be a local problem. LOL. The brackets I used for the word 'sometimes' are gone also. Whatever. Maybe someone could tell me if the brackets are missing on their screen? Changed the bracket set and it worked. ??

So, in effect we all agree that BH's and associated EH's most certainly do exist.
In reality, other then for sensationalism purposes, there was never any real doubt.

The observational evidence says they exist as real natural phenomena. The quantum gravity papers that have been linked as a proof that they don't exist as natural phenomena have been misinterpreted by most everybody. Declan tried to explain why but not many were listening. It would be boring but for being pissed off at the nonsense threads started by whoever.

 

[paddoboy]

The observational evidence says they exist as real natural phenomena. The quantum gravity papers that have been linked as a proof that they don't exist as natural phenomena have been misinterpreted by most everybody. Declan tried to explain why but not many were listening. It would be boring but for being pissed off at the nonsense threads started by whoever.

In light of the mysterious disappearance of the "LIKE" button, I would like to submit "I LIKE"

 

[Layman]

Couldn't you say that the light from a distance galaxy would be information that was lost from that galaxy forever? If the galaxy was moving away from ours at close to the speed of light, information about that photon might never be able to make it's way back to that galaxy ever! Anything inside of the visible universe would have to be trapped in the visible universe forever, or information would be lost. Spacetime traveling outside the visible universe would be traveling faster than light relative to our galaxy. Then there would have to be an information barrier at the edge of the visible universe according to quantum mechanics. Nothing could escape it or the information left from it would be lost forever. Oh noes, what have I done?

Then according to quantum mechanics the universe couldn't be larger than the visible universe or quantum mechanics wouldn't be completely valid either. Maybe, it just means scientist put way too much confidence in things like conservation of information... The visible universe itself is like a black hole. Heck, you could go on and say that quantum mechanics says the universe cannot be infinite by this line of reasoning. Then black holes exist, and things get lost in them forever. Quantum mechanics doesn't always guarantee that you will always find your car keys. That type of information can become lost to you forever, and it is possible you will never find them and have to get new ones made.

 

[tashja]

Sorry. Another glitch prevented me from posting.

 

[tashja]

RJBeery: Start with an existing black hole and an event horizon radius R at time T. Say the black hole is being "fed" an infinite series of golf balls, one after the other, which are all stamped numerically such that the current golf ball external to the event horizon is 1.0 * 10^32.

See linked img: http://i1373.photobucket.com/albums/ag380/rjbeery/golfball_black_holes_zps339d1899.png

Now, starting at time T, run the clock backwards to T_past until R_past = R/2. What does the scene look like? Do golf balls with numbers less than 1.0 * 10^32 appear? If they do then there is a time T_crossover such that T_past < T_crossover < T where we would have witnessed the event horizon expand due to matter crossing it. This cannot happen. If the golf balls numbered 1 through 1.0∗1032−1 ever existed then we would theoretically be able to observe them, with perfect instrumentation, forever. But in this thought experiment the black hole at time T is made of nothing but golf balls numbered 1 through (1.0*10^32)-1. This is a contradiction, therefore the event horizon cannot exist.

Prof. Unruh:

There is a difference between seeing and being. The event horizon is the surface such that light travelling along this surface, will take an infinite time to get out from the vicinity of the black hole to infinity (to the outside observer.) To see something, the light has to get out to you from the object. Thus the light by which you see it takes a long time to get out. Does that mean that because the light takes a long time to get out, the event did not happen? Surely not.

Imagine a waterfall where the water at some point flows faster than sound. It again takes sound an infinite time to get out from that point--It travels out very very slowly very near that point. But just because you never heard the fish scream as it fall over that point does it mean that the fish never went over the waterfall?

Imagine one flashed a light at the golf ball and looked at it in the reflected
light. There would be a last flash which you would ever see hit the golf ball.
That flash could be used to define the time at which the golf ball hit the
horizon.

Note that I am accepting the contention that you could see the golf ball

arbitrarily close to the horizon. The problem is that because the light takes

a successively longer time to get out, the light is also red-shifted.

Thus to see it, the light leaving the golf ball would have to have successively higher
frequency, and higher energy.

Very very rapidly, the light would have to have a frequency such that each photon had more energy than the energy (mc^2) in the whole universe. This would itself alter the geometry of the system. I.e, the statement that one could always continue to see the ball is actually
not true.

William G. Unruh | Canadian Institute for|
Physics&Astronomy | Advanced Research |
UBC, Vancouver,BC | Program in Cosmology |
Canada V6T 1Z1 | and Gravity | www.theory.physics.ubc.ca/

 

[Billy T]

I gave you an answer in post 241. Over the path of the falling photon [that's right the photon is in free fall along the null geodesic] it's energy is a constant of the motion. The relativistic energy equation is derived from this. The energy and momentum are constants of the motion for all objects including light along the natural path.

I don't understand this "free fall along the null geodesic" but think you are saying the photon has a definite energy, E, if that could be measured in a reference frame traveling with it (at speed of light) because I don't know what the frequency would be and think the energy is directly proportional to that. If it could be then the "energy mass" of M = E/c^2 is constant too. Certainly if you absorb 10 billon red color photons in a calorimeter your measure a temperature rise less than when the color is green. If measuring the frequency in a frame moving ever closer to the photon speed of travel the "red shift" is transforming the AC electric field into one that is static in time but modulated in space.

I don't know if it helps or is just more confusion, but photons are typically a few dozen centimeters long - a finite number of cycles, so don't even by simple Fourier analysis have a precisely well defined frequency and energy {much less when QM's E[delta T} product > 0 is considered.

I think energy and momentum of a closed system are both constants, including photo traveling thru "empty spaces."

I don't know your answer to the mutual gravitational attraction question I asked. I.e. if two photons are side by side and traveling along two exactly parallel paths only one nanometer apart for length of a million light years, do they merge into a common path? Or oscillate thru each other and follow the average path direction? Or have no effect on the other's path despite being a source of very weak gravity {I think you are saying they do.)? Each by its self is not a "closed system" if they do make weak gravity. Assume they are not in any other gravitational field the might have a gradient making such long parallel paths impossible.

Paddoboy in post 277 said:
"question re a photon adding to gravity/spacetime curvature.
It does, but by an insignificant amount, solely due to its momentum."

So he is suggesting I think that each would curve the space to make the other's path move closer to itself - I.e. they merge if originally "side-by-side" on exactly parallel paths for long enough time in other wise space with no gravity.

I don't know why it is not equally valid to say: "Solely due to its energy" That seems more natural to me as they do have "energy mass" of M = E/c^2.

 

[OnlyMe]

I asked 6 questions in post 234 here: http://www.sciforums.com/threads/qm-gr-black-holes-cannot-exist.142658/page-12#post-3230015
......
Only only me gave some discussion, but not any clear answer. One problem with his discussion suggesting that photons do "warp space time" {make gravity) is he speaks of the photon energy as the mass making this gravity. {but there is no unique "photon energy" E for all frames.)

Billy, I do tend to ramble and my earlier reply was filled with qualifications and hedging...

I did say that I don't believe that photons have an independent gravitational field. Maybe I should have just said they have no gravitational field of their own origin.

If they did we would see photons interfering with each other all over the place. Something that we don't observe.

So two photons with close parallel paths won't merge to a single path over time and/or distance.

This much I thought actually addressed the bulk of what I thought you were driving at. The rest and that which follows was attempting to point out that any answer must depend on from what underlying theoretical basis you begin.

I hedged partly because GR can be interpreted to suggest that photons do contribute to gravity. I think Bruce was on that track.

I also hedged my comment some based on some of the work of Haisch et al which suggests that both inertia and gravity emerge from an interaction between matter and a very short EM wavelength spectrum of the ZPF associated with the zitterbewegung motions of fundamental particles. In that work, partially to address energy density problems and its affect on the cosmological constant, photons, or in the context of their work, the EM spectrum of the ZPF, does not self gravitate. Keep in mind that they are working from the perspective of SED and treat the EM spectrum of the ZPF as real, rather than virtual. Which does not seem unreasonable to me since we know that photons are zipping through all of what we call the vacuum.... But gravity emerges from the interaction of that portion of the EM background and matter. From there came my earlier wording re photons and independent gravitational fields.

I like much of what I see in the Haisch et al approach, even understanding that there are unanswered problems, just like all other QM approaches to gravity, so far.

 

[OnlyMe]

I don't know why it is not equally valid to say: "Solely due to its energy" That seems more natural to me as they do have "energy mass" of M = E/c^2.

I don't believe that the equation E=mc^2 or M=E/c^2 can be applied directly to the photon, other than in the original context of how the mass of an atom is affected by photon absorption/emission events. We can speak of the frequency, wavelength and momentum associated with a photon, but not mass.

Another hedge.., whether you are dealing with GR or quantum gravity, photons are affected by gravity.., and may play some role in the dynamics of curved spacetime.

 

[brucep]

I don't understand this "free fall along the null geodesic" but think you are saying the photon has a definite energy, E, if that could be measured in a reference frame traveling with it (at speed of light) because I don't know what the frequency would be and think the energy is directly proportional to that. If it could be then the "energy mass" of M = E/c^2 is constant too. Certainly if you absorb 10 billon red color photons in a calorimeter your measure a temperature rise less than when the color is green. If measuring the frequency in a frame moving ever closer to the photon speed of travel the "red shift" is transforming the AC electric field into one that is static in time but modulated in space.

I don't know if it helps or is just more confusion, but photons are typically a few dozen centimeters long - a finite number of cycles, so don't even by simple Fourier analysis have a precisely well defined frequency and energy {much less when QM's E[delta T} product > 0 is considered.

I think energy and momentum of a closed system are both constants, including photo traveling thru "empty spaces."

I don't know your answer to the mutual gravitational attraction question I asked. I.e. if two photons are side by side and traveling along two exactly parallel paths only one nanometer apart for length of a million light years, do they merge into a common path? Or oscillate thru each other and follow the average path direction? Or have no effect on the other's path despite being a source of very weak gravity {I think you are saying they do.)? Each by its self is not a "closed system" if they do make weak gravity. Assume they are not in any other gravitational field the might have a gradient making such long parallel paths impossible.

Paddoboy in post 277 said:
"question re a photon adding to gravity/spacetime curvature.
It does, but by an insignificant amount, solely due to its momentum."

So he is suggesting I think that each would curve the space to make the other's path move closer to itself - I.e. they merge if originally "side-by-side" on exactly parallel paths for long enough time in other wise space with no gravity.

I don't know why it is not equally valid to say: "Solely due to its energy" That seems more natural to me as they do have "energy mass" of M = E/c^2.

The null geodesic is the natural path light follows. Lights natural path through curved spacetime [gravity]. Technically the light is in free fall along the natural path. For all objects following the natural path the energy and momentum are constants of the motion [makes sense]. When you derive the relativistic energy equation, for SR, momentum and energy are constants of the motion. That fits your comment referring to a closed system. This closed system would be the photon and the local spacetime curvature. GR is a local theory of gravity. No action at a distance forces to include in the physics. No forces or accelerations to account for just as Newton predicted with his law of inertial motion. So what does this mean? It means the photon energy contributes to the local spacetime curvature as it passes along its natural
path. In this sense it has a very small effect on the local spacetime geometry. You called it a small gravitational field. The energy and momentum are constants of inertial motion for the photon, just like every object in free fall, which verifies there is no exchange of energy with the field as it passes by. FYI: GR predicts the local spacetime curvature is an infinitesimal. Since the tidal field is the first derivative of the spacetime curvature local tidal effects are an infinitesimal. By local I mean a small set of points on the manifold that has some physical extent. I think this is enough for you to get answers to your query.

 

[OnlyMe]

The null geodesic is the natural path light follows. Lights natural path through curved spacetime [gravity]. Technically the light is in free fall along the natural path. For all objects following the natural path the energy and momentum are constants of the motion [makes sense]. When you derive the relativistic energy equation, for SR, momentum and energy are constants of the motion. That fits your comment referring to a closed system. This closed system would be the photon and the local spacetime curvature. GR is a local theory of gravity. No action at a distance forces to include in the physics. No forces or accelerations to account for just as Newton predicted with his law of inertial motion. So what does this mean? It means the photon energy contributes to the local spacetime curvature as it passes along its natural
path. In this sense it has a very small effect on the local spacetime geometry. You called it a small gravitational field. The energy and momentum are constants of inertial motion for the photon, just like every object in free fall, which verifies there is no exchange of energy with the field as it passes by. FYI: GR predicts the local spacetime curvature is an infinitesimal. Since the tidal field is the first derivative of the spacetime curvature local tidal effects are an infinitesimal. By local I mean a small set of points on the manifold that has some physical extent. I think this is enough for you to get answers to your query.

Bruce, though in the past I have been at odds with the way you describe some aspects of GR, I think I am in agreement with the way you explained things here.

I say I think, only because I have too often in the past, misunderstood the underlying intent.

Anyway I give this post a thumbs up.

 

[Layman]

I don't believe that the equation E=mc^2 or M=E/c^2 can be applied directly to the photon, other than in the original context of how the mass of an atom is affected by photon absorption/emission events. We can speak of the frequency, wavelength and momentum associated with a photon, but not mass.

Another hedge.., whether you are dealing with GR or quantum gravity, photons are affected by gravity.., and may play some role in the dynamics of curved spacetime.

I don't think you can apply E=mc^2 to a photon either. A photon is considered a form of energy. If it had zero mass then according to Einsteins equation, it would have zero energy as well. Then finding the mass of a photon would give a non-zero answer. I think E=mc^2 is more of an approximation, and Einstein just pulled it out of a hat. It actually doesn't describe quantum gravity. I mean who really cares about anything that involves E=mc^2 in the equations anyways?

If it was the secret to quantum gravity, I would bet a lot of physicist would be hitting themselves over the head because of that one. Although, it does seem like it would be interesting, because then the mass of the photon would mostly depend on its wavelength, since E = hc/λ for the photon. λ would be the wavelength, h is the Planck constant, and c is the speed of light. Then for the mass of the photon it would be something like, m = h/λc.

With the discovery of the Higgs Boson, it would seem like the wavelength would have a lot to do with the mass of a particle. It is supposed to mean that gravity is created by the Higgs Field. The Higgs Field would determine the mass of a particle by how much drag a particle creates in the field. Then the wavelength could be a determining factor of the amount of drag a particle could create.

It has only been proven that photons are massless to a certain degree of error. The main problem about finding the mass in elementary particles is that equipment isn't accurate enough to detect the force of gravity. Like I said before, the photon could have a small mass that would fit into this degree of error, and if it is truly zero than we would be completely screwed in finding a theory of quantum gravity because frankly zero doesn't really do much for any type of equations.

 

[RJBeery]

Wow Tashja, you got Unruh to respond. I'm pretty impressed with your ability to get experts to chime in! I do have some follow-ups for his comments but they would be wasted in this thread. Perhaps you could give me his address? I would cc you on any correspondence!

 

[brucep]

Bruce, though in the past I have been at odds with the way you describe some aspects of GR, I think I am in agreement with the way you explained things here.

I say I think, only because I have too often in the past, misunderstood the underlying intent.

Anyway I give this post a thumbs up.

The real problem is nobody understands what I'm talking about because it's GR and very few people are familiar with how the theory works. You think you are but you don't have the scholarship since you haven't done the work to find out how it works [the actual physics]. Everything I've written about relativistic physics is based on my studies associated with Einsteins model of gravity. That's the way it will always be in a venue such as a public science forum. No problem but everytime you disagreed with my interpretation you're disagreeing with relativistic physics in general. You've made it clear many times that you don't trust the physics, predictions, experimental results, etc.. My underlying intent was to discuss the physics. All the other bullshit is the result of trying to discuss it where the cranks rule.

 

[brucep]

I don't understand this "free fall along the null geodesic" but think you are saying the photon has a definite energy, E, if that could be measured in a reference frame traveling with it (at speed of light) because I don't know what the frequency would be and think the energy is directly proportional to that. If it could be then the "energy mass" of M = E/c^2 is constant too. Certainly if you absorb 10 billon red color photons in a calorimeter your measure a temperature rise less than when the color is green. If measuring the frequency in a frame moving ever closer to the photon speed of travel the "red shift" is transforming the AC electric field into one that is static in time but modulated in space.

I don't know if it helps or is just more confusion, but photons are typically a few dozen centimeters long - a finite number of cycles, so don't even by simple Fourier analysis have a precisely well defined frequency and energy {much less when QM's E[delta T} product > 0 is considered.

I think energy and momentum of a closed system are both constants, including photo traveling thru "empty spaces."

I don't know your answer to the mutual gravitational attraction question I asked. I.e. if two photons are side by side and traveling along two exactly parallel paths only one nanometer apart for length of a million light years, do they merge into a common path? Or oscillate thru each other and follow the average path direction? Or have no effect on the other's path despite being a source of very weak gravity {I think you are saying they do.)? Each by its self is not a "closed system" if they do make weak gravity. Assume they are not in any other gravitational field the might have a gradient making such long parallel paths impossible.

Paddoboy in post 277 said:
"question re a photon adding to gravity/spacetime curvature.
It does, but by an insignificant amount, solely due to its momentum."

So he is suggesting I think that each would curve the space to make the other's path move closer to itself - I.e. they merge if originally "side-by-side" on exactly parallel paths for long enough time in other wise space with no gravity.

I don't know why it is not equally valid to say: "Solely due to its energy" That seems more natural to me as they do have "energy mass" of M = E/c^2.

Look at section 1.5.2. I used to look at this site twenty years ago.

http://www.physicsguy.com/ftl/html/FTL_part1.html

 

[OnlyMe]

The real problem is nobody understands what I'm talking about because it's GR and very few people are familiar with how the theory works. You think you are but you don't have the scholarship since you haven't done the work to find out how it works [the actual physics]. Everything I've written about relativistic physics is based on my studies associated with Einsteins model of gravity. That's the way it will always be in a venue such as a public science forum. No problem but everytime you disagreed with my interpretation you're disagreeing with relativistic physics in general. You've made it clear many times that you don't trust the physics, predictions, experimental results, etc.. My underlying intent was to discuss the physics. All the other bullshit is the result of trying to discuss it where the cranks rule.

First, you did or did not get that I was saying I was in agreement with the post I was responding to?

Second, when you begin a post with a comment like that in bold above, you start to sound like Farsight, claiming some kind of final unchallengeable authority.

As for the rest.., I don't know what your personal issues are, so I just won't comment further, other than to say,

Learn to take a complement and let it drop at that.

 

[danshawen]

It's too late to say that BHs "cannot form", or "cannot exist" according to QT after they have actually been "observed" in our own galaxy. What I don't understand is that I have even talked to physics teachers who argue about this and by doing so conveniently forget about the existence of Sagittarius A. Stars orbit it. Nothing but radio waves come out of it, or the region it is in.

The situation would be different if this thread posited that a certain type of black hole, within a certain range of mass or constitution could not exist, or that QT was inadequate to describe its dynamics. I might agree with the latter idea, at some level.

OnlyMe, for example, claimed in one of the later posts that photons alone could not constitute the mass/energy of a black hole. The one equation that IS common to both QT and classical physics is E = mc^2, and it means exactly what it says. It was originally derived without giving photons any mass, but they nonetheless possess energy, which amounts to the same thing. If the photons are bounded, as they would be in a black hole, they can even have inertia, the same way they do when they are bound in the energy states of electrons within atoms. In other words, black holes can exist without any matter in their cores or inner shells. It would behave exactly the same in terms of gravitation if all of the mass had been accreted to energy, and all of that mass converted to energy was orbiting in the region of the event horizon. It may not be able to form with this constitution, but almost certainly, it can end in this state. This could even be the reason such beasties slowly (in cosmological time) evaporate in the manner they have been observed to do. This process is also something that we can directly observe happening in the Milky Way.

 

[OnlyMe]

OnlyMe, for example, claimed in one of the later posts that photons alone could not constitute the mass/energy of a black hole. The one equation that IS common to both QT and classical physics is E = mc^2, and it means exactly what it says. It was originally derived without giving photons any mass, but they nonetheless possess energy, which amounts to the same thing.

I don't believe I ever said anything like, "photons alone could not constitute the mass/energy of a black hole.". And in saying this I am not claiming that they could, or event that what we experience as EM radiation or photons, actually exists within an event horizon. I don't know anything about the physical reality or composition of what lies inside the event horizon, of what we call a black hole. This is where old maps used to say, "beyond here there be monsters". Today we temper that by saying beyond here {meaning an event horizon}, is the domain of theory, orr in the context of lay discussion very often imagination.

Ask a varied group of physists what E = mc^2 means and how it applies to physics and you will get many different and differing answers. For myself I will once again through out a paraphrased quote from a still up recovered source,

All mass can be thought of as an expression of energy, but not all energy can be thought of as mass.​

I don't believe that mass and energy are the same thing, though there is an obviously complex relationship between the two.

 

[nimbus]

The null geodesic is the natural path light follows. Lights natural path through curved spacetime [gravity]. Technically the light is in free fall along the natural path. For all objects following the natural path the energy and momentum are constants of the motion [makes sense]. When you derive the relativistic energy equation, for SR, momentum and energy are constants of the motion. That fits your comment referring to a closed system. This closed system would be the photon and the local spacetime curvature. GR is a local theory of gravity. No action at a distance forces to include in the physics. No forces or accelerations to account for just as Newton predicted with his law of inertial motion. So what does this mean? It means the photon energy contributes to the local spacetime curvature as it passes along its natural
path. In this sense it has a very small effect on the local spacetime geometry. You called it a small gravitational field. The energy and momentum are constants of inertial motion for the photon, just like every object in free fall, which verifies there is no exchange of energy with the field as it passes by. FYI: GR predicts the local spacetime curvature is an infinitesimal. Since the tidal field is the first derivative of the spacetime curvature local tidal effects are an infinitesimal. By local I mean a small set of points on the manifold that has some physical extent. I think this is enough for you to get answers to your query.

Brucep
Your post reminded me of this from the MIT site.
Please see pdf for equations, they come out screwy with my cut and paste quote here.

What then is the path of a freely-falling body in the presence of gravity? According to Newtonian physics, the answer is given by solving ~F = m~a = md2~x=dt2 as a differential equation for ~x(t).
Gravity causes acceleration, and one might expect therefore that the maximal-aging result breaks down. A body obviously accelerates in a gravitational field, and so an unaccelerated path cannot be the correct one. However, according to general relativity it still works out that the correct path is the one that maximizes proper time!
It seems astonishing that a result from special relativity carries over directly to general relativity without modification. The key is that, in the paradigm of general relativity, free-fall motion arises not from acceleration but from the effects of spacetime curvature. As we will see, the appearance of acceleration arises naturally from extremal paths in a curved spacetime.
We say "appearance of acceleration" because ordinary acceleration depends on the motion of one's reference frame. In an inertial reference frame in Newtonian gravity, a body moves at a constant velocity if no forces act on it. In Newtonian theory, an inertial reference frame can be extended over all of spacetime. But we have already argued in the first set of notes that there are no global inertial reference frames in curved spacetime.2 Consequently the notion of acceleration is ambiguous! Acceleration depends on frame, and if there are no preferred frames, there is no preferred concept of acceleration.
The ambiguity of acceleration is encapsulated in Einstein's famous Equivalence Principle:3

From… MIT

Pick pdf titled ‘How Gravitational Forces Arise from Curvature.’
Page 1 and 2.

 

[paddoboy]

As far as I'm concerned, EVERY POSTER ignoring the mountain of evidence for BH's, is either willfully ignorant, or simply playing stupid for attention.