(Alpha) General relativity is self-inconsistent

QUOTE:
"andrewgrey: my understanding is that Zanket has been trying to say that General Relativity is self-inconsistent because the Principle of Equivalence does not hold true. My own understanding is that the principle was acknowledged to not hold true for proper gravity many years ago, but this does not destroy General Relativity. Can you offer your opinion on this?"


I think you are speaking of the fact that the Equivalence Principle is only approximately true on ever decreasing inertial frame sizes as your measurement devices get more accurate.

In other words, "proper gravity" is never really like a global constant acceleration in the z direction, for example. This setup is really flat, I believe.

Andrew A. Gray
 
An (approximate) inertial frame centered on the escaping particle cannot straddle the horizon.
What you call an “(approximate) inertial frame”, I will call an “inertial frame”, as Taylor and Wheeler do.

So it appears that your proof needs rewording once again, as it seems that it is always possible to construct an (approximate) inertial frame centered on the escaping particle, but it just cannot straddle the horizon.
My proof doesn’t need rewording. It’s valid as it is. The supporting info in the OP notes that Taylor and Wheeler define “flat spacetime” as a “Region of spacetime in which it is possible to set up a free-float (inertial) reference frame”. Since Y is an inertial frame by definition, and it cannot be set up throughout the spacetime of X, then, as the OP notes, the spacetime cannot be flat throughout X. Then X cannot be an inertial frame. That is, an inertial frame cannot fall through the horizon of a black hole. GR required this conclusion, yet the theory predicts otherwise, contradicting itself, so it is self-inconsistent.

To refute the proof, you would need to show that special relativity (SR), which applies to flat spacetime, does not allow some inertial frames to be set up throughout the flat spacetime of another inertial frame. I don’t see how you have done that, or could do that. You can’t invoke a horizon to do that, because no horizon exists in an inertial frame in SR.
 
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Zanket,

Sorry if I have not made myself clear.

QUOTE:
"Since . . . . . .the spacetime cannot be flat throughout X. Then X cannot be an inertial frame."



I guess I will try this to be a little clearer.

According to GR, there can be no flat inertial frames if gravity is present. If gravity is present, there just are no flat inertial frames, period.

Another way to say this:

If spacetime is flat, then there is no gravity.

Another way to say this:

If gravity is present, then spacetime is not flat.

Another way to say this:

Gravity and flat inertial frames are mutually exclusive.


Do you at least understand why am saying this, Zanket?


Andrew A. Gray
 
Do you at least understand why am saying this, Zanket?
No, I don’t. It sound like you’re going back to the “not exactly flat” argument and trying to say that it explains how GR can be self-inconsistent even though Y cannot be setup throughout X. Is that what you’re saying?

It doesn’t matter that “If gravity is present, then spacetime is not flat”. Clearly it doesn’t matter to Einstein or Thorne. In the OP Thorne says that the equivalence principle “asserts that small, freely falling frames in the presence of gravity are equivalent to inertial frames in the absence of gravity”. Do you agree with Thorne?

It is true that the spacetime in X is not exactly flat, which is the same as saying that the tidal force in X is not completely nonexistent. But that is not a panacea that explains any degree of difference between the behavior in X and the behavior in an inertial frame J in an idealized universe without gravity. Do you agree? It seems that you disagree. If so, please explain. The explanation must refute the equivalence principle.

The definitions of “inertial frame” and “flat spacetime” that I’m using already account for the tidal force (= spacetime curvature) in our real, gravity-endowed universe. As I noted before, the spacetime in an “inertial frame” is not exactly flat by definition. You’re arguing against the definitions of Taylor, Thorne, and Wheeler. Their definitions agree with the equivalence principle.
 
So sorry, Zanket,

I am having trouble following your logic when the definitions are not exactly as they seem to mean. So I will try a little harder here. I think I see what you are saying.

Our problem, as I perceive it, is covering the horizon with outgoing-free-fall coordinates. Perhaps this statement is a better one for me.

I do not think that this can be done. I believe that coordinate singularities would appear. Agreed?

If not, then we must go back and show this.

If you agree, then we must discuss putting an "inertial frame" across these coordinate singularities.

Then, if this can't be done, we must discuss whether this is a contradiction in GR or not.

Agreed?


Andrew A. Gray
 
I am having trouble following your logic when the definitions are not exactly as they seem to mean.
The definitions in the OP are fine, and I’m using them correctly. They explicitly allow the spacetime in an inertial frame to be negligibly curved, and allow flat spacetime to be negligibly curved, thereby incorporating the equivalence principle. The definitions come straight from Taylor, Thorne, and Wheeler. I think it is best to drop an adherence to definitions that do not apply to our real, gravity-endowed universe—Einstein certainly thought so.

Our problem, as I perceive it, is covering the horizon with outgoing-free-fall coordinates. Perhaps this statement is a better one for me.

I do not think that this can be done. I believe that coordinate singularities would appear. Agreed?
Talking about coordinate singularities takes the discussion on a tangent. We’ve already established a physical night vs. day difference between X and an inertial frame J that is wholly above a horizon, which is a far cry from a coordinate singularity, which is not physical. In X, some objects cannot exist, like the lattice of clocks mentioned in the proof in the OP. There is no such limitation in J. That difference between X and J is blatantly inconsistent with the relativity principle stated in the OP.

You’ve already agreed that Y cannot cover all of X initially, due to the horizon. Then the relativity principle allows only one way to refute the OP: show that some inertial frame K cannot cover all of J initially; that is, show that J is still equivalent to X (or negligibly differs). I don’t see how you could do that, since there’s no horizon in J. The “not exactly flat” argument doesn’t help, not only because the definitions I’m using already account for a negligible tidal force (= spacetime curvature), but also because the tidal force could be slightly greater in J than in X.

The OP proves that no theory of gravity can predict black holes and be consistent with the relativity principle.
 
Zanket: The definitions in the OP are not fine. In the OP Thorne says that the equivalence principle “asserts that small, freely falling frames in the presence of gravity are equivalent to inertial frames in the absence of gravity”. The two are not identical. Einstein knew this. Look up Einstein and proper gravity. When you ignore that "negligible curvature" what you end up with is a flat hill. It's a contradiction in terms. I know what andrewgray is saying, and would urge you to really focus on this.
 
OK, Zanket,

Believe it or not, I am beginning to understand what you are saying. And I am even willing to go along with your definitions for the moment. So, in my opinion, the only solution to this problem is to try to construct a set of outgoing-freefall-inertial-frames and see what happens when you try and extend them across the horizon. The ingoing freefall frames are given here:

http://sciforums.com/showpost.php?p=1353875&postcount=3

http://www.modelofreality.org/Sect1_4.html

Instead of the inbound version (1.36) we will use the outbound version. We will have dR=0 (R=constant) for observers who are centered on outbound freefall frames.

$$dR = dt - \frac{sqrt{r/2M}}{(1-2M/r)}dr $$

I will fire up Mathematica and get back to you. Carry on.



Andrew A. Gray
 
Zanket: The definitions in the OP are not fine. In the OP Thorne says that the equivalence principle “asserts that small, freely falling frames in the presence of gravity are equivalent to inertial frames in the absence of gravity”. The two are not identical. Einstein knew this.
Strictly speaking they are not identical, but for all practical purposes they are the same, to emphasize Einstein's choice of words. Why did he choose the word "same" when they are not really the same? Because the difference is truly negligible; it doesn't matter. The difference can be as negligible as desired in principle. The negligible difference does not explain that some elephants can exist in J (an inertial frame that is wholly above a horizon) but not in X (an inertial frame that is falling through a horizon).

The same concept occurs in calculus, which calculates the area under a curve by summing the areas of rectangles below the curve. In the limit of infinitesimally narrow rectangles, the sum of the areas of the rectangles is equivalent to the area under the curve. For finitely narrow rectangles, the difference between the sum of the areas of the rectangles and the area under the curve is not zero but can be as small as desired. Equivalence to 10[sup]7000[/sup] significant digits is possible. X and J can also be equivalent to 10[sup]7000[/sup] significant digits. Why quibble over the remaining difference as if it could explain that some elephants can exist in J but not in X?

Look up Einstein and proper gravity. When you ignore that "negligible curvature" what you end up with is a flat hill. It's a contradiction in terms. I know what andrewgray is saying, and would urge you to really focus on this.
By sticking to GR's predictions and postulates here, my argument is sound. Einstein uses the word "same" when comparing an inertial frame that has no spacetime curvature to an inertial frame that has negligible curvature. Then I can depend on sameness. You're really arguing against Einstein, not me.

The bottom line is, "negligible" means "too small or unimportant to be worth considering", yet you still insist that I consider a negligible curvature. Physics does not redefine the word "negligible". Then I can disregard your point and stick to the dictionary.

If the OP proved only a negligible difference between X and J, then you'd have a point. But the OP proves a giant night vs. day difference between X and J. The OP proves that some elephants that can exist in J cannot exist in X. Namely, no elephant in X can stay at rest with respect to the escaping particle and straddle the horizon. Put a free test particle in J that moves at the same velocity in J as the escaping particle does in X. Unlike in X, there is no problem filling J to the brim with elephants that all stay at rest with respect to the particle. That giant difference between X and J is a clear violation of the relativity principle stated in the OP.

I searched on Einstein and proper gravity. I don’t see anything that refutes my points above. If you disagree, can you provide a link?
 
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Zanket, read this essay Einstein's Gravitational Field by Pete Brown. I can't say it's 100% correct or totally authoritative, but it's interesting how he compares the "Einstein Interpretation" with the "Modern Interpretation". This is his website:

http://www.geocities.com/physics_world/

Here's the essay and the abstract:

http://xxx.lanl.gov/abs/physics/0204044

There exists some confusion, as evidenced in the literature, regarding the nature of the gravitational field in Einstein's General Theory of Relativity. It is argued here the this confusion is a result of a change in interpretation of the gravitational field. Einstein identified the existence of gravity with the inertial motion of accelerating bodies (i.e. bodies in free-fall) whereas contemporary physicists identify the existence of gravity with space-time curvature (i.e. tidal forces). The interpretation of gravity as a curvature in space-time is an interpretation Einstein did not agree with.

Here's an excerpt, note the reference to Thorne:

It is widely assumed that, according to Einstein’s general theory of relativity, gravitation is a curvature in space-time. There is a well accepted definition of space-time curvature. As stated by Thorne:

space-time curvature and tidal gravity are the same thing expressed in different languages, the former in the language of relativity, the latter in the language of Newtonian gravity.

However one of the main tenets of general relativity is the Principle of
Equivalence: A uniform gravitational field is equivalent to a uniformly accelerating frame of reference. This implies that one can create a uniform gravitational field simply by changing one’s frame of reference from an inertial frame of reference to an accelerating frame, which is rather difficult idea to accept. A uniform gravitational field has, by definition, no tidal forces and thus no space-time curvature. Thus according to the interpretation of gravity as a curvature in spacetime a uniform gravitational field becomes a contradiction in terms (i.e. no tidal forces where there are tidal forces). This apparent contradiction is obviously quite confusing and can certainly be misleading. In A brief history of relativity published in the above mentioned issue of Time, Stephen Hawking writes

I still get two or three letters a week telling me Einstein was wrong. Nevertheless, the theory of relativity is now completely accepted by the scientific community, and its predictions have been verified in countless applications. [...] His idea was that mass and energy would warp spacetime in some manner ... Objects like apples or planets would try to move on straight lines through space-time, but their paths would appear bent by a gravitational field because space-time is curved.

Given such a statement by a respected physicists and with a great deal of
experimental data to back up this claim is it reasonable to question this notion of gravity being a curvature in space-time? Is it reasonable to assume that the bent path an object takes when moving through...


Pay special attention to page 20, and google on Synge:

This can be very confusing and is reflected by a comment made by
Synge 40.

... I have never been able to understand this principle…Does it mean that the
effects of a gravitational field are indistinguishable from the effects of an
observer’s acceleration? If so, it is false. In Einstein’s theory, either there is
a gravitational field or there is none, according as the Riemann tensor does
not or does vanish. This is an absolute property; it has nothing to do with any
observers world line … The Principle of Equivalence performed the essential
office of midwife at the birth of general relativity, but, as Einstein remarked,
the infant would never have gone beyond its long clothes had it not been for
Minkowski’s concept [of space-time geometry]. I suggest that the midwife be
buried with appropriate honours and the facts of absolute space-time faced.

A similar example is that of Ray 41.

It is very important to notice that in a freely falling frame we have not
transformed away the gravitational field since the Riemann tensor
(gravitation Û Riemann tensor) will not vanish and we will still measure
relative acceleration ….. The first thing to note about the 1911 version of
the principle of equivalence is that what in 1911 is called a uniform
gravitational field ends up in general relativity not to be a gravitational field
at all – The Riemann tensor is here identically zero. Real gravitational fields
are not uniform since they must fall off as once recedes from gravitating
matter.
 
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I don't know why Einstein said "same" Zanket. But if you apply that "neglible" to your calculus analogy, you'd be doing away with the very curve itself. You equate the curve to a perfectly flat line, and you're left with no area beneath the curve, and no route to understanding what gravity is. The neglible is axiomatic. There's no justification for it. There can be no curve if the small section of it is absolutely the same as that flat line.

Go back to the basics of an inertial reference frame. You're apparently motionless with respect to the visible universe, in a given inertial reference frame. I whoosh past at 1000km/s, and you consider me to be in a different inertial reference frame. If you accelerate quickly to catch me up and travel alongside me, you've changed your inertial reference frame. Any acceleration changes your inertial reference frame. And a continuous gravitational acceleration continuously changes your inertial reference frame. Because that's what the gravity is. It's equivalent to a continuously changing inertial reference frame, and if there is no continuously changing frame there is no gravity. You just can't have a fixed inertial reference frame in a gravity freefall situation. It's a contradiction in terms. So you just can't have an inertial reference frame crossing an event horizon. It's like pointing to a mountain peak that you say straddles the 5000m elevation line, but then you say it's flat. It's ambiguous. Contradictory. It just doesn't make sense.

I'm sorry, I'm not quite clear about those elephants. And I'm sorry to bang on about this. But I do think it's important. As I've said before, I think you do have a point that is generally misunderstood. But IMHO it doesn't invalidate GR. It just challenges the current interpretation, which I believe is not quite the same as the one Einstein eventually held.
 
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Farsight, you CAN have inertial frame in a gravitational field, although you need to limit yourself in a small region of spacetime, and that frame will be accelerating.
 
temur: if the inertial reference frame is accelerating, it's a non-inertial reference frame.

http://id.mind.net/~zona/mstm/physics/mechanics/framesOfReference/framesOfReference.html

If you're in free fall, are you accelerating? To maintain an inertial reference frame, you have to say no. Which would mean gravity is not accelerating you, which means gravity is not a force. Instead you have to say you're accelerating after you've fallen to the surface of the earth. Then you'd consider yourself to be in a non-inertial reference frame even though you're motionless. The only force is coming from the ground beneath your feet preventing your "uniform" motion that would have you falling faster and faster towards the centre of the earth. It all gets a little screwy, and subtle, but IMHO we tend to miss the thing that the gravity actually is - a difference in our local space that cannot be negligible, rather than a magical mysterious action-at-a-distance force.

See my post above where I mentioned Synge.
 
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I guess we are talking the same thing, but it surptises me that you concluded that there is no inertial frame crossing the horizon. If it is in free fall, then it is reference frame. So just drop a small ball from a bit above the horizon and the ball will be in free fall, and it will be in an inertial frame in a short time interval around the moment it crosses the horizon. Only thing is that you have to take very small region of spacetime, which is no problem since in principle you can have a magnifier as strong as you need.
 
temur: if the inertial reference frame is accelerating, it's a non-inertial reference frame.
I agree with temur above. A frame can be inertial, and remain inertial, when it’s gravitationally accelerating. Only when it’s noninertially (nongravitationally) accelerating is it a noninertial frame. The frame of a thrusting rocket is an example of a noninertial frame.

Even in our gravity-endowed universe a frame can remain inertial indefinitely; I disagree that “a continuous gravitational acceleration continuously changes your inertial reference frame”. I think it is obvious from the supporting info in the OP that Einstein, Taylor, Thorne, and Wheeler are all in agreement with me.

Temur says: “So just drop a small ball from a bit above the horizon and the ball will be in free fall, and it will be in an inertial frame in a short time interval around the moment it crosses the horizon.” I say that a frame can in principle remain inertial for a year or more (on a clock affixed to the frame); you just need a sufficiently large black hole. The frame remains inertial for as long as the tidal force in the frame remains negligible for the purposes of your experiments. Taylor and Wheeler also talk about “small” regions and “short” times for an inertial frame, but these are arbitrary words whose precise magnitude depends on the experiment. In one of the examples in their book, they employ an inertial frame that is two million light years long and lasts for at least two million years. A frame in a stable circular orbit can remain inertial for billions of years because the tidal force in it does not increase.

In Pete Brown’s essay, a key point is this: “The main goal of this paper is not to present a new interpretation of gravity. For pedagogical purposes we review an old one, that of Albert Einstein’s”. The paper is just discussing an interpretation. Multiple interpretations can lead to the same predictions; e.g. both tidal force and spacetime curvature. The OP deals with GR’s predictions. It isn’t affected by an interpretation.

Brown says “A uniform gravitational field has, by definition, no tidal forces and thus no space-time curvature. Thus according to the interpretation of gravity as a curvature in spacetime a uniform gravitational field becomes a contradiction in terms (i.e. no tidal forces where there are tidal forces).” See that it’s only an apparent contradiction. Brown says “Given experimental limitations one can establish criteria in which tidal effects may be ignored.” That resolves the apparent contradiction. A uniform gravitational field is better defined as having a negligible tidal force, not no tidal force. A uniform gravitational field is better thought of as a negligibly nonuniform gravitational field. There’s a contradiction only when one rigidly sticks to definitions of a uniform gravitational field, inertial frame, or flat spacetime in which there are “no tidal forces and thus no space-time curvature”.

It seems to me that what happens is this: People first learn SR, which is an idealized environment where there is no tidal force. When they learn GR they stick to SR’s definitions, causing contradictions to be seen. But there’s no requirement that SR’s definitions carry over to GR unchanged. Einstein effectively changed them with the equivalence principle, to allow a negligible tidal force in an inertial frame. That makes sense, since otherwise SR could not even be experimentally confirmed. We might not be having this discussion if it weren’t for the fact that terms like “negligibly nonuniform” and “negligibly curved” get tedious, resulting in the shorthand terms “uniform” and “flat”.

Because that's what the gravity is. It's equivalent to a continuously changing inertial reference frame, and if there is no continuously changing frame there is no gravity.
I disagree. To me and Taylor and Wheeler, gravity is indicated by the tidal force, synonymous with spacetime curvature. A negligible tidal force exists in any inertial frame.

But if you apply that "neglible" to your calculus analogy, you'd be doing away with the very curve itself. You equate the curve to a perfectly flat line, and you're left with no area beneath the curve, and no route to understanding what gravity is. The neglible is axiomatic. There's no justification for it. There can be no curve if the small section of it is absolutely the same as that flat line.
Using negligibly wide rectangles under the curve does not equate the curve to a single perfectly flat line; it equates it to a series of negligibly wide segments of perfectly flat lines, the area under which approximates the area under the curve. The approximation can be as precise as desired. The narrower the rectangles, the more precise the approximation.

Some of the letters on your screen now look like they have curves right? Well, if you could zoom in on them you’d see that they comprise only square blocks, which, by virtue of being negligibly large (to your eyes), can be used to approximate a curve.

If the crew of the International Space Station (ISS) let some object float across the station, it will seem to follow a straight path relative to the station even if the time required is the time it takes for them to complete an orbit around the Earth. The object’s path curves relative to the Earth, but relative to the ISS it is straight.

So you just can't have an inertial reference frame crossing an event horizon.
It wouldn’t matter to the OP even if you were right, because GR predicts that an inertial frame can fall through a horizon, as the supporting info in the OP shows. You’d be arguing against GR, not the OP.

As I've said before, I think you do have a point that is generally misunderstood. But IMHO it doesn't invalidate GR. It just challenges the current interpretation, which I believe is not quite the same as the one Einstein eventually held.
I disagree. The OP does invalidate GR, by showing that it blatantly violates the relativity principle (one way to put it). The OP does not challenge an interpretation. It shows that what GR claims is the same cannot be the same by any stretch of the imagination.
 
If you're in free fall, are you accelerating? To maintain an inertial reference frame, you have to say no.
Relative to the frame you are not accelerating. But the frame is accelerating relative to a source of gravity. If you can’t feel the acceleration, it’s gravitational (inertial) acceleration.

Instead you have to say you're accelerating after you've fallen to the surface of the earth. Then you'd consider yourself to be in a non-inertial reference frame even though you're motionless.
Well, you still noninertially accelerate relative to something falling.

The only force is coming from the ground beneath your feet preventing your "uniform" motion that would have you falling faster and faster towards the centre of the earth.
Yes. On the surface your acceleration is noninertial; you noninertially accelerate, pushed by the electromagnetic force that prevents the Earth from imploding due to gravity. If you continued falling your acceleration would be negligibly nonuniform (“uniform”) over a short time or a short distance. If you were falling toward a sufficiently large object, your acceleration relative to the object could be negligibly nonuniform for years on your clock.
 
Thanks for responding at such length, Zanket. We share much common ground of course, but I feel I'm still not quite communicating my "contradiction in terms" point. This is going to be a long post, but I'd like to start by taking a look at your first paragraph:

I agree with temur above. A frame can be inertial, and remain inertial, when it’s gravitationally accelerating. Only when it’s noninertially (nongravitationally) accelerating is it a noninertial frame. The frame of a thrusting rocket is an example of a noninertial frame.

OK, imagine you and I are in black boxes a million miles apart travelling through space at some steady velocity. You start to fall into a black hole, but you feel no acceleration. You remain in your inertial reference frame, and to you there are no forces at work. There is no "force of gravity". But when I observe you, I see that you're accelerating away from me. To me you're in a non-inertial reference frame, and I claim that the force of gravity is accelerating you. Your paragraph above mixes these two viewpoints. That's the basis of the contradiction in terms.

Even in our gravity-endowed universe a frame can remain inertial indefinitely; I disagree that “a continuous gravitational acceleration continuously changes your inertial reference frame”. I think it is obvious from the supporting info in the OP that Einstein, Taylor, Thorne, and Wheeler are all in agreement with me.
I wouldn't be too sure about Einstein, but yes, I think the other people are in agreement with you. And I think they're missing the trick, especially Thorne when he talks about time travel. Going back to my scenario above, imagine that the black hole was suddenly removed and your freefall ceased. (Please can we ignore any accelerations or "gravity waves" caused by this). Thereafter both you and I would agree that you are back in an inertial reference frame, travelling at some steady but higher velocity. If the black hole reappears briefly to accelerate you some more, we would disagree about your inertial reference frame, then when it disappeared we would agree again. A freefall is like this but smooth and continuous. There's no acceleration in your eyes, no force, no action-at-a-distance, and no gravity. But there is in my eyes. To me your frame is constantly changing, to you it isn't. Not unless you look out the window. (You know, I consider myself an Einstein "fan", but I'm not a fan of reference frames because they don't actually exist).

Temur says: “So just drop a small ball from a bit above the horizon and the ball will be in free fall, and it will be in an inertial frame in a short time interval around the moment it crosses the horizon.” I say that a frame can in principle remain inertial for a year or more (on a clock affixed to the frame); you just need a sufficiently large black hole. The frame remains inertial for as long as the tidal force in the frame remains negligible for the purposes of your experiments. Taylor and Wheeler also talk about “small” regions and “short” times for an inertial frame, but these are arbitrary words whose precise magnitude depends on the experiment. In one of the examples in their book, they employ an inertial frame that is two million light years long and lasts for at least two million years. A frame in a stable circular orbit can remain inertial for billions of years because the tidal force in it does not increase.
I feel happier about circular orbits. Again "neglible" looks like a problem. If it's neglible and yet spans the horizon how can the horizon be inside the frame? You're saying there's no difference here and yet the very presence of the horizon says there is a difference here. How can I explain this? In your freefall inertial reference frame, there was no action at a distance force. The thing that was causing you to accelerate with respect to me has to be local to you and every single atom of your body. Yes, the tidal force might be very small and your acceleration might be changing only slightly. But in my eyes you are definitely accelerating in a given direction. There is a definite non-uniformity and it is local to you, in what you consider to be your "uniform" local frame. The gravity is the non-uniformity. Hence a uniform gravitational field is a uniform non-uniformity, and is contradiction in terms.

In Pete Brown’s essay, a key point is this: “The main goal of this paper is not to present a new interpretation of gravity. For pedagogical purposes we review an old one, that of Albert Einstein’s”. The paper is just discussing an interpretation. Multiple interpretations can lead to the same predictions; e.g. both tidal force and spacetime curvature. The OP deals with GR’s predictions. It isn’t affected by an interpretation.
Noted. I'll reread the OP. Sorry if I've lost the plot somewhere along the line.

Brown says “A uniform gravitational field has, by definition, no tidal forces and thus no space-time curvature. Thus according to the interpretation of gravity as a curvature in spacetime a uniform gravitational field becomes a contradiction in terms (i.e. no tidal forces where there are tidal forces).” See that it’s only an apparent contradiction. Brown says “Given experimental limitations one can establish criteria in which tidal effects may be ignored.” That resolves the apparent contradiction. A uniform gravitational field is better defined as having a negligible tidal force, not no tidal force. A uniform gravitational field is better thought of as a negligibly nonuniform gravitational field. There’s a contradiction only when one rigidly sticks to definitions of a uniform gravitational field, inertial frame, or flat spacetime in which there are “no tidal forces and thus no space-time curvature”.
I agree with this. I say gravity is not "curved spacetime", but instead is a gradient in c across your local frame that you don't notice but I do.

It seems to me that what happens is this: People first learn SR, which is an idealized environment where there is no tidal force. When they learn GR they stick to SR’s definitions, causing contradictions to be seen. But there’s no requirement that SR’s definitions carry over to GR unchanged. Einstein effectively changed them with the equivalence principle, to allow a negligible tidal force in an inertial frame. That makes sense, since otherwise SR could not even be experimentally confirmed. We might not be having this discussion if it weren’t for the fact that terms like “negligibly nonuniform” and “negligibly curved” get tedious, resulting in the shorthand terms “uniform” and “flat”.
Agreed. The frames can become quite confusing. The whole subject can. If it didn't I think we'd have bottomed out gravity a long long time ago.

Farsight: “Because that's what the gravity is. It's equivalent to a continuously changing inertial reference frame, and if there is no continuously changing frame there is no gravity.” I disagree. To me and Taylor and Wheeler, gravity is indicated by the tidal force, synonymous with spacetime curvature. A negligible tidal force exists in any inertial frame.
See my paragraphs above regarding viewpoint. If you stick to your reference frame, there's no force acting on you, no "force of gravity", no gravity. If we shift to my viewpoint, everything changes.

Farsight: But if you apply that "neglible" to your calculus analogy, you'd be doing away with the very curve itself. You equate the curve to a perfectly flat line, and you're left with no area beneath the curve, and no route to understanding what gravity is. The neglible is axiomatic. There's no justification for it. There can be no curve if the small section of it is absolutely the same as that flat line. Using negligibly wide rectangles under the curve does not equate the curve to a single perfectly flat line; it equates it to a series of negligibly wide segments of perfectly flat lines, the area under which approximates the area under the curve. The approximation can be as precise as desired. The narrower the rectangles, the more precise the approximation.
We're talking at cross purposes here. At right angles in fact. I was looking at sections of the curve and saying they can't be flat, because if they were you'd have no curve.

Some of the letters on your screen now look like they have curves right? Well, if you could zoom in on them you’d see that they comprise only square blocks, which, by virtue of being negligibly large (to your eyes), can be used to approximate a curve.
No problem. But if you say they are flat, all the letters end up looking like this _.

If the crew of the International Space Station (ISS) let some object float across the station, it will seem to follow a straight path relative to the station even if the time required is the time it takes for them to complete an orbit around the Earth. The object’s path curves relative to the Earth, but relative to the ISS it is straight.
No problem. It's straight and it's curved. Ain't relativity fun!

Farsight: So you just can't have an inertial reference frame crossing an event horizon. It wouldn’t matter to the OP even if you were right, because GR predicts that an inertial frame can fall through a horizon, as the supporting info in the OP shows. You’d be arguing against GR, not the OP.
I've said elsewhere that I consider this to be a hypothetical situation. You cannot in reality fall through the horizon, because at this point time dilation becomes infinite. My rule of thumb says that infinities don't happen in reality, so we should look to see if the mathematics needs some refinement. Hence I say c goes to zero at the horizon. This means a collapsing star takes forever to collapse, and none have finished collapsing yet. Hence they're frozen stars. The event horizon becomes a wall of eventless "solid space". An eventful and "uniform" local frame simply cannot span it.

Farsight: As I've said before, I think you do have a point that is generally misunderstood. But IMHO it doesn't invalidate GR. It just challenges the current interpretation, which I believe is not quite the same as the one Einstein eventually held. I disagree. The OP does invalidate GR, by showing that it blatantly violates the relativity principle (one way to put it). The OP does not challenge an interpretation. It shows that what GR claims is the same cannot be the same by any stretch of the imagination.
Noted. I've been thinking a shift in interpretation might fix GR singularity infinities and resolve your issue, but like I said I'll reread the OP again. If you'd prefer to agree to differ don't hesitate to say so and I'll leave it at that.
 
I don't mean to interrrupt you and zanket's conversation, but allow me to post some ideas here.

OK, imagine you and I are in black boxes a million miles apart travelling through space at some steady velocity. You start to fall into a black hole, but you feel no acceleration. You remain in your inertial reference frame, and to you there are no forces at work. There is no "force of gravity". But when I observe you, I see that you're accelerating away from me. To me you're in a non-inertial reference frame, and I claim that the force of gravity is accelerating you. Your paragraph above mixes these two viewpoints. That's the basis of the contradiction in terms.
You have to use parallel transport in this case to compare velocities, and to first approximation you will see that zanket's velocity is constant.

I wouldn't be too sure about Einstein, but yes, I think the other people are in agreement with you. And I think they're missing the trick, especially Thorne when he talks about time travel. Going back to my scenario above, imagine that the black hole was suddenly removed and your freefall ceased. (Please can we ignore any accelerations or "gravity waves" caused by this). Thereafter both you and I would agree that you are back in an inertial reference frame, travelling at some steady but higher velocity. If the black hole reappears briefly to accelerate you some more, we would disagree about your inertial reference frame, then when it disappeared we would agree again. A freefall is like this but smooth and continuous. There's no acceleration in your eyes, no force, no action-at-a-distance, and no gravity. But there is in my eyes. To me your frame is constantly changing, to you it isn't. Not unless you look out the window. (You know, I consider myself an Einstein "fan", but I'm not a fan of reference frames because they don't actually exist).
To everybody the frame is constantly changing, i.e., zanket is shifting from one inertial ("free fall") frame to another, but if you consider sufficiently tiny region zanket is staying in one inertial frame.
 
I don't mean to interrrupt you and zanket's conversation, but allow me to post some ideas here.

I'm rather interrupting Zanket's OP, temur. Maybe too much.

You have to use parallel transport in this case to compare velocities, and to first approximation you will see that zanket's velocity is constant.
Did you mean Zanket's acceleration rather than velocity would be approximately constant? I'll assume yes.

To everybody the frame is constantly changing, i.e., zanket is shifting from one inertial ("free fall") frame to another, but if you consider sufficiently tiny region zanket is staying in one inertial frame.

Yes, if we had a very large black hole then freefall Zanket would be staying in what he deems to be an inertial reference frame. The tidal forces would be "negligible", and he could conduct lots of experiments in his black box and wouldn't detect any gravity. The crucial thing is that he cannot consider himself to be in a "uniform gravitational field" without looking outside his black box. He's got to look at me and other things in the Universe to realise that whilst he feels no forces in his inertial reference frame, he's accelerating compared to his previous steady state motion through space - compared to his previous inertial reference frame. Then he realises that despite those null experiments in his black box, his frame is somehow changing. In a sense, the gravitational field is that changing frame.
 
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