General relativity is self-inconsistent

[InfiniteNow]

Some claim that black holes have been observed, so the proof of GR’s self-inconsistency must be wrong. But there is no direct observational evidence of a black hole, and the indirect evidence relies on the validity of GR (as in “if GR is the correct theory of gravity, then the observation indicates a black hole”).

 

Chandra :: Photo Album :: RX J1242-11 :: 18 Feb 04

Chandra :: Photo Album :: Images by Category: Black Holes

 

 

What now?

[snoopy]

To Zanket,

 

I have constantly given you quotes from your own sources that contradicts what you are saying.

 

Yes in 1907 Einstein introduced the principle of equivalence which as you say is at the core of GR.

 

What you do not say is this

 

"In 1912 Einstein published several papers on gravitation. In these he realised that the Lorentz transformations will not hold in this more general setting. Einstein realised that the gravitational field equations were bound to be non-linear and the principle of equivalence appeared only to hold locally "

 

This bit about it only holding under local conditions is key here.

 

Eventually this led Einstein to a new geometry of spacetime and the correct field equations for gravity.

 

So in 1912 Einstein realises that the principle of equivalence which comes from SR and led Einstein into GR only holds in GR under local conditions and not in the "general" way he had initially been hoping for because it does not hold in a general way it cannot be said to hold under GR.

 

Here is a website which explains just that. (the quoted area)

 

General relativity

 

The region of space you are choosing cannot be considered to be local

so therefore the principle of equivalence does not hold in this area.

 

I am not going to bother with rest as I keep on proving what you are saying is palpably false and still you come back with more and more 'falsehoods' and 'half truths'.

 

So like you I dont see the point in continuing the conversation.

 

 

For the last time Goodbye

:)

[andrewgray]

I don’t agree that this Rindler horizon at r=2M is the same as the event horizon of a black hole. The predictions of these two types of horizons differ. For example, whereas an object can pass upward (outward) through a Rindler horizon (it just can’t reach the accelerating observer), GR predicts that an object at an event horizon must fall below the horizon. More on this below.

 

Zanket,

 

I also do not think that an event horizon is the same as a Rindler (or SR hyperbolic) horizon. I was only saying that they were indistinguishable in a small neighborhood of the horizon, where one frame of reference is the local accelerated Schwarzchild coordinates, and the other is some freefall inertial frame.

 

I went on to construct both the ingoing and outgoing freefall coordinates here.

 

Are we now in disagreement about these outgoing freefall coordinates?

 

Are you going to make me actually construct the local hyperbolic coordinates in a small "ε neighborhood" of an event horizon?

 

 

As I recall, what you were saying is that you think an event horizon does not exist at r=2M; you think that only a Rindler horizon exists there, for an observer hovering at r=2M+ε. That is, you think that objects can pass upward through r=2M, in defiance of the widely published predictions to the contrary for GR (and you gave some math to support this). You think that those widely published predictions are the result of a bad interpretation of GR’s math. Do I have all that right?

 

You are close. What I was saying was that both the Rindler and event horizons are horizons because the accelerated coordinate times go to infinity and the accelerated oberservers (both the Schwarzchild & the hyperbolic observers) can never "see" an object enter or leave the horizon in a finite amount of time. However, in the object's own proper time it can actually enter or leave, contrary to traditional Schwarzchild analysis. (Rindler analysis, however, allows objects to enter and leave the horizon trivially since it is flat spacetime.)

 

What I am saying is that you are correct, there must be a local inertial reference frame anywhere in space where there is not a singularity. Even at a Schwarzchild horizon. Your "inconsistency" comes from misinterpretation of the Schwarzchild metric, and it is indeed an inconsistency. So if you believe in the equivalence principle, I hope that you will eventually see that your analysis leads to the conclusion that locally, the GR horizon is just like the SR horizon.

 

Andrew A. Gray

[Guest Zanket]

Chandra :: Photo Album :: RX J1242-11 :: 18 Feb 04

Chandra :: Photo Album :: Images by Category: Black Holes

 

What now?

There’s no direct observational evidence of a black hole at those links. And the indirect evidence depends on the validity of GR. This fact is conveniently buried at the site:

 

From How is it possible to measure the density of a black hole, let alone a supermassive black hole? (boldface mine):

 

The research involves looking at the motions of stars in the centers of galaxies. These motions imply a dark, massive body whose mass can be computed from the speeds of the stars.

And how is the mass computed? By plugging in the speed of the stars into GR. After they use GR to get the mass, they see whether GR predicts that such a mass within such a volume indicates a black hole. If so, then voilà, a black hole is “confirmed”, GR is “confirmed”, and the grant money for more research can keep rolling in.

 

From ARE BLACK HOLES REAL? (boldface mine):

 

The implication of a black hole is based on the idea that it is gravity which makes the gas orbit, and stronger gravity leads to higher speeds. Except for a black hole, no one knows how to cram so much matter into so little volume.

“No one knows” because GR is the only theory of gravity considered in the analysis.

 

I like this quote from Daniel Fischer, editor of an astronomy journal, as told in Black Holes: A Traveler’s Guide: “The biggest unsolved problem [about black holes] is whether there are any black holes in the universe. In my opinion, the evidence so far is slim, and even black-hole aficionados admit this fact (when they believe the public/taxpayers aren’t listening).”

[Guest Zanket]

I have constantly given you quotes from your own sources that contradicts what you are saying.

The only contradictions you’ve shown were those that contradicted my sources.

 

"In 1912 Einstein published several papers on gravitation. In these he realised that the Lorentz transformations will not hold in this more general setting. Einstein realised that the gravitational field equations were bound to be non-linear and the principle of equivalence appeared only to hold locally"

The principle applies locally anywhere, as Einstein says in the OP. It does not apply just wherever you want to unilaterally declare as local. The principle applies in an inertial frame that straddles a horizon; that fits within the range of “anywhere”. An inertial frame is local by definition.

 

The region of space you are choosing cannot be considered to be local

so therefore the principle of equivalence does not hold in this area.

You’ve given nothing to support this but your opinion, and it clearly contradicts multiple sources in the OP, as I’ve noted and you’ve ignored. Who made you the king of what region of space can be considered local? The equivalence principle holds in any inertial frame, just like Einstein says it does, and an inertial frame can exist anywhere in empty spacetime, just like Einstein says it can.

 

I am not going to bother with rest as I keep on proving what you are saying is palpably false and still you come back with more and more 'falsehoods' and 'half truths'.

Yeah, falsehoods like direct quotes from top physicists including Einstein. :)

[Guest Zanket]

Are we now in disagreement about these outgoing freefall coordinates?

No, I think we are in agreement, as far as I am entertaining the notion (I haven’t confirmed your math).

 

Are you going to make me actually construct the local hyperbolic coordinates in a small "ε neighborhood" of an event horizon?

No, not necessary. I took your question too literally it seems.

 

What I was saying was that both the Rindler and event horizons are horizons because the accelerated coordinate times go to infinity and the accelerated oberservers (both the Schwarzchild & the hyperbolic observers) can never "see" an object enter or leave the horizon in a finite amount of time.

Let’s be clear that in the case of an object leaving the horizon (passing upward through it), the accelerated observers never see the object leave the horizon because neither the object nor the object’s light (including its image) can ever reach them.

 

Your "inconsistency" comes from misinterpretation of the Schwarzchild metric, and it is indeed an inconsistency.

I understand your viewpoint here.

 

So if you believe in the equivalence principle, I hope that you will eventually see that your analysis leads to the conclusion that locally, the GR horizon is just like the SR horizon.

This is where we left off before. I’m saying that, even if the GR horizon is just like the SR (Rindler) horizon, GR is still self-inconsistent, and not just due to a misinterpretation. I gave my reasoning in my last post to you, starting at “Where we left off our discussion”. What about that reasoning? To make your case that GR’s is self-consistent mathematically, you need to address and refute that reasoning.

 

There is no problem with a Rindler horizon in SR of course. In GR, however, when a Rindler horizon exists below an observer hovering at some r-coordinate, it’s a real problem. The underlying reason for the difference is because GR predicts that objects can escape to r=infinity, whereas in SR, no object can escape an accelerated observer. A relativistic rocket can in principle always catch up to and pass any inertial object.

 

The only way that GR could avoid being self-inconsistent due its prediction of a Rindler horizon at r=2M, is if it predicted that nothing can escape to r=infinity. In that case the box shown in the OP could be passing by the observer hovering at r=2M+ε, with the particle at rest relative to the box (so that GR does not contradict SR in X), yet without the part of the box that passes upward through the horizon ever reaching r=2M+ε. (Although to visualize this we have to imagine that ε is something greater than zero in the limit.) There would be no problem at least locally in this case because the box and the particle it encloses would eventually fall back below the horizon in their own proper time, just like would happen in SR for an observer in a relativistic rocket, observing the box & particle. But GR is of course not saved by this logic because it does predict that objects above r=2M can escape to r=infinity (and it must predict this, given observations).

 

In a self-inconsistent theory of gravity that incorporates SR, there cannot be a Rindler horizon predicted below a hovering observer, save for at r=0.

[snoopy]

To Zanket,

 

First of all I have to apologise for my rude behaviour to you in a previous post in which I told you not to post. It was meant as a joke but I admit it was in poor taste and ill-judged on my part.

 

I hope you take the apology.

 

However in all other aspects I still believe you are wrong but I dont see your opinions being changed by logical argument and you have ignored previous evidence I supplied that this is the case.

 

 

I will try one more simpler approach to you to prove that local inertial frames

are done away with in GR.

 

This is a link to Wiki on inertial frames

 

Inertial frame of reference - Wikipedia, the free encyclopedia

 

under the general relativity section

 

It says -

 

"GR modifies the distinction between nominally inertial and non-inertial effects by replacing SR's flat euclidean geometry with a non-euclidean metric. In GR the principle of inertia is replaced with the principle of geodesic motion whereby objects move in a way dictated by the curvature of spacetime."

 

(So without the 'principle of inertia' there can be no 'principle of equivalence')

 

It also says -

 

"This phenomenon of geodesic deviation means that inertial frames of reference do not exist globally as they do in Newtonian Mechanics and SR."

 

(This is in direct contradiction of your earlier statements that inertial frames exist anywhere.)

 

It also says -

 

"However GR reduces to SR over sufficiently small regions of spacetime where curvature effects become less important and the earlier inertial frame arguments come back into play. Consequently SR is now described only as a "local" theory. (However this refers to the theory's application rather than its derivation)"

 

This is something I have mentioned before "local" regions of space are regions that can be considered as flat or euclidean. The region of space you have chosen cannot be considered in this way (as I have said before). This however does not make me the "King of Space" as you have jested in a previous post but it does imply that you cannot use SR in this region and are forced by convention to use GR.

 

Incorrectly applying theories will lead you to wrong answers.

 

Your assertion that GR is self-inconsistent is false (I have supplied all the evidence that any fair minded person can follow)

 

Again as the link in Wiki suggests when you have a region of space you can consider to be flat GR reduces to SR and the 'principle of equivalence' comes back into play along with 'inertial frames of reference'. I believe this was the point Einstein was making in your quote from him.

 

I hope in least a small way I have changed your mind that GR is self-inconsistent. However I sincerely doubt that this is possible but I wish you all the luck in the world with your attempts to prove me wrong, as this will definitely be the last time I post to you on this subject.

 

Cheers

:)

[Guest Zanket]

I hope you take the apology.

Yes, thanks.

 

However in all other aspects I still believe you are wrong but I dont see your opinions being changed by logical argument and you have ignored previous evidence I supplied that this is the case.

I haven’t ignored your evidence. I’ve shown that it contradicts the many reputable sources I gave in the OP.

 

"This phenomenon of geodesic deviation means that inertial frames of reference do not exist globally as they do in Newtonian Mechanics and SR."

 

(This is in direct contradiction of your earlier statements that inertial frames exist anywhere.)

You are misinterpreting that; it isn’t such a contradiction. “Anywhere” does not mean “global”. In SR, a single inertial frame can encompass the entire idealized, gravity-free universe in that theory; that is, its size can be global. In GR, the size of an inertial frame is limited to a region of spacetime throughout which the tidal force (synonymous with spacetime curvature) is negligible for the purposes of a given experiment. But in GR an inertial frame, with its sufficiently small size, can still exist anywhere in empty spacetime. In Einstein’s verbatim statement of the equivalence principle in the OP he uses both the words “small” and “anywhere”—he’s not contradicting himself.

 

This is something I have mentioned before "local" regions of space are regions that can be considered as flat or euclidean. The region of space you have chosen cannot be considered in this way (as I have said before).

You have been claiming that an inertial frame cannot straddle a horizon. But you’ve given nothing but your opinion to support this, and I’ve given lots of references to the contrary. For example, look at what Taylor and Wheeler say, given in the OP starting with “Our old, comfy, free-float (inertial) frame carries us unharmed to the center of a black hole”.

 

This however does not make me the "King of Space" as you have jested in a previous post but it does imply that you cannot use SR in this region and are forced by convention to use GR.

To make your case that an inertial frame cannot straddle a horizon, you would need to refute Einstein, Taylor, Thorne, and Wheeler, just to start. Your opinion does not suffice. In GR, SR applies locally everywhere in empty spacetime. There are no exceptions in a region that straddles a horizon, unless GR is self-inconsistent. No exception is noted in the equivalence principle, which tells us under what conditions SR applies in GR. You will not find such an exception noted anywhere in the reputable annals of physics.

 

Look at the definition of an inertial frame as given in the OP. You have claimed that the tidal force in a region straddling a horizon must be significant (not negligible), hence an inertial frame cannot straddle a horizon. But the tidal force can be calculated in GR, and for a sufficiently large black hole, the tidal force that GR predicts for a region straddling a horizon can shown to be arbitrarily small. That explains references like this one from Pickover that I gave in the OP: “The tidal forces at the horizon of this gigantic black hole would be weaker than those produced by Earth, which are already impossible for us to feel”.

 

Again as the link in Wiki suggests when you have a region of space you can consider to be flat GR reduces to SR and the 'principle of equivalence' comes back into play along with 'inertial frames of reference'. I believe this was the point Einstein was making in your quote from him.

The equivalence principle is always “in play” in GR. It tells us under what conditions SR applies in GR. The principle is not one of SR, as you seem to imply.

 

I hope in least a small way I have changed your mind that GR is self-inconsistent. However I sincerely doubt that this is possible but I wish you all the luck in the world with your attempts to prove me wrong, as this will definitely be the last time I post to you on this subject.

Thanks on the wish. I hope you understand why I cannot be convinced by an argument that comprises nothing more than an opinion that directly contradicts not only the many references I have given in the OP, but also any decent book on GR, which will tell you that nothing strange need happen as you fall through a horizon (your frame can stay inertial).

[andrewgray]

Zanket,

 

Sorry if I did not address all of your logic from the first post. I had to see where we stood after you had some time to digest everything. I will try to address all of your logic (I'm not ignoring), but I need to go just one step at a time.

 

I think we are in agreement [about the outgoing coordinates], as far as I am entertaining the notion (I haven’t confirmed your math).

 

So you agree that objects can escape an event horizon in their own proper time, but not observable by accelerated Schwarzchild observers as their time coordinates go to infinity?

 

 

Andrew A. Gray

[Guest Zanket]

So you agree that objects can escape an event horizon in their own proper time, but not observable by accelerated Schwarzchild observers as their time coordinates go to infinity?

Yes, I'm entertaining that notion. I agree for the sake of our argument.

 

However, since the accelerated Schwarzschild observer at r=2M+ε has a Rindler horizon at r=2M, it isn't saying much to say that an object can escape r=2M in its own proper time, for it can escape to no higher r-coordinate. And I'm ignoring for now the fact that the acceleration required to stay at r=2M is infinite.

[andrewgray]

Ok, then. I will let you reconsider this at any time if you like. But for now we will go on the assumption that traditional Schwarzchild analysis is incorrect and that things can escape event horizons in their own proper time.

 

So here is the rest of your logic:

 

Where we left off our discussion, I had made the point that your proof of a Rindler horizon at r=2M, for an observer hovering at r=2M+ε, reinforces the proof in the OP. Even if you disagree with me that an object can pass upward through a Rindler horizon, it doesn’t seem to matter here, because you agreed with me that nothing can reach the accelerating observer for which that horizon exists, from at or below the horizon. That is, you agreed that nothing, not even light, can reach r=2M+ε from r<=2M. I see no difference between that and GR’s prediction that nothing can pass upward through r=2M. If the box in the OP cannot increase its radial position (e.g. the part of the box at r=2M cannot reach r=2M+ε—any higher r-coordinate), then the proof in the OP holds: the box cannot be at rest relative to the particle, in which case GR contradicts SR in X, where GR says that SR applies.

 

Note also that your proof depends upon GR’s equation for the acceleration required for an observer to hover at a given r-coordinate. At r=2M that acceleration is infinite, in which case all objects at r<=2M must fall to r=0. Then the part of the box at r<=2M must fall, which reinforces the proof in the OP. While you have shown that there is a Rindler horizon at r=2M for an observer hovering at r=2M+ε, GR contradicts SR’s prediction that an object can pass upward through r=2M.

 

The bottom line is this: Even if you were right, GR would still contradict SR in X. And when GR contradicts SR in X, GR is self-inconsistent.

 

So let us look at a few of your "sub-statements":

 

That is, you agreed that nothing, not even light, can reach r=2M+ε from r<=2M. I see no difference between that and GR’s prediction that nothing can pass upward through r=2M.

 

This statement depends heavily on which observer you are talking about. In the outgoing freefall inertial frame, an object's world line can intersect the accelerated coordinate r=2M-ε, then intersect r=2M, and then intersect r=2M+ε. This is what it means that "things can escape event horizons in their own proper time".

 

Agreed?

 

At r=2M that acceleration is infinite, in which case all objects at r<=2M must fall to r=0. Then the part of the box at r<=2M must fall . . .

 

Zanket, look again at this Rindler Horizon diagram:

 

 

We see that as we move to the left, the acceleration of the ξ¹ coordinates becomes greater and greater, until finally, at ξ¹ = -1/g , the acceleration becomes infinite.

 

But just because a coordinate acceleration becomes infinite , that doesn't mean too much. Since this is flat spacetime, we are still free to dance around back and forth, in and out, past the ξ¹ = -1/g coordinate world line twice.

 

Agreed?

 

 

 

Andrew A. Gray

[Guest Zanket]

This statement depends heavily on which observer you are talking about. In the outgoing freefall inertial frame, an object's world line can intersect the accelerated coordinate r=2M-ε, then intersect r=2M, and then intersect r=2M+ε. This is what it means that "things can escape event horizons in their own proper time".

I partly disagree. An object can escape a Rindler horizon in its own proper time, but it cannot reach the accelerated observer at r=2M+ε. Then the object's world line cannot intersect r=2M+ε or any higher r-coordinate. The statement "things can escape event horizons in their own proper time" does not mean to me that something can escape to r=infinity; it just means that they can pass upward through the horizon. In this case the accelerated observer is immediately above (arbitrarily close to) the horizon, so an object passing upward through r=2M can reach no higher r-coordinate. The object would be forever stuck “passing upward through” r=2M (and that is ignoring the fact that GR says that the acceleration a material object requires to stay put at r=2M is infinite).

 

Here you say: “A [Rindler] horizon is not a place that objects are prohibited from passing thru, a horizon is a place from which accelerated observers outrun light and simply never see objects pass!” And I agree. The accelerated observer at r=2M+ε outruns anything—even light—passing upward through r=2M. Then something passing upward through r=2M cannot reach r=2M+ε, which is the next higher r-coordinate.

 

Consider a relativistic rocket in an idealized, gravity-free universe. Let wherever the rocket is be r=0. An object that tries to catch up to the rocket starting from a distance 1/a below the rocket (as measured in a rocket's MCIF) will never reach it, hence never reach r=0, even though its movement is unimpeded.

 

But just because a coordinate acceleration becomes infinite , that doesn't mean too much. Since this is flat spacetime, we are still free to dance around back and forth, in and out, past the ξ¹ = -1/g coordinate world line twice.

I agree that this is true for a Rindler horizon. But we’re talking about more than just a Rindler horizon here. When I say that the acceleration required to stay put at r=2M is infinite, I’m not talking about a Rindler horizon at r=2M, I’m talking about GR’s prediction. In your proof that shows that there is a Rindler horizon at r=2M for an observer hovering at r=2M+ε, you depended upon GR’s equation that shows that an infinite acceleration is required to stay put at r=2M. No object passing upward through the Rindler horizon at r=2M can reach any higher r-coordinate, because the accelerated observer is at r=2M+ε. The object would be forever stuck “passing upward through” r=2M. But GR does not allow that. The properties of a Rindler horizon at r=2M do not override GR’s prediction for the acceleration required to hover at r=2M, which is derived from Einstein’s field equations; that prediction is not a misinterpretation.

[andrewgray]

OK, Zanket. Then are we back to where you do not believe that the outgoing freefall coordinates are correct?

 

Remember, we had this diagram for the outgoing Schwarzchild freefall coordinates:

 

 

The R=constant observers cross from r<2M and reach r=∞ in their own proper time. Schwarzchild time t becomes infinite in the mean time, though. The discussion was here if you remember:

 

Outgoing FreeFall inertial coordinates in Schwarzchild Geometry

 

 

Andrew A. Gray

[andrewgray]

It was just exactly the time reversal of the infalling coordinates:

 

 

The R=constant observers fall right through the horizon, but the Schwarzchild time goes to infinity before it happens.

 

Andrew

[Guest Zanket]

Oops, I added something to my last post after yours. See mine at “Consider a relativistic rocket...”

 

I’ve never completely understood your diagrams! That’s why I broke out the relativistic rocket equations and did some of my own analysis in a spreadsheet, and I suggested that you try to show me using those equations. I entertained your claim that r=2M is really just a Rindler horizon. In that case an object can pass outward through it; it just can’t reach the accelerated observer. But when you showed your proof that shows that the accelerated observer is at r=2M+ε, I realized that an object passing upward through r=2M could not reach any higher r-coordinate. My position evolved during our discussion, as I came to better understand a Rindler horizon and how it applies in this case.

 

It seems that you already agreed with me, regardless of your diagrams, when you said “A [Rindler] horizon is not a place that objects are prohibited from passing thru, a horizon is a place from which accelerated observers outrun light and simply never see objects pass!” An object doesn’t pass the accelerated observer in its own proper time in some way that the accelerated observer never sees. Rather the object never catches up to the accelerated observer; the accelerated observer remains always ahead of the object along the object’s axis of motion.

 

Do you agree that in an idealized, gravity-free universe, a relativistic rocket can be deemed to be always at d=0, where d is distance measured in the frame of an inertial frame momentarily comoving with the rocket (MCIF)? Do you agree that an object can pass upward through the Rindler horizon at d=-1/a? Do you agree that the object can never reach d=0?

[andrewgray]

OK, Zanket,

 

Now I see where we are becoming confused.

 

An accelerated hyperbolic coordinate system really has two parts, shown as red and blue in this diagram:

 

 

We have projected the world line onto the x-axis (really x¹-axis) to emphasize that the path of the hyperbolic observer remains one-dimensional.

 

First the red part.

 

At t=-∞, the hyperbolic observer is moving to the left, towards the horizon, and the acceleration is causing him to slow down and stop when both times are zero. During this portion of the transformation, the hyperbolic observer is not outrunning light that is shined to the right.

 

Next the blue part.

 

As t→+∞, the hyperbolic observer starts moving to the right, away from the horizon, and the acceleration is causing him to start outrunning light that is shined to the right.

 

You are only imagining the blue part of the coordinate transformation, and not the red. During the red part, the observer is moving at almost the speed of light towards the horizon. This red part of the transformation is when it is possible to "leave" the horizon.

 

The blue part of the transformation is when it is possible to "enter" the horizon.

 

Do you see that the hyperbolic observer's red and blue paths are time symmetric and one is time reversed from the other?

 

There is a direct analogy with the accelerated Schwarzchild coordinate transformation as well.

 

Imagine an object falling and entering a Schwarzchild event horizon. In the object's own proper time, you probably agree that the object can "enter" the horizon, right?

 

Well, if you time reverse this occurrence, you have the equations for the object "leaving" the horizon. If there was no mishap entering, then there will be no mishap leaving. It is completely time symmetrical.

 

So the complete analogy to the Schwarzchild coordinates would be an object ejected upwards through the horizon as the r=2M coordinate was moving towards the object (red phase), and then ..

 

The object "stops" and falls back "into" the r=2M coordinate in a perfect time reversal (blue phase). The Schwarzchild observer sees neither crossing. And not because the r=2M coordinate is "outrunning the light signal" during the "red" phase. It is because the accelerated Schwarzchild time coordinate comes from minus infinity during the red phase.

 

 

Comments?

 

 

Andrew A. Gray

[Guest Zanket]

At t=-∞, the hyperbolic observer is moving to the left, towards the horizon, and the acceleration is causing him to slow down and stop when both times are zero.

...

Do you see that the hyperbolic observer's red and blue paths are time symmetric and one is time reversed from the other?

I’m sorry, I don’t see what you’re seeing in those diagrams. (I do try!) I can’t grasp the concept of the accelerated observer moving relative to the horizon. The horizon is a fixed distance below the accelerated observer, as measured in the frame of an MCIF of it. That distance is 1/a.

 

I’d really like you to make your case using the relativistic rocket equations, which are derived from the page in Gravitation where you scanned the diagram above. The equations are far simpler than the diagram to me, and I can easily accept or reject what you’re saying using them. What I am seeing is that they disagree with you. So why not try to prove me wrong using them? I can’t accept what you’re saying unless I can confirm it with those equations.

 

Imagine an object falling and entering a Schwarzchild event horizon. In the object's own proper time, you probably agree that the object can "enter" the horizon, right?

Yes, I agree.

 

Well, if you time reverse this occurrence, you have the equations for the object "leaving" the horizon. If there was no mishap entering, then there will be no mishap leaving. It is completely time symmetrical.

I disagree that the equations are completely time symmetrical. An object that falls past the accelerated observer and falls through the Rindler horizon in its own proper time can subsequently pass upward through the horizon in its own proper time, but can never reach the accelerated observer to rise past that observer. (I can confirm that using the relativistic rocket equations.) If the object could do that, the horizon would not be a horizon. Why call it a horizon if it has no affect whatsoever on the behavior of the object?

 

So the complete analogy to the Schwarzchild coordinates would be an object ejected upwards through the horizon as the r=2M coordinate was moving towards the object (red phase), and then ..

 

The object "stops" and falls back "into" the r=2M coordinate in a perfect time reversal (blue phase). The Schwarzchild observer sees neither crossing.

I can agree that the object’s fall is time symmetrical with its rise when it first rose upward through the horizon, in its own proper time. But I disagree that its rise can be time symmetrical with its fall when it first fell past the accelerated observer and fell through the horizon.

 

I’m confused as to what your position is at this point. It seems to me that you have contradicted yourself. Can you restate your position in clear terms? It seems to me that your position is that GR is mathematically self-consistent because you think that the box shown in the OP can be at rest with respect to the particle. You think the box can pass upward through the horizon, even though you’ve shown a proof that there is a Rindler horizon at r=2M for an observer hovering at r=2M+ε. Do I have that right?

 

My position is that your proof shows that the box cannot pass upward through the Rindler horizon at r=2M. If it could, then a particle of the box could pass upward through the horizon to reach the observer hovering at r=2M+ε, in which case the horizon would not satisfy the definition of a Rindler horizon.

 

Some questions I asked above were designed to elicit your position: Do you agree that in an idealized, gravity-free universe, a relativistic rocket can be deemed to be always at d=0, where d is distance measured in the frame of an inertial frame momentarily comoving with the rocket (MCIF)? Do you agree that an object can pass upward through the Rindler horizon at d=-1/a? Do you agree that the object can never reach d=0?

 

Also I have not seen you explain how an object could reach a higher r-coordinate from r<=2M when GR predicts that the acceleration required just to remain at r=2M is infinite. You depended on that equation in your proof.