(Alpha) General relativity is self-inconsistent

In the real world, all instruments have limited precision, so a small enough tidal force is not detectable, and is therefore negligible for the purpose of the measurement.
In the world of "in principle", perfect instruments will register zero tidal force in perfectly flat space, which thus has negligible tidal force.
I agree. But that’s the opposite of what you said to which I replied. You said “The essential problem with your argument is that your thought experiment is defined in such a way that no matter how small the tidal acceleration is, it is never negligible for the purpose of your particular experiment”. That’s saying that that no matter how small the tidal acceleration is, it is always significant for the purpose of my experiment. But all my experiment does is measure the tidal force. And the value of a measurement does not affect the measurement of that value.

True enough, for the purpose of this discussion... but your thought experiment is not only a measurement of the tidal force.
Your thought experiment is specifically set up to require a measurable tidal force.
My thought experiment requires a measurable tidal force only because that shows that the spacetime in X cannot be flat, which is not an experiment but rather only a conclusion. If a thought experiment makes its case by showing that the tidal force is above a certain value, that does not prove that the thought experiment does more than just measure the tidal force.

The particle's position and motion can not be specified in X unless the tidal force is known.
What is your evidence? It is given that the particle is in X and escaping from the black hole; that’s all the information about its position and motion that is needed for the thought experiment to make its case. GR allows the particle to be in X and be escaping from the black hole; the supporting info in the OP shows that.

In that case, Y can not cover the same region of space-time as X.
You have tried to define Y using three self-inconsistent conditions: there is no inertial frame in which the particle is at rest that covers the same region of space-time as X. To see why, you only need to consider the distance from the particle to the event horizon in the particle's inertial rest frame... in that frame, the event horizon is infinitely distant from the particle.
The self-inconsistency you describe is GR’s fault, not mine. According to Taylor and Wheeler, GR predicts that Y can cover the same region of spacetime as X. A self-inconsistency of GR cannot be used to refute a proof of that self-inconsistency. More on this below.

Since relative acceleration depends on both distance and time, which are certainly dependent on the chosen reference frame, it is not at all clear that tidal acceleration is not frame dependent.
Applying some maths will be necessary to know for sure, I think.
No math is necessary. Do you agree that a strain gauge can show the tidal force on an object? Do you think the value shown on the gauge is affected by your velocity relative to the gauge? (That is analogous to asking if the ocean tides are altered by your velocity relative to the Earth.) The particles’ relative acceleration is measured in the frame of one of the particles, just like their relative velocity would be.

You misread Taylor and Wheeler. They do not imply that it is possible to set up any inertial frame in a region of spacetime that can be considered flat.
They do imply that. It would be nice if they were more specific. I know I’m right because flat spacetime is the arena of SR, and in SR any inertial frame can be set up in any other inertial frame. In SR, the frame of any free test particle in an inertial frame J is likewise inertial when extended throughout J.

According to the literal meaning of the quoted extract, it is sufficient that one inertial frame exists across a given region of space-time for that region to be considered flat. (Don't read too much into that, either... that introductory paragraph needs to be understood in the context of the rest of the text.)
So, if X is inertial, then (according to the extract) the space-time in X can be considered flat, regardless of whether and other frame in that space-time is inertial or not.
You’re taking Taylor and Wheeler too literally. If we took them literally, then an inertial frame could contain a black hole, because they don’t specify that the tidal force must be negligible throughout the frame, but rather only in the frame, which could be taken to be just in a small part of it. (Thorne is likewise not specific enough.) To make your case, you need to show that the frame of a free test particle in an inertial frame J can be noninertial when extended throughout J. But to do that you would need to refute SR.

Without doing the maths, I can't be sure, but here's what I suspect:
During the time covered by X, the rod is (close enough to) at rest w.r.t. the escaping particle. It is also close enough to at rest w.r.t. the event horizon - ie in X, the particle, the rod, and the EH all seem to be moving at c.
Yes, they all seem to be moving at c in X, or close to it. Nevertheless, the particle is escaping, and the rod is at rest w.r.t. the particle and straddles the horizon. And the tidal force in X is negligible since X is an inertial frame. Then the rod must be passing outward through the horizon; it must be escaping too. But GR doesn’t allow that, and so the rod cannot exist. If there is any contradiction there then that’s GR’s fault, not mine.

At a later time, however, the outer end of the rod will be slowed down enough for the EH to catch up and pass it - unless the rod breaks. I don't know how strong the force acting on the rod will be. I don't know if it's negligible or huge, so I don't know if breakage is likely or not.
I'm fuzzy on the precise details here... like I said, a mathematical analysis would be good.
If the “outer end of the rod will be slowed down enough for the EH to catch up and pass it”, then the rod does not stay at rest w.r.t. to the escaping particle, and does not exist as given. That would make my case. Because Y is inertial, the rod, or a latticework of synchronized clocks, can stay at rest w.r.t. to the escaping particle. If you were right then Y could not exist in X, and then the spacetime in X could not be flat, and then GR would be self-inconsistent.

In X, the escaping particle is moving at effectively the same speed as the horizon. I'm not sure if it will be negligibly less than c, or if the negligible curvature allows it to be moving negligibly faster than c. Maths...
All that matters is that the particle is escaping, and GR allows that. The rod, or a latticework of synchronized clocks, is given to stay at rest w.r.t. to the escaping particle, and GR allows that in Y, an inertial frame. If the curvature of spacetime is negligible, which it is because X is an inertial frame, then the tidal force on the rod is negligible, because spacetime curvature = tidal force. Then the rod would be passing outward through the horizon, which GR does not allow, and so the rod cannot exist. Hence GR is self-inconsistent. If you find some inconsistency regarding the velocity of the particle, that’s GR’s fault, not mine.

The rod is negligibly stretched (I think) in X.
In the hovering observer's frame it is clearly either stretched considerably, or undergoes tremendous tidal acceleration, or both.
The rod cannot be at once both negligibly stretched and tremendously stretched by a tidal force. A strain gauge on the rod cannot show two values at once.

The situation isn't symmetrical. I don't see that the same logic should apply in each direction. I think that the box frame is inertial, while the rod's rest frame is not.
The situation is symmetrical. Please provide evidence to the contrary. The lack of evidence to the contrary proves my point that the situation is symmetrical. To prove your point you need to show something asymmetrical about the situation.

Perhaps if you crunched the numbers you'd discover otherwise. Or not.
Feel free to crunch them and present your results.
I gave the equation for the tidal force, direct from Taylor and Wheeler, and velocity was not an input. (In the link I gave you, Taylor and Wheeler are clear that the equation I gave you is the one and only equation for spacetime curvature = tidal force in Schwarzschild geometry.) Then my case is made that the tidal force is not dependent on velocity. It is impossible to crunch numbers with velocity on an equation that does not take velocity as an input.

Notice that the units of the curvature factor are neither force nor acceleration.
I don't know how to extract tidal acceleration from the equations for the Schwarzachild metric, but it's clear that it's not simply the curvature factor.
It’s a factor; it’s dimensionless, unitless, like the Lorentz factor for time dilation and length contraction. It tells you the degree of the tidal force relative to the tidal force given by the equation using other inputs. Do you think the Lorentz factor does not indicate time dilation just because it’s unitless?

I don't know how to use the Schwarzschild metric, but I have the impression that it's based around a single spherically symmetrical massive body in an otherwise empty universe...
Yes. Also the object is uncharged and nonrotating.

... so I'd guess that the dr term in the equation is the distance from the center of that body.
In the language of the metric, you're not looking for a velocity term, you're looking for dr, dt, and d(phi), I think.
I don’t know what you’re getting at here. The curvature factor that I gave is all you need to know the degree of tidal force in Schwarzschild geometry. The incorporation of the curvature factor in the Schwarzschild metric is the sole difference between the Schwarzschild metric and the metric for flat spacetime, the metric of SR, in which the tidal force is nonexistent. This link says: “The tidal force goes as M / r^3 at distance r from a black hole of mass M.” You won’t find any reputable reference that supports a claim that tidal force is frame or velocity dependent. Nor would that make sense, since a strain gauge can show the degree of tidal force, and shows the same value to observers in all frames.
 
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OK, I think we've got too many threads going at once... Let's take one bit at a time, if that's OK.

Is X Inertial?
I agree. But that’s the opposite of what you said to which I replied. You said “The essential problem with your argument is that your thought experiment is defined in such a way that no matter how small the tidal acceleration is, it is never negligible for the purpose of your particular experiment”. That’s saying that that no matter how small the tidal acceleration is, it is always significant for the purpose of my experiment. But all my experiment does is measure the tidal force.
It's not the opposite of what I said - it reinforces it.

Look:
In the real world, all instruments have limited precision, so a small enough tidal force is not detectable, and is therefore negligible for the purpose of the measurement.
Your experiment uses unlimited precision.

In the world of "in principle", perfect instruments will register zero tidal force in perfectly flat space, which thus has negligible tidal force.
Your experiment is explicitly set in space that is not perfectly flat.

Put those two together. The conclusion is that the tidal force in your experiment is never negligible.

And the value of a measurement does not affect the measurement of that value.
I'm not sure what you mean by this, or how it is relevant.
In your experiment, the "value of the measurement" is always non-zero, and hence never negligible for the purpose of that experiment.

Perhaps it would be useful to pin down exactly how your experiment measures the value of the tidal force?

Could you conduct the measurement in perfectly flat space to get a zero reading?

My thought experiment requires a measurable tidal force only because that shows that the spacetime in X cannot be flat, which is not an experiment but rather only a conclusion.
By setting up your experiment across the event horizon andincluding an escaping particle, you've made it a condition of your experiment that the spacetime in X is not flat. This is in conflict with your assertion that X is inertial.

You can't have it both ways - you can't say "let X be inertial", and then set conditions so that X is not inertial.

What is your evidence? It is given that the particle is in X and escaping from the black hole; that’s all the information about its position and motion that is needed for the thought experiment to make its case. GR allows the particle to be in X and be escaping from the black hole; the supporting info in the OP shows that.
It's not enough to know that it can be done... you need the precision to be able to do it. If you don't have enough precision to know the tidal force, then you don't have enough precision to place the particle.

You're trying to have it both ways - you want the tidal force to be negligible, but you want to place the particle so precisely that you can't neglect the tidal force.
 
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Can Y cover the same spacetime as X?

zanket said:
The self-inconsistency you describe is GR’s fault, not mine. According to Taylor and Wheeler, GR predicts that Y can cover the same region of spacetime as X.
Taylor and Wheeler predict no such thing.

They do imply that. It would be nice if they were more specific.
Perhaps if you worked through the text?

I know I’m right because flat spacetime is the arena of SR, and in SR any inertial frame can be set up in any other inertial frame. In SR, the frame of any free test particle in an inertial frame J is likewise inertial when extended throughout J.
But the spacetime in question is not quite flat... As soon as you include that escaping particle, you've gone beyond the realm of SR.

You’re taking Taylor and Wheeler too literally. If we took them literally, then an inertial frame could contain a black hole, because they don’t specify that the tidal force must be negligible throughout the frame, but rather only in the frame, which could be taken to be just in a small part of it. (Thorne is likewise not specific enough.)
Like I said, it needs to be read in context with the rest of the text. It's not the Bible, it's a textbook.

To make your case, you need to show that the frame of a free test particle in an inertial frame J can be noninertial when extended throughout J. But to do that you would need to refute SR.
...or include just a little curvature :)


More to come...
 
Zanket,

This is Andrew Gray. I have spent many years studying the Schwarzchild geometry and GR, and the answer to your paradox jumped right out to me. Your question is very similar to one that I asked myself a couple of years ago.

Cut and paste this link into a browser and zoom in all the way: (newbies can't link (edit: changed to links by Pete))

Horizon.gif

The entire chapter is here if you want:

Section 1.5


As you will see, the diagram is of a numerically integrated solution to a laboratory frame that spans the horizon and emits light pulses in all directions just prior to dropping through the horizon. This will map out the geometry in a very insightful way.

Now on to your paradox. I agree with you on your setup. To get a local Lorentz frame, one must only specify to what degree of accuracy is demanded, and the size of the local Lorentz frame is fixed from that.

So let's just say that we set up a local Lorentz frame that is a cubic meter, and let it drop through a horizon. And lets just say that when the center of the cube is at the horizon, then there exists a place in the outer part of the cube where a particle going radially outward at .999999c can escape the black hole.

I believe that you have set everything up in a very well defined way so far, using definitions that are correct.

This particle moving at .999999c defines the Y local Lorentz frame, and initially it overlaps X. Let's arbitrarily set the accuracy so that X and Y initially cover the same space.

Another particle, below the horizon, traveling at .999999c in the same direction (according to X) stays equidistant from the particle outside the horizon (according to X) initially, to within the accuracy demanded! This can be shown rigorously, as Lemaitre's coordinates allow an exact solution. If you do not believe me, then I might be convinced to show you mathematically if you do not see from this logic.

Now comes the kicker:

The particle outside the horizon, and part of the local Lorentz frame Y, simply leave and become disjoint from the local Lorentz frame X before any paradox can occur!



Andrew A. Gray
 
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Relative to X, the event horizon will be moving with the speed of light . So Y would be moving faster than light.
 
Relative to X, the event horizon will be moving with the speed of light . So Y would be moving faster than light.

temur,

Y is presumed to be moving at .999999c in X. And the horizon is moving at c in X because there exists a light geodesic that is "stuck right at the horizon", not leaving. So the horizon is "in between the two particles" which are moving at .999999c in X. But we know that the horizon will never catch up to the outside particle, as it has the necessary escape velocity.

Again, to within the accuracy of the local Lorentz frames, the outside particle will leave the local Lorentz frame X before the horizon can "catch up to it".

"So just put the outside particle closer to the horizon", I can just hear you say. Well, then the particle will have to be going at say, .99999999999999999c, to escape capture, and the horizon still will not catch up before the outside particle is "gone".

Here is the story about the Lemaitre coordinate system falling with the Lab:


Section 1.4


Zoom in if necessary.


Andrew A. Gray
 
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I have spent many years studying the Schwarzchild geometry and GR, ...
Welcome to sciforums Andrew. It is nice to have someone with your level of knowledge here, especially since you can put things in English.

To get a local Lorentz frame, one must only specify to what degree of accuracy is demanded, and the size of the local Lorentz frame is fixed from that.
Agreed.

I believe that you have set everything up in a very well defined way so far, using definitions that are correct.
Nice to have confirmation on that, thanks.

Another particle, below the horizon, traveling at .999999c in the same direction (according to X) stays equidistant from the particle outside the horizon (according to X) initially, to within the accuracy demanded!
Agreed.

If you do not believe me, then I might be convinced to show you mathematically if you do not see from this logic.
I almost always prefer logic and examples in these discussions.

Now comes the kicker:

The particle outside the horizon, and part of the local Lorentz frame Y, simply leave and become disjoint from the local Lorentz frame X before any paradox can occur!
I don’t see how this resolves a key point of the paradox, which is:

zanket said:
But GR predicts that nothing may pass outward through a horizon. Then Y cannot extend below the horizon (if only because otherwise a latticework of synchronized clocks, that stays at rest with respect to Y and is spread throughout Y, would be passing outward through the horizon), and so the spacetime cannot be flat throughout X.
Taylor and Wheeler confirm that the lattice described can exist in an inertial frame, like Y. In fact that is a valid way to verify that a frame is inertial.

You said:

andrewgray said:
Let's arbitrarily set the accuracy so that X and Y initially cover the same space.
But GR does not allow X and Y to initially cover the same space. X straddles the horizon. If X and Y initially covered the same space then Y would initially straddle the horizon too. But if Y straddled the horizon for even a moment, the lattice described would be passing outward through the horizon in that moment, which GR does not allow.

Consider: The tidal force (= spacetime curvature) in Y is initially negligible by the definition of an inertial frame. Y is moving outward, away from the black hole, so the tidal force in Y becomes progressively weaker. Then Y keeps its status as an inertial frame indefinitely. Then the lattice stays essentially at rest with the escaping particle indefinitely. The particle stays surrounded by the lattice, which is spread throughout Y. When the escaping particle is at a great distance from the black hole, so too will the lattice be, still surrounding the particle.

Then how do you explain that Y, including the lattice, can initially straddle the horizon (so it can initially cover the same space as X), without violating GR’s prediction that nothing may pass outward through a horizon?
 
Welcome to sciforums Andrew. It is nice to have someone with your level of knowledge here, especially since you can put things in English.

I appreciate that Zanket. This forum seems like a much friendlier place to communicate because of all the sane rules for alpha proposals.

Now, back to our topic.

But GR does not allow X and Y to initially cover the same space. X straddles the horizon. If X and Y initially covered the same space then Y would initially straddle the horizon too. But if Y straddled the horizon for even a moment, the lattice described would be passing outward through the horizon in that moment, which GR does not allow.

OK, let's take a look at what is happening in our example so we can get "a feel for it". Let's say that when our 1 cubic meter Lorentz frame X is halfway into the horizon, a moving 1 cubic meter Lorentz frame Y magically appears going .999999c relative to X. Let's say that it a "rigid grid" with clocks at each cm[sup]3[/sup] mark (Thought experiment only of course).

This means that the two frames overlap each other for about 3 nanoseconds according to X. Now the horizon is moving at exactly c, so that means that X sees the horizon gaining on the grid at approximately .000001c. That means that while our two frames overlap , the horizon gains 1 micrometer on the grid, or perhaps a ½ micrometer since it appeared at the halfway point. By agreement, this still will be within the agreed tolerances for the local Lorentz frames. The difference between the grid point where a particle escapes and the particle itself is below the tolerance by agreement.
Then the lattice stays essentially at rest with the escaping particle indefinitely.

No, the lattice stays essentially at rest with the escaping particle for about 3 nanoseconds. After that, the two diverge, the particle getting outside the agreed upon tolerances for local Lorentz frames.





Andrew A. Gray
 
Hi Andrew,
I think zanket would be interested into what physically happens to a rod straddling the horizon and moving at 0.999999c relative to X during those 3 nanoseconds.

I'm interested too. I think I get it, and tried to explain earlier in the thread... but I haven't got a great grasp of GR.
 
This forum seems like a much friendlier place to communicate because of all the sane rules for alpha proposals.
Kudos to Pete for that. He came up with the rules.

OK, let's take a look at what is happening in our example so we can get "a feel for it". Let's say that when our 1 cubic meter Lorentz frame X is halfway into the horizon, a moving 1 cubic meter Lorentz frame Y magically appears going .999999c relative to X. Let's say that it a "rigid grid" with clocks at each cm[sup]3[/sup] mark (Thought experiment only of course).

This means that the two frames overlap each other for about 3 nanoseconds according to X. Now the horizon is moving at exactly c, so that means that X sees the horizon gaining on the grid at approximately .000001c. That means that while our two frames overlap , the horizon gains 1 micrometer on the grid, or perhaps a ½ micrometer since it appeared at the halfway point. By agreement, this still will be within the agreed tolerances for the local Lorentz frames. The difference between the grid point where a particle escapes and the particle itself is below the tolerance by agreement.
...
No, the lattice stays essentially at rest with the escaping particle for about 3 nanoseconds. After that, the two diverge, the particle getting outside the agreed upon tolerances for local Lorentz frames.
In the original post (OP), it is given that the escaping particle stays at rest with respect to Y; that is, Y is the escaping particle’s inertial frame (a “Lorentz frame”). I will assume that is the case for your example as well. Then your grid is equivalent to my lattice. If the grid does not stay essentially at rest with the escaping particle indefinitely, then Y is not an inertial frame. Do you agree that in an inertial frame, a free test object initially at rest with respect to the frame stays at rest with respect to the frame (and not just for 3 nanoseconds)? The grid is a free test object. How do you explain that Y is an inertial frame, yet a free test object in Y that is initially at rest with respect to the frame does not stay at rest with respect to the frame?

Put the escaping particle at the center of one of those cm[sup]3[/sup] gaps in the grid, which is initially at rest with respect to the particle. Tell me, what force moves the particle away from that position relative to the grid? If Y is an inertial frame, then when the particle is at a great distance from the black hole, it will still be at that position relative to the grid, still at the center of that gap in the grid. And that would mean that the grid escaped the black hole too, in which case the grid, which initially straddled the horizon, had to pass outward through the horizon. But GR doesn’t allow that.

In the OP, Taylor and Wheeler define an inertial frame as “a reference frame in which a free test particle initially at rest remains at rest”. No free test particle that is in Y and below the horizon can stay at rest with respect to Y. (In any such particle’s frame, the escaping particle recedes at a relativistic velocity.) Then Y cannot be an inertial frame and straddle the horizon. Can you refute that?

BTW, I do follow your explanation. It seems that you are correct that the horizon will gain on the grid, because—as you point out—in X the grid is moving outward slower than the horizon is. But the explanation must be self-consistent. It isn’t self-consistent because Y cannot be an inertial frame and straddle the horizon. The self-inconsistency is not the fault of your explanation. It’s a self-inconsistency of GR, the theory on which your explanation depends. (Specifically, you depend on GR’s prediction that Y can cover the same space as X initially. But GR also predicts that Y cannot be an inertial frame and straddle the horizon, in which case Y cannot cover the same space as X initially.) Unless of course you can show that I’m wrong.
 
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Pete said:
I think zanket would be interested into what physically happens to a rod straddling the horizon and moving at 0.999999c relative to X during those 3 nanoseconds.

I'm interested too. I think I get it, and tried to explain earlier in the thread... but I haven't got a great grasp of GR.

The free-falling observer X is not aware of crossing over the horizon. His spacetime is smooth and his coordinates are fine.

The observer Y that is going outward at .999999c relative to X probably had to expend a lot of energy to get into that situation, but once there and in "free fall", he too does not sense the horizon.

If the gravitational tidal forces are relatively small at the horizon (they do not crush things), then both observers feel fine near the horizon. The lattice that spans the horizon likewise "feels" nothing out of the ordinary. That is, it "feels" similar to being just outside the horizon, or similar to being just inside the horizon.

Anyway, back to Zanket.

zanket said:
If the grid does not stay essentially at rest with the escaping particle indefinitely, then Y is not an inertial frame. Do you agree that in an inertial frame, a free test object initially at rest with respect to the frame stays at rest with respect to the frame (and not just for 3 nanoseconds)? The grid is a free test object. How do you explain that Y is an inertial frame, yet a free test object in Y that is initially at rest with respect to the frame does not stay at rest with respect to the frame?

Bernard Schutz does the best job explaining this in A First Course in General Relativity:

modelofreality.org/inertial.jpg

(cut & paste into browser, zoom in)

The key thing that he says is:

"Since any nonuniformity is, in principle, detectable, a frame can only be regarded mathematically as inertial in a vanishingly small region".

So in this case Zanket, the particle would only stay near to a vanishly small inertial frame grid point for a longer and longer time.

Tell me, what force moves the particle away from that position relative to the grid?

The whole theory of GR is based on gravity not being a force. The two separate because their free-falling geodesics diverge due to spacetime curvature caused by gravity.

Then Y cannot be an inertial frame and straddle the horizon. Can you refute that?

Once again, technically, there cannot be a finite sized inertial frame where spacetime is not flat. Only practically, and to the limits agreed upon by the experimenters. So technically, an inertial frame Y cannot really straddle anything where gravity is present.


Andrew A. Gray
 
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The whole theory of GR is based on gravity not being a force.
Agreed. No force exists to move the particle away from its initial position relative to the grid. With no force to move it, the particle stays at rest relative to the grid.

The two separate because their free-falling geodesics diverge due to spacetime curvature caused by gravity.
The tidal force (= spacetime curvature) in Y is negligible according to the definition of an inertial frame. The tidal force in Y negligibly stretches and squeezes the grid, as described by Thorne in the OP. That does not cause the particle to significantly move away from its initial position relative to the grid.

Observations show that the Milky Way is moving toward the Andromeda galaxy. Then the tidal force imparted by Andromeda on the Milky Way grows continuously stronger. Yet the North Star has indicated north on Earth with good accuracy for at least a thousand years. For at least a thousand years, the free-falling geodesics of the Sun and the North Star have not significantly diverged due to the spacetime curvature caused by Andromeda. Those two stars have remained essentially at rest relative to each other all that time, despite the growing tidal force imparted by Andromeda. Do you disagree?

By the definition of negligible, a negligible tidal force cannot cause a significant effect. It certainly does not force free test particles to move away from each other at a relativistic velocity, as I noted here:

zanket said:
In the OP, Taylor and Wheeler define an inertial frame as “a reference frame in which a free test particle initially at rest remains at rest”. No free test particle that is in Y and below the horizon can stay at rest with respect to Y. (In any such particle’s frame, the escaping particle recedes at a relativistic velocity.) Then Y cannot be an inertial frame and straddle the horizon. Can you refute that?

That was a key question; I don’t see how your explanation has refuted it. The tidal force in the International Space Station (ISS) is presumably negligible for the purposes of the vast majority of the experiments that are conducted there. Yet I have never read or seen that a free object in the ISS must move at a relativistic velocity relative to the station; rather, I’ve seen objects in the ISS floating essentially at rest with respect to it.

Do you agree that the grid + particle can be considered to be a single object N (like how we consider a bunch of stars to be the Milky Way)? Do you agree that the negligible tidal force in Y will not significantly change the shape of N? Do you agree that if the particle moves a great distance away from the grid, the shape of N has significantly changed?

How could the tidal force in Y move the part of the grid that is initially above the escaping particle (i.e. further away from the black hole) to a position below the particle? Why wouldn’t it move that part of the grid and the particle in lockstep? How does the tidal force know to move that part of the grid, but not the particle too? And if that part of the grid stays above the particle, how does the grid not escape from the black hole too? Do you think the negligible tidal force in Y tears the grid apart?

andrewgray said:
Once again, technically, there cannot be a finite sized inertial frame where spacetime is not flat. Only practically, and to the limits agreed upon by the experimenters. So technically, an inertial frame Y cannot really straddle anything where gravity is present.
I call this the “not exactly flat” argument. It is true that the spacetime in Y is not exactly flat, which is the same as saying that the tidal force in Y is not completely nonexistent. But that is not a panacea that explains any degree of difference between the behavior in Y and the behavior in an inertial frame J in an idealized universe without gravity. Do you agree? It explains only a negligible such difference; Einstein is very clear about that in his statement of the equivalence principle in the OP.

In the OP Thorne says “small, freely falling frames in the presence of gravity are equivalent to inertial frames in the absence of gravity”. That makes it clear that we need not be concerned with “technically”; it does not refute my argument. Moreover, the definitions for an inertial frame and flat spacetime in the OP explicitly allow a negligible tidal force—they apply to a gravity-endowed universe. So, while a finite sized inertial frame requires flat spacetime, “flat” does not necessarily mean perfectly flat; by definition the spacetime can be negligibly curved. By accounting for a nonzero tidal force in their definitions of an inertial frame and flat spacetime (and Thorne does likewise in his definition of an inertial frame), Taylor and Wheeler override the “technically” argument you make, and justly so, because the equivalence principle tells us that the argument has a very limited application.

I have shown that the difference between the behaviors in J and Y is radical when Y straddles the horizon, and that radical difference is not reduced in the slightest no matter how much the tidal force is reduced in Y (by, say, reducing the size of Y). J and Y cannot be considered equivalent when Y straddles the horizon. Then Y cannot be considered to be an inertial frame when it straddles the horizon.

In the OP Thorne says “All small, inertial (freely falling) reference frames in our real, gravity-endowed Universe must be "created equal"; none can be preferred over any other in the eyes of the laws of physics”. Therefore, to refute me using the “not exactly flat” argument, you would need to show that a free object in the ISS must move at a relativistic velocity relative to the station, just as the escaping particle in Y must move at a relativistic velocity in the frame of any free test particle that is in Y and below the horizon. I don’t see how you have done that, or could do that.

According to GR, X can be arbitrarily large and still have a tidal force throughout it that is 10[sup]7000[/sup] times smaller than the tidal force throughout your room now, for a sufficiently large black hole. Regardless, in the frame of any free test particle that is in Y and below the horizon, the escaping particle must recede at a relativistic velocity according to GR, even if those particles are initially an infinitesimal distance apart. Do you agree that this behavior contradicts what you (I assume it’s you) said here?: “Put a green pea in front of your face and let it go. If you are in an inertial frame, it will stay where you let it go.” Or do you think that the fact that the spacetime in a finite sized inertial frame in our gravity-endowed universe is not exactly flat explains that the meaning of “stay” can include “recede at a relativistic velocity”, so that your frame can still be deemed inertial when the pea shoots away at a relativistic velocity?

Let the pea be escaping from the black hole, initially located just above the horizon. Put your face in front of it when you are just below the horizon. You cannot stay at rest with respect to the pea, or anywhere close to it. According to GR, the pea must recede at a relativistic velocity in your frame, because the horizon is receding outward at c in your frame and the pea moves outward away from the horizon.

Do you still think that Y—the escaping particle’s frame—can be an inertial frame and straddle the horizon?
 
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Relative to X, the event horizon will be moving with the speed of light . So Y would be moving faster than light.
Yes, in X the horizon is moving outward at c, the speed of light. Y is the escaping particle’s frame. The particle is in both X and Y initially. In X the particle is moving outward faster than the horizon is, otherwise it would not be escaping. Then it would seem that Y is moving faster than c in X. In no inertial frame that is wholly above the horizon is it possible for a free particle to exist such that not even light can gain on it. (Like light that an observer in X shines radially outward toward the particle, starting the moment the horizon is crossed; such light stays at the horizon indefinitely according to GR.) X is an inertial frame with properties not shared by other inertial frames, in defiance of Einstein’s own equivalence principle, and Thorne, who in the OP says “All small, inertial (freely falling) reference frames in our real, gravity-endowed Universe must be "created equal"; none can be preferred over any other in the eyes of the laws of physics”.

If there is any problem with that, then it’s GR’s fault, not mine. The OP shows that GR predicts that Y can initially cover the same region of spacetime as X does, and shows that GR also predicts that Y cannot extend below the horizon, in which case Y cannot initially cover the same region of spacetime as X does. That’s a self-inconsistency. It is okay for me to set up a thought experiment in which Y initially covers the same region of spacetime as X does, because GR allows that. The self-inconsistency itself cannot be used to prevent me from showing it.
 
Once again, technically, there cannot be a finite sized inertial frame where spacetime is not flat. Only practically, and to the limits agreed upon by the experimenters. So technically, an inertial frame Y cannot really straddle anything where gravity is present.

Andrew---I tried to make this point to zanket many moons ago. Good luck:)
 
Let the pea be escaping from the black hole, initially located just above the horizon. Put your face in front of it when you are just below the horizon. You cannot stay at rest with respect to the pea, or anywhere close to it. According to GR, the pea must recede at a relativistic velocity in your frame, because the horizon is receding outward at c in your frame and the pea moves outward away from the horizon.

Do you still think that Y—the escaping particle’s frame—can be an inertial frame and straddle the horizon?

OK, Zanket, I do see what you are saying. And I agree. An (approximate) inertial frame centered on the escaping particle cannot straddle the horizon.

So any (approximate) inertial frame centered on an escaping particle would have to be limited to space outside the horizon.

It seems like all (approximate) inertial frames that straddle the horizon must be ingoing.

OK, I think I've got that. So back to your proof.

Since the spacetime throughout an inertial frame is flat, it must be possible to set up an inertial frame Y that extends throughout X and in which a free test particle, that is above the horizon and moves away from the black hole indefinitely, stays at rest.

In this case:
Spacetime thoughtout an (approximate) inertial frame is (approximately) flat.

And we have just seen that it is impossible to set up an (approximate) inertial frame Y, centered on an escaping particle, that extends throughout X (straddles the horizon).

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.


Andrew A. Gray
 
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?

Zanket, please correct me if I've got this wrong, and apologies in advance if so.
 
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