Guest Zanket Posted July 20, 2007 Report Posted July 20, 2007 The green light does not forbid you from going through. If for some reason you don't go through, at the most you are annoying the driver behind you but you aren't violating the green light.This doesn’t show a problem with the proof in the OP. If a theory said: The green light does not forbid you from going through.The green light forbids you from going through. Then the theory would be inconsistent. You really don't realize how this makes it meaningless to say that the box straddles the horizon. You also fail to realize the difference between saying "Some birds can fly." and SR not forbiding birds from flying, and also that between saying "No birds can fly." and GR predicting that a particular bird cannot fly.You’ve shown no problem with the proof in the OP here. You should not just allude to refutations; you should make them. Do you agree that the following theory about birds is inconsistent?: Some birds can fly.No birds can fly. If you disagree, then I say that you are illogical. GR says: The box and the particle can be at rest relative to each other.The box and the particle cannot be at rest relative to each other. Do you see any contradiction there? Do you disagree that GR implies both of these statements? GR implies the first statement by saying that SR applies in X. GR implies the second statement by saying that objects above r=2M (like the particle) can escape to r=infinity, and that nothing (like the box) can pass outward through the horizon. Likewise, SR describes dynamics of bodies in terms of their masses but makes no assumption about these having the effect called gavity. It therefore does not forbid the apple from remaining where you let go of it. GR does, by saying that the frame at rest with the ground isn't an inertial one, as it would be according to SR. Why is this not inconsistent?How is the question relevant to the discussion? The box and the particle are both freely falling in an inertial frame. There is no ground in the proof in the OP. Let the escaping particle let go of an apple. Does it remain where the particle let go of it (i.e. does it remain at rest relative to the particle)? If the apple spans the horizon then it (the whole apple) cannot remain at rest relative to the particle, according to GR, even when the apple is wholly within X, an inertial frame. But GR also says that SR applies in X, and SR says that the apple can remain at rest relative to the particle. Then GR is self-inconsistent. In a region of spacetime that straddles the event horizon, which directions are spacelike and which are timelike?It doesn’t matter. Regardless what GR says beyond the following statements that it implies ...: The box and the particle can be at rest relative to each other.The box and the particle cannot be at rest relative to each other. ... the theory would still be inconsistent. Once a theory contradicts itself, no additional statement it makes reverses that. To refute me, you need to show that one or both of those statements are not implied by GR. About the equivalence principle, in the OP Thorne says “Most importantly, the principle of relativity must be true: 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.” X is an inertial frame. You’re implying that because GR says that some directions in X are spacelike and some directions are timelike, SR can work differently in X than it does in an inertial frame Y that is wholly above the horizon. But no, GR would be inconsistent in that case, because it says that SR works the same in X as it does in Y. GR cannot have it both ways. The self-inconsistency of GR cannot be used to refute a proof of that.
andrewgray Posted July 21, 2007 Report Posted July 21, 2007 Zanket, These diagrams are not inconsistent with the rocket equations. Here are the rocket and the light that the rocket is "outrunning": The "rocket equation" is represented by the green path, and the pursuing light represented by the purple path. Notice that the light's R coordinate is constantly decreasing. However, as the rocket heads toward the speed of light, the rocket and the light will head toward being equidistant in the inertial frame. So the accelerated R coordinate of the light can be forever decreasing, but the separation distance between it and the rocket will remain essentially the same forever according to the inertial observer. That is, the rocket stays at R=1, the pursuing light heads for R=-∞, but after a while, the rocket and the light stay essentially equidistant in the inertial frame forever. That is, decreasing R coordinates, that in themselves are accelerating to a point exceeding c, are not a a good measure of inertial frame separations. That is (further), just because the light's R coordinate is decreasing without limit, that doesn't mean that the light is receding from the rocket. This was also shown to be the same for the Schwarzchild coordinates in a small neighborhood near the horizon. Andrew A. Gray
Qfwfq Posted July 23, 2007 Report Posted July 23, 2007 You’ve shown no problem with the proof in the OP here. You should not just allude to refutations; you should make them.I've been trying to lead you toward it but there seems to be no hope. You can't even see the distintion in the simple logic that I discuss. According to town law, people can sit on the park benches, and many people do. Mr. and Mrs. Smith are very law abiding people, they would never sit on those benches if law said that you can't. Mr. and Mrs. Smith don't allow their five year old son to sit on those park benches, considering some recent local events. Are Mr. and Mrs. Smith self inconsistent? How can they be both perfectly law abiding and concerned about little Johnny, at the same time? GR implies the first statement by saying that SR applies in X. GR implies the second statement by saying that objects above r=2M (like the particle) can escape to r=infinity, and that nothing (like the box) can pass outward through the horizon.And where does GR imply the first, in the same case in which it implies the second? Only in your misunderstanding of it. If I went further about the specific case of near the event horizon, you would mistake it for criticism of the equivalence principle on the basis of the "not exactly flat" argument. How is the question relevant to the discussion? The box and the particle are both freely falling in an inertial frame. There is no ground in the proof in the OP.I wasn't talking about the OP. Let the escaping particle let go of an apple. Does it remain where the particle let go of it (i.e. does it remain at rest relative to the particle)?Of course, it implies that the apple was escaping too, which means the apple must have been ouside the event horizon. You’re implying that because GR says that some directions in X are spacelike and some directions are timelike, SR can work differently in X than it does in an inertial frame Y that is wholly above the horizon. But no, GR would be inconsistent in that case, because it says that SR works the same in X as it does in Y. GR cannot have it both ways. The self-inconsistency of GR cannot be used to refute a proof of that.No, not because GR says. SR says that some directions are spacelike and some directions are timelike. I asked you which, in a spacetime region that straddles the event horizon; GR, via the Schwarzschild metric, give a hint about which. And you haven't answered.
Guest Zanket Posted July 25, 2007 Report Posted July 25, 2007 These diagrams are not inconsistent with the rocket equations.Please show your work using the relativistic rocket equations. When I use those equations, I see that a photon cannot pass upward through a Rindler horizon and reach the accelerated observer for who that horizon exists. My results agree with those at the Rindler Horizon site. You agreed that there’s a Rindler horizon at r=2M for an observer hovering at r=2M+ε, where ε is arbitrarily small. The Rindler horizon is at a distance 1/a below the hovering observer in that observer’s frame, and 1/a is arbitrarily small for an observer hovering at r=2M+ε, since GR’s equation for the acceleration required to hover at a given r-coordinate approaches infinity as r decreases to approach 2M. In the frame of a freely falling observer dropped from rest at r=2M+ε (at rest with respect to the observer hovering there), the distance in geometric units to the observer hovering there is: d = (sqrt(1 + (a * t)^2) - 1) / a The variables are defined at the relativistic rocket site. In the frame of the freely falling observer, the distance in geometric units to a photon emitted radially upward at t=0 from the Rindler horizon is: d_photon = t - (1 / a) The photon travels t light years in t years, starting from 1/a light years below the freely falling observer, all in that observer’s frame. You can plug these formulas into a spreadsheet to see that, regardless of the values of the variables a and t, d_photon is always less than d. For a given a, d_photon asymptotes to d as t increases, just like the Rindler Horizon site shows it does: This shows mathematically that according to GR nothing—not even light—can reach r=2M+ε from r=2M (i.e. nothing can pass outward through r=2M), in which case the box in the OP cannot be at rest relative to the escaping particle. My results agree with the widely published predictions of GR, and contradict the claim you have been making. Can you refute my results using the relativistic rocket equations? There’s an easier way to see that the relativistic rocket equations agree with me, and disagree with you. In the hovering observer’s frame, time is stopped at the Rindler horizon at r=2M, due to gravitational time dilation (the crew of the rocket experiences the equivalent of a uniform gravitational field, according to the equivalence principle; the relativistic rocket site confirms). Then a photon emitted radially upward from r=2M stays at r=2M in the hovering observer’s frame, just like the Rindler Horizon site says: “In other words, the event of the emission of the flash of light is always in the present, and always a distance of 1/a behind the ship. As far as the passengers on the ship are concerned, the flash of light isn't getting any closer.”
Erasmus00 Posted July 25, 2007 Report Posted July 25, 2007 SR allows but does not require two bodies each anywhere in a common inertial frame to be at rest relative to each other. GR forbids that in X I added a very important bold statement to the above. SR does NOT require a specific two objects to be at rest relative to each other. Hence, the fact that GR does not allow it isn't really a problem. Consider- we all agree that near a black hole's event horizon the metric can be expanded and locally described by a rindler horizon. The rindler horizon IS a flat spacetime- so we have no contradiction. -Will
andrewgray Posted July 25, 2007 Report Posted July 25, 2007 OK, One last, last attempt. Let's let a=1 for ultimate simplicity. The relativistic rocket equations in their simplest form: [math]t=\sinh (T)[/math][math]x=\cosh (T)[/math] yields hyperbolic motion from your diagram: [math]x^2-t^2 = 1[/math] We need the simplest forms. The coordinates (R,T) used by the accelerated rocket frame are given by: [math]t = R \sinh(T)[/math][math]x= R \cosh(T)[/math] So the rocket remains at R=1 with an acceleration of 1. These coordinates satisfy the flat metric: [math]-d \tau^2 = -R^2dT^2 + dR^2 + dy^2 + dz^2[/math] in the area to the right of the horizon. Now consider two balls spanning the horizon (R=0), and imagine a "box around them". We wish one ball to start approx ε inside the horizon, and one approx ε outside the horizon at x=0. The balls are traveling just under the speed of light. The balls actually start separated by 1.99999999999999998ε. Question 1: Can the two balls remain at a distance 1.99999999999999998ε as the outside ball gets to very large R, say R=1,000,000? (i.e., for as long as desired). Question 2: What is happening to the R coordinate of the other ball as this happens? Well, the answer to question 1 is obvious. The outside ball makes it to R=1,000,000 when [math]R\sinh T = \frac{R \cosh T}{.99999999999999999} - \epsilon [/math] or when [math]1,000,000\sinh T = \frac{1,000,000 \cosh T}{.99999999999999999} - \epsilon [/math] If we let ε=.01, then the outside ball reaches R=1,000,000 when T=18.4 and when t=4.9x1013 The two balls can obviously remain at a distance of 1.99999999999999998ε for as long as desired, and the outside ball can get to any large value of R desired. (Just increase the velocity enough to make it to any R, but below the speed of light). Question 2 is trickier, and this question is the cause of all of the confusion. We want to extend the hyperbolic coordinates below the horizon, even though they are singular at the horizon (just like the Schwarzchild coordinates are). The criteria used is that they must satisfy the Schwarzchild metric reduced to first order: [math]-d \tau^2 = R^2dT^2 - dR^2 + dy^2 + dz^2[/math] Where we have calculated that the dT2 and dR2 coefficients change sign like the Schwarzchild metric, reduced to first order. So for the region below the horizon, the coordinates become: [math]t = -R \cosh(T)[/math][math]x= -R \sinh(T)[/math] These coordinates still satisfy the flat spacetime metric [math]-d \tau^2 = R^2dT^2 - dR^2 + dy^2 + dz^2[/math] So when the outside ball is at R=1,000,000 , what R coordinate does the inside ball have? We have: [math]t=4.9x10^{13}[/math] [math]x =4.9x10^{13} (.99999999999999999) - .01 [/math] [math]t^2-x^2=R^2[/math] Thus, the inside ball is at R = -1,010,000 when the outside ball is at R=1,000,000, and yet they remain equidistant. The conclusion? The two balls can remain equidistant as the outside ball gets out to infinity, no problem. The horizon remains inbetween, no problem. ΔR=2,010,000, yet the balls remain 1.99999999999999998ε apart. The balls remain equidistant as the outside ball escapes to infinity, and as the inside ball has ever decreasing R, as required by the equivalence principle, keeping GR consistent. In this case, the inside ball does not have to "come out of the horizon" to remain equidistant from the outside ball. Andrew A. Gray
Guest Zanket Posted July 25, 2007 Report Posted July 25, 2007 According to town law, people can sit on the park benches, and many people do. Mr. and Mrs. Smith are very law abiding people, they would never sit on those benches if law said that you can't. Mr. and Mrs. Smith don't allow their five year old son to sit on those park benches, considering some recent local events. Are Mr. and Mrs. Smith self inconsistent?No, but this is a false analogy, so it's irrelevant. It would be a true analogy to the OP if the law said that you both can and cannot sit on the benches. And where does GR imply the first, in the same case in which it implies the second? Only in your misunderstanding of it.GR implies the first in the same case it implies the second because both apply to the same events in X. You’ve shown no misunderstanding of mine. I wasn't talking about the OP.False analogies to the OP are irrelevant to this discussion. Of course, it implies that the apple was escaping too, which means the apple must have been ouside the event horizon.The apple need not be wholly outside of the event horizon according to GR, because GR says that SR applies to X, an inertial frame that extends below the horizon. The apple can be anywhere in X and be at rest relative to the escaping particle, according to SR. No, not because GR says. SR says that some directions are spacelike and some directions are timelike. I asked you which, in a spacetime region that straddles the event horizon; GR, via the Schwarzschild metric, give a hint about which. And you haven't answered.Then I rephrase to: You’re implying that because SR says that some directions in X are spacelike and some directions are timelike, SR can work differently in X than it does in an inertial frame Y that is wholly above the horizon. But GR would be inconsistent in that case, because it says (via the equivalence principle) that SR works the same in X as it does in Y. GR cannot have it both ways. The self-inconsistency of GR cannot be used to refute a proof of that. If you want to lock the thread or ban me or whatever because I don’t answer a question that clearly cannot make your case (i.e. an irrelevant question), go ahead. I insist that people show relevancy before I answer their question, if ask for it. Explain to me how you can possibly make your case with the answer to your question, and if your explanation convinces me that you can possibly make your case with the answer, I’ll answer the question.
Guest Zanket Posted July 25, 2007 Report Posted July 25, 2007 I added a very important bold statement to the above. SR does NOT require a specific two objects to be at rest relative to each other. Hence, the fact that GR does not allow it isn't really a problem.You should not quote me and change what I said. Make it your own statement instead. “Allows but does not require” contradicts “disallows”. SR need not require the box and the escaping particle to be at rest relative to each other, for GR to be self-consistent by both allowing and disallowing those to be at rest relative to each other. Consider- we all agree that near a black hole's event horizon the metric can be expanded and locally described by a rindler horizon. The rindler horizon IS a flat spacetime- so we have no contradiction.This logic simply ignores what makes GR self-inconsistent. GR says that the spacetime in X is both flat and not flat. GR says that the spacetime in X is flat because X is an inertial frame and the spacetime in an inertial frame is flat. But if the spacetime in X was flat then the box and the escaping particle could be at rest relative to each other. GR doesn’t allow that, so the spacetime in X cannot be flat.
Guest Zanket Posted July 26, 2007 Report Posted July 26, 2007 The balls remain equidistant as the outside ball escapes to infinity, and as the inside ball has ever decreasing R, as required by the equivalence principle, keeping GR consistent. In this case, the inside ball does not have to "come out of the horizon" to remain equidistant from the outside ball.Let’s be clear that you have not shown here that something can pass outward through r=2M, which is what you have claimed before. I agree with you here and I gave my own example above, where a relativistic rocket decelerates alongside an indefinitely long rod in an idealized, gravity-free universe. But you’re not refuting the OP here. To refute the OP you need to show that GR does not disallow anything to pass outward through the horizon. That is, you need to show that GR does not predict black holes. Only when GR does not predict black holes can it be consistent with what else it says. It seems that you think you have shown that the box and the escaping particle in the OP can be at rest relative to each other even as the box does not pass outward through the horizon. But you have not shown that. Step back to see the absurdity of what you’re suggesting. Do you really think that the box and the escaping particle can be at rest relative to each other even as the particle passes Earth, where the Earth is at r=1 million light years? Can you see that the particle would need to have exited the box by that point? The size of the box is limited by the size of X. The particle can exit the box only by moving relative to it. Your logic works only when the balls share an inertial frame forever, and when the distance between them as either measures is arbitrarily large (that's why I used an indefinitely long rod in my example). But the particle is escaping to r=infinity in our real, gravity-endowed universe, so it won’t share X forever, or any other inertial frame that contains the box. That’s a mistake in your logic. For the box and the particle to eventually no longer share an inertial frame while both remain freely falling, they must move relative to each other in X. To refute me you need to use the relativistic rocket equations to show that something can pass outward through r=2M, so that the particle can be at rest relative to the box as the particle escapes to r=infinity (in which case it will remain inside the box and the box will also escape to r=infinity). Can you do that?
InfiniteNow Posted July 26, 2007 Report Posted July 26, 2007 I wasn't talking about the OP. False analogies to the OP are irrelevant to this discussion. Wow, dude. Talk about missing a point. :)
Guest Zanket Posted July 26, 2007 Report Posted July 26, 2007 Wow, dude. Talk about missing a point. :)Like what? A false analogy was made, followed by the question "Why is this not inconsistent?" That cannot refute the OP.
Erasmus00 Posted July 26, 2007 Report Posted July 26, 2007 This logic simply ignores what makes GR self-inconsistent. GR says that the spacetime in X is both flat and not flat. GR says that the spacetime in X is flat because X is an inertial frame and the spacetime in an inertial frame is flat. But if the spacetime in X was flat then the box and the escaping particle could be at rest relative to each other. GR doesn’t allow that, so the spacetime in X cannot be flat. GR says that the sapcetime in X has to be flat. Many have shown this spacetime IS a Rindler metric- which IS a flat spacetime. End of story. This isn't ignoring your claimed inconsistency. Again, a recap of your argument1. SR is flat spacetime2. GR should reduce to SR locally3. Near a black hole, GR does not reduce to SR. You fail at 3, because again, near a black hole GR reduces to a rindler metric which IS a flat spacetime. -Will
Guest Zanket Posted July 26, 2007 Report Posted July 26, 2007 GR says that the sapcetime in X has to be flat. Many have shown this spacetime IS a Rindler metric- which IS a flat spacetime. End of story. This isn't ignoring your claimed inconsistency.Not end of story, because you do ignore the fact that GR also says that the spacetime in X cannot be flat. You ignore the fact that GR does not allow the box and the particle to be at rest relative to each other, which it must allow since SR allows that and GR says that SR applies in X. (If GR allowed the box and the particle to be at rest relative to each other then r=2M could not be a horizon like GR says it is, for then the box would have to pass outward through it.) A self-inconsistent theory contradicts itself. You can't refute me by simply ignoring the contradiction I show and claiming that you're not ignoring it. Again, a recap of your argument1. SR is flat spacetime2. GR should reduce to SR locally3. Near a black hole, GR does not reduce to SRThat is not my argument. Here is a summary of my argument:1. SR is flat spacetime2. GR reduces to SR locally3. In a local region straddling the horizon of a black hole, GR, according to its own predictions, both reduces and does not reduce to SR Here is an analogy to the argument you are making. Let a theory about birds make the following statements: Robins can fly.Robins cannot fly. If I pointed out that this theory is inconsistent, you would say “It’s not, because it says that robins can fly. End of story”. See a problem there?
Erasmus00 Posted July 26, 2007 Report Posted July 26, 2007 3. In a local region straddling the horizon of a black hole, GR, according to its own predictions, both reduces and does not reduce to SR You haven't shown this. Where does my proof go wrong- 1. A Rindler metric is an SR metric2. Near an event horizon, GR locally mathematically reduces to a Rindler metric3. Therefore, locally GR reduces to SR. For your point to be true (GR reduces to AND does not reduce to SR) you need to show that a rindler metric is BOTH SR and NOT SR. Good luck. Also Here is an analogy to the argument you are making. Let a theory about birds make the following statements: Robins can fly. Robins cannot fly. If I pointed out that this theory is inconsistent, you would say “It’s not, because it says that robins can fly. End of story”. See a problem there? Your analogy is flawed. The better analogy would be "Robins can fly in some situations and can't fly in others. " Followed by "in one situation, Robins can't fly." -Will
Guest Zanket Posted July 26, 2007 Report Posted July 26, 2007 You haven't shown this. Where does my proof go wrong- 1. A Rindler metric is an SR metric2. Near an event horizon, GR locally mathematically reduces to a Rindler metric3. Therefore, locally GR reduces to SR.There’s nothing amiss here. But it’s not a proof that refutes the OP, because it doesn’t address the inconsistency that I show. For your point to be true (GR reduces to AND does not reduce to SR) you need to show that a rindler metric is BOTH SR and NOT SR. Good luck.No. Does anything prevent GR from also postulating that SR does not apply locally near a horizon? No, because GR can say whatever it wants. GR could postulate that, although it would then be self-inconsistent. Then I have proven that to show that GR reduces to AND does not reduce to SR, I need not show that a rindler metric is BOTH SR and NOT SR. Instead I could just show that GR also postulates that SR does not apply locally near a horizon, if GR postulated that. No good luck needed. Your analogy is flawed. The better analogy would be "Robins can fly in some situations and can't fly in others. " Followed by "in one situation, Robins can't fly."No, the analogy was perfect. GR says that the spacetime in X is both flat for the purposes of all situations (by saying that SR applies in X) and always not flat for the purposes of all situations (by disallowing the box and the particle to be at rest relative to each other, as SR would allow). In your “refutation” you say that the spacetime in X is flat and simply ignore that GR also says that the spacetime in X is not flat. Since you think the spacetime in X is flat, you must agree that the box and the escaping particle can be at rest relative to each other. Right? Then you would need good luck to refute me, because if they were at rest relative to each other then the box would be passing outward through the horizon, in contradiction to the definition of a horizon. You’re just supporting my case.
Qfwfq Posted July 26, 2007 Report Posted July 26, 2007 No, but this is a false analogy, so it's irrelevant. It would be a true analogy to the OP if the law said that you both can and cannot sit on the benches.You do not claim that SR says both "you can" and "you can't" so it isn't a false analogy. You are skating on very, very thin ice. GR is "abiding of" SR (which says "you can") but it is also "concerned about Johnny" (it says "you can't"); the analogy is good and GR doesn't violate SR. GR implies the first in the same case it implies the second because both apply to the same events in X. You’ve shown no misunderstanding of mine.Now the exact sense in which GR is "abiding of" SR can only be expressed with differential geometry and you have been showing you don't understand it, you quote "Einstein's words" the way the people in Babette's Feast quote the words of the Parson. That isn't scientific debate. ...because GR says that SR applies to X, an inertial frame that extends below the horizon.GR does not say this at all. If GR did say that SR applies exactly to a spacetime region partly with [imath]r<r_S[/imath] and partly with [imath]r>r_S[/imath], then the locus of points with [imath]r=r_S[/imath] would have no special status at all. You’re implying that because SR says that some directions in X are spacelike and some directions are timelike, SR can work differently in X than it does in an inertial frame Y that is wholly above the horizon. But GR would be inconsistent in that case, because it says (via the equivalence principle) that SR works the same in X as it does in Y.No, I don't imply that SR "works differently", I simply say that SR distinguishes spacelike, timelike and null directions. What is an event horizon? Any spacelike submanifold is an event horizon. Can you observe an event that will happen tomorrow? Why is the locus of points with [imath]r=r_S[/imath] called the event horizon? To see why, you would have to understand the yet outstanding question, instead of claiming it irrelevant. ...question that clearly cannot make your case (i.e. an irrelevant question)...Easy to say "clearly", without even understanding it...I insist that people show relevancy before I answer their question, if ask for it.this is no excuse...Explain to me how you can possibly make your case with the answer to your question, and if your explanation convinces me that you can possibly make your case with the answer, I’ll answer the question.I could explain once we have the answer but, once we do, there wouldn't be much point in you supplying it, would there? Hint: Going from [imath]r>r_S[/imath] to [imath]r<r_S[/imath], the dr and dt elements of the Schwarzschild metric trade signs, which means that dr and dt trade roles. Does this help any?
Guest Zanket Posted July 26, 2007 Report Posted July 26, 2007 You are skating on very, very thin ice. GR is "abiding of" SR (which says "you can") but it is also "concerned about Johnny" (it says "you can't"); the analogy is good and GR doesn't violate SR.GR does violate SR. It says that SR applies in X, but then does not allow SR to work as expected in X, by not allowing the box and the particle to be at rest relative to each other. The ice is thick, not thin. Now the exact sense in which GR is "abiding of" SR can only be expressed with differential geometry and you have been showing you don't understand it, you quote "Einstein's words" the way the people in Babette's Feast quote the words of the Parson. That isn't scientific debate.No, it can also be expressed in terms of widely published predictions, which can be supported without differential geometry. (For example, above I showed mathematically that GR predicts that nothing can pass outward through a horizon, by using the relativistic rocket equations.) Einstein’s words in the equivalence principle are relevant to this discussion, and they explicitly refute you. This is a scientific debate. GR does not say this at all. If GR did say that SR applies exactly to a spacetime region partly with [imath]r<r_S[/imath] and partly with [imath]r>r_S[/imath], then the locus of points with [imath]r=r_S[/imath] would have no special status at all.GR does say this, in the equivalence principle. GR makes no exceptions for where SR applies locally in empty spacetime. The equivalence principle says “anywhere”. Einstein did not redefine the word “anywhere” to mean “anywhere except in a region straddling a horizon”. Also search the OP for “Here are confirmations”. Taylor, Thorne, and Wheeler disagree with you. Note that both Andrew and Erasmus agree that SR applies in X, disagreeing with you. The ice you are skating on broke. You're flailing in the water. Easy to say "clearly", without even understanding it...this is no excuse...I could explain once we have the answer but, once we do, there wouldn't be much point in you supplying it, would there?Why not just explain where you’re going with the timelike and spacelike reasoning? It’s reasonable for me to ask when it seems clear that you cannot use it to refute me. I don’t like going down dead end paths, paths that are already refuted; that takes time and leads people to not see the forest for the trees. I suspect your reasoning has something to do with your erroneous belief that “GR does not say this at all”. Hint: Going from [imath]r>r_S[/imath] to [imath]r<r_S[/imath], the dr and dt elements of the Schwarzschild metric trade signs, which means that dr and dt trade roles. Does this help any?It doesn’t help refute the OP. GR says that SR works in X the same as it does in any other inertial frame. The equivalence principle is clear about that. Nothing you can point out about X will refute that.
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