andrewgray Posted June 26, 2007 Report Posted June 26, 2007 Zanket, Again, sorry if I am not stating things clearly. First, It’s not “just like”, at least not when the acceleration at r=2M is given by GR’s equation that I gave above. The infinite acceleration at r=2M applies to all material objects, even those in free fall. The infinite acceleration at s=0 applies to only the rocket. Let's look at that acceleration again. [math]a = \frac{M}{r^2 \sqrt{1-2M/r}}[/math] Again, we are in a small neighborhood, so we only need this expression to first order. [math]a \;\;=\;\; \frac{M}{r^2 \sqrt{1-2M/r}}[/math] [math]\;\;\;= \;\frac{M}{(R/2+2M)^2\sqrt{\frac{R^2}{4M}}}[/math] [math]\;\;\;=\; \frac{M}{(R^2/4 + 2RM +4M^2)\frac{R}{\sqrt{4M}}} [/math] [math]\;\;\; \approx \; \frac{M}{4M^2\frac{R}{\sqrt{4M}}} \text{ where we drop terms O(2) and higher}[/math] [math]\;\;\; = \; \frac{1}{\sqrt{4M}R} [/math] [math]\;\;\; = \; \frac{1}{R} \text{ Since we have set 4M=1}[/math] Zanket, in the small neighborhood around r=2M, this acceleration is "just like", since the coordinate acceleration in the vicinity of the Rindler Horizon is 1/s. Everything that I can see is just the same. Continuing, if we make one more transformation: [math]T=R \sinh(t)[/math][math]X=R \cosh(t)[/math][math]Y=y[/math] Then the Schwarzchild metric in the small neighborhood of r=2M becomes: [math]-d\tau^2 = -dT^2 + dX^2 + dY^2[/math] and we see no limitations on object travel in the vicinity. Zanket, Both equations show that material objects near r=2M are free to go whereever. The contradiction that you have found is here: * All material objects at r=2M must fall. * Material objects at r=2M need not fall. One of these statements is incorrect, and it is the first one.The first one is incorrect because of incorrect application of GR principles to the Schwarzchild Geometry, not because GR's first principles are incorrect. To me, this shows that GR is at least consistent. It does not necessarily show that GR is correct. I would be willing to have that discussion with you if you like. Andrew A. Gray
Guest Zanket Posted June 26, 2007 Report Posted June 26, 2007 Both equations show that material objects near r=2M are free to go whereever.But I’m not talking about near r=2M. I’m talking about at r=2M. To figure out your thinking, I need answers to these questions to start: Do you agree that GR predicts that the gravitational acceleration at r=2M is infinite? Do you think a relativistic rocket at r=2M can increase its radial position? sanctus 1
andrewgray Posted June 26, 2007 Report Posted June 26, 2007 Sorry, Zanket. I seems like in our discussion almost every word needs a precise definition like in a legal court case. Yes, by "near r=2M", I mean in a neighborhood of a point in an inertial frame whose worldline is intersecting the coordinate worldline r=2M. "At r=2M" is an oxymoron. r=2M is not spacelike. It is only a coordinate. For example, I can imagine a "light-speed" coordinate system with a coordinate velocity at the speed of light in some inertial frame (t,x). (t,x)->(t',x') t'=tx'=ct This "lightspeed" coordinate system is a valid coordinate system and it can be used to do physics, but it is impossible to be at x'=1. This is also an oxymoron. However, a worldline for some point in the inertial frame can intersect the "light-speed" coordinate worldline x'=1, and be "near" to it. Do you agree that GR predicts that the gravitational acceleration at r=2M is infinite? No. Accelerations are only properly defined for timelike (for objects) worldlines. The worldline r=2M is a null (for light) worldline. The coordinate acceleration for this is infinite. The analogy would be that in our "light-speed coordinate frame", the coordinate velocities of objects can be at the speed of light. But this is meaningless since we are not talking about velocities relative to an inertial frame. Do you think a relativistic rocket at r=2M can increase its radial position? Again, "relativistic rocket at r=2M" is an oxymoron. You are stuck in using the accelerated Schwarzchild coordinate system as your thinking base. Your thinking base should be from some inertial frame. The laws of physics only hold in inertial frames. Since we have shown that the Rindler Horizon and the Gravitational Horizon are exactly the same "near" r=2M (see definition above), why do you get that an object can cross outwards through the worldline s=0, but you do not get that an object can cross outwards through the worldline R=0? Anyway, the correct way of saying this would be: A rocket whose worldline intersects the worldline r=2M-h can subsequently intersect the worldline r=2M, then "be at" r=2M+h, then "be at" r=2M+{some large value}. Of course, Schwarzchild observers would not live long enough to see this happen. Andrew A. Gray
Guest Zanket Posted June 27, 2007 Report Posted June 27, 2007 Again, "relativistic rocket at r=2M" is an oxymoron.I disagree that there is anything amiss with the idea of a “rocket at r=2M”. In GR, every location in empty spacetime has an r-coordinate associated with it. A rocket can certainly be at a location in empty spacetime whose r-coordinate is 2M. You are stuck in using the accelerated Schwarzchild coordinate system as your thinking base. Your thinking base should be from some inertial frame.My thinking base is from an inertial frame, one that is straddling r=2M. In the OP the inertial frame X is falling through the horizon. Just put a rocket at r=2M in that frame. The relativistic rocket equations apply to X. The laws of physics only hold in inertial frames.Yes, like in X. Since we have shown that the Rindler Horizon and the Gravitational Horizon are exactly the same "near" r=2M (see definition above), why do you get that an object can cross outwards through the worldline s=0, but you do not get that an object can cross outwards through the worldline R=0?I do not “get that” because there is a key difference between the two situations. No rocket can be at s=0, because its acceleration is always finite, whereas a rocket can be at r=2M. At r=2M, the rocket’s finite acceleration cannot overcome the infinite gravitational acceleration there to allow the rocket to reach a higher r-coordinate. If not even a rocket can do that, then neither can any other material object. It seems you’re thinking of rockets that successively hover at ever lower r-coordinates above r=2M. You’re thinking that no rocket can be at r=2M, because for the rocket to hover there its acceleration would need to be infinite, which is impossible. But there is no requirement in this discussion that a rocket hover at some r-coordinate; that’s just an analogy you arbitrarily adopted. The rocket could be drifting downward to r=2M (even when its proper acceleration is constant), or it could just turn its engine off and free fall to r=2M, and then turn its engine back on. Think of X, an inertial frame falling through r=2M. The frame is gravitationally accelerating with respect to r; the frame’s downward r-coordinate velocity is increasing. Let a relativistic rocket be in X, accelerating directly upwards. The rocket is accelerating relative to X. The rocket’s r-coordinate can be decreasing even as its distance d increases with respect to X. If the rocket reaches r=2M, it can never again increase its r-coordinate. But as long as X remains inertial (i.e. as long as the tidal force in X remains negligible) and the rocket is in it, the relativistic rocket equations return valid results for the rocket with respect to X. If you still disagree with me, these are key questions: Do you think the radius of a body to which the Schwarzschild metric applies (a spherically symmetric, uncharged nonrotating body) can hold steady at r=2M? If so, is the gravitational acceleration at its surface finite? Do you think that the minimum radius of such a body is r=2M+ε? sanctus 1
andrewgray Posted June 28, 2007 Report Posted June 28, 2007 I disagree that there is anything amiss with the idea of a “rocket at r=2M”. In GR, every location in empty spacetime has an r-coordinate associated with it. A rocket can certainly be at a location in empty spacetime whose r-coordinate is 2M. Zanket, Definitions have gotten to us again. "Rocket at r=2M" implies to me that the rocket is fixed at constant r=2M, and stays there. This is impossible. Rather, say "rocket crossing r=2M". Agreed? My thinking base is from an inertial frame, one that is straddling r=2M. In the OP the inertial frame X is falling through the horizon. Just put a rocket at r=2M in that frame. The relativistic rocket equations apply to X. I do not understand what this means. Perhaps you mean that some interior point in X is crossing the worldline r=2M. And, can rocket equations apply to an inertial frame? Do you mean to some point in X? . . . there is a key difference between the two situations. . . Zanket, I have shown mathematically that the two types of horizons are identical in a small neighborhood "near" r=2M. (see definition). So I really don't understand how there can be any "differences". Do you think the radius of a body to which the Schwarzschild metric applies (a spherically symmetric, uncharged nonrotating body) can hold steady at r=2M? If so, is the gravitational acceleration at its surface finite? Do you think that the minimum radius of such a body is r=2M+ε? No, no object can hold steady at r=2M. However, a spherical body can have a radius that expands "outward" through the coordinate worldline r=2M. But again, nothing can hold steady along the coordinate worldline r=2M. The coordinate worldline r=2M moves at lightspeed relative to an inertial frame. So the smallest value for a constant r spherical body would be r=2M+ε. But this does not say that all bodies whose surface worldlines intersect r=2M must then go on and collapse. This is the mistake in traditional Schwarzchild analysis, which by the way, is what your insightful analysis has flushed out. Thank you for doing so. "Near the gravitational horizon", the horizon looks like this: where we have used [math]r=\frac{R^2}{2}+2M[/math] and here we are not to scale. Clearly, anything can pass upward through r=2M, but cannot remain fixed at r=2M. An object passing upward through the horizon is exactly the time reversal of the object falling through the horizon. Perhaps you disagree with the Rindler diagram, or perhaps you cannot understand it. But there it is. This is the flat spacetime diagram of a neighborhood "near" r=2M. Andrew A. Gray P.S. And of course, the Schwarzchild observer would not live long enough to see the crossing.
Guest Zanket Posted June 28, 2007 Report Posted June 28, 2007 Definitions have gotten to us again. "Rocket at r=2M" implies to me that the rocket is fixed at constant r=2M, and stays there. This is impossible. Rather, say "rocket crossing r=2M". Agreed?OK, that’s fine with me. My thinking base is from an inertial frame, one that is straddling r=2M. In the OP the inertial frame X is falling through the horizon. Just put a rocket at r=2M in that frame. The relativistic rocket equations apply to X.I do not understand what this means. Perhaps you mean that some interior point in X is crossing the worldline r=2M. And, can rocket equations apply to an inertial frame? Do you mean to some point in X?The relativistic rocket equations apply to a rocket in an inertial frame. Within X, an inertial frame that is falling through r=2M, the rocket can be crossing r=2M. Zanket, I have shown mathematically that the two types of horizons are identical in a small neighborhood "near" r=2M. (see definition). So I really don't understand how there can be any "differences".Your math can be correct even when GR is self-inconsistent. Your math does not prove that GR is self-consistent; you’re implying otherwise, but that’s a mistake in logic. When I say there is a difference, I’m talking about in the context of more of GR than just your analysis. GR’s self-inconsistency overrides “identical”. I think you’ll see the difference soon. It’s a matter of me figuring out a way to convey it in a way that you can appreciate. That’s the purpose of my line of questioning to you. Once you have seen that your viewpoint cannot be supported by logic (i.e. it’s illogical), you can go back to understand how the math was deceptive. That said, I am always open to being wrong. No, no object can hold steady at r=2M.Agreed. However, a spherical body can have a radius that expands "outward" through the coordinate worldline r=2M.When such an expanding body’s radius is crossing r=1.99M, what is the gravitational acceleration at its surface? Rocks on the surface are not weightless as the body expands, right? If so, then the surface must have a gravitational acceleration, a specific value for g, right? What is it? Explain why whatever positive value for g you could possibly say is already returned by GR’s equation for gravitational acceleration, for an input r>2M. The equation returns every possible positive rational value for an r>2M. Why should the same value for g apply to two different radii, when presumably gravity always strengthens as r decreases? You’re implying that gravity can weaken as r decreases. Do you agree? Note that GR’s equation is unaffected by whether the body is expanding or contracting; that property is not an input to the equation. The equation works even when the body’s radius is not holding steady. But again, nothing can hold steady along the coordinate worldline r=2M. The coordinate worldline r=2M moves at lightspeed relative to an inertial frame.Agreed to both. In what direction? (Not a rhetorical question.) I say, in agreement with texts on GR, that it moves outward at lightspeed. That explains why no material object can ever attain a higher r-coordinate once it crosses inward through r=2M. To increase its radial position, the object would have to move faster than c relative to the inertial frame, which is impossible. This is the mistake in traditional Schwarzchild analysis, which by the way, is what your insightful analysis has flushed out. Thank you for doing so.You’re welcome. I think there’s yet more for you to see.
andrewgray Posted June 28, 2007 Report Posted June 28, 2007 That’s the purpose of my line of questioning to you. Once you have seen that your viewpoint cannot be supported by logic (i.e. it’s illogical), you can go back to understand how the math was deceptive. Zanket, I would hope that my mind is open enough that if you show me the logic, I will believe. However, so far it appears to me that fundamental GR equivalence principles remain consistent in the neighborhood of a horizon. I did not mean to imply that this derivation shows GR's consistency everywhere. Perhaps you can find other inconsistencies. This was an inconsistency, though not quite what you had hoped for. It was an inconsistency in Schwarzchild analysis, not GR fundamentals. So in a way you were correct. When such an expanding body’s radius is crossing r=1.99M, what is the gravitational acceleration at its surface? Rocks on the surface are not weightless as the body expands, right? If so, then the surface must have a gravitational acceleration, a specific value for g, right? What is it? Zanket, can you imagine a collapsing gravitational body crossing inward past r=1.99M? Well, rocks on the surface feel some sort of finite acceleration during this collapse, right? I do not think that you doubt this. Well, then, it would be possible to time reverse this process and have the surface emerge past r=2M with the same finite accelerations in the opposite direction. Agreed? Explain why whatever positive value for g you could possibly say is already returned by GR’s equation for gravitational acceleration, for an input r>2M. The equation returns every possible positive rational value for an r>2M. Why should the same value for g apply to two different radii, when presumably gravity always strengthens as r decreases? You’re implying that gravity can weaken as r decreases. Do you agree? No, the acceleration increases as r decreases. However, you are again confused with the fact that none of these collapses/expansions are done with r=constant. Again, during a collapse inward, the gravitational acceleration increases as the surface goes inward past r=2M, the acceleration remaining finite. Just time reverse this process, and there you go. In what direction? (Not a rhetorical question.) I say, in agreement with texts on GR, that it moves outward at lightspeed. This is where I think you are getting tripped up. If something feels a constant acceleration, then sometime in the past, there was an inertial frame where the motion stopped. For example, if I feel a constant outward acceleration of 4 m/s², and my velocity is 32 m/s outward, then 8 seconds ago I was stopped, and 16 seconds ago I was going 32 m/s inward! This is the same behavior relative to inertial frames of constant acceleration r and s coordinates near horizons. That explains why no material object can ever attain a higher r-coordinate once it crosses inward through r=2M. To increase its radial position, the object would have to move faster than c relative to the inertial frame, which is impossible. From these last two posts, I can see that you do not fully understand the Rindler coordinate diagrams. These diagrams show that (according to GR and SR) the motion into and out of the horizon is time reversal symmetric, and there is no problem moving out past r=2M. Just imagine the object going in past r=2M, then time reverse the worldline of the object and you then can imagine it coming out. Accelerations remained finite going in (no argument there), so accelerations remain finite coming out. Andrew A. Gray
Guest Zanket Posted June 28, 2007 Report Posted June 28, 2007 These diagrams show that (according to GR and SR) the motion into and out of the horizon is time reversal symmetric, and there is no problem moving out past r=2M. Just imagine the object going in past r=2M, then time reverse the worldline of the object and you then can imagine it coming out. Accelerations remained finite going in (no argument there), so accelerations remain finite coming out.I understand that you think the motion is time reversal symmetric. I have understood that you are showing that with those diagrams. I’m focusing on showing you that the motion is not time reversal symmetric. Both situations can be predicted by GR when it is self-inconsistent. Even if your math and logic is correct that shows that the motion is time reversal symmetric, it does not refute my case that GR is self-inconsistent, because a self-inconsistent theory can say “the motion is time reversal symmetric” and “the motion is not time reversal symmetric”. It can say both, where those are both supported by math. To refute me, you need to directly refute the logic I give to show a contradiction. Zanket, can you imagine a collapsing gravitational body crossing inward past r=1.99M? Well, rocks on the surface feel some sort of finite acceleration during this collapse, right? I do not think that you doubt this.If the collapse is slow enough that the surface is not freely falling, then yes, the rocks feel an acceleration. Well, then, it would be possible to time reverse this process and have the surface emerge past r=2M with the same finite accelerations in the opposite direction. Agreed?No, I disagree that the surface can cross outward through r=2M. I will show this to be impossible (according to GR) in hopefully the clearest way yet. For some input r between r=2M and r=2M+ε, GR’s equation for gravitational acceleration returns a value that is higher than any given multiple of any given finite value for the acceleration the rocks are feeling. (The acceleration the equation returns approaches infinity as r approaches 2M from above.) Then regardless what acceleration the rocks are feeling, the expanding body’s surface would confront an opposing gravitational acceleration that reduces its upward r-coordinate velocity to zero before it reaches r=2M+ε. This logic shows that the process is not time reversible. The surface of a body can fall through r=2M, but it cannot cross outward through r=2M. From page B-14 of the book Exploring Black Holes by Taylor and Wheeler: The rain observer launches light pulses toward the center and away from the center. We shall find that inside the horizon, light shot out the front and light shot out the back both move inward. Hence all particles, shot out the front or out the back (necessarily with speeds less than that of light), must also move inward. These facts are important because the only true proof that a horizon exists is the demonstration that worldlines can run through it only in the inward direction, not outward.Taylor and Wheeler show this mathematically. How do you explain that your math contradicts them? Sorry I edited this post so much. I wanted to plug a hole in my logic.
andrewgray Posted June 29, 2007 Report Posted June 29, 2007 OK, Zanket, I get what you are saying. I think that you are saying that my analysis is probably correct, but why do all these other scientists disagree with me? In other words, I must explain why the traditional Schwarzchild analysis is incorrect, including the logic from Taylor and Wheeler. Agreed? Andrew A. Gray
Guest Zanket Posted June 29, 2007 Report Posted June 29, 2007 I get what you are saying. I think that you are saying that my analysis is probably correct, but why do all these other scientists disagree with me?Yes, including scientists like me. :) In other words, I must explain why the traditional Schwarzchild analysis is incorrect, including the logic from Taylor and Wheeler.Yes, if you want to refute my claim that GR is self-inconsistent. Earlier in our discussion I had thought that black holes might be just an interpretation. But Taylor and Wheeler show it mathematically (unfortunately too much too to put here). And I’ve shown logic that tells how to prove mathematically that the box in the OP cannot pass outward through r=2M: either by using the Rindler horizon at r=2M for an observer hovering at r=2M+ε (to show that a photon crossing r=2M can’t reach r=2M+ε), or by using GR’s equation for gravitational acceleration (to show that a material object crossing r=2M can’t reach r=2M+ε). sanctus 1
andrewgray Posted July 3, 2007 Report Posted July 3, 2007 Zanket, To prove that the traditional sub-horizon Schwarzchild analysis is incorrect, We will first concentrate on proving that the acceleration of geodesics leaving the horizon is finite. OK, consider an object hurled upward from r=2M+ε, it moves upward until it reaches rmax, then falls back to r=2M+ε, like geodesics A and B in the following diagram: We take for fact that there are no discontinuities in the geometry near r=2M. If you dispute this, then we must digress. As we take [math]Limit_{\epsilon \right 0}[/math], the object experiences r values that get closer and closer to r=2M. So what is the acceleration that the object "feels" as the object travels along geodesics that originate closer and closer to r=2M? Are you ready for this? The acceleration that the object feels along these geodesics is zero. These are "freefall" geodesics that originate near r=2M, and by definition, the acceleration felt along these geodesics is zero. The way we accomplish this is that from r=(2M+ε/2) to r=(2M+ε), we accelerate the object up to the geodesic velocity, then we allow the object to "free fall" up to rmax and back down again. The object feels no acceleration after r=2M+ε. Therefore, since we are taking the spacetime near r=2M to be continuous, and since the [math]Limit_{\epsilon \right 0}\;\text{(acceleration) = 0}[/math], then the actual value of the acceleration felt by a geodesic that originates at r=2M is also zero. Zanket, unless you want to argue discontinuities, then you must accept this due to mathematical theorems on continuity. The acceleration felt going out from r=2M is zero, and the acceleration felt going into r=2M is zero. Andrew A. Gray
Guest Zanket Posted July 3, 2007 Report Posted July 3, 2007 Are you suggesting that Taylor and Wheeler made a mistake in their math? Do you disagree that they show mathematically that the Schwarzschild metric predicts that all photons below r=2M fall? If not then it would seem you are helping me make my case that GR is self-inconsistent. The OP already notes that, according to SR, the box in the OP should be able to pass outward through r=2M. Other than showing that T&W (and others) made a mistake in their math, I don’t see how you could “prove that the traditional sub-horizon Schwarzchild analysis is incorrect”. Remember, a self-inconsistent theory can both agree and disagree with your analysis. A self-inconsistent theory contradicts itself.
andrewgray Posted July 5, 2007 Report Posted July 5, 2007 Yes, they made a mistake. I do not have Taylor and Wheeler's book, but I assume that their logic is the same as found in Misner, Thorne, and Wheeler's Gravitation http://www.amazon.com/Gravitation-Physics-Kip-S-Thorne/dp/0716703440/ref=pd_bbs_2/102-3323562-0106554?ie=UTF8&s=books&qid=1183593003&sr=8-2. They base their logic on the Schwarzchild metric: These Schwarzchild coordinates are accelerated coordinates, and are not inertial. At r=2M, there is a coordinate singularity. The term (1-2M/r) goes to zero, and its reciprocal goes to infinity. However, the curvature tensor is well behaved at r=2M, so spacetime is well-behaved and continuous. Also notice that for r<2M, the coordinates are actually improper. The first two terms of the Schwarzchild metric, coefficients of dt² and dr², change sign. According to John Wheeler, (he actually explained this to me himself), t becomes the spacelike variable, and r becomes timelike for r<2M. Thus, according to their logic, since r is timelike for r<2M, it must strictly be unidirectional. So they conclude that r must be only decreasing for r<2M. Their derivation (see Gravitation pages 595-615 and 843-844) goes something like this: Notice that since they have used -e2Φ as the coefficient of dt², that they have assumed that t is the timelike variable from the very start, since -e2Φ is presumably always negative. Thus, the solution for r<2M is thus suspicious since this circumstance contradicts the initial assumption. Changing which variable is the timelike variable changes the Einstein equations. So, for non-vaccuum situations, their solutions may be invalid for r<2M, since the timelike variable would be associated with the density. Trr would then be the density, and not Ttt. However, if the stress tensor is identically zero, then the solution may still be valid. However, there is suspicion since the initial assumption involves assuming t is the time variable. -e2Φ will have to become positive and thus Φ will be imaginary. Close examination of the Einstein equations will be necessary. If r is the time variable, is r²dΩ² justified as a space differential? Is the angular space differential independent of the spacial variable t for all time r<2M? It would be much better to solve the equations with an inertial set of coordinates so these problems with their suspicions do not appear. I will be looking into this, and will get back to you. Feel free to check this scenario yourself if you like. I can share with you some other improper accelerated coordinate experiences that I have dealt with, though. Consider a coordinate system rotating about the z-axis. Here is a case where we have used accelerating coordinates, and there are regions of spacetime where we cannot correctly identify the timelike variable by the signature of the metric! Since -(1-ω²ρ²) becomes positive for ρ>1/ω, the coefficient of dt² becomes positive, and t appears to not be the timelike coordinate. Additionally, for dτ to be non-imaginary, dφ must be negative. φ is unidirectional and not the time variable! Using the same logic as Taylor and Wheeler, then φ must be the time variable! But it clearly is not. This is flat spacetime! Therefore, using the signature of the metric to identify the timelike coordinate for accelerated coordinates does not work! Additionally, if a space variable becomes unidirectional, this does not necessarily mean that it is the time variable for accelerated coordinates! I will almost guarantee you that this is their mistake! Andrew A. Gray
Guest Zanket Posted July 5, 2007 Report Posted July 5, 2007 I don’t get all of that, since I’m not as mathematical as you. However, I can say that T&W use what they call a “rain-frame metric”, the metric for an inertial observer falling to r=0. They show a plot similar to this one by Penrose (click on it for more info): You’re suggesting that the light cones do not become vertical at r=2M. You’d be suggesting that a lot of peoples’ math is erroneous. I can’t rule that out; after all, the OP is about a flaw of GR. Math can be correct but incorrectly applied, and I think that’s what you’re suggesting happened. I’ve shown in two ways that the box in the OP cannot pass outward through the horizon according to GR, which makes my case. I don’t see that you’ve refuted either one. Going back to your post #62, for an inertial object crossing upward through r=2M, rmax will be 2M. That’s because the gravitational acceleration between r=2M and r=2M+ε is arbitrarily high—higher than any finite value imaginable. The higher the gravitational acceleration on a rising inertial object, the smaller rmax will be, right? An arbitrarily high gravitational acceleration on an object prevents it from increasing its radial position. Even if you say that rmax can be as high as 2M+ε, that’s not good enough to refute me, because the box in the OP needs to be able to stay at rest relative to the particle indefinitely for GR to be able to be self-consistent, and the particle is escaping to r=infinity. Do you think GR’s equation for gravitational acceleration is incorrectly derived from the Schwarzschild metric? Do you agree that the equation predicts that an object dropped from rest at r=2M+ε passes an observer hovering at r=2M in the limit at c in the limit? Time reverse the process to see that an object rising at c in the limit past an observer hovering at r=2M in the limit comes to rest at r=2M+ε. Then no material object passing upward through r=2M will pass r=2M+ε. (And ε is zero in the limit.) Do you disagree?
Qfwfq Posted July 5, 2007 Report Posted July 5, 2007 I'm not exactly sure which one you mean, by "GR’s equation for gravitational acceleration". Can you supply it in LaTeX?
Guest Zanket Posted July 5, 2007 Report Posted July 5, 2007 I'm not exactly sure which one you mean, by "GR’s equation for gravitational acceleration". Can you supply it in LaTeX?Its reciprocal is in LaTeX on the third line on the left of this proof. It returns the initial acceleration that an observer fixed at a given r-coordinate measures for an object dropped from rest there, or the acceleration the observer feels there. a = M / (r^2 * sqrt(1 - (2M / r))) Where a is the acceleration in geometric units, M is the mass of the body in geometric units, r is the r-coordinate (circumference of a great circle centered on the body, divided by (2 * pi)).
Recommended Posts