Zanket Posted November 8, 2006 Share Posted November 8, 2006 So, if, from Eq. (17) we get a finite, nonzero radius when we plug in v = c (for the photon), there will be areas that require a v > c to escape from, i.e. Black Holes. You are assuming that eq. 17 is valid. But section 2 of my paper shows that GR contradicts itself about that equation. You have to refute that inconsistency. You can't just ignore it. Section 2 has a relevant reader-author comment: Reader: Had you followed the rules of general relativity for computing spacetime curvature, there would be no inconsistency. [in your case, it'd be "Had you used eq. 17 for calculating escape velocity ...] Author: The finding that black holes are precluded was found by means general relativity allows. A conflicting result by alternate means does not resolve the inconsistency—it is the inconsistency. Eq. 17 applies to photons as well. That's why books say that not even a photon can escape from at or below the Schwarzschild radius. Link to comment Share on other sites More sharing options...

Espen Posted November 9, 2006 Share Posted November 9, 2006 I'm afraid that section 2 of your paper shows no contradiction of GR whatsoever. The reason for that I've already pointed out: you're completely ignoring the photon! In fact, I searched your paper for the word photon, but it doesn't occur even once! So of course it's easy to show an inconsistency in GR when you're only considering part of the theory! I'll repeat my self. You're only talking about massive particles, and then you're right, their velocities will always asymptote c. For a photon, however, it's velocity will always equal exactly c. Link to comment Share on other sites More sharing options...

Zanket Posted November 9, 2006 Share Posted November 9, 2006 I need not show a flaw of GR in multiple ways. One way is enough. I show an inconsistency of GR regarding massive particles. Then I need not be concerned with photons. You're only talking about massive particles, and then you're right, their velocities will always asymptote c. This disagrees with GR, namely eq. 17. Then you must agree that GR is inconsistent, or else be illogical. Link to comment Share on other sites More sharing options...

Espen Posted November 9, 2006 Share Posted November 9, 2006 Well, technically Eq. (17) only applies for radii greater than the Schwarzschild radius. The equation was derived using coordinates that are comoving and at rest relative to the object falling. Such frames cannot readily exist within the Schwarzschild radius. For one thing, the proper time (i.e. the watch following the falling object) would show imaginary time! Furthermore, at the Schwarzschild radius the time and space coordinates are interchanged: r becomes a time coordinate, and t becomes a space coordinate. Well, this is just a tidbit of what there is to know about Schwarzschild geometry and black holes. This is a subject with much to learn, and I know I have far more to learn myself. What we've talked about here in general, and Eq. (17) in particular, are just smidgens of the whole story. I'm afraid your paper is far too lacking in extent, depth and understanding of the subject to be taken even remotely serious, I'm sorry to say... Link to comment Share on other sites More sharing options...

Zanket Posted November 9, 2006 Share Posted November 9, 2006 Well, technically Eq. (17) only applies for radii greater than the Schwarzschild radius. Section 2 shows that v always asymptotes to c, i.e. as long as the particle falls. That includes below the Schwarzschild radius. You agreed with that, saying "their velocities will always asymptote c". If eq. 17 agreed with section 2 and you, then it would apply both above and below the Schwarzschild radius, and always return v < c. I'm afraid your paper is far too lacking in extent, depth and understanding of the subject to be taken even remotely serious, I'm sorry to say... That's a common refrain, but neither you nor anyone else has proven a problem with it. Subjective impressions are scientifically meaningless. A theory can be refuted by as little as an arrow pointing to a division-by-zero error. Link to comment Share on other sites More sharing options...

Espen Posted November 10, 2006 Share Posted November 10, 2006 Section 2 shows that v always asymptotes to c, i.e. as long as the particle falls. That includes below the Schwarzschild radius. You agreed with that, saying "their velocities will always asymptote c". If eq. 17 agreed with section 2 and you, then it would apply both above and below the Schwarzschild radius, and always return v < c. Indeed v of a massive particle always asymptotes c, locally. Now, Eq. (17) was derived using the metric from Schwarzschild's exterior solution, a metric with a special set of coordinates that exhibit a coordinate singularity at the Schwarzschild radius. Furthermore, like I've said, below the Schwarzschild radius the proper time of the falling object will become imaginary, which means that the analysis breaks down, as well as Eq. (17). Trust me, I know this is hard to grasp, but that doesn't make it any less real. A theory can be refuted by as little as an arrow pointing to a division-by-zero error. Perhaps a new theory under development, but in order to refute a well tested, well scrutinized and well established theory you need more than an incomplete understanding of it. You know, the more I discuss this with you, the more I begin to realize the futility of convincing you. I believe James "The Amazing" Randi said it the best: No amount of contrary evidence will ever un-convince the true believer. Link to comment Share on other sites More sharing options...

Ragib Posted November 11, 2006 Share Posted November 11, 2006 Zanket, your theory may appear more approachable to a general audience, such as 14 year olds like myself, rather than one comprised of only experienced physicists, if your were to rewrite it in SI units. If possible, could you send me a paper that has all the equations in SI units rather than Plank units, that would be a great help. Natural Units are good for equations, but nothing more. In deriving equations Natural units are frowned upon by most, as it provides less understanding of the formulae. Only after derivation are natural units used, to simplify calculations and nothing else. A flaw may be found with much less effort if SI units were used. Now, I have not completely analysised your paper yet, which I will as soon as possible, but it seems you have a circular argument going on. You seemed to critcise Espens comment on assuming equation 17 is correct, which you said you have shown an inconsistancy in. However, in a seperable section you use your assumption that equation 17 is not correct to prove another flaw. Now i quote your paper: "The universe always existed. There are no singularities." "Presumably the universe endlessly cycles between expansion and contraction; otherwise the galaxies would already be infinitely spread apart from one another because the universe always existed." Would you please, using your theories equations, tell us to what extent it expands then contracts? Not to mention, having existed for an infinite amount of time, shouldn't the second law of thermodynamics result in a heat death of the universe already, or the universe reached thermal eqilibrium? I can not remember the quote exa ctly, or by who it was from, but someone once said that "If your theory disagrees with observances, well even sometimes experimentalists get it wrong. But if your theory is incompatible with the Second law of thermodynamics, I can offer you no hope". Quite true I have to say. I'll get back to you later once i've read it, it may take me up to a week to fully analyse it though. Please do not dismiss my comments merely on the factor of my age, you will find I know just as much Relativity as you do. To All: As we have not found an error yet, do not dismiss this theory yet. Even if the probablity of its physical correctness is not very high, if we do find an error it may show us something new we have not considered before. Just like Eienstein's criticism of Heisenbergs Uncertainty Principle lead to a deeper understanding of the Principle itself. Link to comment Share on other sites More sharing options...

D H Posted November 11, 2006 Share Posted November 11, 2006 Equation 17 is from special relativity. It follows from the Minkowski metric for flat spacetime, [math]ds^2 = -c^2dt^2 + dr^2 + r^2d\theta^2 + r^2\sin^2\theta d\phi^2[/math] Quoting from the paper, Reader: Special relativity does not apply to curved spacetime. Author: It is used only in flat spacetime herein. It is indeed used, but that use is incorrect. From the paper, Let a test particle fall radially from rest at infinity toward a large point mass ... The Minkowski metric does not apply in such a case. The mass curves spacetime. The governing equations in the case of a point mass [math]M[/math] with zero angular momentum are the Schwarzschild metric, [math]ds^2 = -c^2\left(1-\frac{2GM}{c^2 r}\right)dt^2 + \left(1-\frac{2GM}{c^2 r}\right)^{-1} dr^2 + r^2d\theta^2 + r^2\sin^2\theta d\phi^2[/math] A theory can be refuted by as little as an arrow pointing to a division-by-zero error. or by a misapplication of the Minkowski metric when the Schwarzschild metric should have been used instead. This thread belongs in "Speculations" with all of the other "I found a flaw in relativity" conjectures. Link to comment Share on other sites More sharing options...

Zanket Posted November 11, 2006 Share Posted November 11, 2006 Indeed v of a massive particle always asymptotes c, locally. Yes, locally, but locally everywhere, so v asymptotes to c as long as the particle falls, even in the local region that straddles the Schwarzschild radius, and local regions below that. This was inferred by means GR allows. There is a uniform gravitational field that straddles the Schwarzschild radius. The equivalence principle demands that v always asymptotes to c within that local region. Now, Eq. (17) was derived using the metric from Schwarzschild's exterior solution, a metric with a special set of coordinates that exhibit a coordinate singularity at the Schwarzschild radius. Furthermore, like I've said, below the Schwarzschild radius the proper time of the falling object will become imaginary, which means that the analysis breaks down, as well as Eq. (17). This is all irrelevant. What is relevant is that eq. 17 contradicts something that can be inferred by means GR allows. Section 2 already covers this with this reader-author comment: Reader: The velocity v asymptotes to c only as low as the Schwarzschild radius, at the event horizon. Author: The analysis above shows that v always asymptotes to c; i.e. as long as the particle falls. Since this was inferred by means general relativity allows, the theory cannot demand otherwise without being inconsistent. Trust me, I know this is hard to grasp, but that doesn't make it any less real. I’m aware of what GR says about the region at and below the Schwarzschild radius. The paper is concerned with GR’s inconsistency about that, which you are ignoring. Perhaps a new theory under development, but in order to refute a well tested, well scrutinized and well established theory you need more than an incomplete understanding of it. No, even the most venerable theory can be refuted by an arrow pointing to a division-by-zero error. Can you prove your viewpoint logically or mathematically? I didn’t think so. You know, the more I discuss this with you, the more I begin to realize the futility of convincing you. I believe James "The Amazing" Randi said it the best: No amount of contrary evidence will ever un-convince the true believer. This is a common tactic: If I disagree then call me stubborn. I am refuting your points with logic and reason. You will convince me only by incontrovertibly refuting me in the same manner. Link to comment Share on other sites More sharing options...

D H Posted November 11, 2006 Share Posted November 11, 2006 This is all irrelevant. What is relevant is that eq. 17 contradicts something that can be inferred by means GR allows. I’m aware of what GR says about the region at and below the Schwarzschild radius. The paper is concerned with GR’s inconsistency about that, which you are ignoring. Apparently you are not aware. The Schwarzschild metric applies outside the Schwarzschild radius. Equation 17 does not apply, from which your whole argument fails. Link to comment Share on other sites More sharing options...

Zanket Posted November 11, 2006 Share Posted November 11, 2006 Natural Units are good for equations, but nothing more. In deriving equations Natural units are frowned upon by most, as it provides less understanding of the formulae. Only after derivation are natural units used, to simplify calculations and nothing else. A flaw may be found with much less effort if SI units were used. I disagree. Since geometric units make simpler equations and hence simpler calculations, and are convertible to SI units with no ambiguity (and vice versa), there’s little reason to use SI units. No errors creep into derivations by the use of geometric units. Simpler equations and calculations make flaws easier to see. Geometric units are common in texts about relativity, as noted in the link in the Definitions section of my paper. (The link gives you the conversion factors.) For example, Taylor and Wheeler use geometric units throughout their book Exploring Black Holes, for which my paper includes an online reference. See page 2-13 for their take on this topic. Now, I have not completely analysised your paper yet, which I will as soon as possible, but it seems you have a circular argument going on. You seemed to critcise Espens comment on assuming equation 17 is correct, which you said you have shown an inconsistancy in. However, in a seperable section you use your assumption that equation 17 is not correct to prove another flaw. Section 2 shows that eq. 17 from the reference (GR's equation for escape velocity, equivalent to eq. 4 in my paper, so hereafter I reference eq. 4 to avoid confusion) contradicts a prediction that can be inferred by means GR allows. The equation is inconsistent with section 1, thereby shown to be incorrect. Section 4 shows that Einstein's eq. 8 is derivable from eq. 4, hence eq. 8 is incorrect as well. There's no circular argument there. Can you be more specific? Would you please, using your theories equations, tell us to what extent [the universe] expands then contracts? I decline. Keep in mind that cosmology models are based on incomplete observational information. Not to mention, having existed for an infinite amount of time, shouldn't the second law of thermodynamics result in a heat death of the universe already, or the universe reached thermal eqilibrium? I can not remember the quote exa ctly, or by who it was from, but someone once said that "If your theory disagrees with observances, well even sometimes experimentalists get it wrong. But if your theory is incompatible with the Second law of thermodynamics, I can offer you no hope". Quite true I have to say. If this was a problem for me, then it would have been a problem as well for Einstein, who for years thought the universe was likely static, having always existed. In my model, the universe can oscillate between arbitrarily dense (maximum contraction) and arbitrarily sparse (maximum expansion) states. So there’s no requirement that it reach thermal equilibrium or result in heat death. Please do not dismiss my comments merely on the factor of my age, you will find I know just as much Relativity as you do. I’ll treat you the same as anyone. Link to comment Share on other sites More sharing options...

Zanket Posted November 11, 2006 Share Posted November 11, 2006 From the paper, Let a test particle fall radially from rest at infinity toward a large point mass ... The Minkowski metric does not apply in such a case. Section 2 doesn’t use SR for the particle’s whole fall, but rather only in each local region. The following conclusion ...: Then v always asymptotes to c; i.e. as long as the particle falls. ... is based on the following logic ...: The particle always falls through a uniform gravitational field (a succession of them, each applying only locally) since a nonuniform gravitational field is everywhere uniform locally. ... which is not a global usage of SR. This thread belongs in "Speculations" with all of the other "I found a flaw in relativity" conjectures. Good thing you’re not a mod then, or else there might be the same rush to judgment here as in other forums. The Schwarzschild metric applies outside the Schwarzschild radius. The Schwarzschild metric applies outside the Schwarzschild object, including at and below the Schwarzschild radius. Equation 17 does not apply, from which your whole argument fails. Eq. 17 applies only as low as the Schwarzschild radius, but that is irrelevant. What is relevant is that a prediction inferred by means GR allows contradicts eq. 17. The following conclusion in section 2 ...: Then v always asymptotes to c; i.e. as long as the particle falls. ... is not dependent on eq. 17, and contradicts eq. 17. When v always asymptotes to c, it asymptotes to c even in the local region that straddles the Schwarzschild radius, hence it is less than c at the Schwarzschild radius, in contradiction with eq. 17. As I noted to Espen above, section 2 already covers this with this reader-author comment: Reader: The velocity v asymptotes to c only as low as the Schwarzschild radius, at the event horizon. Author: The analysis above shows that v always asymptotes to c; i.e. as long as the particle falls. Since this was inferred by means general relativity allows, the theory cannot demand otherwise without being inconsistent. I have a feeling you’ll ignore that again though. Link to comment Share on other sites More sharing options...

Espen Posted November 11, 2006 Share Posted November 11, 2006 No, even the most venerable theory can be refuted by an arrow pointing to a division-by-zero error. Can you prove your viewpoint logically or mathematically? I didn’t think so. Using the coordinate system that gives Eq. (17) it follows that the infalling particle with it's proper time, T, at a distance r from the center will observe a time t on a watch at rest relative to the center given by t = T/sqrt(1-R/r) where R is the Schwarzschild radius. Now, observe what happens when your infalling particle reaches the Schwarzschild radius, you get a division by zero. Now, I'm pretty confident that you've already decided that you've shown an incosistency in GR, regardless of what the facts may be. So, I'm curious as to see how you'll deal with this division-by-zero error I've pointed to (which you yourself claims can be the end of any theory). Will you dismiss it as irrelevant again? Or will you ignore it and refer to your website and say that you've shown an inconsistency? Regardless, I doubt you'll even attempt to solve this inconsistency! Link to comment Share on other sites More sharing options...

D H Posted November 11, 2006 Share Posted November 11, 2006 Zanket, This deserves to be put into "Speculations" because it is a crackpot conjecture. Good thing you’re not a mod then, or else there might be the same rush to judgment here as in other forums. Mods on other forums are a little less forgiving of crackpot notions. Section 2 doesn’t use SR for the particle’s whole fall, but rather only in each local region. Yes, it does. You are using equation 17, which assumes the Minkowski metric over all space. This is an invalid assumption in the presence of a gravitational source. The following conclusion ... Then v always asymptotes to c; i.e. as long as the particle falls. is based on inappropriately applying the Minkowski metric to all space. I have a feeling you’ll ignore that again though. All your paper does is show an inconsistency between special and general relativity. BFD. Why do you think Einstein called one theory "special" (i.e., true only under special circumstances) and the other, "general"? General relativity yields special relativity in the limit [math]M\to0\text{\ or\ }r\to\infty[/math]. I have a feeling you’ll ignore that again though. Link to comment Share on other sites More sharing options...

Zanket Posted November 12, 2006 Share Posted November 12, 2006 Now, observe what happens when your infalling particle reaches the Schwarzschild radius, you get a division by zero. That was a good attempt. A division-by-zero error leads to logical problems, such as the one that generates the absurdity that 2 = 1 here. The site says “In practice, division by a term in any algebraic argument requires an explicit assumption that the term is not zero or a justification that the term can never be zero”. GR satisfies that requirement for your example by preventing the particle from reaching the Schwarzschild radius in the frame measuring t. In that frame the particle moves ever closer to but never reaches the Schwarzschild radius, as any decent text on black holes will tell you. So there will never be a division by zero in that example. So, I'm curious as to see how you'll deal with this division-by-zero error I've pointed to (which you yourself claims can be the end of any theory). Will you dismiss it as irrelevant again? I meet all challenges head on. It is possible for something to be irrelevant you know. Link to comment Share on other sites More sharing options...

Zanket Posted November 12, 2006 Share Posted November 12, 2006 This deserves to be put into "Speculations" because it is a crackpot conjecture. I’m the crackpot, yet you cannot show one problem with my paper that I cannot refute. Hmm. Yes, it does. You are using equation 17, which assumes the Minkowski metric over all space. This is an invalid assumption in the presence of a gravitational source. There may have been some confusion above. Epsen referenced eq. 17 from my online reference #7 from Taylor and Wheeler; that’s GR’s equation for escape velocity, and GR’s equation for v in section 2. You are referring to eq. 17 in section 8 in my paper. The equivalence principle lets eq. 17 be used in the presence of a gravitational source, as long as it’s used only locally (as given by the definition of “local” in my paper). Review Einstein’s equivalence principle from here: The outcome of any local non-gravitational experiment in a laboratory moving in an inertial frame of reference is independent of the velocity of the laboratory, or its location in spacetime. What is an inertial frame? It’s a local frame in free fall; i.e. one in the presence of a gravitational source. SR, including eq. 17 from my paper, can be used in an inertial frame. And that’s the only way section 2 uses it. How else do you think SR can be experimentally confirmed, when every laboratory is in the presence of a gravitational source? According to your logic, SR cannot be experimentally confirmed, let alone applied at all. All your paper does is show an inconsistency between special and general relativity. No, it shows an inconsistency between two predictions of GR for the same situation. Section 2 shows, inferred by means GR allows, that v always asymptotes to c even in a local region that straddles the Schwarzschild radius. That disagrees with GR’s eq. 4 in my paper (the same as eq. 17 in my online reference). Link to comment Share on other sites More sharing options...

Ragib Posted November 12, 2006 Share Posted November 12, 2006 Zanket, sure even for Eienstein he believed in a static universe. But we are not dealing that. We are dealing with his theory, which DOES predict a non-static universe. And who ever said Zankets a crackpot, seriosuly think before you say that, you havent found anything wrong yet. Link to comment Share on other sites More sharing options...

Ragib Posted November 12, 2006 Share Posted November 12, 2006 God damn i wish Matt grime was still here, he would do this in a second... Link to comment Share on other sites More sharing options...

D H Posted November 12, 2006 Share Posted November 12, 2006 And who ever said Zankets a crackpot, seriosuly think before you say that, you havent found anything wrong yet. Yes, we have. He misapplies the equivalence principle and proceeds from this point to assume a uniform gravitational field everywhere. According to the equivalence principle, the crew of a relativistic rocket experiences the equivalent of a uniform gravitational field. The equivalence principal only applies at a point. I do not see a single tensor equation in the paper. The equations of motion in general relativity are tensor equations. I do see a lot of hand-waving and misapplications of various equations. The free-fall velocity will asymptote to c, but this occurs at the Schwarzschild radius, not at r = 0. Link to comment Share on other sites More sharing options...

Zanket Posted November 12, 2006 Share Posted November 12, 2006 Zanket, sure even for Eienstein he believed in a static universe. But we are not dealing that. We are dealing with his theory, which DOES predict a non-static universe. What is your point? My model allows for a non-static universe too. My point was that Einstein believed for years that the universe was likely static, yet he apparently saw no issue between that and the second law of thermodynamics. Then there probably was no issue. Link to comment Share on other sites More sharing options...

Zanket Posted November 12, 2006 Share Posted November 12, 2006 He misapplies the equivalence principle ... You have demonstrated that you don’t understand it. You think SR is inapplicable in the in the presence of a gravitational source, i.e. inapplicable period. How do you explain the experimental confirmation of SR then? The equivalence principle is what allows SR to be used in the presence of a gravitational source. ... and proceeds from this point to assume a uniform gravitational field everywhere. According to the equivalence principle, the crew of a relativistic rocket experiences the equivalent of a uniform gravitational field. That quote does not imply a uniform gravitational field everywhere. The quote is implied by any decent text on the equivalence principle. The equivalence principal only applies at a point. That’s no more an issue for me than it is for all the texts on the equivalence principle that show it applied to larger regions. The reader-author comment in my introduction addresses this issue and shows that Feynman, Taylor, Thorne, and Wheeler all apply the equivalence principle to larger regions. I do not see a single tensor equation in the paper. The equations of motion in general relativity are tensor equations. Can you prove logically or mathematically that tensor equations are required? I didn’t think so. A reader-author comment in section 5 addresses this: Reader: Without new field equations, your theory is worthless. Author: Field equations are not required. Using only the new metric and the principle of extremal aging, a principle of special relativity, one can make falsifiable predictions. I do see a lot of hand-waving and misapplications of various equations. You’re the one doing the handwaving. Your "I do not see a single tensor equation in the paper" is a prime example. You've not shown a misapplication of any equation. The free-fall velocity will asymptote to c, but this occurs at the Schwarzschild radius, not at r = 0. Section 2 shows that v always asymptotes to c even in the local region that straddles the Schwarzschild radius, or else the equivalence principle is invalidated. Then v is less than c at the Schwarzschild radius, and this was inferred by means GR allows. You haven’t refuted that; you've only insisted otherwise without a basis. Link to comment Share on other sites More sharing options...

Espen Posted November 13, 2006 Share Posted November 13, 2006 I finally feel we're getting somewhere, Zanket. First you're of course right. In that frame any time intervall observed on the falling watch will eventually become infinite, i.e. the falling watch will appear to reach the horizon, but never fall in. So, we agree that frame of reference depends on what you see. Following that I hope we also agree that choise of coordinate system depends on what results you get? Then you're alleged inconsistency is easily explained by GR as a mere bad choice of coordinates. Let me try to illustrate more easily the situation. The best way to illustrate is to imagine lightcones falling towards the center. As you now, massive particles only exists inside the lightcone. Now, if we use the same coordinates used to derive the almost infamous Eq. (17) we get that the infalling lightcones collapse at the Schwarzschild radius, i.e. the coordinate system and any equations resulting from the use of it breaks down at the Schwarzschild radius, including Eq. (17). Your claim is that Eq. (17) holds even here, whereas I've just shown, by means general relativity allows, that it doesn't. What you have to do to avoid lightcones from collapsing, and thus being able to make sense of what happens inside the Schwarzschild radius is to transform to different coordinates. Some famous examples of such coordinates are the ingoing Eddington-Finkelstein coordinates and the Kruskal-Szekeres analytical extension of the Schwarzschild spacetime properly named Kruskal-Szekeres coordinates. It is thus quite clear that, by means general relativity allows, Eq. (17) only applies outside the Schwarzschild radius, and that different coordinates are required to cover more of the Schwarzschild space-time manifold. Your alleged inconsistency is thereby explained and accounted for, by means general relativity allows. Finally I'd like to bring to your attention something I noticed in your article, the following statement: Likewise, Taylor and Wheeler place no limit on the size of an inertial frame. Here you're referring to pages GL-4 and 2-4, sect. 3 (your ref. #6). I then refer you to page 1-14, sect. 8 in that same book. The sections begins with the following sentence: In practice there are limits on the space and time extent of the free-float (inertial) frame. I just thought you should know if there's a typo or something. Link to comment Share on other sites More sharing options...

Ragib Posted November 13, 2006 Share Posted November 13, 2006 Eienstein did see the issue of Entropy and Thermal Eqilibrium. He never claimed his theory could explain this, he merely thought it an improvement to current theories of gravity. Once it was shown by Hubble the universe is indeed expanding, he had that problem fixed. Explain why these issues are irrelevant to your theory? These may prove it falsifiable, which every good theory should do. Link to comment Share on other sites More sharing options...

Zanket Posted November 13, 2006 Share Posted November 13, 2006 Now, if we use the same coordinates used to derive the almost infamous Eq. (17) we get that the infalling lightcones collapse at the Schwarzschild radius, i.e. the coordinate system and any equations resulting from the use of it breaks down at the Schwarzschild radius, including Eq. (17). To avoid confusion I won’t refer to eq. 17 from my online reference; I’ll refer to eq. 4 in my paper, the same equation. Notice the circular logic in your statement. You’re using eq. 4 to show that eq. 4 doesn’t apply below the Schwarzschild radius. Really it’s by interpretation that eq. 4 shows that v (of section 2) cannot be measured at and below the Schwarzschild radius, because otherwise SR would be violated. In other words, someone looked at eq. 4 and said, “Holy crap, SR will be violated unless we assume that no object can be fixed at an altitude at and below the Schwarzschild radius, disallowing v from being measured”. And so it was assumed that all objects must fall there. Eq. 4 is still valid there; it just has no practical purpose. Your claim is that Eq. (17) holds even here, whereas I've just shown, by means general relativity allows, that it doesn't. No, that is not my claim. Section 2 shows that eq. 4 is invalid. That’s because it is inferable by means GR allows (and not by eq. 4) that v is less than c there, rather than v not applying there as eq. 4 indicates. Because GR makes contradicting predictions about the same situation, it is inconsistent. It is thus quite clear that, by means general relativity allows, Eq. (17) only applies outside the Schwarzschild radius, and that different coordinates are required to cover more of the Schwarzschild space-time manifold. Your alleged inconsistency is thereby explained and accounted for, by means general relativity allows. You are looking at only one side of the inconsistency, which is to say you’re ignoring the inconsistency. (That’s what everyone does.) Suppose a theory about birds contains these statements: Most birds can fly. No birds can fly. If I pointed out that these statements contradict each other, using your reasoning one would say either “Zanket, you’re wrong because the theory clearly says that most birds can fly” or “Zanket, you’re wrong because the theory clearly says that no birds can fly”. Section 2 already has two reader-author comments that address this: Reader: The velocity v asymptotes to c only as low as the Schwarzschild radius, at the event horizon. Author: The analysis above shows that v always asymptotes to c; i.e. as long as the particle falls. Since this was inferred by means general relativity allows, the theory cannot demand otherwise without being inconsistent. Reader: At and below an event horizon no object can be fixed at an altitude, so your analysis fails. Author: That puts the cart before the horse. When v always asymptotes to c then so does escape velocity, in which case escape velocity is always less than c and then there are no event horizons. Since this was inferred by means general relativity allows, the theory cannot demand otherwise without being inconsistent. Finally I'd like to bring to your attention something I noticed in your article, the following statement: Likewise, Taylor and Wheeler place no limit on the size of an inertial frame. Here you're referring to pages GL-4 and 2-4, sect. 3 (your ref. #6). I then refer you to page 1-14, sect. 8 in that same book. The sections begins with the following sentence: In practice there are limits on the space and time extent of the free-float (inertial) frame. I just thought you should know if there's a typo or something. The distinction is made by “in practice”. If you look more closely you’ll see that there is no contradiction between their statement and mine. First, look at their definition on pg. GL-4 for a free-float (inertial) frame, which I shorten to “a frame with respect to which the tidal force can be neglected for the purposes of a given experiment”. Notice that no limit on the size of the frame is placed. Next, see sample problem 2 on pg. 1-4, where they “assume that a single free-float frame can stretch all the way from Sun to Andromeda”, a distance of two million light years. Finally, read the text of section 8 starting on pg. 1-14. It should become clear that “in practice” means “for the purposes of a given experiment”. For one experiment, the inertial frame may be no larger than a breadbox before the tidal force becomes significant. For another experiment, the inertial frame may be a billion light years long, with a negligible tidal force throughout. In practice (for any given experiment), there is a limit on the size of the frame. But by definition (for experiments in general), there is no such limit. Link to comment Share on other sites More sharing options...

Zanket Posted November 13, 2006 Share Posted November 13, 2006 Eienstein did see the issue of Entropy and Thermal Eqilibrium. He never claimed his theory could explain this, he merely thought it an improvement to current theories of gravity. Once it was shown by Hubble the universe is indeed expanding, he had that problem fixed. Explain why these issues are irrelevant to your theory? These may prove it falsifiable, which every good theory should do. Can you prove with references that this is an issue? Let's be clear that you say this is an issue with any universe that always exists. I've read extensively on cosmology, and the only issue with entropy that I've seen is for a universe that always oscillates between a big bang and a big crunch. Is that the one you're talking about? Harrison says this issue is highly speculative. Link to comment Share on other sites More sharing options...

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