Any coordinate system in GR?
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From: Igor on 27 Aug 2006 13:31 Koobee Wublee wrote: > Tom Roberts wrote: > > Koobee Wublee wrote: > > > > ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2 > > > ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2 > > > > > > K is an integration constant chosen to fit Newtonian result. It is > > > ** K = 2 G M / c^2 > > > For the record, Mr. Bielawski is claiming the above two metrics are the > > > same despite one manifests a black hole and other one not. > > > > This is wrong, and both exhibit a black hole. You forgot to specify the > > regions of validity of the coordinates: in the first one the horizon is > > at r=0 (note the metric components are singular there); the black hole > > is the region -K<r<0. While it is labeled "r", that coordinate does > > indeed have perfectly valid negative values inside the horizon, and r is > > timelike there. From the last term it is clear that r is not "radius" in > > this first line element. > > Oh, no. Not you too. I should not be surprised from the master of > word salad himself. > > You should read the following equations are mutually EXCLUSIVE of each > other. Only one of them can exist at the same time in one world. The > valid range for r for both equations are (r >= 0). And that's where it all goes wrong. Again you need to learn to transform a domain. Also you need to understand that just because you call it r, it may not represent an actual polar radial coordinate. >At (r = 0), we have > the very center of the gravitating object. > > ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2 > ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2 > > > [...] > > > > These two line elements do indeed correspond to the same metric, > > projected onto different coordinates. <shrug> > > Your claim is forever caste in stone (the equivalence in cyberspsace). So is your ignorance and unwillingness to learn.
From: Igor on 27 Aug 2006 13:49 Koobee Wublee wrote: > Igor wrote: > > Koobee Wublee wrote: > > > > ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2 > > > ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2 > > > > > > K is an integration constant chosen to fit Newtonian result. It is > > > > > > ** K = 2 G M / c^2 > > > > > > For the record, Mr. Bielawski is claiming the above two metrics are the > > > same despite one manifests a black hole and other one not. > > > > That's just plain wrong. The black hole is still there. > > The first metric does not manifest a black hole. Show me how you get a > black hole from > > ** (1 + 2 G M / c^2 / r = 0) where (r >= 0) Actually r >= -K, and it is not a valid polar radial coordinate. You just keep failing to transform the domain properly. > > You just > > refuse to see it. Math has rules and disobeying them is no excuse. > > Come back when you've truly learned some of them, especially > > transforming a domain. > > Hey, this is very simple math. If you cannot do this, you need to > consider going back to junior high school to brush up on your basic > algebra. It is very simple math and apparently, you've failed to understand it. What part of transforming a domain are you still failing to grasp? > > > > > However, since Schwarzschild Metric is much simpler than > > > > > Schwarzschild's original solution, Schwarzschild Metric is embraced by > > > > > the physics communities today. > > > > > > > > It's embraced because it's the same. > > > > > > You embraced it because of your denial of faulty GR. > > > > You've never shown anything to be faulty, except for your understanding > > of the math. > > I have shown them through out many posts in the past year or so. Every > post is backed up by rigorous mathematical analysis. No they have not been. Some of us have repeatedly pointed out where you went wrong, but you keep making the same stupid mistakes. We have no responsibility for you unwillingness to learn. > > > You rejected it because of your denial of faulty GR. You have 90 years > > > of fun playing with Voodoo Mathematics that gives rise to GR. It is > > > time to tear it down. > > > > You have a hell of a lot of nerve calling something voodoo when you > > can't even play by the proper rules to begin with. > > Oh, I play within the rules of mathematics. Since Mr. Bielawski fails > at simple numeric operation, he needs to go back to kindergarten. If > you still have not learnt proper arithmetic, maybe you will study them > with my 2.5 year old twins in the future. As far as you go, you need > to go back to junior high to study the basic algebra. Argument from ignorance will never work. > > Who said all the solutions are the same? Nobody. You need to > > understand the distinction between the concepts of same and equivalent. > > And brushing up on transforming a domain wouldn't hurt either. > > Each metric is merely a unique and independent solution to a set of > differential equations called Einstein Field Equations. These are > merely differential equations. Since you, Mr. Bielawski, and Dr. > Roberts are all claiming all solutions are the same. To be fair, you > are saying all solutions to a set of differential equations are the > same. Nobody ever said they were the same. Look at them. They have different forms, because you are using a different variable. But one is reachable from the other by way of a coordinate transformation, so the two are said to be EQUIVALENT. But when you perform a transformation, you also need to transform the domain. You keep failing to do that and then you call us uneducated. And the fact that you refuse to recognise your own little mistakes makes it worse. A true scholar will recognise his mistakes and learn from them, but you don't want to do that. > > > As multiple solutions to the vacuum field equations are discovered, > > > there are actually an infinite number of them. With infinite number of > > > solutions, it is shaking the very foundation of GR and SR. The house > > > of cards will soon inevitably collapse. However, refusing to give up > > > GR and to comfort themselves in false sense of security, they choose to > > > embrace Voodoo Mathematics. In doing so, they blindly claim all > > > solutions are indeed the same regardless manifesting black holes, > > > constant expanding universe, accelerated expanding universe. VOODOO > > > MATHEMATICS REPRESENTS THE ACHIEVEMENT IN PHYSICS DURING THE LAST 100 > > > YEARS. It is very sad that these clowns are regarded as experts in > > > their field. > > > > Try to keep up. There's only one Schwarzschild solution. If you can > > find any others by merely transforming the coordinates, it won't ever > > count as an independent solution. How can it? > > All solutions to a set of differential equations are related but > independent of each other. Why is it a surprise to be able to obtain > them through a theorem I showed you guys? What about Birkoff's theorem? > > But you keep claiming that it is. And that's just plain wrong. > > The rules of mathematics show so. Which you keep showing that you're completely ignorant of. Learn how to transform domains. > > You can go ahead and believe > > what you want to, even it's wrong. I dare you to find one more > > solution that satisfies the conditions of Schwarzschild and that is > > truly independent of his original solution. > > What do you mean by the conditions of Schwarzschild? Do you > Schwarzschild Metric as discovered by Hilbert or Schwarzschild's > original metric? They're equivalent also. > > Birkoff proved that it can't be done. > > Birkoff assumed that Schwarzschild Metric is the only solution to > derive that theorem. His assumption is blatantly wrong. His theorem > is thus total rubbish. He assumed no such thing. But he proved that there are no spherically symmetric solutions that are not equivalent to Schwarzschild.
From: Igor on 27 Aug 2006 13:55 Koobee Wublee wrote: > JanPB wrote: > > Koobee Wublee wrote: > > >> ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2 > >> ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2 > >> > >> K is an integration constant chosen to fit Newtonian result. It is > >> > >> ** K = 2 G M / c^2 > > > > At r = -2GM/c^2. Your restriction r>0 is wrong, it should read > > r>-2GM/c^2. > > > > The reason for it this is that the first form of the metric is obtained > > from the second (for which r>0) by means of the coordinate change: > > How do you know which of the two is the first form? And why? > > > ...where you are recycling the letter "r" in the first form where I > > have used "u". Look at the formula r=u+K. It's patently obvious that > > when r changes between 0 and infinity then u changes between -K and > > infinity. Tell me, is this not true? > > No. Still ignoring the transformation of the domain? If so, you may be hopeless.
From: Igor on 27 Aug 2006 13:56 Koobee Wublee wrote: > JanPB wrote: > > Koobee Wublee wrote: > >> ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2 > >> ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2 > >> > >> K is an integration constant chosen to fit Newtonian result. It is > >> > >> ** K = 2 G M / c^2 > >> > >> How do you know which of the two is the first form? And why? > > > > If one starts - as is usual - with a spherically symmetric > > (t,r,theta,phi)-space with r>0, etc., then any spherically symmetric > > metric can be locally written in the two forms we had before (not the > > two forms you wrote above - the two forms with the sign switch). One > > solves the Einstein equations which describe a metric (uniquely, > > because the equations happen to be ODEs in this case) described by > > components which happen to blow up at r=2m in that basis. This is your > > second equation. > > Of course, r >=0. This is spherically symmetric polar coordinate. > > > If one started instead with a spherically symmetric > > (t,r,theta,phi)-space with labelling set so that r>-K (K - some number) > > then you'd end up with the solution that looks like your first > > equation. Same thing, different chart. > > Is there any reason for you to avoid my simple question? > > > > > ...where you are recycling the letter "r" in the first form where I > > > > have used "u". Look at the formula r=u+K. It's patently obvious that > > > > when r changes between 0 and infinity then u changes between -K and > > > > infinity. Tell me, is this not true? > > > > > > No. > > > > Are you sure? This is aritmetic now. One more time: assuming r=u+K if r > > has values between 0 and infinity then u has values between what and > > what? > > Very sure. In both equations, r >= 0. Now I know you're hopeless.
From: I.Vecchi on 27 Aug 2006 14:49
Daryl McCullough wrote: > I.Vecchi says... > > >The question is whether the requirement of geodesic completeness (which > >in this setting is obtained by shifting trouble to infinity) is > >appriopriate, i.e. physically relevant. > > The answer is definitely "yes". Consider an observer in freefall near the > event horizon. Using his local coordinates, there is *nothing* to prevent > him from reaching and passing the event horizon in a finite amount of > proper time, because for him, spacetime near the event horizon is > approximately *flat*. Yes, but the problem is to define what is the event horizont for him. What does it mean PHYSICALLY that he has crossed the event horizon? Let me go back for a moment to the Agata and Bruno example ([1]). For Agata, Bruno never crosses the horizon . She cannot associate a time T_o to Bruno stepping over the horizon. For an external observer Bruno NEVER crosses the horizon. Let's consider Bruno perspective then. You are right that nothing prevents him to reach the horizon, the problem is the the horizon's position in his perspective becomes arbitrary. If his proper time at the horizon had a physically determined value he should be able to look at his clock just before talking the plunge (e.g. switching off the rocket that keeps him over the horizon) calculate and say: "In ten minutes I will have crossed the horizont", and that would mean that in his proper time 10 minutes would pass and then he'd look at his clock and know he's over . The problem is that in order to calculate that proper time he needs to know on which KS chart he is. Now, as we have seen ([2]), the correspondence between the KS chart and the Schwarzschild chart is undetermined, so he will not be able to do that. The question is then, is there any chart independent way (i.e. independent from the space-time measurement model that defines a chart) to calculate Bruno's proper time across the horizon? And my tentative answer is, no there is not. Actually, it seems to me that different charts correspond to different proper times that may be attributed to Bruno and that there is no way to associate a unique chart to Bruno based on the Schwarzschild chart alone (i.e. based on information that is available this side of the horizont). >Anything other than demanding geodesic completeness > would violate the equivalence principle, I think. Would it? I would like to inspect a rigorous proof that it would, and examine the role that proper time plays in it. I am not saying that no such proof exists, though indeed i doubt it, but it seems to me that the arguments being thrown around are riddled with implicit assumption that may or may not be warranted. It may be worthwhile to make them explicit and scrutinise them. Besides, note that neither KS is geodesically complete. It's just that in it the only obstruction to geodesic completeness is the curvature singularity at the origin. That does not violate the equivalence principle , as far as I know. > >I surmise that there are geodesically incomplete extensions, > >corresponding to hybrid solutions which may be physically relevant. > > > >I am not keen on throwing away interesting extensions/solutions just > >because they do not not comply with some arbitrary uniqueness > >criterion. > > I don't think that there is anything arbitrary about the criteria. Good for you. Cheers, IV PS I am reading a paper by Christian Fronsdal (http://arxiv.org/abs/gr-qc/0508048) built around "an observer who is aware of a limited portion of space", where "the horizon recedes as it is approached and has no physical reality". It's good deconstructivist stuff, highlighting some of the issues we are discussing. [1] http://groups.google.com/group/sci.physics.relativity/msg/9ca04b30b02cc1ef [2] http://groups.google.com/group/sci.physics.relativity/msg/8a5ffdd9c5944461 |