Any coordinate system in GR?
in [Math]
|
Prev: Prime numbers
Next: Am I a crank?
From: carlip-nospam on 25 Aug 2006 18:20 In sci.physics.relativity I.Vecchi <tttito(a)gmail.com> wrote: > Tom Roberts ha scritto: [...] >> The Kruskal-Szerkes coordinates show that these two extensions are >> merely two aspects of the complete (inextensible) manifold. > Your construct relies on the arbitrary duplication of the the horizon, > inventing two copies of what is actually a single > physical/observational domain. It remains a fact that the extension > across the horizon is not unique and that each extension yields a > distinct physical object. The Kruskal-Szerkes extension is the unique maximal analytic extension of the Schwarzschild exterior geometry. Which part do you want to give up? Steve Carlip
From: Koobee Wublee on 25 Aug 2006 18:54 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). 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).
From: Koobee Wublee on 25 Aug 2006 19:12 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) > 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. > > > > 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. > > 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. > 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. > > 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? > But you keep claiming that it is. And that's just plain wrong. The rules of mathematics show so. > 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? > 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.
From: Sorcerer on 25 Aug 2006 20:44 "Koobee Wublee" <koobee.wublee(a)gmail.com> wrote in message news:1156547523.877362.164340(a)m73g2000cwd.googlegroups.com... | 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) | | > 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. | | > > > > 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. | | > > 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. | | > 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. | | > > 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? | | > But you keep claiming that it is. And that's just plain wrong. | | The rules of mathematics show so. | | > 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? | | > 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. For the record, http://www.androcles01.pwp.blueyonder.co.uk/DominoEffect.GIF Hence anyone using c has to be a lunatic with a peanut for a neuron. Androcles
From: JanPB on 25 Aug 2006 23:14
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) 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: r = u + K t = t theta = theta phi = phi ....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? > > 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. These are *tensor* PDEs, not scalar. The particular form of the equations therefore changes depending on the coordinate system used. The solutions of these _particular_ equations (the actual PDEs you solve) are then *components* of the tensor you seek. That tensor in the spherically symmetric vacuum case is unique although - obviously - a particular coordinate representation will look different and will have correspondingly transformed domain compared to some other coordinate representation. > > 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. To a *scalar* PDE, yes (with technical qualifications). But you have here a non-zero-rank *tensor* PDE which means for any given coordinate system it's a PDE for *tensor components* in that system. These components are going to look different in different coordinates but the *tensor*, that is the actual *solution* they describe, is the same. Of course it's possible to have multiple *tensor solutions* in general but this is not what happens in the spherically symmetric vacuum case. > > 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? I assume he meant spherically symmetric and vacuum. > > 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. What utter nonsense. Birkhoff only assumed the two things I mentioned: spherical symmetry and vacuum (and signature 2 and other obviosities). You don't assume what you want to prove for goodness sake. You simply write down the Einstein equation and out comes the *unique* tensor because in this case this equation reduces to an ODE, so the solution must be unique once the boundary conditions are pinned down. It's an elementary fact about ODEs in general (look up Picard-Lindeloef theorem). -- Jan Bielawski |