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From: Igor on 25 Aug 2006 13:56 Koobee Wublee wrote: > JanPB wrote: > > Koobee Wublee wrote: > > > > Schwarzschild found a unique solution to the differential equations of > > > Einstein Field Equations in free space. Hilbert found another one that > > > he called it Schwarzschild Metric. Recently, Mr. Rahman also a > > > contributor of this newsgroup presented another solution. The > > > spacetime with this metric is > > > > > > ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2 > > > > > > The metric above indeed is another solution which anyone can easily > > > verify because its simplicity. Notice Rahman's metric and > > > Schwarzschild's original metric do not manifest black holes. > > > > No, the metric above is equal to Schwarzschild's metric. The form above > > is obtained by a coordinate change from the original one, hence the > > metric remains the same (tensors do not change under coordinate > > changes). > > ** 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. 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. > > > 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. > > > Mr. Bielawski and Igor have not > > > understood Schwarzschild's original paper and choose to blindly reject > > > Schwarzschild's original solution and others. > > > > There is nothing to reject. One can _prove_ Schwarzschild's metric is > > unique. Off the horizon it follows immediately from the particular form > > of the Einstein equation in the spherically symmetric case (which is > > what we have) and the uniqueness of the extension over the horizon is > > slightly more involved but it follows from a similar argument. > > 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. > > > As multiple solutions to the vacuum field equations are discovered, > > > there are actually an infinite number of them. > > > > Yes, and 2+2=5. > > Only a Voodoo Mathematician like yourself would claim it to be so. > Furthermore, extending the analogy of you claiming all differential > equations having the same solution, we have the following quadraic > equation. > > ** x^2 - 3 x + 2 = 0 > > Solving for x, we get (x = 1) or (x = 2). According your silly claim, > all solutions are the same. Thus, 1 = 2. 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.
From: Tom Roberts on 25 Aug 2006 14:53 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. Looking just at the last term (solid angle), it is clear that r=-K in the first line element corresponds to r=0 in the second, and r=0 in the first corresponds to r=K in the second. These two line elements do indeed correspond to the same metric, projected onto different coordinates. <shrug> Tom Roberts
From: I.Vecchi on 25 Aug 2006 16:14 Tom Roberts ha scritto: > I.Vecchi wrote: > > I would say that there are at least two distinct ways to prolong the > > solution across the horizon, which yield respectively the black hole > > and the white hole solution, describing two different physical > > phenomena. They correspond to two distinct choices of the > > Eddington-Finkelstein coordinates at the horizon, i.e. to crossing the > > horizon on future-directed or on past-directed curves. The extended > > solution depends on how space-time is extended beyond the horizon. > > 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. A black hole is not a white hole. Period. IV
From: Edward Green on 25 Aug 2006 17:51 Bill Hobba wrote: > "Edward Green" <spamspamspam3(a)netzero.com> wrote in message > news:1156222614.472337.73970(a)p79g2000cwp.googlegroups.com... > > Bill Hobba wrote: > > > >> "Edward Green" <spamspamspam3(a)netzero.com> wrote ... > > > > <stretching the manifold vs. compressing the coordinate system> > > > >> First why is one unphysical and the other not? > > > > Well, that depends on whether it is "physical" to stretch the manifold. > > Since it is not a material then obviously it is a 'word salad's with no > meaning like saying stew smells like isomorphism. You are merely being insulting now. Obviously a manifold is a mathematical abstraction. Obviously any word normally applied to material objects tentatively when applied to a mathematical abstraction should be understood to refer to a putative property of the abstraction analogous to the named material property. I could equally well scoff that since a manifold is not a material then it makes no sense assign "curvature" to it. And yet sense can be so made, as you know. Of course, in the continuing verbal self-defense your minimax strategy makes necessary (proving me maximally stupid based on minimal evidence), I hasten to add that I am _not_ arguing that since a meaning can be assigned to "curvature" meaning must also be assignable to the strain or "stretching" of a manifold. I am merely indicating that such a putative identification is not the silly category error you imply. In fact, the idea will have meaning to the extent we assign meaning to it. Modeling the rubber band by a one dimensional manifold, we will find it natural to assign a pointwise property to the model called "strain". Is this part of the minimal structure of a differential manifold, or something added? I don't know. Is there something analagous to strain assigned to the manifold which is an element of the model of GR? I don't know. Since I know that I don't understand the model, I don't know that there must be a property of the model of GR plausibly described as a local strain state of spacetime. I am not confident however that there is no such property is possible merely because Bill Hobba asserts it, because Bill Hobba denigrates me merely for describing a putative property via a word applicable to material objects, as if such alone were sufficient proof of idiocy, when Bill Hobba in fact knows that mathematics freely borrows terms from the quotidian world. Your counter argument is falacious. > > In a simple minded analogy, the one dimensional manifold might be > > represented by a rubber band: stretching the rubber band is a > > physically distinct situation from compressing the coordinates -- one > > might be able to determine the state of strain by local measurements, > > or at least determine the relative strain for two locations. In GR the > > "stretch" might correspond to a gravitational field vs. flat spacetime. > > Not _everything_ is an artifact of the coordinate system. > > You were on the right track above - stretching is not a property of the > space-time manifold. That's better: a blunt assertion. Maybe you are right, but you will forgive me if I don't assign your claim 100% Bayesian confidence just at this moment. You have after all given me no reason at all to believe you beyond your blunt assertion. (Leave off the implicit insults in the interest of brevity. Thanks.)
From: Tom Roberts on 25 Aug 2006 18:17
I.Vecchi wrote: > Tom Roberts ha scritto: >> I.Vecchi wrote: >>> I would say that there are at least two distinct ways to prolong the >>> solution across the horizon, which yield respectively the black hole >>> and the white hole solution, describing two different physical >>> phenomena. They correspond to two distinct choices of the >>> Eddington-Finkelstein coordinates at the horizon, i.e. to crossing the >>> horizon on future-directed or on past-directed curves. The extended >>> solution depends on how space-time is extended beyond the horizon. >> 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, Sure. As is easily seen in a Kruskal diagram, and as I said. Here you seem to be in violent agreement with what I said (but later you get it wrong). It is arbitrary which portion of the horizon you consider, but as both are contained in the manifold it is only the analyst who is subject to this choice, not the manifold itself (or the physical system for which it is a model). > inventing two copies of what is actually a single > physical/observational domain. Not really. Those regions of the complete (inextensible) manifold are DIFFERENT. That is, considered from a given point in spacetime near but outside the horizon, the past and future horizons are NOT "the same". This difference between past and future is true in ANY spacetime, of course. That is, for any point in any manifold of GR, every locus in the past lightcone of the point is disjoint from every locus in the future lightcone of the point. This is true in everyday life -- just think about how different is your ability to observe events in the past from events in the future. > It remains a fact that the extension > across the horizon is not unique and that each extension yields a > distinct physical object. Not "physical object" but rather region of the manifold. Basically you happened to choose E-F coordinates that do not cover the manifold, but both sets of E-F coordinates cover the exterior region; which set of E-F coordinates you choose will determine into which region of the manifold you can extend. BTW neither choice includes other regions of the complete (inextensible) manifold, but Kruskal-Szerkes coordinates include them all. > A black hole is not a white hole. Period. Sure. And both are contained in the Kruskal diagram, and in the complete (inextensible) manifold. Have you never looked at a Kruskal diagram? -- you seem rather unknowledgeable about basic aspects of this manifold (yes, singular -- you are discussing different regions of a single manifold). Any reasonably modern textbook on GR will have it. Tom Roberts |