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From: Ken S. Tucker on 25 Aug 2006 02:34 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). > > > 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. > > > 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. > > > As multiple solutions to the vacuum field equations are discovered, > > there are actually an infinite number of them. > > Yes, and 2+2=5. > > > With infinite number of > > solutions, it is shaking the very foundation of GR and SR. > > Sure. My boots are all torn already. > > Give us a break. > > -- > Jan Bielawski Jan, I think you should study the original paper Ed Green cited, there is no such thing as an event horizon or Black-Hole's. BTW, Dr. Loinger and I discussed this at length. In short, the original Schwarzschild Solution has been bastardized and mis-understood for simplicity. The bastardized version became popularized and embraced by astronomers who now see BH's under their beds. Regards Ken S. Tucker
From: I.Vecchi on 25 Aug 2006 08:35 JanPB ha scritto: > 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). > > > 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. > > > 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. 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. IV
From: Tom Roberts on 25 Aug 2006 11:35 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. Tom Roberts
From: Koobee Wublee on 25 Aug 2006 12:29 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. > > 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. > > 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. > > 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. You really need to go back to kindergarten and study above basic mathematics instead of embracing Voodoo Mathematics. This is the best advice I can give you. > > With infinite number of > > solutions, it is shaking the very foundation of GR and SR. > > Sure. My boots are all torn already. > > Give us a break. If you want a break, you can find a hole on the ground and hide your head in it, just like an ostrich. In the meantime, the house of cards called GR and SR, where all pieces are bonded together through Voodoo Mathematics, is coming down.
From: Igor on 25 Aug 2006 13:43
Koobee Wublee wrote: > Edward Green wrote: > > Igor wrote: > > > > The metric will always compensate to make up for the difference. > > > > So you claim. I believe I have found the answer to my own question (as > > usual), and while I did not state the requirement 100% correctly, I was > > on the right track. I happened upon the correct formulation in > > Schwarzschild's 1916 paper on the field of a mass point: > > > > "The field equations ... have the fundamental property that > > they preserve their form under the substitution of other > > arbitrary variables in lieu of x1,x2, x3, x4, as long as the > > determinant of the substitution is equal to 1." > > > > http://arxiv.org/PS_cache/physics/pdf/9905/9905030.pdf > > This is an excellent article. It is very easy to read and understand. > Thanks to Schwarzschild's ingenuity. > > > If the old coordinates were orthogonal and the new coordinates remained > > orthogonal (which requirement I did not state) and we kept the same > > units, mine would be a sufficient condition to meet the above > > requirement on the named determinate. In general we have more latitude > > than that, _but_ we are not free to chose the new coordinates > > arbitrarily, even if they are invertible in a neighborhood. The unit of > > _volume_ at least must be preserved. In the case of a one-dimensional > > manifold this reduces to the requirement to keep the same units. > > 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. > > However, since Schwarzschild Metric is much simpler than > Schwarzschild's original solution, Schwarzschild Metric is embraced by > the physics communities today. Mr. Bielawski and Igor have not > understood Schwarzschild's original paper and choose to blindly reject > Schwarzschild's original solution and others. > > 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? But you keep claiming that it is. And that's just plain wrong. 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. Birkoff proved that it can't be done. |