Re: PROOF: Schwarzschild Radius r=2*G*M/c^2 is wrong

On Sep 9, 9:18 pm, Koobee Wublee <koobee.wub...@xxxxxxxxx> wrote:
On Sep 9, 8:27 pm, Eric Gisse wrote:

On Sun, 09 Sep 200, Koobee Wublee wrote:
I take this back. The following solution is only static and
spherically symmetric but not necessarily asymptotically flat.

You can't change the geometry through a coordinate transformation! How
many times do you need to be told?

I am not changing the geometry. All the mathematics I have shown is
based on the invariance in the geometry. <shrug>

Eric - remember Koobee has his own private terminology for everything.

Schchwarzschild is asymptotically equivalent to Minkowski space.

Yes, but this does not mean all the static and the spherically
symmetric solutions are asymptotically flat. <shrug>

There exists a coordinate transformation between the two "different"
manifolds. Thus, this one is asymptotically flat.

Since there are infinite choices of coordinate systems, there exist
infinite coordinate transformations of the same invariant geometry.
However, this is not what I am addressing. I need to establish a
choice of coordinate system before I can proceed to derive the
Christoffel symbols of the second kind. Thus, the choice of
coordinate system must be fixed at this point. Then, we see the
Riemann curvature tensor, the Ricci curvature tensor, the field
equations, and finally the metrics (solutions). In doing so, the
choice of coordinate system must remain the same throughout each
process.

You phrase it extraordinarily clumsily and in a typical amateurish
overcomplicated fashion but agreed so far.

Since there are an infinite number of solutions to the same
choice of coordinate system,

No, this is false in _this particular case_. In general yes, an
equation may have many solutions but the peculiarity of the
spherically symmetric vacuum problem is that it happens to lead to
equations which are guaranteed to have unique solution by virtue of
the general class of differential equations to which they happen to
belong.

we must conclude that each metric
(solution) describes a unique universe.

It would be so if there were multiple solutions corresponding to the
same coordinate system. But there is only one. I repeat that this is
not a generic situation but a peculiarity of the spherically symmetric
vacuum problem.

Also, you can always find
solutions that are static and spherically symmetric but not
asymptotically flat. Thus, this invalidates Birkhoff's theorem in
which you have bet all your chips on. <shrug>

Name one.

Just like before, you cannot explain why you are so fixated on the
Schwarzschild metric as the unique solution.

IT IS NOT UNIQUE - that is the WHOLE point. We have been telling this
to you for years now!

You have being bullsh*tting for years now. It is time to act yourself
and not one of these henchmen of the Church of Einstein to enforce
your Gestapo-like marshal thoughts.

It's a mathematical statement you are making, a false one, and it can
be resolved 100% and without throwing silly tantrums invoking "Church
of Einstein" and similar idiocies.

It would not work in an
intellectual society. Do we even reside in an intellectual society
with all you henchmen of the Church of Einstein spreading false
gospels?

Again, this is mathematics. Save your metaphors for a different
newsgroup.

[cut]

--
Jan Bielawski

.

Relevant Pages

  • Re: Are There Unresolver Foundational Issues With GR
    ... 2 different metrics using the same coordinate system represent 2 ... as the mathematics that represent them is very different. ... The geometry cannot be determined, shaped, modified, or annihilated. ...
    (sci.physics.relativity)
  • Re: PROOF: Schwarzschild Radius r=2*G*M/c^2 is wrong
    ... spherically symmetric but not necessarily asymptotically flat. ... I am not changing the geometry. ... coordinate system must be fixed at this point. ... Riemann curvature tensor, the Ricci curvature tensor, the field ...
    (sci.physics.relativity)
  • Re: PROOF: Schwarzschild Radius r=2*G*M/c^2 is wrong
    ... spherically symmetric but not necessarily asymptotically flat. ... I am not changing the geometry. ... coordinate system must be fixed at this point. ...
    (sci.physics.relativity)
  • Re: For everyone who thinks they can revolutionize science...
    ... The metric of spacetime does not have a valid Lagrangian. ... In order to describe this invariant geometry ... using a particular set of coordinate system, ... The Riemann curvature tensor is NOT a 4x4x4x4 matrix, ...
    (sci.physics.relativity)
  • Re: Derivation of Lorentz Transformation (ctd. from sci.physics.research)
    ... >> transforms to x' = c t' in the other coordinate system and vice versa. ... > there is therefore no reason to assume they hold in general for events ... ready to use mathematics to express physics related ... As long as you don't aknowledge your errors and the ...
    (sci.physics.relativity)