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

On Sun, 09 Sep 2007 21:18:25 -0700, Koobee Wublee
<koobee.wublee@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>

You aren't?

Calculate the half dozen independent Riemann components and let M-->
0. Then tell me what you get.

Do be sure to use the right definition for Riemann this time.


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>

With the added assumption of a time-like Killing vector at infinity,
it sure does! Welcome to Birkhoff's theorem - enjoy your stay.


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.

Says who?

Do the phrases "chain rule" and "change of variables" not mean
anything to you?

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.

Would this be the same Riemann tensor you wrote down that had the
incorrect index placement in several of the components that you seem
to think can be represented by a matrix, or would this be the actual
Riemann tensor?

Anyways, I'll tell you what. If you can show me even one case in which
changing the coordinate system changes the physical content of a
tensor, I'll admit your right.


Since there are an infinite number of solutions to the same
choice of coordinate system, we must conclude that each metric
(solution) describes a unique universe.

Then why can I always construct a coordinate transformation that
relates these "unique" universes?

I can do it EVERY SINGLE TIME, as I have shown you EVERY SINGLE TIME.
You have been unable to find a solution that I cannot do this to.


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>

If you can find me one solution that conforms to the _explicit_
conditions of Birkhoff's theorem [not just the conditions you think
are relevant] but is not asymptotically flat I'll admit you are right.

[...]
.

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