On May 31, 2006, at 3:07 PM, Jack Sarfatti wrote:

Naively speaking, the Kerr ring singularity has radius a, so that when a
m, it's naked.
One naively thinks that the Kerr singularity shrinks to a spatial point
singularity when a - 0, but the limit point is peculiar as we see by
comparing the Penrose diagrams

http://www.gothosenterprises.com/bla...s/penrose2.gif

The Kerr singularity is TIMELIKE, i.e. vertical jagged line when a =/= 0

At the actual limit a = 0 there is a qualitative jump to the horizontal
SPACELIKE singularity.

Of course from the POV of any observer OUTSIDE the SSS event horizon,
the SPACELIKE singularity is INFERRED to be a SPATIAL POINT at the
"center" of the spherical surface 2m. However, that is not what is
experienced by the INSIDE observers - actually they are crushed out of
existence before they can measure anything. This is an example of Lenny
Susskind's "black hole complementarity."

http://www.astro.northwestern.edu/ra...e_Schwpar.html

On May 31, 2006, at 11:20 AM, Paul Zielinski wrote:

Jack Sarfatti wrote:

On May 31, 2006, at 9:51 AM, Paul Zielinski wrote:

Jack Sarfatti wrote:


On May 31, 2006, at 1:17 AM, Paul Zielinski wrote:

Jack Sarfatti wrote:

Read any text book Paul. We look for asymptotic flat solutions as a
matter of PHYSICS.


Of course I understand that Newtonian correspondence requires that the
field of a spherically
symmetric active mass be asymptotically flat far from the *source*.

But if there is no source, and no Ricci curvature anywhere, what does
it mean to demand asymptotic
flatness far from the source?


The idea is that such vacuum solutions exist!

All kinds of solutions exist for R_uv = 0. Why pick asymptotically flat
solutions when there is no material source?

Because they model elementary particles as Bohm hidden variables if e^2
~ G*m^2. For example Burinski's model, except I make G much larger
perhaps from the extra space dimension idea.


I think the correct GR answer is that the Bianchi identities ensure that
any local perturbation of the Weyl curvature at a point
or in a region of spacetime -- whatever its origin -- results in Weyl
curvature in the rest of the spacetime that diminishes
asymptotically to zero at large distances from the perturbation.

You must have missed my

The Bianchi identities lead to

Guv^;v = 0

This is only for the zero torsion metric connection.

However, in classical vacuum

Guv = 0

Because

R = 0

Ruv = 0

Therefore, why do you need the Bianchi identities in that case?

The Bianchi identities are only important when you have Ricci curvature!

We are here talking about a soliton (geon) of pure astigmatic tidal
stretch-squeeze Weyl conformal curvature without any "gravity lens"
focusing (defocusing) volume contracting (expanding) Ricci curvature
from zero point energy or ordinary matter/radiation.

This will hold even if there is no active mass and no Ricci curvature
anywhere -- just a local perturbation of the Weyl curvature.

If Guv = 0 globally, why should Guv^;v = 0 matter?


What you are really saying here is that if the Weyl curvature is
somehow perturbed from zero at a given point, or locally within a given
region, then the surrounding vacuum will also exhibit Weyl curvature
that diminishes as you recede from the perturbation, and fall off
asymptotically as you approach infinite separation from the perturbation.
I suppose all this follows from the Bianchi identities.

singularity, and we are left with a pure vacuum solution, what does
"far from the source" mean?


In the case of collapsing matter there is an event horizon at 2m that
is the "source." - obviously.

Source of what?

The event horizon is a geometrodynamic structure! It's a dimple in the
geometrodynamic field where the light rays are not allowed to escape to
future infinity where r 2m.


Because it is a null surface with an associated congruence of null
geodesics that traps outgoing light rays.

Because what? Yes it is a null surface. I see no logic in your sentence.


There is no curvature singularity at 2m in the standard model -- only a
coordinate defect (in Schwarzschild coordinates). So how can an event
horizon act as a source of the SSS field?

Your statement is confused. The spacelike singularity is inside the
event horizon. You are asking nonsense questions.


But even if there is no event horizon, there can still be a curvature
singularity -- which is the situation in the Schwarzschild solution.

WRONG! A naked singularity can only happen in the Kerr solution for a
m. Cosmic censorship works in the SSS case. There is always an event
horizon in the maximally extended SSS vacuum solution of Ruv = 0, i.e.
zero Ricci curvature globally.

The Chapline-Laughlin dark star is different of course because it has
Ricci curvature induced by zero point energy at the event horizon and in
the interior.

Ruv = 0 only works when the zero point energy is precisely zero!


On the other hand, according to the singularity theorems if there is an
event horizon it must enclose a curvature singularity.
So with or without an event horizon, the singularity is the actual
source of Weyl curvature in the vacuum region.

So the most you can say here is that to an outside observer, who cannot
see the curvature singularity, the event horizon "looks" like a
gravitational source although that is not to say that it actually is a
source.

Is that what you meant?

I mean more than that. I mean that real matter is made this way - this
is Wheeler's whole idea of "Mass without mass."

Wheeler's "Geometrodynamica" as Bohm's hidden variables with pilot QUBIT
waves:

Mass without mass i.e. pure Weyl curvature, which when coarse-grained
has emergent Ricci curvature.

Charge without charge, i.e. quantized electric flux through tiny
wormholes from ODLRO vacuum coherence, i.e. monopole defect with Dirac
string. The Dirac string is also the Penrose "belt" to make "spinors".

Volume without volume i.e. quantized area + holography as in my archive
paper.