Subject: Gravity Fields of Dark Energy Exotic Vacuum Domain Walls.

http://disc.server.com/discussion.cgi?disc=68326;article=2819;

is the color version of the work below.

On Aug 22, 2006, at 3:54 PM, Paul Zielinski wrote:

The cosmic string and vacuum domain wall solutions are globally flat outside the sources.

Is this accurate? NO! Not for the vacuum wall at least.

The exotic vacuum ODLRO domain wall is an off-mass-shell virtual dark energy effect different from the on-mass-shell slab of ordinary matter.

"The gravitational field of a domain wall is rather unusual and is very different from that of an ordinary massive plane."

http://www.novelconceptsinc.com/picts/calculators-slab-thermal-resistance.gif

http://www.mrao.cam.ac.uk/ppeuc/astronomy/papers/axenides/node2.html

&

"Cosmic Strings and Other Topological Defects"

A. Vilenkin & E. Shellard 13.3

"the gravitational field of the wall is repulsive" p.378

"The Einstein equations [with vacuum ODLRO wall source] have no static solutions having planar and reflectional symmetry."

I. Vacuum domain wall

The vacuum wall is in the y-z plane and the source tensor Tuv has a Dirac delta function &(x). The metric solution can be represented as

ds^2 = (1 - g|x|/2c^2)^2(cdt)^2 + dx^2 + (1 - g|x|/2c^2)^2e^2(gt/c)(dy^2 + dz^2)

"The x-t part is a 1 + 1 Rindler metric describing a flat space in the frame of reference of a uniformly accelerated observer."

This is LNIF non-geodesic observer because the acceleration of a LIF geodesic observer is by construction zero. That is the geodesic is the straightest world line in space-time. It need not be "straight" in the projected spacelike hypersurface.

Note that the statement of flatness is only in the 1 + 1 x-t slice not in the whole 3 + 1 manifold so that Zielinski may be misunderstanding what "flatness" means in this problem.

"The observer at x = 0 will see (geodesic) test particles moving away from the wall with acceleration g in agreement with the Newtonian analysis."

When we see cannon balls in free fall almost parabolic orbit

http://www.zonalibre.org/blog/diversovariable/archives/baron-munchausen.jpg

We are LNIF observers on the non-geodesic world line of the rotating rigid Earth's surface whose center of mass is on a geodesic round the Sun at a focus of the slightly precessing elliptical orbit of that center. The cannon ball is geodesic in Einstein's theory, though not in Newton's where a gravity force is acting on the cannon ball. There is a bait and switch here as we flip paradigms to explain the same phenomenon in almost opposite ways. The word "geodesic" changes meaning. However, there is zero acceleration on "geodesics" in both Newton's and Einstein's informal language and in both cases the geodesic is the straightest possible path. Newton's space + time is flat Euclidean, Einstein's space-time is curved.

Look at the 1 + 1 Rindler metric slice warp factor more closely

(1 - g|x|/2c^2)^2 = 1 - g|x|/c^2 + (g|x|/2c^2)^2

The effective Newtonian potential per unit test mass is

V(Newton) = -g|x| + (g|x|/2c)^2

The "Newtonian" gravity force on the Einsteinian "geodesic" test particle in the POV of the LNIF Rindler uniformly accelerated observer is

- dV(Newton)/d|x| = + g - g^2|x|/2c^2

Note that Zielinski did misunderstand. The first term on the RHS is an approximately uniform repulsion seen by the LNIF observer looking at a test particle on a geodesic in this curved space-time. This approximation is only good when g|x|/c^2 << 1. The second term will give curvature to the spacetime in the Einstein description and it is an attraction. Note the balance at

+g - g^2|x|/2c^2 = 0

i.e.

1 = g|x|/2c^2

which is an infinite red shift horizon.

Note that the non-geodesic Rindler observer needs to apply a NON-GRAVITY FORCE, e.g. firing a rocket engine to maintain his privileged frame status in which the metric field is represented.

The total 3 + 1 is not globally flat as Zielinski wrongly said. Indeed, the x = constant slices form a 2 + 1 DeSitter space of positive zero point energy with negative pressure. This DeSitter space has constant positive curvature /\. It is not at all flat with a vanishing curvature tensor.

ds*^2 = dt^2 - e^gt/c(dy^2 + dz^2_

Note that the COSMOLOGICAL CONSTANT in the 2 + 1 slice here is

/\ = g^2/c^4

Note that the DARK ZERO POINT STRESS TENSOR of the Vacuum ODLRO "Wall" is

Tuv = Ttt&(x)diag(1,0,1,1)

Ttt = wall tension in the y-z plane

In the weak field limit the effective GR Einstein-corrected Poisson gravity equation is

Grad^2V(Newton) = 4piG(Ttt - Txx - Tyy - Tzz) = 4piGTtt(1 - 0 - 1 - 1) = -4piGTtt

i.e. repulsive for positive Ttt = energy density that has dominating negative pressure here by a factor of 2. The broken rotational symmetry has changed the usual pressure factor of 3 to 2.

1 + 3w = -1

w = - 2/3

in this problem

Note also that GTtt/c^4 = /\ = g^2/c^4

i.e. Ttt(Wall) = g^2/G

## Tuesday, August 22, 2006

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