Vilenkin's deSitter Dark Energy Vacuum Bubble

Look Paul, I explained the whole thing correctly earlier today.
Look at the metric from POV of the geodesic observers.
The exotic dark energy "quintessent" w = -2/3 vacuum ODLRO plane wall source

Tuv ~ (c^4/\/G)&(x)diag(1,0,1,1)

signature 1 -1,-1,-1

1 + 3w = 1 - 0 - 1 - 1 = -1

w = - 2/3

note that

c^4/\/G = c^4g^2/c^4G = g^2/G

The LIF geodesic observers INSIDE THE DYNAMIC DESITTER VACUUM 2D
SPHERICAL SHELL see it collapsing at constant objective proper
acceleration g from infinity down to finite area /\^-1 and re-expanding
back out to infinity.

/\^-1 = c^4/g^2 is the minimum area of the vacuum spherical shell at t =
0 when it stops and reverses.

The zero proper invariant acceleration geodesic observers see the 3 + 1
Minkowski metric

ds^2 = c^2dt*^2 - dx*^2 - dy*^2 - dz*^2

Now the LNIF Rindler non-geodesic properly accelerating observers at g
along x =/= x* see an infinite plane wall instead of the dynamic
DeSitter spherical shell.
These Rindler observers must do work in order to exist. They need to
fire rocket engines precisely unlike the LIF geodesic observers who get
a free ride.

The interior 3 + 1 space-time is GLOBALLY FLAT, i.e. R^uvwl = 0 for ALL
observers LNIF & LIF.

However, the LNIF Rindler observers see a DeSitter 2 + 1 transverse
slice (t-y-z) and a Rindler longitudinal slice (t-x) where from the LNIF
the geodesic test particles appear to be in a Newtonian force field

V(Newton) = -gx + (1/2)g^2x^2/c^2

f(Newton) = -dV/dx = + g(repulsive) - g^2x/c^2 attractive

With geodesic test particle event horizon(s) at

1 + V(Newton)/c^2 = 0

1 - gx/c^2 + (1/2)g^2x^2/c^3 = 0

The metric you wrote only covers the interior of the DeSitter spherical
shell not the exterior.

On Aug 24, 2006, at 5:54 PM, Paul Zielinski wrote:

Let me re-word this.

Flat everywhere but at x = 0 does not *just* mean flat "in a 1-1 x-t slice".
It means the spacetime *as a whole* is locally Riemann flat *everywhere*
except at x = 0."


Vilenkin makes it quite clear that the domain wall spacetime *as a
whole* is Riemann flat (R^u_vwl = 0) everywhere *except* on the wall
itself. If it is only flat on x-t surfaces then how could R^u_vwl = 0
everywhere outside of the wall?

However, you are right that in Vilenkin's 1983 solution each y-z, t
hypersurface is a (2 + 1) de Sitter space embedded in a 4D Riemann-flat
spacetime, which implies constant positive curvature along any y-z, t
hypersurface.

The point is that the "Rindler" metric on the x, t surfaces is a
Riemann-flat metric on those surfaces that results in real gravitational
acceleration away from the wall along x. That means that this "Rindler"
metric does not actually represent
a Rindler frame in a Minkowski spacetime, but instead represents a real
geometric tilt of the metric along the x direction. According to the
equivalence principle, the observable effects are the same either way;
but the two cases are nevertheless not the same from a theoretic
standpoint, since a coordinate tranformation cannot change the geometric
relationship between test particle geodesics and the wall.

This is a Red Herring. No one ever said otherwise.

So while you have a valid point that this domain wall solution is not as
simple as it looks at first glance, my argument still works in this
example, since the effect of the flat x-t metric is to curve the actual
geometry of the test particle geodesics and produce real gravitational
acceleration away from the source along the x direction,

No, you keep misunderstanding this. The geodesic test particles in
APPEAR to be in a conservative force field f(Newton)

V(Newton) = -gx + (1/2)g^2x^2/c^2

f(Newton) = -dV/dx = + g(repulsive) - g^2x/c^2 attractive

relative to the LNIF non-geodesic Rindler observers with constant proper
g from firing rockets. It's exactly like the free falling cannon ball
geodesic relative to the non-geodesic Earth in Einstein's POV - switch
"geodesic" to get Newton's POV for same problem.
In any case the objective 3 + 1 Riemann curvature tensor is zero for
everyone, but not the 2 + 1 DeSitter slice that for the LNIF guys looks
curved i.e. /\. The LIF guys do not slice it in same way so one cannot
directly compare these subspaces. What everyone agrees is that 3 + 1
curvature tensor vanishes.


which a coordinate transformation alone cannot do. So this is not
actually a Minkowski spacetime, as you claim. Vilenkin doesn't actually
say that it is; he only says "the metric is Minkowski", meaning that it
is formally of the Minkowski type; just as when he
says that the x-t metric is a "Rindler metric", by which he means that
it is *formally* the same as a metric that represents a Rindler frame in
an actual Minkowski spacetime. So I think it is you, and not Vilenkin,
who is confused here.

I think you are making a meaningless verbal distinction without any
physical meaning.

You always get this effect locally in Riemann normal coordinates in a
*curved* spacetime, since then the metric assumes its normal form [-1,
1, 1, 1] -- but in the curved case obviously this doesn't mean that you
are actually in a Minkowski spacetime. The situation here is very
similar, even though the net Riemann curvature of the spacetime outside
the wall is zero. Here you have a real flat geometric metric gradient
that remains constant, while the net *coordinate* gradient goes to zero
in any free-fall frame. That does not mean that Vilenkin's solution is a
Minkowski spacetime. It simply means that
the metric assumes its *normal form* in any free fall frame in the
Riemann-flat region.

Sorry I don't understand what point you are making.

If the 4th rank curvature tensor = 0 at all points in a 4D simply
connected space-time region then that is sufficient to say that the
geodesic observers have metric

ds^2 = (cdt)^2 - dx^2 - dy^2 - dz^2

and that any curvilinear metric guv(x) in this manifold describes a set
of non-geodesic local observers using non-gravity forces to maintain the
given curvilinear representation i.e.

ds^2 = gu'v'(x)dx^u'dx^v' (LNIF') = (cdt)^2 - dx^2 - dy^2 - dz^2 (LIF)

R^u'v'w'l'(x') = R^uvwl(x) = 0 INVARIANT VANISHING OF ALL THE TENSOR
COMPONENTS FOR ALL OBSERVERS.

In his 1981 paper on the weak field solution there is no mention of /\,
and Vilenkin writes a linearized field equation

(?^2 - &_t^2) h_uv = 16?G (T_uv - 1/2?_uv T)

subject to harmonic coordinates. So why he jumps straight to an exact
solution with /\ =/= 0 in the 1983 paper without even writing a new
field equation is still a mystery to me

Z.