Thursday, August 24, 2006

Through the Fun House Mirror

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)


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

(Δ^2 - &_t^2) h_uv = 16πG (T_uv - 1/2η_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


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