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Black Hole as Wormhole?

by PeterDonis
Tags: black, hole, wormhole
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PeterDonis
#1
Feb16-13, 05:52 PM
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A recent thread on Hacker News led me to this paper by Poplawski:

http://arxiv.org/abs/0902.1994

The abstract says:

"We consider the radial geodesic motion of a massive particle into a black hole in isotropic coordinates, which represents the exterior region of an Einstein-Rosen bridge (wormhole). The particle enters the interior region, which is regular and physically equivalent to the asymptotically flat exterior of a white hole, and the particle's proper time extends to infinity. Since the radial motion into a wormhole after passing the event horizon is physically different from the motion into a Schwarzschild black hole, Einstein-Rosen and Schwarzschild black holes are different, physical realizations of general relativity. Yet for distant observers, both solutions are indistinguishable. We show that timelike geodesics in the field of a wormhole are complete because the expansion scalar in the Raychaudhuri equation has a discontinuity at the horizon, and because the Einstein-Rosen bridge is represented by the Kruskal diagram with Rindler's elliptic identification of the two antipodal future event horizons. These results suggest that observed astrophysical black holes may be Einstein-Rosen bridges, each with a new universe inside that formed simultaneously with the black hole. Accordingly, our own Universe may be the interior of a black hole existing inside another universe."

Has anyone else come across this paper before? It looks questionable to me because of the known properties of isotropic coordinates, but perhaps I'm missing something.

(Note: I see from searching PF that other papers by Poplawski have been commented on, but I haven't seen anything about this one specifically. This one doesn't seem to rely on his views about torsion, which are crucial in his other papers.)
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DaleSpam
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Feb17-13, 07:15 AM
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Quote Quote by PeterDonis View Post
Has anyone else come across this paper before? It looks questionable to me because of the known properties of isotropic coordinates, but perhaps I'm missing something.
I just pulled it up today, and I found his comments quite odd. Do you know what he means by "is Galilean for r->∞". Also, I don't get that the form of the metric is unchanged under that substitution.
Bill_K
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Feb17-13, 10:31 AM
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the Einstein-Rosen bridge is represented by the Kruskal diagram with Rindler's elliptic identification of the two antipodal future event horizons.
It's not clear to me that this can be done without causing a problem at the origin of the Kruskal diagram. This is like saying, take Minkowski space and identify the past and future null cones.

PeterDonis
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Feb17-13, 11:26 AM
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Black Hole as Wormhole?

Quote Quote by DaleSpam View Post
Do you know what he means by "is Galilean for r->∞".
I think he means to say "Minkowski at infinity", i.e., asymptotically flat. I can't be sure, though.
PeterDonis
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Feb17-13, 11:29 AM
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Quote Quote by Bill_K View Post
It's not clear to me that this can be done without causing a problem at the origin of the Kruskal diagram. This is like saying, take Minkowski space and identify the past and future null cones.
This looked fishy to me too, but I don't have access to the Rindler paper that he references for this, so I can't check the original source. I was hoping someone here might, or might at least have heard something about this "elliptic identification" thing.
PeterDonis
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Feb17-13, 11:54 AM
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Quote Quote by Bill_K View Post
It's not clear to me that this can be done without causing a problem at the origin of the Kruskal diagram.
It also appears that the author is not claiming that his solution is a vacuum solution; he appears to be putting an infinitely dense sheet of lightlike radiation on the "southwest to northeast" diagonal, and saying that that is what allows the "elliptic identification" to work.
Mordred
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Feb17-13, 12:26 PM
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Naty had a thread in cosmology section with his torsion model. The thread was Can torsion avoid the black hole singularity. Or domething like that lol. I pulled the link to the paper. Its one I've been looking at. I still have to study the OP paper.

http://arxiv.org/abs/1007.0587
DaleSpam
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Feb17-13, 12:34 PM
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Quote Quote by PeterDonis View Post
I think he means to say "Minkowski at infinity", i.e., asymptotically flat. I can't be sure, though.
OK, I just calculated all of the curvature components and (unsurprisingly) they are all indeed 0 in the limit as r-> ∞, but several are infinite in the limit as r->0. Also, I still cannot get that the substitution leaves the form of the metric unchanged. I think he is simply wrong at the very beginning.
PeterDonis
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Feb17-13, 01:29 PM
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Quote Quote by DaleSpam View Post
Also, I still cannot get that the substitution leaves the form of the metric unchanged.
His notation for isotropic coordinates is not the one I'm used to, but that part looks OK to me; he's just stating a known fact about isotropic coordinates, that they double cover the region exterior to the horizon. MTW has an exercise that demonstrates the same substitution; I'll write it in the notation used there, which is the one I'm used to (Poplawski's notation has [itex]r_g = 2M[/itex]). Start with:

[tex]ds^2 = \frac{\left( 1 - M/2r \right)^2}{\left( 1 + M/2r \right)^2} dt^2 - \left( 1 + \frac{M}{2r} \right)^4 \left( dr^2 + r^2 d\Omega^2 \right)[/tex]

Substitute [itex]r = M^2 / 4R[/itex]; that means [itex]dr = - (M^2 / 4R^2) dR[/itex], and [itex]M / 2r = 2R / M[/itex]. This gives:

[tex]ds^2 = \frac{\left( 1 - 2R/M \right)^2}{\left(1 + 2R/M \right)^2} dt^2 - \left( 1 + \frac{2R}{M}\right)^4 \left( \frac{M^4}{16 R^4} dR^2 + \frac{M^4}{16 R^2} d\Omega^2 \right)[/tex]

Refactor the terms:

[tex]ds^2 = \frac{\left( M / 2R \right)^2}{\left( M / 2R \right)^2} \frac{\left( 1 - 2R/M \right)^2}{\left(1 + 2R/M \right)^2} dt^2 - \left( 1 + \frac{2R}{M}\right)^4 \frac{M^4}{16 R^4} \left( dR^2 + R^2 d\Omega^2 \right)[/tex]

[tex]ds^2 = \frac{\left( M/2R - 1 \right)^2}{\left(M/2R + 1 \right)^2} dt^2 - \left( 1 + \frac{2R}{M}\right)^4 \left( \frac{M}{2 R} \right)^4 \left( dR^2 + R^2 d\Omega^2 \right)[/tex]

[tex]ds^2 = \frac{\left( 1 - M/2R \right)^2}{\left(1 + M/2R \right)^2} dt^2 - \left( 1 + \frac{M}{2R} \right)^4 \left( dR^2 + R^2 d\Omega^2 \right)[/tex]

So he's right that the metric is formally unchanged by the substitution; but he claims that by putting an infinitely dense sheet of lightlike radiation at the horizon, he can construct a solution where the patch 0 < r < M/2 is a *different* region than the patch M/2 < r < infinity, so a timelike geodesic that falls in to r = M/2 can then "fall" back outward from r = M/2 to smaller and smaller values of r, which correspond to larger and larger values of R (the transformed radial coordinate), so it takes an infinite amount of proper time to reach r = 0 (which corresponds to R = infinity). I'm not sure the infinitely dense sheet thing works, though.
DaleSpam
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Feb17-13, 03:00 PM
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Quote Quote by PeterDonis View Post
that means [itex]dr = - (M^2 / 4R^2) dR[/itex]
Oops, that is the (obvious) part that I missed. I made the substitution for all of the r terms, but not the dr term.
PeterDonis
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Feb17-13, 05:16 PM
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Quote Quote by DaleSpam View Post
Oops, that is the (obvious) part that I missed. I made the substitution for all of the r terms, but not the dr term.
I think that's the expected "gotcha" in the MTW exercise; I missed it too the first time I tried it.


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