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neginf
Aug25-11, 11:37 AM
I've read that the universe is Euclidean and have also read that space is bent by gravity. Descriptions of geometry near black holes almost sounds like hyperbolic geometry.

1. Is this so?
2. If it is, does it mean we're in a universe that is Euclidean overall but has non Euclidean regions ?

phinds
Aug25-11, 12:09 PM
I've read that the universe is Euclidean and have also read that space is bent by gravity. Descriptions of geometry near black holes almost sounds like hyperbolic geometry.

1. Is this so?
2. If it is, does it mean we're in a universe that is Euclidean overall but has non Euclidean regions ?

I think that's exactly right, although I believe that it is NOT absolutely known that the U is perfectly flat, just that it is flat to within our current ability to measure it and assumed to be very likely perfectly flat.

WannabeNewton
Aug25-11, 12:37 PM
A spherically symmetric, static black hole is, well, just that. If you look at the schwarzchild metric, the r^{2}d\Omega ^{2} term indicates that the geometry outside the black hole is spherically symmetric as well. Where did you read that it was hyperbolic by the way? If it was hyperbolic then it wouldn't have \frac{\partial }{\partial \phi } as a killing field. Even for a kerr black hole, which isn't spherically symmetric, the geometry outside the black hole isn't hyperbolic.

neginf
Aug25-11, 01:04 PM
I didn't read that it was hyperbolic, just misinterpreted it. Although I can't remember where, I read that if something was close enough to a black hole to fall in, from a certain distance away, that thing would appear never to fall all the way, just fall ever more slowly.

That reminded me of something I learned a long time ago about the upper half plane the hyperbolic metric. If I remember right, a disc moving towards the real line would just decrease in size and never quite get there.

WannabeNewton
Aug25-11, 01:22 PM
While it is true that if you start on the y axis, then any other point on the upper half plane would have an infinite distance from the original point for that metric, the story of observes getting closer and closer to the EH but never actually reaching it is another story. While it does seem to take infinite coordinate distance to get to the EH for the schwarzchild black hole, remember that coordinate related quantities are secondary to geometric quantities. If you compute the proper distance, instead of the coordinate distance, you will see that it is finite. Therefore, in your frame you will fall past the EH.

Chalnoth
Aug26-11, 12:34 AM
I think that's exactly right, although I believe that it is NOT absolutely known that the U is perfectly flat, just that it is flat to within our current ability to measure it and assumed to be very likely perfectly flat.
I don't think that's generally assumed. I'm pretty sure most cosmologists assume the opposite: that it is very likely to be either slightly open or slightly closed, with closed being preferred. Though in many models it is very likely so incredibly flat that we'd never be able to measure the overall curvature.

cepheid
Aug26-11, 12:48 AM
I don't think that's generally assumed. I'm pretty sure most cosmologists assume the opposite: that it is very likely to be either slightly open or slightly closed, with closed being preferred. Though in many models it is very likely so incredibly flat that we'd never be able to measure the overall curvature.

Why would "slightly not flat" be assumed when the theoretical motivation would seem to be for \Omega_\textrm{tot} = 1 ? Is this assumption of slightly closed because inflation is thought to have produced something very very close to flat but not necessary exactly so?

Chalnoth
Aug26-11, 12:52 AM
Why would "slightly not flat" be assumed when the theoretical motivation would seem to be for \Omega_\textrm{tot} = 1 ? Is this assumption of slightly closed because inflation is thought to have produced something very very close to flat but not necessary exactly so?
Yes to the second question. But there's also the point that this is a geometrical factor that would require infinite fine-tuning to be made identically flat.

phinds
Aug26-11, 06:51 AM
Yes to the second question. But there's also the point that this is a geometrical factor that would require infinite fine-tuning to be made identically flat.

Interesting point. Thanks.

I believe I had read, and was following this in my logic (apparently incorrectly, from what you are saying, which I had not thought about), that the consensus was that it would be an amazing coincidence for it to be VERY close to 1 but not actually 1, out of all the possible values.

Chalnoth
Aug26-11, 08:01 AM
Interesting point. Thanks.

I believe I had read, and was following this in my logic (apparently incorrectly, from what you are saying, which I had not thought about), that the consensus was that it would be an amazing coincidence for it to be VERY close to 1 but not actually 1, out of all the possible values.
I've never heard that argument. Especially since we do have a mechanism to drive the total density fraction exponentially-close to one.

TrickyDicky
Aug26-11, 11:37 AM
I believe I had read, and was following this in my logic (apparently incorrectly, from what you are saying, which I had not thought about), that the consensus was that it would be an amazing coincidence for it to be VERY close to 1 but not actually 1, out of all the possible values.

You might have read it in pop-science journals, I recall having read something like that there.
The real fact is rather the opposite, it would be an amazing coincidence for it to be exactly 1 instead of the infinity of values close to 1 either above or below.
Besides the only spacial geometry that can never be empirically proved is the flat limit case as it could always be suspected that the fact one measures a flat space is due to the lack of precision of the measuring tools to measure a very tiny curvature so that it goes undetected. That is not the case with the open and closed universes that could actually be shown to be the real ones thru experiment.