Is the Universe Flat or Curved According to Planck Probe Data?

[Chalnoth]

Planck is designed to probe the high energy [gamma] spectrum. It is well suited for exploring the very early universe where high energy events were common.

I'm sorry, but I believe you're thinking of Fermi. Planck is primarily a CMB instrument, and probes radiation from 30GHz to 857GHz, which is in the millimeter wave range (wavelengths from 10mm to 0.35mm).

 

[Ich]

Cosmological observations only constrain the local geometry of the universe, because we can only observe our local neighborhood.

...with "local" meaning out to z=1089. Our observable universe is flat.
We can't know what's behind the observable universe. Further, topology is not really constrained by curvature, too.

The Schwarzschild solution is spatially flat

No, it's curved like http://en.wikipedia.org/wiki/Schwarzschild_metric#Flamm.27s_paraboloid".

 

[bapowell]

No, it's curved like http://en.wikipedia.org/wiki/Schwarzschild_metric#Flamm.27s_paraboloid".

By flat, I mean R = 0.

 

[Ich]

By flat, I mean R = 0 .

Yes. It isn't flat.

 

[caspiansea3]

I recently came across this paper:
How flat can you get? A model comparison perspective on the curvature of the Universe
http://arxiv.org/abs/0901.3354"

In it the authors say "We show that, given current data, the probability that the Universe is spatially infinite lies between 67% and 98%". By spatially infitite do they mean that the curvature is exactly zero (omega = 0)?

 

[skippy1729]

I recently came across this paper:
How flat can you get? A model comparison perspective on the curvature of the Universe
http://arxiv.org/abs/0901.3354"

In it the authors say "We show that, given current data, the probability that the Universe is spatially infinite lies between 67% and 98%". By spatially infitite do they mean that the curvature is exactly zero (omega = 0)?

All classical physical measurements have a range or nonzero tolerance.

Omega = 1.01 +/- 0.02 means that Omega = 1 is POSSIBLE - universe might be infinite

Omega = 1.01 +/- 0.005 means that Omega = 1 is IMPOSSIBLE - universe is finite

This is primary school reasoning. Their probabilities are rubbish.

Skippy

 

[Ich]

By spatially infitite do they mean that the curvature is exactly zero (omega = 0)?

Zero or negative.

 

[George Jones]

Please correct a layman if I’m wrong, but the WMAP CMB-measurements on 'flatness' deals with the Density parameter (Ω), resulting in a Closed, Open or Flat (local) universe, right?

Consequently, if the (local) universe turns out to be perfectly flat (in terms of Ω), it could still be spatially curved, in the shape of a torus, right...?

Or, did I miss something crucial...

Yes and no. Flatness refers to the intrinsic curvature of a three-dimension spatial hypersurface that results when when the time coordinate is held constant.

Yes, if \Omega =1, a spatial hypersurface could have the same topology as S^1 \times S^1 \times S^1 instead of the usual topology of \mathbb{R}^3. A universive like this, howver, doesn't satisfy the the cosmological principle.

No, in this context, the intrinsic curvature of S^1 \times S^1 \times S^1 is not non-zero, i.e., the intrinsic curvature is zero.

In your mind's eye, you "see" a torus as curved, but this is misleading. You are visualizing extrinsic (not intrinsic) curvature in some abstract higher-dimensional embedding space. Even if the spatial sections of a flat universe have topology \mathbb{R}^3, their extrinsic curavture non-zero. Counterintuitive.

 

[George Jones]

By flat, I mean R = 0.

I am confused. What is R? The Ricci scalar? If so, what has this to do with spatial curvature? Spatial is the curvature is the curvature of a three-dimensional hypersurface that results when a timelike curvature is held constant. The hypersurface curvature is the curvature associated with the hypersurface metric that is induced by the sapcetime metric.

Of course, the hypersurace depends on the choice of timelike coordinate. For example, as Ich has noted, the t coordinate from standard Schwarazschild coodinates for the Schwarzschild metric gives rise to non-flat spatial hypersurfaces, while the time coordinate from Painleve-Gullstrand coordinates for the Schwarzschild metric gives rise to flat spatial hypersurfaces.

 

[slogan120]

If you guys could help me clarify this point up to now that would be great.

Spatial vs Spacetime curvature

If its correct to say that space is flat, and space time is curved, does that not imply only time is curved? Or is this a misinterpretation of information?

I would ask more about the implications of this fact, but I want to check that statement first.
Please also don't use acronyms as they can lead to minutes of confusion and frustration when they only cost secs to type.

 

[Chalnoth]

If its correct to say that space is flat, and space time is curved, does that not imply only time is curved? Or is this a misinterpretation of information?

The tensor describing the curvature can, with a particular choice of coordinates, be separated into a time-time component, space-time components, and space-space components. With the coordinates usually used for cosmology, the space-space components all come out to zero on cosmological scales, while the space-time and time-time components remain nonzero.

A visual interpretation of this result can be to simply consider a flat rubber sheet that is expanding with time. The time-time and space-time curvature components describe this expansion. The space-space components are zero because it's flat.