Curvature and spatial closure of the universe.

[Dr. Jekyll]

Everytime I get into this subject I get confused. All these [math]\Omega_j[/math] and [math]\Lambda_k[/math]. Hmm...now, where was I?

 

Oh here, let us take the latest WMAP-data. Maybe I got some decimals wrong (disregard that), but let us assume a density parameter as [math] \Omega_{tot}=1.003 \pm 0.010[/math]

 

I'm a bit drunk atm, , but [math]\Omega_{tot}[/math] is a sum of different ratios (density ratios), also including a possible cosmological constant - [math]\Lambda_{something}[/math]?

 

Now, assume, hypotetically, that [math]\Omega_{tot}=1.0000000...[/math]. As I get it, that only implies that the universe will expand on and on. However, the universe CAN still be closed.

 

Which parameters measured, determine if the universe is spatially closed or open?

 

Back in the mid 90s, when I studied physics, it was thought that it all depends on gravity vs. mass. But, now we all know that shit is wrong due to dark energy.

 

So, can someone (Martin?) elaborate on this subject?

Edited August 13, 2009 by Dr. Jekyll

[Martin]

Doctor, there is an auxilliary piece of notation which even tho drunk you should start getting used to, which is [math]\Omega_k[/math]. It exists merely to simplify some formulas and is defined by

 

Oh here, let us take the latest WMAP-data. Maybe I got some decimals wrong (disregard that), but let us assume a density parameter as [math]\Omega_k = 1 - \Omega_{tot}[/math]

 

So in the case you offered as an example, suppose in fact that

[math] \Omega_{tot} = 1.003[/math]

Then an equivalent way to say this would be

[math]\Omega_k = - 0.003[/math]

 

In that case the standard cosmo model say that the universe is spatially closed and that space has the overall geometry of a 3D hypersphere and it tells you how to compute the RADIUS OF CURVATURE of space which is the radius which the hypersphere would have if you imagine immersing it in 4D Euclidean space.

 

The model doesn't say that such a surrounding 4D space exists, it assumes that whatever exists at this moment is concentrated in the 3D hypersphere. We have no evidence of anything outside or around it. But it still can have a radius of curvature. That would affect things like the sum of angles of large triangles, and how long it would take to go all the way around if you could freeze expansion. The circumference is of course 2pi times the radius.

And the radius of curvature tells you the volume. The formula for a 3D hypersphere is 2 pi^2 R^3,

 

To compute the radius of curvature is easy, it is just the Hubble radius divided by the square root of the absolute value of Omega_k.

 

In the example, the absolute value of Omega_k is just 0.003.

 

And the Hubble radius c/H is something like 13.7 billion lightyears.

So you divide that by the square root of 0.003.

 

Drunk or not you should be able to do this:D. What do you get?

 

bTW if Omega_tot is exactly one, then Omega_k is exactly zero. And then the radius of curvature is infinite. Which describes spatial flatness. There are exotic topologies, toroids, that are spatial flat and still spatial closed, and smartassers will mention them now and then, but the basic picture of spatial flat is spatial infinite. Euclidean space.

 

I hope it turns out Omega_tot is just a tiny bit greater than one, and then we have a simple hypersphere spatial closed case, and we never have to hear about toroids again. It will the simple surface-of-balloon picture just raised from 2D to 3D.

 

Tell me what you get for the RoC and I'll see if I agree

--------------------------

 

For more information google "Komatsu cosmology wmap" and you will get things like this:

http://arxiv.org/abs/0803.0547

and look in Table 2 on page 4 and you will see the 95% errorbar for Omega_k

 

and you will also see how they calculate a 95% lower bound on the radius of curvature, using the highest Omega_k, the high end of their errorbar because that gives the smallest RoC.

Edited August 14, 2009 by Martin

[Dr. Jekyll]

Thanks for the reply Martin! Since I'm a bit drunk again, let us divide: (13.7e9/sqrt(0.003))/1e9 = 250 billion light years.

 

How is [math]\Omega_k[/math] or [math]\Omega_{tot}[/math] estimated? I don't remeber the details, and weeding through the Nasa articles to refresh is a bit tedious. [math]\Omega_{tot}=\Omega_m+\Omega_r+\Omega_k+\Omega_{\Lambda}[/math] correct?

 

I mean, does WMAP data give a [math]\Omega_{tot}[/math] (loosely speaking), and then we can derive [math]\Omega_k[/math]. Or, is [math]\Omega_m[/math], [math]\Omega_k[/math], etc. measured and they sum up to give a value of [math]\Omega_{tot}[/math]? I have assumed the latter.

[Widdekind]

So, from Friedmann's Equations, and the "density over-contrast" [math]\Omega_{k}[/math], you can calculate the Curvature Radius of the Cosmos to be about 300 billion light-years.

 

Now, according to Wikipedia, the combined mass of the Observable Universe is of order 10⁵⁵ kg — equivalent to a Schwarzschild Radius ("Geometric Mass") of order 10²⁸ m, or 1000 billion light-years.

 

And, this negelects all the "Level I Parallel Universes" which could quite conceivably exist, w/in our own normal Universe, out past & beyond our own finite visible horizon (Scientific American reports -- Majestic Universe ; History Channel The Universe -- Parallel Universes (TV)).

 

Furthermore, Physicist Stephen Hawking has famously shown that the Big Bang is consistent w/ the catastrophic explosion of a "reverse Black Hole", albeit on a "vastly larger scale" than anything humans could conceivably see in this visible Universe (National Geographic Channel Naked Science -- Hawking's Universe (TV)).

 

So, how could it be, that the Cosmos's Curvature Radius is actually less than the Schwarzschild Radius of the "mega-massive hyper-Black Hole" that the Cosmos could have come from ??

[Martin]

Thanks for the reply Martin! Since I'm a bit drunk again, let us divide: (13.7e9/sqrt(0.003))/1e9 = 250 billion light years.

 

How is [math]\Omega_k[/math] or [math]\Omega_{tot}[/math] estimated? I don't remeber the details, and weeding through the Nasa articles to refresh is a bit tedious. [math]\Omega_{tot}=\Omega_m+\Omega_r+\Omega_k+\Omega_{\Lambda}[/math] correct?

 

I mean, does WMAP data give a [math]\Omega_{tot}[/math] (loosely speaking), and then we can derive [math]\Omega_k[/math]. Or, is [math]\Omega_m[/math], [math]\Omega_k[/math], etc. measured and they sum up to give a value of [math]\Omega_{tot}[/math]? I have assumed the latter.

 

Jeykl and Widdekind, thanks for both your responses and for continuing the thread in a quantitative manner. Sorry not to have responded earlier.

 

Jeykl, drunk or sober you clearly know how to calculate.

 

What WMAP reports is a 95% confidence interval for Omega_k.

 

You ask does the actual measuring give you Omega_k or does it give you Omega_tot. But these are logically the same because Omega_k is defined to be

1 - Omega_tot.

 

The same data analysis proceedure that gives a value for one gives a value for the other at the same uncertainty level. I am not expert enough to tell you why they have the alternate notation in the first place. Represents the same information. Apparent redundant notation.

 

Jeykl, you ask how spatial curvature can be measured. they use several methods and check them against each other. One way is very simple to understand and is by galaxy counts. As you may know methods of successive approximation involve an element of (not quite) circularity

 

You believe near flatness and if just assume exact flat for starters, then you can set up a conversion of redshift to present distance and make a catalog of how many galaxies by distance. At the present moment.

Now if there is zero spatial curvature then the volume of an R-ball should increase as R^3.

And we assume largescale uniform distribution of galaxies. So the number within an R-ball should increase as R^3.

Counting galaxies is a way of measuring volume.

 

But if there is positive spatial curvature then the volume of space within distance R from us, the R-ball, will NOT increase as R^3. It will increase more slowly.

 

By analogy think of the 2D case on surface of usual sphere. The area within an R-disk does not increase as fast as R^2.

 

By counting galaxies, and measuring the volume of an R-ball, and seeing how it depends on increasing radius, one can gauge the curvature.

 

Now, to do successive approximations, one has to go back and CORRECT one's original formula for cranking out the present distance to a galaxy from its redshift. But fortunately this will change things only a little bit. The count within a given R distance will be slightly different, and one can get a corrected estimate of the curvature. But basically we're done.

 

That's one way.

 

Now I should go fetch that 95% confidence interval.

 

BTW for background context, google "Komatsu cosmology" and you get this UTexas page:

http://gyudon.as.utexas.edu/~komatsu/

It has some latest news about Planck satellite (the next thing after wmap. It gives an idea what Eichi Komatsu is like.

Then google "Komatsu wmap" and you get this NASA page:

http://lambda.gsfc.nasa.gov/product/map/dr3/map_bibliography.cfm

The one by Komatsu et al is the wmap paper relevant to cosmo.

In that paper the confidence interval is given on page 4.

−0.0179 < Ω_k < 0.0081

 

The greatest positive curvature is the case where Omega_tot = 1.0179

 

And that case is the smallest radius of curvature. So you get a lower bound of 13.7/sqrt(0.0179) billion lightyears.

This comes to 102.4 billion lightyears.

 

However some more recent data suggests that the figure for the Hubble radius should be 13.4 rather than 13.7. There is always some revision going on.

So we can say that the lower bound estimate is roughly 100 billion. The radius of curvature is at least that, and it might be much bigger. Indeed space could be flat, with infinite radius of curvature, not curved at all. We don't really know very much yet.

 

On page 4 it is better to use the column labeled WMAP + BAO + SN because that is based on wmap data pooled with other project's data, so is more reliable.

Edited August 26, 2009 by Martin

[Widdekind]

Rudolf v.B. Rucker (The Fourth Dimension, pg. 211) depicts a (2+0)D visualization of Spacetime, in an isotropic Universe "marred" (my word) by one mega-massive object:

 

 

Thus, 2D Flatland Space is curved, "through Hyperspace", into a shape like a lightbulb.

 

 

 

Now, if you imagine concentrating more & more mass into that mega-massive "central mass", perhaps that lightbulb-shaped Space would "shrink into itself" something like this:

 

 

And, running time in reverse, perhaps you could explain the Big Bang ???

 

Note that the radius of the "throat" stretching down towards the mega-mass approaches the Schwarzschild Radius of the primordial "reverse Black Hole"; whereas, the radius of the ballooning Space beyond that "throat" would be the Radius of Curvature of the Universe.

[Widdekind]

If the Hubble Constant

 

[math]H = \frac{1}{R} \frac{dR}{dt} = 75 \; km \; sec^{-1} \; Mpc^{-1}[/math]

 

and if the current Curvature Radius

 

[math]R \approx 100 \; Gpc[/math]

 

then can one conclude, that the "velocity of Spacetime outwards through Hyperspace" is

 

[math]H \; R \approx 25\; c[/math]

 

? That is, the "Hyper-velocity of Spacetime, thru Hyperspace, is ~25 times the Speed-of-Light ('Mach 25')" ??