|Jun17-08, 08:23 AM||#1|
Radius of curvature
Is the radius of curvature of a Universe-model constant through-out the Universes' lifetime? Or does this have to be adjusted depending on the redshift we are looking at?
|Jun18-08, 06:01 PM||#2|
Niles maybe I can take a shot at that. If you take the universe as a balloon with galaxys on the skin and enlarge it the radius changes and varies with the distance the light has traveled. That is a 3D solution to a 4D problem. Now there is no radius as there is no center as such but you can say to the center of my brain the radius changes with the red shift. Please don't curse me as I do think simple thoughts.
|Jun18-08, 06:25 PM||#3|
If the true value of Omega (at the present moment) is 1.01 then the average radius of curvature (using the standard cosmology model) is about 130 billion lightyears.
The radius of curvature, if it is finite, is increasing at the rate of 1/140 of a percent every million years. That is, very slowly.
A perfectly flat universe would have infinite radius of curvature. We cannot exclude this case. When Omega is measured, you get an error bar of uncertainty, around 1.01. If, for example, it turned out that Omega was exactly 1.0, then the universe would be spatially flat and have infinite radius of curvature. It's possible. The data is not good enough to rule that out.
But if you choose to imagine that the universe is NOT exactly spatially flat, and that it has a slight overall positive curvature (like with the balloon analog) then spatially it is the 3D analog of a 2D sphere surface. And then, with the kind of best estimate 1.01 figure for Omega the radius of curvature will turn out to be around 130 billion.
That means the circumference is about 800 billion. If you could instantaneously travel 800 billion lightyears then at this very moment you could go off in a straight line and end up back in the same place. In practice such a circuit would be impossible because the universe is all the while expanding and you can only travel finite velocity.
In case you want to do the numbers. the presentday Hubble distance is c/H = 14 billion lightyears.
You start with the current Hubble parameter H and you calculate that c/H
Then you divide that by sqrt(Omega - 1)
Taking Omega to be 1.01, that is sqrt( 0.01) = 0.1
So you are dividing the Hubble distance by about 0.1. That should give 140 billion
but when I include a little more accuracy and do it more carefully, I get closer to 130 billion.
(the best fit Omega is more like 1.011, the presentday Hubble distance is not quite 14 billion etc etc)
|Jun19-08, 02:08 AM||#4|
Radius of curvature
Thanks all - very nice replies indeed!
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