Edgar, if Omega=1 the universe wouldn't have the geometry of a hypersphere, 3D space would be "flat"--it would be more like a "hyperplane". Only if Omega is greater than 1 would it have the positive curvature of a hypersphere (and if Omega is less than 1 space would have a hyperbolic geometry with negative curvature, whose 2D analogue looks something like a saddle--see the 3 basic types of geometry shown at http://www.astro.ucla.edu/~wright/cosmo_03.htm )
Also, the Friedmann-Lemaître-Robertson-Walker metric which Omega appears in makes an important simplifying assumption: instead of a bunch of localized stars and galaxies, it treats all of space as being filled with a "perfect fluid", and it assumes there is a way to define simultaneity in a way that results in a bunch of 3D slices where the density of the fluid is perfectly uniform in each slice (though it decreases from earlier slices to later slices as the universe expands). So even if you decide to define your "present" in terms of such a slicing, it's obvious that in the real universe there are variations in density of matter, so this doesn't give us a guide as to what the correct definition of simultaneity would be in the neighborhood of a given localized clump of matter. In particular, there are many different coordinate systems with different definitions of simultaneity that are used by physicists to describe the neighborhood of a black hole--Schwarzschild coordinates, Eddington-Finkelstein coordinates, and Kruskal-Szekeres coordinates being some of the most commonly-used ones--so what experiment would you propose for deciding which definition of simultaneity is the "correct" one? Jesse On Thu, Jan 30, 2014 at 5:51 PM, Edgar L. Owen <[email protected]> wrote: > Liz, > > Good question. Give me the formula to get the radius of a 4-dimensional > hypersphere from the curvature and I'll tell you. I asked for this already > and Brent gave me a formula that seems to make some extraneous assumptions. > The problem is that Omega doesn't simply seem to be the curvature in the > ordinary sense of hyperspherical geometry so some sort of initial > conversion of Omega needs to be made first and I don't know what that must > be... > > Edgar > > > > On Thursday, January 30, 2014 3:37:20 PM UTC-5, Liz R wrote: >> >> On 31 January 2014 04:03, Edgar L. Owen <[email protected]> wrote: >> >>> Richard, >>> >>> I've already answered this same questions on multiple occasions. >>> >> >> :-) >> >>> >>> There isn't any direct mathematical relationship so far as I can see >>> though we should be able to compute p-time from Omega, the curvature of the >>> universe. >>> >> >> Omega = 1 (http://en.wikipedia.org/wiki/Curvature_of_the_universe) to a >> very good approximation, according to the latest measurements - so what's >> p-time? >> >> -- > You received this message because you are subscribed to the Google Groups > "Everything List" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to [email protected]. > To post to this group, send email to [email protected]. > Visit this group at http://groups.google.com/group/everything-list. > For more options, visit https://groups.google.com/groups/opt_out. > -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To post to this group, send email to [email protected]. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/groups/opt_out.
