The cosmic microwave background that we observe uniformly around us is usually explained by assuming that our universe is the surface of a four dimensional sphere. That way the uniformity makes sense since there is no center. My question is if this is true then what is the explanation that describes the fact that the farther we look into space, the further we look back in time. I can't perfectly picture this and see how it would coexist. Help me out.

  • $\begingroup$ Here is a crazy thing. I was always a believer that it is wrong to think of the space as a non-compact thing. That is just unimaginable. But then I started comparing space with the surface of the earth. Now, you can imagine that a non-compact thing. So, why should the space be one? I can't. So, I started asking myself, what if the space is not a non-compact R^3 but a compact set the surface of a 4-sphere like you guys are talking about. Now, that is compact. And I am happy to see that you have other reasons to claim that it is. But now I have food for your thoughts. What if the whole thing goe $\endgroup$ – user116034 Sep 9 '14 at 2:54
  • $\begingroup$ s on. The four dimension including time as the fourth is a surface of a 5-sphere? And it goes on and on and on? Think about it. It should make sense. And I fancy a way to beat speed of light by finding the fifth dimension. lol!-Nitin (Mathematician and not Astronomer) P.S.: I will love to hear what you have to say to that. $\endgroup$ – user116034 Sep 9 '14 at 2:54
  • $\begingroup$ @dotancohen Oh, shoot. I didn't check the numbers after "user". My bad; deleting comment. $\endgroup$ – HDE 226868 Sep 30 '14 at 14:59

The surface of the 4-dimensional ball (called 3-sphere) is a slice through the universe as a whole for a fixed cosmic time. This slice describes just three spatial dimensions. The observable universe is a tiny part of this 3-sphere; hence it looks flat (3-dimensional Euclidean space) up to measurement precision (about 0.4% at the moment).

Adding time makes the universe 4-dimensional. The observable part is similar to a 3+1-dimensional Minkowski space-time. The universe as a whole may be a de Sitter space-time. A de Sitter space-time is the analogon of a sphere ( = surface of a ball), embedded in a Minkowski space, instead of an Euclidean space, but it's not literally a sphere.

If time would be taken as an additional spatial dimension, the de Sitter space would ressemble a hyperboloid of revolution, if embedded in a 5-dimensional hyperspace. The difference to an Euclidean space is due to the different definition of the distance: In a 4-dimensional Euclidean space the distance between two points is defined by $l = \sqrt{\Delta x^2+\Delta y^2+\Delta z^2+\Delta t^2};$ for a 3+1-dimensional Minkowski space it's $l = \sqrt{\Delta x^2+\Delta y^2+\Delta z^2-\Delta t^2}.$ For simplicity the speed of light has been set to $1$.

This model of the universe as a whole can hold, if it actually originated (almost) as a (0-dimensional) singularity (a point); but our horizon is restricted to the observable part, so everything beyond is a theoretical model; other theoretical models could be defined in a way to be similar in the observable part of the universe, but different far beyond, see e.g. this Planck paper.

  • $\begingroup$ But isn't it necessary for us to view our universe as the surface of a 4-D sphere in order to explain the CMBR appearing uniformly in all directions? That is the only way we would be seeing this background radiation in all directions, uniformly regardless of our position in space. $\endgroup$ – user3138766 Apr 16 '14 at 2:10
  • $\begingroup$ And by the way does that mean the surface of the 3 sphere is 3 dimensional? $\endgroup$ – user3138766 Apr 16 '14 at 2:12
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    $\begingroup$ A 3-sphere can be seen as the surface of a 4-ball. The 4-ball is 4-dimensional; its surface, the 3-sphere, is 3-dimensional, and has no surface. $\endgroup$ – Gerald Apr 16 '14 at 2:14
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    $\begingroup$ That's pretty intense! So our universe itself would be this 3-D surface of a 4-D sphere or some other shape possibly? Can you see how hard this is to visualize, I'm trying to see it in my head and the question that keeps popping up is, if the surface is 3-D then how can it be a surface? $\endgroup$ – user3138766 Apr 16 '14 at 2:16
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    $\begingroup$ Almost perfect isotropy of the CMB is explained in current models by inflation, eliminating amost any primoridial heterogeneities. -- Thinking about the universe as the surface of a 4-ball is probably the easiest way. But no significant difference to the infinite Euclidean model could be detected thus far; hence any non-trivial topology (including the 3-sphere) - if present - must most likely be beyond the horizon of the recombination (CMB). $\endgroup$ – Gerald Apr 16 '14 at 2:52

I believe your confusion is from combining two popular simplifications of our Universe. As we look further away we see further back in time because of the finite speed of light. So these distant objects are also evolving with us, but that light hasn't hit Earth yet.

It might help to know that the observable universe may only be 14 Gyr old, but its radius is 46 Gly, not 14 Gly. If the speed of light was infinite, then we wouldn't be observing back in time as we look at more distant objects.

  • $\begingroup$ Could you please expand a bit on the latter paragraph of your answer? I realize where you're going with it, but I'm not sure all the readers will, and judging by the lack of upvotes, that seems to be the case. Cheers! $\endgroup$ – TildalWave Sep 10 '14 at 4:43

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