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According to recent measurements our observable universe is roughly 93 billion light years in diameter; also it appears (according to WMAP measurements) that spacetime is flat.

Supposing space is infinite.

It seems to me that it isn't outside logical possibility that there is another observable universe completely outside of our observational range and so far away it has no appreciable effect on the curvature of our universe.

Note, I'm not using the word universe here as everything in space.

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  • $\begingroup$ Perhaps you might be interested in Tegmark's Parallel Universes, arXiv:astro-ph/0302131, in particular Level I: Regions beyond our cosmic horizon ("the uncontroversial cosmological concordance model"). $\endgroup$
    – Řídící
    Commented May 13, 2013 at 22:22
  • $\begingroup$ I'm not clear on what's being asked. How can an "observable universe" be "completely outside of our observational range?" Doesn't that mean it's observable and unobservable, which would be a contadiction? Why do you only care about this other region's effect on the curvature of our region? All interactions propagate at a maximum speed of c, including gravity, which is what propagates curvature effects (gravity is curvature). To the extent that I understand what's being asked, I think the answer is yes, cosmological event horizons exist, and 2 observers can be in disjoint observable regions. $\endgroup$
    – user4552
    Commented May 13, 2013 at 22:29
  • $\begingroup$ @BenCrowell (and Mozibur) I presume that te OP means (non-intersecting) Hubble volumes? $\endgroup$
    – Řídící
    Commented May 13, 2013 at 22:31
  • $\begingroup$ @Crowell: It's difficult to express exactly what I'm saying given I'm not aware of current terminology here. It would perhaps be easier with a diagram. I think my title says it a lot better than the body of the question. $\endgroup$ Commented May 13, 2013 at 22:38
  • $\begingroup$ @Gugg: I don't think that interpretation makes sense, since the OP uses the word "observable," and the Hubble volume is smaller than the observable volume: physics.stackexchange.com/q/12819/4552 $\endgroup$
    – user4552
    Commented May 13, 2013 at 22:38

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As noted above in comments, I'm not competely sure I understand the question. But anyway, I'll give it a shot.

The answer is model-dependent. The standard cosmological model at the moment is the Lambda-CDM model. This model has various parameters. Depending on these parameters, the spatial curvature can be positive, negative, or zero. Observation puts (model-dependent) bounds on the spatial curvature: What is the curvature of the universe? Given these bounds, we can put (model-dependent) bounds on the size of the universe: Size of the universe . We then find that the universe is much larger than our own observable region. Therefore the (model-dependent) answer to the question (as I construe it) is yes: there are other regions of the universe in which observers would have observable regions that don't intersect our observable region (and probably never will, given the acceleration of expansion).

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  • $\begingroup$ this is what I was looking for. $\endgroup$ Commented May 13, 2013 at 22:55
  • $\begingroup$ It should be noted that in a flat infinite universe, every point is the center of an observable universe. Even my observable universe may be shifted a few thousands of km from that of other users in the site. That might even be true for some large but finite non flat universes. $\endgroup$
    – Pere
    Commented May 16, 2022 at 18:35

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