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I watch this movie:

How big is the universe https://www.youtube.com/watch?v=b34gwCKIM50#t=50m00s

I can't understand where is there triangle with 180 degrees.

What exactly is this red spot on the cosmic microwave background?

Is it a base of triangle? How does it appear that this spot is smooth surface but not a concavity?

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  • $\begingroup$ The triangle is defined by the two green laser beams and the map of the CMB. Its lives ordinary 3D space so the sum of all the angles is 180 degrees. The red spot represents a hotter than average region on the sky of the CMB. Yes it is the base of the discussed triangle. It looks smooth because the map has poor resolution. With Planck (sci.esa.int/planck/…) it would look more clumpy. $\endgroup$
    – chris
    Commented Apr 5, 2014 at 18:41
  • $\begingroup$ I'm confused. How can a projection onto a flat surface and 2 laser beams not give a triangle with 180° angles? $\endgroup$
    – LDC3
    Commented Apr 5, 2014 at 19:04
  • $\begingroup$ To me, the red spot under observation appears to be 2 spots merged together. He states that the distance to the hot spot is known so that the angles can be calculated. If it is 2 spots, they could be at different distances. Besides, I thought the CMB was analyzed at the greatest distances observable. $\endgroup$
    – LDC3
    Commented Apr 5, 2014 at 19:25
  • $\begingroup$ I completelly understand the movie, but this moment and further. $\endgroup$
    – user1346
    Commented Apr 6, 2014 at 8:25
  • $\begingroup$ The video seems to refer to the first acoustic peak of the baryon-photon plasma during recombination. The corresponding angle is about 1 degree in the sky, indicating a flat universe. I'm not yet able to understand and explain it in a simple way. $\endgroup$
    – Gerald
    Commented Apr 6, 2014 at 22:41

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There are features in the CMB, the distance and size of which can be derived from cosmological models (6-parameter Lambda CDM model). The expected size of the features can be predicted in terms of angles in the sky, depending on how curved large-scale space-time is. (The features are a statistical property of tiny temperature fluctuations in the CMB indicating the largest scale of causal connection - kind of sound waves - in the baryon-photon plasma before recombination 380,000 years after the Big Bang, called first acoustic peak of the CMB power spectrum. The angle should be about 1 degree for a flat space-time, in accordance with observation.) Things aren't quite straightforward, e.g. due to the way the universe has been expanding, therefore it's done with simulations.

Angles look larger in a positively curved space-time (e.g. a sphere) than in a flat space-time (e.g. a plane) than in a negatively curved space-time (e.g. a saddle).

The curvature of space-time follows the same principles as curvature of 2-dimensional surfaces. A triangle in an Euclidean plane has an angle sum of 180 degrees, while a triangle on a sphere has an angle sum of more than 180 degrees. Hence an object of a fixed size and a fixed distance looks larger (takes a larger angle) on a sphere than on a plane.

Here a sketch of a triangle on a sphere (may be interpreted as an observer looking to an object) and a triangle of the same size - meaning the same lengths of the sides in this case - in a plane:

enter image description here

(More on non-Euclidean geometry.)

If the observable universe is positively curved, it could be a fragment of the surface of a 3-sphere resp. 3+1-de Sitter space-time. A 3-sphere is finite. If the observable universe is flat (curvature 0), it could be a fragment of an infinite Euclidean space resp. 4-dimensional Minkowski space-time. That's plausible, but not the only possible extrapolation beyond the observable universe. An almost flat observable universe could e.g. be a fragment of a huge 3-spherical universe or of a huge 3-torus. Planck data don't indicate (in a significant way) that kind of structure outside the observable universe, but extrapolations become less valid the further we go beyond the observable universe.

According to the current (Lambda-CDM) model of the universe together with the best-fit parameters, the universe is flat within the limits of measurement. Hence it's plausible (by extrapolation), but not necessary, that the universe is either huge or even infinite.

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  • $\begingroup$ It would be equally true to say that if the universe negatively curved, it could be a fragment of Minkowski spacetime; 'could be' has quite a bit of wiggle room. But the FLRW family of solutions is so much more populous than the special cases of dS/Minkowski/AdS spacetimes that I think it it is somewhat misleading to single them out, except perhaps to say that they're all maximally symmetric. Being lambdavacua, all three are observationally ruled out, since our universe contains non-negligible stress-energy besides the cosmological constant, and so they cannot be "possible extrapolations." $\endgroup$
    – Stan Liou
    Commented May 9, 2014 at 5:29
  • $\begingroup$ @StanLiou If we take the large-scale mass-energy of the observable universe as equally distributed, couldn't the resulting stress-energy tensor be re-interpreted as part of the metric, like Lambda can be re-interpreted as vacuum energy? The question has been about CMB features, and their appearent magnification/reduction by the large-scale curvature of spacetime. I've no better idea, how to keep the answer simple enough to answer the question essentially. Otherwise we've to deal with arbitrarily curved manifolds which are almost flat in the large-scale observable universe, due to mass-energy. $\endgroup$
    – Gerald
    Commented May 9, 2014 at 12:18
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    $\begingroup$ No, because any stress-energy term that's proportional to the metric tensor is going to be a perfect fluid with pressure equal to the negative of its energy density, which is not the case for the non-Λ matter in our universe. Pressure and energy density intermix between local inertial frames, and this special equality allows it to be invariant. ... The logic about extrapolating works with regard to spatial curvature only, which is the only kind you've explicitly considered in the answer. Although I'd note that the links you gave assume a more general subclass of FLRW than dS/M/AdS trio anyway. $\endgroup$
    – Stan Liou
    Commented May 9, 2014 at 20:20

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