Timeline for Is de Sitter space with non-zero curvature an acceptable model for the universe?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 17 at 8:59 | comment | added | OON | @P.C.Spaniel by the way the universe may have nonzero $k$ but very large scale factor i.e. curvature radius much larger than the observable universe radius. Then it will look the same regardless of the true $k$. Inflationary stage more or less guarantees that this will be the case | |
| Apr 17 at 8:52 | comment | added | OON | @P.C.Spaniel Yep, you are right. The de Sitter is special in the sense that such preferred slice is not unique but for other FRLW spacetimes this is not the case | |
| Apr 14 at 22:35 | comment | added | P. C. Spaniel | I arrived many years late, but I think maybe the question (or at least the question I have now) is: If the value of $k=-1,0,1$ depends on the coordinates, which are a gauge freedom of GR, then why do we say that we have measured experimentally that $k=0$? We can just change coordinates and make it be whatever we want. I think my tentative answer is that the matter content of the universe fixes a prefered slicing for us (namely, the slicing in which the CMB is homogeneous) and that slicing is flat ($k=0$). | |
| Sep 11, 2017 at 16:16 | comment | added | Frederic Thomas | Thank you very much! Isn't it amazing ? This is the reason I've been asking so insistingly. When I posted my principal question I would never thought that. Still I can hardly believe it. | |
| Sep 11, 2017 at 11:30 | comment | added | OON | @FredericThomas Ok, look. You can't percieve the whole space at once. You can only measure something locally (like light travelled through the whole universe but you can only detect it when it arrives to you) Then if you live in de Sitter you may interprete the same observations as coming from universe with $k=0$ or $k=\pm 1$. You can't distinguish them! | |
| Sep 11, 2017 at 10:52 | comment | added | Frederic Thomas | It's very difficult to express my key question: Is it possible on a place (moving correspondingly to achieve the desired effect i.e. k unequal zero), but of course not with v>c, (this is what I mean with causality)) inside of our universe to perceive our universe with another k than the one (k=0) we observe on earth ? | |
| Sep 11, 2017 at 9:39 | comment | added | OON | @FredericThomas What do you mean by "respecting causality"? Note that in all such coordinate systems of dS $t=\operatorname{const}$ is spacelike, $\vec{x}=\operatorname{const}$ is timelike. Absolute future of any point at $t=t_0$ (defined by the future lightcone with origin at that point) lies in $t>t_0$. | |
| Sep 11, 2017 at 9:10 | vote | accept | Frederic Thomas | ||
| Sep 11, 2017 at 9:10 | comment | added | Frederic Thomas | In the meantime I had the time to consult the literature. I understand the subject now much better. Just a last question: With a 4d change of a coordinate system with is compatible with GR (respecting causality etc.) will not be able to change from a universe from k=0 to another one k=+1 or -1 ? | |
| Aug 28, 2017 at 19:21 | comment | added | OON | @FredericThomas Well, "seems to be rather strange" is not really a good argument, isn't it? You talk about how 3d curvature change under 4d coordinate transformations. 3d curvature is not a 4d tensor. Probably what is so "strange" about all that - it's just an example that in the general relativity our beloved 3d is just an artificial auxiliary construct, just some imaginary slicing of the spacetime. Concerning matter density, you can always include cosmological constant term into the matter part like dark energy. So de Sitter spacetime is not really empty in a certain sense. | |
| Aug 28, 2017 at 7:54 | comment | added | Frederic Thomas | Thank you for the update. Actually, it still does not hit the heart of my main question, so let me be a bit provocative: It seems to be (at least for me) that changing the coordinates I can change the metrics from e.g. k=-1, to k=0 or k=+1. That's seems to be rather strange. Also I understood that the LFRW metric is based on some non-zero matter density which is apparently not the case for the de Sitter space. | |
| Aug 25, 2017 at 21:03 | comment | added | OON | @FredericThomas I tried to answer your question in the new version. Comment if something still is not understood | |
| Aug 25, 2017 at 21:02 | history | edited | OON | CC BY-SA 3.0 |
added 2194 characters in body
|
| Aug 24, 2017 at 22:15 | comment | added | Frederic Thomas | Thank you very much for this answer. I was already suspecting something like this. However, now I've even more questions. The de Sitter space is a solution of the Einstein's equations. So everything is already fixed. By which condition is the spacelike cut fixed, by more than was is already given by the field equations? Can you comment on this more ? Thanks | |
| Aug 22, 2017 at 14:00 | history | answered | OON | CC BY-SA 3.0 |