Timeline for Would a closed universe with special relativity violate causality? Does the universe have to be simply connected?
Current License: CC BY-SA 3.0
14 events
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| Oct 2, 2016 at 16:03 | comment | added | Jim Conant | Jeffrey Weeks wrote an expository note about this as well: jstor.org/stable/… | |
| Oct 2, 2016 at 15:46 | answer | added | Zurab Silagadze | timeline score: 3 | |
| Apr 2, 2013 at 20:52 | comment | added | Brian Rushton | Thanks you, this answers my question completely (especially the linked article!). | |
| Apr 2, 2013 at 18:48 | comment | added | Ricardo Andrade | Aaron's comment seems to fully resolve the question. Stated another way, the two "reference frames" that the question mentions are not equivalent, in the sense that there exists no isometry of $S^1\times\mathbb{R}$ (with the obvious Lorentz metric) which takes one of the geodesics (the stationary Earth twin) to the other one (the twin on the spaceship). This is just a consequence of the fact that the only isometries of $\mathbb{R}^2$ (i.e. elements of the 2-dimensional Lorentz group) which descend to a diffeomorphism of $S^1\times\mathbb{R}$ must be translations in $\mathbb{R}^2$. | |
| Apr 2, 2013 at 18:28 | comment | added | Igor Khavkine | A very old question. The answer, as Aaron suggests, is standard. See, for instance, dx.doi.org/10.1119/1.16373. | |
| Apr 2, 2013 at 18:22 | history | edited | Brian Rushton | CC BY-SA 3.0 |
Clarification
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| Apr 2, 2013 at 18:02 | comment | added | Ryan Budney | There's no relationship between a universe being `closed' and it being simply connected. And your edit about "poorly behaved" quadratic forms doesn't make sense to me. | |
| Apr 2, 2013 at 17:10 | comment | added | Aaron Bergman | Was a little too quick there -- there are closed spacelike curves; just not the first things that come to mind. | |
| Apr 2, 2013 at 16:56 | comment | added | Aaron Bergman | This should probably be closed because it's pretty standard. First of all, you don't really mean closed geodesics (which would be examples of closed timelike curves, ie, time travel). You want a universe with some foliation of spacelike compact surfaces. The easiest example is a cylinder: S^1 x R. For the twin paradox to hold, you would need the Lorentz group to act isometrically on this space for a given flat metric. But, it's easy to see that the full Lorentz group won't act here. In fact, the choice of a metric defines for you a distinguished frame, solving the 'paradox'. | |
| Apr 2, 2013 at 16:43 | comment | added | Steven Landsburg | If two twins follow different paths to the same point, their odometers will in general differ. So will their clocks. Why is either of these paradoxical? | |
| Apr 2, 2013 at 16:30 | comment | added | Misha | Brian: I could not quite understand the geometric question: What do you mean by "poorly behaved"? Are you asking if nontrivial fundamental group of the space-time implies existence of a closed geodesic, provided that the space-time is geodesically complete and locally-flat? Then the answer is positive. | |
| Apr 2, 2013 at 16:21 | history | edited | Brian Rushton | CC BY-SA 3.0 |
Added small change about Lorenz group
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| Apr 2, 2013 at 16:16 | history | edited | Brian Rushton | CC BY-SA 3.0 |
Reformulating mathematically
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| Apr 2, 2013 at 16:06 | history | asked | Brian Rushton | CC BY-SA 3.0 |