Timeline for Time-separation function on "globally hyperbolic" spacetimes with everywhere timelike boundary
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16 events
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| Oct 16, 2015 at 12:36 | comment | added | Clemens Sämann | I am not sure if that is what you are after but if the metric is just continuous and the spacetime globally hyperbolic, then maximal curves still exist: Prop. 6.4, p. 23 in link.springer.com/article/10.1007/s00023-015-0425-x (or arxiv.org/abs/1412.2408). | |
| Sep 29, 2015 at 22:42 | comment | added | Pedro Lauridsen Ribeiro | Hence, one has to see if the available results on causality for continuous and / or $C^1$ metrics suffice to yield a characterization of maximal causal curves on the double, and use reflection across the boundary to complete the job. | |
| Sep 29, 2015 at 22:41 | comment | added | Pedro Lauridsen Ribeiro | I believe the loss of causal convexity in the double is not such a big deal, though. If you have a (say, locally Lipschitz) causal curve in the double linking two points of the original manifold, you can reflect back the pieces in the other half. The result, it seems to me, is another locally Lipschitz curve (probably with more breaks) with the same Lorentzian arc length since reflection across the boundary in the double is an isometry. The big problem is really the lack of regularity of the double's metric across the boundary. | |
| Sep 29, 2015 at 22:26 | comment | added | Umberto Lupo | Of course! The problem is that above I repeatedly wrote "causally convex" when I was really thinking "causally compatible". But then again I'm not sure why I ever thought causal compatibility could help in the first place. Sorry for the waste of time. | |
| Sep 29, 2015 at 22:18 | comment | added | Pedro Lauridsen Ribeiro | Think, for instance, of the causal diamond $U=\{(t,x)\in M\ |\ |t|<1-|x|\}$ in the 2-dim. half-Minkowski space-time $M=\{(t,x)\ |\ x\geq 0\}$. It is obviously geodesically convex and causally convex, but its image in the full 2-dim. Minkowski space-time (which is the double of $M$) is not causally convex since one can slightly deform any timelike curve with a boundary segment into a timelike curve whose corresponding segment has its interior inside the other half of the double. | |
| Sep 29, 2015 at 22:12 | comment | added | Umberto Lupo | I'm not sure why you say that "the image of U won't be causally convex, no matter how small, precisely because the boundary is timelike". My intuition would be that e.g. taking the double of the spacetime $\{ (t,x) : x \geq 0\}$ (or of similar subspacetimes of higher dimensional Minkowski spacetimes) would not cause the issue you raise. Though I of course agree that generically there would be issues. | |
| Sep 29, 2015 at 22:04 | comment | added | Pedro Lauridsen Ribeiro | Well, you can use the double of the space-time manifold, but in this case one has two problems: (1) the metric in the double extending the original metric is at best continuous (it is at least $C^1$ if the boundary is totally geodesic), and (2) even if it is smooth, the image of $U$ won't be causally convex, no matter how small, precisely because the boundary is timelike. It is possible to generalize results from causality theory to space-times with rough metrics, but I'm not sure how much of the theory survives with metrics this rough (specially if the boundary is not totally geodesic). | |
| Sep 29, 2015 at 21:56 | comment | added | Umberto Lupo | I also have the feeling that things could work if it holds that, locally around every point $p$ of the boundary, there exists a neighbourhood $U$ and an isometric embedding of $U$ into a spacetime without boundary $N$, such that $U$ is mapped to a causally convex subset of $N$. But this hardly seems to me like the most helpful of criteria! | |
| Sep 29, 2015 at 21:47 | comment | added | Umberto Lupo | Good point, @Pedro. Thanks. Now if only one could argue more generally... :-) | |
| Sep 29, 2015 at 20:52 | comment | added | Pedro Lauridsen Ribeiro | I believe the usual characterization still holds if the space-time is strongly causal and the boundary is totally geodesic. In a geodesically and causally convex neighborhood, two causally related points are connected by a unique causal geodesic segment which must be the unique maximal causal segment connecting both points, whose interior either belongs to the interior (if one of the endpoints also is) or to the boundary (if both endpoints also do). See, for instance, Propositions 4.32, 4.33 and Theorem 4.34, pp. 165-166 of Beem-Ehrlich-Easley's "Global Lorentzian Geometry" (2.ed., CRC, 1996). | |
| Sep 29, 2015 at 20:29 | comment | added | Pedro Lauridsen Ribeiro | Ah yes, actually I meant $\gamma(0)=p$ and $\gamma(1)=q$ there, so it fits your context better. Unfortunately, I can no longer edit that comment... | |
| Sep 29, 2015 at 18:57 | comment | added | Umberto Lupo | @Pedro, thanks for the comments. With your second comment, you have hit precisely the technical core of my question, which I was aware of but should have exposed myself in the question! Also, a small thing: I think you perhaps meant "future-directed" in your first comment? | |
| Sep 29, 2015 at 18:10 | comment | added | Pedro Lauridsen Ribeiro | I'm not aware of any characterization of maximal causal curves in space-times with timelike boundary. At least, once again due to the fact that segments of maximal causal curves are also maximal, one has that any segment with nonvoid interior belonging either to the interior or to the boundary should be a causal (pre)geodesic in the submanifold to which it belongs. However, in principle there is nothing preventing the intersection of the curve with (say) the boundary from happening in a pretty nasty subset of parameters in $[0,1]$, so it's hard to say more without a more detailed analysis. | |
| Sep 29, 2015 at 18:00 | comment | added | Pedro Lauridsen Ribeiro | Your question essentially reduces to the following: in a time-oriented Lorentzian manifold $(M,g)$ with timelike boundary, given two chronologically related points $q\ll p$, is it possible to have a maximal, past-directed causal curve $\gamma:[0,1]\rightarrow M$ such that $\gamma(0)=q$, $\gamma(1)=p$ and $\gamma|_{[\epsilon,1]}$ is a(n achronal) null geodesic for some $\epsilon\in(0,1)$? It is clear that strict monotonicity of $(*)$ holds iff the answer is negative, since any segment of $\gamma$ must be maximal (non-strict monotonicity always holds by the reverse triangle inequality). | |
| Sep 29, 2015 at 12:46 | history | edited | Umberto Lupo | CC BY-SA 3.0 |
Changed "geodesic" to "smooth geodesic" in last sentence of second-to-last paragraph.
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| Sep 10, 2015 at 17:15 | history | asked | Umberto Lupo | CC BY-SA 3.0 |