Some questions about causal structure of space-time. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T01:47:22Z http://mathoverflow.net/feeds/question/47086 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47086/some-questions-about-causal-structure-of-space-time Some questions about causal structure of space-time. Anirbit 2010-11-23T12:13:51Z 2010-11-23T13:09:12Z <ul> <li><p>Let $(\hat{M},\hat{g})$ be the conformal compactification of a space-time $(M,g)$. Let $I^+$ be the conformal null infinity and $J^{-}(I^+)$ be its causal past. Then the spacetime will be called "asymptotically strongly predictable" if there exists a subset $\hat{V}$ of $\hat{M}$ such that $(\hat{V},\hat{g})$ is globally hyperbolic and contains the closure in $\hat{M}$ of $M \cap J^{-}(I^+)$. Define as a "black hole" the region $B=M /\ J^{-}(I^+)$. </p> <ol> <li>I think from the above it should imply that there cannot be a causal curve which starts in the region $B$ and enters $M \cap \hat{V}$. Is this correct and if yes then how does one prove it? </li> </ol></li> </ul> <p>(In specific examples what happens is that such a curve cannot remain always causal if it has to do such a journey. But I guess this not a good way of thinking since the restriction is purely topological) </p> <ol> <li><p>Under the above conditions when does it also follow that $(M \cap \hat{V},\hat{g})$ and/or $(M\cap \hat{V},g)$ is globally hyperbolic? </p></li> <li><p>What is the meaning of "trace of a terminally indecomposable past set" ? Is there any special significance if a compact set happens to be the trace of a terminally indecomposable past set of a maximal future directed time-like curve? (Reading around I get the impression that such compact sets somehow specify physically reasonable initial conditions)</p></li> </ol> http://mathoverflow.net/questions/47086/some-questions-about-causal-structure-of-space-time/47091#47091 Answer by Willie Wong for Some questions about causal structure of space-time. Willie Wong 2010-11-23T13:09:12Z 2010-11-23T13:09:12Z <p>1st statement is false. You need to add "future directed", else a past causal curve can of course leave the black hole region, since time-reversed, it is just a white-hole. For the correct statement, the proof is immediate following the usual causal relations of Penrose and Kronheimer: if $\exists$ such a causal curve, there must $\exists p,q$ such that $p\in B$ and $q\in \hat{V}$ with $q \in J^+(p)$. Which is equivalent to $p \in J^-(q)$. Now using that the causal relations form an ordering, you have that </p> <p>$$p \in J^-(q), q\in J^-(I^+) \implies p \in J^-(I^+)$$</p> <p>contradicting the assumption that $p\in B$. (As a side remark, this is very, very basic stuff. If the question were just about this fact, I would've voted to close. You should read E. H. Kronheimer and R. Penrose (1967). On the structure of causal spaces. Mathematical Proceedings of the Cambridge Philosophical Society, 63.)</p> <p>(Second side remark: I don't think the word topological means what you think it means. Pure topology is not enough to contrain causal structure, beyond some trivial things about Euler characteristic and completely compact space-times.)</p> <hr> <p>For the second question (labeled 1 again by the Markdown software), it is important to remember that the causal structure of space-time is conformally invariant, and that global hyperbolicity is an <strong>intrinsic notion</strong> rather than extrinsic. To say it another way, you need to remember that </p> <ol> <li><p>If $(V,g)$ is a globally hyperbolic space-time domain, then for any conformal change of metric $g\to \hat{g}$, $(V,\hat{g})$ is also globally hyperbolic. (This is immediate using the Cauchy surface definition, since conformal changes preserve causal relations; using the compact causal diamond definition, you need to remember that continuous functions attain its maxima and minima on a compact set.) </p></li> <li><p>If $(V,g)$ is a globally hyperbolic space-time domain, and let $\phi$ be an isometric embedding of $(V,g)\to (M,\hat{g})$, then, the restricion $(\phi(V),\hat{g})$ is also globally hyperbolic. </p></li> </ol> <p>So, $\hat V$ embeds into $\hat V \cap M$ isometrically by the identity map. And $(\hat V, \hat g)$ is globally hyperbolic, so by the above two points its image must be globally hyperbolic. (By the way, the above two observations are trivial consequences of the definitions for global hyperbolicity. You should be able to prove them yourself.)</p> <hr> <p>For your third question (as a side remark: the question is actually pretty poorly posed. If you are going to ask about terminology, please at least provide a reference on which paper it is in which you found the phrase that is confusing you; my crystal ball tells me that you are asking about Christodoulou's 1999 CQG paper "On the global initial value problem and the issue of singularities", but most other people won't have a convenient divination device in their office.), it helps to note that the original phrase is about</p> <blockquote> <p>the trace of a terminally indecomposable past set on a space-like hypersurface.</p> </blockquote> <p>And as such it means the same thing generally meant when you take the trace of <em>any</em> object onto a hypersurface: as a set you want to consider the intersection of whatever object you are talking about with the hypersurface, and you want to inherit any geometric structure by pushing forward with the restriction operator. </p> <p>In context the condition that "the trace of a TIP on a space-like hypersurface is compact" means that the intersection of the TIP with the space-like hypersurface is a compact set in the induced topology of the hypersurface. </p> <p>The intuition is that in Minkowski space (or any small perturbations of it), the only TIP associated to maximal time-like geodesics is the whole space-time. And in particular, its trace on any Cauchy hypersurface <strong>cannot</strong> be compact. (For maximal time-like curves, you can get particle horizons if the curves become asymptotically null, but it is easy to see that those TIPs correspond to a half-space in Minkowski space that lies below a null hyperplane, so the same conclusion holds.) In general, if you have a compactification of space-time with a future time-like infinity (the terminus of all maximally extended, complete future time-like curve), then the TIP associated to it must have non-compact trace on any Cauchy hypersurface. </p> <p>So in fact, I think your parenthetical impression is completely wrong. That a TIP has compact trace should suggest to you that the maximal time-like curve your are looking at is incomplete, which should suggest to you that its terminus is a "singularity". (You should compare this picture with the usual picture in the Penrose singularity theorem.) Which is why Christodoulou formulated his version of weak cosmic censorship conjecture in terms of such sets. (It would also do you good to work through the paper of Geroch, Kronheimer and Penrose, "Ideal Points in Space-time", Proc. Roy. Soc. London (1972). )</p>