6
$\begingroup$

It is well-known that, in globally hyperbolic spacetimes, the time separation function $\tau$ (aka Lorentzian distance function) enjoys the following property: fix a point $p$ and a point $q \in I^-(p)$. Then if $\gamma : [0,1] \to M$ is a past-directed causal curve starting at $q$, the function \begin{equation} \lambda \mapsto \tau(p,\gamma(\lambda)) \qquad (*) \end{equation} is strictly increasing. A proof of this fact follows simply by noting that, in this context: (a) a maximal causal curve between $p$ and any point in $I^-(p)$ exists and is necessarily a smooth timelike geodesic; (b) due to the reverse triangle inequality, one only needs to worry about the case where $\gamma$ is an achronal null geodesic segment. The conclusion then easily follows (see e.g. the proof of Proposition 3.1 in http://arxiv.org/abs/1010.5032).

Suppose one were interested in studying (time-oriented) spacetimes with smooth boundary and suppose also that the boundary is everywhere timelike. Then one could fairly straightforwardly come up with a notion of global hyperbolicity for such spacetimes: namely strong causality plus, as usual, compactness of the sets $J^-(x) \cap J^+(y)$ (where causal and chronological curves are defined in the same way as in the case with no boundaries, say using piecewise smooth curves). This has in fact been already done and the causal theory of such spacetimes was studied by D. A. Solis in his PhD thesis (University of Miami, 2006). See also Section 2.2 in http://arxiv.org/abs/0808.3233. It turns out that globally hyperbolic spacetimes with timelike boundary are causally simple, the time-separation function is continuous, and there exist maximal continuous causal curves between any two causally related points.

To get a flavour of the kind of result that does not still hold in this category, notice that it is no longer true that if $q \in J^-(p) \setminus I^-(p)$ then every causal curve from $p$ to $q$ is an achronal null (pre)geodesic. Consider for instance (1+2)-dimensional Minkowski space with a timelike cylinder removed, a point $q$ on the boundary and a point $p$ (either on the boundary or in the interior) "behind" the cylinder. It is clear that for some such $p$ there are causal but not timelike curves to $q$, but also that none of these curves are null geodesics. On the other hand, one can prove the weaker result that if $\gamma : [0,1] \to M$ is a causal past-directed curve from $p$ to $q$ with $\gamma(0,1) \subset \mathrm{Int} \, M$ then either $q \in I^-(p)$ or $\gamma$ is a smooth null (pre)geodesic. Similarly, the "smooth geodesic" bit in remark (a) in my first paragraph clearly fails to hold here.

Having said all that (!): can anyone think of an example showing that the function defined by $(*)$ should not be expected to be strictly increasing if one allows for timelike boundaries (but under the assumption of "global hyperbolicity")? I should say that I don't necessarily believe that a counterexample exists; but it is clear IMHO that the strategy outlined in my first paragraph does not straightforwardly adapt, and I am not aware of any alternative methods.

$\endgroup$
  • 1
    $\begingroup$ 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). $\endgroup$ – Pedro Lauridsen Ribeiro Sep 29 '15 at 18:00
  • 1
    $\begingroup$ 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. $\endgroup$ – Pedro Lauridsen Ribeiro Sep 29 '15 at 18:10
  • 1
    $\begingroup$ @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? $\endgroup$ – Umberto Lupo Sep 29 '15 at 18:57
  • 1
    $\begingroup$ 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). $\endgroup$ – Pedro Lauridsen Ribeiro Sep 29 '15 at 20:52
  • 1
    $\begingroup$ 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! $\endgroup$ – Umberto Lupo Sep 29 '15 at 21:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.