Take a globally hyperbolic topologically trivial spacetime $M \cong \mathbb{R} \times \Sigma$, $\Sigma \cong \mathbb{R}^{(n-1)}$. Given $p, q \in M$, such that there exists a future-directed null geodesic $\ell$ between $p$ and $q$, is this equivalent to the condition that $p \nearrow q$, an horismos relation ($q$ is on $p$'s lightcone), ie $p \leq q$ and $p \not \ll q$?

This is fairly obviously not true for say, totally vicious spacetimes, where $p \ll p$ for all points, ie every point has a closed timelike curve (there isn't even any horismos to be on), and for a globally hyperbolic example, the Minkowski cylinder $\Sigma = S$, where a null geodesic will connect to a point in $p$'s own lightcone after one turn. On the other hand, this is certainly true of Minkowski space, as well as any spacetime related to it by a Weyl transform.

This would be equivalent to proving that, if $q \in I^+(p)$, then there is no null geodesics linking $p$ to $q$ which, given the properties of globally hyperbolic spacetimes, means that there is a maximizing timelike geodesic linking the two points. If $q = \exp_p(v)$ for some $v$, this would be correct (since $\exp_p I^+(0, \mathbb{R}^n) = I^+(p, M)$), but that would be assuming that $\text{Im}(\exp_p) = M$ for such a spacetime, which I am not sure is correct even for such a benign example.

Is such a thing true and if so how to show it?

  • 2
    $\begingroup$ Can you add a good quickstart guide to what a "horismos" is, for the befuddled non-GRists out there? $\endgroup$ Sep 12 '19 at 11:42
  • 1
    $\begingroup$ Please also explain “totally vicious spacetimes”. $\endgroup$
    – G. Smith
    Sep 12 '19 at 16:19
  • 2
    $\begingroup$ And could you explain notation like $\not \ll$ in this context? $\endgroup$
    – user4552
    Sep 12 '19 at 19:28

This is not true in general. As an example take a spherically symmetric (regular) ultracompact object's spacetime. That is, an spherical symmetric object that is compact enough to fit inside its own lightring, but without a horizon. The upshot is that we end up with a spacetime that is Schwarzschild outside some radius smaller than the lightring, and has some regular matter filled region inside such that the spacetime remains topologically trivial.

Expressed in Schwarzshild coordinates, a line with constant spacial coordinates on the lightring radius will be timelike. It is immediately clear that there are pairs of points on this line that are also connected by a null geodesic going around the lightring.


A lightlike geodesic does not have maximal Lorentzian length if it contains a pair of conjugate points. So if the lightlike geodesic $\sigma\colon [0,1]\to M$ contains a pair of conjugate points in its interior, the endpoints $p$ and $q$ are connected by a timelike curve and so are not horismotically related. Thus you are asking if it is true that a topologically trivial spacetime can have lightlike geodesic with conjugate points. The answer is affirmative, it is sufficient to consider a Riemannian manifold $(\Sigma, h)$, where $\Sigma$ has topology $\mathbb{R}^2$ (the higher dimensional case is analogous), that admits geodesics with conjugate points. Then your direct product spacetime has also lightlike geodesics with conjugate points. As for $(\Sigma, h)$ consider a flat 2 dimensional world and now bump the landscape by introducing a kind of mountain. Intuitively you can adjust the new geometry so as to have conjugate points (you can connect points on opposite sides of the mountain using different paths, so at least you have cut points that are sufficient for your purpose of spoiling length-minimality of geodesic in the Riemannian base and Lorentzian length-maximality of causal geodesics in the product).

  • $\begingroup$ Quite a coincidence since all of these questions came about from reading GR synchronization, which included a few papers of your own! $\endgroup$
    – Slereah
    Oct 5 '19 at 8:32

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.