Given two points in Lorentzian spacetime $p,q\in M$, is it true that there is only a unique null geodesic (up to affine reparametrization) that connects that the two points?

On the one hand, it seems that the answer is "yes", since null geodesics are obtained from the geodesic equation which have unique answers given initial data and all I need to fix is the endpoint $q$ if initial data starts at point $p$. On the other hand, I can imagine a case of, say, gravitational lensing, where null rays from a source (point $p$) behind a gravitational lens are bent along two different null directions and reach the Earth (point $q$), which seems to imply that the answer is "no".

I am trying to understand the physics so I am avoiding full dive into uniqueness proof right now, unless that's the only way to go.

Edit: I should have been more careful in excluding the standard counterexamples, such as those with spherical spatial topology. If I have to choose a sufficiently good restriction, it would be to let $M$ be Schwarzschild spacetime --- in particular, the exterior of the Schwarzschild black holes or spherical stars. Naively, I expect this to be non-unique but I am not sure if the solution is best dealt with using caustics and stuff used for singularity theorems.

  • $\begingroup$ Have you heard of Einstein crosses and Einstein rings? $\endgroup$ – Qmechanic Feb 25 '19 at 21:19
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    $\begingroup$ Not even close. For example, if space is spherical there are infinitely many null geodesics between opposite poles. $\endgroup$ – knzhou Feb 25 '19 at 22:52
  • $\begingroup$ @Qmechanic Typically the light arriving to us from an Einstein ring or cross, will not have been emitted from the same spacetime event. Differently put, the Shapiro delay for different images in an Einstein cross will differ. $\endgroup$ – mmeent Feb 26 '19 at 13:32
  • $\begingroup$ Right, not necessarily emitted from the same spacetime event. $\endgroup$ – Qmechanic Feb 26 '19 at 17:55

No, it is not. Even in the Schwartzschild metric, for example, given two point that only differ for the time coordinate, you can easily find two geodesics, one being the rotation around the BH, the other one being having an initial velocity pointing outwards.

It is true that the geodesic equations have one and only one solution for a given "initial data", but when you're saying $p$ and $q$ you are only taking about coordinates, not velocities, thus aren't setting proper "boundary conditions"

  • $\begingroup$ What is D&D enthusiast? :-) $\endgroup$ – magma Feb 26 '19 at 9:23
  • $\begingroup$ It should mean that I love to play Dungeons and Dragons! I'm not sure if it was supposed to be in the bio, I put it there anyway $\endgroup$ – Mauro Giliberti Feb 27 '19 at 8:39
  • $\begingroup$ It is perfectly ok Mauro I just did not recognise the acronym $\endgroup$ – magma Feb 27 '19 at 18:01
  • $\begingroup$ I'm relieved, haha. I'm still kind of a newbie $\endgroup$ – Mauro Giliberti Feb 28 '19 at 14:40
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    $\begingroup$ @MauroGiliberti so is it unique if I fix the null vector tangent to null geodesic $\gamma$ at $p$? If I understand correctly, what you are saying is that I have not specified sufficient initial data to fix uniquely the geodesics if I only have $\gamma(p)$ and $\gamma(q)$. Maybe you could incorporate this aspect in your answer? I think I almost get it. $\endgroup$ – Everiana Mar 15 '19 at 3:46

The geodesic equation is second order , so you need an initial position and an initial first derivative in order to get an unique solution. If you have an initial and final position then you may get an unique solution if the two positions are very close to each other , in practice when you can ignore curvature effects. In other words, you approximate space time locally as a flat space-time so you can join any two events by a unique geodesic.

For some manifolds like the sphere, this is not possible at all. Even if two Points are close together, there are always at least two geodesics connecting them.


Given two points in a Lorentzian spacetime $p, q\in M$, there will generically be no null geodesic connecting the two points. A null geodesic will only exist of $q$ is an element of the lightcone of $p$, $L_p$.

In a sufficiently small neighbourhood of $p$, $L_p$ is actually a cone, and for any $q\in L_p$ inside this neighbourhood the null geodesic connecting $p$ and $q$ will be unique. However, further away from $p$ it is possible that $L_p$ self-intersects. These intersections are known as the "caustics" of $p$. If $q$ lies on these caustics, then there are two (or more) null geodesics connecting $p$ and $q$.

In special cases it is even possible for there to exist $p$ and $q$ for which there are infinitely many null geodesics connecting $p$ and $q$. As an easy example, consider the Lorentzian spacetime $\mathbb{R}\times S^2$ (with the obvious Lorentizan metric). By symmetry it is easy to see that if $p$ and $q$ are on opposite poles of the $S^2$ and $q\in L_p$, then there are infinitely many null geodesics connecting $p$ and $q$.

In general however, note that since the caustics of $p$ are self-intersections of $L_p$, the form at most a measure zero subset of $L_p$. Hence, for a generic point $q\in L_p$ there will be a unique geodesic connecting $p$ and $q$.


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