So of course when stuff falls into black holes, the geodesic for anything ends in that singularity. However, isn't it technically true that a light ray that originates from the sun and then hits the Earth and then the absorption of that light ray by matter count as the end of that geodesic? So singularities aren't the only places geodesics end right because light rays can also terminate on absorption with other matter right?


4 Answers 4


Do not confuse geodesics (the abstract geometrical things) with the paths of particles. Particles follow geodesics, but the fact that a particle is created at some point and destroyed at another doesn't mean that the geodesic starts and ends.

For example, on the surface of the Earth a great circle is a geodesic. A plane may take off, follow a great circle for some time, and then land again. The path of the plane follows the geodesic, but is only a part of the geodesic, not the whole.


This doesn't seem to be so because geodesics are an intrinsic property of space time and not simply the path of a light ray or quantum. Indeed we might say that the light is no longer traveling along a geodesic, but that says nothing about the existence of the geodesic itself. Except in regions of powerful gravitation, the geodesics of space time will not be altered, even if individual particles, quanta, etc. are.

  • $\begingroup$ Your word choice "property" implies it, but: It should be noted that geodesics are not anything that "exists" as a separate entity, even though we may paint lines on paper to visualize that property. Very similar to longitudes and latitudes on a globe, indeed, which do not cause the ocean to flow backwards in time somewhere in the Pacific ;-). $\endgroup$ Commented Mar 5 at 16:15
  • $\begingroup$ @Peter-ReinstateMonica Good point! Geodesics exist as an intrinsic mathematical property of space-time, however, they cannot be separated out apart from it. We cannot fill a test tube with some "geodesic stuff". $\endgroup$ Commented Mar 5 at 17:07

First let me clarify an important point about black holes. In the case of realistic, rotating black holes, it is not at all true that all incident bodies and light rays necessarily end up in the singularity. No, that's a common, widespread misrepresentation. In other words, most geodesic orbits do not meet the singularity. Neither timelike geodesics nor lightlike geodesics. Translated into concrete terms, this means that if a test body or spacecraft enters the event horizon, it is not at all necessarily destined to enter the singularities and be destroyed. It depends on the properties of Kerr spacetime (and Kerr-Newman spacetime), on what kind of geodesics they form. In fact, careful analyses show that very specific conditions must be met for geodesics to end up in the singularity; ergo, the vast majority of possible geodesic and other orbits avoid the ring singularity.

Why is it so common that all incident bodies are necessarily destroyed in the singularity? Probably because of the Schwarzschild spacetime, the non-rotating, static black hole. That's really the case. But the Kerr spacetime, the rotating black hole, is radically different, what I just wrote is true. Scientists have known these things since the late 1960s, yet the misinterpretation persists today. Here is one of the key articles, from 1968, that revealed how geodesics really approach and avoid the singularity. (Chapter 3 of the article deals with these issues)

Carter paper about Kerr spacetime

Now let's get to answering the question. Geodesics (for both material bodies and photons) are the orbits in a given spacetime that are freely moving. So, if they are only affected by gravity, i.e. the curvature of spacetime, they move on geodesics. If they are subject to any other force, interacting with other material bodies, they will be deviated from the geodesic. In other words, if a ray of light is reflected by a surface or absorbed by a body, its motion will be diverted from the geodesic path it was on and it will be on a different world line. Geodetic paths are properties of spacetime itself, not the real life paths of bodies or photons. So the beam of light in question only moves geodesically until it reaches the point where it interacts with bodies. The theoretically calculated geodesic path continues and does not end at the reflecting or absorbing body. But these bodies prevent the beam of light from continuing along that geodesic.

The geodesics do not describe what specific accidents happen to real bodies and photons. They describe the trajectories that spacetime theoretically imposes on things that move freely, obeying only the curvature of spacetime, without obstacles.

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    $\begingroup$ Not my field, but doesn't the mass inflation instability kill any possibility that Kerr is a realistic description past the inner horizon? $\endgroup$
    – Sten
    Commented Mar 5 at 7:42
  • $\begingroup$ In the case of realistic rotating black holes the inner (Cauchy) horizon becomes singular, meaning that all geodesics that enter the black hole will hit this singularity. $\endgroup$
    – TimRias
    Commented Mar 5 at 11:23
  • $\begingroup$ No. In the Kerr and Kerr-Newman spacetimes the only true curvature singularity is the ring singularity, which I wrote about avoiding. Neither the outer nor the inner horizon is a curvature singularity. What spacetime is it that you claim it is? @TimRias $\endgroup$ Commented Mar 5 at 13:06
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    $\begingroup$ The inner horizons of Kerr and Kerr-Newman are known to be unstable, and become singular with even the slightest perturbation. The interior of Kerr most certainly is not a "realistic" solution, but rather a pathological outlier due to its symmetries. $\endgroup$
    – TimRias
    Commented Mar 5 at 13:27
  • $\begingroup$ Can you point me to a source for this? I note that the calculation way of perturbations of Kerr spacetime is far from clear and solid. $\endgroup$ Commented Mar 5 at 13:38

To give an analogy around what geodesics are. On maps of the Earth, there are lines that are also called geodesics, in fact, that's where the spacetime geodesics get their name from.

enter image description here
Geodesic representing the shortest flight path (purple) vs the apparent shortest path (red)

Geodesics represent the shortest path between two points in real world. When drawn on a flat map, they will appear curved. This is because the maps you would draw them on represent - in case of the Earth - a sphere projected on a flat piece of paper. The line will not be straight on the map, because there is a distortion between the map and reality.

When talking about spacetime, this is similar - because of a mass changing the shape of spacetime, what would be considered a straight path is actually a detour.

Now while this analogy does not work for black holes, it does work for your question - if following a geodesic you hit a brick wall or an ocean, it does not mean the geodesic ends there, you just can't follow it any more.

  • $\begingroup$ Maybe a slight rewording: a geodesic is the shortest path between two points, full stop. A geodesic on a sphere may not correspond to a straight line on a map projection. $\endgroup$
    – chepner
    Commented Mar 6 at 18:17
  • $\begingroup$ @chepner The OP notes that the lines corresponding to geodesics will not be straight when drawn on a flat map. $\endgroup$ Commented Mar 6 at 22:42
  • $\begingroup$ @chepner I reworded it a bit. $\endgroup$ Commented Mar 6 at 23:53

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