I have known for a very long time that light cannot exit a black hole. I can even understand some of the simpler reasonning about it, such as escape velocity, or space geometry inside the black hole.

But I have a consistency problem. I learned in school (very long ago) that the path followed by light is independent of its orientation on this path. This seems confirmed by the fact that light follows geodesics. Possibly I misunderstand what that means or imply.

It the path is independent of travel orientation, any path in is necessarily a path out. Thus, if no light can escape a black hole, I would expect that no light can enter it.

Yet, I often read that it does (though we cannot see it happen).

So, what do I misunderstand?

I have also some problem reconciling that with some descriptions of the event horizon as a rather quiet place when its radius is very large. But that is just a side remark.

Possibly related question: What is the reasoning behind the idea that light cannot escape from a black hole?


5 Answers 5


What you learned is correct. More simply, it's a consequence of the "time reversal symmetry" of most of fundamental physics. This symmetry is still present in general relativity. But, it's obscured by the standard system of coordinates. When you transform these coordinates into the Kruskal coordinate system, you not only have a black hole, you also have a white hole, which is a region that nothing can go into.

IN this system, a geodesic falling into a black hole also came out of the white hole, and when you reverse time, then you swap the black hole and the white hole, and the picture is identical.

  • $\begingroup$ Thanks. To be very honest, Kruskal coordinates is beyond me. Is there a chance of explaining this in a somewhat more naive way. $\endgroup$
    – babou
    Jun 14, 2015 at 22:25
  • $\begingroup$ @babou: the key detail is that the black hole solution also contains a white hole, and that reversing the path also flips the white hole and the black hole, and you end up with the same path. $\endgroup$ Jun 15, 2015 at 15:11
  • $\begingroup$ I understood as much about the white hole. But it means little to me physically (other than time inversion view). I am wondering whether we are talking of the same thing. The reversal I had in minf (and learned in school) is that, if you put a mirror on the path of the photon, orthogonal to that path, it will go back where it came from. It is a reversal without inverting time (unless mirrors do that to photons). What I get from your answer is that the geodesics are time oriented, so that reversing path in space only is a different matter than "going backward" on a geodesic. Am I correct? $\endgroup$
    – babou
    Jun 15, 2015 at 16:07
  • $\begingroup$ However, there are no white holes and light does not leave :/ $\endgroup$
    – user65208
    Feb 26, 2018 at 8:44
  • $\begingroup$ @dgrat: yes, and if you can solve the arrow of time problem in a clear and comprehensive way, I will personally applaud for you at your nobel prize ceromony. Until then, it's useful to discuss how pure GR has a time-inversion symmetry. $\endgroup$ Feb 26, 2018 at 17:53

In Newtonian point-particle mechanics, time reversal is equivalent to simultaneously bouncing every particle in the universe off of a mirror, so that all velocities negate. But in more complicated theories, the operation of "time reversal" has more complicated effects. For example, in classical electromagnetism, magnetic fields reverse direction under time reversal, and in quantum mechanics, wave functions get complex-conjugated. If you just reverse particle velocities in these theories, you won't replay the movie backwards.

General relativity is another one of those "more complicated theories." It's true that in GR, light rays retrace their paths backwards under time reversal, but "time reversal" does not just mean negating all point-particle trajectories and bouncing all light rays off mirrors; you need to time-reverse the underlying metric as well, and that's not always a very intuitive operation. For example, a rotating black hole drags nearby space along with it, and under time reversal, the direction of frame dragging gets reversed.

In turns out that outside a (Schwarzchild) black hole, the metric is indeed time-reversal invariant, so you can just bounce light rays off mirrors and they'll retrace their paths. But inside the black hole, the roles of space and time in some sense switch places. Going toward the center of a black hole is like going forward in time: you always do it. The time-reversed black hole metric is that of a white hole, which is exactly like a black hole except everything always falls out of it. If you think about it this way, a black hole is not a place - it's a time: the infinitely far future. And a white hole is the infinitely far past at the exact same place.

Here's one more way to think about it: the operation of "bouncing everything off mirrors" and reversing all particle velocities without performing time reversal (i.e. without switching the direction of magnetic fields, etc.) corresponds to the rather strange mathematical operation of negating infinitesimal displacements $d{\bf x}$ without negating position ${\bf x}$ or time $t$, so that all velocities ${\bf v} = d{\bf x}/dt$ negate. In GR, this would correspond to negating the spacelike components of four-velocities $U^\mu$ but not the timelike components. But inside of a black hole the vector $\partial_r$ is actually timelike and $\partial_t$ is spacelike (this is the more precise version of my statement "space and time switch places" above). So if you "bounce a light ray off a mirror" then $\partial_r$ does not change sign, so the radial velocity $dr/d\tau$ doesn't either, and the light ray actually keeps falling into the black hole despite having been (mathematically) "reflected off the mirror!"


In a comment, the asker said: Is there a chance of explaining this in a somewhat more naive way. How about this:

Imagine we live on the ocean. The sea-surface is "space". And as you were taught, for any path on the surface of the ocean there will be a reverse path.

Now imagine the ocean is flowing in some places. In fact let's imagine a giant waterfall. As you get closer to the waterfall, the water is moving faster and faster towards the cliff. There will be a point of no escape. When the water is flowing towards the cliff at the same speed as your maximum ship-speed, the best you can do is point yourself away from the cliff at top speed, and hold in one spot. If you get any closer to the cliff the water will drag you backwards, even at top speed.

Relativity says that space itself can bend, move, or "flow". A black hole is like the waterfall. The event horizon is the line where space is "flowing" into the black hole at the speed of light. A photon of light at the even horizon, pointing outwards, will stay exactly on the event horizon. A photon inside the event horizon will get dragged backwards, into the hole.

The idea that any light path is reversible is normally correct - except when gravity is so intense that time and space get bent to match the speed of light. A light path is reversible, except no light path can reverse in time. A black hole twists space-time so severely that the directions into-the-hole and out-of-the-hole become time directions. Any path into the hole is a path towards the future, and any path out of the hole is a path towards the past. Light paths can only point into the future. If a path touched the event horizon, the path can only point inwards.

So the path is still reversible, except the entire future of the reverse path is inside the hole.

  • $\begingroup$ I'm sorry Alsee, but Einstein described a gravitational field as a place where space is "neither homogeneous nor isotropic". He rejected Gullstrand–Painlevé coordinates for good reason - we do not live ins some Chicken-Little world where space is falling down. The waterfall analogy is a "lies to children" I'm afraid. $\endgroup$ Apr 25, 2019 at 17:36
  • $\begingroup$ @JohnDuffield I do not dispute that your critique of the analogy was accurate. However I question how useful it was. The asker explicitly requested a simpler "more naive" answer. I believe the asker will find my imperfect-analogy to be more useful and more understandable than a reference to "Gullstrand–Painlevé coordinates". $\endgroup$
    – Alsee
    Apr 25, 2019 at 18:55
  • $\begingroup$ the crucial point is that the waterfall analogy is wrong. Light doesn't behave like a fish in a waterfall fish. See The Speed of Light Everywhere the Same? by PhysicsFAQ editor Don Koks. He says “Einstein talked about the speed of light changing in his new theory”. And this: "light speeds up as it ascends from floor to ceiling". In a strong gravitational field, it speeds up all the more. $\endgroup$ Apr 26, 2019 at 7:41

To illustrate what I believe is the answer, here are some examples making use of a spacecraft at three locations:

  • outside the event horizon,
  • at the event horizon, and
  • inside the event horizon.

The black hole is non rotating, and contains a singularity at its center. The spacecraft has a mirror which will reflect incoming photons back along the same path from which they arrived. The incoming photons are created from a source at a substantial distance from the black hole, and are aimed directly toward the spacecraft. The source of photons, spacecraft, and black hole center are colinear. The spacecraft has a powerful engine which can provide infinite acceleration.

Example 1: The spacecraft is maintaining position at a distance of 1.5 Schwarzschild Radius. For a 10 Solar Mass black hole, the spacecraft will need to accelerate at 100 billion g (see https://www.spacetimetravel.org/expeditionsl/expeditionsl.html) to maintain a constant position at 1.5 Schwarzschild Radius. When a photon arrives from the photon source it will be blue shifted to higher energy, the photon will reflect off the mirror back to where it was created. On it's journey back the photon will be red shifted back to it's original wavelength.

Example 2: The spacecraft is maintaining position at the Schwarzschild Radius. This is not actually possible, because the spacecraft would have to be under an infinite acceleration to maintain position at the event horizon. However, laying that aside, when a photon arrives it would be blue shifted by an infinite amount, and would therefore have infinite energy. The infinite energy photon, would reflect off the mirror and on it's journey back to it's source, it would be red-shifted back to it's original wavelength. So, a photon with infinite energy could escape from the black hole's Schwarzschild Radius.

Example 3: The spacecraft is again maintaining position at the Schwarzschild Radius. The spacecraft emits a normal (finite energy) photon in the outward direction. In this case the photon would be red-shifted by a factor of infinity over 0 distance. So this normal photon would not escape the black hole.

Example 4: The spacecraft is maintaining position inside the Schwarzschild Radius. In this case, similar to Examples 1 and 2, it seems that the photon that was emitted outside the black hole would enter the black hole, reflect off the mirror and return back to it's source outside the black hole. Obviously this is wrong. There are several problems with this scenario as described, which are:

  • a) The photon can not leave the black hole.
  • b) In order for the spacecraft to maintain position at 1.5 Rs required an acceleration of 100 billion g, and to maintain position at Rs required infinite g. Extrapolating it seems that more than infinite g would be required to maintain position inside Rs; which doesn't make sense.
  • c) Similar to (b) the energy of the photon arriving at the stationary spacecraft inside the Schwarzschild Radius would have greater than infinite energy. Also, doesn't make sense.

Fortunately black hole theory doesn't allow for stationary objects within the Schwarzschild Radius. Any object inside the Schwarzschild Radius has an inward motion which will bring the object to the singularity, where it will be absorbed. No amount of thrust from our spacecraft's motor can counter this inward motion of the spacecraft toward the singularity.

Now, suppose the photon enters the black hole and reflects off the mirror of the spacecraft. Remember the spacecraft & mirror are moving inward toward the singularity. The photon would reflect back on it's path. However, from the point of view of a hypothetical stationary observer within the black hole, the reflected photon would have lost energy due to Doppler shift. And the reflected photon, now having less energy, would not have sufficient energy to leave the black hole.

  • $\begingroup$ Thank you for your answer. From your example 2, I gather that photons reach blackholes with infinite energy at the Schwarzschild Radius (and you seem to imply even more inside that radius). I am wondering whether that makes sense at all, independently of my original question. Where would that energy come from? $\endgroup$
    – babou
    Oct 10, 2018 at 7:44

I learned in school (very long ago) that the path followed by light is independent of its orientation on this path.

There's 2 things going on there. One is that our knowledge has changed since you went to school (or more correctly when the textbook author went to school), and the other is that these statements tend to hold true only at human timescales and dimensions.

Most of physics as we know it breaks down* at the extremely small and extremely large end of the spectrum. As we do not relate well to either the quantum or astronomical ends of the scale neither do the truisms.

  • or we realize that our perception is a case where most of the full formula can be ignored. Newtonian vs. relativistic acceleration is a good example.

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