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This whole question and subquestions are based on the assumption that light rays on the event horizon are normal to the event horizon, so my apologies if this is not correct

In A Brief History of Time, in the frist one or two pages of chapter 7, "Black holes ain't so black", Hawking states the following fact: the paths of light rays on the event horizon could never approach one another.

He concludes that if they did, they had to run into each other sometime and thus they could (or have to?) fall into the black hole, thus they could not have been on the event horizon in the first place per the definition of an event horizon.

So he states that light rays on the event horizon have to run in parallel or away from each other as not to run into each other.

I have a few problems in understanding this:

  1. What does it mean for two light rays to run in parallel on an event horizon? (And what does it mean for two light rays to run away from each other)

I thought that light rays the event horizon was like a sphere, so a light ray on the event horizon would have to be a normal to this sphere as not to be falling into the black hole. Would this mean that the light stays stationary on the event horizon? This can't be true since the speed of light is a universal constant. I think the curving of spacetime comes in here, but I don't know how.

And: Would two parallel light rays on the event horizon be two normals at different positions on the event horizon (sphere)?

  1. Why do two light rays fall into the black hole if they run into each other?

If my assumption about parallel light rays being normals to the event horizon is correct, then I guess that non-parellel light rays would fall into the black hole because not all of the speed (c) of at least one of the light rays is pointed in the outward/normal direction, so gravity is stronger than this outward direction speed vector, so a ray falls into the black hole. But that would also happen if there was just one light ray not running into another light ray that was not a normal to the event horizon, right?

I've already read this question: What does Hawking mean by “Light rays that form the edge of the event horizon could never approach one another”?, but I did not understand the explanation in terms of null congruences

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  • $\begingroup$ At first I thought light rays on the event horizon were orbiting the black hole, but this is the photon sphere if I'm not mistaken, because light rays inside the photon sphere normal to this sphere can still escape it because then all of their speed is in the direction outside of this sphere. $\endgroup$ Jan 3 '19 at 14:43
  • $\begingroup$ @BenCrowell A Brief History of Time, Chapter 7, probably first or second page depending on edition $\endgroup$ Jan 3 '19 at 17:49
  • $\begingroup$ @BenCrowell I added a detail to the argument about running into each other. I also think he's talking about two or more light rays already at the event horizon and collide at a later time if they don't run parallel. So not about light rays from outside the event horizon hitting light rays exactly on the event horizon. $\endgroup$ Jan 3 '19 at 17:58
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Here's the relevant passage from A Brief History of Time:

It means that the boundary of the black hole, the event horizon, is formed by the light rays that just fail to escape from the black hole, hovering forever just on the edge (Fig. 7.1). It is a bit like running away from the police and just managing to keep one step ahead but not being able to get clear away!

Suddenly I realized that the paths of these light rays could never approach one another. If they did they must eventually run into one another. It would be like meeting someone else running away from the police in the opposite direction - you would both be caught! (Or, in this case, fall into a black hole.) But if these light rays were swallowed up by the black hole, then they could not have been on the boundary of the black hole. So the paths of light rays in the event horizon had always to be moving parallel to, or away from,each other. Another way of seeing this is that the event horizon, the boundary of the black hole, is like the edge of a shadow - the shadow of impending doom. If you look at the shadow cast by a source at a great distance, such as the sun, you will see that the rays of light in the edge are not approaching each other.

If the rays of light that form the event horizon, the boundary of the black hole, can never approach each other, the area of the event horizon might stay the same or increase with time, but it could never decrease because that would mean that at least some of the rays of light in the boundary would have to be approaching each other.

With this context, we can see that Hawking isn't referring to just any light rays that intersect the event horizon, he's referring to light rays that form the horizon, i.e., each such ray is one that lies completely inside the horizon. This is a very special subset of the rays that intersect the horizon.

What does it mean for two light rays to run in parallel on an event horizon?

In Riemannian or semi-Riemannian geometry, parallelism is not as clearly defined a concept as it is in flat space or flat spacetime. Basically we can only define this concept for geodesics that are close together, approaching to within a distance that is small compared to the scale set by the curvature. If they cross at a point, then they are not parallel at that point. If nearby geodesics are parallel, then this means the same thing as in flat spacetime, because locally, spacetime is flat.

I thought that light rays the event horizon was like a sphere, so a light ray on the event horizon would have to be a normal to this sphere as not to be falling into the black hole.

Topologically it's a 3-cylinder. Just as you can make a 2-cylinder by sliding a circle along an axis perpendicular to the plane of the circle, you can make a 3-cylinder by sliding a sphere. Here the axis we're sliding along is timelike. The rays he's talking about are not normal to the cylinder, they're parallel to it.

Would this mean that the light stays stationary on the event horizon? This can't be true since the speed of light is a universal constant.

Any observer at the event horizon will measure the rays to be moving at $c$. There are two common ways to visualize this, either with tipping of light cones or with Penrose diagrams. I have nonmathematical presentations in both styles in my book Relativity for Poets. See sections 11.4 and 11.5.

Why do two light rays fall into the black hole if they run into each other?

He's not really talking about a physical collision like two planes colliding and falling out of the sky. He's just saying that if these rays were colliding, then the black hole would be losing area, which is what would happen if the event horizon were shrinking.

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  • $\begingroup$ Thanks for the answer and the pdf. Also sorry for not stating more clearly that I indeed meant light rays that form the event horizon. $\endgroup$ Jan 3 '19 at 23:44
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I have a few problems in understanding this:

Since when I first read Hawking's book, years ago, I always asked myself what could understand anyone who was not amply familiar by itself with that matter. Apparently Hawking believed that not using formulae and recurring to silly metaphors, like

It would be like meeting someone else running away from the police in the opposite direction - you would both be caught!

would help the general reader. I believe the opposite. But this is an opinion, and I know that opinions are considered OT on SE. So let me enter the argument.

$\let\th=\vartheta \let\phi=\varphi$ The main fact that Hawking is leaving out in that "explanation" is that his "paths" are spacetime paths. As @BenCrowell remarked, the event horizon is a 3-cylinder. I hope you may be familiar with some kinds of coordinates used for Schwarzschild black holes. I'll refer to Kruskal-Szekeres', to me the best suited for our purpose (incidentally Hawking draws several diagrams referring to Eddington-Finkelstein coordinates, which are far from intuitive - and obvously says nothing about that).

K-S metric is $$ds^2 = 4\,{r_0^3 \over r}\,e^{-r/r_0}(du^2 - dv^2) + r^2 d\th^2 + r^2 \sin^2\!\th\,d\phi^2$$ where $r$ is a function of $u^2-v^2$ I don't write. You only need to know that on the event horizon ($r=r_0$) $u$ and $v$ satisfy $u=\pm v$, so that only one is needed. I'll keep $v$, which is timelike (you can see that from the metric). Since on the horizon $r=r_0$, a constant, you can verify Ben's statement: $v\in\Bbb R$ and $\th$, $\phi$ are spherical coordinates on a 2-sphere of radius $r_0$. The event horizon is a $\Bbb R\times S^2$ manifold.

Note that when Hawking speaks about horizon's area he is referring to the area of the spacelike section, i.e. $4\pi\,r_0^2$.

What about lightlike geodesics? They must satisfy $ds^2=0$. The term in $du^2-dv^2$ vanishes because of the constraint between $u$ and $v$. The remaining part in $d\th^2$, $d\phi^2$ is definite positive, so that it may be 0 only if $d\th=d\phi=0$. So we see that lightlike geodesics on the horizon have $\th$ and $\phi$ constants - a fixed point on the sphere. In other words, lightlike geodesics on the horizon are generatrices of the 3-cylinder.

Now Hawking's statement makes sense: two generatrices are obviously parallel (in spacetime). If they could get nearer, the sphere's area would decrease. This isn't the full story, but I hope what I wrote did help you.

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  • $\begingroup$ I am not familiar with Schwarzschild black holes, nor do I know what all those equations are saying. Where should I start getting into the maths? I tried looking up Kruskal-Szekeres coordinates, but the wikipedia article says they're defined in terms of Schwarzschild coordinates, which I also do not know. I initially wanted to ask you in chat, but I don't know how to do that. $\endgroup$ Jan 6 '19 at 17:32
  • $\begingroup$ @TheCodingWombat Just a short reply - at this time of night I'm tired and not too able to address hard matters. The main difficulty everyone trying to answer questions - here in SE, but not only - encounters is that he/she's not informed about the level of knowledge of questioner. Sometimes it can be deduced from kind and style of answer, but not always. So I'm sorry if I didn't match your wants, in a matter intrinsically hard in itself. $\endgroup$
    – Elio Fabri
    Jan 6 '19 at 20:50
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    $\begingroup$ I don't even know how to chat, but IMO it would not be the right thing, since your problem is widely felt and a public discussion would of much wider interest - I believe. Unfortunately it appears to me that SE is not well suited for such issues. Anyhow I'll try to say more - perhaps tomorrow. Stay tuned. $\endgroup$
    – Elio Fabri
    Jan 6 '19 at 20:51
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Maybe one must not think of light rays, a classical concept and hopefully parallel light rays are well derived mathematically as discussed in the link you give, but rather of photons.

Photons are elementary particles and light ( and its geometrical representation into light rays) emerges from the superposition of the complex wavefunction of each individual photon with the zillions that compose the classical electromagnetic wave. see here and links for an explanation.

If a photon is on the horizon, it will either keep on going down towards the singularity, or it will be trapped to be going around as in the photon sphere.

If a photon is not going in "parallel" with the orbit of another photon, their paths will intersect, by definition of parallel. There is a photon photon interaction, and depending on its energy it can either create particles ( high energy) where they will be swallowed by the black hole or emerge as part of the hawking radiation, or might just scatter off each other elastically, changing directions, with very high probability of one going down into the black hole and the other leave and escape if it has enough energy, again being part of the hawking radiation. Thus over time, the only trapped photons on the horizon have to be in parallel orbits around the black hole.

If all photons are in parallel trajectories, the light ray will also have to be in parallel trajectories, as it is emergent from the zillions of photons.

This argument of course holds for the photon sphere too . This is a hand waving argument, if you do not want to go into the mathematics of the subject given in the link you have.

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  • $\begingroup$ "It will either keep on going down" - do you mean toward the singularity, so toward the center of the black hole? What does going around the black hole on the event horizon mean? If it can go around with the speed of light and not get pulled in by gravity, doesn't that mean that a photon already over the event horizon and thus in the black hole that moves in a normal direction to the surface of the black hole escape because that has a greater outward velocity than the photons going round? $\endgroup$ Jan 3 '19 at 17:55
  • $\begingroup$ Also, does this mean that the oribiting photons have to lie on a geodesic of the event horizon, and thus the only way for them to run in parallel, would be that they "follow" each other, or thus that their paths coincide? Or can they also follow shorter circles around the black hole? $\endgroup$ Jan 3 '19 at 18:07
  • $\begingroup$ @TheCodingWombat I explicitly stated that it is a handwaving argument. The hypothesis / statement is that light rays are parallel. ,I take it to photons, that is all.. Geodesics are a matter of calculations, mentioned at the link of a similar question by the OP. Photons can be on the same path, but as point particles theoretically an infinity of geodesics is available, they just should not cross. and interact. $\endgroup$
    – anna v
    Jan 3 '19 at 19:09
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    $\begingroup$ @annav. I know you like to reduce everything to elementary particles. I have the opposite view. GR is a classical theory and knows nothing of photons. Moreover I find dangerous to bring up a concept so liable to misunderstandings. Photons are quite different things from what most layman would think. They aren't tiny particles of light, like minuscule grains of sand. Photons are quantum particles whose properties are very far from common sense. $\endgroup$
    – Elio Fabri
    Jan 4 '19 at 17:05
  • $\begingroup$ @annav. You wrote that the classical description of light (be it waves or rays) emerges from the superposition of the complex wavefunction of each individual photon with the zillions that compose the classical electromagnetic wave. And then? How can one understand a lightike geodesic along this way? Surely we may not say that a single photon follows a lightlike geodesics in spacetime. $\endgroup$
    – Elio Fabri
    Jan 4 '19 at 17:05
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We can't say light at the event horizon is a light... Because light have some of its distinct properties.. After entering into the horizon boundary its properties totally gets changed... What i think is that... It is beyond our thinking about the state of light rays at event horizon... They might be travelling with greater speed more than C... Escaping of light is not a simple phenomenon.. light cannot be skipped from the darkened end... But can be swallowed by some matter that we call black hole... What I think is that black holes absorbs the light and split it into its band of colours... Thus it is easy for the light to get absorbed by something ... A pure light can't be absorbed directly by the black hole because its rate of occuring gravitational collapse is much smaller than speed of light to pass with in it... .. So bending of space time, split the light and thus gets absorbed.... And their is nothing idle in the universe.. So there is nothing like parallel....so light at event horizon may be travelling in either parallel or non parallel directions... There will be no effect of their motions as light photons of different colours (splitted light) can easily merge and again disperse....

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  • $\begingroup$ This answer contains a number of quite unclear statements, starting with “they might be travelling at speed more than $c$”?? $\endgroup$ Mar 1 '20 at 18:15

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