On the other hand, light follows a light-like trajectory and, therefore, propagates along the event horizon, since the horizon of a black hole  is a light-like surface.

There are very nice lecture notes by Sean Carroll (http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll_contents.html , find on the top-left the link to an arXiv.org post for the full lectures in .pdf) where you can find an illustrative analysis of the lighthcones of a probe near the horizon in different coordinate systems (http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll7.html). There, the Eddington-Finkelstein coordinates are constructed, in which it is clear that an infalling observer won't even notice that it crosses the horizon.

http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll_contents.html

http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll7.html

Thanks for the answers.

One more question though regarding "light propagating along the event horizon":

As far as I remember then light orbits are stable at 1.5 r_S (Schwarzschild radius) ... not at 1 r_S.

Is the light propagating at all atop of the horizon?

A couple clarifications here:

The curvature at the horizon of a black hole depends on the mass of the black hole. For larger black holes the curvature can be arbitrarily small - it occurs at very large r, and the curvature decreases there, but a particle there cannot escape.

A photon may "hover" at the horizon if it is directed radially away from the singularity. Locally, it travels at the speed of light but an inertial reference frame at that point is "falling" into the hole at the same speed relative to infinity. Inside the black hole horizon, photons directed either away from the center or toward it both move nearer the center, one reaching it before the other. For a light like orbit of the hole, part of the photon speed must be around the hole and part fighting the local free fall, so such an orbit must be further out than the horizon.

``Hover'' can't be given physical meaining, since a massless particle doesn't have a rest frame. What does require care is, whether we're talking of the freely falling observer, or the observer at infinity. From the point of view of the observer at infinity, the light cone becomes tangent to the event horizon. Therefore it's a light-like surface and light, sent from infinity, will move along this surface.

Thanks. I agree, obviously. Let's see, if this is better: A light ray, inbound from the exterior, intersects the horizon at one point. *Beyond* that point it, also, follows the light cone-more precisely the ``reflected'' direction, sloppily expressed. The light cone is parallel to the event horizon there, so it does travel along the event horizon, with respect to the observer at infinity. (Similarly, a light ray from the interior, assuming that could be defined, would intersect the horizon at one point and follow the light cone that's, also, parallel to the event horizon-consistent with statement that light can't get out.  I think this does provide a check of sorts. From the point of view of the infalling observer, on the other hand, things may be very different. 

Even though the event horizon is a lightlike surface I don't agree that photons can travel along it. However, they might remain at r=2m if outwardly directed and maintaining fixed angular coordinates. No circular orbits of the BH are possible for any particles within the photon sphere at r=3m (as Bartosz rightly said). From the perspective of any external observer, photons grind to a halt at the horizon, irrespective of their arrival direction. This is another reason why photons cannot explore the horizon - besides, it takes infalling photons an infinite time to reach the horizon according to any externally located clock.

But the content of the statement that a surface is light-like *does* mean that photons can travel on it-else the statement doesn't make sense. And this is consistent with the fact that, following such a trajectory, they would reach infinity in infinite time, as measured by the observer at infinity. 

Agree completely. But that's just my point: that photon trajectories lie on the event horizon.

Robert, I didn't refer to stationary observers and said nothing to imply that the geometry breaks down at the horizon.

In science the phenomena that exist in nature usually have precedence over purely hypothetical objects. It is improbable that this universe is host to stationary eternal black holes, and non-stationary astrophysical black holes formed by gravitational collapse only give rise to event horizons in an asymptotic sense owing to gravitational time dilation. This is an important and often forgotten/overlooked point - especially since we have known for decades that black holes radiate, precluding event horizon formation in infinite external time through near-horizon evaporation processes that terminate in finite external time. It is only because many people are now so "hung up" on using horizon-penetrating coordinates to investigate maximally extended spacetimes that the non-existent black hole information paradox is currently considered a genuine problem.

If light is to escape a Schwarzschild BH from just outside the horizon it must initially travel in a very tight, radially directed, outwardly pointing cone. For photons at r=3m the cone will have widened up sufficiently that it occupies a full hemisphere, thereby allowing circular orbits - consistent with the presence of the photon sphere at this radius. Clearly, as radius decreases from 3m to 2m the cone narrows monotonically without any abrupt discontinuity at r=2m (thus prohibiting photons from roaming around on the horizon).

Robert, I've encountered plots like this before but a major drawback of Penrose diagrams is that they gloss over the infinite time dilation at the event horizon, and the fact that it takes infinite time for a trapped surface to form according to any external observer. I base this on the 1939 paper describing dynamical collapse by Oppenheimer & Snyder in which the event horizon only forms asymptotically. Precious few dynamical solutions of the field equations have ever been found. If the BH evaporates before external observers witness particles reaching an event horizon, in what sense does the event horizon exist at all for any particle? Surely there can be no loss of information.

I think we agree on photon trajectories. I didn't mean to imply that the cones disappear completely at the horizon, just that the opening angle reaches zero at r=2m so that only radial photons can remain there. The description of the event horizon as a lightlike surface can seem a bit odd but I think is because we are more accustomed to thinking of particles and their worldlines. If photons were sheet-like as opposed to point-like then it would make sense that the horizon is a surface on which a 'particle sheet' must travel at the speed of light to remain there.

The description of evaporation is outside general relativity-it's not described by the classical equations of motion. It seems compatible with general relativity, if the black hole is large, since, then, the rate is low, spacetime changes slowly and can be described by ``patching'' solutions to Einstein's equations and the black hole spacetimes,  a classical notion, remains valid.  So the two are distinct. The statement that the event horizon is a light-like surface, doesn't mean that only extended objects can propagate on on it-it means that all curves on it are light-like. Therefore the worldline of a particle that moves on such a surface is light-like and this does correspond to the propagation of a  physical particle, namely a massless particle, that's not a ghost.

Robert, acts of visual observation are irrelevant to this discussion. What should not be ignored is gravitational time dilation. Consider the diagonal black line marked horizon in your diagram and some parallel line located below it. Your diagram fails to give any hint that there is infinite time dilation at the horizon but not on the parallel line below it. Penrose diagrams give the impression that the surface of a collapsing star can cross the horizon quite easily while the observer's proper time remains finite. If you really believe this then you've misunderstood what the famous work of Oppenheimer & Snyder showed for the all-important dynamical situation. Although it does not address BH evaporation, the Kruskal diagram at least makes it clear that there is a time barrier at r=2m for a stationary BH.

Robert, time dilation is absolute in the sense that the ratio of the proper time of one test clock to the proper time of another test clock is perfectly well-defined by the spacetime geometry and the motions of the clocks. There's no need to drag communication by light or other signals into any of this. It is not that infalling particles cannot ever cross the event horizon (or even plummet into a singularity) it is that there are global constraints on when they can do so. Gravity increasingly retards their passage of time and hence motion as the horizon is approached. There is no gravitational time dilation in the Rindler spacetime so the analogy is flawed. Oppenheimer & Snyder studied the dynamical collapse of a spherically symmetric pressure-free dust cloud. It is now known that analytical methods cannot handle more complex situations. Therefore, as their results have not been challenged, they had the last word on this back in 1939. If you have no further comment to add then neither do I.

Stam, describing the event horizon as a lightlike surface does not literally mean that lightlike worldlines traverse that surface. All it means is that the event horizon is a surface where a collection of radially propagating photons can remain. Photons cannot orbit the BH at radii below 3m (which is why this special radius is known as the photon sphere). These things can be gleaned with a bit of algebra from the Schwarzschild metric.

A light-like surface can only mean that the light-like worldlines make it up. Of course the not so trivial question is, whether such worldlines can become patched to worldlines that come from elsewhere in the spacetime-and here one needs to think a bit more and study the Penrose diagram. Then one recognizes that, to obtain a worldline that belongs  on the event horizon, on must start from a very special point. So the incoming photon, does cross the horizon and ends up in the singularity, in general. 

"A light-like surface can only mean that the light-like worldlines make it up."

Intuition might suggest so but there is an important distinction in this context between a surface and a worldline. Fortunately, the metric clarifies what is meant by the terminology and does resolve the ambiguity here.

The question is put incorrectly. In reality the event horizon is not formed at a colapse of a heavy spherical body. The general relativity is based on a hypothesis, that the Riemannian geometry is the most general type of the space-time geometry. It is not so. If one constructs a space-time geometry on the basis of a metric approach, when the geometry is a monistic conception, described by the world function and only by the world function,  there is a lot of non-Riemannian geometries. Constructing the general relativity on the basis of these non-Riemannian geometries, one obtains extended general relativity (EGR). In the EGR the collapse of a heavy body  stops, because near the center of the body  the antigravitation is induced. These antigravitation stops the collapse, and the event horizon is not formed. These results are published:  "Induced antigravitation in the extended general relativity". Gravitation and Cosmology, 2012, Vol. 18, No. 2, pp. 107–112,( 2012). DOI: 10.1134/S0202289312020089 and "General relativity extended to non-Riemannian space-time geometry." Electronic Journal of Theoretical Physics 11, No. 31 (2014) pp 177-202, Available also at http://arXiv.org/abs/0910.3582v7

If you consider influence of matter on the space-time geometry, one needs to consider all possible geometries at the beginning of the consideration. Consideration of "good models" is not reasonable. One should consider correct models, but not good ones. Besides, my way of description is coordinateless. It is much better, than the  case, when one cannot construct a space-time geometry in the coordinateless way. It means simply that your knowledge of a geometry is not sufficient.

All proponents of black holes assign, quite unwittingly, to their black holes, an escape velocity and no escape velocity simultaneously at the same place (at the 'event horizon'). Since it  is impossible for anything except a fantasy to have and not have an escape velocity simultaneously at the same place, the black hole is a fantasy. See Section V of the following:

General Relativity: In Acknowledgement Of Professor Gerardus ‘t Hooft, Nobel Laureate,

http://viXra.org/abs/1409.0072

I   think, that it is useless to discuss any conception   or any calculation, if they are deduced on the basis of wrong mathematical formalism. At least, I am not going to waste time for such a discussion. I shall point out only, why formalism of the Riemannian geometry is not available for discussion of black holes. At first, the Riemannian geometry is inconsistent. For instance, if in an Euclidean plane, where geometry is constructed according to rules of the Riemannian geometry construction, one makes a hole, then the plane with a hole is not embedded isometricly ito the plane without the hole.. Mathematicians know this fact, but they say: "We shall concider onle convex set of points."  But it is only a detail. The main problem consist in the fact, that mathematicians accept only axiomatizable geometries, because they know only one method of the geometry construction: deduction of a geometrical statements from axioms.

They do not accept construction of a generalized geometry as a result of deformation of the Euclidean geometry, when the obtained geometry is multivariant,  generally speaking. (it means that the equivalence relation is intransitive). The do so, although axiomatizability or nonaxomatizability is an attribute of the method of the geometry construction, but not an attribute of the GEOMETR ITSELF. As a result the set of Riemannian geometries form a small part of all possible space-time geometries. A use of this small part of all possible geometrie for constructing the General relativity is a mistake (a use inadequate mathematical formalism). 

Yuri Rylov, Certainly the pseudo-Riemannian geometry of Einstein's Relativity Theory is nonsense. It bears no relation to reality. Mathematics is only as good as its axioms, and mathematics is not physics. Mathematics is just a tool for physics and engineering, and very little of what the pure mathematicians play with is used for solving physical problems.

Furthermore, the pseudo-Riemannian geometry of General Relativity is not even consistently applied within its own ambit by Einstein and his followers, and so it leads to contradictions, both mathematical and also in terms of its alleged physical meaning. This is on top of the contradictory physical principles employed by Einstein at the outset.

Now one does not have to utilise any mathematics at all to prove that black holes are nonsense. As I pointed out in my previous post, all proponents of black holes assert that their black holes have and and do not have an escape velocity simultaneously at the same place. But this is impossible! So the black hole is nonsense.

In General Relativity gravity is not a force, because it is spacetime curvature. All proponents of the black hole assert that all their black holes have a finite (non-zero) mass. This mass they say is concentrated in the black hole's 'singularity' where density is infinite and spacetime curvature is infinite. Thus they assert that a finite mass produces infinite gravity. That too is nonsense - no finite mass can produce infinite anything, let alone infinite gravity and infinite density. So the black hole is nonsense.

Einstein and his followers assert that the material sources of his 'gravitational field' are both present and absent by the very same mathematical constraint (energy-momentum tensor = 0). That too is impossible! So the black hole is nonsense.

Similar simple arguments prove that big bang creationism is also nonsense.

A little mathematics is sufficient to prove that General Relativity violates the usual conservation of energy and momentum for a closed system, so it is in conflict with a vast array of experiments.

There are two things any physical theory must satisfy: (1) logical consistency, (2) correspondence with experiment. General Relativity fails on both counts.

All this and more is covered in detail in the following:

General Relativity: In Acknowledgement Of Professor Gerardus ‘t Hooft, Nobel Laureate,

http://viXra.org/abs/1409.0072

Further examples of Einstein's incompetence and his mysticism are his claims that elementary charges are massless 'wormholes' and that electrons and protons are composed of two wormholes, one for each of their respective masses and one for each of their respective charges. Einstein asserted that electrons and protons are each a "two bridge problem". Einstein's bridge is now most often called a 'wormhole'. The wormhole nonsense is starkly revealed in the following:

Wormholes: Science Fiction or Pure Fantasy?,

https://www.youtube.com/watch?v=16_GuYobDZ4

(transcript, with citations: http://vixra.org/abs/1410.0073)

I shall not discuss considerations, based on incorrect space-time geometry. Incorrectness lies in the fact, that most part of possible space-time geometries is ignored. If these ignored geonetries are taken into accont, the black holes connot exist, because the induced antigravitation stops a collapse of a heavy body. See publications:  "Induced antigravitation in the extended general relativity ". Gravitation and Cosmology, 2012, Vol. 18, No. 2, pp. 107–112,( 2012). DOI: 10.1134/S0202289312020089 and  "General relativity extended to non-Riemannian space-time geometry."  http://arXiv.org/abs/0910.3582v7

Please read my publication on the subject. Why? I am author in Physical Review, 2) derivation of singularity is nice-looking and astonishingly short.

Article Moving into Black Hole: is there a wall?

Dear Dmitri, could you present an electronic version of your paper and its title? The reference to Physical Revew is insufficient to find the paper.