# Queuing by the event horizon

I understand that an observer outside the event horizon (EH) of a black hole (BH) will not see anything disappearing from outside the EH - only the effects of the time dilution near the EH.

Assume that a large number of objects are on its way towards the EH and the last objects can observe all the others. It is right to assume that the all – from the last observer’s viewpoint – cross the EH at the same time?

A natural follow up issue is the pattern the observer sees thereafter, i.e. inside the EH.

• Um, could you try to word it better please? Are you asking the shape which the objects pass the event horizon? Or if you can see anything pass the event horizon? – C. Jordan Jul 19 at 11:40
• Comments are not for extended discussion; this conversation has been moved to chat. – ACuriousMind Jul 21 at 12:32

General relativity is based on the equivalence principle, and one way of stating the equivalence principle is that in small enough regions of spacetime, the effects of gravity become undetectable for free-falling observers. So if your cloud of observers is small, or if we restrict our attention to a small part of it, then nothing special happens as far as they're concerned. The ones in back continue to see the ones in front. However, if A passes the horizon before B, then any light emitted by A after crossing the horizon will be detected by B after B has crossed the horizon as well.

Personally, I find it extremely difficult to reason about this sort of thing unless I draw a type of diagram called a Penrose diagram. I have a simple, nonmathematical explanation of Penrose diagrams in this book: http://www.lightandmatter.com/poets/ . See section 11.5.

You can both ask and answer this question by introducing a coordinate choice suited to an observer freely falling towards the horizon. A good choice is Lemaitre coordinates; see https://en.wikipedia.org/wiki/Lema%C3%AEtre_coordinates

To clarify the question, by "observe" do you mean "receive light signals from" or do you mean "deduce, according to an analysis carried out in such-and-such a coordinate system"? I will comment briefly on both questions

1. Receipt of light signals. From outside the horizon one will never receive light signals from something right at the horizon. However, one can receive signals from each object as it approaches the horizon. An analysis in Schwarzschild coordinates shows that the intensity of such signals falls exponentially as the object approaches the horizon, and so does the frequency. So the object will be seen to "wink off" in a rapid exponential decay, not unlike the way the light pulse from an atom decays exponentially when it undergoes a transition to the ground state, or the way a capacitor discharges through a resistor. The point relevant to your question is that you do not have to wait for an infinite time to go by, even in Schwarzschild coordinates, in order to say that the object has become undetectable owing to its approach to a horizon. Comparing to the case of ordinary spontaneous emission from an atom, or exponential decay of a capacitor, one might say the capacitor never arrives fully at the discharged condition, but in practice that is not a very insightful thing to say. The more important point is that exponential decay is, for all practical purposes, decay in a finite time. Similarly, an observer far from a black hole sees each object disappear in a finite time.

For an observer in free fall nearer to the horizon the analysis will change a bit but I think qualitatively the same behaviour will be seen, for a long line of objects. So the answer is that the line of objects wink out one after the other, not all at the same time. However, if the set of objects is not very much extended in space (to be precise, the spatial distances are small compared to the smallest local radius of curvature of spacetime) then the local inertial frame of one of them becomes a good approximation for treating the whole problem, and then it becomes just like everyday experience; each object can see all the others steadily (see answer by Ben Crowell).

1. "Observe" in the sense of "deduce". In Lemaitre coordinates (and also in other coordinates such as Kruskal) the various worldlines do not all cross the horizon at the same event. Therefore the objects are not observed to reach the horizon at the same time.

Finally, a confession. I have not done any mathematical analysis before writing this answer. I think it is correct but if I have missed something then please someone point this out and I will update the answer.

• "objects are not observed to reach the horizon at the same time" - Here is the explanation by Kevin Brown: "One common question is whether a man falling (feet first) through an even horizon of a black hole would see his feet pass through the event horizon below him. As should be apparent from the schematics above, this kind of question is based on a misunderstanding. Everything that falls into a black hole falls in at the same local time, although spatially separated, just as everything in our city is going to enter tomorrow at the same time." - mathpages.com/rr/s7-03/7-03.htm – safesphere Aug 31 at 4:35
• @safesphere this comment by Brown is wrong, because the horizon is a null surface not a spacelike surface. – Andrew Steane Sep 1 at 14:16
• Kevin has never been wrong, unlike people like us on this site :) This is why MathPages is so famous. He never said the horizon was spacelike. For instance, consider a timeslice of r=2M inside the horizon (where r is coordinate time). It is the same for everything ever crossing to the inside. Now, on the outside, you never see anything crossing the horizon regardless of the frame, because of the infinite time dilation. Thus, if you fall, you see the object in front of you being outside until the moment you cross. So you cross at the same proper time outside and appear at the same time inside. – safesphere Sep 1 at 14:34
• Even in the mainstream view, Ben's statement that "nothing special happens at the horizon" is an urban legend. An infinite time dilation happens at the horizon in any reference frame, not only in the Schwarzschild coordinates. If you fall a light year behind me, in your frame I'd still be there before the horizon even when you are only $1mm$ away. Clearly, in your frame, I can't cross before you. Our timelike separation outside becomes null at the horizon and spacelike on the inside. – safesphere Sep 1 at 15:37
• @safesphere At the horizon the metric is $\eta_{\mu\nu}$ (the Minkowski metric) in Riemann normal coordinates (also called geodesic coordinates, local inertial frame, etc.) – Andrew Steane Sep 1 at 17:24

Observers falling into a black hole never observe themselves, or indeed anything falling ahead of them to cross an event horizon. Instead they see the horizon retreat before them. They only ever meet the horizon at the moment they reach the singularity, when the horizon has shrunk to a point.

I go through the details in my answer to Taking selfies while falling, would you be able to notice a horizon before hitting a singularity? It happens because although inside the horizon light cannot move outwards, you can fall inwards fast enough to catch up with the light being shone outwards from an object ahead of you. So from your (falling) frame it appears that light is still moving outwards towards you.

The key point is that your statement:

I understand that an observer outside the event horizon (EH) of a black hole (BH) will not see anything disappearing from outside the EH - only the effects of the time dilution near the EH.

is true only for observers hovering outside the horizon. It does not apply to observers falling freely through the horizon, so your last observer falling freely into the black hole would not observe a pileup at the horizon.

I'll reproduce the diagram from the answer I linked above that shows the trajectories of two observers falling into a black hole:

Note that the two trajectories do not intersect. They cross the horizon at different times and they hit the singularity at different times. This shows only two infalling observers, but it would apply equally to your train of many observers. That would simply add more lines to the diagram, none of which would intersect and all of which would cross the horizon and hit the singularity at different points in spacetime.

I understand that an observer outside the event horizon (EH) of a black hole (BH) will not see anything disappearing from outside the EH - only the effects of the time dilution near the EH.

He'll see gamma ray bursts. People tend to forget about gamma ray bursts, but they're important. It was the detection of gamma-ray bursts, by satellites checking for Russian nuclear explosions, that reawakened interest in general relativity in the 1960s.

Assume that a large number of objects are on its way towards the EH and the last objects can observe all the others. It is right to assume that the all – from the last observer’s viewpoint – cross the EH at the same time?

No. Not at all. See Einstein's 1939 paper on a stationary system with spherical symmetry consisting of many gravitating masses. He said this: “g44 = (1 – μ/2r / 1 + μ/2r)² vanishes for r = μ/2. This means that a clock kept at this place would go at the rate zero. Further it is easy to show that both light rays and material particles take an infinitely long time (measured in “coordinate time”) in order to reach the point r = μ/2 when originating from a point r > μ/2”. Unfortunately Einstein missed a trick here, in that falling bodies don't slow down. They fall faster and faster. As to why, take a look at this: "As a simple geometric consideration shows, the curvature of light rays occurs only in spaces where the speed of light is spatially variable”. That's Einstein again. So your observers are falling faster and faster towards the black hole because the speed of light is getting lower and lower. If this continued, eventually they'd be falling faster than the local speed of light. That can't happen because of the wave nature of matter. So something else has to happen. A gamma ray burst. Your observer sees all the other infalling observers erupting into gamma rays bursts one after the other. BOOM! BOOM! BOOM! See Friedwardt Winterberg's 2001 paper gamma ray bursters and Lorentzian relativity. This was the original "firewall" paper. It's reference 87 in an apologia for firewalls.

A natural follow up issue is the pattern the observer sees thereafter, i.e. inside the EH.

He doesn't see anything. He's long dead. I know people say he doesn't observe anything unusual as he crosses the event horizon, but that's based on a version of general relativity that contradicts Einstein. See this image from page 848 of Gravitation by Misner Thorne and Wheeler:

Look at drawing (a) on the left. The vertical dashed line denotes the event horizon. The curve on the right of the vertical dashed line denotes the path of the infalling body outside the event horizon. It gets closer and closer to the event horizon as the time t increases. Note though that the chart is chopped off at the top, and the infalling body somehow crosses the event horizon at time t = infinity. It somehow jumps over the end of time. Then the curve on the left of the vertical dashed line denotes the path of the infalling body inside the event horizon. The drawing (b) on the right shows Kruskal-Szekeres coordinates where the t = infinity has been swept under the carpet by the use of "tortoise" coordinates. That's where seconds last for a longer and longer time, eventually lasting for an infinite time. Amazingly, people take this sort of thing seriously.

• So just to be clear, you're saying that everything falling into a black hole explodes at the point where it would exceed the velocity of light measured in the Schwarzschild coordinates? And that all those relativity textbooks saying that a freely falling observer notices nothing special as they cross the horizon are wrong? – John Rennie Jul 31 at 5:42