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There are a lot of questions about crossing the EH (event horizon) of a black hole on this site.

Some of them suggest, that when you cross the horizon, nothing special happens, you don't even notice crossing the horizon, and some suggest that it is even impossible to detect the horizon locally.

Nothing special happens to the observer as they cross the event horizon.

Falling into a black hole

In your co-ordinate system you will notice nothing unusual.

What do you feel when crossing the event horizon?

There will be no discontinuity in behaviour at the event horizon.

Taking selfies while falling, would you be able to notice a horizon before hitting a singularity?

Now there are others, who suggest that inside the horizon, everything, including light must move towards the singularity, the singularity becomes a point in time (future).

So inside the horizon even a light ray directed outwards actually moves inwards not outwards.

How does light behave within a black hole's event horizon?

https://arxiv.org/abs/2002.01135

Is the event horizon locally detectable?

Based on the first one, when two astronauts cross the EH together, their walkie talkie (or radio) could keep working.

Based on the second one, this is not so clear. Obviously, outside the horizon, the radio still works, because EM waves from the sender still spread spherically, and would still reach the receiver. But once you cross the horizon, the curvature becomes so extreme, that the escape velocity exceeds the speed of light. Thus, EM waves would not spread spherically anymore, but only towards the singularity. Based on this, the EM waves from the sender might not be able to reach the receiver anymore, this the radio stops working when crossing the EH.

Just to make it clear, I am asking about two astronauts, co-moving, falling in together, and will the radio stop working between the two of them?

Question:

Does the radio (between two co-moving astronauts) stop working when crossing the event horizon?

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  • $\begingroup$ Is there radial distance between astronauts, or tangential? $\endgroup$
    – stuffu
    Commented Jul 11, 2020 at 20:21
  • $\begingroup$ @stuffu tangential. $\endgroup$
    – Árpád Szendrei
    Commented Jul 11, 2020 at 21:42
  • $\begingroup$ Simon's animations are misleading due to a poor choice of coordinates. They show the singularity as an object in space while instead it is a moment in time that does not exist anywhere in space. There is no inward and outward directions inside a black hole. They instead refer to the future and past. So radio waves go $360^o$ in all directions, but none of these directions point toward or away from the singularity. Please also note that the inner Schwarzschild spacetime does not exist in reality. It is eternal and cannot be created by a star collapse. $\endgroup$
    – safesphere
    Commented Jul 19, 2020 at 6:20
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    $\begingroup$ @safesphere " the inner Schwarzschild spacetime does not exist in reality. It is eternal and cannot be created by a star collapse." Can you please elaborate on this one? $\endgroup$
    – Árpád Szendrei
    Commented Jul 19, 2020 at 15:59
  • $\begingroup$ In a star collapse, both inside and outside, the coordinate time $t$ Is timelike for all values of $t$ trough infinity. This means that $t$ inside is never spacelike, so the inner spacetime is never Schwarzschild (in which $t$ is spacelike). This is the deep meaning of the fact that the Schwarzschild solution is eternal. The inner spacetime is either Schwarzschild forever or not Schwarzschild forever. It cannot be both. See this question (not the answers) for details: math.stackexchange.com/questions/3513195 $\endgroup$
    – safesphere
    Commented Jul 19, 2020 at 17:53

8 Answers 8

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The infalling observer who is free falling with negative escape velocity v=-c√(rs/r) will receive redshifted signals from the far away observer all the way down to the singularity (if he falls in with less than the escape velocity the signal he receives might as well be blueshifted).

The far away observer will receive redshifted signals from the infalling observer until the end of time, although the last signal he receives at the end of eternity will be the infinitely redshifted signal the infalling observer sent when he crossed the horizon.

All the signals the infalling observer sends after he crossed the horizon will not make it out since their dr/dt<0 inside the horizon (and dr/dt=0 for an outgoing signal right at the horizon).

In this simulation of a freefalling observer (red) who emits a signal (36 photons with 10° separation, the photons are depicted green) at r=rs/2 (t=0.8619286) in Raindrop coordinates you see that the radially inward directed photons move faster towards the singularity than the free falling observer, and the outward directed ones slower.

Edit: to adress the question in the comment I updated the animation to show a second observer who crosses the horizon with a delay of Δt=0.1GM/c³ and also emits a signal when he is between the horizon and the singularity to make it obvious that both observers catch each others signal.

So two free fallers are able to exchange light signals if their separation is not too large; if you send a signal right after you crossed the horizon it might not reach an observer that is right before the singularity (and vice versa), but an observer close below rs/2 can communicate with an observer close above rs/2.

However, the observer above rs/2 will receive the signal only when he himself has already fallen below the radius where the lower observer was when he emitted the signal (the signal directed at him still travels inwards, but slower than himself), while the lower observer will be overtaken by the radially inward directed photon emitted by the higher observer:

Two observers inside a black hole sending signals to each other, animation

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  • $\begingroup$ @Yukterez very nice answer, but minor comment: please use mathjax for mathematical formulas. $\endgroup$
    – Prof. Legolasov
    Commented Jul 12, 2020 at 6:20
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    $\begingroup$ @Yukterez SE policy says to prefer gender-neutral pronouns when writing or editing posts, so this was rolled back to the revision that added them. I also merged in your updates on the images. If the pronouns make the answer unclear, please try to revise it according to the gender-neutral requirement. $\endgroup$
    – tpg2114
    Commented Jul 13, 2020 at 11:31
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    $\begingroup$ @Gendergaga your answer was good! Please don’t delete it. The fact that safesphere disagrees with a post should not be taken too seriously. He disagrees with a lot of good answers and doesn’t write his own answers $\endgroup$
    – Dale
    Commented Jul 21, 2020 at 17:01
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    $\begingroup$ @Dale - the original answer is backed up at archive.is/wEtJF#selection-1455.41-1455.50 but I will write a new answer and this times I'll use better coordinates and better pronouns, just wait a little until it is finished $\endgroup$
    – Yukterez
    Commented Jul 21, 2020 at 17:23
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    $\begingroup$ The latest edit turned this answer into not-an-answer, so I've rolled it back. If you think your answer is incorrect and you don't believe it should be here anymore, you're welcome to delete it. Deletion is the proper way to deal with a post that you want removed; editing it into a non-answer is not. (Note that there was an earlier issue about mass deletion, but this is separate. Deleting an occasional answer because you honestly think it's wrong and not fixable is fine, even if deleting a larger number of posts all in one go would not be.) $\endgroup$
    – David Z
    Commented Jul 21, 2020 at 19:24
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Does the radio (between two co-moving astronauts) stop working when crossing the event horizon?

Assuming that the black hole is massive enough that there are negligible tidal effects at the horizon then their radios would continue to work and their conversation would carry on without a pause.

Now there are others, who suggest that inside the horizon, everything, including light must move towards the singularity, the singularity becomes a point in time (future).

This is true also. There is no contradiction between the two claims. Because the astronauts are also falling in towards the singularity it is not necessary for light to go outward in order to go from one astronaut to the other. If you draw the worldlines of the communications you will find that indeed they never go outward.

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    $\begingroup$ Thank you so much! $\endgroup$
    – Árpád Szendrei
    Commented Jul 12, 2020 at 15:43
  • $\begingroup$ Your assumption is wrong, my comments didn’t cause Simon to delete his answer or change his user ID to the name of a popular German book. See the edit history of the answer and the statement in his profile: physics.stackexchange.com/users/24093/gendergaga $\endgroup$
    – safesphere
    Commented Jul 22, 2020 at 22:19
  • $\begingroup$ It isn’t an assumption. In revision 24 he explicitly said “Safesphere's comment ... convinced me that my answer was wrong.” $\endgroup$
    – Dale
    Commented Jul 25, 2020 at 13:20
  • $\begingroup$ Do you seriously have no idea of what’s going on? He has changed his ID and deleted the revision 22 for the reasons now clearly stated in his profile. He was suspended and the answer was forcefully undeleted by SE. Then he was looking for any excuse to delete is again. Now he is deleting all his answers. $\endgroup$
    – safesphere
    Commented Aug 4, 2020 at 0:59
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Thus, EM waves would not spread spherically anymore, but only towards the singularity

This is not true because inside the horizon "radially inward" is the only possible radial direction but the metric is still spherically symmetric, so the restriction is only abut the radial coordinate.

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If the astronauts are both infalling yet one (A) is closer to the singularity than the other (B), then their communication is already in trouble as they approach the horizon. Their different position in the gravitational field means that B is slowed relative to A. As the one further in crosses the horizon, his radio messages to his partner fade out -- smoothly, not suddenly -- because all their energy is lost. I don't think communication can be regained even after A crosses the horizon.

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According to "The Large scale structure of space-time" by Steven W. Hawking and George F.R. Ellis. cited by Neil deGrasse Tyson in the following short video, if you fall into a black hole, you will end-up in an entire new universe. They mentions that the outside horizon of our universe has the same properties as the event horizon of a black hole.

The point is if light cannot escape a black hole, it is because it's liberation speed is higher than the speed of light, but our universe is expanding at a very high speed and light cannot escape it because it is expanding faster than the speed of light.

So, if both astronauts didn't set up a way to prevent them to be separated, they might well end up separated by a growing distance which would prevent their communication from working.

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Let's first consider two astronauts hovering near the event horizon of a black hole. When they communicate, using laser beams, they notice that they must aim the beams almost straight up, and that there is almost no delay, the beam is very fast. (In their accelerating frame the speed of light somewhere above them is billion cees)

An outside observer would say that it takes a very long time for the information to travel between the astronauts.

Next the astronauts turn off the rocket motors that made the hovering possible. Now things become normal for them. So it takes a normal time for the beam to traverse the distance.

Astronauts' very fast communication was outside observer's very slow communication, and now the communication slows down even more. The outside observer knows that the astronauts must be communicating extremely slowly now.

So it must be so that falling to the center of a black hole takes extremely long time according to an outside observer. That way the astronauts can have a quite long conversation while falling.

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    $\begingroup$ But they can say very little during thiss... loongg ....... cooonnnnvvvvvveeeeeeerrrrrrrsssssssssaaaaaaaaaaattttttttttttttttttttttttttttiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo $\endgroup$
    – Peter - Reinstate Monica
    Commented Jul 12, 2020 at 14:34
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    $\begingroup$ @Peter-ReinstateMonica go on... $\endgroup$
    – Michael
    Commented Jul 12, 2020 at 23:01
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This is my second answer. Don't worry, it starts diverging from my first answer at some point.

Let's first consider two astronauts hovering near the event horizon of a black hole. When they communicate, using laser beams, they notice that they must aim the beams almost straight up, and that there is almost no delay, the beam is very fast. (In their accelerating frame the speed of light somewhere above them is billion cees)

An outside observer would say that it takes a very long time for the information to travel between the astronauts.

Next the astronauts turn off the rocket motors that made the hovering possible. Now things become normal for them. So it takes a normal time for the beam to traverse the distance.

Astronauts' very fast communication was outside observer's very slow communication, and now the communication slows down even more. The outside observer knows that the astronauts must be communicating extremely slowly now.

When the rocket motors were turned off, four things happened according to an outside observer:

  1. Speed of communication jumped from very slow to extremely slow
  2. The speed of the brains of the astronauts did not change
  3. The speed of communication started accelerating
  4. The speed of the brains of the astronauts started accelerating

All this happened because the astronauts were moving very fast when they were hovering next to the event horizon - which is often said to be moving at the speed of light. When they turned the motors off they started decelerating, and the ticking rate of their clocks and the communication speed started accelerating.

So, an outside observer says that radio waves can travel between the astronauts many times during the fall, because the radio waves travel quite fast between the astronauts, much faster than between astronauts that hover near event horizon.

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  • $\begingroup$ Don't infalling observers tick slower as seen by an outside observer? In particular, the outside observer sees a finite number of ticks whereas the infalling observer sees a greater number. $\endgroup$
    – Criticizing Israel not allowed
    Commented Jul 13, 2020 at 10:21
  • $\begingroup$ @user253751 Sure. Outside observer spends infinite time seeing those ticks that occur above the horizon, because it takes an infinite time for light to travel from horizon to outside observer, which my outside observer knows. My outside observer is an outside commentator, her job is to comment the events inside and outside the event horizon. $\endgroup$
    – stuffu
    Commented Jul 13, 2020 at 16:40
  • $\begingroup$ Right, so the outside observer sees the speed of communication slow down as the observer falls closer to the event horizon, right? $\endgroup$
    – Criticizing Israel not allowed
    Commented Jul 13, 2020 at 19:37
  • $\begingroup$ @user253751 Yes. $\endgroup$
    – stuffu
    Commented Jul 13, 2020 at 20:01
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The radios would not work because the gravity gradient would rip the radios ( and the astronauts) into small pieces.

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    $\begingroup$ I was going to say that this would depend on the radius of the event horizon, but after looking up the largest known black hole (TON 618, event horizon radius = 1300 AU), I read that it puts out energy equal to about 140 trillion times that of our sun. Even getting near the event horizon is going to be problematic. $\endgroup$
    – EvilSnack
    Commented Jul 13, 2020 at 15:16
  • $\begingroup$ How does a black hole put out energy? $\endgroup$
    – R.W. Bird
    Commented Jul 14, 2020 at 12:44
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    $\begingroup$ The black hole doesn't, but its accretion disk does. All of the in-falling matter (already moving at a significant fraction of c) collides with other matter undergoing the same fate, and this releases energy, some of which propagates outward. $\endgroup$
    – EvilSnack
    Commented Jul 19, 2020 at 19:50

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