Why do light rays emitted at the event horizon stay there in a black hole?

[ChemGuy]

In another post someone postulated that if two observers, at different distances, fall into a black hole two things happen:

1. When the first passes the event horizon all information flow between the two stops.

2. When the second passes the event horizon information flow would resume.

I don't think I agree with the either postulate. In the first case the observer on the inside would continue to "see" the second observer. The second observer would, of course "see" nothing. In the second case I think the situation would be exactly the same. Just because the second observer is not inside the event horizon does not mean that light can go from No. 1 to No. 2. Wouldn't the photon "leave" No.1 but never reach No. 2? Or would No. 2 eventually run into the photon?

 

[George Jones]

In another post someone postulated that if two observers, at different distances, fall into a black hole two things happen:

1. When the first passes the event horizon all information flow between the two stops.

2. When the second passes the event horizon information flow would resume.

I don't think I agree with the either postulate. In the first case the observer on the inside would continue to "see" the second observer. The second observer would, of course "see" nothing. In the second case I think the situation would be exactly the same. Just because the second observer is not inside the event horizon does not mean that light can go from No. 1 to No. 2. Wouldn't the photon "leave" No.1 but never reach No. 2? Or would No. 2 eventually run into the photon?

From looking at my crude attempt at a (Kruskal-Szekeres) spacetime diagram, it appears that the singualrity (not the event horizon) is the only bar to communication between 1 and 2.

Let A be the event at which 1's worldline intersects the event horizon and B be the event at which 2's worldline intersects the event horizon. At A, 1 can send a (light)signal that 2 receives at B. At B, 2 can send a light signal that, if the singularity is far enough in the future, 1 receives inside the even horizon.

 

[ChemGuy]

I think I understand but after 1 passes A he can not send a signal that reaches 2. One is in side the event horizen and 2 is outside.

At B can 1 send a signal that will reach 2?

For B I agree that at some point time slowes down so much that a signal from 2 will never reach 1.

 

[George Jones]

I think I understand but after 1 passes A he can not send a signal that reaches 2.

Yes he can.

One is in side the event horizen and 2 is outside.

B will receive the signal after he, too, is inside the event horizon, providing the singularity is far enought in the future that it doen't intercept the signal first. Since A is inside when he sends the signal, and B receives the signal while also inside, the signal doesn't have to cross the event horizon.

At B can 1 send a signal that will reach 2?

No. B is not on 1's worldline, so 1 cannot send a signal, to 2 or otherwise, at B.

Spacetime diagrams are crucial for understanding these concepts.

 

[Wallace]

I think I understand but after 1 passes A he can not send a signal that reaches 2. One is in side the event horizen and 2 is outside.

If observer 2 stayed outside the event horizon then they would never receive the signal, but if they cross the horizon then they can. The event horizon is a barrier that defines where light cones from that point in all end up in the central singularity in the future. In fact at the exact point of the event horizon, the 'outgoing' light rays are in fact stationary i.e. they remain at a fixed spatial co-ordinate forever. Go the smallest bit closer the singularity and the 'outgoing' light rays will bend over and point towards the singularity (no escape!), go the smalled bit further away and the outgoing rays point back out to infinity (you can escape!)*. But any signal sent at precisely the event horizon will in a sense remain there for all time. As subsequent observers plunge through the event horizon they can receive the signal, regardless of where they were when it was sent.

*Note: the exact 'shape' of the light cone paths depends on the co-ordinates you are using Schwarschild, Eddington-Finkelstein, Kruskal etc but the underlying message is that same, light rays emitted at the event horizon stay there!

 

[ChemGuy]

Thanks guys.

 

[Chris Hillman]

In another post someone postulated that if two observers, at different distances, fall into a black hole two things happen:

1. When the first passes the event horizon all information flow between the two stops.

2. When the second passes the event horizon information flow would resume.

I don't think I agree with the either postulate.

Well, you were right to object, since both statements are quite wrong; that's not what gtr says at all. Where did you read this? Is it possible that you simply misunderstood something?

Rather than trying to explain without pictures what is wrong with these two statements (but you can try http://www.math.ucr.edu/home/baez/RelWWW/history.html) , since a picture is worth a thousand words here I will simply recommend that you consult some of the fine textbooks which explain "block diagrams" (also called "conformal diagrams" or "Carter-Penrose diagrams") which are often used to exhibit the "global structure" of specific spacetime models, such as the Schwarzschild vacuum, Reissner-Nordstrom electrovacuum, or Kerr vacuum solutions of the Einstein field equation (EFE).

Wallace, you wrote "light rays emitted at the event horizon stay there", but you should have said "outwardly and radially moving laser pulses emitted by some observer as he falls past r=2m remain at r=2m" (in the Schwarzschild vacuum solution). See the execellent pictures in MTW if you don't see why I objected.

Chris Hillman