# Schwarzschild solution to Einstein's equation and singularities

In the book D'Inverno - Introducing Einstein's relativity, in the non-rotating black holes part, the following statement has been stated.

In Eddington-Finkelstein coordinates at $r=2m$ the radially outgoing photons "stay where they are".

What is the meaning of this statement? Does it mean that photons have come to rest, but that doesn't make sense because photons always travel at a speed $c$?

• D'Inverno uses scare quotes because $dr/dt$ is not a velocity, it's just the rate of change of a coordinate, or a "coordinate velocity." – Ben Crowell Sep 1 '17 at 3:45

## 1 Answer

The important point is that "stopped" depends on the coordinates. The usual coordinates $(t, r)$ used for black holes are good for someone standing far away from the black hole but they say weird things about what happens at the horizon.

What the quote says is that if a photon is at $r = 2M$ (the horizon) and not moving sideways, then its $r$ coordinate will stay constant as $t$ advances. But one of the basic principles of general relativity is that special relativity always holds locally. In simpler words, you always observe light near you to be moving at $c$. But if the light is far away from you (say, you are safely in Earth while the light is at the black hole), these coordinates are just numbers, and the "velocity" as calculated in them may not be very physical. To put it another way, $dr/dt$ is only the speed of the photon if it is very far away from the black hole.

But the $(t,r)$ coordinates kind of fail at the horizon. Like I said before, the quote says that certain photons have $dr/dt = 0$: the photon is "glued" to the horizon. But we know that only someone who is right next to the photon will always observe it moving at $c$.

So consider the following logic: we know that the photon stays at the horizon, because its $r$ coordinate is constant (this is what "stopped" in quotes means). Also, someone in free fall will always observe light near them to travel at $c$. So think what happens if you fall through the horizon: the light must move at $c$ away from you, and since it and the horizon stay together, the horizon also moves away from you at the speed of light.

• But then that would mean that I will see that at the horizon the photons have stopped. I don't understand when you said that I will see it recede away from me at the speed of light. Can you be a bit more clearer – Khushal Sep 1 '17 at 11:59
• @Aniket I've added more detail, please tell me what you think. – Javier Sep 1 '17 at 14:35
• I understood your argument and it seems logical. You are saying that the horizon moves away from me at the speed of light. I guess there is no way to verify this but this seems to be the most logical answer. – Khushal Sep 5 '17 at 13:17
• @Aniket experimentally verifying this would be pretty hard, yes, but it's not just a logical argument: it's what the theory says, the same theory that predicts the existence of black holes. – Javier Sep 5 '17 at 13:34