Why is there a definite boundary of a black hole that marks point of no return?

Question

Why is there an Event horizon of a black hole?

Why is it when something enters the horizon, it is inexorably pulled inside the black hole but not so if it hovers outside the horizon? How do we understand this?

Answer 1

I'm guessing that you want an intuitive explanation rather than a mathematical proof that nothing can escape a black hole. If you want the maths then see my answer to Why is a black hole black?

I think it is possible to give some intuition as to why a horizon exists, but I'm going to have to make an analogy and the analogy is going to seem strange at first. But bear with me because there is madness in my method.

Suppose you start accelerating with an acceleration $a$, and at the same time I shine a laser towards you from a distance $c^2/a$ behind you. Remarkably, as long as you keep accelerating the light from the laser will never reach you. I go into the maths behind this in this answer to the question What is the proper way to explain the twin paradox? but to give the headline, although you can never travel faster than light (in my frame) your speed tends asymptotically towards $c$ fast enough that the light ray travelling at $c$ takes an infinite time to reach you.

This means that as long as you remain at the constant acceleration $a$ then in your rest frame there is an event horizon a distance $c^2/a$ behind you. That is, nothing from behind that horizon, including light, can ever reach you. If I'm allowed a bit of maths then the geometry of spacetime in your (accelerating) rest frame is given by the Rindler metric:

$$ \mathrm ds^2 = -\left(1 + \frac{a\,x}{c^2} \right)^2 c^2\mathrm dt^2 +\mathrm dx^2 $$

and you can immediately see that this becomes singular when $1 + a\,x/c^2 = 0$ i.e. when $x = -c^2/a$. This is the horizon, and it's why I started saying I would shine the laser from a distance $c^2/a$ behind you.

The point of all this is that it is basically what happens with a black hole. We've established that in the rest frame of an accelerating observer a horizon exists, and if you are hovering a fixed distance from a black hole then you are an accelerating observer so as a result a horizon exists. It might seem a bit odd to say you are accelerating when your distance from the black hole remains constant, but the key point is that general relativity does not distinguish between inertial and gravitational acceleration. Both are treated in the same way. I go into this in more detail in How can you accelerate without moving?

So my point is that the event horizon in a black hole is closely analogous to the Rindler horizon that exists for an accelerating observer. They aren't identical because the acceleration for observers hovering above a black hole depends on distance so unlike the accelerating observer it isn't constant. However the analogies are close enough to make one more important point.

Although a constantly accelerating observer sees a horizon the horizon only exists for them. For me standing on the Earth shining the laser at you no horizon exists. All I see is the laser closing on you but never quite catching you. And I would see exactly the same thing if you were hovering above the black hole and I were falling freely into it.

Suppose we start at the same distance from the black hole, and you remain there hovering while I fall inwards, then at the exact moment I cross the horizon I shine my laser at you. What I see is much the same as when I shine the laser at an accelerating observer. In my frame the light heads away from me at velocity $c$ in your direction, but even if I had infinite time available (which I don't because a messy death awaits me at the singularity) I would never observe the light to reach you.

And finally, for me falling inwards no horizon exists, just as no horizon exists when I'm on Earth shining the laser at you accelerating away. A freely falling observer never observes themselves to cross a horizon. Instead they see an apparent horizon that retreats before them and they reach it only at the moment they hit the singularity.

This answer ended up longer than I expected, and I fear my simple analogy might have ended up complicated. But if you've managed to stay with me the main point I want you to take away is that the horizon exists because of the acceleration, and for the non-accelerating observer no horizon exists. It is because there are observers for whom no horizon exists that the event horizon is just a coordinate singularity, not a true singularity like the one at the centre of the black hole.