I have read this question:

so looking toward the singularity, you see the horizon retreating from you as you fall in - even after you've already crossed the horizon.

Are black holes naked singularities for an observer within the event horizon?

And this one:

A falling observer does not experience passing through an event horizon as you describe.

Thought experiment - would you notice if you fell into a black hole?

As far as I understand, the event horizon is not a physical object, but a boundary between regions of space. Now the first answer says that you can see the horizon retreat in front of you, but the second one says (and some other answers on this site) that you can't even notice the horizon, not even when you cross it (if the BH is large and tidal forces are not relevant). Now the latter seems logical, since it is just a boundary, and it is not noticeable. Then how can an infalling observer see something to retreat in front of it, if you can't even notice it? And how can you cross it if you see it retreat in front if you?


  1. How can the infalling observer see the event horizon to retreat in front after the observer crossed it (and can't even notice it)?

1 Answer 1


The first answer is incorrect. The author may have confused the event horizon with the so-called antihorizon:

The antihorizon is the approximate location of matter that fell into the hole earlier, including the matter that originally formed it, though it is so redshifted that all you really see is blackness. The antihorizon is what you see when you look at a black hole from outside, and if you fall in while facing downward, it continues to be what you see ahead of you even after you cross the event horizon.

After you cross the event horizon, you can see it behind you, in the sense that light emitted from it will reach you. You can't use this to determine when you crossed the event horizon because the same thing happens when you cross any other outward-moving null surface.

  • $\begingroup$ Thank you so much! $\endgroup$ Dec 4, 2021 at 4:39
  • $\begingroup$ @safesphere Kruskal coordinates have a "formal Lorentz boost" symmetry $(T,X)\mapsto (γ(T-βX), γ(X-βT))$. If you start with a diagram like this one and boost it by a large amount, the matter ends up so close to the line $T+X=0$ that it may as well be taken to be on it. Compactifying then gives you the Penrose diagram in the answer. I added the word "approximate" to the answer. $\endgroup$
    – benrg
    Dec 4, 2021 at 6:39

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