I've heard it mentioned many times that "nothing special" happens for an infalling observer who crosses the event horizon of a black hole, but I've never been completely satisfied with that statement. I've been trying to actually visualize what an infalling observer would see (from various angles), and I'd like to know if I understand things correctly.
Suppose we travel near a sufficiently large black hole (say, the one in the center of the Milky Way) so that we could neglect tidal effects near the horizon, and suppose that it's an ideal Schwarzschild black hole. Suppose that I'm falling perfectly radially inward, and you remain at a safe distance (stationary with respect to the black hole).
1) If I'm looking inward as I fall, the event horizon will always appear to be "in front" of me, even after I've crossed the event horizon relative to you, and will continue to be "in front" of me right until I'm crushed by the singularity. This makes perfect sense (and correct me if it's wrong), but it's the following case with which I'm having the most difficulty:
2) If I'm looking back at you (an outside observer), what effects, if any, would I observe? My reasoning was: as I approach the event horizon, a "cosmological horizon" begins to close in around me, beyond which I can no longer observe the universe. At the point when I cross the event horizon relative to you, my cosmological horizon will have "engulfed" you, since I'm effectively traveling faster than light relative to you. Therefore, for me, the universe would redshift out of observability when I cross the event horizon.
Is the above correct? I guess it can't be, if all sources say that "nothing special happens"… but I don't completely understand why. Or is the following more accurate:
2a) If I'm looking back at you, I will continue to observe you even after crossing the event horizon (until my demise at the singularity), since the light emitted by you went into the event horizon along with me, even though I can no longer communicate with you (but I have no way of knowing it). And if this is true, would you appear at all red- or blue-shifted to me? Also, if this is true, do I still have a rapidly-collapsing cosmological horizon around me (even though I can't observe it)?
Best Answer
The answer to this question is addressed, for the case of a simple Schwarzschild black hole, by Taylor & Wheeler in their book called "Exploring black holes" (Addison, Wesley, Longman, 2000, pages B20-B24). There is a huge difference to what would be perceived by a "shell" observer that arranges to be stationary just outside the event horizon compared with an observer that is free-falling into the black hole.
Your question supposes that an observer is falling into the black hole. That observer will be able to see the outside universe from inside the event horizon. The light will be blue-shifted and it will be distorted/aberrated in such a way that as the observer approaches the singularity, the light from the rest of the universe is pushed outward (i.e. the viewing angle with respect to the fall-line becomes larger) into a halo and finally an intense ring of blue-shifted radiation that goes all around the sky, with blackness both in front of the observer and behind. Nothing special happens to the observer as they cross the event horizon.
A "shell" observer, using almost-infinite rocket power to hover just above the event horizon would see the whole universe compressed to a small, intense, blue-shifted dot overhead.
Of course there cannot be "shell" observers inside the event horizon since everything is compelled to move towards the singularity.
NB: This all just assumes classical GR theory. For further information and animations you could look at Andrew Hamilton's set of resources.