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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)?

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you are still able to see rays from outside the event horizon after you fall, they just become severely blue-shifted – lurscher Jul 23 '12 at 17:57
I don't understand your 1) question - since the event horizon is invisible (there is nothing special going on at the event horizon) then in what sense is it visible "in front of me" as you pass it. – FrankH Jul 23 '12 at 18:06
@FrankH, it would be "visible" in the sense that I would see a black circular region in front of me, no matter how close to the singularity I get. Is that not correct? – Dmitry Brant Jul 23 '12 at 18:25
I think I agree with your statement in 2. As you are looking back at the outside observer, you would see the observer rushing away from you at an increased velocity. Eventually, the observer will be beyond your observable universe (like you said, you are moving away faster than the observers light can reach you.) – Michael Jul 23 '12 at 18:33
Your eye would be pulled apart from the rest of your body before you reached the event horizon. ? – Bingo Jan 16 '13 at 10:08

3 Answers 3

The answer is actually a little werider than you imply. The proper way to look at this problem is to look at the paths of light rays intersecting your position as you get closer and closer to the black hole (since these are reversible, they can represent you sending a signal out, or you observing distant stars).

When you are far away, the black hole will get gradually larger as you approach. The closer and closer you get, a bigger and bigger fraction of the sky will appear dark, until it fills the whole horizon. You are still not inside the event horizon, however. As you move closer, the event horzon will fill a larger and larger portion of the sky behind you, until the whole universe will appear as a little dot on the horizon directly away from the hole, and then will disappear as you cross the horizon.

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This disagrees with the account given by Taylor and Wheeler in their excellent book on Black Holes. Only a point that was originally directly behind you remains directly behind you. The rest is distorted outward and ultimately becomes an infinitely thin ring splitting the sky in two at the singularity and with an infinite blue shift. – Rob Jeffries Jun 1 at 7:47
@RobJeffries: I don't quite get what you're saying (what I said was mostly just limited to the apparent shape of the horizon), but the singularity is definitely not visible outside the horizon. – Jerry Schirmer Jun 1 at 14:55
Its the bit about the whole universe appearing as a dot and then disappearing. The "dot" part is I think only true if you can somehow arrange to be stationary just outside the event horizon. If falling into the black hole nothing special happens at the event horizon and you can still see the outside universe, and it actually distorts outwards, tending towards being a blueshifted, intense ring as the singularity is approached. – Rob Jeffries Jun 1 at 15:35

There is a relatively new theory (2012) called the firewall theory, that says that at the event horizon there is a huge "wall of fire" as such. This is because quantum entangled particles that cross the horizon (or one half of a pair of entangled particles) becomes tricky and starts breaking laws like the monogamy of entanglement. So a group of physicists thought that there must be a mechanism that breaks entanglement at the event horizon to avoid this paradox. In order to do this there would need to be a huge amount of energy at the event horizon. So for any observer, their journey might stop at the event horizon because they would simply burn up or be ripped to atoms. The firewall theory is still just a theory though and a lot of people don't like it (eg: Stephen Hawking) but, just a thought :)

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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.

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