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It has been said time and again, that an observer who falls into a black hole will not notice anything special. Is this really true?

There is of course the problem with the tidal forces, but I assume that these can be lowered beyond any limit by making the black hole big enough, right?

But what if I take a cigar-shaped spaceship and let it free-fall into a black hole? There will be a moment, where the head is already inside, but the tail is still outside. This of course requires some increase in gravity along the axis of the spaceship, but this increase can be so small, that an observer won't have to worry about it, but still the event horizon will be inside the spaceship.

Now the obeserver could send a light beam from the head to the tail of his spaceship and have it reflected there. Since the observer is in free fall and tidal forces are minimal, I would expect that the beam would just bounce back. This would be in line with the statement, that he does not observe anything special.

However that cannot be true, because he has just sent a beam of light across the event horizon.

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If I remember correctly the bigger the black hole is, the tidal forces that the body feels are weaker, and the smaller it is they are stronger, but in all reality if you are falling in the black hole you are doomed one way or another :D – dingo_d Dec 28 '13 at 22:43

The "nothing special" is a local property and applies only to those who are small enough. In extreme situations like black holes, the usual allowed area of locality becomes much smaller due to the extreme curvature. A cigar-shaped spaceship is not local as it spans some radial distance.

In addition, from an outside observer's point of view, there is never a point when the spaceship is straddling the horizon. That is because from an outsider's POV, infalling material gets "stuck" to the horizon in a black hole.

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Are you saying, that nothing straddling the even horizon can be small enough to be considered "local"? If so, then there is something special for an observer free-falling into a black hole! – Martin Drautzburg Dec 28 '13 at 22:21
@MartinDrautzburg Well, the other thing is that from the observer's point of view, the true event horizon always lies below him. When he passes the externally visible event horizon, nothing special happens. – Manishearth Dec 28 '13 at 22:26
I didn't know that for the falling observer the event horizon is always below him. That shomewhat shoots this thought experiment, doesn't it? Mind elaborating a bit and posting this as an answer? – Martin Drautzburg Dec 28 '13 at 23:17
@MartinDrautzburg from here "In terms of visual appearance, observers who fall into the hole perceive the black region constituting the horizon as lying at some apparent distance below them, and never experience crossing this visual horizon". The falling observer does cross a "point of no return", but that point is nothing special otherwise. But from their point of view (moving frame), light particles ahead of them have a different point of no return. – Manishearth Dec 28 '13 at 23:20
I'm a bit fuzzy on this though, it would be best if someone more experienced answered :) – Manishearth Dec 28 '13 at 23:21

To observe "nothing special" near a black hole, you would have to be staring very intently away from it. If you could see it as more than a "missing" point source, you could see the entire universe wrapped around it in perfect Einstein rings. The closer you approached the black hole, the more it would intrude into your remaining field of vision. Around 1.5 Schwarzschild radii, the only stable orbit available to light is a perfect circle, meaning the rest of the universe would appear to be piling up onto a perfectly flat infinite nothing. Below the photon sphere, there are no stable orbits. The black hole would be clearly visible as a gigantic missing thing curling around you — only light from a shrinking circle above could reach you.

Robert Nemiroff has simulated the effects of some high-gravity environments:

Virtual Trips to Black Holes and Neutron Stars

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Just a hunch, but I am guessing that the time it would take for the light beam to reach the tail of the ship would be longer than it takes for the tail to pass through the event horizon.

Keep in mind that "nothing special" is a massive understatement. It's not like nothing is happening when you are that close to a black hole :) But, relative to what goes on outside of the horizon, nothing else happens (at first).

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