3. Won't it take forever for you to fall in? Won't it take forever for the black hole to even form?

On my worldline as I fall into the black hole, it turns out that the Schwarzschild coordinate called t goes to infinity when I go through the event horizon. That doesn't correspond to anyone's proper time, though; it's just a coordinate called t. In fact, inside the event horizon, t is actually a *spatial* direction, and the future corresponds instead to decreasing r. It's only outside the black hole that t even points in a direction of increasing time. In any case, this doesn't indicate that I take forever to fall in, since the proper time involved is actually finite.

At large distances t *does* approach the proper time of someone who is at rest with respect to the black hole. But there isn't any non-arbitrary sense in which you can call t at smaller r values "the proper time of a distant observer," since in general relativity there is no coordinate-independent way to say that two distant events are happening "at the same time." The proper time of any observer is only defined locally.

A more physical sense in which it might be said that things take forever to fall in is provided by looking at the paths of emerging light rays. The event horizon is what, in relativity parlance, is called a "lightlike surface"; light rays can remain there. For an ideal Schwarzschild hole (which I am considering in this paragraph) the horizon lasts forever, so the light can stay there without escaping. (If you wonder how this is reconciled with the fact that light has to travel at the constant speed c-- well, the horizon *is* traveling at c! Relative speeds in GR are also only unambiguously defined locally, and if you're at the event horizon you are necessarily falling in; it comes at you at the speed of light.) Light beams aimed directly outward from just outside the horizon don't escape to large distances until late values of t. For someone at a large distance from the black hole and approximately at rest with respect to it, the coordinate t does correspond well to proper time.

So if you, watching from a safe distance, attempt to witness my fall into the hole, you'll see me fall more and more slowly as the light delay increases. You'll never see me actually *get to* the event horizon. My watch, to you, will tick more and more slowly, but will never reach the time that I see as I fall into the black hole. Notice that this is really an optical effect caused by the paths of the light rays.

This is also true for the dying star itself. If you attempt to witness the black hole's formation, you'll see the star collapse more and more slowly, never precisely reaching the Schwarzschild radius.

Now, this led early on to an image of a black hole as a strange sort of suspended-animation object, a "frozen star" with immobilized falling debris and gedankenexperiment astronauts hanging above it in eternally slowing precipitation. This is, however, not what you'd see. The reason is that as things get closer to the event horizon, they also get *dimmer*. Light from them is redshifted and dimmed, and if one considers that light is actually made up of discrete photons, the time of escape of *the last photon* is actually finite, and not very large. So things would wink out as they got close, including the dying star, and the name "black hole" is justified.

As an example, take the eight-solar-mass black hole I mentioned before. If you start timing from the moment the you see the object half a Schwarzschild radius away from the event horizon, the light will dim exponentially from that point on with a characteristic time of about 0.2 milliseconds, and the time of the last photon is about a hundredth of a second later. The times scale proportionally to the mass of the black hole. If I jump into a black hole, I don't remain visible for long.

Also, if I jump in, I won't hit the surface of the "frozen star." It goes through the event horizon at another point in spacetime from where/when I do.

(Some have pointed out that I really go through the event horizon a little earlier than a naive calculation would imply. The reason is that my addition to the black hole increases its mass, and therefore moves the event horizon out around me at finite Schwarzschild t coordinate. This really doesn't change the situation with regard to whether an external observer sees me go through, since the event horizon is still lightlike; light emitted at the event horizon or within it will never escape to large distances, and light emitted just outside it will take a long time to get to an observer, timed, say, from when the observer saw me pass the point half a Schwarzschild radius outside the hole.)

All this is not to imply that the black hole can't also be used for temporal tricks much like the "twin paradox" mentioned elsewhere in this FAQ. Suppose that I don't fall into the black hole-- instead, I stop and wait at a constant r value just outside the event horizon, burning tremendous amounts of rocket fuel and somehow withstanding the huge gravitational force that would result. If I then return home, I'll have aged less than you. In this case, general relativity can say something about the difference in proper time experienced by the two of us, because our ages can be compared *locally* at the start and end of the journey.