Answers posted by Matt McIrvin of Harvard University, last updated 02-FEB-1995. Text in this file is NOT copyrighted by RJN.

Contents:

1. What is a black hole, really?

2. What happens to you if you fall in?

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

4. Will you see the universe end?

5. What about Hawking radiation?
Won't the black hole evaporate before you get there?

6. How does the gravity get out of the black
hole?

7. Where did you get that information?

What such a solution really looks like is a "metric," which is a kind of generalization of the Pythagorean formula that gives the length of a line segment in the plane. The metric is a formula that may be used to obtain the "length" of a curve in spacetime. In the case of a curve corresponding to the motion of an object as time passes (a "timelike worldline,") the "length" computed by the metric is actually the elapsed time experienced by an object with that motion. The actual formula depends on the coordinates chosen in which to express things, but it may be transformed into various coordinate systems without affecting anything physical, like the spacetime curvature. Schwarzschild expressed his metric in terms of coordinates which, at large distances from the object, resembled spherical coordinates with an extra coordinate t for time. Another coordinate, called r, functioned as a radial coordinate at large distances; out there it just gave the distance to the massive object.

Now, at small radii, the solution began to act strangely. There was a "singularity" at the center, r=0, where the curvature of spacetime was infinite. Surrounding that was a region where the "radial" direction of decreasing r was actually a direction in *time* rather than in space. Anything in that region, including light, would be obligated to fall toward the singularity, to be crushed as tidal forces diverged. This was separated from the rest of the universe by a place where Schwarzschild's coordinates blew up, though nothing was wrong with the curvature of spacetime there. (This was called the Schwarzschild radius. Later, other coordinate systems were discovered in which the blow-up didn't happen; it was an artifact of the coordinates, a little like the problem of defining the longitude of the North Pole. The physically important thing about the Schwarzschild radius was not the coordinate problem, but the fact that within it the direction into the hole became a direction in time.)

Nobody really worried about this at the time, because there was no known object that was dense enough for that inner region to actually be outside it, so for all known cases, this odd part of the solution would not apply. Arthur Stanley Eddington considered the possibility of a dying star collapsing to such a density, but rejected it as aesthetically unpleasant and proposed that some new physics must intervene. In 1939, Oppenheimer and Snyder finally took seriously the possibility that stars a few times more massive than the sun might be doomed to collapse to such a state at the end of their lives.

Once the star gets smaller than the place where Schwarzschild's coordinates fail (called the Schwarzschild radius for an uncharged, nonrotating object, or the event horizon) there's no way it can avoid collapsing further. It has to collapse all the way to a singularity for the same reason that you can't keep from moving into the future! Nothing else that goes into that region afterward can avoid it either, at least in this simple case. The event horizon is a point of no return.

In 1971 John Archibald Wheeler named such a thing a black hole, since light could not escape from it. Astronomers have many candidate objects they think are probably black holes, on the basis of several kinds of evidence (typically they are dark objects whose large mass can be deduced from their gravitational effects on other objects, and which sometimes emit X-rays, presumably from infalling matter). But the properties of black holes I'll talk about here are entirely theoretical. They're based on general relativity, which is a theory that seems supported by available evidence.

For ordinary black holes of a few solar masses, there are actually large tidal forces well outside the event horizon, so I probably wouldn't even make it into the hole alive and unstretched. For a black hole of 8 solar masses, for instance, the value of r at which tides become fatal is about 400 km, and the Schwarzschild radius is just 24 km. But tidal stresses are proportional to M/r^3. Therefore the fatal r goes as the cube root of the mass, whereas the Schwarzschild radius of the black hole is proportional to the mass. So for black holes larger than about 1000 solar masses I could probably fall in alive, and for still larger ones I might not even notice the tidal forces until I'm through the horizon and doomed.

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.

That, at least, is the story for an uncharged, nonrotating black hole. For charged or rotating holes, the story is different. Such holes can contain, in the idealized solutions, "timelike wormholes" which serve as gateways to otherwise disconnected regions-- effectively, different universes. Instead of hitting the singularity, I can go through the wormhole. But at the entrance to the wormhole, which acts as a kind of inner event horizon, an infinite speed-up effect actually does occur. If I fall into the wormhole I see the entire history of the universe outside play itself out to the end. Even worse, as the picture speeds up the light gets blueshifted and more energetic, so that as I pass into the wormhole an "infinite blueshift" happens which fries me with hard radiation. There is apparently good reason to believe that the infinite blueshift would imperil the wormhole itself, replacing it with a singularity no less pernicious than the one I've managed to miss. In any case it would render wormhole travel an undertaking of questionable practicality.

Short answer: No, it won't. This demands some elaboration.

From thermodynamic arguments Stephen Hawking realized that a black hole should have a nonzero temperature, and ought therefore to emit blackbody radiation. He eventually figured out a quantum- mechanical mechanism for this. Suffice it to say that black holes should very, very slowly lose mass through radiation, a loss which accelerates as the hole gets smaller and eventually evaporates completely in a burst of radiation. This happens in a finite time according to an outside observer.

But I just said that an outside observer would *never* observe an object actually entering the horizon! If I jump in, will you see the black hole evaporate out from under me, leaving me intact but marooned in the very distant future from gravitational time dilation?

You won't, and the reason is that the discussion above only applies to a black hole that is not shrinking to nil from evaporation. Remember that the apparent slowing of my fall is due to the paths of outgoing light rays near the event horizon. If the black hole *does* evaporate, the delay in escaping light caused by proximity to the event horizon can only last as long as the event horizon does! Consider your external view of me as I fall in.

If the black hole is eternal, events happening to me (by my watch) closer and closer to the time I fall through happen divergingly later according to you (supposing that your vision is somehow not limited by the discreteness of photons, or the redshift).

If the black hole is mortal, you'll instead see those events happen closer and closer to the time the black hole evaporates. Extrapolating, you would calculate my time of passage through the event horizon as the exact moment the hole disappears! (Of course, even if you could see me, the image would be drowned out by all the radiation from the evaporating hole.) I won't experience that cataclysm myself, though; I'll be through the horizon, leaving only my light behind. As far as I'm concerned, my grisly fate is unaffected by the evaporation.

All of this assumes you can see me at all, of course. In practice the time of the last photon would have long been past. Besides, there's the brilliant background of Hawking radiation to see through as the hole shrinks to nothing.

(Due to considerations I won't go into here, some physicists think that the black hole won't disappear completely, that a remnant hole will be left behind. Current physics can't really decide the question, any more than it can decide what really happens at the singularity. If someone ever figures out quantum gravity, maybe that will provide an answer.)

Often this question is phrased in terms of gravitons, the hypothetical quanta of spacetime distortion. If things like gravity correspond to the exchange of "particles" like gravitons, how can they get out of the event horizon to do their job?

Gravitons don't exist in general relativity, because GR is not a quantum theory. They might be part of a theory of quantum gravity when it is completely developed, but even then it might not be best to describe gravitational attraction as produced by virtual gravitons. See the FAQ on virtual particles for a discussion of this.

Nevertheless, the question in this form is still worth asking, because black holes *can* have static electric fields, and we know that these may be described in terms of virtual photons. So how do the virtual photons get out of the event horizon? Well, for one thing, they can come from the charged matter prior to collapse, just like classical effects. In addition, however, virtual particles aren't confined to the interiors of light cones: they can go faster than light! Consequently the event horizon, which is really just a surface that moves at the speed of light, presents no barrier.

I couldn't use these virtual photons after falling into the hole to communicate with you outside the hole; nor could I escape from the hole by somehow turning myself into virtual particles. The reason is that virtual particles don't carry any *information* outside the light cone. See the FAQ on virtual particles for details.

Information about evaporation and wormholes came from Robert Wald's _General Relativity_ (Chicago: University of Chicago Press, 1984). The famous conformal diagram of an evaporating hole on page 413 has resolved several arguments on sci.physics (though its veracity is in question).

Steven Weinberg's _Gravitation and Cosmology_ (New York: John Wiley and Sons, 1972) provided me with the historical dates. It discusses some properties of the Schwarzschild solution in chapter 8 and describes gravitational collapse in chapter 11.

Posted to sci.astro frequently asked questions by Michael McIrvin.

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