Item 14. Some Frequently Asked Questions About Black Holes updated 24-MAR-1993 ------------------------------------------------- original by Matt McIrvin 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? 1. What is a black hole, really? In 1916, when general relativity was new, Karl Schwarzschild worked out a useful solution to the Einstein equation describing the evolution of spacetime. This solution, a possible shape of spacetime, would describe the effects of gravity *outside* any spherically symmetric, uncharged, nonrotating object (and would serve approximately to describe even slowly rotating objects like the Earth or Sun). It worked in much the same way that you can treat the Earth as a point mass for purposes of Newtonian gravity if all you want to do is describe gravity *outside* the Earth's surface. 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 other coordinate systems without affecting anything physical, like the spacetime curvature. Schwarzschild expressed his metric in terms of coordinates which, at large distances from the black hole, resembled spherical coordinates with an extra coordinate t for time. Another coordinate, called r, functioned as a radial coordinate at large distances; it just gave the distance to the black hole. 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. 2. What happens to you if you fall in? Suppose that, possessing a proper spacecraft and a self-destructive urge, I decide to go black-hole jumping and head for an uncharged, nonrotating ("Schwarzschild") black hole. In this and other kinds of hole, I won't, before I fall in, be able to see anything within the event horizon. But there's nothing *locally* special about the event horizon; when I get there it won't seem like a particularly unusual place, except that I will see strange optical distortions of the sky around me from all the bending of light that goes on. But as soon as I fall through, I'm doomed. No bungee will help me, since bungees can't keep Sunday from turning into Monday. I have to hit the singularity eventually, and before I get there there will be enormous tidal forces-- forces due to the curvature of spacetime-- which will squash me and my spaceship in some directions and stretch them in another until I look like a piece of spaghetti. At the singularity all of present physics is mute as to what will happen, but I won't care. I'll be dead. 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. 3. Won't it take forever for you to fall in? Won't it take forever for the black hole to even form? Not in any useful sense. The time I experience before I hit the event horizon, and even until I hit the singularity-- the "proper time" calculated by using Schwarzschild's metric on my worldline -- is finite. The same goes for the collapsing star; if I somehow stood on the surface of the star as it became a black hole, I would experience the star's demise in a finite time. Now, 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 anybody'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. 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 can only be unambiguously defined locally, and if you're at the event horizon in the process of 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 illusion 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 object is about 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 from me just outside it hangs around near there for a long external time, no matter where it is.) 4. Will you see the Universe end? If an external observer sees me slow down asymptotically, it might seem reasonable that I'd see the universe speed up asymptotically-- that I'd see the universe end in a spectacular flash as I went through the horizon. This isn't the case, though. What an external observer sees depends on what light does after I emit it. What I see, however, depends on what light does before it gets to me. And there's no way that light from future events far away can get to me. Faraway events in the arbitrarily distant future never end up on my "past light-cone," the surface made of light rays that get to me at a given time. 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. 5. What about Hawking radiation? Won't the black hole evaporate before you get there? (First, a caveat: Not a lot is really understood about evaporating black holes. The following is largely deduced from information in Wald's GR text, but what really happens-- especially when the black hole gets very small-- is unclear. So take the following with a grain of salt.) 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 "slowdown" effect is an illusion caused by 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.) 6. How does the gravity get out of the black hole? Purely in terms of general relativity, there is no problem here. The gravity doesn't have to get out of the black hole. General relativity is a local theory, which means that the field at a certain point in spacetime is determined entirely by things going on at places that can communicate with it at speeds less than or equal to c. If a star collapses into a black hole, the gravitational field outside the black hole may be calculated entirely from the properties of the star and its external gravitational field *before* it becomes a black hole. Just as the light registering late stages in my fall takes longer and longer to get out to you at a large distance, the gravitational consequences of events late in the star's collapse take longer and longer to ripple out to the world at large. In this sense the black hole *is* a kind of "frozen star": the gravitational field is a fossil field. The same is true of the electromagnetic field that a black hole may possess. 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? First of all, it's important to realize that gravitons are not as yet even part of a complete theory, let alone seen experimentally. They don't exist in general relativity, which is a non-quantum theory. When fields are described as mediated by particles, that's quantum theory, and nobody has figured out how to construct a quantum theory of gravity. Even if such a theory is someday built, it may not involve "virtual particles" in the same way other theories do. In quantum electrodynamics, the static forces between particles are described as resulting from the exchange of "virtual photons," but the virtual photons only appear when one expresses QED in terms of a quantum- mechanical approximation method called perturbation theory. It currently looks like this kind of perturbation theory doesn't work properly when applied to quantum gravity. So although quantum gravity may well involve "real gravitons" (quantized gravitational waves), it may well not involve "virtual gravitons" as carriers of static gravitational forces. 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? The answer is that 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. That is a tricky subject for another (future?) FAQ entry. Suffice it to say that the reasons virtual particles don't provide an escape hatch for a black hole are the same as the reasons they can't be used for faster-than-light travel or communication. 7. Where did you get that information? The numbers concerning fatal radii, dimming, and the time of the last photon came from Misner, Thorne, and Wheeler's _Gravitation_ (San Francisco: W. H. Freeman & Co., 1973), pp. 860-862 and 872-873. Chapters 32 and 33 (IMHO, the best part of the book) contain nice descriptions of some of the phenomena I've described. 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.