[Physics FAQ] - [Copyright]

Updated 1995 by MM.
Original by Matt McIrvin.

What happens to you if you fall into a black hole?

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/r3.  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.

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.

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.

Will you see the universe end?

If an external observer sees me slow down asymptotically as I fall, 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.

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 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.)

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.