Sticky Black Holes

Sticky Black Holes

Outside a Schwarzschild black hole (or any non-rotating star or planet, or other spherically symmetric object) we can use Schwarzschild coordinates. Far from the hole, the time coordinate of Schwarzschild coordinates -- "Schwarzschild time" -- corresponds to "wall clock time".

A photon at the event horizon takes infinite time to get from the horizon to a point far away from the hole. In other words, it never gets out. In Schwarzschild coordinates, the velocity of light doesn't depend on what direction it's going in -- light paths are reversible, just as they are in the inertial frames of special relativity. Going or coming, traversing a particular path takes the same amount of time in either direction. So, it also must take infinite time for a photon to fall down to the event horizon from far away. In other words, in Schwarzschild coordinates, a photon -- or anything else -- can never fall into a black hole, because it can never actually get to the event horizon. This is most unexpected!

[NOTE -- Before I continue, I should mention that the horizon is typically analyzed using Kruskal-Szekeres coordinates, which are nonsingular at the horizon. However, they also don't map so nicely into our "real world" coordinates far from the hole, which is why I wanted to come to an understanding of what's going on with the Schwarzschild coordinates directly.]

"You Can't Get There from Here"

Any standard text on general relativity will provide a rather complicated proof that it takes infinite Schwarzschild time to fall to the event horizon. Here's a demonstration of something simpler which leads to the same conclusion.

Let's look at the fastest thing around, a photon, as it heads directly toward the event horizon of a Schwarzschild black hole.

The black hole has mass M (measured in meters. The conversion factor from kg to meters is roughly 10^-27) and is non-rotating and uncharged. The Schwarzschild metric which describes the field around the hole, expressed as a line element, is

(1)

where the coordinates are (t, r,

) and d

is defined as

.

We can simplify this quite a bit. First, let's assume the mass of the hole is 1/2, which, by the way, is a bit more than 100 Earth masses. Then all those 2M terms turn into 1's. Now we're looking at a black hole which is 2 meters across, but the results will apply to a hole of any size. Next, let's look at the region just outside the event horizon:

(2)

and, substituting 1 for 2M and 2M+

for r, we have:

(3)

Now the metric close to the horizon looks like this:

(4)

The photon, being a photon, has no proper velocity. But we can find a tangent to its path. Since it travels a null geodesic, the proper length of any such tangent vector must be zero, so, up to a multiplicative constant, we can see by inspection of the metric that a tangent to its path as it travels straight into the hole must look like this:

(5)

Now, I want to find out when it'll make it to the horizon, expressed in the Schwarzschild coordinates. For that, we want the three-velocity, which is just the spatial part of the 4-velocity divided by the time component of the 4-velocity (rate at which its space coordinate is changing, divided by the rate at which its time coordinate is changing):

(6)

The 3-velocity is just the ordinary time derivative of the radius coordinate, so we have

(7)

The photon slows down as it gets close to the horizon, and finally stops if it gets to the horizon.

The solution to that equation is the exponential function, so we must have

(8)

and

never reaches zero, and the photon never gets to the horizon! (And so it never actually stops, either.)

Why you Can Get There from Here, After All

But this is all in Schwarzschild coordinates. In a particle's own frame of reference, it actually gets to the horizon in a finite amount of proper time.

So what's really going on? If you fall into a black hole, do you just "hibernate" until the end of time, falling more and more slowly? Do you get a ringside seat when the Last Trump is blown, as you sit in your spaceship, suspended forever just above the event horizon?

Well, it's a little hard to be sure when we're discussing simultaneity at a distance in curved space, but the answer seems to be a clear "NO". You actually fall through the horizon and on into the singularity more or less instantly. It appears that the Schwarzschild time coordinate is very misleading in this case. The time it takes a photon to come up out of the hole from a point just above the horizon grows arbitrarily as the starting point approaches the horizon -- that's a fact. And a photon which falls toward the hole, but bounces off a mirror just above the horizon and comes back out, takes a very long time to make the round trip -- that's a fact. BUT is the time it takes to go down (almost) to the horizon really the same as the time it takes it to come back "up the hill" again? The coordinate system says so. As far as I can tell, however, the correct answer is a resounding "no" -- the photon goes down the hill a lot faster than it comes back up the hill.

Note carefully, however, that this alleged difference in the time it takes to go down versus the time it takes to come up cannot be measured or detected in any way. All we can say for sure is that something very strange is going on, and a mental model in which the photon goes faster one way than the other seems to fit nicely with it.

There is also a gedanken experiment we can do which will make the "fast in -- slow out" scenario seem very plausible. Let's drop a mirror into the hole. Then we'll wait a while, to give it time to get close to the horizon. Finally we'll send a photon in after it, and wait for it to be reflected from the mirror. If the mirror's really sticking just above the horizon, going slower and slower, then the photon should zoom in, bounce off the mirror, and come back out where we can detect it. Right? And that should happen, no matter how long we wait before we send in the photon. If that's actually the case, it's proof that the mirror never gets to the horizon!

Well, guess what -- it's not the case. If the mirror is "too close" to the horizon when the photon is sent in, the photon will not catch up with it before it gets to the horizon. In that case, even though the mirror may seem to be hanging forever above the horizon according to the equations, it's impossible to interact with it in any way: even something traveling toward it at the speed of light never gets to it. This is rather surprising, unless -- as we already suspected -- the mirror had actually zoomed right on through the horizon, leaving behind a cloud of "slow" photons which will still be struggling up out of the hole for the rest of Eternity.

The proof that the photon can't always catch up to the mirror is reasonably straightforward, if we take the same approach we took in the first section, and use a simplified metric close to the horizon.

A Proof that the Photon Can't Always Catch the Mirror

We again assume we're looking at a black hole of mass 1/2 (or roughly 100 Earth masses), and we're concerned only with the region just outside the event horizon. In this area, if

= r-1, then the Schwarzschild metric (1) reduces to (4), which we repeat here:

(9)

If we drop the mirror and then immediately fire a photon at it, of course the photon will catch up with the mirror and bounce off. But we're not concerned with that case. We're concerned only with the case where we wait until the mirror has already approached within

of the horizon before we release the photon.

If the photon is to catch up with the mirror in this case, then it, too, must eventually be close to the horizon, but not as close as the mirror. So, let's define

(10) 0 <

₁ <

₂ << 1

where the mirror is at distance

₁ and the photon is at distance

₂ from the horizon. We'll also set our clock to 0 at this moment. Then at time 0, the 4-velocity of the mirror can be written as

(11)

where a and b are both functions of time, and are both > 0.

We want the 3 velocity of the mirror. It's

(12)

This must be between 0 and the 3-velocity of a photon. But the 3-velocity of a photon approaching the horizon is just ε, so we must have that, for all values of t,

(13)

Since the photon is at ε₂ at time 0, we know that the equation of motion of the photon, which takes the form of (8) close to the horizon, must be

. To find the point where the photon and the mirror intersect, we need the equation of motion of the mirror, too.

For that, we will need to set up the equation of geodesic and integrate it, which unfortunately is a little messy. First, though, we can observe that the squared magnitude of the 4-velocity is always -1, so we have

(14)

or

(15)

which implies that a and b must be cosh(u) and sinh(u), respectively, for some function u. Integrating the geodesic equation for the mirror (see Lemma, below) shows that

(16)

where

(17)

and we find that the 4-velocity of the mirror is

(18)

Then the 3-velocity of the mirror is

(19)

Cross-multiplying to separate terms and integrating, we get

(20)

With a simple change of variables and the help of an integral table, we find this is

(21)

Exponentiating and inverting, we get

(22)

and finally, isolating

_f and renaming T to t, we get the equation we really want, which is

(23)

When the photon intercepts the mirror, this must match the position of the photon, which we obtained up above. So at interception we have:

(24)

Multiplying through by cosh^-2(K)

cosh²(t/2 + K)/

₂, we get:

(25)

Expanding cosh²(t/2 + K) into exponentials:

(26)

Isolating the exponentials and multiplying them out:

(27)

But:

(28)

so by converting this to an inequality we can eliminate t. Then, multiplying across by

and expanding cosh as exponentials, we have

(29)

and hence, if

(30)

then there can be no solution, and the photon never catches up with the mirror.

Lemma: Derivation of the 4-Velocity of the Mirror

The 4-velocity must be parallel transported by itself -- or, in other words, the particle doesn't "turn" relative to the direction it's trying to go in as it moves. In terms of equations, the covariant derivative of the 4-velocity with respect to itself is zero. Now we need to evaluate that derivative, set it to zero, and solve the resulting equations.

The covariant derivative is just the ordinary directional derivative plus the amount we gain as a result of changes in the "values" of the basis vectors, and that amount depends on the metric.

The metric is diagonal and depends only on

, and we're only concerned with the first two terms. Inverting a diagonal matrix is easy, and the partial derivatives we need are pretty simple. Here are all the terms and partials from the metric which we'll need:

(L.1)

The only partials which are nonzero have three ones or two zeros and a one. That, plus the fact that the metric is diagonal, means only a few of the Christoffel symbols are nonzero:

(L.2)