Sticky Black Holes |

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!

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

(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

(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

(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

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.

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

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

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.

(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 <

where the mirror is at distance

(11)

where

We want the

(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 ε

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

(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

(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

(25)

Expanding cosh

(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 byand 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.

The covariant derivative is just the ordinary directional derivative

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)

(L.3)

(L.4)

Now we need to solve a system of two equations:

(L.5)

I'm going to assume that the form of the velocity is

(L.6)

where

(L.7)

Starting with vV

(L.8)

so

(L.9)

and

(L.10)

Next, we consider

(L.11)

so

(L.12)

and

(L.13)

At this point we could rewrite

(L.14)

and

(L.15)

where

(L.16)

which depends on the 3-velocity of the mirror at point ε

(L.17)