Natural Science Forum / Physics / Relativity / February 2006

Yes. But note that while that clearly is valid for paths between a
specified pair of points in spacetime, the singularity is not a point in
spacetime. Indeed, for Schw. spacetime there are an infinite number of
limit points to which the rocket could travel (given it has sufficient
thrust capability), so it is not intuitively obvious which path
maximizes elapsed proper time. But one could guess there are only 3
likely candidates: on the light cone headed inward, on the light cone
headed outward, and the geodesic determined at the horizon; actual
computations show that indeed the geodesic path maximizes elapsed proper
time between horizon and the limit point of the singularity.

Yes. Nobody seriously expects GR to remain valid in the neighborhood of
a singularity. Indeed, the notion "singularity" is usually thought to be
an artifact of GR and its classical roots.

Short answer: at the speed of light. A longer answer is given below.

There can be different types of horizons, but I'll just discuss the
usual one for Schw. spacetime, the event horizon. It is defined as the
limiting locus in 3-space from which a timelike object with arbitrarily
large acceleration cannot escape to spatial infinity. Relative to any
locally-inertial frame (which must necessarily be infalling), the
horizon always is moving at the speed of light. This is not limited to
Schw. spacetime, but for Schw. spacetime the horizon is stationary with
respect to a distant observer even though it is moving at c relative to
a local observer.  When something falls into the black hole, its horizon
expands with the local speed of light.

I can describe this in more detail for the specific case of a
spherically symmetric situation, using Birkhoff's theorem. Imagine a
spherically symmetric black hole with "mass" M, so initially its horizon
is at r=2M (I'll use Schw. coords. throughout). Consider a spherically
symmetric thin shell of mass m falling into the black hole from spatial
infinity. Once the shell is at or inside r=2(M+m), the horizon is at
r=2(M+m). Earlier, when the shell was far outside that value, the
horizon was approximately at r=2M. And as the shell approaches r=2(M+m)
from outside, the horizon expands from inside at the local speed of
light, meeting the shell precisely at r=2(M+m). An observer hovering
between r=2M and 2(M+m) would make local measurements consistent with a
black hole with horizon at r=2M until the shell passes her, at which
time she is inexorably pulled into the black hole. Note this observer
can still hover between the time the horizon passes her (headed outward)
and the time the shell passes her (headed inward); during this time she
could even head outward, but cannot possibly escape. Moreover, there is
no local experiment she could perform to learn of the horizon's approach
or passing -- the horizon is a global phenomenon with no definitive
local characteristics that are measurable (she could learn of the
shell's approach via normal means such as radar). For the diabolical
case of the shell being radiation, that intrepid observer cannot
possibly learn of the shell's approach until it reaches her, at which
time she is doomed.

Computing precisely how the horizon gets from r~2M to r=2(M+m) is
problematical because it depends on one's choice of time coordinate
(unlike Schw. spacetime there is no timelike Killing vector here).

Tom Roberts    tjroberts@lucent.com