Dr. Photon wrote:
TR wrote
Instead of dropping pebbles, imagine
dropping robotic spaceships each
programmed to keep a constant
radar-measured distance from the previous
one; let the first keep its
engines off so it falls freely.
Ok, communicating spaceships I accept. However, I am still concerned
with what they are measuring and how. If both time and space are
getting warped, I am loath to use a "time-of-flight" to measure a
distance. If you argue that the space-time is locally flat when
considered on a small enough length scale so that this is valid, then
there are many dangers I'm not sure how to deal with when propagating
this up the sequence of spaceships, which is a far from locally flat
sequence. Maybe you know how to do this properly, but to make it
clearer to me I will argue the following:
Hovering above the BH is a system of stationary "calibrated"
spaceship
markers. Each is hovering 1m apart according to some "standard ruler"
that everyone accepts.
First, I should point out that no such ruler exists. Even in ordinary
special relativity, a distance measurement is only meaningful if you
specify the frame of reference in which it is made.
There are only a finite number of these
markers, as the distance to the EH is finite. When a spaceprobe is
dropped from the "mother" spaceship, it sends out a pulse every time
it passes a marker. At the first marker it sends a pulse to the
mother
ship, which then drops a second ship. This second ship can only pass
by the 1m mark when it has received a pulse from the lower ship which
has passed the 2m marker. This is continued as needed, with each
dropship not allowed to pass its marker until it has received a
signal
that the next marker has been passed.
In this system, AFAIK, the time taken for light to travel between
markers is longer and longer the closer to the EH you are. This will
cause the whole chain to back up, and the mother ship will let out
the
next ship at longer and longer intervals. Hence, in this system, the
rate of drop at the mother ship falls to zero??? This is true, even
if
the distance between markers approaches zero?
In the system you suggest, the rate of drop at the mother ship will
indeed fall toward zero. But the system doesn't maintain a constant
distance between the falling probes.
From the point of view of the first probe, which is freely falling, the
markers are moving upward at some velocity v. The distance between
adjacent markers is 1m * sqrt(1-v^2/c^2), less than 1m. But by the
time the light from the probe reaches the next marker up, the marker
will have moved by a factor of v * the time it took the light to travel
the distance. This time will be equal to the distance divided by c-v,
or 1m * sqrt(1-v^2/c^2) / (c-v). The resulting distance between the
first and second probes, in the first probe's frame of reference, is 1m
* sqrt[(c+v)/(c-v)]. As the first probe approaches the event horizon,
the relative velocity between it and the hovering markers approaches c,
and the second probe will lag behind the first by a greater and greater
distance. Eventually, the first probe will have long fallen past the
event horizon, but the second probe will be hovering above the event
horizon, getting further and further from the first probe, as it awaits
the last few signals.
If we have the case where the EH expands,
The event horizon is always moving upward at the speed of light by
definition. Infalling masses affect the surface area of the event
horizon, but the effect of a small infalling mass will be insignificant
and can be safely ignored.
and the first ship passes
through the EH in a finite amount of mothership time,
Once again, this is a meaningless statement, unless you carefully
specify a definition of simultaneity. By choosing your definition of
simultaneity, you can choose whether something passes the event horizon
in a finite amount of outside time or not.
then the whole
chain stops as nobody receives a signal to move ahead. My contention
is that this fall to zero rate happens smoothly (asymptotically if an
ideal stationary EH is not allowed to swallow the first ship, with
arbitrarily small distances between markers), and over a considerable
amount of mothership time. This can also be appreciated by
considering
the signal from the lowest dropship skipping all the intermediates
until it gets to the mothership - this received rate slows to zero
(doesn't it???).
Is this equivalent to your system? If not why not? In the system I
describe, I find it very hard to believe that the rate goes to "c",
and then the whole system crashes.
BTW: when you say that the time to cross the horizon is finite, on
what does that time depend? Is it longer for a less massive falling
object? Is it proportional to the ratio of masses of falling object
to
BH mass? Or is it proportional to BH mass alone? Maybe it doesn't
depend on anything much?
It depends on the initial position and velocity of the object, and the
mass, and to a smaller extent, the charge and angular momentum of the
black hole. The mass of the infalling object doesn't make a
difference. In principle, there's the possibility that the space-time
geometry of the black hole and its event horizon would be affected by
the mass, but you'd need a ridiculously large mass to do this. For our
purposes, any effect of the object on the black hole can be very safely
ignored. Aside from that, the mass can't possibly matter; remember
that in a vacuum, feathers and stones fall at the same rate.
If the time does depend on the masses, then I choose a case where the
apparent time is longest,
What do you mean by "apparent time" here?
in order to exaggerate the effect I am
discussing. Ideally I would like an "infinite time" case, as it is
clear to me that the mothership cannot let out an infinite number of
dropships all spaced 1m apart.
Also:
JB wrote
The light has problems catching up to it.
And so we see the main problem; since the ship is accelerating, its
distance from the point the light was emitted is changing.
So do you agree that the *apparent* speed of light (=total
distance/total time) does go to zero???
Assuming we're measuring it over a short enough distance to neglect the
effect of space-time curvature, and that we are making all measurements
in the same inertial frame of reference, the answer is a definite no.
The total distance in your calculation must include the additional
distance the light has to travel due to the upward acceleration of the
ship receiving it.
Over long distances, such as what would happen if the spaceship waited
far away from the black hole, space-time curvature can play tricks on
you and give you various speeds for light if you attempt to measure it
this way. But if distances are long, and gravitation is significant,
there's no such thing as an inertial frame, anyway.
.
