| Topic: |
Science > Physics |
| User: |
"Dr. Photon" |
| Date: |
07 Jan 2005 02:02:23 PM |
| Object: |
What is the distance to a black hole? |
JB wrote
Another possibility comes to mind - about "optical illusions"
again.
The mass actually does fall through the horizon, and keeps on
pulling
the line behind it, you only don't *see* it falling through, due to
the light taking such a long time to get to you. So the winch keeps
letting out line.
That's essentially what happens.
Eek! That was the last one I expected, and I was almost embarrassed I
posted it. It seemed to have so many things wrong with it. Unless you
agree that the line keeps getting let out, but only asymptotically
slowly?
I don't really like this answer though, as it is not
clear to me what happens to the "infinitesimal tension" at the
winch.
If the line is taut, there will be a small tension all along the
line,
and the forces are balanced. But I'm not sure if that's what you're
asking. Could you clarify?"
If the line were totally massless, and the winch had no inertia, there
would be no tension at the winch. But I would like an infinitesimal
tension at the winch, just to have a communication with the mass and
make sure the line is "just taut", but does not affect the mass' rate
of fall. In this case, I can sense at what speed the mass is going at,
reference everything back to the spaceship, and I am only interested
in "proper" units at the spaceship. There are thus no ambiguities, and
all my statements are "according to an observer on the spaceship". I
strongly presume this is not original, but so long as it works. If the
mass were to go below the event horizon, I would lose the tension
signal and things become hard to interpret (meaningless?). I accept
that the tension signal takes a finite time to propagate.
The way it seems to me is that the time dilation seems to act a bit
like syrup (!). If you drop a pebble off a tall building, initially it
speeds up. But then it drops into some water, which slows its rate off
fall. At the bottom of the water there is some syrup which slows it
further. And then there is molten tar which slows it further, ... etc.
So if the pebble is on the end of a fishing line, its rate of fall
will "really" slow down as measured by whoever is holding the line. If
you drop pebbles sequentially, obviously no pebble will overtake the
one in front, but the distances between them will vary as they
encounter different viscosities. According to the guy holding the
fishing line, the pebble slows asymptotically to zero. I'll take a
look at the maths over the weekend, but I don't guarantee anything...
I'm not afraid of an equation, but the correct application and
interpretation can go horribly wrong (see previous, and not just
me!!!).
I was intersted to read on the net that before the term black hole was
introduced, they were called "frozen stars", because objects falling
in appear frozen above the horizon. So I guess I am clarifying that
this appearance is "real" in more ways than one - they actually are
frozen (according to you). And that the appearance ("ghost" image) is
pretty accurate of the current state of the object, as the object
hasn't moved much in the time it takes the light to get to you. Which
struck me as a bit bizarre at first - an object freely falling in the
strongest gravitational field possible *actually* moves at almost zero
velocity (according to you).
If the above is correct, and maybe even if it isn't, I am still left
with a puzzle - the speed of light still looks like it slows down. If
we agree that the length of line let out is finite, and the time taken
by light to go from the spaceship to the mass and back is increasingly
long, then the average velocity of the light will apparently go to
zero. This is valid "spaceship distance" and "spaceship time"!!! If
someone set out in a podule to the mass and back to the ship, he would
have a different story to tell, but that is not my point here.
Also, I don't believe the line has to break above the event horizon.
If light can escape, then surely a tension signal can also? If someone
with amazingly powerful thrusters could escape, then an amazingly
strong rope could pull him? tension <=> thrust? At the EH (or below?),
the tension may well be infinite, and this is how the line knows it
has crossed it! Actually, below the horizon is totally weird, a "high
up" segment of the line can "talk" to a segment below it, but the
segment below cannot talk to the one above it??? (because of
past/future considerations). I don't see how atomic bonds can work in
such circumstances, so the line is atomised (particleised??? :) ).
I also don't quite get the "point of last recall". Assuming that the
EH does not expand to meet the infalling mass, then according to the
spaceship the time for the mass to cross the EH is infinite. If the
ship then sends a light signal to the mass, as the light will always
travel through any point of space faster than the infalling mass, it
is bound to catch up on it, as it has an infinite amount of time to do
so, and no matter what the head-start is. Even though the light would
also apparently take an infinite amount of time to cross the EH, it
still has to catch up with the mass (which takes a "longer" infinity
to cross the EH???). In a more real case, the event horizon will
expand to meet the mass in a finite time, so in this case only there
is a "point of last recall". What do you think? According to an
infalling mass there is always a "point of last recall", but in the
first case above is at the even horizon and corresponds to an infinity
of external time?
cheers,
BR
.
|
|
| User: "Jim Black" |
|
| Title: Re: What is the distance to a black hole? |
08 Jan 2005 05:53:34 PM |
|
|
Dr. Photon wrote:
JB wrote
Another possibility comes to mind -
about "optical illusions" again.
The mass actually does fall through
the horizon, and keeps on pulling
the line behind it, you only don't
*see* it falling through, due to
the light taking such a long time
to get to you. So the winch keeps
letting out line.
That's essentially what happens.
Eek! That was the last one I expected, and I was almost embarrassed I
posted it. It seemed to have so many things wrong with it. Unless you
agree that the line keeps getting let out, but only asymptotically
slowly?
No, unless there is significant tension on the line, it will be let out
faster and faster, at a speed approaching that of light.
I don't really like this answer though, as it is not
clear to me what happens to the
"infinitesimal tension" at the winch.
If the line is taut, there will be
a small tension all along the line,
and the forces are balanced. But
I'm not sure if that's what you're
asking. Could you clarify?"
Thinking about this again, there may be situations in which the forces
are unbalanced. If the tension were being produced by the inertia of
the line as it is accelerated from being at rest with respect to the
spaceship to falling with the rest of the line, the tension of the
whole line wouldn't change instantly, but the changes in tension would
proceed down the line at no more than the speed of sound in the line.
This would introduce additional complications, including the line being
stretched greatly as the speed at it was being let out approached the
speed of sound. So, for simplicity, I'm going to assume instead that
the line is subject to a small constant tension. In ordinary
circumstances, such a small tension would have little effect if the
mass was at the bottom of the line, but if the mass was cut off, the
line would start accelerating upward.
If the line were totally massless, and the winch had no inertia,
there
would be no tension at the winch. But I would like an infinitesimal
tension at the winch, just to have a communication with the mass and
make sure the line is "just taut", but does not affect the mass' rate
of fall. In this case, I can sense at what speed the mass is going
at,
reference everything back to the spaceship, and I am only interested
in "proper" units at the spaceship. There are thus no ambiguities,
and
all my statements are "according to an observer on the spaceship". I
strongly presume this is not original, but so long as it works. If
the
mass were to go below the event horizon, I would lose the tension
signal and things become hard to interpret (meaningless?). I accept
that the tension signal takes a finite time to propagate.
Well, if the mass were to go below the event horizon, you wouldn't
suddenly lose tension on the line. But you do lose the ability to use
the tension to tell whether or not the mass is still there. If we
attach a device to the mass that snips the line after it has gone below
the event horizon, the part of the line just above the mass will start
accelerating upward immediately. But the release of tension travels
upwards at the speed of sound in the line, and can never overtake the
event horizon, which is moving upwards at the speed of light. So
someone at the spaceship would have no way of telling that this had
happened.
The way it seems to me is that the time dilation seems to act a bit
like syrup (!). If you drop a pebble off a tall building, initially
it
speeds up. But then it drops into some water, which slows its rate
off
fall. At the bottom of the water there is some syrup which slows it
further. And then there is molten tar which slows it further, ...
etc.
So if the pebble is on the end of a fishing line, its rate of fall
will "really" slow down as measured by whoever is holding the line.
If
you drop pebbles sequentially, obviously no pebble will overtake the
one in front, but the distances between them will vary as they
encounter different viscosities. According to the guy holding the
fishing line, the pebble slows asymptotically to zero. I'll take a
look at the maths over the weekend, but I don't guarantee anything...
I'm not afraid of an equation, but the correct application and
interpretation can go horribly wrong (see previous, and not just
me!!!).
In a sense, gravitational time dilation isn't even real, but is just a
useful mathematical construct. Consider what it means in detail. It
tells you, if you start two clocks at the same time, and stop them at
the same time, what the ratio between the amount of time shown on the
clocks will be. The problem is that, if you try to do this in
practice, there is no simple way of starting the two clocks at exactly
the same time, because a signal from one to the other can travel no
faster than the speed of light. You have to define the pairs of
starting and ending times to be simultaneous. The amount of time
dilation you measure depends on how your definition. Within certain
limits, one definition of simultaneity is just as good as any other.
However, some definitions are more useful than others.
I was intersted to read on the net that before the term black hole
was
introduced, they were called "frozen stars", because objects falling
in appear frozen above the horizon. So I guess I am clarifying that
this appearance is "real" in more ways than one - they actually are
frozen (according to you). And that the appearance ("ghost" image) is
pretty accurate of the current state of the object,
You see the problem? Which events you say are happening to the object
currently depend on how you think about it, how you define
simultaneity.
as the object
hasn't moved much in the time it takes the light to get to you. Which
struck me as a bit bizarre at first - an object freely falling in the
strongest gravitational field possible *actually* moves at almost
zero
velocity (according to you).
The term "frozen stars" was used at a time when black holes were not
understood as well as they are now. It was some time before physicists
figured out how to deal mathematically with the event horizon, and
realized that things actually can fall through. The name "frozen
stars" has fallen out of use because it can be misleading.
If the above is correct, and maybe even if it isn't, I am still left
with a puzzle - the speed of light still looks like it slows down. If
we agree that the length of line let out is finite, and the time
taken
by light to go from the spaceship to the mass and back is
increasingly
long, then the average velocity of the light will apparently go to
zero. This is valid "spaceship distance" and "spaceship time"!!! If
someone set out in a podule to the mass and back to the ship, he
would
have a different story to tell, but that is not my point here.
Look at it from the perspective of the falling line; the signal travels
at light speed, but the spaceship is accelerating, and soon is
travelling up the line at near light speed. 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. If you
watch from somewhere nowhere near the black hole, you may not need to
accelerate anymore, but you have another problem; the curvature of
space-time itself is playing with the distance.
Also, I don't believe the line has to break above the event horizon.
If light can escape, then surely a tension signal can also? If
someone
with amazingly powerful thrusters could escape, then an amazingly
strong rope could pull him? tension <=> thrust? At the EH (or
below?),
the tension may well be infinite, and this is how the line knows it
has crossed it! Actually, below the horizon is totally weird, a "high
up" segment of the line can "talk" to a segment below it, but the
segment below cannot talk to the one above it??? (because of
past/future considerations). I don't see how atomic bonds can work in
such circumstances, so the line is atomised (particleised??? :) ).
If the line is freely falling, and we're dealing with a short length of
line so that tides aren't significant, there won't be much tension at
all. If you keep the line from freely falling, there will be a lot of
tension. If you don't let the line out at all, the line must tear at
some point above the event horizon. This point can be made arbitrarily
close to the event horizon by making the line stronger. But a line
that is being held up cannot extend down the event horizon, because the
event horizon is moving upwards at the speed of light, and a winch line
can't move that fast.
I also don't quite get the "point of last recall". Assuming that the
EH does not expand to meet the infalling mass, then according to the
spaceship the time for the mass to cross the EH is infinite. If the
ship then sends a light signal to the mass, as the light will always
travel through any point of space faster than the infalling mass, it
is bound to catch up on it, as it has an infinite amount of time to
do
so, and no matter what the head-start is. Even though the light would
also apparently take an infinite amount of time to cross the EH, it
still has to catch up with the mass (which takes a "longer" infinity
to cross the EH???). In a more real case, the event horizon will
expand to meet the mass in a finite time, so in this case only there
is a "point of last recall". What do you think? According to an
infalling mass there is always a "point of last recall", but in the
first case above is at the even horizon and corresponds to an
infinity
of external time?
It's simultaneity again. As long as the guy in the spaceship can never
receive a light signal to the contrary, he is free, if he wishes, to
define that what is happening to the infalling mass "right now" is that
it is near the event horizon. That does not mean that he can send a
rescue mission to the mass where is "right now." The rescue mission
cannot arrive at the mass "right now"; it has obey the speed laws
getting there, and by that time it will be too late. It is because the
guy in the spaceship can't send a signal to the object as it crosses
the horizon, and yet he hasn't received a signal confirming it has
crossed, that he can define that even to be happening "right now,"
forever.
.
|
|
|
|
|
Related Articles |
|
|
