| Topic: |
Science > Physics |
| User: |
"Dr. Photon" |
| Date: |
06 Jan 2005 07:41:45 AM |
| Object: |
What is the distance to a black hole? |
TR wrote
"Well, "short" is a matter of details. For instance, in a flat
manifold
and a ship accelerating at 1 g, that "short" distance is about 1
lightyear(!)."
Doh! This is something new I have a problem with now. If you have a
spaceship and test mass floating together in a flat manifold, and then
the ship accelerates away and leaves the mass behind (reeling out a
line as necessary), I don't see why the line will ever break. In this
case, the mass is at rest, as is the line. As the line is being fed
out as necessary, it does not experience an "F=ma" that it would feel
if it was actually accelerating the mass. The only tension I see is an
infinitesimal one at the winch, just enough to signal that more line
is needed, but not enough to produce any noticeable acceleration of
the mass. Basically I am only using the line as a metre stick, so I
can tell at the ship how far away the mass is (because I know how much
line has been fed out). The line (in the sense of "rope") has ideal
properties such as maximum tensile strength, minimum mass, and minimum
elasticity (as far as possible, I accept that such things like
infinite differential acceleration would break the line though).
Back to the main question:
I don't like the pebble dropping analogy. The pebbles get further
apart as they fall, while the line does not. The line is fed out at a
rate corresponding to the rate of fall of the *first* pebble only. If
you were on top of a building, and dropped a mass off, in my view the
line would be fed out faster until the mass hit the ground, at which
time the line would stop being fed out. You now know the distance to
the ground. At all times, the tension at the winch is only
infinitesimal, just to give you a signal of whether the mass is
falling or not.
In the event horizon case, as the distance is finite and the time to
get there infinite (according to the spaceship), then the rate of fall
asymptotically goes to zero. Anything else imples an infinite distance
(surely??? :) ). Hence the winch will slow down asymptotically to
zero, even though the spaceship is nominally accelerating like mad. It
would appear that the mass has a "soft landing" onto the event
horizon, a finite distance away. The tension at the winch is still
only infinitesimal, ie, you are in no sense impeding the fall of the
mass by "hanging on" to it.
According to the spaceship, then, the event horizon is similar to the
surface of a planet - you need to put on your thrusters in order to
hold your position above its surface, but at some stage, the mass
rests on that surface so you don't need to let out more line. In the
planet case, the mass is supported by contact repulsion from the
matter of the planet, while in the BH case the mass is "supported" by
time dilation.
On the other hand, if we say that an equivalent situation is that the
spaceship is simply accelerating away from the mass, then it looks as
if the line will have to be fed out faster and faster forever. But
this implies infinite distance, so it can't really be equivalent??? It
is only equivalent up until the point where the time dilation factor
becomes greater than the local acceleration factor (according to the
spaceship!). Maybe it is approximately equivalent in weak
gravitational fields (small time dilation), but not valid when
approaching an event horizon (large time dilation) (???).
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. I don't really like this answer though, as it is not
clear to me what happens to the "infinitesimal tension" at the winch.
At the moment I go for the "finite distance" and "soft landing"
answers, bizarre though they may seem.
BR.
p.s. apologies, but my "reply" function only posts my replies as new
sub-threads.
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| User: "Tom Roberts" |
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| Title: Re: What is the distance to a black hole? |
09 Jan 2005 10:29:45 AM |
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Dr. Photon wrote:
If you have a
spaceship and test mass floating together in a flat manifold, and then
the ship accelerates away and leaves the mass behind (reeling out a
line as necessary), I don't see why the line will ever break.
It won't. But this is a very different physical situation than the
original one of a hovering spaceship in Schwarzschild spacetime. The
ship will have to put out line at an increasing rate asymptotically
aproaching c, and the only stress on the line is right at the ship.
Back to the main question:
I don't like the pebble dropping analogy. The pebbles get further
apart as they fall, while the line does not. The line is fed out at a
rate corresponding to the rate of fall of the *first* pebble only.
You are making unphysical asumptions, such as: the line is completely
massless, and the line is infinitely resistant to internal stress. I
don't think anything makes sense with such assumptions, and relaxing
them makes this resistant to analysis....
If
you were on top of a building, and dropped a mass off, in my view the
line would be fed out faster until the mass hit the ground, at which
time the line would stop being fed out. You now know the distance to
the ground. At all times, the tension at the winch is only
infinitesimal, just to give you a signal of whether the mass is
falling or not.
That is a rather naive dscription, and cannot possibly hold in the
curved manifold near a black hole.
In the event horizon case, as the distance is finite and the time to
get there infinite (according to the spaceship), then the rate of fall
asymptotically goes to zero.
You cannot conclude that, because the curvature of the manifold is
important. The pebble analogy shows the ship can keep putting out line
forever.
[... elaboration on that error]
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.
This is closer, except once the mass passes the horizon it cannot keep
pulling on the line. This is part of the reason your assumption of
massless line is naive and impossible.
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. Eventually
the second one will lose radar communication with the first and this
program will fail -- this occurs within a finite proper time of the
second spaceship. Ditto for the third, fourth, .... The analogy to your
line is that the line breaks; this occurs outside the horizon for a line
of any finite strength.
Tom Roberts tjroberts@lucent.com
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| User: "Jim Black" |
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| Title: Re: What is the distance to a black hole? |
09 Jan 2005 12:21:56 PM |
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Tom Roberts 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.
Eventually
the second one will lose radar communication with the first and this
program will fail -- this occurs within a finite proper time of the
second spaceship. Ditto for the third, fourth, .... The analogy to
your
line is that the line breaks; this occurs outside the horizon for a
line
of any finite strength.
If all the spaceships are allowed to fall through the horizon, even
when the first ship goes below the event horizon, its signal will still
get to the second ship, once it too is below the event horizon. There
is obviously a delay in the signal, but if the distance between the
ships is short, the delay will not be much more than the expected delay
due to the finite speed of light. It's only if the ships are
programmed to try to stop falling through the event horizon at some
point that the line of communication must break there.
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| User: "Tom Roberts" |
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| Title: Re: What is the distance to a black hole? |
09 Jan 2005 06:16:54 PM |
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Jim Black wrote:
Tom Roberts 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.
Eventually
the second one will lose radar communication with the first and
this program will fail -- this occurs within a finite proper time
of the second spaceship. Ditto for the third, fourth, .... The
analogy to your
line is that the line breaks; this occurs outside the horizon for a
line of any finite strength.
If all the spaceships are allowed to fall through the horizon, even
when the first ship goes below the event horizon, its signal will still
get to the second ship, once it too is below the event horizon.
Yes, where it happens depends on the details, but communication is
inevitably lost. In particular, after passing the horizon if the first
ship reflects N radar pulses from the second ship, in general the second
will not see all N pulses before intersecting the singularity. And also,
when the second ship receives a given radar echo from the first, the
second is deeper inside the horizon than the first one was when the
pulse reflected (which is why it cannot see them all).
Tom Roberts tjroberts@lucent.com
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| User: "Ahmed Ouahi, Architect" |
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| Title: Re: What is the distance to a black hole? |
10 Jan 2005 09:47:45 AM |
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........... ...For the time being, mostly the system of the
triangulation would be the most appropriate along that matter. A definitely
with the specific technique especially based on the mathematics in a general
and on the geometry in a particular. However, along the triangulation, when
you would know the length of one side of a determined a triangle and the
angles of a two corners, a definitely, you could arrive to determine all its
other measures without moving nowhere or trying anything else, whatsoever.
........... ...However, do take an example to see more a clear. For
instance, do take a two persons who wanted to know the distance to planets,
or more simply how far it is - a specific detail- a specific planet.
Therefore, along that matter, would be a certainly the system of a
triangulation the most appropriate, as along which, the first thing those
persons, they do have to specify a certain distance between each other, or a
more a simple would be, that they would specify a define mark somewhere for
one person and they would specify an other mark somewhere else, where the
other person should be placed. However, we do not have to miss, that the
both of them, they do have to fix or to look to the planet, that they have
had specified together along their volunty to know its distance to where
they are placed themeselves, however, at exactly and specifically at the
same moment....... ...
............ ...Therefore, in the meantime, a simply you do have to
think about a line which would connect all the three principals of this what
those two persons are doing - both of them and any of the planet des have
had decided - along a triangle form. Therefore, in that case, would be the
measurement of the length of the baseline between those two elements and the
angles of their two corners and the rest which it has had to be only
calculated. However, all this, should be a simply done by also knowing, that
the interior angles of the triangle systematically accomplishes the 180
degrees, therefore, when you do know the sum of the two of the angles, you
would be allowed to determine the third along a simple calculation, however,
when you do precisely know the shape of the triangle and also the lenght of
one side, it would make you to determine the lenght of the other side,
simply as that, a definitely as a matter a
fact!!!!!!!!!!!!!!!!!!................ ...
--
Ahmed Ouahi, Architect
Best Regards!
"Tom Roberts" <> kirjoitti viestissä
news:WPjEd.13198$by5.12797@newssvr19.news.prodigy.com...
Jim Black wrote:
Tom Roberts 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.
Eventually
the second one will lose radar communication with the first and
this program will fail -- this occurs within a finite proper time
of the second spaceship. Ditto for the third, fourth, .... The
analogy to your
line is that the line breaks; this occurs outside the horizon for a
line of any finite strength.
If all the spaceships are allowed to fall through the horizon, even
when the first ship goes below the event horizon, its signal will still
get to the second ship, once it too is below the event horizon.
Yes, where it happens depends on the details, but communication is
inevitably lost. In particular, after passing the horizon if the first
ship reflects N radar pulses from the second ship, in general the second
will not see all N pulses before intersecting the singularity. And also,
when the second ship receives a given radar echo from the first, the
second is deeper inside the horizon than the first one was when the
pulse reflected (which is why it cannot see them all).
Tom Roberts
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| User: "Jim Black" |
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| Title: Re: What is the distance to a black hole? |
07 Jan 2005 12:11:32 AM |
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Dr. Photon wrote:
TR wrote
"Well, "short" is a matter of details. For instance, in a flat
manifold
and a ship accelerating at 1 g, that "short" distance is about 1
lightyear(!)."
Doh! This is something new I have
a problem with now. If you have a
spaceship and test mass floating
together in a flat manifold, and then
the ship accelerates away and leaves
the mass behind (reeling out a
line as necessary), I don't see why
the line will ever break. In this
case, the mass is at rest, as is
the line. As the line is being fed
out as necessary, it does not
experience an "F=ma" that it would feel
if it was actually accelerating the
mass. The only tension I see is an
infinitesimal one at the winch, just
enough to signal that more line
is needed, but not enough to produce
any noticeable acceleration of
the mass.
If the line is being let out faster and faster, it indeed doesn't have
to break. In the situation referenced above, the line is being dragged
along with the ship.
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.
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?
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