Re: the black fishing hole
- From: Gerry Quinn <gerryq@xxxxxxxxx>
- Date: Sat, 16 Sep 2006 20:04:32 +0000 (UTC)
In article <Pine.LNX.4.64.0609150954390.18547
@zeno3.math.washington.edu>, tessel@xxxxxx says...
On Thu, 14 Sep 2006, Gerry Quinn wrote
(concerning a thought experiment in
which a static observer, who uses his rocket engine to hover outside the
event horizon of a Schwarzschild hole, slowly lowers a weight on a
radially oriented string toward the horizon, or a similar thought
experiment):
From his perspective, the dangling object will never reach the event
horizon.
However, the event horizon is only a finite distance away, and he can
easily pay out more string than this.
You asserted that "the event horizon is only a finite distance away", but
it's already been pointed out several times in the past few weeks why this
assumption is naive. And the same point must have been made dozens of
times here in the past few years!
I don't recall this being "pointed out", but if I had I would have
dismissed it as obfuscation. For an observer at a distance - such as
Zef, who is after all the active party in this thought experiment - the
position of the event horizon is certainly well-defined. And the
Schwarzschild metric is defined in terms of a radius. If the radial
coordinate becomes timelike, the length of string will be redefined
along with it, and will remain finite. The use of the word "distance"
for the difference between two radial coordinates is not particularly
problematic.
Furthermore, under any interpretation, the distance to a point
arbitrarily close to the event horizon, where there is no need for
ambiguity in distinguishing space from time, is well-defined and tends
toward a finite limit. There is no good reason to think of the
distance to the event horizon as other than this limit. So *of course*
Zef can pay out more string than this limit. Remember, this thought
experiment concerns the observations of Zef. What goes on at or below
the event horizon has no relevance to him, as he will never observe it.
If he does, he will find that the string goes slack,
Well, the string will break at some point before the weight reaches the
horizon, because the tension in the string diverges, because of the fact
about acceleration required for a bit of matter to hover outside the hole
as noted above.
It is illuminating to consider what happens if he pays out string
slowly, in order to maintain tension at all times, or at least
periodically.
I thought that's precisely the thought experiment which we've been
discussing, except that of course that in a quasistatic thought experiment
the tension increases -monotonically-.
Actually the original poster referred to "lowering an object below the
event horizon", and wondered whether a string of mere finite length
would leave the object dangling above it. Of course this question
indicates a fundamental confusion about the distance to the event
horizon. If a viewpoint that engenders such confusion has indeed been
promulgated here, perhaps that explains it.
This string is very strong with a huge elastic constant, so that we can
ignore its changes in length, and any variations in tension will travel
up and down it at almost the speed of light. Let us say the object is
dangling twenty feet above the event horizon, and he lets out six
inches, and waits for the tension to be restored.
Something is wrong here. In both Newtonian gravitation and in gtr, the
tension in the string should monotonically increase as the weight is
slowly lowered. The computation works the same way in both theories:
consider the situation at a given time and work out the total tension in
the string at that time. (In the physics literature, in a discussion of
this situation, one might say that the dangling weight is "quasistatic".)
When I say 'restored' I mean that the fact of tension is restored; that
the dangling object is supported by a tensile force in the string.
There seems no need to concern ourselves with its value.
The line goes slack, a signal represented by an absense of tension in
the string travels down at the speed of light, the object drops six
inches and is pulled up short, then an impulse of restored tension
travels up again, and Zef observes that the string is once again taut,
so he can pay out another six inches.
Are you talking about letting the object fall freely for a bit, then
stopping it by -jerking- (and then pulling with constant magnitude) with
the string? This is obviously a more complicated thought experiment, and
jerking the string certainly won't help avoid snapping it. I strongly
suggest you first work out the tension for the quasistatic thought
experiment, comparing Newtonian gravitation and gtr.
I think it's rather simple and intuitive to deal with the transition
between two well-defined static states. Clearly, mileage varies
regarding what is considered intuitive.
The problem for Zef is gravitational time dilation.
Citation? (Second request.)
I don't recall a previous request. But leaving that aside...
Are you reading a paper by someone whom you believe claims that the
standard analysis in gtr textbooks is -wrong-, or what? If so, is it
possible that you misunderstood?
I'm not reading anybody's paper, nor have I consulted a textbook to
find out what the "standard analysis" is. The question of what happens
when someone lowers a weight on a string towards the event horizon of a
static black hole is trivially simple, given the approach I have
outlined, which admittedly took me a few moments thought to come up
with.
So I can give no citation, except the post to which you are responding.
[Note that I made a mistake in the math, which I have corrected in a
follow-up post.]
What happens in the case where he pays out line continuously, always
maintaining tension in the string, should now be obvious. He will have
to pay it out slower and slower to maintain tension.
Maintain tension? If you mean, "maintain constant tension", of course he
can't do that unless he pays out the string faster and faster, according
to either Newtonian gravitation or gtr! (That is, he'd have to pay out
the string at just the right rate to maintain constant tension, so the
dangling weight would be neither quasistatic or freefalling at any time;
it would be in a "controlled fall".)
When I say "maintain tension" I mean just that. Yes, the weight of the
mass will increase as it is lowered. I assume that his fishing
apparatus can support any finite weight.
But your assertion that paying out the string faster and faster will
maintain constant tension is simply wrong. The effects of
gravitational time dilation will eventually overcome the effects of
increased weight. Even if he maintains constant tension for, as you
say, a controlled fall, the amount of string to be paid out per second
will eventually fall to zero. He will still asymptotically approach
the same total length of paid out string if he maintains any tension at
all. This asymptotically approached length is precisely the distance
to the event horizon, as measured by Zef, in the time-honoured fashion
of dropping a weighted string.
The total length paid out will asymptotically approach the distance
from Zef to the event horizon, but will never reach it.
In gtr, the string will break before the weight reaches the horizon (this
should be obvious from the result noted above about the magnitude of the
outward radial acceleration required for a test particle to remain
static). Up to the point where that happens the length paid out will of
course be finite. But if you try to compute it you should see why in
curved spacetime, "length of the string" requires some assumptions in how
you chose to define this notion, which is the point you forgot above!
If you find that the use of curved spacetime confuses the issue, I
suggest you consider transforming the problem into Zef's coordinates,
in which spacetime is flat. Matters are quite simple then. The amount
of string paid out is indeed finite. It also approaches a fixed value.
Am I to infer from the above paragraph that you think it will increase
without bound so long as the string resists breaking? Certainly that's
the impression that the OP seems to have got.
As for "length of the string", this obviously means its proper length -
as measured by the number of knots along it or in some such manner. To
put it another way, it is the length paid out by Zef. We assume that
under the conditions of the experiment it is so strong as to be
effectively inextensible until such time as it breaks and the gedanken
is no longer valid. In flat or curved spacetime, this proper length is
a well-defined value, admitting no ambiguity.
Mostly likely you will integrate the spatial line element for the static
exterior Schwarzschild chart. This is the "pedometer distance", since it
gives the distance which would be reported by an ant equipped with a
pedometer who crawls along the string from the observer to the dangling
weight. (So this is defined in terms of a quasistatic observer.)
Yes, that is the proper length. There's no need to concern ourselves
with any other for the purposes of this problem.
But if you define the "radar distance", or "optical diameter distance",
etc., you should expect to obtain slightly different results!
It shouldn't be either surprising or dismaying that you should have
multiple competing definitions of distance "in the large" in any curved
spacetime. In any case, all reasonable definitions of "length of the
string" will yield a finite length, until the string breaks, and of course
they all agree in the Newtonian limit.
I'm at a loss to know why I would need to consider alternative
definitions of length. The question concerned the length paid out by
Zef.
(Confirming all this is another good exercise.)
You -could- consider a more complicated thought experiment, but it makes
good sense to work out the details of the simplest one first!
[Snipped: very complicated argument involving Taylor Series of degree 7
and such like.]
My approach using gravitational time dilation gives qualitative answers
instantly, and also gives an easy approach to showing how the string
must be paid out slower and slower to maintain tension. The first
version was wrong - corrected version below:
The length of time (as seen by an observer at infinity) for an impulse
to travel between u and R (dropping all constants) is:
integral (u to R) ( dx / sqrt( 1 - M / x ) )
That's a function of u, which we need to integrate from r to R:
integral (r to R) ( integral (u to R) ( dx / sqrt( 1 - M / x ) ) du
...to get the relative time needed to drop the object from R to r while
maintaining tension in the line.
Is there anything wrong with the above?
- Gerry Quinn
.
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