Fun with black holes
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From: Tom Roberts on 13 May 2010 11:22 Darwin123 wrote: > [... approximately correct description, with minor errors in details > (e.g. a rotating black hole has one horizon and an ergosphere, the > outer boundary of which is not a horizon but is related)] See: K.Thorne, _Black_Holes_&_Time_Warps_. He discusses in detail just about everything brought up in this thread, plus a lot more. It is a popular, non-mathematical book that I think you will enjoy. Tom Roberts
From: eon on 13 May 2010 14:40 On May 13, 5:13 pm, Tom Roberts <tjrob...(a)sbcglobal.net> wrote: > Edward Green wrote: > > Consider an ordinary Schwarzschild black hole: > > (1) Let an object be dropped from an arbitrary value of the radial > > coordinate above the event horizon. > > (2) Is there ever a case where the object has not reached the event > > horizon where it is _not_ possible to reverse its trajectory and > > escape back to its starting point, by means of a sufficiently powerful > > thrust? > > No. The event horizon is the place where this is no longer possible. > > Note: the context is GR; this is all a gedanken, so computations can be carried > out to arbitrary precision. you are contradicted by evaporation and maxwell equations, which says that what falls in must get out > > (3) Assuming the answer to (2) is "no", let T be the greatest lower > > bound on the time to make a return trip. > > But there is no useful GLB (i.e. T=0). This should be obvious as an > infinitesimally-short trip is always possible. > > Here's a question related to what I think you wanted to ask: > > Consider an arbitrary location outside the horizon and an upward impulse of any > appropriate magnitude (including arbitrarily large), and consider a radial path > that is dropped from rest at that location, receives the upward impulse at its > closest approach to the horizon, and is in freefall back up to its starting > point which it reaches at rest. Given an arbitrarily large value T (> 0), is > there such a path that takes time T to return to its starting point (as measured > by a clock at rest at that point)? > > The answer is: Yes. > > Note that as T increases, the magnitude of the required impulse increases, and > the distance of closest approach to the horizon decreases. > > > in what sense does a distant > > object ever finish falling into the event horizon? > > A distant object never "falls in", because it is "distant". > > An object that is simply dropped from outside the horizon will fall through the > horizon and intersect the singularity inside, in a finite proper time after > being dropped. But a distant observer [#] will never observe it reach the horizon. > > The distant observer can see it approach arbitrarily close to > the horizon, but never actually reach it. As impossible, any light or whatever will start orbiting far before any event horizon, note, without falling in it !! so you cannot see nor expect to see anything at all !!! the object is > observed to approach the horizon [@], the intensity of light > reaching the distant observer approaches zero, regardless of > whether the object emits the light or reflects illuminating > light (but emitted light is redshifted more than reflected > light). Here for simplicity I'm considering the object, > observer, and any illuminating light to always be located on > a single radial line, with observer uppermost. > > [#] The "distant" observer can be arbitrarily close to the horizon, > but must hover motionless relative to the black hole; the falling > object is between observer and black hole. > > [@] Note I did not say "as the object approaches the horizon" > -- the difference is important because it is ambiguous to extend > simultaneity from observer to near the horizon; there's no such > ambiguity in discussing what the observer observes. > > Tom Roberts
From: BURT on 13 May 2010 15:19 On May 13, 11:40 am, eon <ynes9...(a)techemail.com> wrote: > On May 13, 5:13 pm, Tom Roberts <tjrob...(a)sbcglobal.net> wrote: > > > Edward Green wrote: > > > Consider an ordinary Schwarzschild black hole: > > > (1) Let an object be dropped from an arbitrary value of the radial > > > coordinate above the event horizon. > > > (2) Is there ever a case where the object has not reached the event > > > horizon where it is _not_ possible to reverse its trajectory and > > > escape back to its starting point, by means of a sufficiently powerful > > > thrust? > > > No. The event horizon is the place where this is no longer possible. > > > Note: the context is GR; this is all a gedanken, so computations can be carried > > out to arbitrary precision. > > you are contradicted by evaporation > and maxwell equations, which says > that what falls in must get out > > > > > > > > (3) Assuming the answer to (2) is "no", let T be the greatest lower > > > bound on the time to make a return trip. > > > But there is no useful GLB (i.e. T=0). This should be obvious as an > > infinitesimally-short trip is always possible. > > > Here's a question related to what I think you wanted to ask: > > > Consider an arbitrary location outside the horizon and an upward impulse of any > > appropriate magnitude (including arbitrarily large), and consider a radial path > > that is dropped from rest at that location, receives the upward impulse at its > > closest approach to the horizon, and is in freefall back up to its starting > > point which it reaches at rest. Given an arbitrarily large value T (> 0), is > > there such a path that takes time T to return to its starting point (as measured > > by a clock at rest at that point)? > > > The answer is: Yes. > > > Note that as T increases, the magnitude of the required impulse increases, and > > the distance of closest approach to the horizon decreases. > > > > in what sense does a distant > > > object ever finish falling into the event horizon? > > > A distant object never "falls in", because it is "distant". > > > An object that is simply dropped from outside the horizon will fall through the > > horizon and intersect the singularity inside, in a finite proper time after > > being dropped. But a distant observer [#] will never observe it reach the horizon. > > > The distant observer can see it approach arbitrarily close to > > the horizon, but never actually reach it. As > > impossible, any light or whatever will > start orbiting far before any event horizon, > > note, without falling in it !! > > so you cannot see nor expect to > see anything at all !!! > > the object is > > > > > observed to approach the horizon [@], the intensity of light > > reaching the distant observer approaches zero, regardless of > > whether the object emits the light or reflects illuminating > > light (but emitted light is redshifted more than reflected > > light). Here for simplicity I'm considering the object, > > observer, and any illuminating light to always be located on > > a single radial line, with observer uppermost. > > > [#] The "distant" observer can be arbitrarily close to the horizon, > > but must hover motionless relative to the black hole; the falling > > object is between observer and black hole. > > > [@] Note I did not say "as the object approaches the horizon" > > -- the difference is important because it is ambiguous to extend > > simultaneity from observer to near the horizon; there's no such > > ambiguity in discussing what the observer observes. > > > Tom Roberts- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text - Light is never overcome by gravity.
From: Edward Green on 13 May 2010 18:03 On May 13, 11:13 am, Tom Roberts <tjrob...(a)sbcglobal.net> wrote: > Edward Green wrote: > > Consider an ordinary Schwarzschild black hole: > > (1) Let an object be dropped from an arbitrary value of the radial > > coordinate above the event horizon. > > (2) Is there ever a case where the object has not reached the event > > horizon where it is _not_ possible to reverse its trajectory and > > escape back to its starting point, by means of a sufficiently powerful > > thrust? > > No. The event horizon is the place where this is no longer possible. > > Note: the context is GR; this is all a gedanken, so computations can be carried > out to arbitrary precision. Yes. > > (3) Assuming the answer to (2) is "no", let T be the greatest lower > > bound on the time to make a return trip. > > But there is no useful GLB (i.e. T=0). This should be obvious as an > infinitesimally-short trip is always possible. By "return trip" I really meant to write "round trip": even so, it's not obvious to me why an infinitesimally-short return trip is always possible, unless you mean in terms of proper time of the initially infalling object. But no matter: > Here's a question related to what I think you wanted to ask: > > Consider an arbitrary location outside the horizon and an upward impulse of any > appropriate magnitude (including arbitrarily large), and consider a radial path > that is dropped from rest at that location, receives the upward impulse at its > closest approach to the horizon, and is in freefall back up to its starting > point which it reaches at rest. Given an arbitrarily large value T (> 0), is > there such a path that takes time T to return to its starting point (as measured > by a clock at rest at that point)? > > The answer is: Yes. I accept your reformulation. I like its symmetry, in fact: the object "bounces". > Note that as T increases, the magnitude of the required impulse increases, and > the distance of closest approach to the horizon decreases. > > > in what sense does a distant > > object ever finish falling into the event horizon? > > A distant object never "falls in", because it is "distant". I assume you had some serious point to make here, but that sounds like a play on words. Would it changes things if I had written "initially distant"? Is the point "once distant, always 'distant' "? > An object that is simply dropped from outside the horizon will fall through the > horizon and intersect the singularity inside, in a finite proper time after > being dropped. But a distant observer [#] will never observe it reach the horizon. > > The distant observer can see it approach arbitrarily close to > the horizon, but never actually reach it. As the object is > observed to approach the horizon [@], the intensity of light > reaching the distant observer approaches zero, regardless of > whether the object emits the light or reflects illuminating > light (but emitted light is redshifted more than reflected > light). Here for simplicity I'm considering the object, > observer, and any illuminating light to always be located on > a single radial line, with observer uppermost. > > [#] The "distant" observer can be arbitrarily close to the horizon, > but must hover motionless relative to the black hole; the falling > object is between observer and black hole. > > [@] Note I did not say "as the object approaches the horizon" > -- the difference is important because it is ambiguous to extend > simultaneity from observer to near the horizon; there's no such > ambiguity in discussing what the observer observes. Hmm... I think you mean here "what the observer sees". As Timo N. writes, an observer in SR is really an infinite collection of clocks and rods (or something close to that that makes sense, if I failed to translate adequately). But evidently no similar concept makes its way through to GR. Anyway, to eliminate the ambiguity of simultaneity from the observer to near the horizon, I inserted the round trip. There is no ambiguity in the time required to make a round trip. I wanted to make the round trip as short as possible, which I guess means accelerating the outgoing object to as near light speed as possible. I think the round trip time still diverges, but this way we can't say it would have failed to diverge if we had only hurried up on the return leg, rather than allowing a leisurely elastic bounce. Now I know this is going to annoy you, but I still adhere to my contrarian viewpoint that the object never falls in in the life of this universe, the finite proper time not withstanding. I believe time more or less comes to a halt for the infalling object, and this resolves the apparent contradiction. The fact that the symmetric retrieval time (your T) becomes unbounded trumps the finite proper time argument for me, anyway.
From: eric gisse on 13 May 2010 19:50
Edward Green wrote: [...] > Now I know this is going to annoy you, but I still adhere to my > contrarian viewpoint that the object never falls in in the life of > this universe, the finite proper time not withstanding. I believe time > more or less comes to a halt for the infalling object, and this > resolves the apparent contradiction. The fact that the symmetric > retrieval time (your T) becomes unbounded trumps the finite proper > time argument for me, anyway. For fucks sake. Stop comparing things that can't be compared. Time is not an invariant. An observer watching something fall into a black hole will never see it happen. The thing actually falling in does it in finite time according to its' clock. |