Fun with black holes
in [Relativity]
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From: eric gisse on 13 May 2010 19:50 eon wrote: [...] I wonder if the dyslexic troll posts under a different name here.
From: Edward Green on 13 May 2010 19:52 On May 12, 8:35 pm, Darwin123 <drosen0...(a)yahoo.com> wrote: > On May 12, 5:50 pm, Edward Green <spamspamsp...(a)netzero.com> wrote: <...> > I think there is an implicit question of how the black hole can > form at all in the lifetime of the fiducial observer. By fiducial > observer, I mean an observer far away from the event horizon. I think > that it isn't possible for a single object to fall past the event > horizon because of the asymmetry of the fall. If objects are falling > in the black hole from all directions, the situation changes. > An object can't fall into the event horizon in a finite time > unless that black hole is expanding. If the event horizon is moving > away from the center of the black hole, then the event horizon can > overtake the object even if the object seems "frozen" near the event > horizon. The event horizon can grow if there is more than one falling > object. > An example of this occurring is in the spherical collapse of a > massive star. During the collapse, there is always a small amount of > material that seems to be falling very slowly as seen by the fiducial > observer (i.e., far away observer). To the fiducial observer, a small > amount of material may seem frozen at the edge of the black hole. > However, material is coming toward the black hole center in all > directions. The combined gravity of material near the event horizon > makes the event horizon grow, as seen by a fiducial observer. Thus, > the material near the black hole will be engulfed by the growing event > horizon. > If the star collapses asymmetrically, then other strange things can > happen. A naked singularity can form. A spinning black hole can have > strange phenomena with two horizons. However, the spherically > symmetrical collapse has been solved in general relativity. Black > holes can form under this condition. > The above is my conjecture based on what I read. If someone knows > general relativity very well and knows where I made a mistake, I will > be glad to hear from him. The question is really about how black holes > can form at all in a finite time. Well, you seem to have answered it, qualitatively, and with Tom Roberts' imprimatur: stuff doesn't reach the horizon, the horizon rises to meet the stuff. If that's what the equations of GR predict for a spherically symmetric collapse, I will accept it. I still have trouble accepting that a gravitationally insignificant test particle is ever going to finish falling into an already extant black hole, on its own. I wonder if I own Kip Thorne's book. It is possible. I must mount a search. :-)
From: Tom Roberts on 14 May 2010 00:04 Edward Green wrote: > 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, Let your object drop by 1 meter and then return (via sufficiently powerful thrust). Then let it drop by 1 mm and then return. Then a micron.... You did not specify how far down it has to go. But once it reaches the horizon, there is no possible amount of thrust that can make it return. >>> 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. The play on words is YOURS, not mine. > Would it changes things if I had written "initially > distant"? Yes, of course. Then the object could fall in towards the black hole and no longer be "distant". > Is the point "once distant, always 'distant' "? You said "distant OBJECT [emphasis mine]", not "distant point". If the object approaches the horizon then it is no longer distant. So it is not "once distant, always distant", but rather "once it is not distant then it is not distant". I keep repeating this to just about everyone around here: you must be more precise in word and thought. >> 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". Here "sees" = "observes". > 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). That is a reference frame, not an observer. In most discussions of SR, an observer is analogous to a (single) human; for instance, an observer needs assistants [*] pre-positioned along the trajectory of a moving object to measure either its length (if it is a ruler), or its tick rate (if it is a clock). A reference frame inherently has all the necessary assistants. [*] each equipped with a clock synchronized to the observer's clock and knowledge of their position, in the observer's frame. > But evidently no similar concept makes its way > through to GR. It does, as long as one only attempts to cover a "small" region of spacetime, where "small" depends on the curvature of the region. > 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. Yes. But you left the ambiguity of how far down the trip goes. > 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. "All physics is local" [Einstein]. The distant observer, and her coordinates, have nothing to do with what happens near the horizon. Use ANY coordinates that are valid in a neighborhood of the horizon, and you'll compute that a falling object goes right through the horizon. There is no "contradiction", it's just that not every observer can observe everything. Tom Roberts
From: eon on 14 May 2010 05:47 On May 14, 6:04 am, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote: > Edward Green wrote: > > 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, > > Let your object drop by 1 meter and then return (via sufficiently powerful > thrust). Then let it drop by 1 mm and then return. Then a micron.... You did not > specify how far down it has to go. > > But once it reaches the horizon, there is no possible amount of thrust that can > make it return. not true, before event horizon you must have infinite angular velocity, from the flush, which by integration gets even more infinite position these two quantities would easy cancel a black hole
From: Daryl McCullough on 14 May 2010 06:51
Tom Roberts says... > >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. > > >> (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. How I interpreted Ed's question is this: Let's fix the initial r-coordinate of the object, r_0. Let r_closest be the smallest value of the r-coordinate. Then, in fact, there is a minimal time t_min(r_0,r_closest) for an object to go from r_0 to r_closest and back again. And, I think Ed's claim is that limit r_closest --> r_schwarzschild of t_min(r_0, r_closest) = infinity This is true. If you keep r_0 fixed and decrease r_closest, then the round trip time increases without bound. I'm not sure exactly what is supposed to follow from it, however. -- Daryl McCullough Ithaca, NY |