Fun with black holes
in [Relativity]
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From: Edward Green on 12 May 2010 19:36 On May 12, 11:26 am, eric gisse <jowr.pi.nos...(a)gmail.com> wrote: <...> > Mazeltov! You have re-discovered the basic fact that an observer outside the > event horizon of a black hole can never see an object pass through the event > horizon in finite amounts of proper time. I thought the round trip aspect gave more meat to the observation. If the object merely falls in, there is always the suspicion that something funny is going on with our sense of time, given the observation of a distant object near the horizon. If we get the object back again... and as fast as possible, and we _still_ find the time lapse goes to infinity, it begins to look more convincing that the object never reaches the horizon.
From: Darwin123 on 12 May 2010 20:35 On May 12, 5:50 pm, Edward Green <spamspamsp...(a)netzero.com> 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? > > (3) Assuming the answer to (2) is "no", let T be the greatest lower > bound on the time to make a return trip. > > (4) Does T tend to infinity as the distance from the event horizon > tends to zero? > > (5) Assuming the answer to (4) is "yes", in what sense does a distant > object ever finish falling into the event horizon? 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.
From: eric gisse on 12 May 2010 13:25 Edward Green wrote: > On May 12, 11:26 am, eric gisse <jowr.pi.nos...(a)gmail.com> wrote: > > <...> > >> Mazeltov! You have re-discovered the basic fact that an observer outside >> the event horizon of a black hole can never see an object pass through >> the event horizon in finite amounts of proper time. > > I thought the round trip aspect gave more meat to the observation. If > the object merely falls in, there is always the suspicion that > something funny is going on with our sense of time, given the > observation of a distant object near the horizon. If we get the object > back again... and as fast as possible, and we _still_ find the time > lapse goes to infinity, it begins to look more convincing that the > object never reaches the horizon. Only according to external observers. Calculate the proper time elapsed for an object going past the horizon. It passes _and reaches the singularity_ in a finite, and rather small, amount of time.
From: dlzc on 12 May 2010 23:14 Dear Darwin123: On May 12, 5:35 pm, Darwin123 <drosen0...(a)yahoo.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. There is no problem. You proceed to strain at *the light signals* that have to struggle *out of the gravity well* to get to said observer. > I think that it isn't possible for a single > object to fall past the event horizon because > of the asymmetry of the fall. There is no problem, there is no asymmetry. If you were a barrel rider, heading for Niagara Falls, and you could only signal via water waves, at some point, your water waves take infinitely long to get to someone upstream. > An object can't fall into the event > horizon in a finite time .... "can't be *seen* to cross the event horizon in finite time". .... > 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. Happening now in the LHC, so you'd better try and understand it. The light signals, and how long they take to get to exterior observers, does not keep the infaller from crossing the event horizon. David A. Smith
From: Tom Roberts on 13 May 2010 11:13
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. 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 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 |