Fun with black holes
in [Relativity]
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From: Tom Roberts on 14 May 2010 09:44 eon wrote: > On May 14, 6:04 am, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote: >> 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 You are confused: 1. I was discussing an object on a straight radial path, so its angular velocity (wrt the black hole) is zero. 2. There is no "flush" (whatever that is supposed to mean). 3. Integration does not work that way. 4. One cannot "cancel" a black hole, because it is a geometrical property of the manifold. Tom Roberts
From: Edward Green on 14 May 2010 19:27 On May 14, 6:51 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > 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 Yes. > 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. Consider an ensemble of rocket probes dropped into a black hole. These probes have the property that some of them will fire their rockets off at a random r_closest and return to us, others will fall in without firing their rockets. No matter how long we wait, some fraction of the ensemble (all dropped from the same height at the same time) will continue to trickle back home. Therefore there is no time we can wait until we can say "OK, by now, the remaining probes have all crossed the event horizon, and no more will return. This strongly suggests to me that the probes never cross the event horizon, in a sense that doesn't depend on ambiguities in simultaneity and time dilation. But you need the continued stream of returning probes to see this clearly. (We probably do need to assume symmetrical round trips to make this argument). Thorne, in his book referenced earlier in this thread, essentially says several times in scenarios 'OK, by now the probe has fallen in', or words to that effect, although we continue to receive red-shifted signals back from it. I question this usage. Now, if the event horizon rises to meet the probe -- which must have some gravitational mass, however slight -- then that's different. Then the probe really will be eventually absorbed. In this case we would observe a cut off in our return distribution of rockets, and could conclude "by now the remainder have fallen inside the horizon".
From: Edward Green on 14 May 2010 20:58 On May 14, 6:51 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: <...> BTW; thanks for continuing to demonstrate that one can disagree without being disagreeable.
From: Daryl McCullough on 14 May 2010 21:08 Edward Green says... >Consider an ensemble of rocket probes dropped into a black hole. These >probes have the property that some of them will fire their rockets off >at a random r_closest and return to us, others will fall in without >firing their rockets. > >No matter how long we wait, some fraction of the ensemble (all dropped >from the same height at the same time) will continue to trickle back >home. Therefore there is no time we can wait until we can say "OK, by >now, the remaining probes have all crossed the event horizon, and no >more will return. This strongly suggests to me that the probes never >cross the event horizon, in a sense that doesn't depend on ambiguities >in simultaneity and time dilation. But you need the continued stream >of returning probes to see this clearly. But there is the analogous situation with the rocket undergoing constant proper acceleration. You have a huge mothership that is accelerating continually. Then, in the accelerated coordinate system of the mothership, there will be a distance, L, below the mothership such that no signal sent from farther away than that will ever reach the mothership. This is the Rindler event horizon. You can then release a bunch of rocket probes, which will drop behind the mothership and get arbitrarily close to the Rindler horizon. Then the probe will accelerate and return to the mothership. The time required for the round trip goes to infinity as the probes get closer and closer to the Rindler horizon. But in this case, we know that life goes on past the Rindler horizon. Because the horizon is just an artifact of using accelerated coordinates. Nothing changes when something crosses the horizon, except that there is a point at which it becomes impossible to catch up to the mothership (unless the mothership stops accelerating). -- Daryl McCullough Ithaca, NY
From: eon on 15 May 2010 05:10
On May 14, 3:44 pm, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote: > eon wrote: > > On May 14, 6:04 am, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote: > >> 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 > > You are confused: > 1. I was discussing an object on a straight radial path, so its angular > velocity (wrt the black hole) is zero. fluid dynamics says you have a sink, so a radial path is unreal > 2. There is no "flush" (whatever that is supposed to mean). how not, you have a sink, with angular velocity component must be much higher than its radial i so this in a move sold by a Brian, or was it Stephan or something > 3. Integration does not work that way. how not? > 4. One cannot "cancel" a black hole, because it is a geometrical property > of the manifold. is not it an object, but only a property of an object ??? how come? > > Tom Roberts thanks |