Fun with black holes
in [Relativity]
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From: BURT on 24 May 2010 21:53 On May 24, 6:47 pm, Edward Green <spamspamsp...(a)netzero.com> wrote: > On May 24, 2:18 am, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote: > > > Edward Green wrote: > > > Finite proper time either means it actually > > > falls in, or actually freezes. > > > No. Finite proper time means a finite time to a comoving observer. An object > > that crosses the horizon in a finite proper time after a given point on its > > trajectory outside the horizon definitely DOES fall into the BH, according to a > > comoving observer. This is not "debatable". > > > You seem to be hung up on the notion that any observer ought to be able to > > observe all aspects of "reality" -- that's quite naive. > > I remind you of the class of trajectories in SR of unbounded (but > always finite) acceleration which accumulate only finite proper time > on their trip to infinity. How do I know something similar is not > happening here? In some sense the free fall acceleration becomes > unbounded here also, because to stop it would require unbounded > acceleration. > > However, I thought you agreed with me that for a massive test > particle, the horizon will rise to meet it, which renders the question > moot. The end of time applies to all forms of time. Proper time ends at the event horizon. Mitch Raemsch
From: Tom Roberts on 25 May 2010 02:10 BURT wrote: > Black holes are disproven on the basis that they violate motion laws > and energy laws. No, they don't. But you must use the last of motion and energy of GR, because it is the applicable theory. > Falling violates SR while energy for light can go > infinite. SR is COMPLETELY IRRELEVANT here. As is your silly claim about light energy. Rather than wasting your time posting nonsense to the net, why don't you get a good textbook and STUDY? Tom Roberts
From: Tom Roberts on 25 May 2010 02:14 Edward Green wrote: > I suspect neither of us knows enough to answer this definitively, but > analyticity suggests to me that if the spacetime curvature immediately > adjacent to the horizon is vibrating, then so is the horizon. There is a theorem in GR that in a given manifold, the area of the union of all black holes' horizons cannot decrease. This greatly limits the ability for an horizon to "vibrate". > And BTW, the above quote also supports my contention that black holes > can have hair, they just tend to radiate it away. The theorems refer > to _stationary_ black holes, not excited ones. An excited black hole > is obviously not solely characterized by mass, charge and angular > momentum. Yes. Tom Roberts
From: Tom Roberts on 31 May 2010 13:15 Greg Neill wrote: > The remote > observer watches the falling object, and as it approaches the > horizon there is a 'last detectable photon' from the object > and a 'last gravitational wave' from the settling of the > manifold. No. An observer can never know whether one more will appear if she waits a bit longer. This is similar to the impossibility of finding the "largest integer less than infinity". At some point in time the observer can note that it's been a while since the last detected photon, and she can extrapolate the fall-off and GUESS no more will come for quite a while, but she can never KNOW that no more will come. > What I gather is that after the romote observer detects the > last photon and wave he will not see any further adjustment > in the position of the BH with respect to the center of mass > of the system as that was defined defined by the object + BH > prior to their meeting. A distant observer cannot observe the horizon in any way without sending probes down near to it, such as remote-controlled rockets or light rays. Assuming GR is correct, the observer can COMPUTE where the horizon ought to be, based on measurement she makes. Tom Roberts
From: Greg Neill on 31 May 2010 18:36
Tom Roberts wrote: > Greg Neill wrote: >> The remote >> observer watches the falling object, and as it approaches the >> horizon there is a 'last detectable photon' from the object >> and a 'last gravitational wave' from the settling of the >> manifold. > > No. An observer can never know whether one more will appear if she waits a bit > longer. This is similar to the impossibility of finding the "largest integer > less than infinity". The number of photons emitted by the object crossing the event horizon is finite, so the exponential decay is only an approximation and doesn't continue forever. This is somewhat like the analysis of the collapse of a star to a black hole, wherein there is also a 'last photon' effect in the dimming of the system as it passes below its own Schwarzchild radius. There the luminosity goes something like L ~ exp(-t/(3*sqrt(3)*M)), and the "last photon" emerges in time t ~ (M/Msun)*10^-3 seconds after the collapsing star starts dimming. I'm taking these approximations from Pickover's book (Black Holes: A Traveller's Guide), but I'm pretty sure that MTW covers this, too. > > At some point in time the observer can note that it's been a while since the > last detected photon, and she can extrapolate the fall-off and GUESS no more > will come for quite a while, but she can never KNOW that no more will come. Even so, I think that the extinction rate will be sufficiently fast so that one can define a practical point in time (for the observer) that can be considered 'lights out', much like for second order systems we can consider all the excitement to be over after five time constants. I may be mistaken, but I think that the extinction rate will make the 'time constant' in this case much shorter than the other time measures of interest in the problem. > > >> What I gather is that after the romote observer detects the >> last photon and wave he will not see any further adjustment >> in the position of the BH with respect to the center of mass >> of the system as that was defined defined by the object + BH >> prior to their meeting. > > A distant observer cannot observe the horizon in any way without sending probes > down near to it, such as remote-controlled rockets or light rays. Assuming GR is > correct, the observer can COMPUTE where the horizon ought to be, based on > measurement she makes. > I don't argue with what you say, but I don't see how it addresses my query. One shouldn't have to observe the horizon in order to determine where the center of mass of the black hole lies -- it should be obvious from the motions of test bodies or 'gravitational compass' triangulations. My query was about whether or not the BH's motion ceases to be affected (accelerated) by the interaction with the colliding object at the same time as it 'appears' to have become unobservable (by photon or gravitational wave emissions). Presumably infalling object becomes unobservable as a separate mass center at some point in time closely associated with the photon extinction. Does the BH then cease to behave as though it's participating in a two body interaction immediately? If so, the horizon, as insubstantial as it is, behaves like a solid surface of impact as far as the outside observer can tell (in terms of transferring conserved quantities like momentum). |