Fun with black holes
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From: BURT on 31 May 2010 19:08 On May 31, 3:36 pm, "Greg Neill" <gneil...(a)MOVEsympatico.ca> wrote: > 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). Light's redshift violates laws of energy for a black hole. At the event horizon light energy infinitely red shifting leaving infinitely small energy violates energy laws. Infinite blueshift violates the same law. Mitch Raemsch
From: Greg Neill on 31 May 2010 20:59 BURT wrote: [snip drivel] Your comments are not lucid, nor desired.
From: Tom Roberts on 1 Jun 2010 20:10 Greg Neill wrote: > 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. Hmmm. The exponential decay is the distribution of the number of emitted of protons over time -- it is STATISTICAL. Still, the distant observer can never KNOW when the last one reaches him. A good guess can probably be made. > 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). As I said before, I don't think this query is well defined. Phrasing it all in terms of what the distant observer observes is straightforward; I don't know the answer to that question. I doubt anyone does, unless they have simulated it with a numerical solution to the field equation. Tom Roberts
From: BURT on 1 Jun 2010 21:16 On Jun 1, 5:10 pm, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote: > Greg Neill wrote: > > 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. > > Hmmm. The exponential decay is the distribution of the number of emitted of > protons over time -- it is STATISTICAL. Then I guess you qualify as a statistic. Mitch Raemsch > Still, the distant observer can never > KNOW when the last one reaches him. A good guess can probably be made. > > > 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). > > As I said before, I don't think this query is well defined. Phrasing it all in > terms of what the distant observer observes is straightforward; I don't know the > answer to that question. I doubt anyone does, unless they have simulated it with > a numerical solution to the field equation. > > Tom Roberts- Hide quoted text - > > - Show quoted text -
From: Greg Neill on 1 Jun 2010 23:32
Tom Roberts wrote: > Greg Neill wrote: >> 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. > > Hmmm. The exponential decay is the distribution of the number of emitted of > protons over time -- it is STATISTICAL. Still, the distant observer can never > KNOW when the last one reaches him. A good guess can probably be made. > > >> 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). > > As I said before, I don't think this query is well defined. Phrasing it all in > terms of what the distant observer observes is straightforward; I don't know the > answer to that question. I doubt anyone does, unless they have simulated it with > a numerical solution to the field equation. Fair enough. Perhaps the answer can be teased out of some of the numerical simulations that have been done for the merging of black holes. I'll have to check the literature. I do seem to recall that, at least for some period of time, such a merger exhibits a continuous but oblate horizon. Whether or not one would be justified in surmising the continuing existance of separate mass concentrations (singularities in the case of two merging black holes) out of sight behind such an horizon is perhaps problematical. |