Fun with black holes
in [Relativity]
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From: Greg Neill on 19 May 2010 08:54 Tom Roberts wrote: > Greg Neill wrote: > [My context is GR.] > > Let me assume no other massive objects exist, that the black hole is > Schwarzschild, that the object is pointlike, and it falls straight in. Then the > "gravitational compass" points in a fixed direction, as the c-o-m does not move > (conservation of momentum, which applies because of the asymptotically flat > manifold). Note that the "gravitational compass" is insensitive to the passage > of the gravitational waves generated by the infall. Okay, sounds reasonable. > > At least I think all that is true. The caveat is that > conservation of momentum only applies APPROXIMATELY > here, but I think the approximation is extremely good > and better than required. All that I say here is an > educated guess based on knowledge of the properties of > black holes and the structure of GR, not the result of > any rigorous computation. > > >> The question that arrises is, at what point in time, if any, >> with the observed center of mass coincide with the center >> of the black hole? > > Using the distant observer's time and observations at her location: As the > object is observed to approach the horizon, the asymmetry of the horizon + > object will quickly get redshifted to unobservability. That asymmetry is related > to the observed difference between the center of the black hole and the c-o-m. Okay, I can go along with emitted gravitational waves being redshifted, although I'm not sure about how redshift applies to what should be the shape of space around the BH and infalling mass (for instance, say for example that it happens to be another BH of similar mass...). > > >> That is, when will the external observer >> conclude that the infalling mass has met its fate at the >> singularity? > > That is a COMPLETELY different question. An unanswerable one. The relaxing of > the horizon to a sphere has nothing to do with the infalling object intersecting > the singularity. Indeed they occur at different locations in the manifold, and > the horizon is not causally connected to any point inside the horizon (including > all points near the singularity). > > Indeed there is no point in the entire manifold that is > causally connected to the limit point of the object's > trajectory intersecting the singularity. > > The distant observer cannot observe the infalling object cross the horizon (much > less intersecting the singularity inside) -- the relevant data get redshifted to > unobservability. So this applies to the net gravitational field as well as to mundane means of observation then. A distant observer will "see" the relaxation of the horizon to a sphere occur in a very short time indeed, much as the quick extinction of light arriving from an object that falls in. After that, the BH just "looks" like a slightly bigger (more massive) BH. Presumably the collision of the BH and object will then appear like a perfectly inelastic collision as far as the observer can see. There won't be any wobbling about of the horizon around the center of mass while the object follows its inevitable trajectory to the singularity. Would this also necessarily follow for an object meeting the horizon on a non radial path, the angular momentum being transferred almost instantly to the spin of the BH?
From: Edward Green on 21 May 2010 17:24 On May 16, 11:58 pm, eric gisse <jowr.pi.nos...(a)gmail.com> wrote: > Edward Green wrote: > > [...] > > What's the point of this? You are rehashing a well known point that > observers outside a black hole's event horizon do not see the transit of an > object through the event horizon in finite observer time. > > It doesn't matter how you rephrase the question, the answer is always going > to be the same. The point is that the answer sounds like some kind of trick, whereas I am proposing that the object never "really" penetrates the horizon, for suitable values of "really". Unless, that is, as something I have read leads me to believe, the horizon eventually rises to meet the infalling mass. Thereafter the black hole rings out like a bell until the perturbation to its horizon has been absorbed. Black holes _can_ have hair, they just tend to go bald when they have it.
From: Darwin123 on 21 May 2010 17:42 On May 12, 11:14 pm, dlzc <dl...(a)cox.net> wrote: > 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. Theoretically, there could be a large asymmetry. According to some theories, that is how a "naked singularity" forms. A naked singularity is different from a black hole, if it exists. I am not saying that I am sure that naked singularities exist. I mean that some physicists have predicted that such an object can form if the collapse of the star is highly asymmetric. If the matter falls in symmetrically, even with spin, a black hole forms. Look up "naked singularity". >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. That analogy would work fine in a Galilean invariant universe. In a Galilean invariant universe, the water flows at a constant rate up to the edge of the Falls together with the boat. The boat always moves with the water. In a relativistic universe, the boat slows down as it approaches the edge. So the boat ends up hanging at the edge of the falls, without ever going over. > > 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". > > ... Okay. However, that doesn't really answer the question. The problem is connecting the observations of the falling observer with the observations of the far away observer.
From: eric gisse on 21 May 2010 18:41 Edward Green wrote: > On May 16, 11:58 pm, eric gisse <jowr.pi.nos...(a)gmail.com> wrote: >> Edward Green wrote: >> >> [...] >> >> What's the point of this? You are rehashing a well known point that >> observers outside a black hole's event horizon do not see the transit of >> an object through the event horizon in finite observer time. >> >> It doesn't matter how you rephrase the question, the answer is always >> going to be the same. > > The point is that the answer sounds like some kind of trick, whereas I > am proposing that the object never "really" penetrates the horizon, > for suitable values of "really". According to the object actually doing it, it does (which is the only answer that matters). According to the external observer, it doesn't. This doesn't need to be rehashed. > Unless, that is, as something I have > read leads me to believe, the horizon eventually rises to meet the > infalling mass. Uuuuuuhhhhhhh no. The horizon is a static fixture of the manifold. > Thereafter the black hole rings out like a bell until > the perturbation to its horizon has been absorbed. No again. There is no analysis anywhere which supports this. > > Black holes _can_ have hair No again. > , they just tend to go bald when they have > it.
From: BURT on 21 May 2010 18:54
On May 21, 3:41 pm, eric gisse <jowr.pi.nos...(a)gmail.com> wrote: > Edward Green wrote: > > On May 16, 11:58 pm, eric gisse <jowr.pi.nos...(a)gmail.com> wrote: > >> Edward Green wrote: > > >> [...] > > >> What's the point of this? You are rehashing a well known point that > >> observers outside a black hole's event horizon do not see the transit of > >> an object through the event horizon in finite observer time. > > >> It doesn't matter how you rephrase the question, the answer is always > >> going to be the same. > > > The point is that the answer sounds like some kind of trick, whereas I > > am proposing that the object never "really" penetrates the horizon, > > for suitable values of "really". > > According to the object actually doing it, it does (which is the only answer > that matters). According to the external observer, it doesn't. This doesn't > need to be rehashed. > > > Unless, that is, as something I have > > read leads me to believe, the horizon eventually rises to meet the > > infalling mass. > > Uuuuuuhhhhhhh no. The horizon is a static fixture of the manifold. > > > Thereafter the black hole rings out like a bell until > > the perturbation to its horizon has been absorbed. > > No again. There is no analysis anywhere which supports this. > > > > > Black holes _can_ have hair > > No again. > > > > > , they just tend to go bald when they have > > it.- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text - The cause of round curved geometry is the center of energy density. Gravity gives a geometric center to space and energy. Mitch Raemsch |