Fun with black holes
in [Relativity]
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From: eric gisse on 24 May 2010 03:35 Tom Roberts wrote: > eric gisse wrote: >> Edward Green wrote: >>> 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. > > Not for the case of an object falling into a black hole from far away. > > >>> 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. > > Yes, there is. It's well known. But one must be careful in defining terms, > because a distant observer cannot observe it. I saw a nice discussion of this in MTW, and by all means it seems like the perturbations don't exist because there is no means of observing them since everything is redshifted into oblivion. > > I believe it is discussed (in general terms) in the book > by Kip Thorne I referenced earlier in this thread. MTW as well, but I had a different interpretation. > > > Tom Roberts
From: eric gisse on 24 May 2010 03:51 Tom Roberts wrote: > eric gisse wrote: >> Once the matter is redshifted into oblivion it is a part of the black >> hole as far as external observers are concerned. > > Yes. > > >> As for the horizon _growing_ before the matter reaches it, nonsense. The >> horizon will only grow in response to further input of mass-energy. > > No. > > Here's proof: Consider a Schw. black hole of "mass" M, and an infalling > thin spherical shell of mass m that is initially far outside the horizon. > Initially the horizon is essentially at r=M, but everywhere outside the > shell the geometry is that of a black hole with "mass" M+m [#]. Completely agree so far... > As the > shell approaches the horizon, the horizon expands outward with local speed > c , and the horizon and shell meet when both reach r=M+m together [#]. This is not at all obvious to me, though the _result_ has to be as stated. > > [#] Birkhoff's theorem. > > Heuristically this can be understood by considering spatial points between > r=M and r=M+m, as the shell approaches. At some time before the shell > actually reaches r=M+m, each such point must be inside the horizon, > because an emitted light ray cannot reach r=M+m before the shell reaches > r=M+m. We know [#] that when the shell reaches r=M+m the horizon must be > at r=M+m. I buy this without any difficulty, though it is not something I have thought about before in this fashion. What isn't clear to me is why the horizon must approach the shell at c > > Yes, I am speaking a bit loosely here. > > >> The event horizon is isn't a membrane. It is not a material surface. It >> is not 'there'. Nothing special happens when something passes through it, >> other than the certainty of never leaving the black hole. > > Yes to all that. But the horizon is a geometrical locus, and when > spacetime is foliated into space and time, the spatial locus at a given > time can change over time. > > > Tom Roberts
From: Tom Roberts on 24 May 2010 05:18 eric gisse wrote: > Tom Roberts wrote: >> Here's proof: Consider a Schw. black hole of "mass" M, and an infalling >> thin spherical shell of mass m that is initially far outside the horizon. >> Initially the horizon is essentially at r=M, but everywhere outside the >> shell the geometry is that of a black hole with "mass" M+m [#]. > > Completely agree so far... > >> As the >> shell approaches the horizon, the horizon expands outward with local speed >> c , and the horizon and shell meet when both reach r=M+m together [#]. > > This is not at all obvious to me, though the _result_ has to be as stated. When the shell is far away, the horizon is at r=M; when the shell is at r=M+m, the horizon is at r=M+m. Everything is continuous, so the horizon MUST move outward smoothly, and reach r=M+m precisely when the shell reaches r=M+m. >> Heuristically this can be understood by considering spatial points between >> r=M and r=M+m, as the shell approaches. At some time before the shell >> actually reaches r=M+m, each such point must be inside the horizon, >> because an emitted light ray cannot reach r=M+m before the shell reaches >> r=M+m. We know [#] that when the shell reaches r=M+m the horizon must be >> at r=M+m. > > I buy this without any difficulty, though it is not something I have thought > about before in this fashion. > What isn't clear to me is why the horizon must approach the shell at c From continuity, the horizon must smoothly expand from its initial r=M to r=M+m when the shell reaches r=M+m. The horizon is always moving with local speed c. Don't forget that any locally-inertial frame valid at the horizon is necessarily infalling; Schw. coordinates aren't valid there. Tom Roberts
From: Greg Neill on 24 May 2010 09:19 Tom Roberts wrote: > Greg Neill wrote: >> 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. > > Not really. A distant observer has no way to "see" the shape of the horizon. the > horizon of a black hole is an abstract geometrical locus at which no timelike > object can escape to spatial infinity. To determine the location of the horizon > requires non-local measurements, such as tracking light rays that approach it, > or dropping powerful rockets that attempt to blast back out, and see which ones > make it and which ones don't. > > When someone says that the asymmetry in the horizon disappears quickly, this is > no sort of measurement, this is determined by examining an appropriate solution > to the field equation (or an approximation to one, such as in computer simulation). That's fine. For the purposes of discussion we can assume the appropriate Gedanken tools for observation, or assume that the mathematics of GR is correct and 'observe' by that means. > > >> Presumably the collision of the BH and object will then >> appear like a perfectly inelastic collision as far as >> the observer can see. > > Yes. > > >> There won't be any wobbling about >> of the horizon around the center of mass > > The distant observer cannot observe this. What such an observer could see is the > gravitational radiation emitted as the horizon relaxes to spherical shape (in > the case being discussed), but the simple "gravitational compass" cannot do so. Again, we can assume any instrumentation desired, of any theoretically possible precision and accuracy. The 'compass' was just an example. > > >> while the object >> follows its inevitable trajectory to the singularity. > > The distant observer cannot attempt to ascribe anything to the object "after" > its trajectory enters the horizon. I put "after" in quotes, because to this > observer the object NEVER actually reaches the horizon. > > So your use of "while" is unwarranted. Okay, let me try to phrase it in a different way. 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. The observer can see nothing of the object beyond that instant on his clock. Yet his clock continues to run, and he can continue to observe the BH even if nothing interesting is happening. 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. From the remote observer's point of view, the center of mass will have 'jumped' to coincide with the center of the black hole as soon as the infalling object becomes undetectable. The motion of the BH in the observer's frame of reference will betray no further evidence of the collision -- all the 'transfer' of conserved quantities appears to take place before or effectively coincident with the detection of the last photon. > >> 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? > > Infalling objects on non-radial paths into a black hole impart angular momentum > to the BH. "Almost instantly" is unwarranted, as discussed above. I think 'almost instantly' is apt since the extinction rate of the means of observation of the infalling object is apparently very rapid indeed from the remote observer's point of view. So an observer equipped to measure the spin of a black hole (by whatever means that might be accomplished -- frame dragging, say) would register that the increase in spin rate for the black hole essentially coincides with the observational disappearance of the infalling object (last detectable photon).
From: BURT on 24 May 2010 13:35
On May 24, 2:18 am, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote: > eric gisse wrote: > > Tom Roberts wrote: > >> Here's proof: Consider a Schw. black hole of "mass" M, and an infalling > >> thin spherical shell of mass m that is initially far outside the horizon. > >> Initially the horizon is essentially at r=M, but everywhere outside the > >> shell the geometry is that of a black hole with "mass" M+m [#]. > > > Completely agree so far... > > >> As the > >> shell approaches the horizon, the horizon expands outward with local speed > >> c , and the horizon and shell meet when both reach r=M+m together [#]. > > > This is not at all obvious to me, though the _result_ has to be as stated. > > When the shell is far away, the horizon is at r=M; when the shell is at r=M+m, > the horizon is at r=M+m. Everything is continuous, so the horizon MUST move > outward smoothly, and reach r=M+m precisely when the shell reaches r=M+m. > > >> Heuristically this can be understood by considering spatial points between > >> r=M and r=M+m, as the shell approaches. At some time before the shell > >> actually reaches r=M+m, each such point must be inside the horizon, > >> because an emitted light ray cannot reach r=M+m before the shell reaches > >> r=M+m. We know [#] that when the shell reaches r=M+m the horizon must be > >> at r=M+m. > > > I buy this without any difficulty, though it is not something I have thought > > about before in this fashion. > > What isn't clear to me is why the horizon must approach the shell at c > > From continuity, the horizon must smoothly expand from its initial r=M to r=M+m > when the shell reaches r=M+m. The horizon is always moving with local speed c. > Don't forget that any locally-inertial frame valid at the horizon is necessarily > infalling; Schw. coordinates aren't valid there. > > Tom Roberts Black holes are disproven on the basis that they violate motion laws and energy laws. Falling violates SR while energy for light can go infinite. Mitch Raemsch |