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Natural Science Forum / Physics / Research / August 2008Time evolution in GR
Question from a GR student: There is a classic question that students ask in GR that I havent been able to find a really good explanation for. Simply put, if a particle as observed from a distant observer would appear to never reach the event horizon of a BH how then could blackholes really ever form? The only thing I've been able to convince myself is that an actual blackhole in the purest sense would never be seen in finite time from a distant observer. If an observer would take the plunge, he could see it... but of corse looking back on the universe would see the "end of time." I think this means that only "approximations" to black holes would exist in the observable universe and this doesnt really seem too bad. The one piece of the puzzle that I know I don't have is that I do not understand how the horizon is formed in the creation of a blackhole. I know this is an active topic of study for numerical relativists... but I can't even find a really good qualitative discussion... any help? thanks, Shawn Thus spake mathkills <mathkills@ureach.com> >Question from a GR student: > [quoted text clipped - 17 lines] >but I can't even find a really good qualitative discussion... any >help? This was considered a puzzle in the early days of gtr. The calculation that a black hole can form was performed by Oppenheimer and Snyder in 1939. If a star consisting of neutron-degenerate matter has mass greater than the Tolman-Oppenheimer-Volkoff limit (thought to be about 1.5-3.0 solar masses, but uncertain because equations of state for extremely dense matter are not well known), then it will continue to collapse. When the density of matter is sufficient the event horizon forms outside the matter contained in the black hole. Regards  SignatureCharles Francis moderator sci.physics.foundations. charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and braces) http://www.teleconnection.info/rqg/MainIndex > This was considered a puzzle in the early days of gtr. The calculation > that a black hole can form was performed by Oppenheimer and Snyder in [quoted text clipped - 4 lines] > When the density of matter is sufficient the event horizon forms outside > the matter contained in the black hole. Does an event horizon form in the sense that there is some kind of discontinuity that comes into being? Or is it a mathematical artefact defined by the escape velocity = c? SignatureDirk http://www.transcendence.me.uk/ - Transcendence UK http://www.theconsensus.org/ - A UK political party http://www.onetribe.me.uk/wordpress/?cat=5 - Our podcasts on weird stuff Thus spake Dirk Bruere at NeoPax <dirk.bruere@gmail.com> >> This was considered a puzzle in the early days of gtr. The >>calculation [quoted text clipped - 9 lines] >discontinuity that comes into being? Or is it a mathematical artefact >defined by the escape velocity = c? The event horizon is a coordinate singularity, which comes into being when there is sufficient matter inside its radius. Regards SignatureCharles Francis moderator sci.physics.foundations. charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and braces) http://www.teleconnection.info/rqg/MainIndex >> Does an event horizon form in the sense that there is some kind of >> discontinuity that comes into being? Or is it a mathematical artefact >> defined by the escape velocity = c? In general, there is no LOCAL discontinuity at an event horizon. Indeed, there is no local measurement that can discover the event horizon of any of the known black-hole manifolds; only a global analysis can do so. There is a "discontinuity" at a horizon that is invisible to local observers but which can be analyzed by global methods: outside the horizon, outward-directed light rays move outward, but inside the horizon, outward-directed light rays move inward. This is invisible to local observers because they are necessarily infalling at a rate such that relative to them the outward-directed light ray moves outward, even though it is actually moving inward (as ascertained by a global analyst). I use the word "analyst" to refer to someone outside the manifold who can examine any aspect of the manifold at her leisure (to her the entire manifold, including all its history, is fixed). An "observer" is an inhabitant of the manifold, and can necessarily only make observations of local phenomena. > The event horizon is a coordinate singularity, which comes into being > when there is sufficient matter inside its radius. Only for poorly-chosen coordinates :-). And it does not actually "come into being" as you suggest, at least for some simple physical situations. In general, an event horizon is a locus consisting of a closed trapped surface, inside of which no timelike or null object can escape to spatial infinity. Such a surface is always moving with speed c relative to any locally-inertial frame, and poorly-chosen coordinates often have a coordinate singularity there, but there are always coordinate charts in a neighborhood of each point of the horizon that have no coordinate singularity there (in general this may not cover the entire horizon, but a finite number of such charts can cover the entire horizon). As I described earlier in the thread with subject "Time evolution in GR", for a spherically-symmetric collapse of a matter shell, the horizon initially forms at the center when there is no mass there at all -- the mass of a thin infalling shell of mass M meets the horizon at r=2M, so the horizon forms before the infalling shell even reaches that value of r. This makes perfect sense -- for any position of the infalling shell, a given point is inside the horizon if no light ray emitted from that point would reach r=2M before the infalling shell. So every point with r<2M must be inside the horizon before the shell reaches r=2M (how much before depends on its location and the rate of the shell's infall). Here "point" is in 4d. Tom Roberts > The event horizon is a coordinate singularity, Yes. > which comes into being when there is sufficient matter inside its radius. No. By definition, a coordinate singularity does not come into existence as a result of a physical process. It comes into existence when we choose to use singular coordinates, and vanishes when we choose to use well-behaved coordinates. Steve Carlip I just realized I misread the post I was responding to... What I *should* have said is: > > The event horizon is a coordinate singularity, > Yes. Really, the event horizon is a physical set of events, which is independent of the choice of coordinates. Specifically, it is the boundary beyond which no signal can reach infinity. An observer inside the event horizon can never send a signal that goes beyond the horizon; this is a statement that has no reference to coordinates. The coordinate singularity at the event horizon -- in a particular set of coordinates -- is an artifact of a poor choice of coordinates. For example, the horizon of a Schwarzschild black hole is singular in Schwarzschild coordinates, but is not singular in Kruskal-Szekeres or Painleve-Gullstrand coordinates. > > which comes into being when there is sufficient matter inside its radius. > No. By definition, a coordinate singularity does not come into > existence as a result of a physical process. It comes into existence > when we choose to use singular coordinates, and vanishes when > we choose to use well-behaved coordinates. The *event horizon* does come into existence when enough energy is inside a small enough region. The coordinate singularity comes into existence when someone decides to use bad coordinates. To return to the original question: I think the basic problem comes from the relativity of simultaneity, the source of much confusion in relativity. The place to start is to note that there is no preferred time coordinate in relativity (special or general), that is, no single special method of synchronizing clocks. This means that an event can take a finite time in one coordinate system and an infinite time in another. To understand what "really" happens, you need to look at coordinate- independent quantities. Does an event horizon "really" form? In classical general relativity, at least, the answer is clearly yes. That is, if you look at a spacetime containing, for example, a collapsing star, there are early time slices on which there is no event horizon, and later time slices on which there is an event horizon. (A time slice just means a spacelike hypersurface, that is, a three- dimensional space on which all events are spacelike separated, i.e., none is to the future or past of another. In particular, one can choose a time coordinate such that a time slice consists entirely of points at a fixed time. But the existence of a slice, and its geometric properties -- e.g., does it contain a piece of an event horizon? -- are coordinate-independent features.) It is also true that an observer who remains outside the event horizon will never "see" the horizon form. This is a matter of definition -- by definition, a signal originating inside or at the horizon can never reach an external observer. But this doesn't mean the horizon doesn't form, unless you want to take the solipsistic position that only those things you or your friends can see are real. (Tonight, take a flashlight, point it toward an empty area in the sky, and briefly flick it on. You can never see the light again -- it's moving away from you at the speed of light, and you can never catch up. If you really chose an empty section of the sky, no one else will ever see it, either. Does this mean it has disappeared from the Universe?) What an observer will see, although not quite an event horizon, is a very rapid fade to black. Light from a collapsing object is so strongly red-shifted that it very quickly becomes undetectable. Suppose, for instance, that an observer is watching a few-Solar-mass star collapse, and is looking in the visible light portion of the spectrum. After about a microsecond, no photon in this spectrum can possibly reach the observer: even if the entire mass of the star were converted to a single photon, it would be red-shifted to an energy below visible light. The observer could switch to looking at longer wavelengths, but for any wavelength she chose, the emitted light would very quickly be red- shifted to invisibility. This is not quite the same as the formation of an event horizon. While it does not happen in general relativity (except in very contrived circumstances), one could imagine a different theory in which a star halts its collapse at a position just outside the location that an event horizon would form. Emitted light would still be so red-shifted that it would be invisible for all practical purposes. But while such a theory could conceivably be constructed, it seems quite artificial and ad hoc. In particular, for a large enough collapsing body, the density of matter at the location where a horizon forms is not especially high, and a freely falling observer will see nothing special happening there. Steve Carlip > The event horizon is a coordinate singularity, [[...]] Some particular coordinates may be singular at the event horizon, but that's a property of those particular coordinates, not of coordinates in general. That is, in general the event horizon is *not* a coordinate singularity. For example, let's look at Schwarzschild spacetime. Here the Schwarzschild time coordinate is singular at the event horizon. But there are other coordinates [e.g., the Eddington-Finkelstein time coordinate, not to mention Kruskal-Szekeres (u,v) coordinates] which are nonsingular on the event horizon, and indeed everwhere near the event horizon. Signature-- "Jonathan Thornburg [remove -animal to reply]" <J.Thornburg@soton.ac-zebra.uk> t <= 31.Aug.2008: School of Mathematics, U of Southampton, England t >= 1.Sep.2008: Dept of Astronomy, Indiana University, Bloomington, USA "Washing one's hands of the conflict between the powerful and the powerless means to side with the powerful, not to be neutral." -- quote by Freire / poster by Oxfam > Question from a GR student: > [quoted text clipped - 17 lines] > but I can't even find a really good qualitative discussion... any > help? This is very much Achilles and the Tortoise. At the Event Horizon time does indeed stand still but only for an instant. In fact total time to fall into a Black Hole is finite, very much so. The original paradox said. Achilles runs the distance the totoise is away. Then he runs the distance to the tortoise, etc. etc. He never catches the tortoise. To get a handle integrate 1/x^2 through zero. 1/ x^x is infinite at zero yet the integral is still finite. - Ian Parker Correction Int(1/√x) = 2√x This is 0 at x=0 - Ian Parker mathkills a écrit: >The one piece of the puzzle that I know I don't have is that I do not >understand how the horizon is formed in the creation of a blackhole. >I know this is an active topic of study for numerical relativists... >but I can't even find a really good qualitative discussion... any >help? Did you read this? http://www.mathpages.com/rr/s7-02/7-02.htm > There is a classic question that students ask in GR that I havent been > able to find a really good explanation for. Simply put, if a > particle as observed from a distant observer would appear to never > reach the event horizon of a BH how then could blackholes really ever > form? The extended solution essentially amounts to tacking on time after eternity. So, the particle falls in after forever never ends. You'll get replies to the effect that this was considered a puzzle that was "solved" a long while ago. However, there is an entierly different layer of "puzzle" on top of the puzzle which was NOT solved: the "eternal" black hole is a purely classical object. There are none -- at least if you believe what quantum theory tells you. Instead, it evaporates. Any solution with a horizon and "trapped surface", in fact, evaporates. This leads to a flat-out contradiction (a.k.a. the information paradox). The only clear solution is that there is, in fact, no trapped surface at all, but something that just looks like one for a long time and can be approximated classically by one. Hawking, himself, since 2004 (as have I since way back, in the 1980's) has basically come out on that side of the issue, essentially saying that nothing ever falls in because by the time it gets there, it's all evaporated away. In recent times -- meaning just in the last year or so -- I've seen a journal reference published which actually draws up the Penrose diagram for an evaporating black hole, INCLUDING the "last moment" event, where it's gone poof and completely evaporated away. Needless to say, the diagram is pretty contorted, almost as much as that for an eternal black hole. Someone else may be able to supply the reference, otherwise I'll look in my cache of papers and see if I can't get a reference. On a somewhat related note, the idea's come to my mind in recent times that since the black hole is the solution respecting spherical symmetry and time symmetry, then it is basically just a reduction of the ONE dimensional fibre bundle with base space R: (coordinate r), gauge group SU(2): (coordinates, lambda, mu, nu); and gauge group E(1): (coordinate t). The reduction SU(2) -> S_2 from the 3-sphere S_3 = SU(2) to the 2-sphere S_2 can be effected by (lambda, mu, nu) -> (2 lambda, mu + nu), if the 3-sphere coordinates are given by (cos(theta) cos(mu), cos(theta) sin(mu), sin(theta) cos(nu), sin(theta) sin(nu)). That means that the black hole is just Euclidean space E(1):r, with a gauge group, locally equivalent to SU(3)xU(1) = U(3). So, you can solve it just by writing the gauge group bundle metric for U(3) x E(1) over E(1). The extension to the maximal solution is from the restricted Schwarzshild solution E(1): (r > 0), which corresponds to the interval (0, infinity) to negative values of r E(1): (-infinity, infinity). That gives you the basis for the Kruskal solution. One can add in the charges for any gauge group by tacking on more symmetry groups. Thus, for electroweak (i.e. U(2)), you have U(3) x U(2) x E(1). For electroweak + color, you have U(3) x S(U(2) x U(3)) x E(1). The same ideas apply for the axial solutions. Here, the base space is E(2): (r, theta), corresponding to the half-plane (r > 0), and the symmetries are axial: U(1): (psi), and temporal E(1): (t). One can thus represent the solutions and charged solutions as bundle metrics over U(1) x U(1) x E(2), and charged metrics (e.g. for U(2) electroweak) as U(2) x U(1) x U(1) x E(2). This gives you a generalization of the charged rotating black hole, with both hypercharge (Y, for the U(1)_Y part of U(2)) and isospin (I for the SU(2)_I part of U(2); thus giving electrical charge as Q = Y + I_3). > if a > particle as observed from a distant observer would appear to never > reach the event horizon of a BH how then could blackholes really ever > form? Well, perhaps they were there from the beginning of the universe. After all, this is true for the standard black hole manifolds (Schwarzschild, Kerr, ...). Why should primordial black holes be less likely than primordial quarks and gluons?.... > The only thing I've been able to convince myself is that an actual > blackhole in the purest sense would never be seen in finite time from [quoted text clipped - 3 lines] > holes would exist in the observable universe and this doesnt really > seem too bad. Hmmm. To an outside observer, a forming black hole and a fully-developed black hole are indistinguishable. For instance, once a spherically symmetric object is smaller than 3 times its Schw. radius, no timelike object can stably orbit at or within that radius, regardless of whether or not it actually collapses to smaller than its Schw. radius. This is rigorous for a spherically symmetric object. I'm pretty sure that similar results apply to asymmetric and/or rotating objects. > The one piece of the puzzle that I know I don't have is that I do not > understand how the horizon is formed in the creation of a blackhole. For a spherically symmetric collapse, Birkhoff's theorem permits us to describe things accurately and well. [I use units with G=c=1.] Imagine a spherical shell of matter of mass M that is symmetric around spatial point P, and is collapsing inward toward P in a spherically symmetric manner. Let there be a distant observer O relative to which P does not move, and let the shell be transparent so O can observe light sources arranged along the line from P to O. When the shell is far from P (i.e. its inner radius is >>M), clearly no horizon is present and all light sources can be seen by O. And once the shell is wholly inside radius 2M, no source inside that radius is visible to O. By imagining a series of thin shells each of mass M, it should be clear that the horizon is initially absent, but at some time when the inner radius of the first shell has not yet reached P, a horizon will form at P and expand outward at the (local) speed of light [#]. The horizon will expand, meeting the outer surface of the first shell at a radius 2M from P, and the horizon remains there while the shell continues collapsing. For later shells this also applies, with M being the total mass inside radius 2M. Clearly a given light source a distance L from P will become invisible to O for light it emits at the time the total mass inside L is L/2 -- that is when the horizon has expanded to engulf the source. [#] The horizon initially forms at P when a light pulse emitted from P would lose the race to r=2M to the outside of the first shell of mass M. And the horizon reaches a given radius when a light pulse emitted from that radius would lose the same race. Horizon and outer surface of the first shell reach r=2M simultaneously. Tom Roberts |
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