
#19
Apr812, 11:33 AM

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The apparent horizon is "where gravitational time dilation goes to zero", though I would prefer to describe it as the location where outgoing light rays stay at the same radius, since "time dilation" is observer dependent but the paths of light rays are not. That is, the apparent horizon is locally defined; you can tell it's there just by looking at the paths of light rays locally. The apparent horizon forms when the surface of the star is just falling inward past the Schwarzschild radius corresponding to the star's total mass; before that, there is *no* apparent horizon anywhere. The absolute horizon is the boundary between the region of spacetime that cannot send signals "to infinity" and the rest of the spacetime. That is, the absolute horizon is globally defined; you have to know the entire future paths of light rays to know where it is. This boundary coincides with the apparent horizon once it forms (for the idealized case of a black hole that is static and "eternal" once formed), but the absolute horizon forms at r = 0 before that and expands outward at "the speed of light" (which is a bit of a misleading way to state it, since as more of the star's matter passes inward the outgoing null surface moves outward more and more slowly) until it meets the surface of the star at just the point where the apparent horizon forms. 



#20
Apr812, 12:20 PM

P: 14

Does the mass per volume remain constant with the expansion of the event horizon? What is interesting is that earths gravity reduces from its maximum when inside the earth linearly. So inside this object we can have the gravitational force reduce by r/a*GravHole? However an object reaching the speed of light at the surface would require infinite force. Yet gravity is with respect to the center of the object as it decreases by 1/r^2. I wonder how its possible for an object to experience infinite force from the gravitational law when it reaches the event horizon? I suppose for this to be possible it would require for the density at the surface to be infinite but the gravitational law only takes into accound mass. Would this suggest that there is infinite mass located at the edge of the event horizon? Or maybe that the distance to the center of the black hole becomes zero due to lorentz contraction. And I believe it would be certainly in error to assume a black hole to have infinite mass. And to assume it has infinite density is the same as saying it has infinite mass. But it certainly has limited mass other wise there would be no limit to its gravity. I guess I like the best the concept that on approach of the black hole the lorentz contraction closes you distance to the center of the black hole to zero. 



#21
Apr812, 12:56 PM

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[tex]a = \frac{M}{r^{2} \sqrt{1  \frac{2M}{r}}}[/tex] This obviously increases without bound as r > 2M, i.e., as the horizon is approached. However, this formula only applies to an observer who is "hovering" at a constant radius; but it is impossible for any massive object (i.e., anything except light or some other massless radiation) to "hover" exactly at the horizon, because it would have to move at the speed of light to do so. 



#22
Apr812, 01:33 PM

P: 16

OK, so a black hole forms when the gravity of a set amount of mass has expelled enough energy that it can no longer support itself. When it collapses it forms a singularity. The singularity is infintely dense and infinetly small. How did a set amount of mass form something infinite? The gravitational signature exist of something with a set mass but if we examine the process in reverse wouldnt we end up with something infinetly large at the limit of density approaching 0? I know that reversing the process is impossible.




#23
Apr812, 09:11 PM

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However, that's not the whole story, because as previous posts in this thread have pointed out, once the singularity is formed at r = 0, the matter in the object that originally collapsed to form it ceases to exist. All that remains is the infinite spacetime curvature at r = 0. The reason that curvature persists is that once it forms, it is selfperpetuating; the vacuum Einstein Field Equation (vacuum because, as I said just now, there is no matter left from the original collapsing object) says that that configuration of spacetime curvature will remain the same for all time. 



#24
Apr912, 02:12 PM

P: 3,967

Just for fun here is the interior Schwarzschild solution expressed as a ratio of proper time (d[itex]\tau[/itex]) to coordinate time (dt): [tex] \frac{d\tau }{dt} = \frac{3}{2}\sqrt{1\frac{R_s}{R}}\frac{1}{2}\sqrt{1\frac{R_s r^{2}}{R^{3}}}[/tex] where r is the location of the clock whose proper time we are interested in and R is the radius of the surface of the dust sphere. Now when R = 9/8 Rs and the clock is at the centre (r=0): [tex] \frac{d\tau }{dt} = \frac{3}{2}\sqrt{1\frac{8}{9}}\frac{1}{2}\sqrt{1\frac{0}{R^{3}}} = 0[/tex] while for clock at the surface of the same dust sphere, R = 9/8 Rs = r, so: [tex] \frac{d\tau }{dt} = \frac{3}{2}\sqrt{1\frac{8}{9}}\frac{1}{2}\sqrt{1\frac{R_s (9/8 R_s)^2}{(9/8R_s)^{3}}} = \frac{1}{2}\frac{1}{2}\sqrt{1\frac{8}{9}} = \frac{1}{3}[/tex] Now let's say the dust sphere continues to collapse so that R = 16/15 Rs then the relative proper time of the clock at the centre (r=0) is: [tex] \frac{d\tau }{dt} = \frac{3}{2}\sqrt{1\frac{15}{16}}\frac{1}{2}\sqrt{1\frac{0}{R^{3}}} = \frac{3}{8}\frac{1}{2} =  \frac{1}{8} [/tex] so now, the clock at the centre is running backwards relative to clocks outside the dust sphere! How this affects how the dust continues to collapse is anyones guess. 



#25
Apr912, 02:36 PM

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#26
Apr912, 04:41 PM

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#27
Apr912, 04:45 PM

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#28
Apr912, 05:36 PM

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(Technically, there is a mathematical solution, the "maximally extended Schwarzschild spacetime", which describes a black hole that is not formed by a collapsing object; but that's a mathematical solution only. As far as I know, nobody believes that it is physically reasonable.) (1) From the standpoint of someone far away from the collapsing object, and from the black hole that forms from it, there is still "mass" present; it does not disappear when the collapsing object vanishes into the singularity. By "mass" I mean that, for example, if you were orbiting the original object before it collapsed (i.e., you were far enough away that you were outside the original object to begin with), your orbit would not change during the collapse and the formation of the black hole (this is in the idealized case of a spherically symmetric collapse with no radiation emitted by the collapsing matter); from the orbital parameters alone you would not even be able to tell that the collapse had happened. So from a global perspective the whole process does "conserve energy". (2) Locally, however, there is indeed an issue with energy conservation when the collapsing matter vanishes into the singularity. However, strictly speaking, physicists do not think that this prediction of GR is what actually happens; rather, they take this prediction of GR, that a singularity of infinite density and infinite spacetime curvature forms, to be an indication that this situation is beyond GR's domain of validity. That is why you will see physicists saying that GR "breaks down" at the singularity, and that we need some new theory, presumably a theory of quantum gravity, to tell us what actually happens to the collapsed object. We would then want that new theory to explain how local energy conservation is maintained. 



#29
Apr912, 05:52 PM

P: 55

@PeterDonis: Thanks for the detailed response. I wonder about one more thing: if there's no energy and no "mass" in the hole, then from where does the energy that powers Hawking Radiation originate?




#30
Apr912, 06:48 PM

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As far as "where the energy comes from", the best simple answer is probably that it comes from the curvature of spacetime; as the hole radiates, its mass slowly decreases (for example, if you were in orbit about the hole, you would see your orbital parameters slowly changing to reflect a slowly decreasing mass), so the spacetime around the hole slowly becomes less curved. However, there is a *lot* more lurking here, enough for multiple threads (and there have been plenty on these forums already). 



#31
Apr912, 08:00 PM

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#32
Apr912, 08:29 PM

P: 3,967

<EDIT> Having given it yet more thought, the t coordinate in the interior solution is the same as the t coordinate in the exterior solution (ie the reference coordinate clock at infinity) because the 2 solutions join smoothly at the boundary. For example, when the dust sphere has a radius of 9/8 Rs and the test clock is at the surface of the dust sphere so that r = R = 9/8 Rs the proper time rate according to the interior solution is: [tex]\frac{d\tau }{dt} = \frac{3}{2}\sqrt{1\frac{8}{9}}\frac{1}{2}\sqrt{1\frac{ (9/8)^2}{(9/8)^{3}}} = \frac{1}{2}\frac{1}{2}\sqrt{1\frac{8}{9}} = \frac{1}{3}[/tex] while the proper time rate according to the exterior solution is: [tex]\frac{d\tau }{dt} = \sqrt{1\frac{8}{9}} = \frac{1}{3}[/tex] The solutions are in agreement, so the proper times are relative to the reference Schwarzschild coordinate clock at infinity in both cases. Perhaps I misunderstand what you are getting at? 



#33
Apr912, 08:36 PM

P: 1,555

That does not, as some often do, imply, that space becomes time and vice versa. It is only the coordinates that change their mapping. 



#34
Apr912, 08:50 PM

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#35
Apr912, 09:32 PM

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Prior to the actual collapse and formation of the singularity at the center why would the clocks there be more dilated than at the surface?? Thanks 



#36
Apr912, 09:48 PM

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That said, your general intuition here is correct. As the object collapses, the density at its center increases, and the "time dilation" at its center becomes more extreme, relative to an observer far away (another way to look at it would be to say that the "potential well" inside the object becomes deeper). See next comment. Also note that, strictly speaking, "acceleration due to gravity" only applies to a *static* observer, one who is "hovering" at a constant radial coordinate r. Once the object has collapsed inside its Schwarzschild radius, there are no such observers in the interior vacuum region inside the event horizon (i.e., outside the surface of the collapsing object but still below the horizon). (I'm not sure if there can still be static observers inside the collapsing object once it has collapsed inside the horizon; I think not, but I have not looked at the math in detail.) 


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