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Are Finkelstein/Kruskal interior black hole solution compatible with Einstein's GR? 
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#109
Dec612, 09:27 AM

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PF Gold
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If we take GR as it has been validated by experiment, and use that theory, without any changes, to make physical predictions about black holes, we find that it predicts that horizons and singularities form, and objects can fall in past the horizons and be destroyed in the singularities. That's not a matter of "interpretation"; it's a matter of using the theory as it's been validated, with the same math and the same rules for translating the math into physical observables, and extending it into a regime where there is no direct experimental validation. When you make the claim that "Einstein's GR says that nothing can ever reach the horizon", you are taking the theory, GR, as it has been validated by experiment, and *changing the rules* for how it is used to make physical predictions in a regime where there is no direct experimental data. The theory, as it has been validated by experiment, uses proper time and other invariants, not coordinate time and other coordinatedependent quantities, to make physical predictions. Proper time and all other invariants are finite at and below the horizon; the fact that coordinate time goes to infinity at the horizon is irrelevant, because the theory as it's been validated by experiment does not assign any physical meaning to coordinate time. By making coordinate time privileged for a particular scenario, black holes, you are changing the theory; the theory you are using is no longer GR, but "GR with a special patch for this situation". It's true that, since we have no direct experimental evidence in this situation, there is no way to experimentally distinguish GR from your "GR with a patch". But that doesn't mean your "GR with a patch" is the same theory as GR. It isn't. All it means is that there is no experimental test we currently know of that distinguishes your theory, "GR with a patch", from GR. 


#110
Dec612, 11:18 AM

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#111
Dec712, 01:23 PM

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#112
Dec712, 01:28 PM

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#113
Dec712, 01:48 PM

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#114
Dec712, 01:59 PM

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The form of the metric in the generalized SC is given by their eq 2. The coordinate time as a function of r for a radial freefalling object is given by eq 12. The proper time as a function of r is given by eq 18. They also give explicit expressions for the free function B for standard Schwarzschild coordinates, EddingtonFinkelstein coordinates, and also for PainleveGullstrand coordinates. 


#115
Dec712, 02:21 PM

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#116
Dec712, 03:04 PM

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PF Gold
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It then becomes necessary to posit a physically plausible origin, e.g. OS collapse, that really does remove the white hole region. 


#117
Dec712, 03:10 PM

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#118
Dec712, 03:40 PM

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I agree, it does not allow you to cover the maximally extended spacetime using their equations. In that sense it is not truly "generalized", but it is generalized enough to easily calculate the quantities of interest by harrylin using a wide variety of coordinates over regions I and II.



#119
Dec712, 10:52 PM

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As a quick check, I computed the explicit formula from equation 18 for the proper time to fall for a Lemaitre observer (who falls "from rest at infinity"), from radius r to the singularity at r = 0: [tex]\tau ( r ) = \frac{1}{\sqrt{2M}} r^{\frac{3}{2}}[/tex] This matches what is given in MTW, although they write it in normalized form, which actually looks neater: [tex]\frac{\tau}{2M} = \left( \frac{r}{2M} \right)^{\frac{3}{2}}[/tex] To get the proper time to the horizon, just subtract 2M from the RHS in the first formula, or 1 from the RHS in the second (to get [itex]\tau / 2M[/itex] to the horizon). 


#120
Dec712, 11:04 PM

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And just to add some quick numbers based on the formula in my last post (which isn't strictly correct for an object that starts from rest at finite r, but which will be *less* than the time for falling from rest any finite r, and the error gets smaller as r gets larger): if we plug in 2M for the Sun (about 3 km), and start from the radius of the actual Sun (about 700,000 km), we have:
[tex]\frac{r}{2M} = \frac{700000}{3} = 233333[/tex] [tex]\frac{\tau}{2M} = ( 233333 )^{\frac{3}{2}} = 112710467[/tex] Multiplying by 2M (i.e., 3), and converting from km to seconds by dividing by c (299792), we get 1128 seconds. The time to the horizon is only 10 microseconds smaller (since that's 3 km divided by c). Again, this is a lower bound (since a Lemaitre observer is moving inward at "escape velocity" at any finite r); the actual proper time to fall from rest at r = 233333M will be larger. If we run the same calculation for the million solar mass black hole at the center of the Milky Way, and start from the same value of r / 2M (which will equate to 233 billion km or about 1556 AU, about 13 times the distance to the Voyager spacecraft but still only about 10^5 light year, so very close by interstellar standards), the result for [itex]\tau / 2M[/itex] remains the same, and we just scale [itex]\tau[/itex] up by a factor of a million; so the time to the singularity would be 1128 million seconds or about 36.4 years, and the time to the horizon would be about 10 seconds shorter. Finally, for a billion solar mass black hole, such as the ones that are thought to be at the centers of quasars, if we start from the same value of r / 2M, we will be about 10^2 light year away when we start; it will take 36,400 years to fall to the singularity, and the time to the horizon will be about 2 hours 47 minutes shorter. Again, all of these times are lower bounds; I suspect the actual numbers for a fall from rest at finite r will be significantly higher even for such a high r / 2M. 


#121
Dec812, 07:57 AM

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@harrylin, were there other numbers you wanted? I am not sure what significance you will assign to them. 


#122
Dec812, 10:29 AM

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Let me go through the mathematics (different from the paper, but I think it's correct). If we consider only radial motion, then the path of a free particle obeys:
We don't actually need to solve the equations to know qualitatively what the solutions look like:
There's nothing at all surprising about these results, except for one thing: Nowhere in the equations does the initial value of [itex]r[/itex] come into play. Which means that there are solutions to the equations of motion in which a particle starts off below the event horizon, and then emerges from the event horizon, and either escapes to infinity or reaches a maximum height and plunges back into the event horizon. How do people exclude these possibilities? 


#123
Dec812, 10:34 AM

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If you look at these solutions, as I said above, in a chart that covers the exterior plus the black hole interior, you will see that the portion that "starts below the event horizon" is *not* covered; but you have to look at the actual coordinates to see this, not just the effective potential (which is basically what you're looking at). 


#124
Dec812, 02:23 PM

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The textbook treatments are all so dry that I skipped over them rather lightly. I suppose you'd want to read up on "time orientable manifolds" if you wanted the formal description of how to do this. (Wald would have this). Informally, lets start with assuming one knows how to construct light cones. Note that one has to be careful about this inside the event horizon if one is using Schwarzschild coordiantes! It's easy enough to determine the two lightlike geodesics that pass through a point, and draw the cone shape that light marks out. But if one draws a point P, one needs to realize that the Lorentz interval between P and P+dt is spacelike inside the event horizon. Which implies that the correct "shading" of the light cone to determine its "inside" region does not include the point P+dt inside the event horizon  given the convention that we "shade" the light cone so that the inside (shaded) region only contains timelike worldlines. Basically, we know that P+dr and Pdr are both timelike intervals inside the event horizons, so both of those are in the "shaded" region, and P+dt and Pdt are not in the shaded region. So, onece we've got the easy part done, shading the light cone correctly so that it only contains timelike worldlines, we still need to determine past vs future. As far as I know, the only way to do this is by convention, given that physics is time reversible. So you pick some external observer, and say that as the Schwarzschild t increases at large R, that that is the future. Then you need to splice all the light cones together in a consistent manner. This is the tricky part. There's really only two choices inside the horizon in Schwarzschild coordinates though  r increasing and r decreasing. It turns out that in the black hole region it's r decreasing, in the white holde region it's r increasing. Its probably easy to demonstrate this by using KS coordinates, where the light cones always point in the same direction , than it is to demonstrate in Schwarzschild coordinates (where they rotate). You'll probably need some nonsingluar coordinate system to convicingly handle the transition over the horizon in any event. 


#125
Dec812, 09:25 PM

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I just realized that I had left out a factor of 2/3 in the formulas I posted for proper time. The formulas should be
[tex]\tau ( r ) = \frac{2}{3 \sqrt{2M}} r^{\frac{3}{2}}[/tex] or in normalized form: [tex]\frac{\tau}{2M} = \frac{2}{3} \left( \frac{r}{2M} \right)^{\frac{3}{2}}[/tex] These are proper times to the singularity; to get proper times to the horizon, subtract 4/3 M from the first formula or 2/3 from the second. All of the times I posted should similarly be multiplied by 2/3, so the correct results are: M = 1 Sun Time from r = 1 solar radius = 233333 M to singularity: 752 s or about 12.5 minutes (7 microseconds shorter to horizon) M = 1 million Suns Time from r = 233333 M to singularity: 24.3 years (7 seconds shorter to horizon) M = 1 billion Suns Time from r = 233333 M to singularity: 24,300 years (1 hour 51 minutes shorter to horizon) 


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