Are Finkelstein/Kruskal "interior black hole solution" compatible with Einstein's GR?

by harrylin
Tags: eep, finkelstein, kruskal, rindler coordinates
Physics
PF Gold
P: 5,517
 Quote by harrylin To my knowledge Einstein's GR as I defined here is our best current theory that hasn't been invalidated by experiment.
As far as I can tell, you are defining "Einstein's GR" in such a way that your claim that nothing can ever actually reach a BH horizon is part of the theory. That means what you are calling "Einstein's GR" is *not* the best current theory that hasn't been invalidated by experiment.

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 coordinate-dependent 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.
Mentor
P: 16,484
 Quote by harrylin I will insist on discussing numbers - as I also tried (but without insisting on it) in this thread.
What numbers are you interested in?
P: 3,178
 Quote by PeterDonis When you make the claim that "Einstein's GR says that nothing can ever reach the horizon", [..]
That's surprising as I'm not aware of having made such a claim; however I asked questions on that topic (I checked the quoted part with Google, but only found Peter's remark here).
 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 [...] your "GR with a patch" [..]
To my knowledge it is Einstein theory as formulated by him that has been put to the test, and that without any patch; but that is a different topic, not belonging to this discussion. Note also that, obviously, his theory is fully his and certainly not mine.
 Quote by DaleSpam What numbers are you interested in?
A simple example of a rocket with a clock in the front and in the back that is falling into a black hole with full description incl. distant time t1 according to Schwarzschild and Finkelstein (r,τ,t,t1) would probably be interesting for many people; I supposed that such examples are available in the literature, but perhaps that isn't the case. So, that's for a next discussion.
P: 3,178
 Quote by PeterDonis [..] what is it that applies to the coordinates of Adam but not Eve, *and* to those of Eve' but not Adam'? I haven't seen an answer yet.
As the discussion continues in the other thread I replied there although I don't suppose to have all the answers; I'm among those who ask questions about black holes. Anyway, thanks for your participation.
Physics
PF Gold
P: 5,517
 Quote by harrylin To my knowledge it is Einstein theory as formulated by him that has been put to the test
Actually it is the theory of GR as originally formulated by Einstein and refined by physicists for almost a century now, that has been put to the test. The Einstein Field Equation, which is what was originally published by Einstein, is unchanged, yes, but Einstein obviously did not know a lot of things about the consequences of the EFE that we know today, and some of the things he apparently believed about those consequences have turned out not to be true. [Edit: perhaps "solutions and their properties" would be a better word than "consequences".]

 Quote by harrylin A simple example of a rocket with a clock in the front and in the back that is falling into a black hole with full description incl. distant time t1 according to Schwarzschild and Finkelstein (r,τ,t,t1) would probably be interesting for many people; I supposed that such examples are available in the literature, but perhaps that isn't the case. So, that's for a next discussion.
I'll await another thread on this specific topic.
Mentor
P: 16,484
 Quote by harrylin A simple example of a rocket with a clock in the front and in the back that is falling into a black hole with full description incl. distant time t1 according to Schwarzschild and Finkelstein (r,τ,t,t1) would probably be interesting for many people; I supposed that such examples are available in the literature, but perhaps that isn't the case. So, that's for a next discussion.
The easiest way I know of for this is to use the generalized Schwarzschild coordinates as presented here: http://arxiv.org/abs/gr-qc/0311038

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 free-falling 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, Eddington-Finkelstein coordinates, and also for Painleve-Gullstrand coordinates.
P: 3,178
 Quote by DaleSpam The easiest way I know of for this is to use the generalized Schwarzschild coordinates as presented here: http://arxiv.org/abs/gr-qc/0311038 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 free-falling 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, Eddington-Finkelstein coordinates, and also for Painleve-Gullstrand coordinates.
Nice - that's constructive. Thanks.
PF Gold
P: 4,862
 Quote by DaleSpam The easiest way I know of for this is to use the generalized Schwarzschild coordinates as presented here: http://arxiv.org/abs/gr-qc/0311038 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 free-falling 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, Eddington-Finkelstein coordinates, and also for Painleve-Gullstrand coordinates.
One observation about this paper is the authors suggest you can 'hide' the white hole issue by using this family of coordinates, and avoiding the corresponding Kruskal family. Not really, IMO, because you can easily show there exist timelike paths beginning and ending on the SC radius, encompassing finite proper time (in standard SC coordinates, the beginning and end t coordinates would be -∞ and +∞, despite finite clock time along the path). The existence of such a timelike path leads immediately to the question of what happened before the beginning of the total path of finite proper time. This leads directly into the white hole region.

It then becomes necessary to posit a physically plausible origin, e.g. O-S collapse, that really does remove the white hole region.
Physics
PF Gold
P: 5,517
 Quote by PAllen One observation about this paper is the authors suggest you can 'hide' the white hole issue by using this family of coordinates, and avoiding the corresponding Kruskal family. Not really, IMO
I agree, and on a quick reading the easiest way to show this would be to construct a similar generalized coordinate chart that, instead of covering regions I and II (exterior and black hole interior) would cover regions IV and I (white hole interior and exterior). I think that can be done just by changing the sign of the du dr term in their generalized line element.
 Mentor P: 16,484 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.
Physics
PF Gold
P: 5,517
 Quote by DaleSpam The proper time as a function of r is given by eq 18.
A key thing to note about this equation is that, when you combine it with equation 12 (since the first term on the RHS of equation 18 is the coordinate time u(r), which is given by equation 12), B cancels out. In other words, the proper time for a radially infalling geodesic, as a function of r, is *independent* of B. That means it's the *same* for *all* of the charts that are included in the family described by this generalized line element.

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:

$$\tau ( r ) = \frac{1}{\sqrt{2M}} r^{\frac{3}{2}}$$

This matches what is given in MTW, although they write it in normalized form, which actually looks neater:

$$\frac{\tau}{2M} = \left( \frac{r}{2M} \right)^{\frac{3}{2}}$$

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 $\tau / 2M$ to the horizon).
 Physics Sci Advisor PF Gold P: 5,517 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: $$\frac{r}{2M} = \frac{700000}{3} = 233333$$ $$\frac{\tau}{2M} = ( 233333 )^{\frac{3}{2}} = 112710467$$ 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 $\tau / 2M$ remains the same, and we just scale $\tau$ 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.
Mentor
P: 16,484
 Quote by PeterDonis A key thing to note about this equation is that, when you combine it with equation 12 (since the first term on the RHS of equation 18 is the coordinate time u(r), which is given by equation 12), B cancels out. In other words, the proper time for a radially infalling geodesic, as a function of r, is *independent* of B. That means it's the *same* for *all* of the charts that are included in the family described by this generalized line element.
That is good to know. Of course, it is exactly what you would expect, but still, the confirmation is good.

@harrylin, were there other numbers you wanted? I am not sure what significance you will assign to them.
P: 1,657
 Quote by PAllen One observation about this paper is the authors suggest you can 'hide' the white hole issue by using this family of coordinates, and avoiding the corresponding Kruskal family.
Perhaps I should start a separate topic for White Holes, but I really don't understand why there is a white hole region.

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:
• $\dfrac{dt}{d\tau} =Q^{-1} K$
• $Q c^2(\dfrac{dt}{d\tau})^2 - Q^{-1} (\dfrac{dr}{d\tau})^2 = c^2$
where $Q = 1 - \dfrac{2GM}{c^2 r}$, and where $K$ is a constant of the motion. Plugging the first equation into the second gives:
• $(cK)^2 - (\dfrac{dr}{d\tau})^2 = Q c^2= c^2-\dfrac{2GM}{r}$
which can be rearranged to look like a problem in Newtonian physics:
• $E = \dfrac{1}{2} m v^2 - \dfrac{GMm}{r}$
where m is the mass of the particle, and where $E = \dfrac{1}{2} mc^2(K^2 - 1)$, and where $v = \dfrac{dr}{d\tau}$

We don't actually need to solve the equations to know qualitatively what the solutions look like:
1. If $E>0$, and $v > 0$ then the particle will escape from the black hole out to infinity.
2. If $v < 0$ then regardless of the sign of $E$ the particle will in a finite amount of proper time reach the singularity.
3. If $E<0$, and $v > 0$ then the particle will rise temporarily, reach a maximum height, turn around and fall toward the black hole, reaching the singularity in a finite amount of time.

There's nothing at all surprising about these results, except for one thing: Nowhere in the equations does the initial value of $r$ 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?
Physics
PF Gold
P: 5,517
 Quote by stevendaryl [*]If $E<0$, and $v > 0$ then the particle will rise temporarily, reach a maximum height, turn around and fall toward the black hole, reaching the singularity in a finite amount of time.
Work this possibility backwards; you will see that the particle's geodesic, in a chart that covers the exterior and the black hole interior, ends at a finite proper time in the past, at a point where all physical invariants are finite. So where did it come from?

 Quote by stevendaryl Which means that there are solutions to the equations of motion in which a particle starts off below the event horizon
But the event horizon you already know about, which is more precisely called the *future* horizon, is an outgoing null surface; nothing can escape from it. So these solutions that start off "below the event horizon" can't be starting off below that horizon; they must be starting off below *another* horizon, the *past* horizon, which is an *ingoing* null surface, so particles can escape but no particle can go from the outside back in.

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).

 Quote by stevendaryl How do people exclude these possibilities?
They aren't excluded; they're the possibilities that answer your question, by telling us that there must be a white hole region in the maximally extended spacetime.
Emeritus
P: 7,443
 Quote by stevendaryl Perhaps I should start a separate topic for White Holes, but I really don't understand why there is a white hole region.
How do we tell the past from the future?

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 light-like 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 P-dr are both timelike intervals inside the event horizons, so both of those are in the "shaded" region, and P+dt and P-dt 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 non-singluar coordinate system to convicingly handle the transition over the horizon in any event.
 Physics Sci Advisor PF Gold P: 5,517 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 $$\tau ( r ) = \frac{2}{3 \sqrt{2M}} r^{\frac{3}{2}}$$ or in normalized form: $$\frac{\tau}{2M} = \frac{2}{3} \left( \frac{r}{2M} \right)^{\frac{3}{2}}$$ 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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