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

PeterDonis

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.

DaleSpam

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.

stevendaryl

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:

• [itex]\dfrac{dt}{d\tau} =Q^{-1} K[/itex]
• [itex]Q c^2(\dfrac{dt}{d\tau})^2 - Q^{-1} (\dfrac{dr}{d\tau})^2 = c^2[/itex]

where [itex]Q = 1 - \dfrac{2GM}{c^2 r}[/itex], and where [itex]K[/itex] is a constant of the motion. Plugging the first equation into the second gives:

• [itex](cK)^2 - (\dfrac{dr}{d\tau})^2 = Q c^2= c^2-\dfrac{2GM}{r}[/itex]

which can be rearranged to look like a problem in Newtonian physics:

• [itex]E = \dfrac{1}{2} m v^2 - \dfrac{GMm}{r}[/itex]

where m is the mass of the particle, and where [itex]E = \dfrac{1}{2} mc^2(K^2 - 1)[/itex], and where [itex]v = \dfrac{dr}{d\tau}[/itex]

We don't actually need to solve the equations to know qualitatively what the solutions look like:

1. If [itex]E>0[/itex], and [itex]v > 0[/itex] then the particle will escape from the black hole out to infinity.
2. If [itex]v < 0[/itex] then regardless of the sign of [itex]E[/itex] the particle will in a finite amount of proper time reach the singularity.
3. If [itex]E<0[/itex], and [itex]v > 0[/itex] 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 [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?

PeterDonis

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?

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).

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.

pervect

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.

PeterDonis

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)