Are Finkelstein/Kruskal Black Hole Solutions Compatible with Einstein's GR?

[Dale]

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.

 

[stevendaryl]

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?

 

[PeterDonis]

[*]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?

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).

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.

 

[pervect]

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, let's 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

\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)