Haelfix said:
2) The white hole horizon is conceptually really bizarre...
Since nothing is allowed to get in, that means that 'test' particles traveling in orbits around the white hole horizon (more precisely the particle horizon) will accumulate there, and there will be a severe blue shift when viewed from infinity. This blue sheet is a sort of classical instability, and it is argued that it leads to gravitational collapse, and thus there is likely a singularity in the future as well!
Exactly what the white hole is is a little mysterious to me. It seems that there is a sense in which there is no difference in the spacetime geometry of a black hole and a white hole; the difference is simply initial conditions of the test particles traveling in that geometry.
Let me explain why I think that.
To simplify, let's talk about purely radial motion, so we can treat the Schwarzschild geometry as if there were only one spatial dimension. Let Q be the Schwarzschild factor defined by: Q \equiv 1 - \frac{2GM}{c^2 r}. Let \tau be proper time. Let U^\mu \equiv \frac{\partial x^\mu}{\partial \tau}. Then for a test particle of mass m moving along a radial timelike geodesic, we have the following conserved quantities:
- K \equiv m c Q U^t. This is sort of the "momentum" in the t-direction.
- H \equiv \frac{m c^2}{2} Q (U^t)^2 - \frac{m (U^r)^2}{2Q}. This is actually \frac{mc^2}{2} \frac{ds^2}{d\tau^2}, so it's just equal to \frac{mc^2}{2}.
Putting these together gives an equation for U^r:
\frac{m}{2} (U^r)^2 - \frac{GMm}{r} = \mathcal{E}
where \mathcal{E} = \frac{K^2}{2 mc^2} - \frac{mc^2}{2}
I wrote it in this way so that you can immediately see that it's just the energy equation for a test particle moving under Newtonian gravity. So without any mathematics, we can immediately guess the qualitative behavior:
If \mathcal{E} < 0, and initially, U^r > 0, then the test particle will rise to some maximum height: r_{max} = \frac{GMm}{|\mathcal{E}|}, and then will fall back to annihilation at r=0 in a finite amount of (proper) time. The interesting case is r_{max} > \frac{2GM}{c^2} \equiv r_S, where r_S is the black hole's Schwarzschild radius. In that case, this scenario represents a particle rising from below the event horizon and then turning around and falling back through the event horizon.
That seems to contradict the fact that nothing can escape from the event horizon, but to see why it doesn't, you have to see what the time coordinate t is doing: In the time period between the particle rising out of the event horizon and falling back into the event horizon, only a finite amount of proper time passes, but an infinite amount of coordinate time passes. In the far past, t \rightarrow -\infty, the particle arises from the event horizon, and in the far future, t \rightarrow +\infty, the particle sinks below the event horizon. The time period while the particle is rising up to the event horizon, and the time period while the particle is falling below the event horizon is not covered by the coordinate t (well, you can still have a t coordinate there, but its connection to the t coordinate above the horizon is broken by the event horizon). So from the point of view of someone far from the black hole, using the t coordinate for time, nothing ever crosses the event horizon (in either direction) for any finite value of t.
Going back to the test particle, we can identify the various parts of the Schwarzschild geometry:
- During the time that the particle is rising below the event horizon, the particle is traveling through Region IV, the white hole interior.
- During the time that the particle is above the event horizon, the particle is traveling through Region I, the black hole exterior.
- During the time that the particle is falling below the event horizon, the particle is traveling through Region II, the black hole interior.
(A fourth region, Region III, is not visited by the test particle, but is a black hole exterior like Region I).
The point is that nothing about the local geometry of spacetime changes in going from Region IV (the white hole interior) to Region II (the black hole interior). The only difference is the sign of \frac{dr}{d\tau}. So the difference between a black hole and a white hole is simply the initial conditions of the test particle. So it's not that the particle is repelled by the white hole and is attracted by the black hole. It's true by definition that:
- If the test particle is below the event horizon and \frac{dr}{d\tau} > 0, then the particle is in the black hole interior.
- If the test particle is below the event horizon and \frac{dr}{d\tau} < 0, then the particle is in the white hole interior.
As for the exterior, the same region, Region I, serves as the exterior of the white hole and the black hole. The same event horizon looks like a white hole in the far past t \rightarrow -\infty, because the test particle is rising from it, and looks like a black hole in the far future t \rightarrow +\infty, because the test particle is falling toward it. (For a realistic black hole formed from the collapsed of a star, there is no event horizon in the limit t \rightarrow -\infty, so there is no corresponding white hole.)
Here are some puzzles having to do with the test particles:
- In the case of many one test particles instead of just one, do all the particles have the same sign of \frac{dr}{d\tau}? They are all falling in the black hole interior, and all rising in the white hole interior. Why aren't there some particles that are rising, while other particles are falling? It turns out that there is a simple answer to this question. If you have \frac{dr}{d\tau} > 0, you can make it \frac{dr}{d\tau} < 0 by reparametrizing: \tau \rightarrow -\tau. So it's possible to arrange it so that all particles have the same sign of \frac{dr}{d\tau}.
- A followup to the first puzzle: If you just arbitrarily flip the sign of \tau for a test particle, it makes no difference, since they have no internal state. But if instead you don't have a test particle, but a physical object, such as a clock or a human being, then flipping the sign of proper time means reversing the usual progression of states. The clock will start running backwards, and the human will start getting younger, instead of older. That's not technically a contradiction, because the laws of physics are reversible, so it's possible for a human to age backwards. But it's a violation of the law of increasing entropy. So if it happens to be the case (and it sure seems to be) that all processes in the universe have the same thermodynamic arrow of time--entropy increases as proper time increases--then puzzle number 1 switches to the question of why is there a universal thermodynamic arrow of time? This boils down to the question: Why was entropy lower in the far past? General Relativity doesn't answer this question. (I'm not sure what does
- Another complication is to include test particles that don't move on geodesics, because of non-gravitational forces. How does that affect the picture?
