Questions for shminux

Since the question is explicitly for me...

General Relativity, like Quantum Mechanics, tends to get weird and counter-intuitive. This is probably one of those cases. The answer is a somewhat disappointing "yes and no".

Specifically, "yes": since the required escape velocity depends on the direction of escape, in general there can be configurations where light emitted in some directions ends up back in the "center" (not a great term, but should do for now), while the light emitted in other directions can escape. This is true even for the regular Schwarzschild back hole. For example, close to the event horizon, you have to shine your flash light nearly straight outward for it to escape, otherwise it bends around and goes in. There are some nice ray tracing simulations of this effect online, and I recall doing one of those as a part of my Masters thesis.

Which brings me to the "no" part: if some light can escape in some directions, its source is not inside the event horizon. Though an event horizon is certainly required in order for some light to not be able to escape (unless, of course, you shine your light directly into the ground, which is not very interesting).

To summarize, in the first approximation your proposed "asymmetric black hole" is actually your garden variety spherical black hole. The story gets more complicated if you want it to have a preferred axis along which escape is harder than along other directions. For example, a rotating (Kerr) black hole does provide the desired effect, because light emitted in the equatorial direction (and slightly tipped in the direction of rotation) can escape, while the light emitted along the axis from the same source might not. This is a sort of "slingshot effect", assisted by the frame dragging of the black hole.

Again, this is not the end of the question, if you dig a bit. For example, why can't a non-rotating black hole form from a large cylinder in a way that there is a preferred axial direction? Were I to ask this on a quiz, the answer I would expect would be the no-hair theorem, which guarantees that all non-rotating and uncharged stationary black holes of the same mass are identical (and so the black hole left by such a cylinder would be identical to a spherical black hole left by a collapsing spherical cloud of dust of the same mass) . The qualifier "stationary" means that the theorem applies to the black hole remnant, i.e. what remains after the dust settles, so to speak.

This is perfectly correct, but not very illuminating, because it does not answer the question "why?". To see how this happens, consider such a cylinder as it gets heavy enough to appreciably bend the spacetime. The parts of it along the axis farther away from the center will feel "more gravity" and so will compress the rest of the cylinder to a more spherical shape. It requires a bit of calculation, but it is possible to show that relativity poses a limit to the tensile strength of any material just enough to prevent such a cylinder from supporting itself, once it bends spacetime enough to prevent some outward-going light from escaping. This is formally known as the Dominant energy condition and is, in essence, due to the requirement of locality: no interactions between the constituents of the cylinder in question can locally propagate faster than light.

I realize that this is a lot to digest, and I might have missed your point entirely, so please feel free to comment.

I'm not sure why Less Wrong is the appropriate place for these kind of questions instead of, say, Reddit's Ask Science. You'd probably get half a dozen physicists jumping over themselves to answer your question over there, no?

I think this would be better as a reply to wherever shiminux said that he was happy and qualified to answer such questions, or as a PM to him. As a discussion post, I class it as noise.

The rod as you describe it could not exist as a stable object, unless I'm missing something. Why would it not collapse under its own gravity?

There are other examples of black holes that lack spherical symmetry. For instance, rotating black holes, described by the Kerr metric, are axially symmetric but not spherically symmetric. Spherical symmetry is not generic; it is a special case.

A long rod (a cylinder) could have a large escape velocity in the direction of its main axis. From its end, to the "infinity". Larger than the speed of light. While the perpendicular escape velocity is lesser than the speed of light.

Is it obvious that this is in fact possible?

That said, there is no requirement for event horizons to be spherically symmetric; that's just the simplest solution.

Edit to add: The question bugged me, so I played around a bit with the classical integrals. I'm provisionally convinced that, for a cylinder of nonzero thickness, you can indeed find some combination of density, length, and thickness such that the escape velocity is greater than c longitudinally from the endcap, but not perpendicularly from the midpoint.

Leaving aside the question of space-time curvature at various places near/around the rod, there is an obvious answer:

If an event horizon forms around some part of the rod, say near one of the ends, that part of the rod is no longer connected to the remainder, and the rod now exists entirely outside of the black hole.

As for whether or not a black hole is symmetric, the answer (in the real universe) is almost certainly no. There's no requirement for symmetry in regard to event horizons (assuming that by 'symmetric', you mean 'spherical'.)

-dentin