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.
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
Even if it was so, the cylinder would exist for a while. It would not be crushed instantly.
But the tensile strength needed, is almost arbitrary small.
I guess I should have made my conclusions explicit:
Classically, the escape velocity is independent of the direction of emission, because the gravitational force is potential (unlike, say, magnetism or friction). In GR the situation is more complicated because of the potential capture by an event horizon.
Light always escapes, regardless of direction (assuming your cylinder is transparent), if there is no horizon close by. In other words, the only time a ray of light can be captured is when it dips under the event horizon. This is basically the definition of the event horizon.
The anisotropic light ray capture happens when light is emitted close to the horizon, such as some light rays are bent hard enough to go through the horizon.
Quite independently of all this, any attempt to create a cylinder such as you describe will fail because it will collapse onto itself before you can make it heavy enough.
Classically, the escape velocity is independent of the direction of emission, because the gravitational force is potential (unlike, say, magnetism or friction). In GR the situation is more complicated because of the potential capture by an event horizon.
In a real world, the escape speed from the system (Earth)-(any black hole) is heavily dependent from the direction you choose to escape. In the direction from Earth to the black hole, it is greater then c. While in the opposite direction it is only the well known 11+ km per second.
What bothers me further. For an observer, far away on the other side of the super massive black hole in the Center, we are sometimes behind the event horizon, sometimes we are not. True?
In a real world, the escape speed from the system (Earth)-(any black hole) is heavily dependent from the direction you choose to escape. In the direction from Earth to the black hole, it is greater then c. While in the opposite direction it is only the well known 11+ km per second.
That cannot be right. For example, in the Earth-Sun system the escape velocity from the Earth's surface is about 11.2km/s (to escape the Earth), but this only gets you to an orbit around the Sun. You need to accelerate to about 42.1 km/s to escape the solar system (neglecting the effects from other planets), regardless of the direction of travel.
For an observer, far away on the other side of the super massive black hole in the Center, we are sometimes behind the event horizon, sometimes we are not. True?
No, not true. Once you are behind the event horizon, you only have moments to live until you run into the singularity, and you can certainly never get out (barring FTL travel). I suspect that I misunderstand your setup, that's why we are having difficulties.
You need to accelerate to about 42.1 km/s to escape the solar system
Sure, but I said "the Earth". Never the less you may include the Sun, okay. It is 40+ km per second then, to escape the system: a black hole -- our planet, in one direction. You can't in the other, you will stumble into a black hole in other direction.
My point was, we have different escape velocities for different directions, from one point. Don't we?
No, it's the same velocity regardless of direction, because the escape velocity is determined by the potential energy, which is just a number for each point and is direction-independent.
From here, you can escape the system planet Earth-SupermassiveBlackHole in almost every direction easy. But not even the light will escape this system if it goes from here toward the SMBH.
From the same point, much different escape velocities, dependent of the escape direction.
What do I miss?
Just like classically light gets consumed by the ground if you aim it wrong, in GR light gets consumed by the black hole if it gets close enough to the horizon (1.5x the horizon radius for a non-rotating black hole). If you aim it better, it misses the black hole and escapes to infinity.
Yes. And a rock flown 1000 km per second will not escape in one direction, it will escape in other.
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 don't know about Thomas, but I sure as heck would prefer to discuss physics on LW (if possible, i.e. if appropriately knowledgeable people are present here) rather than an ordinary forum. Less inferential distance, more common assumptions (e.g. that we're not interested in guessing passwords), etc.
See also here (second bullet). (This post now joins the list of examples.)
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.
This reaction is one reason I have personally never made a post like that. While I find the topic fascinating, I can see how many others would not (and the quick downvote supports your point). It is also not directly relevant to the AGI research or to applied rationality. However, I conjecture that there is a significant overlap between the group of people who are interested in Physics and those interested in the subject matter more relevant to LW, so I'm happy to talk about my area of expertise as long as there is enough interest.
Additionally, Physics in particular and natural sciences in general teach us one essential trait for an aspiring rationalist: humility. What's out there and what is possible is so much richer than our imagination, which is a good thing to keep in mind when considering deep and hard problems, like the FAI.
I am happy to see math and physics puzzles and curiosities on Discussion, if the writeup is good enough. (In this case, it is not.) I am very unhappy to see these "are you smart enough?" gotcha posts that are Thomas' speciality. And calling you out like this was just creepy.
I am very unhappy to see these "are you smart enough?" gotcha posts
Huh, I did not get that vibe at all.
And calling you out like this was just creepy.
To be fair, he asked me privately first if this is OK.
As a discussion post, I class it as noise.
I class it as somewhere between noise and actual negative signal.
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.
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.
I find this hard to believe, actually. Granted, I have not done the calculation, but in my mind the gravitational potential profile is such that the ends are always at a higher potential than the middle, so the classical escape velocity should be less from the ends. Unless the cylinder is of non-uniform density, which would be a totally different problem.
Oh, and classically the escape velocity, which is just sqrt(2*potential), is independent of the initial direction, provided the trajectory does not hit the ground.
For a cylinder of zero thickness, you are correct. The trick is to have nonzero thickness. This doesn't (to first order) affect the gravitational attraction at the endpoints, but it limits how close you can get to the axis. Even if you allow a particle to enter the cylinder, it will then no longer feel the attraction from the outer parts. It therefore seems to me that you can choose a density and radius such that there is never an event horizon as you go perpendicularly in, but a length such that there is one at the endpoints.
I have no answer for the argument that the escape velocity is independent of direction; but is it possible that this result was derived for a sphere? At any rate it may be that, if I got around to playing around with numbers, I would find that there was no solution as I described above.
For a cylinder of zero thickness, you are correct. The trick is to have nonzero thickness.
Imagine a long cylinder cut into segments. Instead of moving along the line, you merely have to take the segment at one end and move it over to the other end. Because of the superposition principle, the change in potential is merely the change cause by moving the segment from one end to the other.
Now, when does moving the segment increase the potential, and when does moving the segment decrease the potential? Can we identify a maximum?
when does moving the segment increase the potential, and when does moving the segment decrease the potential?
That's a good approach. If the segment is closer to the point of interest after it is moved, the potential well is deeper. Which tells you that it is deepest at the center.
I sit corrected; my argument that as you enter the cylinder you can ignore the outer parts breaks down.
is it possible that this result was derived for a sphere?
This is quite general for isolated potential systems. Total energy of a particle: E=KE+PE. E=0 at infinity. KE=1/2mv^2, PE=mP where P is the gravitational potential (volume integral of -G*rho/r dV). Escape velocity corresponds to E=0, so we get v=sqrt(-2P), regardless of anything else.
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
It is NOT a black hole at all in one direction, it is in the other.
The same for pragmatist. THAT was the question.
And that was my point: you can't have that scenario. If you have an event horizon, it's closed. There is no 'other direction'.
-dentin
There's no requirement for symmetry in regard to event horizons (assuming that by 'symmetric', you mean 'spherical'.)
Surprisingly, in 3 spatial dimensions there are very few possible stationary configurations, and they are all quite symmetric, as I mentioned in my other reply. Once you consider extra dimensions, however, the space of possible stationary black hole configurations gets pretty large in a hurry.
I was primarily referring to rotation, which pretty much all 'real' black holes would have, where as I understand it the event horizon is distorted from spherical. I can't really think of any other configurations like that, but I don't know relativity that well either.
-dentin
the event horizon is distorted from spherical
I suppose we use the term "spherical" in different ways. I mean it in the topological sense, to say "shaped like a ball, or a cube or a pear", as opposed to something with holes, like a donut. I guess you take it to mean "a perfect sphere", where it looks exactly the same from all directions (has spherical symmetry). In that sense, yes, there are just two kinds of stationary of black holes, the Schwarzschild one (non-rotating) and the Kerr one (rotating). There are no other kinds in 3D. However, in 4D there are also toroidal black holes, string-like and some irregularly shaped ones, too.
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.
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
Even if it was so, the cylinder would exist for a while. It would not be crushed instantly.
But the tensile strength needed, is almost arbitrary small.
As mister shminux mentioned somewhere, he is happy and qualified to answer questions in the field of the Relativity. Here is mine:
A long rod (a cylinder) could have a large escape velocity in the direction of its main axe. 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 this rod then an asymmetric black hole?