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.
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.
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.
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.
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?
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.
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.
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.