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.
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.
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 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.
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?
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.
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.