11 ms·

Not quite -- in your reference frame, you cross the blackhole's event horizon and go inside, but an observer (who is outside the event horizon) will only see yo

by xtacy 11y ago

Not quite -- in your reference frame, you cross the blackhole's event horizon and go inside, but an observer (who is outside the event horizon) will only see you reach the event horizon and slowly "fade away."

nonbel 11y ago
Thanks, so what is the deal with the four-velocity always equaling c^2? Ie (maybe it is totally wrong): c^2 = (v_t)^2 - (v_x)^2 - (v_y)^2 - (v_z)^2 where v_i indicates speed along that dimension. I thought this was why all the weird time dilation effects occur, velocity along time v_t ->0 as velocity through space increases. You seem to be saying that from the traveler's perspective they are not accelerating faster and faster towards the black hole. As I said, maybe my understanding is just totally incorrect here. Edit: Actually, I was really confused (this is all based on looking it up the other day). According to the above, v_t would need to increase as eg v_x increases. This is just wrong, but I'll leave it here in case someone wants to explain it correctly.
- raattgift 11y ago
  This is asking an essential Special Relativity (SR) question into a conversation about an object where gravitation is not just relevant but central, so I'll also give you an a modern Special Relativity answer cast in such a way to follow with a couple of comments about "c" in General Relativity. Below you'll note that I'll treat GR strictly as geometry. Your equation is some algebraic mangling of the Minkowski line-element, which we can write as ds^2 = -cdt^2 + dx^2 + dy^2 + dz^2 in Cartesian coordinates with a -+++ metric signature. That's the interval in flat spacetime (aka Minkowski spacetime, although in both cases sometimes "time" is omitted when a reader will not be confused). The conversion constant "c" there is the sole free parameter of the Poincare group, which is the local isometry group of Minkowski spacetime (i.e., it applies each point in spacetime), which includes translations, Lorentz boosts and rotations on the three spacelike axes (x, y, and z in the form above) and a unidirectional translation on the timelike axis (t). The parameter corresponds (in SR) to the speed of a massless particle, and light is assumed to be massless. It's a postulate of Special Relativity that all non-gravitational physics is invariant under the Poincare group, and that's baked into the Standard Model, for instance. However, in GR "c" corresponds to the surface of a nonempty, convex, open cone of tangent vectors at some point p on the (curved) manifold; light at point p is constrained to that surface exactly, and massive particles at point p stay inside that boundary. In GR we can only meaningfully compare speed and velocity locally, where that means either in the limit as spacetime intervals go to zero, or equivalently in the local section of the fibre bundle, because the very definition of spacetime curvature means the parallel transport of one vector to another for comparison purposes is path-dependent, and that applies on all four axes, so we cannot even meaningfully compare two clocks (with which we might measure relative velocity) unless they are at the same point in spacetime. However, we can work things out so that a small (but not exactly zero-sized) region of spacetime is treated as flat, and then use Special Relativity or even Newtonian mechanics to discuss the matter content in that region, right down to comparing velocities. This "flattening" can be done quite successfully in a number of ways, and can even be done in a region with significant curvature by using the formalisms of semiclassical gravity. The cost is in artifacts introduced into the non-gravitational content of the region of non-negligbly curved spacetime, most noticeably in terms of differences in particle count and even the interpretation of matter field excitations as particles. This mathematical "flattening away" of the curvature returns us from a causal cone built on a hyperbolization of a series of first-order quasilinear partial differential equations (the Einstein Field Equations) to the Poincare group applying at every point, and thus we return to "c" as being mathematically special, rather than relating to one of perhaps many causal cones (since one can have many hyperbolizations, and thus many timelike-spacelike conversion constants although so far there is only evidence for one).
ars 11y ago
Which is why black holes makes no sense. Nothing can ever fall into them, so they can't ever form, or grow. (Obviously speaking from the POV of Earth.) Since they can't form, and can't grow, how can you see a "supermassive one"?
- Natanael_L 11y ago
  Technically it is only the singularity at the center which doesn't grow when seen from a frame of reference on the outside. But things will still fall in below the event horizon when seen from the outside. A black hole forms when the density (of energy or matter) at some point adds up to a gravity that light can't escape. Colliding two neutron stars would likely instantly cause a high enough density in a volume such that a black hole and its event horizon can form, which would envelop a very large mass in much less than a second.
- lisivka 11y ago
  Black hole has zero gravity at it center. ;-) Moreover, gravity can stop light and other electromagnetic forces, but cannot stop gravity, so time is not stopped.
  - lisivka 11y ago
    WTF? I have a follower-downvoter? Every body has zero gravity forces at their center of mass. We also can clearly see that mass continue to generate gravity forces even after fall to black hole, so time is not stopped, at least for gravity. PS. Every black hole has second event horizon in the center of black hole, and I am very curios is it filled with mass or not. :-/

by xtacy 11y ago

nonbel 11y ago

Thanks, so what is the deal with the four-velocity always equaling c^2? Ie (maybe it is totally wrong): c^2 = (v_t)^2 - (v_x)^2 - (v_y)^2 - (v_z)^2 where v_i indicates speed along that dimension. I thought this was why all the weird time dilation effects occur, velocity along time v_t ->0 as velocity through space increases. You seem to be saying that from the traveler's perspective they are not accelerating faster and faster towards the black hole. As I said, maybe my understanding is just totally incorrect here. Edit: Actually, I was really confused (this is all based on looking it up the other day). According to the above, v_t would need to increase as eg v_x increases. This is just wrong, but I'll leave it here in case someone wants to explain it correctly.

raattgift 11y ago
This is asking an essential Special Relativity (SR) question into a conversation about an object where gravitation is not just relevant but central, so I'll also give you an a modern Special Relativity answer cast in such a way to follow with a couple of comments about "c" in General Relativity. Below you'll note that I'll treat GR strictly as geometry. Your equation is some algebraic mangling of the Minkowski line-element, which we can write as ds^2 = -cdt^2 + dx^2 + dy^2 + dz^2 in Cartesian coordinates with a -+++ metric signature. That's the interval in flat spacetime (aka Minkowski spacetime, although in both cases sometimes "time" is omitted when a reader will not be confused). The conversion constant "c" there is the sole free parameter of the Poincare group, which is the local isometry group of Minkowski spacetime (i.e., it applies each point in spacetime), which includes translations, Lorentz boosts and rotations on the three spacelike axes (x, y, and z in the form above) and a unidirectional translation on the timelike axis (t). The parameter corresponds (in SR) to the speed of a massless particle, and light is assumed to be massless. It's a postulate of Special Relativity that all non-gravitational physics is invariant under the Poincare group, and that's baked into the Standard Model, for instance. However, in GR "c" corresponds to the surface of a nonempty, convex, open cone of tangent vectors at some point p on the (curved) manifold; light at point p is constrained to that surface exactly, and massive particles at point p stay inside that boundary. In GR we can only meaningfully compare speed and velocity locally, where that means either in the limit as spacetime intervals go to zero, or equivalently in the local section of the fibre bundle, because the very definition of spacetime curvature means the parallel transport of one vector to another for comparison purposes is path-dependent, and that applies on all four axes, so we cannot even meaningfully compare two clocks (with which we might measure relative velocity) unless they are at the same point in spacetime. However, we can work things out so that a small (but not exactly zero-sized) region of spacetime is treated as flat, and then use Special Relativity or even Newtonian mechanics to discuss the matter content in that region, right down to comparing velocities. This "flattening" can be done quite successfully in a number of ways, and can even be done in a region with significant curvature by using the formalisms of semiclassical gravity. The cost is in artifacts introduced into the non-gravitational content of the region of non-negligbly curved spacetime, most noticeably in terms of differences in particle count and even the interpretation of matter field excitations as particles. This mathematical "flattening away" of the curvature returns us from a causal cone built on a hyperbolization of a series of first-order quasilinear partial differential equations (the Einstein Field Equations) to the Poincare group applying at every point, and thus we return to "c" as being mathematically special, rather than relating to one of perhaps many causal cones (since one can have many hyperbolizations, and thus many timelike-spacelike conversion constants although so far there is only evidence for one).

ars 11y ago

Which is why black holes makes no sense. Nothing can ever fall into them, so they can't ever form, or grow. (Obviously speaking from the POV of Earth.) Since they can't form, and can't grow, how can you see a "supermassive one"?

Natanael_L 11y ago
Technically it is only the singularity at the center which doesn't grow when seen from a frame of reference on the outside. But things will still fall in below the event horizon when seen from the outside. A black hole forms when the density (of energy or matter) at some point adds up to a gravity that light can't escape. Colliding two neutron stars would likely instantly cause a high enough density in a volume such that a black hole and its event horizon can form, which would envelop a very large mass in much less than a second.
lisivka 11y ago
Black hole has zero gravity at it center. ;-) Moreover, gravity can stop light and other electromagnetic forces, but cannot stop gravity, so time is not stopped.
- lisivka 11y ago
  WTF? I have a follower-downvoter? Every body has zero gravity forces at their center of mass. We also can clearly see that mass continue to generate gravity forces even after fall to black hole, so time is not stopped, at least for gravity. PS. Every black hole has second event horizon in the center of black hole, and I am very curios is it filled with mass or not. :-/