3 ms·

> while a stone thrown up from below the event horizon would cross the event horizon just fine on the way up and on the way down and would return back down succ

> while a stone thrown up from below the event horizon would cross the event horizon just fine on the way up and on the way down and would return back down successfully in its proper time This doesn't agree with my understanding of GR. A stone "thrown up" from within a black hole interior cannot cross the event horizon in any reference frame - it cannot even get closer to it. Look at the future light cones within the black hole interior, e.g. in the illustration at http://en.wikipedia.org/wiki/Eddington–Finkelstein_coordinates http://en.wikipedia.org/wiki/Eddington–Finkelstein_coordinat.... The future light cone of every event within the EH is skewed so far that even light rays directed outwards are drawn closer to the singularity.

VladRussian2 13y ago
the event horizon - Schwarzschild radius - is defined by the specific value of gravitational potential (escape speed equals speed of light) while, given the black hole mass large enough, the gravitational force at the Schwarzschild radius can be as small as we'd like it to be. The various local effects like time dilation, light path curving, etc... are defined by the value of gravitation force, not gravitational potential. The potential defines the fact that anything originating at or below the horizon would never escape completely the gravitational field, i.e. never reach the infinity. The Schwarzschild radius of the mass of the observable Universe is 10B light years. So, several billions years ago, when observable Universe had 10B radius, it would be a black hole (though i think that in less expanded space of the earlier Universe the constants like "c" had different values and thus that Schwarzschild radius was less). We can imagine it in another way - increase 125 times (the observable Universe has 40+B light years radius) the amount of matter, ie. galaxies, stars, etc... inside the 10B radius ball around us, and you'd get the black hole with 10B light years Schwarzschild radius (and i don't think we would ever notice the change - only with time the galaxies's movement will be affected). Obviously, the gravitational field on the surface of that imaginary 10B radius sphere and inside it would increase somewhat (like 125 times on the surface, 125 times 0 pretty much 0) - not even close though to any values to affect space curvature or to prevent anything from crossing it from inside. Of course, anything that would cross it from inside would return back eventually.
- pdonis 13y ago
  given the black hole mass large enough, the gravitational force at the Schwarzschild radius can be as small as we'd like it to be No, it can't; the "acceleration due to gravity", which is the acceleration required to "hover" at a constant altitude above the horizon, diverges as the horizon is approached. What can be made as small as desired by making the hole's mass large enough is tidal gravity at the horizon. several billions years ago, when observable Universe had 10B radius, it would be a black hole No, it wouldn't. A black hole is a stationary spacetime. The spacetime of the universe is not stationary. The spacetime model that describes the universe is very different from the spacetime model that describes a black hole; the fact that you can plug the mass of the observable universe into the Schwarzschild radius formula and get a number out does not mean that number has any physical meaning for the universe.
  - VladRussian2 13y ago
    >No, it wouldn't. A black hole is a stationary spacetime. The spacetime of the universe is not stationary. The spacetime model that describes the universe is very different from the spacetime model that describes a black hole exactly. There is no stationary spacetime in our Universe. Black hole is pretty artificial model where pure mathematical artifacts of singularity at the horizon is taken for the real thing. > the fact that you can plug the mass of the observable universe into the Schwarzschild radius formula and get a number out does not mean that number has any physical meaning for the universe. Taking a big chunk of space and calculating escape speed from its gravitational field - how is that not a physical meaning? At what specific size of the chunk you think it becomes not meaningful?
    - pdonis 13y ago
      There is no stationary spacetime in our Universe. The universe as a whole is not stationary, nor even close to being so; but portions of the universe are very close to being stationary. The solar system, for example. Black holes do not have to be exactly stationary; if they are as close to being stationary as the solar system, that's plenty close enough. Black hole is pretty artificial model The exactly spherically symmetric solution is an idealization, yes; but there are plenty of numerical simulations that show that non-symmetric spacetimes still form event horizons. where pure mathematical artifacts of singularity at the horizon is taken for the real thing. There is no physical singularity at the horizon. Some coordinate charts have a coordinate singularity at the horizon, but that's easily fixed by just using a different coordinate chart. The only physical singularity is at r = 0. Taking a big chunk of space and calculating escape speed from its gravitational field Escape to where? You can't escape from the universe as a whole. The concept of "escape speed" has no meaning for the universe as a whole.
pdonis 13y ago
This doesn't agree with my understanding of GR. Nor mine. You're correct, any object in the interior of the hole, even one that is moving radially outward at the speed of light, gets closer to the singularity (i.e., its r coordinate decreases) with time, which means it can't get closer to the horizon (that would require increasing r).

VladRussian2 13y ago

the event horizon - Schwarzschild radius - is defined by the specific value of gravitational potential (escape speed equals speed of light) while, given the black hole mass large enough, the gravitational force at the Schwarzschild radius can be as small as we'd like it to be. The various local effects like time dilation, light path curving, etc... are defined by the value of gravitation force, not gravitational potential. The potential defines the fact that anything originating at or below the horizon would never escape completely the gravitational field, i.e. never reach the infinity. The Schwarzschild radius of the mass of the observable Universe is 10B light years. So, several billions years ago, when observable Universe had 10B radius, it would be a black hole (though i think that in less expanded space of the earlier Universe the constants like "c" had different values and thus that Schwarzschild radius was less). We can imagine it in another way - increase 125 times (the observable Universe has 40+B light years radius) the amount of matter, ie. galaxies, stars, etc... inside the 10B radius ball around us, and you'd get the black hole with 10B light years Schwarzschild radius (and i don't think we would ever notice the change - only with time the galaxies's movement will be affected). Obviously, the gravitational field on the surface of that imaginary 10B radius sphere and inside it would increase somewhat (like 125 times on the surface, 125 times 0 pretty much 0) - not even close though to any values to affect space curvature or to prevent anything from crossing it from inside. Of course, anything that would cross it from inside would return back eventually.

pdonis 13y ago
given the black hole mass large enough, the gravitational force at the Schwarzschild radius can be as small as we'd like it to be No, it can't; the "acceleration due to gravity", which is the acceleration required to "hover" at a constant altitude above the horizon, diverges as the horizon is approached. What can be made as small as desired by making the hole's mass large enough is tidal gravity at the horizon. several billions years ago, when observable Universe had 10B radius, it would be a black hole No, it wouldn't. A black hole is a stationary spacetime. The spacetime of the universe is not stationary. The spacetime model that describes the universe is very different from the spacetime model that describes a black hole; the fact that you can plug the mass of the observable universe into the Schwarzschild radius formula and get a number out does not mean that number has any physical meaning for the universe.
- VladRussian2 13y ago
  >No, it wouldn't. A black hole is a stationary spacetime. The spacetime of the universe is not stationary. The spacetime model that describes the universe is very different from the spacetime model that describes a black hole exactly. There is no stationary spacetime in our Universe. Black hole is pretty artificial model where pure mathematical artifacts of singularity at the horizon is taken for the real thing. > the fact that you can plug the mass of the observable universe into the Schwarzschild radius formula and get a number out does not mean that number has any physical meaning for the universe. Taking a big chunk of space and calculating escape speed from its gravitational field - how is that not a physical meaning? At what specific size of the chunk you think it becomes not meaningful?
  - pdonis 13y ago
    There is no stationary spacetime in our Universe. The universe as a whole is not stationary, nor even close to being so; but portions of the universe are very close to being stationary. The solar system, for example. Black holes do not have to be exactly stationary; if they are as close to being stationary as the solar system, that's plenty close enough. Black hole is pretty artificial model The exactly spherically symmetric solution is an idealization, yes; but there are plenty of numerical simulations that show that non-symmetric spacetimes still form event horizons. where pure mathematical artifacts of singularity at the horizon is taken for the real thing. There is no physical singularity at the horizon. Some coordinate charts have a coordinate singularity at the horizon, but that's easily fixed by just using a different coordinate chart. The only physical singularity is at r = 0. Taking a big chunk of space and calculating escape speed from its gravitational field Escape to where? You can't escape from the universe as a whole. The concept of "escape speed" has no meaning for the universe as a whole.

pdonis 13y ago

This doesn't agree with my understanding of GR. Nor mine. You're correct, any object in the interior of the hole, even one that is moving radially outward at the speed of light, gets closer to the singularity (i.e., its r coordinate decreases) with time, which means it can't get closer to the horizon (that would require increasing r).