5 ms·

>The event horizon is the "point of no return" beyond which gravity is inescapable. the event horizon is frequently treated as "uncrossable" from inside to out

>The event horizon is the "point of no return" beyond which gravity is inescapable. the event horizon is frequently treated as "uncrossable" from inside to outside which is really different from "inescapable". The later means that whatever speed you have on or below the event horizon, you can't reach infinity. The former is just an impression by an outside observer because in his observation the time has stopped on the event horizon, 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 (the point of "inescapability" is that the stone would always return). While above the horizon, the stone can interact with other stuff there and result of this interaction can be observed outside (doesn't mean on practice by us today or tomorrow :). Many models seems to treat the event horizon as "uncrossable". For example quantum information disappearance - matter goes in, evaporates as Hawking. Yet, just for example, when Hawking radiation decreases the mass of the black hole it causes the shrink of the event horizon and thus whatever "stones"/photons on their way up were stuck (for external observer) in the stopped time of the horizon become free - doesn't mean though that we can observe them on practice as getting out of that gravitational well does take time (again in our time) and redshifts them into oblivion. Like proverbial "the check is in the mail". It is the reason why we can't really observe Hawking radiation until we develop technology to observe light with extremely long wavelength and i just don't have the numbers right now on whether 13B years is enough for the radiation originating right above the event horizon to get out of that well and reach the interstellar space.

millstone 13y ago
> 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.
Steuard 13y ago
Technically speaking, in classical GR if you begin at any point inside the event horizon of a black hole, your entire future light cone is also contained within the event horizon. That means that a black hole event horizon truly is "uncrossable" (as you phrase it), not just "inescapable". (The Newtonian analogy of escape velocities doesn't really capture the details of GR in this case.)

millstone 13y ago

> 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.

Steuard 13y ago

Technically speaking, in classical GR if you begin at any point inside the event horizon of a black hole, your entire future light cone is also contained within the event horizon. That means that a black hole event horizon truly is "uncrossable" (as you phrase it), not just "inescapable". (The Newtonian analogy of escape velocities doesn't really capture the details of GR in this case.)