- #1
Ray Eston Smith Jr
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Escape velocity at particular radius from the gravitational source means the initial speed that an object needs at that raduis in order to coast without limit ("to infinity") without ever falling back to the gravitational source. If the escape velocity at a radius is so large that nothing inside that radius can ever move outside that radius, then the escape velocity at that radius is infinite.
Suppose that a light beam is directed radially outward from a source in the vicinity of a black hole event horizon.
If the light source is exactly on the event horizon, then the beam will become infinitely red-shifted in zero distance. In other words, it won't go anywhere. This is because escape velocity at the event horizon is infinite.
If the light source is a short distance outside the event horizon, then the beam will travel a short distance before it becomes infinitely red-shifted. This is because the escape velocity at that radius is finite but much larger than the speed of light, so that light cannot escape to infinity.
At some distance from the event horizon, the escape velocity will be equal to the speed of light. By definition, this is the Schwarzschild radius. By LaPlace's classical physics derivation, this distance would be 2GM/c-squared. When the light source is at that radius, the beam will travel forever without becoming infinitely red-shifted. If the light source was just inside that radius, the beam would travel a very long, but finite, distance before it became infiinitely red-shifted.
Thus the event horizon is not located at a spatial distance of 2GM/c-squared from the center of mass. It is separated from the center of mass by zero spatial distance and by a time interval of 2GM/c-squared. The center of mass is empty space with null gravity until, after a time interval of 2GM/c-squared (in the free-fall reference frame at the center point), the singularity arrives with all the mass-energy within the spatial Schwarzschild radius falling in on the spatial center point. (That last sentence makes no sense, but it's as close as I can come to visualizing something that happens nowhere after forever.)
Suppose that a light beam is directed radially outward from a source in the vicinity of a black hole event horizon.
If the light source is exactly on the event horizon, then the beam will become infinitely red-shifted in zero distance. In other words, it won't go anywhere. This is because escape velocity at the event horizon is infinite.
If the light source is a short distance outside the event horizon, then the beam will travel a short distance before it becomes infinitely red-shifted. This is because the escape velocity at that radius is finite but much larger than the speed of light, so that light cannot escape to infinity.
At some distance from the event horizon, the escape velocity will be equal to the speed of light. By definition, this is the Schwarzschild radius. By LaPlace's classical physics derivation, this distance would be 2GM/c-squared. When the light source is at that radius, the beam will travel forever without becoming infinitely red-shifted. If the light source was just inside that radius, the beam would travel a very long, but finite, distance before it became infiinitely red-shifted.
Thus the event horizon is not located at a spatial distance of 2GM/c-squared from the center of mass. It is separated from the center of mass by zero spatial distance and by a time interval of 2GM/c-squared. The center of mass is empty space with null gravity until, after a time interval of 2GM/c-squared (in the free-fall reference frame at the center point), the singularity arrives with all the mass-energy within the spatial Schwarzschild radius falling in on the spatial center point. (That last sentence makes no sense, but it's as close as I can come to visualizing something that happens nowhere after forever.)