Please verify that this explanation of black holes is consistent with the mainstream theory. (I have a $10 bet riding on this.)(adsbygoogle = window.adsbygoogle || []).push({});

The Schwarzschild radius is 2GM/C-squared. If the mass M is within R spatial units of the center, then there will be a black hole, with an event horizon having the time-like Schwarzschild radius, in pure time units with no spatial component because the event horizon's radius is within the event horizon, where space-time has only a time component. This means that the spatial radius of a black-hole event horizon is zero. The event horizon is positioned in space in the same mathematical point as the singularity. In other words, the spatial position of the event horizon is at the singularity, not R spacial units out from the singularity. Every point in space, other than the singularity (which is out of space) is outside the event horizon.

This means that an object can approach arbitrarily close (spatially) to the singularity and still be outside the event horizon. Thus the special time-like calculations that apply within the event horizon are not applicable at any point in space (other than the singularity point). So escape velocities at points arbitrarily close spatially to the singularity can still be analyzed like ordinary escape velocities.

At every point within the spatial Schwarzschild radius, the escape velocity is greater than the speed of light. This means that a beam of light originating just inside the spatial Schwarzschild radius and directed radially outward will go a very long, but finite, distance before it becomes infinitely red-shifted. A beam originating just outside the spatial Schwarzschild radius will be more and more red-shifted as it travels, but it will never become infinitely red-shifted. A beam originating very close to the singularity/event horizon will become infinitely red-shifted after a very short, but non-zero, distance.

A rocket ship (with unlimited but finite thrust to overcome gravity, unlimited strength to withstand tidal forces, and piloted by Superman) can approach arbitrarily close, spatially, to a singularity and at any time (or, from the pilot's reference frame, at any time within the finite time interval before his spatial intersection with the event horizon, which is sooner than his temporal intersection with the singularity) the pilot can fire his rockets and accelerate away from the singularity. If he continues to fire his rockets (just enough to slightly overcome the gravity at each point), he will eventually reach the region outside the spatial Schwarzschild radius. In that outer region, the escape velocity is less than the speed of light. If the pilot continues to fire his rockets just enough to overcome gravity, he will eventually reach a point where the escape velocity is less than his current speed. Then the pilot can cut off his rockets and coast forever away from the black hole, always decelerated by the black hole's gravity, but never losing all of his outward velocity. This is how Superman escaped from the black hole of Krypton.

When an object is very close to the singularity/event horizon, it is in a very intense gravitational field, with a very steep gravitational gradient. Thus it experiences strong tidal forces, increasing without limit as the object approaches the singularity/event horizon. (For a very large black hole, the tidal forces at the spatial Schwarzschild radius might be moderate. But remember that the event horizon is not a the spatial Schwarzschild radius, it is at the singularity point, separated from the singularity only by a time interval which is equal to the Schwarzschild radius in time units.) Also, relative to distant observers, the object undergoes a slowing down of time, approaching a stand-still at the singularity/event horizon. I'm not sure whether an observer looking outward from the vincinity of the singularity/event horizon would see red-shifting and time-slowing or blue-shifting and time-speeding (can someone out there tell me?), but I am sure that after returning from near the singularity/event horizon he would find that his distant stationary colleague had experienced a much longer time interval than what he had experienced.

A distant observer (with super infra-red vision) sees all the mass of the black hole between the spatial Schwarzschild radius and the temporal Schwarzschild radius (the event horizon), with most of it "frozen" just outside the event horizon/singularity. The mass never crosses the event horizon, but it doesn't have to cross the event horizon in order to form a black hole. It is a black hole because the mass is inside the spatial Schwarzschild radius, even though the mass never (from a fixed frame of reference) crosses to within the time-like Schwarzschild radius of the event horizon.

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# Black Holes - help needed to settle a bet

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