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Posted by Bruce on February 10, 2002 22:26:52 UTC

There are several perspectives which are very different. I will mention three. The remote observer, the local shell observer, and the remote observer observing from a proper frame. The remote observer is the 'GR bookkeeper who is at rest in flat spacetime far away from M' who will use the ticks on her (you) clock and, coordinate time dt, to observe the astronaut falling radially towards the black hole. The local shell observer at rest on a spherical shell (which encircles the black hole at constant r but outside the event horizon) uses his (your husband) clock, local shell time dtshell, to observe the astronaut as the astronaut crosses over a local spherical shell of constant r. And finally the remote observer (me)observing from a proper frame who uses the proper wristwatch time (dt) of the falling astronaut to observe the falling astronaut motion inside and outside the black hole. This is really a unique perspective because it allows us to observe the falling astronaut even after he is no longer causally connected to our universe (he crossed over the event horizon. Of course this is speculative but the calculations are fundamentally based on Einstein's theory of gravity, general relativity). The following equations are simple to use and they are derived from the Schwarzchild metric which was the first solution to Einstein's field equations and are based on an idealized nonspinning spherically symmetric black hole. M in the equations is mass converted to units of length (meters) where one solar mass = 1477 meters. This means for a solar mass black hole the singularity is at r = 0 and the event horizon is r = 2M (2954 meters) and the photon sphere is r = 3M (4431 meters), etc.. The curvature term in these equations is (1-2M/r).

For you the remote observer, GR bookkeeper,

vradially falling astronaut = dr/dt = -(1-2M/r)(2m/r)1/2

If you substitute 2M for r you will see that the remote observer observes dr/dt -> 0. How can this be? It is all due to how the GR bookkeeper receives news of the falling astronauts progress. As the falling astronaut approaches the event horizon the signal bearing information of his progress is slowed as it climbs out of the gravitational field of the black hole. This slowing of light was predicted by general relativity and confirmed by experiments conducted by Irwin Shapiro. You can read about it by searching 'Shapiro Effect'. Shapiro used the Sun and planets of our solar system for the experiment.

vshell = drshell/dtshell = -(2M/r)1/2

There can be no spherical shell at the event horizon (not enough energy in the universe to keep the shell from falling into the singularity) but your husband can observe that as r -> 2M, drshell/dtshell -> 1. 1 is the local coordinate speed of light in geometric units.

And finally I can observe

vproper = dr/dt = -(2M/r)1/2

Here we can safely substitute 2M for r and show that the proper velocity of the falling astronaut is the speed of light as he crosses the event horizon. You can also integrate this expression for proper velocity and speculate that the average proper velocity of the falling astronaut from r = 2M to r = 0 is (3/2)c. BTW I forgot to mention that the falling astronaut begins his fall from your position as the GR bookkeeper. I mention this because the equations are slightly different for a radially free falling astronaut who begins his fall from a spherical shell or bails out of a spaceship (for instance). The results are the same at the event horizon regardless.
A little more than you asked for but you sound interested.
Regards to you and your husband

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