PeterDonis
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Rena Cray said:Nothing falls into a classical black hole.
Incorrect. See below.
Rena Cray said:It's perfectly fine to use Schwarzschild coordinates on the open set bound by the event horizon for an external observer and the local coordinates of the infalling astronaut.
But it's not fine to then assume that only the open set bounded by the event horizon is accessible to the infalling astronaut, because that's all that's visible to the external observer.
Rena Cray said:In the coordinate system of an external observer, we need to integrate the elapsed time of an infalling test particle from an initial radial displacement to R_s + \delta In nonstandard analysis this is R_s + dR This amount of time is transfinite.
Show your work, please. The integral that gives elapsed proper time on an infalling trajectory for a finite ##r > r_s## to ##r = r_s## is finite; this is a standard textbook problem in GR.
Rena Cray said:But the evaporation time is finite.
I thought you were leaving out Hawking radiation. You should; you need to get the purely classical case right first, before adding quantum complications.
Rena Cray said:For the infalling astronaut, the boundary would not be crossed but recede before him.
Incorrect.
Rena Cray said:If we assume time symmetry, modulo CP, and there are time-like paths for matter to enter or null curves for light to enter, then there are time-like paths for matter to escape and null curves for light to escape
Incorrect. Time symmetry does not require this. It only requires that if there is a "black hole" solution to the field equations, in which objects can fall in but not escape, then there must also be a time-reversed "white hole" solution, in which objects can escape but not fall in. In other words, time symmetry is a property of the solution space of the field equations, not of single solutions considered in isolation.