Perhaps "black holes" should be called "red holes"; after all, as time slows down close to the event horizon, all emitted light becomes red-shifted into the long wave radio spectrum; Mary no longer "sees" Joe, so much as hears him on her radio. Don't count photons; count energy. The number of photons actually goes up dramatically as the frequency goes down. Photons merely count the number of Planck lengths in a single wavelength of "light". Photons are merely the smallest units of a "transaction" in the "transactional interpretation of quantum theory". At 05:17 PM 5/28/2012, Warren Smith wrote:
From: Eugene Salamin <gene_salamin@yahoo.com> If we examine the situation before the black hole has formed, and after it has evaporated, we need not be concerned about the peculiarities of general relativity. Baryonic matter goes in; mostly photons (and gravitons also) come out. This is the "Joe point of view" wherein Joe has been destroyed in the black hole. This must be correct, so what is wrong with the "Mary point of view" in which Joe lingers forever just above the event horizon?
In Joe's rest frame, he emits a finite number of photons before crossing the horizon, and for him, crossing the horizon is not particularly eventful.
--not necessarilly. Joe could emit an infinite number of photons in a finite amount of his time with a finite summed energy (e.g. photon n has energy=1/n^2)... at least if we had magic fake-photons which were point particles not waves so we did not need to worry about their redshifts or antenna effects. Not that this quibble terribly matters. So if photons were "points" not waves and we did not need to worry about their redshifts, then this would enable Joe to stay visible to Mary the whole time.
Mary detects these photons within a finite amount of her proper time, and after the last one, there are no more. Photon number is conserved under propagation and Doppler shift in general relativity. How could Mary test the hypothesis that Joe never actually fell into the black hole? She might send in a beam of light and look for the scattering off of Joe.
--the usual story is that does not work because after some short finite amount of Mary-time, she is unable to emit a photon capable of catching up to Joe. It works until that moment comes, however. Furthermore, after that moment, the light Mary had shined on Joe before, still is in transit toward Joe or back to Mary... hence she would keep on seeing Joe using that old light, forever, except for the fact there are only a finite number of photons in the beam since light is not infinitely divisible. And eventually the last one catches up to Joe, then flies back to Mary, for her last glimpse of Joe.
But on a space-time diagram, when the probe light reaches the event horizon, Joe is already inside. If the light does catch up with and scatter off Joe, the scattered light is trapped inside the horizon, and never gets back to Mary.
Where does the notion of "lingering just outside the horizon" come from?
--it comes from the picture where Joe keeps emitting an infinite number of magic fake-photons constantly, causing Joe to remain visible to Mary eternally, meaning "He never fell in."
Before 1969 or so, the literature often called black hole s"frozen stars" to connot the fact they were "frozen eternally just outside their event horizon." But then Wheeler coined the phrase "black hole" and it stuck -- he laughed off complaints this name was "undignified" and "obscene" :)
In the elementary textbook description for the Schwszschild black hole, we have coordinates r and t that are the usual radius and time far from the black hole.In these coordinates, Joe's orbit r(t) approaches the Schwarzschild radius as t approaches infinity, and so he seems to linger there forever. But just because we name something "t" doesn't make it into a time.
--very true. Indeed t is not time, and r is not length. You can integrate ds along radial or t-directed lines though, to figure out lengths and times... But t really happens to be Mary's time-coordinate, and r for her really is a length coordinate (assuming Mary located at large r value).
Inside the horizon, r becomes timelike, and t becomes spacelike, while on the horizon itself, the metric in these coordinates becomes singular.
The black hole singularity is not so much "at the center" as it is in the future.
--true.
Joe is compelled to fall into the singularity for the same reason that we are all compelled to move into the future, whatever that reason may be.
--true again. As Wheeler once put it (which struck me as rather poetic wording, it has an appealing air of finality) "Time stops. That is the lesson of the Schwarzschild black hole."