From: ram@zedat.fu-berlin.de   
      
   This might be a bit of a speculative question.   
      
     They say that if an object falls into a black hole, it would appear   
     to an observer at a great distance relative to the black hole that   
     the object slows down but never actually reaches the event horizon.   
      
     I'll refrain from pondering the idea that for an object near the   
     event horizon, quantum mechanically due to the uncertainty principle,   
     the probability of being inside the event horizon could already be   
     greater than 0. I'm sure experts have thought about that as well.   
      
     So, where was I? Ah yes, for an outside observer, the object reaches   
     the event horizon only after an infinite amount of time has passed.   
      
     On the other hand, from an outside perspective, the black hole   
     itself could have evaporated after a finite amount of time AFAIK.   
      
     What happens to the object, which from the outside appears   
     to be waiting just short of the event horizon, after the black   
     hole has evaporated from the point of view far from the black hole?   
     Would we find the object outside the event horizon then? - TIA!   
      
   [[Mod. note --   
   A few comments (I suspect the author already knows these things):   
      
   > for an outside observer, the object reaches the event horizon only   
   > after an infinite amount of time has passed   
      
   It's useful to think of the infalling object as sending out continuous   
   light (or radio) signals.   
      
   As the object approaches the event horizon, these signals take longer   
   and longer to reach a (fixed) outside observer, and the signals are more   
   and more redshifted.  This results in the effect the author described,   
   and means that the outside observer never sees any signals emitted on   
   or inside the event horizon.   
      
   It may be useful to rephrase the above-quoted statement as:   
      
      Light/radio signals from the moment the infalling object crosses   
      the event horizon will take infinitely long to reach, and will   
      be infinitely redshifted when they arrive at, a (fixed) outside   
      observer.   
      
   This phrasing makes it clear(er) that the effects we're describing are   
   caused by light/radio signal propagation near the event horizon, rather   
   than the infalling object's motion being in any way arrested.   
      
   Indeed, an observer riding the infalling object itself would measure it   
   passing through the event horizon very quickly; it's "just" that after   
   passing through the event horizon her light/radio signals would never   
   reach the outside observer.   
      
   The time scales are interesting:  For a 1-solar-mass black hole   
   (event horizon is a few kilometers in diameter):   
   * outside observer sees the object's radio signals infinitely redshift   
     and cut off on a time scale of microseconds; the last photon received   
     by the outside observer will be very low-enegy (upon reception) and   
     have originated just before the infalling object crossed the event   
     horizon   
   * infalling observer reaches r=0 singularity in microseconds   
     (but any messages from her about what it's like, will never   
     get to the outside observer)   
   * outside observer sees the black hole evaporate in something   
     on the order of 10^67 years   
      
   As for your actual question, what happens to things inside a black   
   hole when it (eventually) evaporates, it's a good question.  Alas, I   
   don't think we (yet) understand quantum gravity well enough to know   
   the answer. :(  There's a brief introduction to this topic at:   
     https://en.wikipedia.org/wiki/Hawking_radiation#Black_hole_evaporation   
   -- jt]]   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)