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The End of the Universe from the Outside In - as free-faller nears singularity

  1. Aug 22, 2004 #1
    Consider a light source approaching a singularity (in free fall). As the source approaches the singularity, it is approaching infinite gravity, infinite tidal forces, infinite time-slowing, infinte escape velocity, and infinite red-shifting in zero distance (as seen, or not seen, by outside observer). At some finite distance from the singularity (at the Schwarzschild radius), the escape velocity equals the speed of light. A beam from a source at that point will approach infinite red-shifting as its travel distance approaches infinity. A beam from a source closer to the singularity will become infinitely red-shifted after traveling some finite distance.

    Now if I shine a beam outward and it red-shifts to infinity after traveling a finite distance, then if I look at a light source coming from that distant point, I should see it infinitely blue-shifted. In other words, from my reference frame, time has ended outside that distance (the distance where my outgoing beams become infinitely red-shifted). So, as I approach the singularity, I begin to see the end of time (in distant locations) as soon as I cross the Schwarzschild radius. As I get closer and closer to the singularity, I see the end of time closing in on me, closer and closer. At some point, I see the end of the time at the Schwarzshild radius. At that point, if I turned on my retro-rockets, I couldn't escape the Schwarzschild radius, because there would be no universe outside the radius.
  2. jcsd
  3. Aug 23, 2004 #2
    I don't quite get where you're going with the rest, but...
    this is not so. Only at the point of singularity is the incoming light infinitely blue-shifted; at any other point inside the Schw. radius, the blueshift is finite and easily calculable.
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