wangyi said:
Hi, i am confused by the following question:
When an observer is going towards the horizon of a BH from outside, he finds the light from far away so blue shifted that the far future of the far away world presents in front of him. Does that mean when he is crossing the horizon, he has seen everything in the infinite future?
What makes you think this?
To actually calculate the path of the light is somewhat involved, the only way I can think of to do it would be to do a ray trace.
The only website I've seen that talks about this topic has some nice graphics, but doesn't go into the detail of how the calculations were performed.
http://casa.colorado.edu/~ajsh/schw.shtml
(This site has several movies, note that only one of the later movies is a straightforwards dive, see the introduction for details).
http://www.fourmilab.ch/gravitation/orbits/ and textbooks like MTW go into the math of how to calculate photon orbits, but don't present any visual picture of what one would "see" falling into a black hole.
The first website has a quiz section. Look at question #2
2) As you fall freely into a black hole, you see the entire future of the Universe played out before your eyes. True or false?
Click on the blue question mark to be directed to the answer:
Answer to the quiz question 2: False. You do NOT see all the future history of the world played out. Once inside the horizon, you are doomed to hit the singularity in a finite time, and you witness only a finite (in practice rather short) time pass in the outside Universe.
IIn order to watch the history of the Universe unfold, you would have to remain outside the horizon, the Schwarzschild surface. One way to watch all the history of the Universe would be to stay just above the horizon, firing your rockets like crazy just to stay put. The Universe would then appear not only speeded up, but also highly blueshifted (probably roasting you in gamma rays), and concentrated in a tiny piece of the sky just above you.
For a quick "sanity check", let's consider the case where someone is falling straight into a black hole, and is at the photon sphere at 1.5 x the Schwarzschild radius. For a stationary observer at this location, any photons emitted "in front" will fall into the black hole. To figure out what we see, we use the idea of ray tracing.
http://www.siggraph.org/education/materials/HyperGraph/raytrace/rtrace1.htm
We reverse time and ask "where would any visible photons come from"? We want a photon that originates at a distant star, and impacts our location. To do this, we reverse time and trace the outgoing path of a time-reversed photon, and see if it intersects any stars.
The black hole is black, so photons can't start from it. They must come from some object outside the black hole. We will assume that the light we see comes from some distant star, thus we need a reversed-time photon orbit that escapes to infinity.
We can see immediately that for the stationary observer, any such "reversed time" photons must come from behind the observer, not in front (because any photons emitted to the front fall in, any photons that come from the front must have come from the black hole, and the black hole is black).
A falling observer will see things differently due to relatavistic abberation due to his infalling velocity. This will make some of the photons which appeared to fall from "in back" of the stationary observer to be aberrated towards the front. So the black hole will not fill half the sky at the photon sphere for an infalling observer as one might expect. It will still fill a substantial portion of the sky, which will appear black. This seems to tie in reasonably well with the graphics on the website I mentioned earlier.
