Ich said:
Ok, that´s the "simultaneity issue" I brought up. I read the original statement different: It sais that
1. I will fall through and
2. The observer will not see me fall through.
It doesn´t explicitly compare times. But still there´s the other point-->
So after a certain time nobody outside can interact with me. That´s sufficient for me to state that it´s not me there at the horizon but merely my image.
I think you may need to think about this some more.
The person who is falling into the black hole can still receive signals from the outside world. But, once he passes the event horizon, he cannot send signals back.
The question arises - if you have an observer sitting at infinity, sending out pulses once a second, how many of them does the infalling observer see? He can see these pulses even after he crosses the event horizon.
It takes math to really calculate the answer - the only really interesting part of the answer is that he sees a finite number of pulses from the external observer, not an infinite number, before he reaches the black hole.
There is a fairly simple argument to show that the infalling observer can't see an infinite number of pulses, without a lot of math. You have to believe that photons are redshifted when they go "uphill", and blueshifted when they go "downhill". You also have to believe that the tidal forces near a black hole at a given point don't act weirdly as a function of velocity. You also have to believe that the observer hits the singularity in a finite amount of his own proper time.
Given all these assumptions, you can say that from the infalling observers POV, the stretching tidal forces mean that the photons from infinity have to climb "uphill" to reach him. This means that the photons are redshifted.
The point of this is that your mental model must give the following results to be correct:
Photons emitted by the infalling observer don't make it back out
Infalling photons tracing a radial path (ones that fall directly into the black hole) are redshifted when they reach the infalling observer.
The observer reaches both the event horizon (and even the singularity!) in a finite amount of proper time (umm, we're doing a non-rotating black hole here to keep life simple).
Because the infalling photons are redshifted, the observer sees only a finite number of pulses from the outside world before he reaches the event horizon (or before he hits the singularity for that matter). (Not only photons are redshifted, but any time interval is lengthend. If you send pulses out at once a second with a redshift factor of 2, pulses willl arive once every two seconds).
A model that the infalling ovserver never reaches the event horizon usually fails on this point (the finite number of pulses) - because if you think of him never reaching the point, you think that he must receive an unlimited number of pulses. This is wrong.
At the time I lost contact with the universe, I´m as good as through the event horizon - I can´t return (can I?) and nothing from the outside can bring me back. So I will be killed by singularity.
What remains after that time is an image which will fade away and finally be destroyed in the explosion. It´s merely an artifact of curved space time (something like trapped photons) and has nothing to do with me.
Does this make sense?
And how does one calculate that time?
To actually calculate the paths of light and objects, one needs to use a well behaved coordinate system like the Finklestein ingoing coordinates, or Kruskal coordiantes. Of the two, Finklestein coordinates are the easiest to understand. The Finklestein coordinate 'v' is basically just the number of pulses an observer has seen while falling from infinity. Another way of saying this - by consturction, radially infalling light rays have a constant Finklestein coordinate 'v'. This coordinate 'v' is neither a space coordinate, nor a time coordinate, but a 'null' coordinate, becuase it desccribes a light-like path with a constant number.
There's a graph at
http://casa.colorado.edu/~ajsh/collapse.html#kruskal which may be of some help if you study it enough of an infalling black hole in Finklestein coordinates (and in Kruskal coordinates, too).
Given the right coordinates, the job of calculating the trajectories is basically a matter of satisfying the geodesic deviation equations, which are a system of second order partial differential equations.
For specificity we say that the particle follows some parameterized path
x^i(\tau), i.e. t(tau), x(tau), y(tau), z(tau), in the usual cartesian coordinates. In Finklestein coordinates it's actually r(tau), v(tau), theta(tau), phi(tau), where theta and phi are two angles (zero for someone falling straight in, so we can deal only with r(tau) and v(tau) for an observer with no angular momentum.
Then we write down the geodesic deviation equations
\frac{d^2 x^c}{d \tau^2} + \Gamma^a{}_{bc} \frac {d x^b}{d <br />
\tau} \frac {d x^c} {d \tau} = 0
Then we solve these equations.
The \Gamma in the above expression are the Christoffel symbols for the given metric - they can be calculated by hand (or much more easily by programs like GRTensor II) from the metric coefficients. They are basically partial derivatives of the metric coefficients.
Working the problem out in Finklestein coordinates, I get the following equations for what it's worth (I've represented differentiaton with respect to tau by a prime mark. You can see these are second order equations in r'', r', r, v'', v', v, where r and v are the Finklestein coordinates. (r is equivalent to the Schwarzschild radial coordinates).
r'' - 2m/r^2 r'v' +(1/r^2-2m^2/r^3) v' v' = 0
v'' + 1/r^2 v' v' = 0
Oh yeah - you need the boundary conditions to solve the equations - specifically, what are r, r', and v (the radial position, radial velocity, and you can safely set v=0 at tau=0, but you need to specify that that's what you're doing).
