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One of the features of the black hole complementarity is the following :

According to an external observer, the infinite time dilation at the horizon itself makes it appear as if it takes an infinite amount of time to reach the horizon, while the free falling observer reach the horizon in a finite amount of time.

But, because we cannot differentiate between acceleration and gravitation, it may be equivalent to the following "paradox" in Special Relativity.

Le be the plane $z = 0$ be a kind of "horizon".

Let be an observer $O_1$ which is at $z = 1 \space m$ (1 meter), at $t = 0$, and which is moving left (towards decreasing $z$), with a constant speed $(-1) \space m/s$.

Let be a second observer $O_2$, which has the position $z = 1$ m (1 meter), at $t = 0$, but which is moving right (towards increasing $z$), and which is accelerating, in a precise sense (see below).

Divide the initial distance from the observer $O_1$ to the "horizon" $z=0$, into a serie of distance intervals $L_n$, with $L_n = 1/2^n \space m$(meter): $$1/2, 1/4, 1/8,....,..., 1/2^n,...$$

The initial distance (1 meter) between the observer $O_1$ and the horizon is : $$ 1 = \sum^{\infty}_{n=1} L_n$$

For each distance interval $L_n$, the corresponding elapsed time (from observer $O_1$ point of view) needed for observer $O_1$ to reach the end of the interval $L_n$, is $\tau_n = 1/2^n s \space (second)$, because its speed is $-1 \space m/s$.

The total time (from observer $O_1$ point of view), necessary for the observer $O_1$ ,to reach the horizon $z = 0$ is then : $$\tau = \sum^{\infty}_{n=1} \tau_n = \sum^{\infty}_{n=1} 1/2^n = 1 s$$

Now, for each interval $L_n$, we may adjust the speed of the accelerating observer $O_2$, such that, due to the time dilatation, the elapsed time necessary for the observer $O_1$ to travel during the interval $L_n$, from the point of view of observer $O_2$, be :

$$T_n = a_n \tau_n$$, where $a_n$ is a coefficient > 1

From the point of view of observer $O_2$, the time necessary for observer $O_1$ to reach the horizon $z = 0$ is then : $$ T = \sum^{\infty}_{n=1} T_n = \sum^{\infty}_{n=1} a_n \tau_n = \sum^{\infty}_{n=1} a_n/2^n$$

By choosing, for instance, $a_n = 2^{n+\epsilon}$, where $\epsilon >0$, it is easy to see that, from the observer $O_2$ point of view, the observer $O_1$ needs an infinite amount time to reach the horizon $z=0$, while from the $O_1$ point of view, he reaches the horizon in one second!

Did you agree ?

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I thought that a falling object would follow a de-sitter geodesic to an external observer, meaning that it must continue traveling inward at the horizon. I also thought the infinite time dilatation implied that the object is frozen in time as it passes the horizon, not that it stays there forever. But I'm speaking of the formalization for an external observer, the light you see would show someone frozen above the horizon. –  AlanSE Oct 9 '12 at 13:52
Well, if time dilatation is infinite, the object is frozen, but, in my example, the time dilatation, while increasing, is never infinite (at a given external observer time). So the object, from the point of view of the external observer, is never frozen, it moves, but it just don't reach the horizon. –  Trimok Oct 9 '12 at 14:09
I should qualify that I'm not an expert in GR topics, as evidenced by my own questions. I see you describing an asymptotic limit. If, in this formulation, the distance to the horizon is finite, then for this "never crosses" picture hold, then the velocity would have to limit to zero (among other requirements). An inertial object (a "geodesic" if I have terminology right) would always travel inward, and always accelerate to the horizon. Regardless of what happens to the falling observer, the concept of velocity and position still follow their regular rules to the external observer. –  AlanSE Oct 9 '12 at 14:39
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1 Answer

The trouble is that you're choosing the (variable) acceleration of $O_2$ to make $O_1$ "freeze". This isn't a realistic model of someone hovering over a black hole (i.e. at constant acceleration) watching someone else fall in.

Accelerated observers can be handled using just special relativity. Chapter 6 of Gravitation by Misner, Thorne and Wheeler develops this, or for constant acceleration John Baez's article on the relativistic rocket gives a briefer but more easily understod treatment. Finally, the geometry corresponding to an accelerated observer is Rindler geometry. This does contain horizons, but these are cosmological horizons rather than the horizon round a black hole.

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The accelerated observer does see O1 freeze, this is a Rindler horizon, and it's neither black hole (curving away) nor cosmological (curving towards), it's exactly a flat plane. It approximates both horizons in the near horizon limit. –  Ron Maimon Nov 8 '12 at 17:09
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