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 ?
