proper time for someone falling radially into a black hole

The full, very gritty, details were worked out (for a Schwarzschild black hole) at įor the case of the person free-falling from a position $r_0$, then the span of time seen by the falling person when looking back towards $r_0$, before they hit the singularity, is given by Or to put it another way you can contruct a past light cone from any point on the event horizon (or the space-like singularity) and it encompasses only a finite time anywhere outside the black hole. The light beams are always travelling faster, so their $dt/dr$ is always shallower and hence it is possible for them to catch the falling observer as long as the delay in sending the light beam does not exceed a finite critical value. The handwaving reason is that from the perspective of Schwarzschild coordinate time both the falling person and light beams approach the event horizon along an exponential asymptote. The answer in all cases is that there is a finite time available, so no, the infalling person does not look back and see the cosmic aeons flying by. It is also often forgotten that it is not just light reaching them before the falling observer reaches the event horizon, but there is also a little bit more to add in the (proper) time it takes the falling observer to travel from the event horizon to the singularity. The answer of course depends on exactly how far the Earth is from the black hole.

So what will happen to the distances between the masses and what will the clocks show, relative to the others and a clock at infinity? It seems to me that the distances between the masses grows without limit while the proper clocktimes will differ too but not more than the time it takes for light to cover a Schwarzschild radius (which is about 1/100000 seconds for a hole with the mass of the Sun).You are essentially asking the following: if someone falls from the Earth from some way beyond the event horizon of a black hole, how long after they have left can an observer on Earth still signal to them with a light beam? On the back of these masses we have put point-like clocks (which sounds a bit odd but they are not affected by tidal forces). Along the radial distance in your freely falling falling frame that accompanies you, you place poni-like masses with equal distance between them. You start falling freely in a region where the hole's gravity is weak (so it doesn't take an infinite time to fall in. Do the particles end up stretched in space and time? Is the hole maybe evaporated (the evaporation radiation, or Hawking radiation, taking away your last state of being) before you end up anywhere? I'm a bit confused how things in the hole end up. Meanwhile in the universe outside the hole it seems you are stuck on the horizon and in the little proper time that has passed on your watch, the coordinate in the universe has run on speed, so to speak. Still it seems that you can never reach a particle that falls in front of you as it keeps on moving away from you while falling.

It only takes a small amount of time to reach the center, about the time it takes light to cross a Schwarzschild radius distance, which in the case of the Sun is about 1/100000 seconds. Falling freely into a black hole (assuming we survive or being point-sized) all particles in front of you are accelerating away from you in the direction of the singularity which seems flying towards you.

Proper time for someone falling radially into a black hole