4 ms·

It goes like this: We have two observers with a computer. A is outside a black hole. B goes over the event horizon. B is infinitely time dilated seen from A. It

It goes like this: We have two observers with a computer. A is outside a black hole. B goes over the event horizon. B is infinitely time dilated seen from A. It takes forever for B to reach the singularity from A standpoint. B reaches the middle in a finite time. A starts computation. If it halts A sends a result, otherwise it won't. B sees the result in a finite time. If it doesn't, the program didn't halt. If time is discrete, it won't fly I think. This works because there is no smallest time unit in gr. We are working with different types of infinities. A's computational steps take, the further B goes in, less time. Sort of Zeno's paradox. It is easy to map all natural numbers between 0 and 1 on the real line. Just not 1 to 1. There are more problems. How to get the information out and how to survive the divergent blue shift, it is somewhat unclear. B cannot talk back. But still a cool find.

JadeNB 2y ago
> It is easy to map all natural numbers between 0 and 1 on the real line. Just not 1 to 1. For the usual meaning of the term, you certainly can construct a 1-to-1 (that is, injective) map N \to [0, 1] (for example, n \mapsto 10^(-n)); the natural numbers just can't be mapped onto [0, 1] (that is, the map can't be surjective). That's the opposite of the problem we have: it's saying you can losslessly encode a countable amount of information in an uncountable amount of space; but I'm saying conversely that you can't perform an uncountable number of steps in a countably infinite amount of time.
- pino999 2y ago
  But the computation steps are certainly countable, even if there are infinite. The amount of steps in time we can take is uncountable. We can always find a point between two other times.Time dilation does exactly this, making the time a computational step take smaller and smaller.

JadeNB 2y ago

> It is easy to map all natural numbers between 0 and 1 on the real line. Just not 1 to 1. For the usual meaning of the term, you certainly can construct a 1-to-1 (that is, injective) map N \to [0, 1] (for example, n \mapsto 10^(-n)); the natural numbers just can't be mapped onto [0, 1] (that is, the map can't be surjective). That's the opposite of the problem we have: it's saying you can losslessly encode a countable amount of information in an uncountable amount of space; but I'm saying conversely that you can't perform an uncountable number of steps in a countably infinite amount of time.

pino999 2y ago
But the computation steps are certainly countable, even if there are infinite. The amount of steps in time we can take is uncountable. We can always find a point between two other times.Time dilation does exactly this, making the time a computational step take smaller and smaller.