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If such a machine existed in our universe, as it sped up its execution, its parts would approach the speed of light. To continue speeding up its execution, it n
by cvoss 2y ago
If such a machine existed in our universe, as it sped up its execution, its parts would approach the speed of light. To continue speeding up its execution, it necessarily must occupy a smaller and smaller volume, to keep the parts nearer and nearer to each other. Continue further, and you realize that the total information content of the machine can't be sustained at any amount without the thing collapsing itself into a black hole. So, even if the machine, in any sense, "finishes" the computation, the output will live inside a black hole and thus be physically unknowable to us.
I find it so fascinating that fundamental properties about the laws of physics integrate so tightly with what is computable. The standard definition of compatibility involving Turing machines sounds kind of arbitrary, but those machines are representative a truly fundamental concept in our universe: that which is physically knowable, since you can build Turing machines out of the stuff in the universe.
Almost. Turing machines have unbounded tape. But the universe does not contain an unbounded amount of information or space with which to work. Sufficiently large Turing computations, though theoretically finite, are not realizable in our universe. Should such computations be considered decidable or undecidable?
- sqeaky 2y agoI am not sure if such is decidable or not, but your conjecture is decidedly against the spirit of the post. :p
- waveBidder 2y agoYou would probably enjoy Feynman's lectures on computation: https://www.goodreads.com/book/show/17274.Feynman_Lectures_On_Computation https://www.goodreads.com/book/show/17274.Feynman_Lectures_O...
- Quekid5 2y ago> Sufficiently large Turing computations, though theoretically finite, are not realizable in our universe. Should such computations be considered decidable or undecidable? In Practice, I think Undecidable is the only "correct" answer -- because our knowledge is still very incomplete. In Theory/Principle I think Decidable is the "correct" answer because there will very probably be limits that cannot be exceeded. ... but then, to reduce this to a simpler issue: The Halting Problem becomes trivial for any amount of finite state. It just takes a really, really long time (or a huge amount of space) to decide.
- Etherlord87 2y ago> Continue further, and you realize that the total information content of the machine can't be sustained at any amount without the thing collapsing itself into a black hole Non-sequitur here. You never argued why making a machine smaller and smaller makes it into a black hole. As you make circuits denser, you can also make them thinner, therefore avoiding mass increase.
- meroes 2y agoThe parent poster is correct. I’m not a physicist but Nima Arkani Hamed is and he talks about these limits all the time. Going smaller requires more and more energy. Why I’m not 100% sure but it does. At a certain smallness you get a black hole because of this (energy/volume) and more energy only gets you a bigger black hole believe or not. So there is a size limit according to our best theories.
- Etherlord87 2y ago> believe or not I thought this is not how science works. BTW I didn't say the OP is wrong, I said he committed a non-sequitur. Which would mean he's wrong in the sense his reasoning is wrong, but not in the sense his conclusion is false.
- gqcwwjtg 2y agoThe Shannon entropy itself actually is the issue, no matter how small the mass gets. https://en.m.wikipedia.org/wiki/Bekenstein_bound https://en.m.wikipedia.org/wiki/Bekenstein_bound
- m463 2y agomaybe the machine would collapse to a point to compute and rebound back to present results. Maybe call it single-mode computing.