Is it possible to create a wormhole exit without knowing how to do so? If so, how likely is it that there is a wormhole somewhere within listening range?
As for checking the answers, I use the gold standard of reliability: did it work? If it does work, the answer is sent back to the initiating point. If it doesn’t work, send the next answer in the countable answer space back.
If the answer can’t be shown to be in a countable answer space (the countable answer space includes every finite sequence of bits, and therefore is larger than the space of the possible outputs of every Turing Machine), then don’t ask the question. I’m not sure what question you could ask that can’t be answered in a series of bits.
Of course, that means that the first (and probably) last message sent back through time will be some variant of “Do not mess with time” It would take a ballsy engineer indeed to decide that the proper response to trying the solution “Do not mess with time” is to conclude that it failed and send the message “Do not mess with timf”
My very limited understanding is that wormholes only make logical sense with two endpoints. They are, quite literally, a topological feature of space that is a hole in the same sense as a donut has a hole. Except that the donut only has a two dimensional surface, unlike spacetime.
My mostly unfounded assumption is that other time traveling schemes are likely to be similar.
How do you plan to answer the question “did it work?” with an error rate lower than, say, 2^-100? What happens if you accidentally hit the wrong button? No one has ever tested a machine of any sort to that standard of reliability, or even terribly close. And even if you did, you still haven’t done well enough to send a 126 bit message, such as “Do not mess with time” with any reliability.
Is it possible to create a wormhole exit without knowing how to do so? If so, how likely is it that there is a wormhole somewhere within listening range?
As for checking the answers, I use the gold standard of reliability: did it work? If it does work, the answer is sent back to the initiating point. If it doesn’t work, send the next answer in the countable answer space back.
If the answer can’t be shown to be in a countable answer space (the countable answer space includes every finite sequence of bits, and therefore is larger than the space of the possible outputs of every Turing Machine), then don’t ask the question. I’m not sure what question you could ask that can’t be answered in a series of bits.
Of course, that means that the first (and probably) last message sent back through time will be some variant of “Do not mess with time” It would take a ballsy engineer indeed to decide that the proper response to trying the solution “Do not mess with time” is to conclude that it failed and send the message “Do not mess with timf”
My very limited understanding is that wormholes only make logical sense with two endpoints. They are, quite literally, a topological feature of space that is a hole in the same sense as a donut has a hole. Except that the donut only has a two dimensional surface, unlike spacetime.
My mostly unfounded assumption is that other time traveling schemes are likely to be similar.
How do you plan to answer the question “did it work?” with an error rate lower than, say, 2^-100? What happens if you accidentally hit the wrong button? No one has ever tested a machine of any sort to that standard of reliability, or even terribly close. And even if you did, you still haven’t done well enough to send a 126 bit message, such as “Do not mess with time” with any reliability.
I ask the future how they will did it.
I was going to say “bootstraps don’t work that way”, but since the validation happens on the future end, this might actually work.