4 ms·
No, because you can’t have something have that precisely defined of a position.
by moi2388 1y ago
No, because you can’t have something have that precisely defined of a position.
- chii 1y agoyou can - you merely need enough energy to precisely define it (as according to the heisenberg uncertainty principle)! Unless...such an energy exceeds the Schwarzschild radius...
- moi2388 1y agoSo you can’t.
- close04 1y agoYou can't halve a Planck length so you're limited to ~1.6×10^−35.
- lvncelot 1y agoI think current theories break down at less than a Planck length, but they are not constrained to integer multiples of it.
- altruios 1y agoRelativistic speeds can contract the length of any object as measured from an outside observer. If an object the size of 1 Planck length travels fast enough: you won't be able to measure it, as from your position it would be smaller than the Planck length as it passes by. It's not impossible (afaik) for things to be smaller than the Planck length. We just don't have the ability (maybe ever) to measure something smaller than this limit. Now, good luck finding something the size of 1 Planck length, and also to accelerate it to relativistic speeds.
- charcircuit 1y agoBy definition you can if you accept it's continuous.
- moi2388 1y agoNo, because space might be continuous, but that doesn’t mean the uncertainty principle and the Planck limit disappear..
- charcircuit 1y agoThose are only relevant for decoding it.
- auntienomen 1y agoThe Compton wavelength will probably cause trouble for the storage scheme long before gravity becomes a problem.