4 ms·
1bit of information stored at 300K has a mass of 10^-38kg according to [0]. That’s a few orders of magnitude off but almost uncannily close. [0] https://aip.sc
by _Microft 5y ago
1bit of information stored at 300K has a mass of 10^-38kg according to [0]. That’s a few orders of magnitude off but almost uncannily close.
[0] https://aip.scitation.org/doi/10.1063/1.5123794 https://aip.scitation.org/doi/10.1063/1.5123794
- zzt123 5y agoHold up. The article shows a mass difference of 10^-38 for the matter/antimatter states. That’s not a few orders of magnitude off your number, that’s literally your exponent.
- throw0101a 5y agoThis reminds me of a 2004 weblog post by one of the co-creators of ZFS, Jeff Bonwick, on why 128 bits was chosen: > Some customers already have datasets on the order of a petabyte, or 2^50 bytes. Thus the 64-bit capacity limit of 2^64 bytes is only 14 doublings away. Moore's Law for storage predicts that capacity will continue to double every 9-12 months, which means we'll start to hit the 64-bit limit in about a decade. Storage systems tend to live for several decades, so it would be foolish to create a new one without anticipating the needs that will surely arise within its projected lifetime. > If 64 bits isn't enough, the next logical step is 128 bits. That's enough to survive Moore's Law until I'm dead, and after that, it's not my problem. But it does raise the question: what are the theoretical limits to storage capacity? > Although we'd all like Moore's Law to continue forever, quantum mechanics imposes some fundamental limits on the computation rate and information capacity of any physical device. In particular, it has been shown that 1 kilogram of matter confined to 1 liter of space can perform at most 10^51 operations per second on at most 10^31 bits of information [see Seth Lloyd, "Ultimate physical limits to computation." Nature 406, 1047-1054 (2000)]. A fully-populated 128-bit storage pool would contain 2^128 blocks = 2^137 bytes = 2^140 bits; therefore the minimum mass required to hold the bits would be (2^140 bits) / (10^31 bits/kg) = 136 billion kg. > That's a lot of gear. > To operate at the 10^31 bits/kg limit, however, the entire mass of the computer must be in the form of pure energy. By E=mc^2, the rest energy of 136 billion kg is 1.2x10^28 J. The mass of the oceans is about 1.4x10^21 kg. It takes about 4,000 J to raise the temperature of 1 kg of water by 1 degree Celcius, and thus about 400,000 J to heat 1 kg of water from freezing to boiling. The latent heat of vaporization adds another 2 million J/kg. Thus the energy required to boil the oceans is about 2.4x10^6 J/kg * 1.4x10^21 kg = 3.4x10^27 J. Thus, fully populating a 128-bit storage pool would, literally, require more energy than boiling the oceans. * https://blogs.oracle.com/bonwick/128-bit-storage:-are-you-high https://blogs.oracle.com/bonwick/128-bit-storage:-are-you-hi...
- Keyframe 5y agoA fully-populated 128-bit storage pool would contain 2^128 blocks = 2^137 bytes = 2^140 bits; therefore the minimum mass required to hold the bits would be (2^140 bits) / (10^31 bits/kg) = 136 billion kg. 136 billion kg: - ≈ 0.64 × mass of trash produced in the United States in one year ( ≈ 2.36×10^8 sh tn ) - ≈ 0.35 × estimated wet biomass of all humans alive ( ≈ 385 Mt ) - ≈ 1.3 × estimated dry biomass of all humans alive ( 105 Mt ) thanks wolframalpha!
- _Microft 5y agoAnother one: 136 billion kg is the mass of a block of water with a footprint of a square kilometer and a height of 136 meters. Put this way, it does not sound that much anymore, in my opinion.
- arbitrage 5y agomoores law hasn't been accurate in decades, and it was widely regarded as a boondoggle of a throwaway statement for the press well before that, too.