Estimate 4e28 total nucleons in the human body (60Kg of nucleons) it takes about 100 bits to describe the number of nucleons. each nucleon takes about two bits of information to specify type (proton, neutron, antiproton, antineutron)
Figure that a human is about 6′x2′x1′; that’s 1e35x4e34x2e34 plank units. With 8e108 unique locations within that rectangle, each nucleon needs 362 bits to describe location.
Without discussing momentum, it already takes 10^41 bits to describe the location of every nucleon. Any benefit you gain from the distribution of matter not being uniform you lose when describing what the distribution of matter is in a human.
With a discussion of momentum, there is no way to describe an arbitrary nucleon in fixed space, since there is no upper limit to the amount of momentum of a nucleon.
How did you estimate an exabyte for the information content of a human?
How did you estimate an exabyte for the information content of a human?
Googled it. If I had to guess why the number I found was so much smaller, it’s probably because they had a scheme more like “for each molecule, describe it and place it with precision much better than its thermal vibration,” and maybe a few bits to describe temperature.
But yes, even 10^your number of bits will be smaller than 10^^5. Even if we tracked each possible quantum state that’s localized in my body, which would be exponentially more intensive than tracking individual particles, that just means we might have to (but probably not) bump up to a reward of 10^^6 in order to swamp my prior probabilities.
Describe momentum in fixed space. It is acceptable to have an upper limit on momentum, provided that that limit can be arbitrarily high. It is not acceptable for two different vectors to be described differently.
I’m not sure what the smallest unit of angular measure is, or if it is continuous. If it is continuous, there’s no way to describe momentum in finite bits.
An exabyte? Really? 8e18 bits?
(Most values to one sig fig)
Estimate 4e28 total nucleons in the human body (60Kg of nucleons) it takes about 100 bits to describe the number of nucleons. each nucleon takes about two bits of information to specify type (proton, neutron, antiproton, antineutron) Figure that a human is about 6′x2′x1′; that’s 1e35x4e34x2e34 plank units. With 8e108 unique locations within that rectangle, each nucleon needs 362 bits to describe location.
Without discussing momentum, it already takes 10^41 bits to describe the location of every nucleon. Any benefit you gain from the distribution of matter not being uniform you lose when describing what the distribution of matter is in a human.
With a discussion of momentum, there is no way to describe an arbitrary nucleon in fixed space, since there is no upper limit to the amount of momentum of a nucleon.
How did you estimate an exabyte for the information content of a human?
Googled it. If I had to guess why the number I found was so much smaller, it’s probably because they had a scheme more like “for each molecule, describe it and place it with precision much better than its thermal vibration,” and maybe a few bits to describe temperature.
But yes, even 10^your number of bits will be smaller than 10^^5. Even if we tracked each possible quantum state that’s localized in my body, which would be exponentially more intensive than tracking individual particles, that just means we might have to (but probably not) bump up to a reward of 10^^6 in order to swamp my prior probabilities.
Whatever the information content is, unless you have an argument that it’s actually infinite, it’s going to be smaller than 3^^^^3.
Describe momentum in fixed space. It is acceptable to have an upper limit on momentum, provided that that limit can be arbitrarily high. It is not acceptable for two different vectors to be described differently.
I’m not sure what the smallest unit of angular measure is, or if it is continuous. If it is continuous, there’s no way to describe momentum in finite bits.