3 ms·

> Consider that a table has 10^verylargenumber of "spherical" atoms That "verylargenumber" exponent is not actually a large number. It's probably close to 28.

> Consider that a table has 10^verylargenumber of "spherical" atoms That "verylargenumber" exponent is not actually a large number. It's probably close to 28. E.g. for a rough estimate, pretend that the table has a mass of 120kg and is made of only carbon atoms. 120000 gram / (12 gram/mol) = 10000 mol = 6x10^27 < 10^28. On a related note, the estimated number of atoms in the visible universe is around 10^80.

mjburgess 5y ago
So the volume of one atom can be calculated using, c. 30 dp. of pi -- +3 so that errors do not compound. So we need 30 bits/atom to store the volume, for 10^30 atoms -- this is 10^15 petabytes. Suppose we're trying to estimate the shape of the table using the relevant state vector for each atom (ie., radius, coordinate, EM-shell, etc.). Ie., how these atoms interact to produce a structure. Then we need perhaps 10 numbers/atom, so 10^16 petabytes. And consider the computation. The issue is that macroscopic structuring effects, ie., that the table has its tabular shape also exert a top-down effect constraining the structure of the atoms. We'd have to solve that problem through iteration. Ie., we compute a shape from local effects only, then iteratively recompute until increasing orders of scale are included. Suppose we perform 1 operation/atom to compute an interaction, that's 10^30 operations. At 1ns/op, we have 10^21 seconds -- optimistically. That's one scale, now we need to iterate; given the orders of effects are 1nm...1m that's c. 10 scales, say. We're now at 10^22 seconds with 10^16 petabytes of data. And also, since we're running a 10-fold iteration we need more than 10^30 precision, as errors will compound. This is for a table!. As I said, choose your scale, eg., atom to star, and the storage requirments have an essentially physically impossible mass.

mjburgess 5y ago

So the volume of one atom can be calculated using, c. 30 dp. of pi -- +3 so that errors do not compound. So we need 30 bits/atom to store the volume, for 10^30 atoms -- this is 10^15 petabytes. Suppose we're trying to estimate the shape of the table using the relevant state vector for each atom (ie., radius, coordinate, EM-shell, etc.). Ie., how these atoms interact to produce a structure. Then we need perhaps 10 numbers/atom, so 10^16 petabytes. And consider the computation. The issue is that macroscopic structuring effects, ie., that the table has its tabular shape also exert a top-down effect constraining the structure of the atoms. We'd have to solve that problem through iteration. Ie., we compute a shape from local effects only, then iteratively recompute until increasing orders of scale are included. Suppose we perform 1 operation/atom to compute an interaction, that's 10^30 operations. At 1ns/op, we have 10^21 seconds -- optimistically. That's one scale, now we need to iterate; given the orders of effects are 1nm...1m that's c. 10 scales, say. We're now at 10^22 seconds with 10^16 petabytes of data. And also, since we're running a 10-fold iteration we need more than 10^30 precision, as errors will compound. This is for a table!. As I said, choose your scale, eg., atom to star, and the storage requirments have an essentially physically impossible mass.