3 ms·

I'm not sure this is as true as it seems. When creating simulations we do so by defining a computable function (ie., over discrete binary strings) which approx

I'm not sure this is as true as it seems. When creating simulations we do so by defining a computable function (ie., over discrete binary strings) which approximates some non-computable function (ie., over the reals). It is important to note that those non-computable functions are derived, in part, analytically based on theorems which require infinite precision, ie., real numbers. When we define a computable approximation we can only do so by narrowing the simulation to a scale where "the numbers have only a few significant digits". This is really the only thing we can simulate, and even then, we are using mathematical analysis to derive much of it -- which cannot be simulated. If you really wanted to simulate across scales, ie., run a simulation "from atoms to tables", i think we'd be requiring many many many decimal places. It would likely, i think, be physically impossible to do. (Or if not, then atoms to planets would be) -- as the discrete precision required would require a theoretical minimum of mass that would create a black hole. eg., Consider that a table has 10^verylargenumber of "spherical" atoms, and so its aggregate shape would be extremely sensitive to a miscalculation of each atoms shape ... and we would require a precision in PI probably something like 10^-verylargenumber.

wcoenen 5y ago
> 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.

wcoenen 5y ago

> 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.