3 ms·

> This is posed as a computational resource problem, but it strikes me as an information problem. I thought the point of the gas illustration was to show how t

> This is posed as a computational resource problem, but it strikes me as an information problem. I thought the point of the gas illustration was to show how the assumption there's an information problem (i.e. heat, second law) is actually not correct. That it only looks like an information problem and it's really a computation problem. The theory being that if you can compute were the molecules are going to be, from the initial state or from interactions you have already learned from, then the motions don't appear random any more. There are no surprises; you have "decrypted" the apparently random movements. It's just to illustrate the idea, and an immediate objection would be "but we can't know everything to that much detail". That is addressed by a more subtle version of the argument, which says: Although you don't know all the motions precisely, your ability to compute motions from the information you obtained so far gives you progressively increasing knowledge about motions locally or which you recognise as related, and causes "regions of effective coherence" to expand. It's effective coherence not actual coherence, because the molecule motions don't change, only the precision with which you can anticipate some of them as well as relationships between them. What would have appeared random, now with the benefit of some prior information and computation resolves gradually into local clusters of more predictable related motions, even if you don't know every motion accurately. With the result that the effective fluid properties change, so your ability to "swim" through the gas changes. In the 2d closed box model, with perfect balls and perfect interactions (i.e. a mathematically perfect simulation) it's plausible that this may work perfectly. That is, if you have your own "special" ball and it undergoes a number of collisions and you get perfect measurement of those collisions, eventually you end up with enough information to model the contents of the rest of the box. If in that model you can dynamically adjust something about the collisions of your "special" ball, for example changing the ball's shape, mass or radius, it's plausible that can be used to travel anywhere in the box much faster than diffusion, but only if you have the information up to that point and excellent computation - which might be irreducibly hard computation for a reasonably sized box.

a1369209993 6y ago
> an immediate objection would be "but we can't know everything to that much detail". No, the immediate objection would be "but it's physically impossible to know everything to that much detail, because you don't have enough bits of storage[0], and also because of Heisenberg's uncertainty principle". (Both objections are suffient on their own to make this not work except possibly for a homogenous spherical molecule-shaped unphysically-light and -compact hypercomputer in a frictionless vacuum.) 0: That is, it's physically impossible to pack enough bits to describe a cloud of gas onto a storage medium massing significantly less than the entire cloud of gas.
- jlokier 6y ago
  > because you don't have enough bits of storage Inside the computer. That's what makes it a computational reducibility question and not a measurement information-availability objection. (Also: For a twist, assume you have a quantum computer and they are quantum balls.) > Heisenberg's uncertainty principle This raises questions, certainly, but the answers aren't obvious when talking about repeated interactions with the many particles. In the box model, the balls are inevitably entangled with each other at the position-momentum level due to their collisions, even if that entanglement is undetectable in an analogous way to how their motions appear "random" classically. Heisenberg does not apply to each ball independently when they are entangled. In this box model, as your little computer/mind/demon accumulates information-in-principle from many interactions, in addition to classical information it couples to that entangled state, and the independence of Heisenberg limits dissolves because they aren't really independent. (Also: Once you invoke Heisenberg, you've also invoked quantum particles in a box self-interfering. In a box that reduces the amount of information you need to represent a single particle's state to an integer, bounded if the energy is bounded. I'm not sure if that also applies to multiple particles interacting chaotically.) > except possibly for a homogenous spherical molecule in a frictionless vacuum. Well, the model actually is about homogeneous spherical molecules, and vacuum at the molecular level is frictionless, so that's ok :-)
  - a1369209993 6y ago
    > > because you don't have enough bits of storage Edited to clairify, thanks. > the particles are inevitably entangled with each other That increases the information content, from O(N) for N particles to (worst case) O(2^N). Using a quantum computer at best reduces that back down to O(N) qubits. > the model actually is about homogeneous spherical molecules I don't think electron orbitals are spherical enough for that, much less nuclei or polyatomic molecules, but edited anyway.
    - jlokier 6y ago
      > That increases the information content, from O(N) for N particles to (worst case) O(2^N). Classical (model) molecules have infinite bits of information: Their motion parameters are "analogue", which you can think of as real numbers, or infinite precision numbers. For a digital computer, that takes infinite bits unless you know of a constraint upon them. You can't call that O(N). And you can't say they have mass proportional to that kind of information either, because that would be infinite mass. The quantum molecules are in a bounded box. Individual eigenstates are constrained by quantum mechanics in a box into bound states, which are countably enumerable as integers. If you have an upper bound on the energy of the entire box contents, there's a maximum integer required, therefore finite bits to encode an eigenstate. The coefficients associated to each eigenstate in the general wavefunction are complex (and therefore infinite bits), while subject to various constraints, but they are unobservable. Observations select among eigenstates, each of which is represented in finite bits. So is it infinite (like the classical model) or finite? But observation is meaningless in the "special ball is a quantum computer" model. How much information does the "special ball" need from its environment, if it's allowed to entangle with that environment, to outsmart the quantized chaos around itself? Qubits linked to physical measurements and actions are full of paradoxes arising from the mathematics, which makes thought experiments useful. Where does the computation even take place, given that entanglement makes qubits non-local? In the special ball, or in all the balls it's entangled with, affecting them all subtly? I don't think this information question is simple enough to hand-wave as O(N) or O(2^N).

a1369209993 6y ago

> an immediate objection would be "but we can't know everything to that much detail". No, the immediate objection would be "but it's physically impossible to know everything to that much detail, because you don't have enough bits of storage[0], and also because of Heisenberg's uncertainty principle". (Both objections are suffient on their own to make this not work except possibly for a homogenous spherical molecule-shaped unphysically-light and -compact hypercomputer in a frictionless vacuum.) 0: That is, it's physically impossible to pack enough bits to describe a cloud of gas onto a storage medium massing significantly less than the entire cloud of gas.

jlokier 6y ago
> because you don't have enough bits of storage Inside the computer. That's what makes it a computational reducibility question and not a measurement information-availability objection. (Also: For a twist, assume you have a quantum computer and they are quantum balls.) > Heisenberg's uncertainty principle This raises questions, certainly, but the answers aren't obvious when talking about repeated interactions with the many particles. In the box model, the balls are inevitably entangled with each other at the position-momentum level due to their collisions, even if that entanglement is undetectable in an analogous way to how their motions appear "random" classically. Heisenberg does not apply to each ball independently when they are entangled. In this box model, as your little computer/mind/demon accumulates information-in-principle from many interactions, in addition to classical information it couples to that entangled state, and the independence of Heisenberg limits dissolves because they aren't really independent. (Also: Once you invoke Heisenberg, you've also invoked quantum particles in a box self-interfering. In a box that reduces the amount of information you need to represent a single particle's state to an integer, bounded if the energy is bounded. I'm not sure if that also applies to multiple particles interacting chaotically.) > except possibly for a homogenous spherical molecule in a frictionless vacuum. Well, the model actually is about homogeneous spherical molecules, and vacuum at the molecular level is frictionless, so that's ok :-)
- a1369209993 6y ago
  > > because you don't have enough bits of storage Edited to clairify, thanks. > the particles are inevitably entangled with each other That increases the information content, from O(N) for N particles to (worst case) O(2^N). Using a quantum computer at best reduces that back down to O(N) qubits. > the model actually is about homogeneous spherical molecules I don't think electron orbitals are spherical enough for that, much less nuclei or polyatomic molecules, but edited anyway.
  - jlokier 6y ago
    > That increases the information content, from O(N) for N particles to (worst case) O(2^N). Classical (model) molecules have infinite bits of information: Their motion parameters are "analogue", which you can think of as real numbers, or infinite precision numbers. For a digital computer, that takes infinite bits unless you know of a constraint upon them. You can't call that O(N). And you can't say they have mass proportional to that kind of information either, because that would be infinite mass. The quantum molecules are in a bounded box. Individual eigenstates are constrained by quantum mechanics in a box into bound states, which are countably enumerable as integers. If you have an upper bound on the energy of the entire box contents, there's a maximum integer required, therefore finite bits to encode an eigenstate. The coefficients associated to each eigenstate in the general wavefunction are complex (and therefore infinite bits), while subject to various constraints, but they are unobservable. Observations select among eigenstates, each of which is represented in finite bits. So is it infinite (like the classical model) or finite? But observation is meaningless in the "special ball is a quantum computer" model. How much information does the "special ball" need from its environment, if it's allowed to entangle with that environment, to outsmart the quantized chaos around itself? Qubits linked to physical measurements and actions are full of paradoxes arising from the mathematics, which makes thought experiments useful. Where does the computation even take place, given that entanglement makes qubits non-local? In the special ball, or in all the balls it's entangled with, affecting them all subtly? I don't think this information question is simple enough to hand-wave as O(N) or O(2^N).