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Summary of the Ford paradox (which the post takes forever to actually get to): Quantum mechanics has only periodic orbits, but chaotic systems aren't periodic.

Summary of the Ford paradox (which the post takes forever to actually get to): Quantum mechanics has only periodic orbits, but chaotic systems aren't periodic. How do we reconcile this with the existence of, say, double pendulums? My guess: because the abstract double pendulum is a model that only approximates real double pendulums. Analogously, computers based on Turing Machines exist despite Turing Machines requiring an infinite amount of space. --- Also, a minor nitpick: the post seems to equate 'not periodic' or 'chaotic' with 'is random', which I think is misleading. Pseudo-random, I guess, but the output of a simulation of a chaotic system is highly compressible and thus not random in the information theoretic sense.

Fargren 14y ago
Thank you; your minor nitpick doesn't seem minor at all. It was keeping me from noticing I don't see what the problem is. The movement here is deterministic, it's just hard to predict. I don't see the paradox at all.
- codeulike 14y ago
  Exactly. Chaos can be deterministic. Its sensitive dependence on initial conditions (among other things) that makes it chaotic.
  - DougBTX 14y ago
    When simulating a chaotic system, using different rounding for the floating point values will affect the simulation by a small amount, which a chaotic system is liable to amplify. Quantisation of reality seems to be analogous to rounding floats, in the sense that it affects the precision of the measurements. I'd be interested to know whether that affects the "running" of the system like with floats in a simulation, or if it only puts a limit on how well the initial conditions can be defined.
  - arjie 14y ago
    Doesn't the traditional definition _require_ it to be deterministic? I recall James Gleick describing it that way.
  - Kapura 14y ago
    Chaos is deterministic, but I'm always told that some quantum things are NOT deterministic, i.e. a single photon being exposed to two slits in a paper. My understanding is that the opening it goes through is random; how can that be reconciled with deterministic models of the universe?
    - codeulike 14y ago
      Well, thats a different question. http://philosophy.stackexchange.com/questions/1210/in-which-way-does-quantum-mechanics-disprove-determinism http://philosophy.stackexchange.com/questions/1210/in-which-...
vannevar 14y ago
...the output of a simulation of a chaotic system is highly compressible... Is that really true? Losslessly? Do you mean the entire phase space, or the sequence generated on any given run? Because I was under the impression that what distinguishes a chaotic system is that the output cannot be predicted in any way other than running the simulation (the article uses the word 'random', but it might be better to say 'unpredictable'). Compressibility relies on exploiting regularities, and regularity is predictability, isn't it?
- Strilanc 14y ago
  See: Kolmogorov Complexity ( http://en.wikipedia.org/wiki/Kolmogorov_complexity http://en.wikipedia.org/wiki/Kolmogorov_complexity ) The program that simulates the chaotic system is shorter than the unbounded amount of output being generated, and counts as a compression of the output.
  - Gravityloss 14y ago
    Sure, but that is kind of circular. I guess in other words, the more interesting question is this: can we generate hard to compress pseudorandom sequences with simple algorithms - if the algorithm description by itself can not be used as "compression"?
    - MostAwesomeDude 14y ago
      Absolutely. Consider any number which can be expressed as a patterned continued fraction, like pi or e. pi's digits are such that it is very hard to compress, but it is absolutely trivial to generate. (http://en.wikipedia.org/wiki/Continued_fraction http://en.wikipedia.org/wiki/Continued_fraction)
      - Fargren 14y ago
        How is the algorithm that generates the digits of pi a compressed form of pi? It gives me all the information about the number in finite amount of time. In randomness theory, the definition of the most compressed form of a sequence S is, given a computer C, the shortest program with output S in C. So the program that generates the output of the simulation of the pendulums is a compression of it's output. Since it's and infinite sequence that can be compressed into a finite program, it is not random from the perspective of information theory or randomness theory.
Xcelerate 14y ago
> My guess: because the abstract double pendulum is a model that only approximates real double pendulums. This subject is somewhat outside of my area of knowledge, but if I had to guess, this is what resolves the paradox. An abstract mathematical chaotic system is different than a real-life one, and it is my guess that if we performed the experiment, we would find the orbits would be different. QM has withstood every test thrown at it so far, and while it's always possible that a correction to it will be found, I think it will be a very rare event should it happen.

Fargren 14y ago

Thank you; your minor nitpick doesn't seem minor at all. It was keeping me from noticing I don't see what the problem is. The movement here is deterministic, it's just hard to predict. I don't see the paradox at all.

codeulike 14y ago
Exactly. Chaos can be deterministic. Its sensitive dependence on initial conditions (among other things) that makes it chaotic.
- DougBTX 14y ago
  When simulating a chaotic system, using different rounding for the floating point values will affect the simulation by a small amount, which a chaotic system is liable to amplify. Quantisation of reality seems to be analogous to rounding floats, in the sense that it affects the precision of the measurements. I'd be interested to know whether that affects the "running" of the system like with floats in a simulation, or if it only puts a limit on how well the initial conditions can be defined.
- arjie 14y ago
  Doesn't the traditional definition _require_ it to be deterministic? I recall James Gleick describing it that way.
- Kapura 14y ago
  Chaos is deterministic, but I'm always told that some quantum things are NOT deterministic, i.e. a single photon being exposed to two slits in a paper. My understanding is that the opening it goes through is random; how can that be reconciled with deterministic models of the universe?
  - codeulike 14y ago
    Well, thats a different question. http://philosophy.stackexchange.com/questions/1210/in-which-way-does-quantum-mechanics-disprove-determinism http://philosophy.stackexchange.com/questions/1210/in-which-...

vannevar 14y ago

...the output of a simulation of a chaotic system is highly compressible... Is that really true? Losslessly? Do you mean the entire phase space, or the sequence generated on any given run? Because I was under the impression that what distinguishes a chaotic system is that the output cannot be predicted in any way other than running the simulation (the article uses the word 'random', but it might be better to say 'unpredictable'). Compressibility relies on exploiting regularities, and regularity is predictability, isn't it?

Strilanc 14y ago
See: Kolmogorov Complexity ( http://en.wikipedia.org/wiki/Kolmogorov_complexity http://en.wikipedia.org/wiki/Kolmogorov_complexity ) The program that simulates the chaotic system is shorter than the unbounded amount of output being generated, and counts as a compression of the output.
- Gravityloss 14y ago
  Sure, but that is kind of circular. I guess in other words, the more interesting question is this: can we generate hard to compress pseudorandom sequences with simple algorithms - if the algorithm description by itself can not be used as "compression"?
  - MostAwesomeDude 14y ago
    Absolutely. Consider any number which can be expressed as a patterned continued fraction, like pi or e. pi's digits are such that it is very hard to compress, but it is absolutely trivial to generate. (http://en.wikipedia.org/wiki/Continued_fraction http://en.wikipedia.org/wiki/Continued_fraction)
    - Fargren 14y ago
      How is the algorithm that generates the digits of pi a compressed form of pi? It gives me all the information about the number in finite amount of time. In randomness theory, the definition of the most compressed form of a sequence S is, given a computer C, the shortest program with output S in C. So the program that generates the output of the simulation of the pendulums is a compression of it's output. Since it's and infinite sequence that can be compressed into a finite program, it is not random from the perspective of information theory or randomness theory.

Xcelerate 14y ago

> My guess: because the abstract double pendulum is a model that only approximates real double pendulums. This subject is somewhat outside of my area of knowledge, but if I had to guess, this is what resolves the paradox. An abstract mathematical chaotic system is different than a real-life one, and it is my guess that if we performed the experiment, we would find the orbits would be different. QM has withstood every test thrown at it so far, and while it's always possible that a correction to it will be found, I think it will be a very rare event should it happen.