12 ms·

> we conclude that up to 5 per cent of such triples would require an accuracy of smaller than the Planck length in order to produce a time-reversible solution,

> we conclude that up to 5 per cent of such triples would require an accuracy of smaller than the Planck length in order to produce a time-reversible solution, thus rendering them fundamentally unpredictable. Hence, incidentally, classical physics is itself non-deterministic and non-computable.

AnotherGoodName 3y ago
Feels like it's just another way quantum mechanics non determinism bubbles up into the larger world. So not that all that insightful right?
- mjburgess 3y ago
  I don't think it requires any QM to see. I used to have an argument that you could derive non-determinism from the resolution to Zeno's paradox, but i've forgot the steps. Roughly, you need to measure infinitely precisely to give an infinitely precise value to a variable. Via Zeno, continuous time precludes infinite measurement precision (there are no 'moments') and hence infinite precision in measurement. All chaos is doing here is giving us a bound on the actual lack of precision reality has -- I take it, via a zeno-ish argument above, that reality has to have such a bound. Relevantly, none of this requires any reasoning from QM premises or observations. This paper adds-to-the-pile cases where (classical) chaotic systems require measurement beyond-possible spatio-temporal resolution to be deterministic.
  - daxfohl 3y ago
    Classical mechanics does have nondeterministic cases even without appeals to mathematical paradoxes. See https://en.m.wikipedia.org/wiki/Norton%27s_dome https://en.m.wikipedia.org/wiki/Norton%27s_dome as the most well-known example.
    - mjburgess 3y ago
      It's unclear whether that's a concequence of poorly specified premises of the classical mathematical framework. Here, I mean that 'classical reality' as specified in a framework of basic applied mathematics, with no QM premises, produces non-determinism -- via only showing that (classical) chaotic systems exist
    - tsimionescu 3y ago
      Replied elsewhere, but Norton's dome has no real implications for classical mechanics as a full physical theory. It goes away as soon as you add the idea that matter is composed of fixed shape atoms, without even making any assumptions about the shape or size of said atoms - and this was a well-known, if often unstated, assumption even at the time of Newton.
      - btilly 3y ago
        And as soon as you have a rigid 3-body collision, indeterminism comes back. If A hits B before C, you generally get a different result than if A hits C before B. If A hits B and C at the same time, a range of solutions are possible, including both limiting solutions of the previous two variations. The extreme example of this is the indeterminism of a break shot in billiards.
        tsimionescu 3y ago
        Sure, but perfectly rigid bodies are also not compatible with an atomic model of matter, they are just an abstraction used in the maths. If you wanted to model an actual physical experiment, you would model any physical object as two or more rigid bodies connected by springs of an arbitrary small length*. So, anyone using classical mechanics for experimental purposes rather than explorations of the maths would model the collision of three balls in a way where simultaneity is impossible. * This is just like when modeling actual circuits, you model a physical wire as an ideal wire plus a resistor. But we don't say that "classical wire diagrams predict that the current through a wire can be infinite" just because you can draw a circuit with 0 resistance - that circuit modal is simply non-physical.
        btilly 3y ago
        But perfectly rigid atoms are compatible with an atomic model of matter. And in such a model, the indeterminism comes right back. Yes, the idea of perfectly rigid atoms is incompatible with experiment. But so is classical mechanics on the scale where that fact becomes obvious. And so it isn't classical mechanics that is deterministic. You are right that n-body classical mechanics combined with the assumption of a continuous response does become deterministic by the standard existence and uniqueness theorems of ordinary differential equations. But the point still remains, classical mechanics winds up not being deterministic without that extra assumption. And that extra assumption is often omitted in real models of things like elastic collisions.
amai 3y ago
The definition of a Planck length implies Quantum Mechanics. Pure classical physics doesn't have a smallest length scale. Classical mechanics is deterministic.
- codethief 3y ago
  A priori, the Planck length is a a length like any other. It's perfectly fine to talk about that scale in classical physics, without involving QM. > Pure classical physics doesn't have a smallest length scale. Quantum mechanics doesn't have a smallest length scale, either, as far as we know.
  - amai 3y ago
    No, Planck length is not a length like any other. Any other length gets Lorentz contracted, but the Planck length does not. It is a constant in any reference frame. And if you assume hbar->0 as for classical mechanics the Planck length will be zero. So classical mechanics doesn't have a finite Planck length. Quantum mechanics doesn't have a smallest length scale, but Quantum Electrodynamics has, namely the Compton wavelength. And this is what goes into the derivation of the Planck length. So with only classical theory you simply can't derive the Planck length. Classical theory doesn't have a smallest length scale.
    - codethief 3y ago
      > Any other length gets Lorentz contracted, but the Planck length does not. Are you claiming that the value of the constant l_P is the same in every reference frame or that objects of length equal to l_P will have the same length (= l_P) in every reference frame? The former is trivially true, the latter is definitely false unless you assume new physics. See also the discussion in https://physics.stackexchange.com/questions/4094/is-the-planck-length-lorentz-invariant https://physics.stackexchange.com/questions/4094/is-the-plan... > but Quantum Electrodynamics has, namely the Compton wavelength Citation needed. (Besides, the Compton wavelength of what particle exactly?) > So with only classical theory you simply can't derive the Planck length. So classical mechanics doesn't have a finite Planck length. I don't need to derive anything. In classical mechanics I am free to define a constant with value equal to 1.616255(18)×10−35 m and call it the Planck length. Whether or not it has any physical significance is irrelevant. (I mean, questions of physical significance so far have never stopped proponents of the Planck length in QM, either.)
codethief 3y ago
Being unpredictable and being non-deterministic are two completely different things. Classical physics might be the former but it's definitely not the latter.

AnotherGoodName 3y ago

Feels like it's just another way quantum mechanics non determinism bubbles up into the larger world. So not that all that insightful right?

mjburgess 3y ago
I don't think it requires any QM to see. I used to have an argument that you could derive non-determinism from the resolution to Zeno's paradox, but i've forgot the steps. Roughly, you need to measure infinitely precisely to give an infinitely precise value to a variable. Via Zeno, continuous time precludes infinite measurement precision (there are no 'moments') and hence infinite precision in measurement. All chaos is doing here is giving us a bound on the actual lack of precision reality has -- I take it, via a zeno-ish argument above, that reality has to have such a bound. Relevantly, none of this requires any reasoning from QM premises or observations. This paper adds-to-the-pile cases where (classical) chaotic systems require measurement beyond-possible spatio-temporal resolution to be deterministic.
- daxfohl 3y ago
  Classical mechanics does have nondeterministic cases even without appeals to mathematical paradoxes. See https://en.m.wikipedia.org/wiki/Norton%27s_dome https://en.m.wikipedia.org/wiki/Norton%27s_dome as the most well-known example.
  - mjburgess 3y ago
    It's unclear whether that's a concequence of poorly specified premises of the classical mathematical framework. Here, I mean that 'classical reality' as specified in a framework of basic applied mathematics, with no QM premises, produces non-determinism -- via only showing that (classical) chaotic systems exist
  - tsimionescu 3y ago
    Replied elsewhere, but Norton's dome has no real implications for classical mechanics as a full physical theory. It goes away as soon as you add the idea that matter is composed of fixed shape atoms, without even making any assumptions about the shape or size of said atoms - and this was a well-known, if often unstated, assumption even at the time of Newton.
    - btilly 3y ago
      And as soon as you have a rigid 3-body collision, indeterminism comes back. If A hits B before C, you generally get a different result than if A hits C before B. If A hits B and C at the same time, a range of solutions are possible, including both limiting solutions of the previous two variations. The extreme example of this is the indeterminism of a break shot in billiards.
      - tsimionescu 3y ago
        Sure, but perfectly rigid bodies are also not compatible with an atomic model of matter, they are just an abstraction used in the maths. If you wanted to model an actual physical experiment, you would model any physical object as two or more rigid bodies connected by springs of an arbitrary small length*. So, anyone using classical mechanics for experimental purposes rather than explorations of the maths would model the collision of three balls in a way where simultaneity is impossible. * This is just like when modeling actual circuits, you model a physical wire as an ideal wire plus a resistor. But we don't say that "classical wire diagrams predict that the current through a wire can be infinite" just because you can draw a circuit with 0 resistance - that circuit modal is simply non-physical.
        btilly 3y ago
        But perfectly rigid atoms are compatible with an atomic model of matter. And in such a model, the indeterminism comes right back. Yes, the idea of perfectly rigid atoms is incompatible with experiment. But so is classical mechanics on the scale where that fact becomes obvious. And so it isn't classical mechanics that is deterministic. You are right that n-body classical mechanics combined with the assumption of a continuous response does become deterministic by the standard existence and uniqueness theorems of ordinary differential equations. But the point still remains, classical mechanics winds up not being deterministic without that extra assumption. And that extra assumption is often omitted in real models of things like elastic collisions.

amai 3y ago

The definition of a Planck length implies Quantum Mechanics. Pure classical physics doesn't have a smallest length scale. Classical mechanics is deterministic.

codethief 3y ago
A priori, the Planck length is a a length like any other. It's perfectly fine to talk about that scale in classical physics, without involving QM. > Pure classical physics doesn't have a smallest length scale. Quantum mechanics doesn't have a smallest length scale, either, as far as we know.
- amai 3y ago
  No, Planck length is not a length like any other. Any other length gets Lorentz contracted, but the Planck length does not. It is a constant in any reference frame. And if you assume hbar->0 as for classical mechanics the Planck length will be zero. So classical mechanics doesn't have a finite Planck length. Quantum mechanics doesn't have a smallest length scale, but Quantum Electrodynamics has, namely the Compton wavelength. And this is what goes into the derivation of the Planck length. So with only classical theory you simply can't derive the Planck length. Classical theory doesn't have a smallest length scale.
  - codethief 3y ago
    > Any other length gets Lorentz contracted, but the Planck length does not. Are you claiming that the value of the constant l_P is the same in every reference frame or that objects of length equal to l_P will have the same length (= l_P) in every reference frame? The former is trivially true, the latter is definitely false unless you assume new physics. See also the discussion in https://physics.stackexchange.com/questions/4094/is-the-planck-length-lorentz-invariant https://physics.stackexchange.com/questions/4094/is-the-plan... > but Quantum Electrodynamics has, namely the Compton wavelength Citation needed. (Besides, the Compton wavelength of what particle exactly?) > So with only classical theory you simply can't derive the Planck length. So classical mechanics doesn't have a finite Planck length. I don't need to derive anything. In classical mechanics I am free to define a constant with value equal to 1.616255(18)×10−35 m and call it the Planck length. Whether or not it has any physical significance is irrelevant. (I mean, questions of physical significance so far have never stopped proponents of the Planck length in QM, either.)

codethief 3y ago

Being unpredictable and being non-deterministic are two completely different things. Classical physics might be the former but it's definitely not the latter.