5 ms·

So where and how does a jump from nice symmetric reversible equations to turbulent irreversibility happen?

MathMonkeyMan 1y ago
Even three bodies under newtonian gravity can lead to chaotic behavior. The neat part (assuming that the result is valid) is that precisely the equations of fluid dynamics result from their billiard ball models in the limit of many balls and frequent collisions.
- LudwigNagasena 1y ago
  But even millions of bodies under Newtonian gravity lead to reversible behaviour unlike Navier-Stokes.
  - MathMonkeyMan 1y ago
    The Navier-Stokes equations are a set of differential equations. The functions that the equations act upon are functions of time (and space), so the system is perfectly reversible. It's just hard to figure out what the functions are for a set of boundary conditions.
    - bubblyworld 1y ago
      This is not quite right. Time-reversibility means that solutions to your differential equation are invariant under the transformation x(t) -> x(-t). It's pretty easy to verify that is the case for simple differential equations like Newton's law: F = mx''(t) = mx''(-t) since d/dt x(-t) = -x'(-t), and d/dt (-x'(-t)) = x''(-t) Navier-Stokes is only time-reversible if you ignore viscosity, because viscosity is velocity-dependent and you can already see signs of that being a problem in the derivation above (velocity pops out a minus sign under time reversal). From my reading the OP managed to derive viscous flow too, so there really is a break in time-symmetry happening somewhere.
      - MathMonkeyMan 1y ago
        Now I get it, thanks for the explanation. I wonder if "t -> -t" is lost in the Boltzmann step or in the hydrodynamic step.
        doop 1y ago
        It's lost at Boltzmann's "molecular chaos" or "Stosszahlansatz" step. If f(x1,x2) is the two-particle distribution function giving you (hand-wavingly) the probability that you have particles with position and velocity coordinates x1 and others with coordinates x2, then Boltzmann made the simplification that f(x1,x2) = f(x1) * f(x2), ie throwing away all the correlations between particles. This is where the time-asymmetry comes in: you're saying that after two particles collide, they retain no correlation or memory of what they were doing beforehand.
        mannykannot 1y ago
        I assume (on the basis that it has not come up so far in this discussion and my limited further reading) that position-momentum uncertainty offers no justification for throwing away the correlations?
        bubblyworld 1y ago
        The systems we're talking about here are classical, not quantum, so the uncertainty principle isn't really relevant. I think the justification is mainly that it makes the analysis tractable. In physical terms it's simply not true that the interactions are uncorrelated, but you might hope that the correlations are "unimportant" in the long-term. In a really hot gas, for instance, everything is moving so fast in random directions that any correlations that start to arise will quickly get obliterated by chance.
        doop 1y ago
        I don't think it really helps - you're already working in something like a probabilistic formulation. If you want to use a quantum mechanical justification for it then you need to look at some sort of non-unitary evolution. Besides that, I don't think anybody is really arguing that the correlations are actually lost after a collision, just that it's usually a good approximation to treat them as if they are.
    - naasking 1y ago
      > The Navier-Stokes equations are a set of differential equations. The functions that the equations act upon are functions of time (and space), so the system is perfectly reversible. It's hard to take full reversibility seriously given Newton's equations are not actually deterministic. If they're not deterministic, then they can't be fully reversible. Of course maybe these non-deterministic regimes don't actually happen in realistic scenarios (like Norton's Dome), but maybe this is hinting at the fact that we need a better formalism for talking about these questions, and maybe that formalism will not be reversible in a specific, important way.
  - kgwgk 1y ago
    You lose the reversible behavior when you describe the system ignoring almost every degree of freedom.
bubblyworld 1y ago
I've been puzzling about this as well. The best answer I have (as an interested maths geek, not a physicist, caveat lector) is that it sneaks in under the assumption of "molecular chaos", i.e. that interactions of particles are statistically independent of any of their prior interactions. That basically defines an arrow of time right from the get-go, since "prior" is just a choice of direction. It also means that the underlying dynamics is not strictly speaking Newtonian any more (statistically, anyway).
- whatshisface 1y ago
  It comes about when the deterministic collision process is integrated over all the indistinguishable initial states that could lead into an equivalence class of indistinguishable final states. If you set the collision probability to zero it's time reversible even with molecular chaos, and if the particles are highly correlated (like in a polymer) there can still arise an arrow of time when the integral is performed.
  - bubblyworld 1y ago
    Interesting, so if I understand right you are saying that coarse-graining your states can produce an arrow of time on its own? Given some fixed coarse-graining, I can see that entropy would initially increase, since your coarse-graining hides information from you. The longer you evolve the system under this coarse graining the less certain you will be about the micro-states. But I would expect this to eventually reach an equilibrium where you are at "maximum uncertainty" with respect to your coarse graining. Does that sound right at all? And if so, then there must be something else responsible for the global arrow of time, right? > If you set the collision probability to zero it's time reversible even with molecular chaos Is this true for boring reasons? If nothing interacts then you just have a bunch of independent particles in free motion, which is obviously time-reversible. And also obviously satisfies molecular chaos because there are no correlations whatsoever. Maybe I misunderstand the terminology.
    - whatshisface 1y ago
      Yes, almost any coarse graining can lead to an arrow of time unless the map that represents a step forward in time aligns with the coarse graining perfectly. Also, yes, there are often equilibria, but that's the heat death of the universe, and admittedly time is hard to define even macroscopically if everything is at the same temperature. Chaos isn't even necessary, it just gets you there faster. The collisionless case is that way for boring reasons: the map aligns with the coarse graining.
bonvoyage36 1y ago
Strictly speaking, naturally on its own, it doesn't. Detailed equations remain reversible. Even for very big N, typical isolated classical mechanical systems are reversible. However, typical initial conditions imply transitions to equilibrium, or very long stay in it. The reversed process (ending in Poincare return) will happen eventually, but the time is so incredibly long, it can't be verified.
- bonvoyage36 1y ago
  In derivations of the Navier Stokes equations from reversible particle models, the former get their irreversibility from some approximation, e.g. a transition to a less detailed state and a simpler evolution equation for it is made. Often the actual microstate is replaced by some probabilistic description, such as probability density, or some kind of implied average.
rnhmjoj 1y ago
This has been known for a long time: the irreversibility comes from the assumption that the velocities of particles colliding are uncorrelated, or equivalently, that particles loose the "memory" of their complete trajectory between one collision and another. It's called the molecular chaos hypothesis. See https://en.wikipedia.org/wiki/Molecular_chaos https://en.wikipedia.org/wiki/Molecular_chaos

Even three bodies under newtonian gravity can lead to chaotic behavior. The neat part (assuming that the result is valid) is that precisely the equations of fluid dynamics result from their billiard ball models in the limit of many balls and frequent collisions.

LudwigNagasena 1y ago
But even millions of bodies under Newtonian gravity lead to reversible behaviour unlike Navier-Stokes.
- MathMonkeyMan 1y ago
  The Navier-Stokes equations are a set of differential equations. The functions that the equations act upon are functions of time (and space), so the system is perfectly reversible. It's just hard to figure out what the functions are for a set of boundary conditions.
  - bubblyworld 1y ago
    This is not quite right. Time-reversibility means that solutions to your differential equation are invariant under the transformation x(t) -> x(-t). It's pretty easy to verify that is the case for simple differential equations like Newton's law: F = mx''(t) = mx''(-t) since d/dt x(-t) = -x'(-t), and d/dt (-x'(-t)) = x''(-t) Navier-Stokes is only time-reversible if you ignore viscosity, because viscosity is velocity-dependent and you can already see signs of that being a problem in the derivation above (velocity pops out a minus sign under time reversal). From my reading the OP managed to derive viscous flow too, so there really is a break in time-symmetry happening somewhere.
    - MathMonkeyMan 1y ago
      Now I get it, thanks for the explanation. I wonder if "t -> -t" is lost in the Boltzmann step or in the hydrodynamic step.
      - doop 1y ago
        It's lost at Boltzmann's "molecular chaos" or "Stosszahlansatz" step. If f(x1,x2) is the two-particle distribution function giving you (hand-wavingly) the probability that you have particles with position and velocity coordinates x1 and others with coordinates x2, then Boltzmann made the simplification that f(x1,x2) = f(x1) * f(x2), ie throwing away all the correlations between particles. This is where the time-asymmetry comes in: you're saying that after two particles collide, they retain no correlation or memory of what they were doing beforehand.
        mannykannot 1y ago
        I assume (on the basis that it has not come up so far in this discussion and my limited further reading) that position-momentum uncertainty offers no justification for throwing away the correlations?
        bubblyworld 1y ago
        The systems we're talking about here are classical, not quantum, so the uncertainty principle isn't really relevant. I think the justification is mainly that it makes the analysis tractable. In physical terms it's simply not true that the interactions are uncorrelated, but you might hope that the correlations are "unimportant" in the long-term. In a really hot gas, for instance, everything is moving so fast in random directions that any correlations that start to arise will quickly get obliterated by chance.
        doop 1y ago
        I don't think it really helps - you're already working in something like a probabilistic formulation. If you want to use a quantum mechanical justification for it then you need to look at some sort of non-unitary evolution. Besides that, I don't think anybody is really arguing that the correlations are actually lost after a collision, just that it's usually a good approximation to treat them as if they are.
  - naasking 1y ago
    > The Navier-Stokes equations are a set of differential equations. The functions that the equations act upon are functions of time (and space), so the system is perfectly reversible. It's hard to take full reversibility seriously given Newton's equations are not actually deterministic. If they're not deterministic, then they can't be fully reversible. Of course maybe these non-deterministic regimes don't actually happen in realistic scenarios (like Norton's Dome), but maybe this is hinting at the fact that we need a better formalism for talking about these questions, and maybe that formalism will not be reversible in a specific, important way.
- kgwgk 1y ago
  You lose the reversible behavior when you describe the system ignoring almost every degree of freedom.

bubblyworld 1y ago

I've been puzzling about this as well. The best answer I have (as an interested maths geek, not a physicist, caveat lector) is that it sneaks in under the assumption of "molecular chaos", i.e. that interactions of particles are statistically independent of any of their prior interactions. That basically defines an arrow of time right from the get-go, since "prior" is just a choice of direction. It also means that the underlying dynamics is not strictly speaking Newtonian any more (statistically, anyway).

whatshisface 1y ago
It comes about when the deterministic collision process is integrated over all the indistinguishable initial states that could lead into an equivalence class of indistinguishable final states. If you set the collision probability to zero it's time reversible even with molecular chaos, and if the particles are highly correlated (like in a polymer) there can still arise an arrow of time when the integral is performed.
- bubblyworld 1y ago
  Interesting, so if I understand right you are saying that coarse-graining your states can produce an arrow of time on its own? Given some fixed coarse-graining, I can see that entropy would initially increase, since your coarse-graining hides information from you. The longer you evolve the system under this coarse graining the less certain you will be about the micro-states. But I would expect this to eventually reach an equilibrium where you are at "maximum uncertainty" with respect to your coarse graining. Does that sound right at all? And if so, then there must be something else responsible for the global arrow of time, right? > If you set the collision probability to zero it's time reversible even with molecular chaos Is this true for boring reasons? If nothing interacts then you just have a bunch of independent particles in free motion, which is obviously time-reversible. And also obviously satisfies molecular chaos because there are no correlations whatsoever. Maybe I misunderstand the terminology.
  - whatshisface 1y ago
    Yes, almost any coarse graining can lead to an arrow of time unless the map that represents a step forward in time aligns with the coarse graining perfectly. Also, yes, there are often equilibria, but that's the heat death of the universe, and admittedly time is hard to define even macroscopically if everything is at the same temperature. Chaos isn't even necessary, it just gets you there faster. The collisionless case is that way for boring reasons: the map aligns with the coarse graining.

bonvoyage36 1y ago

Strictly speaking, naturally on its own, it doesn't. Detailed equations remain reversible. Even for very big N, typical isolated classical mechanical systems are reversible. However, typical initial conditions imply transitions to equilibrium, or very long stay in it. The reversed process (ending in Poincare return) will happen eventually, but the time is so incredibly long, it can't be verified.

bonvoyage36 1y ago
In derivations of the Navier Stokes equations from reversible particle models, the former get their irreversibility from some approximation, e.g. a transition to a less detailed state and a simpler evolution equation for it is made. Often the actual microstate is replaced by some probabilistic description, such as probability density, or some kind of implied average.

rnhmjoj 1y ago

This has been known for a long time: the irreversibility comes from the assumption that the velocities of particles colliding are uncorrelated, or equivalently, that particles loose the "memory" of their complete trajectory between one collision and another. It's called the molecular chaos hypothesis. See https://en.wikipedia.org/wiki/Molecular_chaos https://en.wikipedia.org/wiki/Molecular_chaos