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Comment by LudwigNagasena | Hacker News Reader

But even millions of bodies under Newtonian gravity lead to reversible behaviour unlike Navier-Stokes.

You lose the reversible behavior when you describe the system ignoring almost every degree of freedom.

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.

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.

Now I get it, thanks for the explanation.

I wonder if "t -> -t" is lost in the Boltzmann step or in the hydrodynamic step.

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.

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?

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.

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.

The former. Diffusion in gases is similar.

equations can be time-symmetric, or invariant re time reversal. What you're describing is equations being invariant re time reversal.

You can call this invariance under time reflection if you like, yeah.

Note that the solutions x(t) are not generally time symmetric. We aren't saying that x(t)=x(-t), we are saying that x(t) is a solution to the differential equation if and only if x(-t) is, which is a weaker statement.

I know what you meant; I've just tried to point out an error in your sentence which pops up sometimes, which may have mislead others. It's all about the time reversal invariance of evolution equations, not solutions.

Oh I see what you mean, it's kinda easy to read my comment as meaning time symmetry. But I do think the phrasing in terms of solutions is correct, provided you interpret it appropriately. As in "is still a solution to the diff eq after transformation" and not "is left unchanged by the transformation".

It's not a good phrasing to express the point, because "solution is invariant under operation O" has an established meaning, that the solution does no change after the operation. What you mean can be properly phrased as "equations are time-reversal invariant".

You've convinced me =)

> 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.

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