71–80 of 106 posts
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> [Quantum Cognition] is a real field with dozen of collaborators and even a textbook. Flat Earth is also a real field, with conferences with hundreds of attendees.
Here's the textbook he wrote, 2nd edition
https://www.cambridge.org/us/universitypress/subjects/psycho...
Looks totally respectable. Why do you feel the need to ridicule it?
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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-asymm…
Earlier quoted context omitted.
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-asymm…
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?
So where and how does a jump from nice symmetric reversible equations to turbulent irreversibility happen?
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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-d…
Sabine Hossenfelder's video on this: https://youtu.be/mxWJJl44UEQ
In my perception Sabine’s quality degraded over the last year or so. Maybe it’s also the topics she covers. I’m not sure why she is getting into fantasies of AGI for example. I liked the skeptical version of her better.
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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-d…
equations can be time-symmetric, or invariant re time reversal. What you're describing is equations being invariant re time reversal.
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.
Earlier quoted context omitted.
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-asymm…
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?
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.
So where and how does a jump from nice symmetric reversible equations to turbulent irreversibility happen?
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.
Earlier quoted context omitted.
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.