4 ms·

I think that for extremely small perturbations the discrepancy will be linear until it reaches a critical threshold, at which point the trajectories diverge exp

I think that for extremely small perturbations the discrepancy will be linear until it reaches a critical threshold, at which point the trajectories diverge exponentially. Basically, you need to get to the point where the path of one atom diverges by ~atomic radius. But until that happens, there is no non-linear mechanism to increase the divergence. For example, if you just consider a billiard ball bouncing between two walls, the divergence in trajectories will only ever grow linearly. But for very small perturbations, the case of collisions between atoms is effectively the same as collisions between walls --- the curvature of the scattering atom can be neglected. After all, the curvature here is a second order term, and when the perturbation is of the order of 10^-68, the non-linear factor will be (10^-68)^2 ~ 10^-136.

pfdietz 5y ago
> I think that for extremely small perturbations the discrepancy will be linear until it reaches a critical threshold An error of epsilon on the angle of a path toward the scatter will cause an error of ~(distance to scatter/size of scatter) epsilon in the angle from the center of the scatterer to the "impact point". So the errors should grow exponentially in the number of scatterings. The (effective) curvature of the scatterer CANNOT be neglected -- it is crucial. Modeling this with large flat reflectors is not appropriate.
- antognini 5y ago
  I spent a bit more time thinking about this, and yes you are correct. Each scattering event increases the discrepancy by ~mean free path / atomic radius, which is about five orders of magnitude. So even starting from a 10^-78 discrepancy, it only takes ~10 collisions to get an order unity discrepancy. I was thinking that the discrepancy due to collisions was negligible compared to the accumulated tidal acceleration, but this is clearly wrong.
  - r_parthasarathy 5y ago
    I'm glad this makes sense; thanks for writing!
  - pfdietz 5y ago
    About three orders of magnitude, not five, but yeah.
r_parthasarathy 5y ago
pfdietz is correct. The key thing is that divergences always multiply, leading to an exponential dependence on number of scattering events. There isn't a transition between a linear and an exponential dependence. I think where you're getting confused is that an exponential can be initially approximated by a linear function (i.e. exp(x) ~= 1 + x), but that doesn't mean that the process is linear. The dominance of the exponential growth is also why the tidal correction, as massive as it seems, gets swamped and ends up not mattering much. -- Raghu Parthasarathy

pfdietz 5y ago

> I think that for extremely small perturbations the discrepancy will be linear until it reaches a critical threshold An error of epsilon on the angle of a path toward the scatter will cause an error of ~(distance to scatter/size of scatter) epsilon in the angle from the center of the scatterer to the "impact point". So the errors should grow exponentially in the number of scatterings. The (effective) curvature of the scatterer CANNOT be neglected -- it is crucial. Modeling this with large flat reflectors is not appropriate.

antognini 5y ago
I spent a bit more time thinking about this, and yes you are correct. Each scattering event increases the discrepancy by ~mean free path / atomic radius, which is about five orders of magnitude. So even starting from a 10^-78 discrepancy, it only takes ~10 collisions to get an order unity discrepancy. I was thinking that the discrepancy due to collisions was negligible compared to the accumulated tidal acceleration, but this is clearly wrong.
- r_parthasarathy 5y ago
  I'm glad this makes sense; thanks for writing!
- pfdietz 5y ago
  About three orders of magnitude, not five, but yeah.

r_parthasarathy 5y ago

pfdietz is correct. The key thing is that divergences always multiply, leading to an exponential dependence on number of scattering events. There isn't a transition between a linear and an exponential dependence. I think where you're getting confused is that an exponential can be initially approximated by a linear function (i.e. exp(x) ~= 1 + x), but that doesn't mean that the process is linear. The dominance of the exponential growth is also why the tidal correction, as massive as it seems, gets swamped and ends up not mattering much. -- Raghu Parthasarathy