4 ms·

Yeah I'm not sure about this math. Midway through the article the author correctly notes that their original argument was wrong. The relevant factor here is n

Yeah I'm not sure about this math. Midway through the article the author correctly notes that their original argument was wrong. The relevant factor here is not the gravitational force, which scales as 1/d^2, but the tidal force, which scales as 1/d^3. But the author claims that after they make this correction, it only changes the result by about 30% or so. But given that they would have to be off by a factor of 1/d (which, in their example is ~4 x 10^-18). Now there's admittedly a non-linear factor here, but even still I have a hard time believing that a change in one of the inputs by 18 orders of magnitude will only result in a 30% change in the end result. Another hint that something is wrong with the math is that the author does not have a term for the width of the balloon, which is a relevant quantity when calculating the tidal force. Even though the math here is dodgy and the example is somewhat artificial, the same idea is super important when accounting for the effect of Galactic tides on the kinematics of stars and clusters in the Galaxy. Tides play a role in disrupting open clusters and dispersing young stars into the Galaxy.

pfdietz 5y ago
The point he made is that because the discrepancy in angle grows exponentially, the change from 1/d^2 to 1/d^3 has only a small effect on the time until the change builds up to O(1).
- antognini 5y ago
  I don't think there's the exponential dependence they claim. I think a more correct derivation would look like this: Let's suppose that we place a 1 g mass at 4 ly. How does this affect the internal dynamics of a helium balloon? The relevant change is the tidal acceleration induced by the 1 g mass: a_tide ~ GMl/r^3 where l is the mean free path in the balloon (l ~ 1 / n sigma), where n is the number density and sigma is the cross section of a helium atom. At room temperature, we have l ~ 4 x 10^-5 m. Now, we want to calculate the change in angle induced by the tidal force across a single mean free path. This is (ignoring factors of order unity) delta phi ~ delta x / l ~ al / v^2 Noting that the mean velocity is v ~ sqrt(kT / m_He), and substituting in the tidal acceleration, we have delta phi ~ GMml^2 / kTr^3 Given these numbers this gives us delta phi ~ 10^-78 At this point it should be pretty clear that this is not going to be a big effect. But let's see how long we would need to follow these interactions until the delta phi had built up to be large enough that a collision that would have happened doesn't. This is: phi_crit ~ r_He / l ~ 10^-6 Now, I think a critical error that the author made is in assuming that the delta phi's build up exponentially. There is a non-linear dependence here, but I don't think it's as fast as described. Consider, for example, an atom bouncing head-on between two stationary atoms. Classically, the atom would continue indefinitely. How long would it be before an atom perturbed by delta phi would miss one of the stationary atoms entirely? There will be a linear increase in delta phi over time, along with a non-linear increase of order ~y / r_He where y is the vertical displacement. But this non-linearity is essentially a cosine, which is very close to zero initially. So the non-linear term will be essentially irrelevant until the linear term has grown to be large enough (i.e., of order unity) that the non-linear cosine term can take over. So it will require of order 10^78 collisions, which given the velocity of the atom of ~1000 m/s, means that it will take ~10^68 years for the atom to miss. Now, due to other non-linearities in the system it would probably in practice be much less than this. But it would certainly be nowhere close to the few microseconds claimed by the author.
  - pfdietz 5y ago
    I think the dependence is clearly exponential. The discrepancy in the scattering angle should grow by a factor of O(mean free path/radius of scatterer) per scattering event.
    - antognini 5y ago
      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.

pfdietz 5y ago

The point he made is that because the discrepancy in angle grows exponentially, the change from 1/d^2 to 1/d^3 has only a small effect on the time until the change builds up to O(1).

antognini 5y ago
I don't think there's the exponential dependence they claim. I think a more correct derivation would look like this: Let's suppose that we place a 1 g mass at 4 ly. How does this affect the internal dynamics of a helium balloon? The relevant change is the tidal acceleration induced by the 1 g mass: a_tide ~ GMl/r^3 where l is the mean free path in the balloon (l ~ 1 / n sigma), where n is the number density and sigma is the cross section of a helium atom. At room temperature, we have l ~ 4 x 10^-5 m. Now, we want to calculate the change in angle induced by the tidal force across a single mean free path. This is (ignoring factors of order unity) delta phi ~ delta x / l ~ al / v^2 Noting that the mean velocity is v ~ sqrt(kT / m_He), and substituting in the tidal acceleration, we have delta phi ~ GMml^2 / kTr^3 Given these numbers this gives us delta phi ~ 10^-78 At this point it should be pretty clear that this is not going to be a big effect. But let's see how long we would need to follow these interactions until the delta phi had built up to be large enough that a collision that would have happened doesn't. This is: phi_crit ~ r_He / l ~ 10^-6 Now, I think a critical error that the author made is in assuming that the delta phi's build up exponentially. There is a non-linear dependence here, but I don't think it's as fast as described. Consider, for example, an atom bouncing head-on between two stationary atoms. Classically, the atom would continue indefinitely. How long would it be before an atom perturbed by delta phi would miss one of the stationary atoms entirely? There will be a linear increase in delta phi over time, along with a non-linear increase of order ~y / r_He where y is the vertical displacement. But this non-linearity is essentially a cosine, which is very close to zero initially. So the non-linear term will be essentially irrelevant until the linear term has grown to be large enough (i.e., of order unity) that the non-linear cosine term can take over. So it will require of order 10^78 collisions, which given the velocity of the atom of ~1000 m/s, means that it will take ~10^68 years for the atom to miss. Now, due to other non-linearities in the system it would probably in practice be much less than this. But it would certainly be nowhere close to the few microseconds claimed by the author.
- pfdietz 5y ago
  I think the dependence is clearly exponential. The discrepancy in the scattering angle should grow by a factor of O(mean free path/radius of scatterer) per scattering event.
  - antognini 5y ago
    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.