3 ms·

I've thought about it, and as far as I can tell, there is no need to construct this dome in such a strange way. A trajectory away from from an unstable equilibr

I've thought about it, and as far as I can tell, there is no need to construct this dome in such a strange way. A trajectory away from from an unstable equilibrium will always start out with zero kinetic energy, and I can't think of any cases where you couldn't stitch together no motion at the equilibrium for t<0 with motion away from it after that to get a solution that obeyed Newton's laws and conservation of energy at all times. That's not a resolution to the philosophical issue, but it could help us come up with one, given that some other systems are so much simpler but have the same property. So, if anyone wants to think about this, you can think about x''(x) = -x^2 just as productively, without the extra complexity.

messe 5y ago
This is a response to part of your comment before you deleted and re-commented. I just thought it was an interesting point to add, and your comment helps give it some context. > [...] I always thought the only necessary initial conditions were position and velocity or position and momentum, a case is presented where that doesn't happen. It's implied in the exposition of Hamiltonian mechanics that those are the only variables that describe a particle at a point in time, and that's how it is treated in computational simulations. There's usually an implicit assumption that the functions you're dealing with are sufficiently smooth (differentiable as much as you want at all points). In Norton's dome, the derivative of the height of the surface (the slope) does not exist at r=0 because d/dr sqrt(r) = 1/(2sqrt(r)).
- whatshisface 5y ago
  What about F = -k x^2/m?
  - messe 5y ago
    What about it? x^2 is a C^\infty function, differentiable and continuous everywhere.
- davrosthedalek 5y ago
  Wait, no, height h propto r^3/2, so dh/dr propto r^1/2. So the slope is well defined, I think. The derivative of the slope isn't, but that's fine.
  - messe 5y ago
    Sorry, you're right. Off-by-one error on my part. The second derivative still doesn't exist though, which means the function isn't smooth.

messe 5y ago

This is a response to part of your comment before you deleted and re-commented. I just thought it was an interesting point to add, and your comment helps give it some context. > [...] I always thought the only necessary initial conditions were position and velocity or position and momentum, a case is presented where that doesn't happen. It's implied in the exposition of Hamiltonian mechanics that those are the only variables that describe a particle at a point in time, and that's how it is treated in computational simulations. There's usually an implicit assumption that the functions you're dealing with are sufficiently smooth (differentiable as much as you want at all points). In Norton's dome, the derivative of the height of the surface (the slope) does not exist at r=0 because d/dr sqrt(r) = 1/(2sqrt(r)).

whatshisface 5y ago
What about F = -k x^2/m?
- messe 5y ago
  What about it? x^2 is a C^\infty function, differentiable and continuous everywhere.
davrosthedalek 5y ago
Wait, no, height h propto r^3/2, so dh/dr propto r^1/2. So the slope is well defined, I think. The derivative of the slope isn't, but that's fine.
- messe 5y ago
  Sorry, you're right. Off-by-one error on my part. The second derivative still doesn't exist though, which means the function isn't smooth.