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Re:Newtonian physics IS deterministic (sorry Norton) (2017) by messe on Sunday August 15, 2021 @05:02PM (#28192514)
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)).

Re:Newtonian physics IS deterministic (sorry Norton) (2017) by davrosthedalek on Sunday August 15, 2021 @06:14PM (#28193031)
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.
Re:Newtonian physics IS deterministic (sorry Norton) (2017) by messe on Sunday August 15, 2021 @07:19PM (#28193478)
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.
Re:Newtonian physics IS deterministic (sorry Norton) (2017) by whatshisface on Sunday August 15, 2021 @05:10PM (#28192563)
What about F = -k x^2/m?
Re:Newtonian physics IS deterministic (sorry Norton) (2017) by messe on Sunday August 15, 2021 @05:54PM (#28192897)
What about it? x^2 is a C^\infty function, differentiable and continuous everywhere.