3 ms·
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 i
by 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 agoWhat about F = -k x^2/m?
- messe 5y agoWhat about it? x^2 is a C^\infty function, differentiable and continuous everywhere.
- davrosthedalek 5y agoWait, 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 agoSorry, 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.