4 ms·

With the time reversal trick, it almost seems like you could say the cause happens in the future and propagates backwards in time? I'm curious that if it doesn

With the time reversal trick, it almost seems like you could say the cause happens in the future and propagates backwards in time? I'm curious that if it doesn't work for a hemispherical dome because of requiring infinite time, does that mean a ball on a hemispherical dome is stable and won't spontaneously roll off?

abdullahkhalids 2y ago
No. What all this shows is that both cases (hemispherical dome or the other one in OP) of a ball in stable equilibrium are outside the domain of applicability of the theory. You should simply not use the theory as is to make predictions about when and how the ball will roll off.
- tsimionescu 2y ago
  Actually, there is nothing mysterious here. The theory, applied correctly, predicts the expected thing: a particle that starts at rest at the apex will never leave it. The other "solution" is not actually a particle at rest (it has a non-0 6th derivative of position, that is the rate of change of the rate of change of its acceleration is 1/6). So it's actually describing a particle whose acceleration will increase at an increasing rate forever, but that just happens to be 0 at time T.
  - abdullahkhalids 2y ago
    Yes. The situation is correctly described by a modified version of standard Newton's theory, in which you start caring about higher derivatives.
- foxglacier 2y ago
  Just in case I'm completely wrong about reality, both cases are unstable equilibrium, aren't they? Equilibrium in that forces balance and zero acceleration but unstable in that any disturbance will lead to it rolling off rather than returning to where it was.
bennythomsson 2y ago
Yes, if it's rolling on a hemispherical dome then it's always been rolling, unless an external force pushed it.
wat10000 2y ago
Time reversible systems make causality fuzzy. There’s nothing special about this example in that respect. You can’t object have ball A hit ball B. A causes B to move. You can then reverse it, end up with an equivalent but time reversed system where B hits A, and B causes A to move.

abdullahkhalids 2y ago

No. What all this shows is that both cases (hemispherical dome or the other one in OP) of a ball in stable equilibrium are outside the domain of applicability of the theory. You should simply not use the theory as is to make predictions about when and how the ball will roll off.

tsimionescu 2y ago
Actually, there is nothing mysterious here. The theory, applied correctly, predicts the expected thing: a particle that starts at rest at the apex will never leave it. The other "solution" is not actually a particle at rest (it has a non-0 6th derivative of position, that is the rate of change of the rate of change of its acceleration is 1/6). So it's actually describing a particle whose acceleration will increase at an increasing rate forever, but that just happens to be 0 at time T.
- abdullahkhalids 2y ago
  Yes. The situation is correctly described by a modified version of standard Newton's theory, in which you start caring about higher derivatives.
foxglacier 2y ago
Just in case I'm completely wrong about reality, both cases are unstable equilibrium, aren't they? Equilibrium in that forces balance and zero acceleration but unstable in that any disturbance will lead to it rolling off rather than returning to where it was.

bennythomsson 2y ago

Yes, if it's rolling on a hemispherical dome then it's always been rolling, unless an external force pushed it.

wat10000 2y ago

Time reversible systems make causality fuzzy. There’s nothing special about this example in that respect. You can’t object have ball A hit ball B. A causes B to move. You can then reverse it, end up with an equivalent but time reversed system where B hits A, and B causes A to move.