6 ms·
The problem is that a three-body system is inherently unstable/chaotic. So even if you run a numerical simulation with the same granularity for a two-body and a
by phreeza 5y ago
The problem is that a three-body system is inherently unstable/chaotic. So even if you run a numerical simulation with the same granularity for a two-body and a three-body system, the three-body simulation will degrade much faster than the two-body system. This is unrelated to the fact that there is a closed form solution, there are many stable, non-chaotic dynamical systems that don't have a closed form solution.
- MontyCarloHall 5y agoConversely, there are also extremely simple closed-form recurrence relations that exhibit chaotic behavior, e.g. the logistic map.
- phreeza 5y agoTrue but a recurrence relationship is not the same as a closed form solution. The differential equation for a three body problem is also very simple.
- MontyCarloHall 5y agoOf course. My point was just that you can evaluate a recurrence relation exactly (i.e. with zero numerical error) and still get chaotic behavior. OP’s mistaken point was that the three body problem’s chaos arises solely from numerical error during simulation, which is untrue.
- acjohnson55 5y agoWould a linear-feedback shift register be an example of a recurrence relation that exhibits chaotic behavior?
- Grustaf 5y agoWell sure. But my point was that for two bodies you don’t need to simulate at all, you can just get the answer for any point in time.
- phreeza 5y agoThe need for numerical simulations is completely unrelated to the predictability and/or stability of the system. If you have a Lyapunov exponent smaller than 0 everywhere, errors don't accumulate and you can simulate numerically for as long as you like.
- Grustaf 5y agoIt's not unrelated, but the implication only goes in one direction. If you have a closed form solution, that implies that you can model it completely forever, but sure the opposite is not true.