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Comment by Aeolun | Hacker News Reader

Huh? If this is the case, wouldn’t a 2 body problem also be practically impossible to calculate?

I mean, you can certainly still predict to a certain (probably high) level of accuracy, but ultimately that motion is also influenced by factors outside your model.

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The question is what happens with a small perturbation and how does it "grow".

For a two body problem, you nudge one of the bodies and it is forever off by a small amount, but your predictions into infinity require only a small adjustment to compensate.

For a three body problem you nudge one of the bodies and only for a very short time do your previous predictions stay true, the change amplifies until nothing you thought might happen before the nudge means anything at all, and a common occurrence is one of the bodies being ejected.

> a common occurrence is one of the bodies being ejected.

So the problem eventually solves itself?

Well, not if the ejected body contains you.

Being ejected from your model is unquestionably a solution of sorts, although it does Raise Questions.

Yes, but but also the problem really is: "given some initial condition space, can we estimate the likelihood of ejection (how many conditions in the condition space eject) in X timeframe"?

> For a two body problem, you nudge one of the bodies and it is forever off by a small amount, but your predictions into infinity require only a small adjustment to compensate.

How does this even work? Where is a place in the universe where there are only two bodies?? Where would the nudge come from if not from a third body?! A ghost?

You're taking it too literally. The 'nudge' means a change in initial conditions, and could result e.g. from measurement uncertainty, not necessarily a literal nudge.

It also doesn't necessarily matter that much if there are more than two bodies, if the gravitational influence of other bodies is small enough, then you can model as a two body problem.

The “nudge” can also come up even with exact starting information. When you calculate the next step and round it to the nearest nanometer, you’ve just nudged the system by enough to eventually make your prediction worthless in the 3-body case.

The relative magnitude of interactions is important. The article specifically mentioned triple star system as an example. Solar system is of course many body, but (a) it has already settled into (sort of) stable configuration (b) there is one massive body and everything else revolves around it. Yes, massive bodies like gas giants have noticeable affect on other bodies, but due to the distances they tend to be in the stable territory.

You do an accounting of the nearby massive objects and their distances from each other, pairs of objects within certain bounds of mass and distance can be treated like there are only those two objects in the universe with the understanding that there will be a small error because of outside influences. It’s an approximation, but in the right circumstances a very good one.

Each planet and the sun can be done like this, ignoring all of the other planets. Each moon and its planet can be considered a two body system ignoring the rest of the moons.

If you just randomly generated a bunch of massive bodies and pressed play, you would have few 2 body systems and a lot of chaos, but that’s a problem that solves itself as the chaos results in either collisions or ejections.

This is sometimes called "infinite sensitivity to initial conditions" or "deterministic chaos".

No. A small error in the initial conditions of a 2-body system produces a small error in the result. In case of a 3-body system, a small error will result in drastically different outcomes. The phase space of a 2-body system is nice and smooth, but a 3-body system’s is more like a fractal.

For a two-body problem the discrepancies between your model and reality increase gradually with time, but it's still possible to predict eclipses decades in advance with a precise albeit imperfect measurement of initial conditions.

With a three-body problem any slight shift causes a wildly different trajectory, bearing no resemblance to the original so your measurements of the initial condition have to be perfect.

No that's not it. The two body problem has a closed, analytical solution, the three body one doesn't, so you need to simulate it. It's a fundamentally different approach.

Both of you are correct.

That's not really the issue. The three body problem does have an analytical solution in the form of a power series, but the problem is that it converges so slowly to be of any practical use.

It is though, if you don’t have a closed form solution you need to use an iterative process to calculate the positions, meaning errors will accrue over time. For a closed form solution that wouldn’t be the case.

Thanks for mentioning the existence of an analytical solution at all though, I wasn't aware of that.

There are iterative methods/systems that stabilize over time. For example, symplectic integrators on tame problems oscillate lightly around the true energy of the system over time. The issue here is the properties of the underlying problem, not the set of solution methods.

> meaning errors will accrue over time

This is not universally true. Error behavior is a function of the particular problem, the algorithm used to approximate its solution, and the properties of input data. A large subtopic of numerical analysis is concerned with this kind of stuff. See [1] or [2] to get a flavor.

[1]: https://en.wikipedia.org/wiki/Numerical_stability

[2]: https://en.wikipedia.org/wiki/Lax_equivalence_theorem

arent eclipses three body problems?

For all practical purposes eclipses are two two-body problems ... Sun vs Earth-Moon pair, and Earth vs Moon. The scale of the Sun is so vastly larger and distances so great from sun to earth there's no need to do 3-body.

If the three had roughly the same mass would be different story.

Better stated, the 2-body problem can be solved with a finite number of standard operations, i.e. a closed-form expression. This solution does not exist for the 3-body problem.

This is not a requirement for a system to lack chaos, nor is it a metric by which we can judge if a system DOES have chaos.

In principle - yes. Except that we can change our frame of reference, and treat the two body problem as a pseudo one body problem (the lab frame becomes the center of mass of one of the bodies). One cannot do this for the three body problem, which gives us at best a pseudo two body problem.

No, the fundamental difference is that the two-body problem can be solved analytically, you can write down a formula. For three bodies and up you only have numerical solutions, simulations, and they will break down over time.

I think this is accurate and I don’t understand why you are being downvoted.

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.

Conversely, there are also extremely simple closed-form recurrence relations that exhibit chaotic behavior, e.g. the logistic map.

True 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.

Of 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.

Would a linear-feedback shift register be an example of a recurrence relation that exhibits chaotic behavior?

Well 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.

The 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.

It'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.

That is not the issue. There are many problems that do not have analytical solutions, but can be approximated numerically to any precision you would like.

The issue with the three-body problem is that it is chaotic, meaning any error will eventually grow to take over the entire solution, making prediction impossible, even in theory. Every chaotic system lacks an analytical solution, but not every system without an analytical solution is chaotic.

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