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I'm new to this "integrable system" stuff, but from what I've read, if there are as many linearly independent constants of motion that are compatible with respect to the poisson brackets as degrees of freedom, then the system is solvable in terms of elementary functions. Is this correct? I get that for each linearly independent constant of motion you can reduce the degree of freedom by one, but I don't understand why the theorem

Theorem (First integrals of the n-body problem) The only linearly independent integrals of the $n$-body problem, which are algebraic with respect to $q$, $p$ and $t$ are the $10$ described above. (http://en.wikipedia.org/wiki/N-body_problem#Three-body_problem)

implies that there is no analytic solution (I think this is synonymous with closed-form solution, and solution in terms of elementary functions). I've been trying to think about it, but I can't reason it, and apparently integrability implies no chaos, which I can't see either.

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    Have you read http://en.wikipedia.org/wiki/Integrable_system ?2012-05-17
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    I had understood that a perfectly elastic simultaneous collision between three bodies could not be solved - momentum and energy are conserved, but this does not determine the subsequent motion.2012-05-17
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    @MarkBennet Can you specify what are the initial conditions of this system which does not determine the subsequent motion?2012-05-17
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    @Xnyyrznaa: I think the idea was that, given an energy E, you could fire three point particles of from the origin in a variety of ways having zero total momentum: particles of equal mass at equal speeds and at $120^o$ angles will illustrate. Reverse one of these motions. How can you tell how the particles emerge?2012-05-17
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    Fun fact: the quantum mechanics 3-body problem *is* solvable!2012-05-17
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    @Mark: while that is true, the set of initial conditions which leads to a three-body simultaneous collision has (at least morally) measure zero, which I don't think is completely the resolution to the question asked.2012-05-18
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    Roughly speaking, the idea is counting the number of degrees of freedom. For each particle you have 6 degrees of freedom (3 for position and 3 for momentum). At three or more particles you have at least 18 degrees and only 10 integrals, so you don't have closed form solutions. How about 2 particles you ask? As it turns out you can cheat a bit there: by working in the centre of mass reference frame, you kill 6 integrals (0 center of mass and 0 linear momentum). However, conservation of momentum means that the two body evolution is only two dimensional! So for the two body problem you really ...2012-05-18
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    ...only have 4 degrees of freedom, which is just enough to kill using the remaining integrals of motion (conservation of angular momentum and conservation of energy). For three and more bodies you don't have such nice reductions.2012-05-18
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    @WillieWong Indeed - I put it in a comment for that reason. It was a one-liner which I remember from a lecture once. There is an article in the Princeton Companion to Mathematics which suggests that Sundman produced a series solution for cases with non-zero angular momentum (which avoids triple collisions).2012-05-18
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    "Integrable by quadrature", which is the classical (Liouville) notion of integrability, does not mean "integrable in terms of elementary functions". It means that you can in principle write down the solution, provided that you are able to compute all antiderivatives and inverses of functions that you happen to come across along the way. But unfortunately not all antiderivatives and inverses of elementary functions are elementary, as you probably know...2012-05-28
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    All of which leads to the curious, and unanswered question, is the solar system stable?2012-07-16
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    @AlexR. in what sense is it solvable? Does it admit a closed form solution or is the solution representable in terms of e.g. quadratures? And, to be precise, by the quantum 3-body problem problem do you mean initial-value problem for TDSE or boundary-value problem for TISE?2017-08-31
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    @Ruslan: the energies of the particles can be derived as generalizations of Lambert's W Function. See here for example: https://en.m.wikipedia.org/wiki/Euler%27s_three-body_problem#Quantum_mechanical_version2017-08-31
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    @AlexR. it's not the _true_ 3-body problem: there the problem is simplified to that of motion of electron (1 body!) in the field of two fixed nuclei.2017-08-31
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    @Ruslan: It was kind of a joke.2017-08-31

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