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This question evolved from a discussion below this answer which explains (among other things) that the total energy of a system offers insight as to the possibility of one (or all) members "escaping".

The total energy would be the sum of the kinetic and potential energies

$$E = \sum_{i=1}^{3}\frac{1}{2}m_i v_i^2 - \sum_{i=1}^{3} \sum_{j>i}^{3} \frac{m_i m_j}{r_{ij}}.$$

Can there be some three body orbit that is energetically unbounded ($E>0$) but where it is still impossible for any of the objects to escape due to do conservation of angular momentum?

Possibly helpful: Equations of motion for the n-body problem

notes:

  1. I'm not asking if there exist orbits that are closed and periodic where escape is impossible for that reason.
  2. I haven't written an expression for angular momentum because there is flexibility about which point it is calculated.

Batominovski's Clarification on the Bounty (as noted by Angela Pretorius in a comment). The energy should be measured with respect to the center-of-mass frame of the system. That is, the condition $$\sum_{i=1}^3m_iv_i=0$$ is enforced.

Based on comments here and my suspicion I've corrected $i \ne j$ to $i > j$ for the potential energy term to avoid double-counting.

uhoh
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    Regarding the link above, I think your post is fine here, although I believe that you might get a quicker answer if you post in the physics forum. – Batominovski May 06 '20 at 06:26
  • @WETutorialSchool thanks, I'm looking for a definitive answer, so quickness is not so important. – uhoh May 06 '20 at 08:51
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    When I was involved in an early version of this project I learned a little about the differences between quasi-periodic, chaotic, and ergodic orbits in various potentials, as a way to describe particles which have enough energy to escape from a trap but don't for some symmetry-related reason. Might be a useful search term. – rob May 06 '20 at 10:31
  • @rob just when I thought I'd settle down and get some work done today you go and show me something shiny :-) – uhoh May 06 '20 at 11:13
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    My suggestion is to post the question at Mathoverflow. – Moishe Kohan May 06 '20 at 19:16
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    It seems unlikely, because as one mass gets very far away, you can give it a large angular momentum by giving it a tiny tangential velocity, which wouldn't significantly affect the other conservation laws. – knzhou May 06 '20 at 20:12
  • @MoisheKohan I'm certainly open to it being migrated there if it improves the changes of a definitive answer. – uhoh May 07 '20 at 00:32
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    Just ask a question at MO and give a link to the MSE post. – Moishe Kohan May 07 '20 at 00:47
  • @MoisheKohan I understand but I'm uncomfortable cross-posting which is strongly discouraged without special circumstances and a moderator's blessing. – uhoh May 07 '20 at 00:55
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    @uhoh: Cross-posting is OK, as long as you state clearly in MO that this is a cross-post and give a link to MSE question. It happened to me in the past: I asked a question first on MSE, then on MO; it was well-received but it turned out to be a duplicate of an earlier MO question, so it was closed for that reason. – Moishe Kohan May 08 '20 at 04:44
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    I should point out that the kinetic energy of a system can be made really high by changing the frame of reference. There are energetically unbounded systems which don't escape from each other, but I guess that you are really asking whether there are energetically unbounded systems which don't escape from a point which has zero velocity in the chosen frame of reference. – Angela Pretorius May 08 '20 at 06:06
  • @Batominovski thank you for your help and adding the +200 bounty, both of which resulted in increased attention. There's no conclusive answer posted yet but I hope that eventually something will be forth coming. – uhoh May 16 '20 at 23:57
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    I voted to close this as unclear. For it is not clear to me what would qualify as an answer. A given configuration either leads to an escape or it doesn't. What has that to do with the conservation of angular momentum? Which is, after all, a consequence of the equations of motion. – Jyrki Lahtonen May 26 '20 at 05:25
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    @JyrkiLahtonen some configurations will not lead to an escape because they are energetically unable, some will not because they are closed. Those are important and interesting distinctions even though both are consequences of the equations of motion. Why would you discriminate specifically against angular momentum and not those other two reasons? Thank you for your comment, but I don't understand why would you take steps to prevent others from answering my question and perhaps yours as well. – uhoh May 26 '20 at 05:32
  • The reason I protest is that you don't make it clear what it means that "no escape" is due to the conservation of angular momentum. How do you hope to achieve the implication "conservation of angular momentum" $\implies$ "no escape"? Or yet in other words: given a non-escaping initial configuration, how can we decide that the non-escape is a due to conservation of angular momentum specifically? Particularly because angular momentum is conserved in all configurations - escape or not. – Jyrki Lahtonen May 26 '20 at 05:40
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    @JyrkiLahtonen okay thank you for the clarification! It is possible that this has been shown already, and an answer to my question might cite such a source where I could discover how this was argued. I don't know if this can be done or not, which is my motivation for asking the question. I appreciate your comments, but I don't know how to address your close vote; I'm not able to answer this in order to demonstrate how it can be answered. – uhoh May 26 '20 at 05:46
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    @JyrkiLahtonen If you are certain that there can neither be a yes or no answer because the question does not make sense mathematically, then perhaps that is the answer to my question! If you post an answer as such and it's well received by others, then I may even click accept and we've all benefitted. But if you are not certain, then why block others from having an opportunity to answer? – uhoh May 26 '20 at 05:46
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    May be you are looking for a configuration where there is an elegant argument involving conservation of angular momentum that leads to the conclusion that no escapes will ever happen? That would be the kind of a question where "we know that this is an answer when wew see it", but there would be no way of proving that no such configuration exists, unless the rules of the game are made very precise. Scratching my head. I won't be sad, if this question escapes closure. But I'm not sure it's a good fit here. – Jyrki Lahtonen May 26 '20 at 05:55

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The answer is no... conservation of angular momentum, by itself, can't be used to prove boundedness of a 3-body system with positive total energy (in the frame where the center of mass is stationary at the origin). For sufficiently large $t$, all escaping bodies (there must be at least 2) will have essentially fixed velocities ${\bf v}_i$ and linearly evolving positions ${\bf x}_i + t {\bf v}_i$. The total angular momentum is $\sum_i \left({\bf x}_i + t{\bf v}_i\right) \times m_i{\bf v}_i = \sum_i m_i{\bf x}_i \times{\bf v}_i$, also a constant. But note that the angular momentum can be changed to any value without changing the total energy, the total momentum, or the center of mass, by adding appropriate offsets to the ${\bf x}_i$. (Keeping the center of mass fixed imposes one vector constraint on these offsets; since at least two bodies are escaping, there is at least one vector degree of freedom remaining.)

In short, conservation of angular momentum doesn't help you because each "escape scenario" belongs to an equivalence class of scenarios (with the same total energy and momentum) that differ only in their angular momenta.

mjqxxxx
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    I am not quite sure how this post answers the question. Could you please elaborate? This post seems to prove only that, when there are at least two escaping bodies, then the total angular momentum can be changed arbitrarily without changing the total energy $E$. The question is whether there exists a bounded system with total energy $E>0$ that is bounded because the conservation of angular momentum. – Batominovski May 13 '20 at 16:47
  • @Batominovski It is unclear to me what the phrase bounded because the angular momentum is conserved means. In particular for the purposes of proving the negation. Logically an answer could be a scenario, where escape is possible if we violate the law of conservation of angular momentum, but I doubt that is the intended meaning :-) – Jyrki Lahtonen May 26 '20 at 05:12
  • (cont'd) I mean, a three-body-system behaves deterministically under Newtonian mechanics. Therefore either there will be an escape or there won't. What has this got to do with due to conservation of angular momentum, which , IIRC, is a consequence of certain symmetries in Newtonian mechanics? – Jyrki Lahtonen May 26 '20 at 05:14
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    @Jyrki Lahtonen, "either there will be an escape or there won't" : whether there will be an escape or not and under what circumstances or assumptions this happens, may in general be a very delicate and interesting mathematical problem, in various different setups. So, in my understanding, the question is very interesting and it is a question which is naturally addressed to mathematicians. I doubt that physics.stackexchange would be a good place to ask this and to get some concrete answer: physicists who can answer such questions are actually mathematicians. – KonKan May 28 '20 at 11:00
  • @KonKan I'm aware (at some level) about the intricacies. And chaotic nature of 3 body systems. I am more concerned about how the asker might decide that any particular example works specifically "due to the conservation of angular momentum". I agree that the problem is mathematically very challenging, but the meaning of the question is not clear. – Jyrki Lahtonen May 28 '20 at 11:20