7 ms·

They would be deterministic. But they are unpredictable. Minuscule fluctuations (coming from influence of some much smaller bodies, that are not in the model, o

by ordu 5y ago

They would be deterministic. But they are unpredictable. Minuscule fluctuations (coming from influence of some much smaller bodies, that are not in the model, or approximations in calculations) can lead to dramatic differences in the outcome. So theoretically speaking, they are deterministic, but practically they are unpredictable.

Aeolun 5y ago
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.
- deleted 5y ago
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- colechristensen 5y ago
  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.
  - stavros 5y ago
    > a common occurrence is one of the bodies being ejected. So the problem eventually solves itself?
    - motoboi 5y ago
      Well, not if the ejected body contains you.
      - TheOtherHobbes 5y ago
        Being ejected from your model is unquestionably a solution of sorts, although it does Raise Questions.
    - dnautics 5y ago
      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"?
  - onlyrealcuzzo 5y ago
    > 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?
    - mannerheim 5y ago
      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.
      - mokus 5y ago
        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.
    - mynegation 5y ago
      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.
    - colechristensen 5y ago
      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.
  - RhysU 5y ago
    This is sometimes called "infinite sensitivity to initial conditions" or "deterministic chaos".
- alephu5 5y ago
  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.
  - Grustaf 5y ago
    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.
    - extropy 5y ago
      Both of you are correct.
    - l332mn 5y ago
      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.
      - Grustaf 5y ago
        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.
        johncolanduoni 5y ago
        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.
- Grustaf 5y ago
  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.
  - bottled_poe 5y ago
    I think this is accurate and I don’t understand why you are being downvoted.
  - 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 ago
      Conversely, there are also extremely simple closed-form recurrence relations that exhibit chaotic behavior, e.g. the logistic map.
      - phreeza 5y ago
        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.
        MontyCarloHall 5y ago
        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.
- sasaf5 5y ago
  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.
  - evanb 5y ago
    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.
- aaaaaaaaaaab 5y ago
  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.
- bsf_ 5y ago
  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.
mjburgess 5y ago
Random doesnt mean non-deterministic anyway. X is random with respect to Y, if knowing Y makes no difference to your predicting that X. QM systems are indeterminate, they are random in the above sense /because/ they are indeterminate. But that isnt what random means.
- elcomet 5y ago
  What you're describing is not randomness, it's independence. It's hard to define randomness. I think non-determinism is better than your definition.
  - mjburgess 5y ago
    It isn't hard to define randomness. It's an epistemic condition on the knowability of Y given X. Non-determinism is an incoherent definition of randomness; classical physical processes are entirely deterministic. The point of a coinflip being random is that it is random with respect to the information both observers of the coinflip have. It isnt random with respect to /any/ piece of information. There are almost no processes which are non-deterministic in this sense. Not enough to bother calling them random; and in physics we do not: the word is indeterminate. Randomness has nothing to do with quantum mechanics; it wasn't invented in the 1920s. It's an epistemic condition. The RANDOM variable X, st. X ~ N(mean, std) provides a random number x -- x isnt random with repect to the outcome which produced x; nor is it random with respect to an index of a vector in which it is contained.
    - elcomet 5y ago
      This does not match the intuitive notion of randomness though. I would say for a given variable to be random, it must not be predictible, given any other variables that humans can know. I don't think it makes sense to say that X is random "with respect to Y", that's just the definition of independence. And a constant variable is independent from all other variables, but it's definitely not random.
      - 5y ago
ninkendo 5y ago
The motion isn’t deterministic if free will exists. Launching a rocket into space decreases earth’s rotation speed ever so slightly, which will have a small impact on the moon’s trajectory due to tidal interactions, and so on.
- athrowaway3z 5y ago
  Please define 'free will' before using it in a sentence about determinism.
  - thaumasiotes 5y ago
    If that was a requirement, discussions of determinism would sound awfully one-sided. ;p
- mxxc 5y ago
  that case is beyond scope of the three body problem
- martincmartin 5y ago
  There are those who believe determinism is not only compatible with free will, but required for free will. https://en.wikipedia.org/wiki/Compatibilism https://en.wikipedia.org/wiki/Compatibilism
simonh 5y ago
It's got nothing to do with perturbations from additional small bodies, the problem is exactly mathematically calculating the outcome for exactly three bodies within reasonable time frames even if you assume perfect ideal knowledge about the system. It turns out this is excruciatingly hard. The n-body problem where you add further bodies, even very small ones, is even harder.
kergonath 5y ago
I think the issue was more with “random” than with “unpredictable”.
deleted 5y ago
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by ordu 5y ago

Aeolun 5y ago

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.

deleted 5y ago
[deleted]
colechristensen 5y ago
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.
- stavros 5y ago
  > a common occurrence is one of the bodies being ejected. So the problem eventually solves itself?
  - motoboi 5y ago
    Well, not if the ejected body contains you.
    - TheOtherHobbes 5y ago
      Being ejected from your model is unquestionably a solution of sorts, although it does Raise Questions.
  - dnautics 5y ago
    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"?
- onlyrealcuzzo 5y ago
  > 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?
  - mannerheim 5y ago
    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.
    - mokus 5y ago
      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.
  - mynegation 5y ago
    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.
  - colechristensen 5y ago
    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.
- RhysU 5y ago
  This is sometimes called "infinite sensitivity to initial conditions" or "deterministic chaos".
alephu5 5y ago
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.
- Grustaf 5y ago
  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.
  - extropy 5y ago
    Both of you are correct.
  - l332mn 5y ago
    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.
    - Grustaf 5y ago
      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.
      - johncolanduoni 5y ago
        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.
Grustaf 5y ago
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.
- bottled_poe 5y ago
  I think this is accurate and I don’t understand why you are being downvoted.
- 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 ago
    Conversely, there are also extremely simple closed-form recurrence relations that exhibit chaotic behavior, e.g. the logistic map.
    - phreeza 5y ago
      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.
      - MontyCarloHall 5y ago
        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.
sasaf5 5y ago
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.
- evanb 5y ago
  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.
aaaaaaaaaaab 5y ago
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.
bsf_ 5y ago
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.

mjburgess 5y ago

Random doesnt mean non-deterministic anyway. X is random with respect to Y, if knowing Y makes no difference to your predicting that X. QM systems are indeterminate, they are random in the above sense /because/ they are indeterminate. But that isnt what random means.

elcomet 5y ago
What you're describing is not randomness, it's independence. It's hard to define randomness. I think non-determinism is better than your definition.
- mjburgess 5y ago
  It isn't hard to define randomness. It's an epistemic condition on the knowability of Y given X. Non-determinism is an incoherent definition of randomness; classical physical processes are entirely deterministic. The point of a coinflip being random is that it is random with respect to the information both observers of the coinflip have. It isnt random with respect to /any/ piece of information. There are almost no processes which are non-deterministic in this sense. Not enough to bother calling them random; and in physics we do not: the word is indeterminate. Randomness has nothing to do with quantum mechanics; it wasn't invented in the 1920s. It's an epistemic condition. The RANDOM variable X, st. X ~ N(mean, std) provides a random number x -- x isnt random with repect to the outcome which produced x; nor is it random with respect to an index of a vector in which it is contained.
  - elcomet 5y ago
    This does not match the intuitive notion of randomness though. I would say for a given variable to be random, it must not be predictible, given any other variables that humans can know. I don't think it makes sense to say that X is random "with respect to Y", that's just the definition of independence. And a constant variable is independent from all other variables, but it's definitely not random.
    - 5y ago

ninkendo 5y ago

The motion isn’t deterministic if free will exists. Launching a rocket into space decreases earth’s rotation speed ever so slightly, which will have a small impact on the moon’s trajectory due to tidal interactions, and so on.

athrowaway3z 5y ago
Please define 'free will' before using it in a sentence about determinism.
- thaumasiotes 5y ago
  If that was a requirement, discussions of determinism would sound awfully one-sided. ;p
mxxc 5y ago
that case is beyond scope of the three body problem
martincmartin 5y ago
There are those who believe determinism is not only compatible with free will, but required for free will. https://en.wikipedia.org/wiki/Compatibilism https://en.wikipedia.org/wiki/Compatibilism

simonh 5y ago

It's got nothing to do with perturbations from additional small bodies, the problem is exactly mathematically calculating the outcome for exactly three bodies within reasonable time frames even if you assume perfect ideal knowledge about the system. It turns out this is excruciatingly hard. The n-body problem where you add further bodies, even very small ones, is even harder.

kergonath 5y ago

I think the issue was more with “random” than with “unpredictable”.

deleted 5y ago

[deleted]