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You think Newtonian physics is non-deterministic? I haven't checked but I think existence and uniqueness for the classical three body problem likely follows fr

You think Newtonian physics is non-deterministic? I haven't checked but I think existence and uniqueness for the classical three body problem likely follows from standard theorems on ODEs. If you can articulate why this is not the case, please do. edit: Yes, see section 2.1 of this term paper: https://sites.math.washington.edu/~morrow/336_13/papers/peter.pdf https://sites.math.washington.edu/~morrow/336_13/papers/pete.... In fact, more is known: the solution exists and is analytic. From Wikipedia: "However, in 1912 the Finnish mathematician Karl Fritiof Sundman proved that there exists an analytic solution to the three-body problem in the form of a power series in terms of powers of t1/3. This series converges for all real t, except for initial conditions corresponding to zero angular momentum. In practice, the latter restriction is insignificant since initial conditions with zero angular momentum are rare, having Lebesgue measure zero. "

deltasixeight 5y ago
I didn't say that. For the three body problem we don't know the closed form solution under Newtonian physics. So if we don't know it, then a closed solution may not even exist. If we don't know whether a closed form solution exists then how can we even know if it's deterministic. You can prove determinism by finding me a solution that's closed and deterministic. My answer is the same as stated: we don't know, and that the claim that we do know, is completely wrong. >In fact, more is known: the solution exists and is analytic These are special cases, under special conditions. A general solution is not actually known. We don't know. We are only assuming that it's deterministic. Also the solution you provided is analytic not closed.
- spekcular 5y ago
  "If we don't know whether a closed form solution exists then how can we even know if it's deterministic. You can prove determinism by finding me a solution that's closed and deterministic." Why would it be necessary to have a closed form solution to know that the solution is deterministic? By the first reference in my previous comment, two identical initial conditions lead to identical time evolutions. That is literally the definition of determinism.
  - deltasixeight 5y ago
    No look at my wording more carefully. You are misinterpreting it. I said we don't know if a "closed form solution" is deterministic because we don't even know if there is a "closed form solution." I also said that you CAN prove determinism to me by finding a closed form solution that is deterministic. I never said you needed to nor did I say that was the only way. Your examples are for special cases. Can you prove it to me that for N-bodies and all possible initial conditions that all solutions are deterministic?
  - deltasixeight 5y ago
    https://news.ycombinator.com/item?id=28184786 https://news.ycombinator.com/item?id=28184786
    - spekcular 5y ago
      The example you quote uses Newton's laws to model a system with ODEs that do not satisfy the hypotheses of the uniqueness and existence theorems. It is a nice example, but it is not relevant here, because the ODEs for the three-body problem do satisfy the hypotheses of those theorems. This particular problem is totally deterministic. "I also said that you CAN prove determinism to me by finding a closed form solution that is deterministic. I never said you needed to nor did I say that was the only way." I agree. One need not give a closed form solution to show it is deterministic. I gave you another way. "Your examples are for special cases. Can you prove it to me that for N-bodies and all possible initial conditions that all solutions are deterministic?" What "special case" is being considered by the term paper I linked above? What are the initial conditions you are worried about, to which the ODE uniqueness and existence theorems do not apply? In other words, where you do think there is a loss of generality? (More precisely, we have a unique solution up until any collision takes place, if it does at all. But this is not a problem with determinism, it is a problem with modeling the bodies as point masses. We could imagine them as hard spheres and then this issue would disappear.)
      - deltasixeight 5y ago
        >What "special case" is being considered by the term paper I linked above. Not referring to that. Referring to the Wikipedia version which immediately stated it's a special case. Anyway you're not focusing on that you're talking about the paper and the paragraph at section 2.1. Your paper assumes the the theorems apply. They don't apply. Why? Because the equations are not Lipschitz continuous when bodies are really really close to each other. They only apply when the bodies are far away from each other which is not always the case for the 3 body scenario. This completely destroys the one solution property and as a result determinism. Again we are both wrong. I said I we didn't know, you said it's deterministic; but now we do in fact know that it is non-deterministic.
- kergonath 5y ago
  > So if we don't know it, then a closed solution may not even exist. If we don't know whether a closed form solution exists then how can we even know if it's deterministic. We can prove that something has some properties even if we don’t know everything about it. Newton’s second law in a many-body system is (in the case of Newtonian gravity) basically a system of independent second-order ODEs. We know a thing or two about these beasts. Anyway, let me try something (without LaTeX, so I apologise in advance for the notation). Let’s think about a many-body system (following Newtonian mechanics, and interacting via Newtonian gravity) from the point of view of its positions-velocities phase space. The solution is not deterministic if there are points (i.e., sets of positions and velocities for all the particles) that are part of at least two trajectories. Let’s assume we are looking at such a point. The forces F and all their derivatives dF/dr are the same along all trajectories, because they depend only on the positions (which are identical). The velocities and the positions are the same because it is the same point in the phase space. The third derivatives of the positions w.r.t time are also identical (d^3r/dt^3 = 1/m dF/dt, and dF/dt = dF/dr dr/dt, and the dr/dt are identical). You can go on to the n-th derivative, n being arbitrary large (d^4 r/dt^4 is a function of only dF/dr, dr/dt, d^2 r/dt^2 and d^2 F/dr^2, all of which are identical along both trajectories, etc). So, it means that both trajectories have the same position in the phase space, and all their derivatives are identical, i.e., both trajectories are identical. So it is deterministic (there is no point in the phase space that belongs to more than one trajectory). It is somewhat related to Liouville’s theorem.
  - deltasixeight 5y ago
    https://news.ycombinator.com/item?id=28184786 https://news.ycombinator.com/item?id=28184786

deltasixeight 5y ago

I didn't say that. For the three body problem we don't know the closed form solution under Newtonian physics. So if we don't know it, then a closed solution may not even exist. If we don't know whether a closed form solution exists then how can we even know if it's deterministic. You can prove determinism by finding me a solution that's closed and deterministic. My answer is the same as stated: we don't know, and that the claim that we do know, is completely wrong. >In fact, more is known: the solution exists and is analytic These are special cases, under special conditions. A general solution is not actually known. We don't know. We are only assuming that it's deterministic. Also the solution you provided is analytic not closed.

spekcular 5y ago
"If we don't know whether a closed form solution exists then how can we even know if it's deterministic. You can prove determinism by finding me a solution that's closed and deterministic." Why would it be necessary to have a closed form solution to know that the solution is deterministic? By the first reference in my previous comment, two identical initial conditions lead to identical time evolutions. That is literally the definition of determinism.
- deltasixeight 5y ago
  No look at my wording more carefully. You are misinterpreting it. I said we don't know if a "closed form solution" is deterministic because we don't even know if there is a "closed form solution." I also said that you CAN prove determinism to me by finding a closed form solution that is deterministic. I never said you needed to nor did I say that was the only way. Your examples are for special cases. Can you prove it to me that for N-bodies and all possible initial conditions that all solutions are deterministic?
- deltasixeight 5y ago
  https://news.ycombinator.com/item?id=28184786 https://news.ycombinator.com/item?id=28184786
  - spekcular 5y ago
    The example you quote uses Newton's laws to model a system with ODEs that do not satisfy the hypotheses of the uniqueness and existence theorems. It is a nice example, but it is not relevant here, because the ODEs for the three-body problem do satisfy the hypotheses of those theorems. This particular problem is totally deterministic. "I also said that you CAN prove determinism to me by finding a closed form solution that is deterministic. I never said you needed to nor did I say that was the only way." I agree. One need not give a closed form solution to show it is deterministic. I gave you another way. "Your examples are for special cases. Can you prove it to me that for N-bodies and all possible initial conditions that all solutions are deterministic?" What "special case" is being considered by the term paper I linked above? What are the initial conditions you are worried about, to which the ODE uniqueness and existence theorems do not apply? In other words, where you do think there is a loss of generality? (More precisely, we have a unique solution up until any collision takes place, if it does at all. But this is not a problem with determinism, it is a problem with modeling the bodies as point masses. We could imagine them as hard spheres and then this issue would disappear.)
    - deltasixeight 5y ago
      >What "special case" is being considered by the term paper I linked above. Not referring to that. Referring to the Wikipedia version which immediately stated it's a special case. Anyway you're not focusing on that you're talking about the paper and the paragraph at section 2.1. Your paper assumes the the theorems apply. They don't apply. Why? Because the equations are not Lipschitz continuous when bodies are really really close to each other. They only apply when the bodies are far away from each other which is not always the case for the 3 body scenario. This completely destroys the one solution property and as a result determinism. Again we are both wrong. I said I we didn't know, you said it's deterministic; but now we do in fact know that it is non-deterministic.
kergonath 5y ago
> So if we don't know it, then a closed solution may not even exist. If we don't know whether a closed form solution exists then how can we even know if it's deterministic. We can prove that something has some properties even if we don’t know everything about it. Newton’s second law in a many-body system is (in the case of Newtonian gravity) basically a system of independent second-order ODEs. We know a thing or two about these beasts. Anyway, let me try something (without LaTeX, so I apologise in advance for the notation). Let’s think about a many-body system (following Newtonian mechanics, and interacting via Newtonian gravity) from the point of view of its positions-velocities phase space. The solution is not deterministic if there are points (i.e., sets of positions and velocities for all the particles) that are part of at least two trajectories. Let’s assume we are looking at such a point. The forces F and all their derivatives dF/dr are the same along all trajectories, because they depend only on the positions (which are identical). The velocities and the positions are the same because it is the same point in the phase space. The third derivatives of the positions w.r.t time are also identical (d^3r/dt^3 = 1/m dF/dt, and dF/dt = dF/dr dr/dt, and the dr/dt are identical). You can go on to the n-th derivative, n being arbitrary large (d^4 r/dt^4 is a function of only dF/dr, dr/dt, d^2 r/dt^2 and d^2 F/dr^2, all of which are identical along both trajectories, etc). So, it means that both trajectories have the same position in the phase space, and all their derivatives are identical, i.e., both trajectories are identical. So it is deterministic (there is no point in the phase space that belongs to more than one trajectory). It is somewhat related to Liouville’s theorem.
- deltasixeight 5y ago
  https://news.ycombinator.com/item?id=28184786 https://news.ycombinator.com/item?id=28184786