# Chaos theory and determinism

My professor in class went a little over chaos theory, and basically said that Newtonian determinism no longer applies, since as time goes to infinity, no matter how close together two initial points are, the distance between them will increase greatly. But why isn't this merely a matter of the imprecision of our measuring instruments? If we can somehow know our initial conditions exactly, wouldn't we still be able to calculate what the system will be like at some time t in the future?

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You got it exactly right, you answered your own question - in principle determinism applies, in practice not so much. –  user566 Feb 11 '11 at 7:11
Not even in principle if you take into account that any measurements physical observers can make will always have some finite error, no matter how small. –  user346 Feb 11 '11 at 7:28
True. Still, you can have an epsilon- delta definition of determinism. To be within epsilon precision of a given final state at some finite future time, you'd have to specify initial state with some precision delta. Of course delta is in practice unattainable. –  user566 Feb 11 '11 at 7:38
Let me suggest the following book about determinism, which is very interesting. It shows that Newtonian mechanics is not deterministic contrary to "popular" belief. But this has nothing to do with chaos. –  Raskolnikov Feb 11 '11 at 10:59
Hi dbrane: actually, it's due to the fact that there is no speed limit in Newtonian mechanics. So something could be incoming from infinity and immediately throw determinism in jeopardy through instantaneous action. That's how far I can remember the argument, probably distorting it. But on the other hand, special relativity does not suffer from the problem. There were other examples of non-determinism, but I forgot them. I'll have a look at the book later. –  Raskolnikov Feb 11 '11 at 13:44
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Well, yes. In a purely mathematical world where you can specify initial conditions exactly, chaotic systems are fully deterministic. It's not like a quantum system with wavefunction collapse, whose evolution can never be specified exactly by the initial conditions.

But in practice, we can never specify (or know) the initial conditions exactly. So there will always be some uncertainty in the initial conditions, and it makes sense to characterize the behavior of a system in terms of its response to this uncertainty. Basically, a chaotic system is one in which any uncertainty in the state at time $t=0$ leads to exponentially larger uncertainties in the state as time goes on, and a non-chaotic system is one in which any initial uncertainty in the state decays away or at least stays steady with time.

In the former (chaotic) case, given that we can't know the initial conditions to infinite precision, there will always be some time after which predictions of the behavior of the system become essentially meaningless - the uncertainty becomes so large that it fills up most of the state space. This is effectively similar to the behavior of a truly non-deterministic (e.g. quantum) system, in that our ability to make predictions about it is limited, so some people call chaotic systems non-deterministic.

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Nice, clear answer! Nitpicky: if chaotic systems have uncertainty growing exponentially in time and non-chaotic have uncertainty constant or less, we've left out situations where the uncertainty grows in time, but less strongly than exponentially. For example, a free classical particle has position uncertainty that is linear in time due to uncertainty in its initial velocity. –  Mark Eichenlaub Feb 11 '11 at 9:11
For a quantitative account, see phy.bris.ac.uk/people/berry_mv/the_papers/Berry174.pdf , where Sir Michael Berry shows how fast a small perturbation (gravity due to a single electron at the edge of the universe) will lead to a complete breakdown of predictability. –  genneth Feb 11 '11 at 9:42
The persons who call chaotic systems non-deterministic are nevertheless wrong and contributing to propagate a misunderstanding. It's better to speak about non-predictable as genneth suggests. –  Raskolnikov Feb 11 '11 at 10:54
No to mention that Quantum Mechanics is also deterministic... –  mbq Feb 11 '11 at 14:10
@David I am a decoherence guy ;-) Nevertheless, you can always extend the system with observer and thus remove the measurement -- and this bigger system will be deterministic. –  mbq Feb 12 '11 at 0:00

I think the following from the wikipedia entry clears up well the terminology:

Chaos theory is a field of study in applied mathematics, with applications in several disciplines including physics, economics, biology, and philosophy. Chaos theory studies the behavior of dynamical systems that are highly sensitive to initial conditions; an effect which is popularly referred to as the butterfly effect. Small differences in initial conditions (such as those due to rounding errors in numerical computation) yield widely diverging outcomes for chaotic systems, rendering long-term prediction impossible in general.1 This happens even though these systems are deterministic, meaning that their future behavior is fully determined by their initial conditions, with no random elements involved.[2] In other words, the deterministic nature of these systems does not make them predictable.[3] This behavior is known as deterministic chaos, or simply chaos.

Bold mine.

Note that chaos results in completely deterministic systems when small errors in initial conditions yield highly divergent solutions. It is the "exactly" in your question that is unattainable, that will be a tiny bit off in chaotic situations,(highly nonlinear response to input parameters) even in computer solutions, because one cannot be more accurate than the computer bits.

Note also that this does not mean that there are no mathematical methods to study the behavior of such systems. There are, and can be predictive in bulk. I would give as an example the study of Tsonis et al who have studied climate with a neural net chaotic model using as inputs bulk behavior of atmospheric and ocean currents.

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Changed to a quote block from italics - hope you don't mind. –  Mark Eichenlaub Feb 11 '11 at 9:53

Classical mechanics is perfectly integrable for two bodies as a closed or isolated system. However, early on it was found that problems existed, where Newton found he could not find a solution for the motion of the planets in a complete form. He made his famous statement that God had to readjust the solar system now and them. Poincare solved the Sweden prize for a solution to the stability of the solar system by demonstrating no such solution existed. This is what opened the door for chaos theory, where Poincare developed methods of separatrices and perturbations on them. For general systems it turns out that Newtonian mechanics is not integrable,

Classical mechanics for systems with three or more bodies can\rq t be solved in closed form. For any $N$ body problem there are 3 equations for the center of mass, $3$ for the momentum, $3$ for the angular momentum and one for the energy. These are $10$ constraints on the problem. An N-body problem has $6N$ degrees of freedom. For $N~=~2$ this means the solution is given by a first integral with degree $2.$ For a three body problem this first integral has degree $8$. This runs into the problem that Galois illustrated which is that any root system with degree $5$ or greater generally have no algebraic roots. First integrals for differential equations are functions which remain constant along a solution to that differential equation. So for $8$ solutions there is some eight order polynomial $p_8(x)~=~\prod_{n=1}^8(x~-~\lambda_n)$, with $8$ distinct roots$\lambda_n$ that are constant along the $8$ solutions. Since $p_8(x)~=~p_5(x)p_3(x)$, a branch of algebra called Galois theory tells us that fifth order polynomials have no general algebraic system for finding its roots, or a set of solutions that are algebraic. This means that any system of degree higher than four are not in general algebraic. At the root of the $N$-body problem Galois theory tells us there is no algebraic solution for $N~\ge~3$.

A problem in classical mechanics is the vanishing denominator problem for three bodies. This has a Hamiltonian $H( J,~\theta)~=~H_0(J,~\theta)~+~\epsilon H_1(J,~\theta)$ for $J~=~(J_1,~J_2)$ and $\theta~=~(\theta_1,~\theta_2)$. Here a generating function written according to the variable $J^\prime$
$$S(J^\prime,~\theta)~=~\theta\cdot J^\prime~+~i\epsilon\sum_{n_1,n_2}{{H_{1,n_1,n_2}}\over{n_1\omega_1(J^\prime)~+~n_2\omega_2(J^\prime)}}e^{ n_1\omega_1(J^\prime)~+~n_2\omega_2(J^\prime)}$$ will be divergent for the resonant condition $n_1\omega_1(J^\prime)~+~n_2\omega_2(J^\prime)~=~0$. This resonance condition has lead many to presume that the solar system can not be stable with resonance conditions. Yet the solar system is replete with near resonance conditions.

The number of resonance conditions that exist on the real line are dense. Within any $\epsilon$ neighborhood there will exist a countably infinite number of possible resonance conditions that correspond to rational numbers. As the orbit of a planet drifts it will pass through these resonance conditions and be chaotically perturbed. It is to be expected that for simple rational numbers, such as $1/12$ for Earth and Jupiter, rather than $1003/12000$ strong resonances occur. For more complex rational numbers it might be expected that the instability will be weaker. In other words if the ratio of frequencies are \lq\lq sufficiently irrational\rq\rq$~$so that $$\Big|{{\omega_1}\over{\omega_2}}~-~{m\over s}\Big|~>~{{k(\epsilon)}\over{s^{2.5}}},~\lim_{\epsilon~\rightarrow~0}k(\epsilon)~\rightarrow~0$$ the orbit is more stable. So an orbit that is removed from a “strong resonance”” condition near a simple rational number will be more stable than an orbit that is near an orbit with a simple rational ratio of frequencies.

This is the basis for Greenberg’s Hamiltonian approach to chaos theory. This is called deterministic for the differential equations are time reverse invariant so the motion of a particle is absolutely determined. However. if you have a slight variation in the initial conditions of that particle is may in general end up arbitrarily far away from its starting point. The tiny variation $\delta z~=~(\delta q,~\delta p)$ becomes amplified by an exponential map $\delta z~\rightarrow~exp(\lambda t)\delta z$, for $\lambda$ the Lyapunov exponent. Any error in the specification of the initial conditions of a body results in the amplification of this error. From an algorithmic perspective a truncation results in numerical overflow errors which grow. So the dynamics of a particle can’t be integrated by computer to arbitrary accuracy into the future, even though nature actually does determine its dynamics.

W. Zurek took this a bit further and considered how quantum fluctuations, where $\delta z$ is set by the Heisenberg uncertainty principle.

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What a great answer. Reminds me that I need to keep coming back to this website. –  Raphael R. Feb 11 '11 at 20:27

I think there is some debate about this, at least according to Steven Strogatz. I paraphrase a part of his lecture that I'm watching: Even systems with perfectly known, perfectly deterministic laws, where all the positions of all the particles, and all the forces, are known, can still be unpredictable. It is such systems that are called chaotic.

This is his view of it. I wish I could provide a link to the video, but it's on my hard drive.

I think this is an important debate to have, if it is still up in the air, because it has a lot of implications for the notion of free will. It's a philosophically deep question.

From my point of view, which is that of a lay person, Strogatz's view seems a bit hard to imagine, but then again, he's the expert.

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An update: I've been watching his subsequent lectures, and he has contradicted himself and gone back to the standard definition of chaos -- that given by David Zaslavsky and anna v. –  Joebevo Mar 16 '12 at 4:32
Everyone talks about that "free will" thing, but I've never heard anyone give a definition of it that satisfies any requirements to a scientifically rigorous statement. So how can we know that determinism clashes with free will, if we don't even know what free will is? –  Thriveth Jun 9 '13 at 21:57
@Thriveth: On subsequent thought, I have to take back what I said about the connection between chaos and free will, since they may not have anything to do with each other. However, I do believe that free will (in some qualified sense) exists, regardless of whether we can define it or not. That's just my position. –  Joebevo Jun 10 '13 at 5:00

I am afraid that your professor is confusing determinismus and computability. Chaotic behaviour is a property of some solutions of a system of non linear ODE.

However any non linear ODE is strictly deterministic - e.g if you know the value of X at time t0, the ODE allows you to compute deterministically X at time t0+dt with an arbitrary accuracy. That's why it is often talked about deterministic chaos.

The difficulties begin when you want to know X at a time T which is no more infinitesimaly close to t0 or in other words when you want to integrate the ODE system for any t. This is only possible numerically for chaotic systems and it is here that the property of exponential divergence (sensibility to initial conditions) kicks in. Indeed the error made in the value of the initial conditions will increase exponentially with time and prevent you to compute accurately the value of the chaotic variable for all times. But this is merely a computability problem, not a determinismus problem.

From the physical point of view, the computability problem will never be solved because of the Heisenberg's uncertainty relation which imposes a non zero value for the uncertainty of the measure of initial conditions. If the uncertainty is equal to the minimum value allowed by QM, it is easy to compute for any chaotic system the time beyond which the uncertainty of the value of the dynamical variable will be of the same order of magnitude as the dynamical variable itself. Beyond this time no chaotic variable can be computed.

For instance the orbital parameters of the Earth which are chaotic cannot be known beyond some 10 millions years. Of course, but this is another question, the unknowability of the orbital parameters is not synonymous with the unknowability of the orbit itself which may be but must not be stable. In our case we were lucky because despite the chaos in the orbital parameters of the Earth, the orbit itself has been stable or quasi stable for at least 4 billions years.

As a caveat one should add that the exponential divergence of orbits is a necessary but non sufficient condition for a system to be chaotic. For instance a variable Y = exp(t) has the property of exponential divergence of nearby trajectories but is computable with an arbitrary accuracy and is not chaotic.

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