# Diffusion in the standard map

Consider the standard map (also known as Chirikov map): $$p_{n+1} = p_n + K \sin(\theta_n) \\ \theta_{n+1} = \theta_n + p_{n+1}$$

I know that the diffusion coefficient according to the Einstein relation is defined through the relation:

$$\sigma^2 = 2Dt$$

being $\sigma$ the variance of the position in the case of the $1$-D random walker. Can you figure out why the diffusion coefficient for the standard map is given by:

$$D = \lim_{n \rightarrow \infty} \frac{<(p_n - p_0)^2>}{n} \quad ?$$

I mean, the $n$ makes sense, but what about the variance, why this is given just in terms of $p$, what happens with the $x$.

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In the more typical situation of particles, position is an integral over the velocity. Writing out the expectation for the mean square displacement you end up with an autocorrelation the velocity. Thus, by the Green-Kubo relations from linear response theory, you get the equation for the diffusion constant. In your case the integrals may be sums, but I expect everything to work out the same nevertheless. –  alarge Aug 9 '14 at 2:33

The following is a passage from Diffusion in the standard map (pdf) by Itzhack Dana and Shmuel Fishman:

A major difficulty in the analysis of chaotic behavior of Hamiltonian systems is the proximity of chaotic and regular orbits on various scales [2]. Thus, for $K \approx 1$, the phase plane of (1) is an intricate mixture of regular and chaotic orbits [3]. For $K \gg 1$ chaotic orbits fill almost the entire phase plane, but "islets of stability", in particular accelerator modes [2, 7], are known to exist for $K$ arbitrarily large. When a chaotic orbit approaches such an islet, it may wander near it in a "regular" fashion. The area of each islet generally decreases when $K$ increases.

Distinctive features of a stochastic or random motion, which allow a statistical description of it, are the rapid decay of correlations and diffusion. The decay of correlations in the chaotic region is related to the local instability, measure Lyapunov characteristic exponent [1, 2, 8]. For $K \gg 1$, the decay of correlations is actually exponential, with the decay exponent proportional to the Lyapunov exponent [8]. The existence of a definite characteristic time for the decay of correlations and the resulting statistical independence generally imply diffusion in the unbounded direction of the action $I$ for $K > K_\mathrm{c}$ [...]. Therefore a definite diffusion coefficient $D$ can be associated with the chaotic motion $$D = \lim_{n\to\infty} \frac{\big\langle(I_n-I_0)^2\big\rangle}{n},$$ where the average is taken over some ensemble of initial positions $(I_0,\theta_0)$ within the chaotic region. For large $K$, where the Lyapunov exponent is large, diffusion was verified unambiguously [2].

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Thank you very much,just to complement and trying to understand completely this point, I'll share you something. I found in this book the following explanation: If we assume $K$ >> $2\pi$, then $p$ will also tipically be large compared to $2\pi$. Thus, we expect $\theta$ to vary very wildly in $[0,2\pi]$. We therefore treat $\theta_n$ as effectively random, uniformly distributed. With these assumptions, the motion in $p$ becomes a random walk with step size $\Delta p_n = K\sin\theta_{n+1}$ –  dapias Aug 11 '14 at 14:02
Yes that makes sense as well, thank you for finding another approach. Good luck. –  Gödel Aug 11 '14 at 19:31