Standard map

How does order dissolve into chaos in a conservative (Hamiltonian) system?

▶ Run the simulationSee the measured result

Units: dimensionless kick strength (Greene 1979: 0.971635 ± 3e-6; MacKay 1983 renormalization: 0.97163540631)

How the lab tests it

Iterate the kicked-rotor map, colour each orbit by its Lyapunov exponent, and measure the chaotic-area fraction as the kick K grows.

What it looks for

the KAM transition — invariant tori break near the critical kick K_c ≈ 0.9716

Standard map K_c, residue criterion & Lyapunov exponent calculator

A rotor, one kick per turn, and a question with no formula behind it. Below some kick strength the momentum of the Chirikov standard map is trapped forever between invariant circles; above it, one continuous path runs to infinity. The strength at which the LAST of those circles dies is 0.971635406, and there is no closed form for it anywhere in science — it is not a root of anything, not a ratio of anything, and every digit ever published came out of a computation. So this page computes it too, instead of quoting it. Greene's 1979 criterion says a circle of irrational winding is dead when the periodic orbits that approximate it stop being stable, and stability is arithmetic: the product of the map's own 2×2 tangent matrices along an orbit has a trace, the residue R = (2 − Tr M)/4 is a quarter of what that trace is missing from 2, and the kick at which R reaches ¼ is what gets bisected. Ten orders of golden-mean convergents up to period 377 land on 0.971635239 ± 1.85e-6 — a bracket that contains MacKay's value — and agree with this lab's own headless recovery to 1.1e-7 without sharing a line of code with it. WHICH circle is an input, and that is the whole point: the rotation number ω = [0; m, m, m, …] = (√(m²+4) − m)/2 is the golden mean only at m = 1, and running the same chain at m = 2 and m = 3 returns 0.957953 and 0.891360. So the claim this world exists to make — that the golden circle is the last one standing — is a number on this page rather than a sentence, and a stored 0.9716 is wrong the moment you ask about a different torus. Four other things here need no iteration at all. At the hyperbolic fixed point the exponent is exact, λ = ln((2+K+√(K(K+4)))/2), and it inverts exactly: a measured λ fixes K = 4·sinh²(λ/2) with no fitting, which is how this lab's own module is calibrated — its two pinned budgets extrapolate onto that formula to 5.6e-14. At K = ½ the multiplier is exactly 2, so λ is exactly ln 2. The two fixed points are one trace each, 2 ∓ K, giving residues ∓K/4 and a genuine landmark at K = 4 where the elliptic one period-doubles — the reason the lab's transport seeds have to avoid it. And the rival is assembled rather than quoted: a pendulum separatrix of half-width 2√K facing another one across a 2π gap in momentum gives π²/4 ≈ 2.467, which is 154% high, and the page reports what coverage the criterion would need to be right instead of repairing it. Three refusals are stated where they belong. This page will not extrapolate its own sequence — Aitken Δ² helps by 3.4× at low order and is worse than the raw answer at high order, with the same code, and an accelerator whose sign you cannot predict is not a measurement. It will not call the residue it lands on a recovery of MacKay's critical residue, because the criterion solved for ¼ and ¼ is already closer to that constant than the arithmetic managed. And it will not quote thresholds for the m ≥ 4 families, where the orbit search loses the branch it was following.