Double Pendulum (Lagrangian Chaos)

Equivalent forms

E = (m_1{+}m_2)gl_1(1-\cos\theta_0) + m_2 gl_2(1-\cos\theta_0)

\ddot\theta_1, \ddot\theta_2 \;\text{from E--L: two coupled nonlinear ODEs}

A two-line Lagrangian whose solutions cannot be written down — the cleanest demonstration that determinism ≠ predictability.

Symbol	Name	SI unit	Dimension	Range
$E$	Release energyoutput Mechanical energy above the hanging state when released from rest at θ₀	J	ML^2T^-2	0 – 200
$m1$	Mass 1 Upper bob mass	kg	M	0.1 – 5
$m2$	Mass 2 Lower bob mass	kg	M	0.1 – 5
$l1$	Length 1 Upper rod length	m	L	0.2 – 2
$l2$	Length 2 Lower rod length	m	L	0.2 – 2
$theta0$	Release angle Initial angle of both rods from vertical (released from rest)	deg	1	5 – 180

Unit systems

Symbol	SI	CGS	Imperial
$E$	J	erg	ft·lbf
$m1$	kg	g	lb
$m2$	kg	g	lb
$l1$	m	cm	ft
$l2$	m	cm	ft
$theta0$	deg	deg	deg

Where it holds

Rigid massless rods, point bobs, frictionless pivots, planar motion. Linearized solution valid only for small angles; the general case must be integrated numerically.

Dimensional analysis

ℒ

\to [ML^{2}T^{-2}]

ml^{2}\theta

̇

^{2} \to [M][L^{2}][T^{-2}]

;

mgl \cos \theta \to [M][LT^{-2}][L] = [ML^{2}T^{-2}]

Every term is an energy.

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Discovery

Euler & Daniel Bernoulli (system); chaos understood after Poincaré · 1738

Euler and Daniel Bernoulli studied linked pendulums in the 1730s and solved the small-angle case (two clean normal modes). The full swinging system resisted everyone — because, as Poincaré's work around 1890 first hinted and 20th-century chaos theory confirmed, it is non-integrable: no formula for its future exists. It became the lab icon of deterministic chaos.

Try this

Two rods, two bobs, zero randomness — yet nudge the start by 0.001° and within seconds the two futures have nothing in common.

Release a double pendulum and its near-identical twin from almost the same angle and watch deterministic chaos tear their trajectories apart.

Research status: stable

Real-world applications

Benchmark system for chaos theory and Lyapunov-exponent estimation
Robotics: a two-link arm is exactly this system plus motors (the 'acrobot')
Human gait and limb-swing modeling
Testing symplectic and variational integrators on a brutally nonlinear target

Common misconceptions

Chaos means randomness — the system is perfectly deterministic; it is prediction that fails, not law
The motion never repeats because energy is lost — even the ideal frictionless system is aperiodic
A better computer could forecast it indefinitely — error doubling is exponential; each digit of precision buys only a fixed extra time slice

Experimental verification

Bench-top double pendulums with optical encoders confirm the equations and the measured Lyapunov exponents; the small-angle normal-mode frequencies match theory to within bearing-friction corrections.

Derivation

Positions:

x_{1} = l_{1}\sin \theta _{1}

,

y_{1} = -l_{1}\cos \theta _{1}

;

x_{2} = x_{1} + l_{2}\sin \theta _{2}

,

y_{2} = y_{1} - l_{2}\cos \theta _{2}

.

T = \tfrac{1}{2}m_{1}

(ẋ

_{1}^{2}

+ẏ

_{1}^{2}) + \tfrac{1}{2}m_{2}

(ẋ

_{2}^{2}

+ẏ

_{2}^{2})

,

V = m_{1}gy_{1} + m_{2}gy_{2}

.

Substituting gives the Lagrangian above; the two Euler–Lagrange equations yield coupled nonlinear ODEs

for \theta

̈

_{1}

,

\theta

̈

_{2}

whose denominators share the factor

2m_{1} + m_{2} - m_{2}\cos (2\theta _{1}-2\theta _{2})

.

Limiting cases

Small angles

(\theta

≪ 1)⟶ Two normal modes, fully predictableLinearized system: in-phase and counter-phase oscillations with fixed frequencies — no chaos.

m_{2} \to 0

⟶ Simple pendulum (upper rod)The lower bob becomes a passive tracer; integrability returns.

Large release angles⟶ Chaotic regime, positive Lyapunov exponentNearby initial conditions diverge exponentially

(\lambda \approx a few /s

for metre-scale rigs).

What if…

What if I release it from

5^{\circ}

instead of

120^{\circ}

?

Near-linear regime: two clean normal modes, no chaos — chaos needs the nonlinearity that large angles unlock.

What if I know the start to 15 decimal places?

You gain only seconds of forecast: with errors doubling every $\sim \tfrac{1}{2} s$ , even machine precision evaporates in about 20 s.

What if I make

m_{1}

≫

m_{2}

?

The upper pendulum barely feels the lower one and swings almost periodically, while the light lower bob still whips around chaotically.

1

Release energy at 120°

Given ·

m1:: 1
m2:: 1
l1:: 1
l2:: 1
theta0:: 120

Find · E

Steps

$\cos 120^{\circ} = -0.5$ , so $(1-\cos \theta _{0}) = 1.5$
Upper-bob chain: $(m_{1}+m_{2})gl_{1}(1.5) = 2\times 9.8\times 1\times 1.5 = 29.4\,\mathrm{J}$
Lower extra drop: $m_{2}gl_{2}(1.5) = 1\times 9.8\times 1\times 1.5 = 14.7\,\mathrm{J}$
$E = 29.4 + 14.7 = 44.1\,\mathrm{J}$ — conserved forever after release

Result ·

E = (m_{1}+m_{2})gl_{1}(1-\cos \theta _{0}) + m_{2}gl_{2}(1-\cos \theta _{0}) = 2\times 9.8\times 1.5 + 9.8\times 1.5 = 44.1\,\mathrm{J}

2

Small-angle normal modes (m₁=m₂, l₁=l₂=l)

Given ·

m1:: 1
m2:: 1
l1:: 1
l2:: 1

Find · Mode frequencies

Steps

Linearize the E–L equations about $\theta _{1}=\theta _{2}=0$
Mode frequencies: $\omega ^{2} = (2 \pm \sqrt{2}) g/l$
$g = 9.8$ , $l = 1$ : $\omega _{-} = \sqrt{0.586\times 9.8} \approx 2.40\,\mathrm{rad/s}$ (in-phase), $\omega _{+} = \sqrt{3.414\times 9.8} \approx 5.78\,\mathrm{rad/s}$ (counter-phase)

Result ·

\omega \pm = \surd ((2\pm \sqrt{2})g/l)

:

\omega _{+} \approx 5.78\,\mathrm{rad/s}

,

\omega _{-} \approx 2.40\,\mathrm{rad/s}

Euler–Lagrange Equation→Simple Pendulum Period→Hamilton's Canonical Equations→Simple Harmonic Motion Period→Principle of Stationary Action→