The Pendulum Phase Portrait and Separatrix

The simple pendulum is the canonical nonlinear conservative system. Its equation of motion $\ddot\theta=-\omega_0^2\sin\theta$ (with $\omega_0^2=g/L$ ) cannot be solved in elementary functions — and yet you can know everything about it without solving anything. Conservation of energy pins every motion to a level curve of one function, and plotting those curves in the $(\theta,p_\theta)$ plane yields the pendulum phase portrait: arguably the most instructive picture in all of mechanics, because it displays every qualitative behaviour of a one-degree-of-freedom system on a single page. Three regimes appear at a glance. Low-energy states rock back and forth (libration), tracing closed loops that shrink, for small amplitude, into the ellipses of the harmonic oscillator. High-energy states whirl over the top forever (rotation), tracing open wavy curves. And separating them is a single critical trajectory — the separatrix — threaded through the unstable equilibrium at the inverted position, on which the pendulum takes literally infinite time to balance on top. Because $\theta$ and $\theta+2\pi$ are the same configuration, the true arena is a cylinder, and this center–saddle–separatrix anatomy turns out to be the universal fingerprint of any system with a periodic potential: Josephson junctions, planetary spin–orbit resonance, charged particles surfing a wave. Mastering this one portrait therefore buys you a template you will reuse across physics — and it is the cleanest stage on which Hamiltonian flow and Liouville's incompressible phase fluid perform, shearing dramatically near the separatrix while conserving every drop of area.

Key takeaways

The planar pendulum has Hamiltonian $H=\frac{p_\theta^2}{2mL^2}-mgL\cos\theta$ ; its level curves $H=E$ are the phase-space trajectories, so the whole portrait is a contour map — no integration required.
For $E<mgL$ the motion is bounded oscillation (libration): closed loops around the stable center at $(0,0)$ . For $E>mgL$ the bob swings over the top (rotation): open curves that wind around the cylindrical phase space.
The critical contour $E=mgL$ is the separatrix, passing through the saddle at $(\pm\pi,0)$ ; on it the period diverges, and the exact libration period is $T=\frac{4}{\omega_0}K(\sin\frac{\theta_0}{2})$ with the complete elliptic integral $K$ .
The separatrix carries the homoclinic orbit $\theta(t)=2\arcsin[\tanh(\omega_0 t)]$ — infinite traversal time, yet each homoclinic loop encloses the finite phase-space area $8mL^2\omega_0$ .
Phase-space area is conserved (Liouville) even as the flow shears violently near the saddle, and the center–saddle–separatrix topology recurs in every periodic-potential system from Josephson junctions to spin–orbit resonance.
Key idea: The level curves of the energy function $H(\theta,p_\theta)$ are the trajectories; the topology of these curves — closed loops (libration), open curves (rotation), and the critical separatrix through the saddle — encodes the entire dynamics without solving the equation of motion.

Where this lives today

The busiest pendulum phase portraits of 2026 are at millikelvin temperatures. The transmon qubit — the workhorse of superconducting quantum processors — is literally a quantized pendulum: the Josephson phase $\varphi$ plays $\theta$ , the junction's $-E_J\cos\varphi$ is the gravitational potential, and the charging energy supplies the kinetic term. The anharmonicity that makes a transmon addressable as a two-level system is the pendulum's period-vs-amplitude effect — the same physics as the $\theta_0^2/16$ of equation (6), reappearing as the nonuniform spacing of quantum levels in the cosine well — and driving a junction past its separatrix (the escape of the "phase particle" into the rotation regime) is exactly the switching event that turns a zero-voltage junction resistive. The same portrait organizes celestial mechanics and plasma physics. Mercury's 3:2 spin–orbit lock and the Moon's synchronous rotation are libration states of a pendulum-like resonance Hamiltonian, with the separatrix deciding capture versus passage; charged particles riding a plasma wave are trapped inside a moving separatrix, bouncing in the troughs — the mechanism behind nonlinear Landau damping; and when weak damping and driving are switched on, the clean homoclinic loop tears into Poincaré's tangle, the standard on-ramp to chaos. One drawing, learned once, keeps paying.

The physical picture

Picture energy as height. The potential $U(\theta)=-mgL\cos\theta$ is a corrugated landscape: valleys at $\theta=0,\pm2\pi,\dots$ where the bob hangs down, hilltops at $\theta=\pm\pi$ where it balances inverted. Roll a marble with too little energy and it rocks in a valley — libration. Give it enough to clear the hilltops and it rolls on forever — rotation. The separatrix is the energy that exactly reaches a hilltop, and a marble arriving at a hilltop with zero speed takes forever to get there. Translate each of those stories into the $(\theta,p_\theta)$ plane and the entire repertoire appears at once:

Figure

The pendulum phase portrait: nested libration loops, open rotation curves, and the red separatrix threading the saddles

Read the portrait like a map. The stable equilibrium (bob hanging down) is an elliptic center: nearby trajectories circle it in closed loops. The unstable equilibrium (bob inverted) is a hyperbolic saddle: trajectories approach along one direction and flee along another, and the red curves through it — the incoming and outgoing branches joined into a loop — form the separatrix. But the flat picture double-counts: $\theta=-\pi$ and $\theta=+\pi$ are the same physical configuration, so the honest phase space is a cylinder. Glue the two edges and the two saddles merge into one, the separatrix becomes a pair of homoclinic loops through that single point, and the "open" rotation curves close up into rings that wind around the cylinder:

Figure

Phase space is a cylinder: gluing θ equals −π to θ equals +π merges the two saddles into one and closes the rotation curves

Finally, think of time as flow. The phase fluid streams fastest where $|p_\theta|$ is largest — the bottom of the swing — and crawls to a near-standstill beside the saddle. A trajectory hugging the separatrix therefore spends almost all of its period loitering near the inverted balance point, which is why the period does not merely grow with amplitude but diverges as the amplitude approaches $180°$ . Even the divergence is gentle — logarithmic in the energy distance from the separatrix — and the curve below is computed, not sketched:

Figure

The period grows with amplitude and diverges logarithmically as the amplitude approaches 180 degrees

From energy contours to the portrait

Everything above becomes quantitative with one function. Start from the Lagrangian $\mathcal{L}=\tfrac12 mL^2\dot\theta^2+mgL\cos\theta$ : the conjugate momentum is $p_\theta=\partial \mathcal{L}/\partial\dot\theta=mL^2\dot\theta$ , and the Legendre transform $H=p_\theta\dot\theta-\mathcal{L}$ hands back the total energy expressed in phase-space coordinates,

H(\theta,p_\theta)=\frac{p_\theta^2}{2mL^2}-mgL\cos\theta \qquad (1)

where: $\theta$ — angle from the downward vertical; $p_\theta=mL^2\dot\theta$ — the angular momentum conjugate to $\theta$ ; $m,L$ — bob mass and rod length; $g$ — gravitational acceleration. Because the constraint is time-independent, $H$ equals the conserved energy $E$ . Note the two anchoring values of the potential: $U(0)=-mgL$ at the bottom, $U(\pm\pi)=+mgL$ at the top, so every trajectory is labelled by an energy $E\in[-mgL,\infty)$ . The Hamiltonian generates the flow through Hamilton's canonical equations — the velocity field that every phase point obeys,

\dot\theta=\frac{\partial H}{\partial p_\theta}=\frac{p_\theta}{mL^2},\qquad \dot p_\theta=-\frac{\partial H}{\partial\theta}=-mgL\sin\theta \qquad (2)

where: $(\dot\theta,\dot p_\theta)$ — the components of the phase velocity at the point $(\theta,p_\theta)$ . The second equation is just the gravitational torque. Setting both components to zero locates the fixed points: $p_\theta=0$ and $\sin\theta=0$ , i.e. the center at $(0,0)$ and the saddle at $(\pm\pi,0)$ — Derivation A classifies them by linearizing this field. Since $H$ is conserved, a trajectory with energy $E$ is confined to the level set $H=E$ ; solving equation (1) for the momentum gives the entire portrait in closed form,

p_\theta(\theta)=\pm\sqrt{2mL^2\,(E+mgL\cos\theta)} \qquad (3)

where: $E$ — the conserved energy labelling the contour; the $\pm$ gives the upper and lower branches. Real solutions require $E+mgL\cos\theta\ge0$ : for $E<mgL$ this restricts $\theta$ to $|\theta|\le\theta_0$ with $\cos\theta_0=-E/mgL$ , producing a closed libration loop with turning points at $\pm\theta_0$ ; for $E>mgL$ it holds for every $\theta$ , producing an open rotation curve. Every curve in the hero figure is this one equation at a different $E$ . The boundary case deserves its own line. Set $E=mgL$ in (3) and use the half-angle identity $1+\cos\theta=2\cos^2(\theta/2)$ : the square root collapses and the separatrix is a pair of cosine arches,

p_\theta^{\,\text{sep}}(\theta)=\pm\,2mL^2\omega_0\cos\frac{\theta}{2} \qquad (4)

where: $\omega_0=\sqrt{g/L}$ — the small-oscillation frequency. The arches meet, with zero momentum, exactly at the saddles $\theta=\pm\pi$ : the separatrix does not cross the saddle, it terminates on it from both sides. Its maximum height $2mL^2\omega_0$ at $\theta=0$ is the momentum of a bob that passes the bottom with barely enough energy to reach the top. How long does one libration take? Integrating $\mathrm{d}t=\mathrm{d}\theta/\dot\theta$ around a loop (Derivation B) produces an answer that is exact for every amplitude,

T_{\text{lib}}(\theta_0)=\frac{4}{\omega_0}\,K\!\left(\sin\frac{\theta_0}{2}\right) \qquad (5)

where: $\theta_0$ — the amplitude (turning angle); $K(k)=\int_0^{\pi/2}\frac{\mathrm{d}\phi}{\sqrt{1-k^2\sin^2\phi}}$ — the complete elliptic integral of the first kind. As $\theta_0\to0$ , $K\to\pi/2$ and $T\to2\pi/\omega_0$ : the harmonic result. As $\theta_0\to\pi$ , the modulus $k\to1$ and $K$ diverges logarithmically — the infinite period of the separatrix, made precise below. For laboratory amplitudes the elliptic integral hides a simple and famous correction. Expanding $K$ for small modulus gives

T_{\text{lib}}\approx\frac{2\pi}{\omega_0}\left(1+\frac{\theta_0^2}{16}+\frac{11\,\theta_0^4}{3072}+\cdots\right) \qquad (6)

where: $\theta_0$ — the amplitude in radians. The leading term is Galileo's (approximate) isochronism; the $\theta_0^2/16$ term is the nonlinearity announcing itself — at $90°$ amplitude the true period already runs $18\%$ long, the effect Huygens battled when pendulum clocks became navigation instruments. On the separatrix itself the trajectory can be written down exactly. Integrating the separatrix relation $\dot\theta=2\omega_0\cos(\theta/2)$ (Derivation C) yields the homoclinic orbit,

\theta_h(t)=2\arcsin\!\big[\tanh(\omega_0 t)\big],\qquad p_\theta(t)=\frac{2mL^2\omega_0}{\cosh(\omega_0 t)} \qquad (7)

where: $\theta_h(t)$ — the trajectory that leaves the inverted position at $t\to-\infty$ and asymptotically regains it as $t\to+\infty$ . The $\tanh$ never reaches $\pm1$ at finite time: the bob spends a finite interval sweeping through the bottom and two infinite tails creeping toward the top. This soliton-shaped pulse in $p_\theta(t)$ — all the action compressed into a few $\omega_0^{-1}$ — is the time-domain face of the separatrix.

Which equation answers which question

you want	use	notes
the energy of a state	$H$ , eq. (1)	conserved; labels the contour
the direction of flow	eq. (2)	clockwise around the center
the trajectory through a point	eq. (3)	closed if $E<mgL$ , open if $E>mgL$
the libration–rotation boundary	eq. (4)	cosine arches through the saddles
the period at any amplitude	eq. (5)	exact, via $K(\sin(\theta_0/2))$
the period at lab amplitudes	eq. (6)	$+\theta_0^2/16$ correction
the motion on the separatrix	eq. (7)	infinite time, finite loop area

Derivation ALocating and classifying the fixed points

We are finding where the phase flow stands still and proving that the two equilibria have opposite characters — one a center that organizes the libration loops, one a saddle that anchors the separatrix. The strategy is the standard one for any portrait: set the velocity field (2) to zero to locate the fixed points, then linearize the flow about each and read the character from the eigenvalues of the linearization. The thing to watch is the sign of the restoring term: a single sign flip in $\sin\theta$ near $\theta=\pi$ converts oscillation into exponential escape, imaginary eigenvalues into real ones.

Step 1. Set the velocity field to zero. From equation (2), $\dot\theta=p_\theta/mL^2=0$ forces $p_\theta=0$ , and $\dot p_\theta=-mgL\sin\theta=0$ forces $\sin\theta=0$ . On the cylinder that leaves exactly two equilibria: $(\theta,p_\theta)=(0,0)$ , the bob hanging at rest, and $(\pi,0)$ , the bob balanced inverted.

Step 2. Linearize about the bottom. Write $\theta=0+\delta\theta$ , $p_\theta=0+\delta p_\theta$ and keep first order, using $\sin\delta\theta\approx\delta\theta$ :

\delta\dot\theta=\frac{\delta p_\theta}{mL^2},\qquad \delta\dot p_\theta=-mgL\,\delta\theta

Combining, $\delta\ddot\theta=-\omega_0^2\,\delta\theta$ — the harmonic oscillator. The eigenvalues of the linearized flow matrix are $\lambda=\pm i\omega_0$ : purely imaginary, so nearby trajectories neither grow nor decay but circulate. This is an elliptic center, and the nested loops around it are the small-amplitude ellipses of the oscillator. Why no spiraling: the eigenvalues have zero real part because the system is conservative — Hamiltonian flow cannot have attractors, so the borderline circulation is exact, protected by energy conservation rather than an accident of linearization.

Step 3. Linearize about the top. Write $\theta=\pi+\delta\theta$ ; now $\sin(\pi+\delta\theta)=-\sin\delta\theta\approx-\delta\theta$ , so the torque term flips sign:

\delta\dot\theta=\frac{\delta p_\theta}{mL^2},\qquad \delta\dot p_\theta=+mgL\,\delta\theta

Now $\delta\ddot\theta=+\omega_0^2\,\delta\theta$ , with eigenvalues $\lambda=\pm\omega_0$ : real, one positive and one negative. Perturbations along the unstable eigendirection grow as $e^{+\omega_0 t}$ , those along the stable one shrink as $e^{-\omega_0 t}$ . This is a hyperbolic saddle, and its stable and unstable manifolds — the directions of approach and escape — are precisely the four separatrix branches meeting at the point. Why the same number appears twice: the growth rate at the saddle equals the oscillation frequency at the center, $\omega_0$ , because both come from the same curvature $|U''|=mgL$ of the cosine potential at its extremes. This coincidence is special to the pendulum, not generic. *Verified: eigenvalue signs match the standard result — stable equilibrium at the bottom, unstable inverted equilibrium (Goldstein Ch. 6; Taylor Ch. 12); the saddle rate $\omega_0$ is the $e$ -folding time constant used in the period-divergence estimate of Derivation B.*

Go deeper · Derivation B · 2 minThe exact period from the energy integral

Derivation BThe exact period from the energy integral

We are deriving the exact libration period (5), extracting from it the small-amplitude expansion (6), and pinning down precisely how the period blows up at the separatrix. The route is the classic one-degree-of-freedom reduction: energy conservation gives $\dot\theta(\theta)$ , the period is the time integral $\oint\mathrm{d}\theta/\dot\theta$ , and a half-angle substitution converts it to the complete elliptic integral $K$ . Watch the modulus $k=\sin(\theta_0/2)$ : every property of the period — isochronism, the $\theta_0^2/16$ correction, the logarithmic divergence — is a standard property of $K(k)$ at small $k$ or at $k\to1$ .

Step 1. Time from energy conservation. For libration with amplitude $\theta_0$ the energy is $E=-mgL\cos\theta_0$ , and (3) gives $\dot\theta=\pm\omega_0\sqrt{2(\cos\theta-\cos\theta_0)}$ . A full period is four quarter-swings from $0$ to $\theta_0$ :

T=\frac{4}{\omega_0}\int_0^{\theta_0}\frac{\mathrm{d}\theta}{\sqrt{2(\cos\theta-\cos\theta_0)}}

The integrand diverges at the turning point $\theta=\theta_0$ , but only as an inverse square root — integrable, exactly as for the harmonic oscillator.

Step 2. Half-angle substitution. Use $\cos\theta=1-2\sin^2(\theta/2)$ in both cosines, so $\cos\theta-\cos\theta_0=2[k^2-\sin^2(\theta/2)]$ with $k\equiv\sin(\theta_0/2)$ , then substitute $\sin(\theta/2)=k\sin\phi$ so that the turning point $\theta=\theta_0$ maps to $\phi=\pi/2$ :

T=\frac{4}{\omega_0}\int_0^{\pi/2}\frac{\mathrm{d}\phi}{\sqrt{1-k^2\sin^2\phi}}=\frac{4}{\omega_0}\,K(k)

Why this substitution: it simultaneously rationalizes the square root and absorbs the turning-point singularity into the smooth upper limit $\phi=\pi/2$ — the standard trick that defines the elliptic modulus, here $k=\sin(\theta_0/2)$ rather than $\theta_0$ itself.

Step 3. Small amplitudes. Expand $K(k)=\frac{\pi}{2}\left[1+\frac{k^2}{4}+\frac{9k^4}{64}+\cdots\right]$ and $k^2=\sin^2(\theta_0/2)=\frac{\theta_0^2}{4}-\frac{\theta_0^4}{48}+\cdots$ ; collecting orders,

T=\frac{2\pi}{\omega_0}\left(1+\frac{\theta_0^2}{16}+\frac{11\,\theta_0^4}{3072}+\cdots\right)

which is equation (6). At $\theta_0=\pi/2$ the series gives $1.1760$ against the exact $2K(\sin45°)/\pi=1.1803$ — already $0.4\%$ shy, a warning that the series is asymptotic in usefulness, not a substitute for $K$ at large swing.

Step 4. The separatrix limit. As $\theta_0\to\pi$ , $k\to1$ , where the elliptic integral has the standard logarithmic asymptote $K(k)\to\ln(4/k')$ with $k'=\sqrt{1-k^2}=\cos(\theta_0/2)$ . Convert to energy: $k'^2=\cos^2(\theta_0/2)=\tfrac12(1+\cos\theta_0)=\tfrac12(1-E/mgL)$ , so with $\epsilon\equiv1-E/mgL$ ,

T\;\sim\;\frac{4}{\omega_0}\ln\frac{4}{\sqrt{\epsilon/2}}\;=\;\frac{2}{\omega_0}\ln\frac{32}{\epsilon}\qquad(E\to mgL^-)

The period grows without bound, but only logarithmically: halving the energy gap adds a fixed increment $(2/\omega_0)\ln2$ to the period, which is why real pendulums released ever closer to inverted show frustratingly modest period gains. Why the rate is set by the saddle: the diverging time is spent within the linear neighbourhood of the saddle, where motion e-folds at the rate $\omega_0$ found in Derivation A — so the coefficient of the logarithm had to be $\propto\omega_0^{-1}$ before any computation. *Verified: numerically, at $\epsilon=0.01$ the asymptote gives $T=5.156\,\mathrm{s}$ against the exact $4K/\omega_0=5.161\,\mathrm{s}$ for $L=1\,\mathrm{m}$ ( $0.09\%$ ); the expansion coefficients $1/16$ and $11/3072$ match Landau–Lifshitz §11; $K(\sin45°)=1.85407$ reproduces the $18.03\%$ correction at $90°$ amplitude.*

Go deeper · Derivation C · 2 minThe homoclinic orbit and the separatrix action

Derivation CThe homoclinic orbit and the separatrix action

We are solving the motion on the separatrix in closed form, confirming that it takes infinite time to traverse, and computing the phase-space area its loop encloses — a finite number, $8mL^2\omega_0$ per homoclinic loop, despite the infinite period. The strategy: on the separatrix the trajectory equation (4) turns Hamilton's first equation into a separable ODE for $\theta(t)$ alone; integrate it, invert, and then evaluate $\oint p_\theta\,\mathrm{d}\theta$ directly. The watch-out is a bookkeeping one: "the separatrix" is two arches, and area statements must say whether they mean one homoclinic loop or the full libration cell they jointly bound.

Step 1. Reduce to a separable equation. On the upper separatrix branch, $p_\theta=2mL^2\omega_0\cos(\theta/2)$ , and $\dot\theta=p_\theta/mL^2$ gives $\dot\theta=2\omega_0\cos(\theta/2)$ . Separate:

\int\frac{\mathrm{d}\theta}{2\cos(\theta/2)}=\int\omega_0\,\mathrm{d}t \quad\Longrightarrow\quad \ln\left|\sec\frac{\theta}{2}+\tan\frac{\theta}{2}\right|=\omega_0 t

choosing $t=0$ at the bottom of the swing, $\theta(0)=0$ .

Step 2. Invert. Exponentiating and converting the secant–tangent combination gives $\sin(\theta/2)=\tanh(\omega_0 t)$ , i.e.

\theta_h(t)=2\arcsin\!\big[\tanh(\omega_0 t)\big]=4\arctan\!\big(e^{\omega_0 t}\big)-\pi

and substituting back, $p_\theta(t)=2mL^2\omega_0\cos(\theta_h/2)=2mL^2\omega_0/\cosh(\omega_0 t)$ — equation (7). As $t\to\pm\infty$ , $\tanh\to\pm1$ and $\theta_h\to\pm\pi$ : the orbit departs from and returns to the same saddle (one point on the cylinder), which is what the prefix homo- in homoclinic records. Why infinite time was inevitable: the saddle is itself a complete trajectory, and distinct trajectories of a smooth flow can never merge in finite time — the separatrix can only approach it asymptotically, at the linear rate $e^{-\omega_0|t|}$ visible in the tails of the tanh.

Step 3. The enclosed action. The area under one homoclinic loop (the upper arch, traversed once around the cylinder) is

\oint p_\theta\,\mathrm{d}\theta=\int_{-\pi}^{\pi}2mL^2\omega_0\cos\frac{\theta}{2}\,\mathrm{d}\theta=2mL^2\omega_0\Big[2\sin\frac{\theta}{2}\Big]_{-\pi}^{\pi}=8mL^2\omega_0

The upper and lower loops together bound the full libration island, of area $16mL^2\omega_0$ . Both numbers are finite: the separatrix encloses finite phase-space area even though the period on it is infinite. A blob of phase fluid straddling the region is sheared without limit — the part inside circulates, the part outside rotates away — yet by Liouville's theorem its area never changes by a single unit. Why this area matters beyond the pendulum: it is the maximum action of any libration orbit, so in semiclassical quantization it counts the bound states trapped in the well ( $N\approx16mL^2\omega_0/2\pi\hbar$ ), and in adiabatic theory it sets the critical action below which slowly-driven systems remain captured in resonance. *Verified: enclosed area $8mL^2\omega_0$ recomputable by direct integration as shown ( $\int_{-\pi}^{\pi}\cos(\theta/2)\,\mathrm{d}\theta=4$ ); the homoclinic solution $\theta=2\arcsin(\tanh\omega_0 t)$ is standard (Taylor Ch. 12) and satisfies $\dot\theta=2\omega_0\cos(\theta/2)$ by direct differentiation.*

Limiting cases

The pendulum interpolates between two familiar idealizations as the energy dial turns:

case	result	meaning
small oscillations, $E\to-mgL$	$\theta(t)\approx\theta_0\cos(\omega_0 t)$ , $T\to2\pi/\omega_0$	loops become harmonic-oscillator ellipses; period amplitude-independent
near the separatrix, $E\to mgL$	$T\sim\frac{2}{\omega_0}\ln\frac{32}{1-E/mgL}$	period diverges, but only logarithmically — the trajectory loiters at the saddle
on the separatrix, $E=mgL$	$\theta(t)=2\arcsin[\tanh(\omega_0 t)]$ , $T\to\infty$	the homoclinic orbit, asymptotic to the inverted balance point
fast rotation, $E\gg mgL$	$\dot\theta\approx\text{const}$ , $p_\theta\approx\pm\sqrt{2mL^2E}$	gravity is a small ripple; phase curves flatten into near-horizontal lines

Common trapThe separatrix is just the biggest possible swing

It is not — and the difference is the whole point. The separatrix is a limiting trajectory that takes infinite time to traverse: the pendulum approaches the inverted point asymptotically, as $e^{-\omega_0 t}$ , and never crosses it in finite time. Nor does the period stay finite and then jump to infinity — it diverges continuously and logarithmically as $E\to mgL$ from either side, so "just below" and "just above" the critical energy both mean extremely slow motion near the top. A second conflation to avoid: phase space here is not a plane. Since $\theta$ is periodic the space is a cylinder; rotation trajectories are genuinely closed loops around it, the two saddles of the flat picture are one point, and the separatrix consists of two homoclinic loops through that single saddle — not four curves crossing at an X. Nothing ever crosses at the saddle: those are four asymptotic branches of distinct trajectories.

Worked examples

1 · A one-metre pendulum released at ninety degrees (confidence builder)

A pendulum of length $L=1.00\,\mathrm{m}$ ( $g=9.80\,\mathrm{m/s^2}$ ) is released from rest at $\theta_0=90°$ . Find the exact period, compare with the small-angle prediction, and check the amplitude expansion.

Step 1. The scales. $\omega_0=\sqrt{g/L}=\sqrt{9.80}=3.130\,\mathrm{rad/s}$ , so the small-angle period is $T_0=2\pi/\omega_0=2.007\,\mathrm{s}$ .

Step 2. The elliptic modulus. $k=\sin(\theta_0/2)=\sin45°=0.7071$ . The complete elliptic integral at this modulus is $K(0.7071)=1.85407$ (the small-angle limit would be $\pi/2=1.5708$ — the integral has grown $18\%$ ).

Step 3. Exact period from equation (5): $T=\dfrac{4K}{\omega_0}=\dfrac{4\times1.85407}{3.130}=\boxed{2.369\,\mathrm{s}}$ , a ratio $T/T_0=2K/\pi=1.1803$ — the true period runs $\mathbf{18.0\%}$ longer than the small-angle formula. A pendulum clock calibrated at tiny amplitude but swung to $90°$ would lose roughly nine minutes per hour.

Step 4. Check the series (6): $\theta_0=\pi/2=1.5708\,\mathrm{rad}$ gives $\theta_0^2/16=0.1542$ and $11\theta_0^4/3072=0.0218$ , so $T/T_0\approx1.1760$ — within $0.4\%$ of exact. The quadratic term alone ( $1.154$ ) would be $2\%$ off: at quarter-turn amplitudes the pendulum is already visibly nonlinear.

2 · Released one degree from inverted (exam level)

The same pendulum is released from rest at $\theta_0=179°$ — one degree short of balancing on top. Find the period exactly and from the near-separatrix asymptote, and comment on how the period scales as the release angle creeps toward $180°$ .

Step 1. Locate the energy relative to the separatrix. $E=-mgL\cos179°=+0.99985\,mgL$ : the energy gap is $\epsilon=1-E/mgL=1+\cos179°=1.523\times10^{-4}$ — the pendulum is within $0.02\%$ of the critical energy.

Step 2. Exact period. $k=\sin(89.5°)=0.9999619$ , so $k'^2=1-k^2=7.615\times10^{-5}$ and $K=6.1278$ . Then $T=\dfrac{4K}{\omega_0}=\dfrac{4\times6.1278}{3.130}=\boxed{7.83\,\mathrm{s}}$ — only $3.9$ times the small-angle period, despite being one degree from an infinite-period orbit.

Step 3. The logarithmic asymptote from Derivation B: $T\approx\dfrac{2}{\omega_0}\ln\dfrac{32}{\epsilon}=\dfrac{2}{3.130}\ln(2.10\times10^{5})=0.639\times12.26=\boxed{7.830\,\mathrm{s}}$ , agreeing with the exact value to $0.002\%$ — at $\epsilon\sim10^{-4}$ the asymptotic regime is fully in charge.

Step 4. The scaling moral. Each factor of $10$ reduction in $\epsilon$ adds only $\Delta T=(2/\omega_0)\ln10=1.47\,\mathrm{s}$ — and since $\epsilon\approx(\pi-\theta_0)^2/2$ , shaving the release angle tenfold closer to inverted cuts $\epsilon$ a hundredfold and adds just $2.94\,\mathrm{s}$ : releasing at $179.9°$ gives $T\approx10.8\,\mathrm{s}$ , at $179.99°$ only $T\approx13.7\,\mathrm{s}$ . To reach even a one-minute period you would need $\epsilon\sim10^{-40}$ — an angular precision of $10^{-18}$ degrees, hopeless against thermal kicks and air currents. Infinite period is a mathematical boundary, never an experimental observation — which is precisely why the separatrix is drawn, not demonstrated.

Origins · 1 minChristiaan Huygens and how this was discovered

Christiaan Huygens

1629–1695 · The Hague; pendulum clocks and the isochronism problem

Huygens turned the pendulum from Galileo's curiosity into the precision technology of its century — and in doing so ran straight into the physics of this article. His 1656 pendulum clock improved timekeeping accuracy roughly a hundredfold, but navigation demanded more, and Huygens discovered the obstruction: a circular-arc pendulum is not isochronous, its period growing with amplitude — the effect quantified by the $\theta_0^2/16$ term of equation (6). His response, published in the magisterial Horologium Oscillatorium (1673), was to prove that the cycloid is the true isochronous curve and to force the bob onto one with curved metal "cheeks," meanwhile inventing the theory of the compound pendulum and the center of oscillation. Two centuries before phase space existed, Huygens had measured its geometry: the widening of the period with amplitude is nothing but the crowding of the libration loops toward the separatrix.

History

1656

— Huygens builds the first pendulum clock, making the pendulum the world's timekeeping standard for the next 270 years.

1673

— Horologium Oscillatorium: Huygens proves the circular pendulum anisochronous and identifies the cycloid as the isochronous path.

1834

— Hamilton recasts dynamics as first-order flow in the coordinates and momenta — the arena in which the portrait lives.

1889

— Poincaré discovers that a perturbed separatrix splits into a homoclinic tangle, the founding mechanism of deterministic chaos.

1962

— Josephson predicts the superconducting junction whose phase obeys the pendulum equation, giving the portrait a solid-state career.

Going further

References. Morin, Introduction to Classical Mechanics, Ch. 5 & Appendix on the pendulum · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., Ch. 6–7 · Landau & Lifshitz, Mechanics, 3rd ed., §11, §16 (finite-amplitude pendulum) · Taylor, Classical Mechanics, Ch. 12 (nonlinear oscillations & phase space) · Strogatz, Nonlinear Dynamics and Chaos, Ch. 6 (the pendulum on the cylinder)

Provenance. Hamiltonian, Hamilton's equations, and fixed-point classification are standard (Goldstein Ch. 6, Taylor Ch. 12). Exact period $T=4K(\sin(\theta_0/2))/\omega_0$ and amplitude expansion $1+\theta_0^2/16+11\theta_0^4/3072$ verifiable against Landau–Lifshitz §11 and standard elliptic-integral references. Separatrix curve, homoclinic solution $\theta=2\arcsin(\tanh\omega_0 t)$ , and enclosed action $8mL^2\omega_0$ recomputable by direct integration. A verifier should recheck the half-angle reduction, the $K(k\to1)$ asymptotics, and the eigenvalue signs $\pm i\omega_0$ vs $\pm\omega_0$ .

Builds on: The Hamiltonian Function · Hamiltonian as Total Energy · Hamilton's Canonical Equations · Phase Portraits and Trajectories · Fixed Points and Stability in Phase Space. Leads to: The Harmonic Oscillator in Phase Space · Liouville's Theorem · Hamiltonian Flow and the Phase Fluid.

Why does the period diverge only logarithmically rather than as a power law?

Because the slowdown happens in the linear neighbourhood of the saddle. Near $(\pm\pi,0)$ the flow is governed by the eigenvalues $\pm\omega_0$ : a trajectory entering at distance $\delta$ from the saddle along the stable direction takes time $\sim\omega_0^{-1}\ln(1/\delta)$ to pass by, since motion there e-folds rather than proceeding at fixed speed. The energy gap sets $\delta\propto\sqrt{\epsilon}$ , so $T\sim(2/\omega_0)\ln(32/\epsilon)$ — the $\ln$ is the fingerprint of exponential dynamics at a hyperbolic point. Power-law divergences require different local structure (e.g. a degenerate, non-quadratic maximum of the potential, where the linearization vanishes). Quick test of understanding: the coefficient $2/\omega_0$ comes from passing two saddle neighbourhoods per period — once at each end of the swing.

Is the separatrix a trajectory a real pendulum can actually be on?

In principle it is a legitimate solution of the equations; in practice it has measure zero and is dynamically unreachable. Worked example 2 shows the futility: even $\epsilon\sim10^{-4}$ in energy buys only $T\approx7.8\,\mathrm{s}$ , and each additional decade of fine-tuning adds a mere $1.47\,\mathrm{s}$ . Thermal fluctuations, air currents, and release jitter set a floor on $\epsilon$ many orders of magnitude above anything interesting. Moreover the separatrix is structurally fragile: any damping destroys it instantly (the saddle connection breaks), and any periodic forcing splits it into a homoclinic tangle. Its importance is not as an achievable motion but as the organizing boundary — the watershed whose mere existence dictates the behaviour of everything near it.

What happens to this portrait when I add damping or driving?

Damping first: the center becomes a spiral sink, the libration loops become inward spirals, and the separatrix ceases to exist as a loop — the saddle keeps its stable and unstable manifolds, but they no longer join; the unstable manifold spirals into the sink. Energy is no longer conserved, so trajectories cut across the old contours. Add periodic driving and the story becomes epochal: the stable and unstable manifolds of the (now periodically flickering) saddle intersect transversally, and one intersection forces infinitely many — Poincaré's homoclinic tangle. Inside the tangle lies a Smale horseshoe and with it deterministic chaos. This is why the damped driven pendulum is the standard first chaotic system: the undamped portrait you learn here is the skeleton the tangle grows on.

Rotation curves never return to the same point on the plane — in what sense is rotation periodic?

On the plane the curve $p_\theta(\theta)$ marches off to $\theta\to\pm\infty$ , which looks aperiodic. But the plane over-describes the system: $\theta=\phi$ and $\theta=\phi+2\pi$ are the same bob position. On the true cylindrical phase space the rotation curve closes after one trip around, and the motion is periodic in every observable sense — $\theta(t)\bmod 2\pi$ , $p_\theta(t)$ , the position of the bob — with period $T_{\text{rot}}=\oint\mathrm{d}\theta/\dot\theta$ over one winding. What distinguishes rotation from libration is not periodicity but topology: a libration loop is contractible on the cylinder (you can shrink it to the center), a rotation loop winds around and cannot be contracted. That winding number is why no continuous deformation of trajectories can convert one into the other without passing through the separatrix.

Why does the elliptic modulus come out as sine of half the amplitude instead of the amplitude itself?

The half-angle is forced by the shape of the cosine potential. The period integrand involves $\cos\theta-\cos\theta_0$ , and the identity $\cos\theta=1-2\sin^2(\theta/2)$ turns this difference into $2[\sin^2(\theta_0/2)-\sin^2(\theta/2)]$ — a difference of squares, which is exactly the form the elliptic substitution $\sin(\theta/2)=k\sin\phi$ rationalizes. The modulus $k=\sin(\theta_0/2)$ then has a clean physical reading: $k^2=(E+mgL)/2mgL$ is the energy above the well bottom measured as a fraction of the well depth $2mgL$ . Libration is $0\le k<1$ , the separatrix is $k=1$ , and rotation corresponds to $1/k<1$ — the standard trick is to transform rotation-regime integrals to modulus $1/k$ .

Liouville says areas are conserved, but trajectories near the saddle separate exponentially. How are both true?

Because stretching in one direction is exactly compensated by squeezing in the conjugate direction. Near the saddle the linearized flow expands the unstable direction as $e^{+\omega_0 t}$ and contracts the stable one as $e^{-\omega_0 t}$ : the product of the two factors is exactly $1$ at every instant, so a small parallelogram of states keeps its area while becoming absurdly elongated. This is shear, not spreading. Note also what this is not: exponential separation at one isolated saddle does not make the pendulum chaotic — the system is integrable, every trajectory is a level curve, and two nearby libration orbits separate only linearly in time (through their period difference). Chaos requires the exponential stretching to be recurrent, which is what the homoclinic tangle of the driven pendulum provides.

Does the number 8mL²ω₀ for the separatrix loop area have any physical use, or is it a curiosity?

It earns its keep in at least three places. Semiclassically, the number of quantum bound states in the cosine well is the libration-cell area over Planck's cell: $N\approx16mL^2\omega_0/(2\pi\hbar)$ — for a transmon qubit this counts the levels below the junction barrier, a design parameter engineers actually tune. In adiabatic theory, the action $\oint p\,\mathrm{d}\theta$ is the conserved quantity when parameters drift slowly, and the separatrix area is the critical action separating orbits that stay trapped in a slowly-modulated resonance from those that escape — the quantitative core of capture-into-resonance calculations in celestial mechanics. And in kinetic plasma theory, the separatrix area of the wave-frame pendulum measures how much phase space a wave can trap, setting the trapped-particle fraction in nonlinear Landau damping.

Where exactly does a quantum pendulum notice the separatrix?

In the spectrum. The quantum pendulum (Mathieu equation) has levels that sit, semiclassically, on quantized contours of the classical portrait. Deep in the well the spacing is nearly uniform, $\hbar\omega_0$ , with the anharmonic compression given by equation (6) translated to level spacings. Near the separatrix energy $E\approx mgL$ the classical period diverges, and since the local level spacing is $\Delta E\approx2\pi\hbar/T(E)$ , the spectrum bunches — a logarithmic pile-up of levels at the separatrix energy. Above it, states come in nearly degenerate clockwise/counterclockwise rotation pairs whose small splitting is the tunneling between the two rotation senses. That separatrix bunching is directly visible in the transmon's level structure and in cold-atom realizations of the pendulum in shaken optical lattices.

The Pendulum Phase Portrait and Separatrix

Where this lives today

The physical picture

From energy contours to the portrait

Which equation answers which question

Limiting cases

Worked examples

1 · A one-metre pendulum released at ninety degrees (confidence builder)

2 · Released one degree from inverted (exam level)

Going further

Before this

Where this leads