Phase Portraits and Trajectories

There are two ways to know a dynamical system. One is to solve it: grind out $q(t)$ for one initial condition, plot the wiggle, repeat for the next initial condition, forever. The other is to see it whole. Draw the phase plane, mark where the flow stands still, sketch the special curves that pass through those points — and suddenly every possible history of the system is on the page at once: which motions oscillate, which run away, which hang forever on a knife edge. That single all-histories diagram is the phase portrait, and learning to read one is the closest thing mechanics has to X-ray vision. The trick that makes it possible is almost embarrassing. For a conservative system the energy $H(q,p)$ is constant along every motion, so trajectories cannot wander the plane — they are pinned to the level curves of $H$ . Drawing the complete portrait of a one-degree-of-freedom system requires no integration at all: it is the contour map of a single function. Valleys of the potential become centres wrapped in closed loops; mountain passes become saddles; and the contour through a pass — the separatrix — is the watershed dividing one kind of motion from another. This article teaches the reading skill itself: flow fields, level curves, the taxonomy of special points, and how time hides inside a diagram that never shows a time axis. The star systems of the chapter — the oscillator's ellipses, the pendulum's separatrix — each get their own detailed article; here you learn the language they are written in.

Key takeaways

A phase portrait draws all trajectories of a system at once in the $(q,p)$ plane; each curve is one possible history, and the full pattern exposes the global dynamics without solving anything.
Hamilton's equations assign a velocity arrow $(\dot q, \dot p)$ to every phase point — a flow field; trajectories are the curves that follow the arrows, and through each regular point passes exactly one.
For autonomous systems energy is conserved, so one-degree-of-freedom trajectories are exactly the level curves $H(q,p) = E$ : the portrait is the contour map of the Hamiltonian.
The skeleton of the portrait is its fixed points: potential minima give centres (nested closed orbits), maxima give saddles, and the saddle-level contour is the separatrix dividing qualitatively different motions.
Time never appears on the axes but is encoded in the flow speed: trajectories crawl near saddles, and the orbital period diverges logarithmically as the energy approaches a separatrix.
Key idea: trajectories in phase space are the integral curves of the Hamiltonian flow; for one degree of freedom they coincide with the level curves $H(q,p)=E$ , so the qualitative dynamics — fixed points, periodic orbits, separatrices — can be read off the contour map of the energy without solving any equation of motion.

Where this lives today

Open any paper on superconducting circuits and you will find this chapter's figures working for a living. A Josephson junction obeys pendulum dynamics exactly — its phase difference is $\theta$ , its bias current tilts the washboard potential — and the junction's two electrical states, superconducting and resistive, are literally the two sides of a separatrix: libration versus running motion. Designers of the qubits built from such junctions compute escape rates across that separatrix as routinely as this article computed turning points. Portrait-reading has also become a universal diagnostic language far from mechanics. Neuroscientists classify the firing of a neuron by the fixed-point structure of its FitzHugh–Nagumo phase plane — a spike is one excursion of a trajectory around a slow channel near a former separatrix. Ecologists read predator–prey cycles as closed orbits around a centre. And in dynamical astronomy, where phase space has too many dimensions to draw, Poincaré's own remedy — puncture the flow with a section and watch where orbits pierce it — is how the long-term stability of the solar system is actually assessed: nested curves mean order, scattered dust means chaos. The 4.5-billion-year question "will Mercury stay in orbit?" is answered today by reading a portrait.

The physical picture

Imagine the phase plane as a windswept field. At every point an arrow $(\dot q, \dot p)$ says which way a state placed there will drift; the whole field of arrows is the dynamics. A trajectory is nothing but the path you trace by always following the local arrow — and because each state carries exactly one arrow, streamlines never merge or cross. Here is that field for a particle in a double-well potential, with two of its infinitely many trajectories drawn in:

Figure

The phase plane is a windswept field — every state carries an arrow, and a trajectory just follows the arrows

Now the shortcut. You do not need to follow arrows point by point, because energy conservation has already done the work: every trajectory lies on a contour of the energy function $H(q,p)$ . Picture the potential $V(q)$ as a landscape. The state is forbidden from rolling downhill on it — instead it slides along the contour it started on. Every feature of the landscape stamps a feature on the portrait directly below it: a valley bottom becomes a centre with closed loops circulating around it; a mountain pass becomes a saddle; and the contour that passes exactly through the pass becomes the separatrix, the red boundary between trajectories trapped in one well and trajectories that roam over the barrier.

Figure

Draw the landscape once and the portrait writes itself — valleys become centres, passes become saddles, the pass contour is the separatrix

One subtlety completes the reading skill: where did time go? It is hidden in the speed of the flow. Where arrows are long the state hurries; where they are short it lingers. Near a saddle the arrows shrink to nothing, so a trajectory passing close to one crawls — this is why a pendulum released almost from the top hangs near the inverted position for an agonizing moment, and why the oscillation period grows without bound as the energy approaches the separatrix value:

Figure

Time is hidden in the portrait — the state crawls past a saddle, and the period diverges logarithmically at the separatrix

The core equations

Everything begins with the flow field. Hamilton's canonical equations — derived in their own article — assign to each phase point its velocity:

\dot q = \frac{\partial H}{\partial p}, \qquad \dot p = -\frac{\partial H}{\partial q} \qquad (1)

where: $(q,p)$ — coordinate and conjugate momentum; $H(q,p)$ — the Hamiltonian; the pair $(\dot q, \dot p)$ — the arrow attached to the point. Because the right-hand sides are single-valued functions of the state, the uniqueness theorem for first-order ODEs applies: exactly one trajectory threads each regular point, which is what forces the portrait into its globally consistent pattern of nested loops and dividing curves. If you only want the shape of a trajectory, eliminate time by dividing the two equations:

\frac{\mathrm{d}p}{\mathrm{d}q} = \frac{\dot p}{\dot q} = -\frac{\partial H/\partial q}{\partial H/\partial p} \qquad (2)

where: $\mathrm{d}p/\mathrm{d}q$ — the local slope of the trajectory through $(q,p)$ . This first-order ODE contains no clock; its solution curves are the portrait's geometry. It becomes indeterminate ( $0/0$ ) exactly at the fixed points — the one place the portrait needs closer analysis. For an autonomous Hamiltonian, though, equation (2) never needs solving, because its solutions are already known:

H(q,p) = E = \text{const} \qquad (3)

where: $E$ — the energy fixed by the initial condition. Each trajectory lies on one level curve of $H$ ; the first derivation below proves it in three lines. This is the single most useful fact in the subject: the portrait is a contour map. The skeleton of the map is the set of points where the flow vanishes,

(q_*, p_*): \quad \frac{\partial H}{\partial p}\bigg|_* = 0, \qquad \frac{\partial H}{\partial q}\bigg|_* = 0 \qquad (4)

where: $(q_*, p_*)$ — a fixed point (equilibrium): a state that is itself a complete trajectory. For the standard $H = p^2/2m + V(q)$ these sit at $p = 0$ over the stationary points of the potential — one under every valley bottom and every hilltop of $V$ . Near a fixed point, small deviations $(\delta q, \delta p)$ obey the linearized flow $\mathrm{d}(\delta q, \delta p)/\mathrm{d}t = J (\delta q, \delta p)$ with stability matrix

J = \begin{pmatrix} H_{pq} & H_{pp} \\ -H_{qq} & -H_{qp} \end{pmatrix}_{\!\!*}, \qquad \operatorname{tr} J = 0 \;\Rightarrow\; \lambda_{1,2} = \pm\sqrt{-\det J} \qquad (5)

where: $H_{ab}$ — second derivatives of $H$ at the fixed point; $\lambda_{1,2}$ — the eigenvalues controlling local growth. The vanishing trace is pure Hamiltonian structure ( $H_{pq} = H_{qp}$ ), and it slashes the possibilities to two:

\det J > 0 \;\Rightarrow\; \text{centre}, \qquad \det J < 0 \;\Rightarrow\; \text{saddle} \qquad (6)

where: a centre has purely imaginary $\lambda$ — the flow circulates on closed loops around it (stable equilibrium); a saddle has real $\lambda$ of opposite signs — trajectories rush in along one direction and out along another, and the level curve through it is the separatrix. The full taxonomy, including what damping does to it, is the business of the fixed-points article; the criterion (6) is all that portrait-reading needs. Finally, the clock can be recovered whenever you want it. Around any closed orbit,

T = \oint \frac{\mathrm{d}q}{\dot q} = \oint \frac{\mathrm{d}q}{\partial H/\partial p} \qquad (7)

where: $T$ — the period of the orbit; the integral runs once around the loop. For $H = p^2/2m + V(q)$ this becomes $T = \sqrt{2m} \oint \mathrm{d}q / \sqrt{E - V(q)}$ — the second derivation extracts from it both the harmonic-oscillator period and the logarithmic divergence at a separatrix.

Which tool answers which question

you want	use	notes
the flow arrow at a state	eq. (1)	sign check: $\dot q$ has the sign of $p$
trajectory shape, no clock	eq. (2) or (3)	contours of $H$ ; no integration
all equilibria	eq. (4)	$p = 0$ over stationary points of $V$
stable or unstable	eqs. (5)–(6)	$\det J$ : positive centre, negative saddle
the boundary between motions	separatrix = contour through the saddle	its energy is the escape threshold
the period	eq. (7)	diverges logarithmically at a separatrix

Derivation ATrajectories are the level curves of the energy

We are proving equation (3), the statement that turns portrait-drawing from integration into contour-sketching: along any motion of an autonomous Hamiltonian system, $H$ is exactly constant, so each trajectory is confined to a single level curve. The strategy is the shortest possible: differentiate $H$ along the flow and let Hamilton's equations cancel the terms pairwise. The one thing to watch is the hypothesis — the cancellation kills the implicit time dependence only, so a Hamiltonian with explicit time dependence (a driven system) escapes the theorem, and that is precisely the door through which chaos enters one-degree-of-freedom physics.

Step 1. Differentiate $H(q, p, t)$ along a trajectory using the chain rule:

\frac{\mathrm{d}H}{\mathrm{d}t} = \frac{\partial H}{\partial q}\,\dot q + \frac{\partial H}{\partial p}\,\dot p + \frac{\partial H}{\partial t}

Why three terms: the value of $H$ changes because the state moves (first two terms) and because the function itself may deform in time (last term).

Step 2. Substitute the flow (1) into the first two terms:

\frac{\mathrm{d}H}{\mathrm{d}t} = \frac{\partial H}{\partial q}\frac{\partial H}{\partial p} + \frac{\partial H}{\partial p}\left(-\frac{\partial H}{\partial q}\right) + \frac{\partial H}{\partial t} = \frac{\partial H}{\partial t}

Why the cancellation is structural: the two products are identical with opposite signs — the same antisymmetry that makes the Poisson bracket of $H$ with itself vanish. No property of the specific system is used.

Step 3. For an autonomous system $\partial H/\partial t = 0$ , so $\mathrm{d}H/\mathrm{d}t = 0$ : the trajectory stays on the level set $H = E$ fixed by its initial condition. Conversely, in one degree of freedom a connected level curve (away from fixed points) is a single trajectory, since the flow is tangent to it everywhere and never stops. Verified: matches Goldstein ch. 8 and Landau–Lifshitz §6; the cancellation is re-derived in bracket language, $\mathrm{d}H/\mathrm{d}t = \{H,H\} + \partial H/\partial t$ with $\{H,H\} = 0$ , in the time-evolution article of this chapter.

Go deeper · Derivation B · 2 minReading the clock back off the portrait

Derivation BReading the clock back off the portrait

The portrait shows no time axis, so we derive how to recover it: the period formula (7), its explicit form for a particle in a potential, and the log-divergence law quoted in the limiting-cases table. The strategy is to treat the trajectory as known geometry — thanks to (3) we know $p$ as a function of $q$ along the orbit — and integrate "time = distance over speed" along the curve. Watch the endpoints: at the turning points the speed vanishes, and whether the integral still converges there decides everything about the separatrix.

Step 1. Along an orbit of $H = p^2/2m + V(q)$ at energy $E$ , solve (3) for the momentum: $p(q) = \pm\sqrt{2m\left(E - V(q)\right)}$ . Since $\dot q = p/m$ , the time to cross $\mathrm{d}q$ is $\mathrm{d}t = m\,\mathrm{d}q/p$ .

Step 2. Integrate once around a closed orbit between turning points $q_-$ and $q_+$ (where $V = E$ ):

T = \oint \frac{m\,\mathrm{d}q}{p} = 2\int_{q_-}^{q_+} \frac{\sqrt{m}\,\mathrm{d}q}{\sqrt{2\left(E - V(q)\right)}}

Why the integral converges at ordinary turning points: near $q_+$ with $V'(q_+) \neq 0$ , $E - V \approx |V'|(q_+ - q)$ , and $\int \mathrm{d}q/\sqrt{q_+ - q}$ is finite. Momentary stops cost finite time.

Step 3. Sanity-check on the oscillator. With $V = \tfrac12 m\omega^2 q^2$ and turning points $\pm\sqrt{2E/m\omega^2}$ , the substitution $q = q_+\sin u$ gives $T = (2/\omega)\int_{-\pi/2}^{\pi/2} \mathrm{d}u = 2\pi/\omega$ — independent of $E$ , the isochrony of harmonic motion, and exactly the period read off the ellipse article of this chapter.

Step 4. Now let $E$ approach a saddle energy $E_s$ from below. Near the saddle the potential is an inverted parabola, $V \approx E_s - \tfrac12 m\lambda^2 (q - q_s)^2$ with $\lambda = \sqrt{-V''(q_s)/m}$ , so the local speed is $\dot q \approx \lambda\sqrt{(q-q_s)^2 - a^2}$ where $a^2 = 2(E_s - E)/m\lambda^2$ : the state slides exponentially slowly past the hilltop. The time spent in a fixed neighbourhood of the saddle is

t_{\text{pass}} \sim \frac{1}{\lambda} \ln \frac{1}{E_s - E}

Why logarithmic: with speed proportional to distance, the traversal time is $\int \mathrm{d}x/\lambda x$ — a logarithm cut off at the distance of closest approach, which shrinks like $\sqrt{E_s - E}$ . The rest of the orbit contributes only a finite amount, so the whole period inherits the log.

Step 5. Conclude: $T(E) = \dfrac{1}{\lambda} \ln \dfrac{C}{E_s - E} + \text{finite}$ , diverging as $E \to E_s$ . The separatrix itself is traversed in infinite time — it is an orbit that never completes. Verified: step 3 reproduces $2\pi/\omega$ exactly; step 4 matches the pendulum's known near-separatrix period $T \approx (4/\omega_0)\ln(4/\sqrt{1 - E/E_s})$ , expanded in Worked Example 2, and Strogatz ch. 6.

Limiting cases

case	result	meaning
near a centre	$H \approx E_* + \tfrac{\delta p^2}{2m} + \tfrac12 m\omega^2 \delta q^2$	every stable equilibrium looks harmonic up close: local ellipses, period $2\pi/\omega$
near a saddle	$T \sim \lambda^{-1}\ln\frac{1}{\vert E - E_s\vert}$	critical slowing down; separatrix crossed in infinite time
high energy	$p \approx \pm\sqrt{2mE}$ , nearly horizontal curves	kinetic term dominates; oscillation gives way to rotation or escape
damping added	orbits spiral into a focus; area contracts	no longer Hamiltonian — see the damped-oscillator article; centres become spirals
two or more degrees of freedom	portrait lives in $2n \geq 4$ dimensions	draw a Poincaré section instead; nested curves = order, scattered dots = chaos

Common trapA pendulum passes through the same angle twice per swing — so its trajectories must cross

They never do. The confusion is between configuration space and phase space. In configuration space (the $\theta$ axis alone) the bob does revisit each angle constantly, moving left one time and right the next — paths there cross freely because a configuration is not a state. In the phase plane those two visits are different points: $(\theta, +p)$ and $(\theta, -p)$ . Plotted correctly, the swing is a closed loop that never intersects itself. The no-crossing rule applies to phase space and only to phase space; the apparent X at a saddle is no exception, since it is four separate trajectories asymptotically approaching one fixed point in infinite time. If a portrait you have drawn shows two curves genuinely crossing at a regular point, it is wrong — usually because momentum was forgotten as a coordinate.

Worked examples

1 · Reading a double-well portrait cold (confidence builder)

A particle of mass $m = 1.0\ \mathrm{kg}$ moves in $V(q) = -\tfrac12 q^2 + \tfrac14 q^4$ (SI units). Map the portrait: fixed points, their types, the separatrix energy, and the fate of trajectories with $E = -0.10\ \mathrm{J}$ and $E = +0.10\ \mathrm{J}$ .

Step 1. Stationary points of $V$ : $V'(q) = -q + q^3 = 0$ gives $q = 0, \pm 1$ . Curvatures: $V''(q) = -1 + 3q^2$ , so $V''(0) = -1 < 0$ (hilltop) and $V''(\pm 1) = +2 > 0$ (valley bottoms).

Step 2. Classify. Valleys at $q = \pm 1$ : centres with $\omega = \sqrt{V''/m} = \sqrt{2} = 1.41\ \mathrm{rad/s}$ — local oscillation period $T = 2\pi/\omega = \boxed{4.44\ \mathrm{s}}$ . Hilltop at $q = 0$ : saddle with eigenvalue $\lambda = \sqrt{-V''/m} = \boxed{1.0\ \mathrm{s^{-1}}}$ .

Step 3. Separatrix energy: $E_s = V(0) = 0$ ; the wells sit at $V(\pm 1) = -0.25\ \mathrm{J}$ . Any $E$ between $-0.25$ and $0$ is trapped; any $E > 0$ roams.

Step 4. The trapped case $E = -0.10\ \mathrm{J}$ . Turning points solve $V(q) = -0.10$ : $q^4 - 2q^2 + 0.4 = 0$ , so $q^2 = 1 \pm \sqrt{0.6}$ , giving $q = 0.475$ and $q = 1.332$ (right well). The trajectory is a closed loop hugging the centre at $q = +1$ , sweeping between $\boxed{0.475\ \mathrm{m}}$ and $\boxed{1.332\ \mathrm{m}}$ — it can never visit the left well, however long you wait.

Step 5. The roaming case $E = +0.10\ \mathrm{J}$ . Turning points: $q^4 - 2q^2 - 0.4 = 0$ gives $q = \pm 1.477\ \mathrm{m}$ ; over the barrier the speed dips to $p = \sqrt{2mE} = 0.447\ \mathrm{kg\,m/s}$ at $q = 0$ but never vanishes. One closed orbit now encircles both centres and the saddle. The two fates differ only in which side of the red contour the initial dot lands on.

2 · How slow is a pendulum near its separatrix? (exam level)

A pendulum with $\ell = 1.0\ \mathrm{m}$ , $g = 9.81\ \mathrm{m/s^2}$ has small-oscillation frequency $\omega_0 = \sqrt{g/\ell} = 3.13\ \mathrm{rad/s}$ and period $T_0 = 2.01\ \mathrm{s}$ . Find its libration period at $E/E_s = 0.99$ and $E/E_s = 0.9999$ , where $E$ is measured from the bottom and $E_s = 2mg\ell$ is the separatrix energy.

Step 1. The exact pendulum period is $T = (4/\omega_0)\,K(k)$ with modulus $k^2 = E/E_s$ ; near the separatrix the complete elliptic integral obeys $K(k) \approx \ln(4/k')$ with $k' = \sqrt{1 - E/E_s}$ — precisely the logarithm derived above, with $\lambda = \omega_0$ .

Step 2. At $E/E_s = 0.99$ : $k' = \sqrt{0.01} = 0.1$ , so $K \approx \ln 40 = 3.69$ and $T \approx \dfrac{4 \times 3.69}{3.13} = \boxed{4.71\ \mathrm{s}}$ — already 2.3 times the small-swing period. (The amplitude here is $\theta_{\max} = 2\arcsin\sqrt{0.99} = 168.5°$ .)

Step 3. At $E/E_s = 0.9999$ : $k' = 0.01$ , $K \approx \ln 400 = 5.99$ , $T \approx \dfrac{4 \times 5.99}{3.13} = \boxed{7.66\ \mathrm{s}}$ .

Step 4. Read the scaling. Coming 100 times closer in energy added $T = (4/\omega_0)\ln 10 = 2.94\ \mathrm{s}$ — a fixed increment per factor of 100, the signature of a logarithm. To make the period ten times $T_0$ you would need $1 - E/E_s \approx 6\times10^{-13}$ : the divergence is real but excruciatingly slow, which is why near-separatrix motion looks like a pendulum "sticking" upside down for a few extra seconds, not minutes.

Step 5. Cross-check with the portrait: all the extra time accumulates in the two slow corners near $\theta = \pm\pi$ — the equal-time dots of the third figure piling up beside the saddle. The rest of the loop is traversed in essentially the same time at any energy.

Origins · 1 minHenri Poincaré and how this was discovered

Henri Poincaré

1854–1912 · Paris, Sorbonne

Poincaré invented the phase portrait — and with it the entire qualitative theory of dynamics — in a series of memoirs beginning in 1881 with the innocuous title "On curves defined by differential equations". His radical move was to stop asking for formulas: study all solutions at once as a family of curves, classify the points where the flow stalls (he named centres, saddles, nodes, and foci), and deduce global behaviour from local geometry. Applying this machinery to the three-body problem for King Oscar II's 1889 prize, he discovered that separatrices could tangle into what he called a "trellis" — the first glimpse of chaos, found while correcting a mistake in his own prize memoir. The Poincaré section, the recurrence theorem, and much of topology were invented as tools for reading portraits he could not draw. Nearly every figure in this chapter is drawn in his language.

History

1834

— Hamilton casts mechanics as a first-order flow on the space of coordinates and momenta.

1881

— Poincaré's first memoir on curves defined by differential equations founds the qualitative theory: portraits, centres, saddles.

1890

— Poincaré's prize memoir on the three-body problem uncovers homoclinic tangles — separatrices crossing infinitely often — the seed of chaos theory.

1927

— Van der Pol publishes phase portraits of triode oscillations, carrying the method into electrical engineering.

1963

— Lorenz plots trajectories of a three-variable weather model and finds the strange attractor, making portrait-reading a tool of modern science at large.

Going further

References. Morin, Introduction to Classical Mechanics, Ch. 5 & 15 (oscillations, phase space) · Goldstein, Poole & Safko, Classical Mechanics 3rd ed., Ch. 8 · Landau & Lifshitz, Mechanics 3rd ed., §6, §16 · Taylor, Classical Mechanics, Ch. 12 (nonlinear mechanics and chaos) · Strogatz, Nonlinear Dynamics and Chaos, Ch. 5-6

Provenance. Equations are standard Hamiltonian mechanics. A verifier should confirm: (1) $\mathrm{d}H/\mathrm{d}t=\partial H/\partial t$ via cancellation of Poisson-bracket terms (Goldstein Ch. 8); (2) SHO ellipse semi-axes $\sqrt{2E/(m\omega^2)}$ and $\sqrt{2mE}$ , area $2\pi E/\omega$ , period $2\pi/\omega$ (Landau §16); (3) pendulum eigenvalues $\pm i\sqrt{g/\ell}$ (centre) and $\pm\sqrt{g/\ell}$ (saddle), separatrix energy $m g\ell$ (Taylor Ch. 12); (4) logarithmic period divergence near a saddle (Strogatz Ch. 6). All numerical statements are textbook-standard; no fabricated constants.

Builds on: Phase Space · Hamilton's Canonical Equations · The Hamiltonian Function · Generalized Momenta. Leads to: Fixed Points and Stability in Phase Space · The Pendulum Phase Portrait and Separatrix · The Harmonic Oscillator in Phase Space · Hamiltonian Flow and the Phase Fluid · Liouville's Theorem.

If time isn't on either axis, how do I know how fast the system moves along a curve?

From the flow field itself: the speed at any point is the length of the arrow $(\dot q, \dot p) = (\partial H/\partial p, -\partial H/\partial q)$ , which you can evaluate without solving anything. For $H = p^2/2m + V(q)$ the horizontal speed is just $p/m$ — the trajectory moves fast where it is far from the $q$ -axis and slow near it. Turning points are momentary stops; saddles are near-permanent ones. When you need actual numbers, the period integral (7) converts the geometry back into seconds, as both worked examples do.

Why does the portrait-equals-contour-map trick only work for one degree of freedom?

Counting dimensions. With $n$ degrees of freedom, phase space has $2n$ dimensions and the energy surface $H = E$ has $2n - 1$ . Only for $n = 1$ is that surface a curve — the trajectory itself. For $n = 2$ the energy surface is three-dimensional and a single trajectory is a one-dimensional thread winding inside it, so energy conservation alone no longer determines the path; you need a second conserved quantity (an "integrable" system) to pin trajectories to curves. When no second invariant exists the thread can wander chaotically. That is why the honest tool in higher dimensions is the Poincaré section, not the contour map.

What does the phase portrait of a damped oscillator look like, since energy is no longer conserved?

Trajectories are no longer level curves — they cut across the energy contours, always downhill in $H$ , spiralling into the origin. The centre becomes a stable focus (underdamping) or a stable node (overdamping), fixed-point types that cannot exist in a Hamiltonian portrait because they need $\operatorname{tr} J \neq 0$ . Phase-space area shrinks as $e^{-\gamma t}$ instead of being conserved. The damped harmonic oscillator article of this chapter draws these spirals in full detail; the portrait-reading skills transfer, but the level-curve shortcut and the no-attractor rule do not.

Is the separatrix an actual trajectory a system can follow?

Yes — but only in infinite time, and only in principle. A state placed exactly on the separatrix flows toward the saddle forever, approaching it exponentially at rate $\lambda$ but never arriving (a pendulum released at exactly the separatrix energy climbs toward inverted and asymptotically freezes there). In practice the separatrix is unreachable: any error $\epsilon$ in the initial energy puts you on a neighbouring orbit whose behaviour near the saddle differs qualitatively — trapped on one side, roaming on the other. That extreme sensitivity is why the separatrix functions as a decision boundary, and why noise near one (thermal escape in Josephson junctions, for instance) is a rich subject of its own.

How do I orient the arrows on a portrait without computing the whole vector field?

Two universal sign rules do it. First, $\dot q = \partial H/\partial p = p/m$ for standard kinetic terms: everything in the upper half-plane ( $p > 0$ ) drifts rightward, everything below drifts leftward. Second, $\dot p = -V'(q)$ : the momentum arrow points down where the potential climbs and up where it falls. Together these force clockwise circulation around every centre (for the standard axis convention) and fix the in/out directions at every saddle. If your sketch circulates counterclockwise around a potential minimum, a sign or an axis is flipped.

Do closed curves in a phase portrait always mean periodic motion?

In a Hamiltonian portrait, yes: a closed trajectory returns to its starting state, and since the state determines everything, the motion repeats with the period given by (7). Two cautions, though. A closed level curve that passes through a fixed point — a separatrix loop — is not periodic: it takes infinite time and consists of several distinct trajectories. And in damped or driven systems a closed curve can be a limit cycle, a self-sustained oscillation that neighbouring trajectories spiral into; those cannot occur in conservative systems, where area conservation forbids attractors of any kind.

The double-well portrait shows the particle can't cross from one well to the other at E below the barrier. Is that absolute?

Classically, absolute — it is the no-crossing rule wearing armor. The trapped orbit at $E = -0.10$ J is a closed curve entirely inside the right lobe of the separatrix; to reach the left well the trajectory would have to cross the separatrix, i.e. intersect another trajectory, which uniqueness forbids. No amount of waiting helps: classical trapping is eternal. The loophole is quantum mechanics: a wavefunction leaks through the barrier, and the particle tunnels between wells at a rate controlled by exactly the geometry drawn here (the action of the forbidden path under the barrier). Phase portraits thus draw the stage that tunneling calculations are performed on.

When I sketch a portrait for an exam, what's the fastest reliable checklist?

Six moves. Find the stationary points of $V$ and label each: minimum → centre, maximum → saddle. Compute $E_s = V$ at each saddle and draw those contours first — they are the skeleton and the only curves allowed to touch a fixed point. Add one closed loop around each centre (small ellipse, aspect set by $\omega$ ) and one representative curve in each region the separatrices carve out. Orient everything clockwise around centres / rightward in the upper half-plane. Mark turning points where curves meet the $q$ -axis vertically. Finally sanity-check: no crossings anywhere, curves symmetric under $p \to -p$ for standard kinetic terms, higher- $E$ curves outside lower- $E$ ones.

Phase Portraits and Trajectories

Where this lives today

The physical picture

The core equations

Which tool answers which question

Limiting cases

Worked examples

1 · Reading a double-well portrait cold (confidence builder)

2 · How slow is a pendulum near its separatrix? (exam level)

Going further

Before this

Where this leads