Fixed Points and Stability in Phase Space

The most famous fixed points in the solar system are the five Lagrange points of the Sun–Earth and Sun–Jupiter systems — equilibria of the rotating-frame Hamiltonian of the three-body problem, classified by exactly the machinery of this article. The collinear points L1–L3 are saddle-type: spacecraft like JWST, parked near Sun–Earth L2 since 2022, must fight the unstable eigenvalue with station-keeping burns every few weeks, a few metres per second of budget per year against an e-folding time of about 23 days. The triangular points L4 and L5, by contrast, are centres (linearly stable whenever the mass ratio is below Routh's threshold $\approx 1/25$), and nature has filled them: over ten thousand Trojan asteroids librate around Jupiter's L4 and L5, and NASA's Lucy spacecraft is currently threading its way toward them, with its first Trojan flyby scheduled for 2027 — a mission flying to visit the neighbourhood of a stable fixed point. The saddle-plus-separatrix motif also runs a machine closer to home. In a tokamak, the poloidal magnetic flux plays the role of a Hamiltonian for the field-line map, and modern "divertor" configurations deliberately shape the coils to place a saddle — the X-point — at the plasma edge; the separatrix through it divides confined field lines from those diverted onto exhaust plates. ITER's entire power-handling strategy, now being assembled toward first plasma in the 2030s, is the engineering of one saddle point and its manifolds. From spacecraft to fusion reactors, "find the fixed points, compute the eigenvalues, respect the separatrix" is a daily professional workflow.

Picture the phase plane as the surface of a flowing fluid. Fixed points are the places where the fluid is perfectly still. Drop a speck of dust nearby and watch what it does: if it circles the still point forever, that point is a centre; if it rushes away along one line while being squeezed inward along another, it is a saddle. The Hamiltonian itself supplies the landscape metaphor: $H(q,p)$ assigns a height to every phase point, and since energy is conserved, trajectories are contour lines of constant height. A centre sits at the bottom of a bowl, wrapped in concentric contours; a saddle sits exactly where two ridges cross, and the two contours passing through it form the separatrix — the watershed of the portrait. The cleanest laboratory for both species at once is a particle in a double-well potential. Each valley bottom is a centre; the hump between them is a saddle; and the energy contour through the saddle is a figure-eight separatrix that divides small oscillations trapped in one well from large orbits that swing over the barrier:

Energy conservation is also the reason certain behaviours are simply absent from this picture. A trajectory that spiralled into a fixed point would have to cross contours of $H$, changing its energy — impossible for a conservative system. Dynamical-systems theory catalogues linear fixed points in a two-dimensional map indexed by the trace and determinant of $J$: spirals and nodes, the equilibria that attract or repel, all require a nonzero trace, which measures how fast phase area grows or shrinks. Hamiltonian mechanics pins the trace to zero, confining every conservative system to a single vertical line of that map:

Think of the eigenvalues as the equilibrium's secret reaction to being disturbed. Real positive eigenvalue: "run away exponentially." Pure imaginary pair: "oscillate, never decide." The character of an equilibrium is encoded entirely in these two numbers $\pm\lambda$ — and because they always come in a plus–minus pair, the verdict is always one of exactly two: turn forever, or escape. The pendulum displays both in one portrait: the hanging position at $\theta=0$ is a centre (a 1.0 m pendulum circulates its phase point at $\omega=\sqrt{g/\ell}=3.13\ \mathrm{rad/s}$), the inverted position at $\theta=\pi$ is a saddle (perturbations grow with e-folding time $1/\lambda = 0.32$ s), and the separatrix between libration and rotation emanates from the saddle. That story gets its own article; here we build the machinery that makes it a two-line calculation.

Everything starts from the canonical equations, which assign a velocity vector to every point of the phase plane. A fixed point is a point where that assignment returns zero, so the search for equilibria is a search for simultaneous zeros of both right-hand sides:

where: $q,p$ — generalized coordinate and conjugate momentum; $H(q,p)$ — the Hamiltonian. Read as a vector field $\mathbf{v}=(\dot q,\dot p)$ on the phase plane, this is the "flow" whose stagnation points we now hunt. Setting both components to zero converts the dynamical condition into a purely geometric one. The requirement $\dot q=\dot p=0$ at a point $(q_0,p_0)$ says that both partial derivatives of $H$ vanish there:

where: $(q_0,p_0)$ — the equilibrium point. Fixed points of the flow are precisely the stationary points of the Hamiltonian function — its minima, maxima, and saddles as a surface over the phase plane. The landscape metaphor is exact: stagnation of the flow happens exactly where the energy landscape is flat. For the workhorse case of a natural Hamiltonian — kinetic plus potential — the two conditions decouple beautifully. The momentum condition fixes $p$, the coordinate condition fixes $q$:

where: $m$ — the mass; $V(q)$ — the potential. Every equilibrium lies on the $q$-axis of the phase plane, at a critical point of the potential — which is why the drop-lines in the double-well figure connect extrema of $V$ straight down to fixed points of the portrait. Near a fixed point, expand. Write the state as the fixed point plus a small displacement $\boldsymbol{\xi}$ and Taylor-expand the velocity field, keeping only the linear term (the constant term vanishes by the definition of a fixed point):

where: $J$ — the Jacobian (stability) matrix of the velocity field evaluated at $(q_0,p_0)$. Solutions of the linear system are superpositions of $e^{\lambda t}$ along the eigendirections of $J$, so the eigenvalues $\lambda$ are growth rates — and the whole classification problem reduces to computing a $2\times2$ matrix. Differentiating the right-hand sides of (1) — the top row from $\dot q=H_p$, the bottom row from $\dot p=-H_q$ — gives the Jacobian in terms of second derivatives of the Hamiltonian:

where: $H_{ab}=\partial^2 H/\partial a\,\partial b$ evaluated at the fixed point. Look at the diagonal: the entries are $+H_{pq}$ and $-H_{qp}$, and for any twice-differentiable Hamiltonian the mixed partials are equal. The trace has cancelled identically — before we choose any particular system:

where: $\operatorname{tr}J$ — the sum of the diagonal entries, equal to the divergence $\nabla\cdot\mathbf{v}$ of the phase-space velocity field. This is Liouville's theorem in its most local form: the phase fluid is incompressible, at every point, for every Hamiltonian. Everything special about conservative fixed points flows from this one line — the first derivation below turns it into the full classification, and the second shows what it forbids.

The derivation above used the special kinetic-plus-potential form, but the eigenvalue structure is universal. With the trace pinned to zero, the characteristic polynomial $\lambda^2-(\operatorname{tr}J)\lambda+\det J=0$ loses its linear term for any one-degree-of-freedom Hamiltonian, and the two eigenvalues are always an exact plus–minus pair determined by a single number:

where: $\det J$ — the determinant of the Jacobian, built from the second derivatives of $H$ at the fixed point. Notice that $\det J$ is exactly the determinant of the Hessian of $H$ — so the flow classification and the landscape classification are the same test. One number now decides the species:

where: the imaginary case means rotation of the displacement vector at fixed amplitude; the real case means exponential escape along the unstable eigendirection. A positive Hessian determinant is a minimum or maximum of the $H$-surface — encircled by closed contours, a centre; a negative one is a saddle of the surface — crossed contours, a saddle of the flow. There is no third species because there is no third sign. For the natural Hamiltonian the determinant is $V''(q_0)/m$, and both rates deserve names because both are measurable:

where: $\omega$ — the small-oscillation angular frequency of libration around a potential minimum; $\lambda$ — the growth rate (inverse e-folding time) of departures from a potential maximum. Same formula, opposite sign under the square root: the frequency at which a marble rattles in a bowl and the rate at which a pencil falls off its tip are two faces of the same curvature. The saddle does one more job: it anchors the separatrix. Since trajectories are level curves of $H$, the special contour through the saddle at $(q_s,0)$ carries the saddle's own energy, and solving $H=E_s$ for $p$ gives its explicit shape:

where: $q_s$ — the position of the potential maximum; $E_s$ — the separatrix energy. Orbits with $E<E_s$ are trapped in one well; orbits with $E>E_s$ pass over the barrier; the separatrix itself is the boundary case that creeps into the saddle in infinite time. For the double well of the hero figure this is the figure-eight $p=\pm\sqrt{q^2-q^4/2}$.

Which quantity answers which question

The classification says a centre's neighbours circulate and a saddle's neighbours flee, but "stable" deserves a precise definition — especially in a world where nothing can ever settle down. The right notion is Lyapunov's: an equilibrium is stable if you can guarantee the system stays within any prescribed tolerance forever, merely by starting it close enough:

where: $\boldsymbol{\xi}(t)$ — the displacement from the fixed point. Note what is not demanded: the orbit need not approach the equilibrium, only remain confined. The stronger property — $\boldsymbol{\xi}(t)\to 0$, called asymptotic stability — is exactly what Derivation B outlawed: a Hamiltonian equilibrium can be Lyapunov stable, but never asymptotically stable. Centres pass the test (11); saddles fail it spectacularly, since an arbitrarily small displacement along the unstable manifold grows like $e^{\lambda t}$. There is a subtlety worth respecting. For a saddle, the linear verdict is final — one genuinely positive eigenvalue guarantees nonlinear instability too. But for a centre the linear verdict is, in general, inconclusive: pure imaginary eigenvalues sit on the knife-edge where neglected nonlinear terms could tip the balance either way. What rescues the conservative case is that we hold a stronger card than linearization — the exact conservation of $H$ itself. That card is the Lagrange–Dirichlet theorem, and it is the third derivation:

Because the whole classification hangs on the local shape of $V$, tuning a physical parameter can rebuild the portrait's skeleton in real time. The generic event in one conservative degree of freedom is curvature passing through zero: $\det J\to 0$, the eigenvalues collide at the origin of the complex plane, and fixed points are created, destroyed, or exchange species. A bead on a rotating hoop shows the classic version — below a critical spin rate the bottom of the hoop is the only stable equilibrium; above it, the bottom turns saddle and two new centres split off symmetrically:

This is a pitchfork bifurcation, and its fingerprints are everywhere in physics: the frequency of oscillation about the surviving equilibrium goes soft ($\Omega\to 0$) as the critical point is approached, exactly like an order parameter's restoring force vanishing at a continuous phase transition. The second worked example below puts real numbers on this hoop. Note also what cannot happen: since the trace is pinned to zero, a conservative fixed point can never cross from stable to unstable by way of a spiral (no Hopf bifurcation without dissipation) — the only doorway between species is $\det J=0$.

For a one-degree-of-freedom natural Hamiltonian, the type of the fixed point at $q_0$ (where $V'(q_0)=0$) is fixed entirely by the curvature $V''(q_0)$:

1 · Full census of a double well (confidence builder)

A particle of mass $m=1.0$ kg moves in $V(q)=\tfrac14 q^4-\tfrac12 q^2$ (SI units: the quartic and quadratic coefficients are $1\ \mathrm{J/m^4}$ and $1\ \mathrm{J/m^2}$). Find every fixed point, classify each, give the oscillation frequency and escape rate, and describe the separatrix quantitatively.

Step 1. Locate the equilibria. Momenta vanish, and $V'(q)=q^3-q=q(q-1)(q+1)=0$ gives three fixed points: $(0,0)$ and $(\pm 1\ \mathrm{m},\,0)$.

Step 2. Compute curvatures: $V''(q)=3q^2-1$, so $V''(\pm 1)=+2\ \mathrm{J/m^2}$ and $V''(0)=-1\ \mathrm{J/m^2}$.

Step 3. Classify. At $q=\pm1$: $\det J=V''/m=+2\ \mathrm{s^{-2}}>0$ — centres, with small-oscillation frequency $\omega=\sqrt{2}=\boxed{1.41\ \mathrm{rad/s}}$ and period $T=2\pi/\omega=4.44$ s. At $q=0$: $\det J=-1\ \mathrm{s^{-2}}<0$ — a saddle with $\lambda=\pm 1.0\ \mathrm{s^{-1}}$: a perturbation off the hump e-folds every $\boxed{1.0\ \mathrm{s}}$.

Step 4. The separatrix. Its energy is the saddle's, $E_s=V(0)=0$, while the wells sit at $V(\pm1)=-0.25$ J. Equation (10) gives $p_{\text{sep}}(q)=\pm\sqrt{q^2-q^4/2}$, a figure-eight reaching out to $q=\pm\sqrt{2}=\pm1.41$ m, widest above the well bottoms where $p_{\text{sep}}(\pm1)=\pm\sqrt{1/2}=\boxed{\pm0.71\ \mathrm{kg\,m/s}}$.

Step 5. One bonus number. The phase area of one separatrix lobe is $\oint p\,\mathrm{d}q=2\int_0^{\sqrt 2}\sqrt{q^2-q^4/2}\,\mathrm{d}q$; the substitution $u=1-q^2/2$ makes the integrand elementary and the area comes out exactly $\boxed{4/3\ \mathrm{J\,s}}$. Every libration orbit trapped in the right well encloses less action than 1.33 J·s — a bound that becomes physical in semiclassical quantization, where it caps the number of quantum states the well can hold.

2 · The bead on a spinning hoop (exam level)

A bead slides without friction on a circular hoop of radius $R=0.20$ m that spins about its vertical diameter at constant rate $\omega$. In the rotating frame the dynamics follows from the effective potential $V_{\mathrm{eff}}(\theta)=-mgR\cos\theta-\tfrac12 m\omega^2R^2\sin^2\theta$, with $\theta$ measured from the downward vertical and kinetic term $p_\theta^2/2mR^2$. Classify all fixed points for $\omega=5.0$ and $\omega=9.0$ rad/s ($g=9.81\ \mathrm{m/s^2}$).

Step 1. Find the equilibria for general $\omega$. Setting $V_{\mathrm{eff}}'=mR\sin\theta\,(g-\omega^2R\cos\theta)=0$ gives $\theta=0$, $\theta=\pi$, and — whenever the equation is solvable — the tilted pair $\cos\theta_0=g/\omega^2R$. The tilted roots exist only for $\omega\ge\omega_c=\sqrt{g/R}=\sqrt{9.81/0.20}=\boxed{7.0\ \mathrm{rad/s}}$.

Step 2. Get the rates from curvature. The coordinate mass is $mR^2$, so $\Omega^2=V_{\mathrm{eff}}''/mR^2$. Differentiating: $V_{\mathrm{eff}}''=mR\left(g\cos\theta-\omega^2R\cos 2\theta\right)$, giving $\Omega^2=g/R-\omega^2$ at $\theta=0$ and, after substituting $\cos\theta_0=g/\omega^2R$, the tidy result $\Omega^2=\omega^2\sin^2\theta_0$ at the tilted points. All rates are independent of the bead's mass, as they must be for gravity-plus-constraint dynamics.

Step 3. Slow spin, $\omega=5.0$ rad/s. Tilted equilibria do not exist ($g/\omega^2R=1.96>1$). At the bottom, $\Omega^2=49.05-25.0=+24.05\ \mathrm{s^{-2}}$ — a centre, oscillating at $\Omega=\boxed{4.9\ \mathrm{rad/s}}$. (Note this is slower than the non-spinning pendulum value $\sqrt{g/R}=7.0$ rad/s: the centrifugal term has already softened the well.) The top $\theta=\pi$ has $V_{\mathrm{eff}}''=mR(-g-\omega^2R)<0$ — a saddle at any spin rate.

Step 4. Fast spin, $\omega=9.0$ rad/s. The bottom flips: $\Omega^2=49.05-81.0=-31.95\ \mathrm{s^{-2}}$, a saddle with escape rate $\lambda=\sqrt{31.95}=\boxed{5.7\ \mathrm{s^{-1}}}$ (e-folding time 0.18 s). The tilted pair appears at $\cos\theta_0=9.81/(81\times0.20)=0.606$, i.e. $\theta_0=\pm\boxed{52.7^\circ}$ — two new centres with $\Omega=\omega\sin\theta_0=9.0\times0.796=\boxed{7.2\ \mathrm{rad/s}}$.

Step 5. Read the bifurcation. Between the two spin rates the portrait's skeleton reorganized exactly as in the pitchfork figure: one centre became a saddle and shed two centres. Approaching $\omega_c$ from below, $\Omega=\sqrt{g/R-\omega^2}\to 0$ — the mode goes soft, the classic warning that a bifurcation is imminent, and the reason this humble bead is the standard classical analogue of a second-order phase transition.

References. Morin, Introduction to Classical Mechanics, Ch. 5 (oscillations) and Hamiltonian sections · Goldstein, Poole & Safko, Classical Mechanics 3rd ed., Ch. 8 · Landau & Lifshitz, Mechanics 3rd ed., §21–§22 · Taylor, Classical Mechanics, Ch. 13 (Hamiltonian mechanics) · Strogatz, Nonlinear Dynamics and Chaos, Ch. 5–6 (linearization and classification) · Arnold, Mathematical Methods of Classical Mechanics, §22 (stability) The key results to keep in mind: the Jacobian of Hamilton's equations has diagonal entries $\pm H_{pq}$ giving zero trace; the eigenvalues for $H=p^2/2m+V$ are $\lambda=\pm\sqrt{-V''/m}$; the small-oscillation frequency is $\omega=\sqrt{V''/m}$; the pendulum has a centre at $\theta=0$ with $\omega=\sqrt{g/\ell}$ and a saddle at $\theta=\pi$. The trace-zero identity is the divergence-free (Liouville) condition, and the classification reduces to the conservative (zero-trace) line of the standard trace-determinant scheme.

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