Exercises — Phase space — trajectories, phase portraits

WHAT the picture shows. A loop that closes on itself means the state $(q,p)$ returns exactly to where it started and repeats — this is periodic (oscillatory) motion. WHY closed ⇒ energy fixed. For a conservative system every trajectory sits on a single contour $H(q,p)=E$. A curve that never leaves one contour means energy is constant. If energy were leaking away (friction), the dot would drift to lower contours and the loop could not close — it would spiral inward. Answer: a stable oscillation with constant total energy.

This is a fixed point (equilibrium point). It is defined by the state never changing: $\dot{q}=0\text{and}\dot{p}=0.$ WHY these two. The flow vector at any point is $(\dot{q},\dot{p})$. "Never moves" means that arrow has zero length — both components vanish. This is the only kind of point where trajectories may touch, because elsewhere the single-valued flow forbids crossings.

• (a) ellipse → undamped oscillator (energy fixed, closed loop). See Harmonic Oscillator.
• (b) inward spiral → damped oscillator (energy bleeding away, dot sinks to origin).
• (c) wavy open line → whirling pendulum — it never turns back, $q$ (angle) keeps increasing, so the curve runs off to the side forever. See Pendulum.

WHAT/WHY — use Hamilton's equations (they turn energy into the velocity of the state point): $\dot{q}=\frac{\partial H}{\partial p}=\frac{p}{m},\dot{p}=-\frac{\partial H}{\partial q}=-kq.$ Plug in $q=1,p=0,m=2,k=8$: $\dot{q}=\frac{0}{2}=0,\dot{p}=-(8)(1)=-8.$ WHAT IT LOOKS LIKE — Figure s01. The blue curve is the energy ellipse that passes through our point. The yellow dot marks the state $(q,p)=(1,0)$ — the rightmost tip of that ellipse, where $q$ is largest and $p=0$. The pink arrow there points straight down with length $8$: momentum is being destroyed by the restoring force, so the dot slides clockwise. This is exactly our computed flow $(0,-8)$.

Answer: $(\dot{q},\dot{p})=(0,-8)$.

$\omega =\sqrt{k/m}=\sqrt{8/2}=2$. Here $\omega$ (angular frequency) is the number of radians of oscillation per second; note $k=m{\omega }^2$, so we can write $H$ either way.

Derive the ellipse (WHY it is an ellipse). Start from energy conservation, $H=E$: $\frac{p^2}{2m}+\frac{1}{2}m{\omega }^2{q}^2=E.$ We want the standard ellipse shape $\frac{q^2}{a^2}+\frac{p^2}{b^2}=1$, because an equation of that form is by definition an ellipse with half-widths $a$ (along $q$) and $b$ (along $p$). To reach it we make the right-hand side equal $1$: divide every term by $E$. $\frac{p^2}{2mE}+\frac{m{\omega }^2{q}^2}{2E}=1.$ Now read off each denominator as a squared half-axis. For the $q$-term, $\frac{m{\omega }^2{q}^2}{2E}=\frac{q^2}{2E/(m{\omega }^2)}$, so $\frac{q^2}{2E/(m{\omega }^2)}+\frac{p^2}{2mE}=1.$ WHY this reading works: in $q^2/a^2$ the number under $q^2$ is $a^2$; whatever sits there, its square root is the half-width. Hence $a=\sqrt{\frac{2E}{m{\omega }^2}},b=\sqrt{2mE}.$ Sanity check on meaning: at $p=0$ all the energy is potential, $\frac{1}{2}m{\omega }^2{q}^2=E$, giving the farthest reach $q=a$ — the max displacement. At $q=0$ all energy is kinetic, $p^2/2m=E$, giving $p=b$ — the max momentum. Now plug numbers ($m=2,\omega =2,E=16$): $a=\sqrt{\frac{2\cdot 16}{2\cdot 2^2}}=\sqrt{\frac{32}{8}}=\sqrt{4}=2,b=\sqrt{2\cdot 2\cdot 16}=\sqrt{64}=8.$ Answer: $\omega =2,a=2,b=8$. See Harmonic Oscillator.

$\dot{q}=\frac{\partial H}{\partial p}=p,\dot{p}=-\frac{\partial H}{\partial q}=-1.$ Energy contours: $E=\frac{p^2}{2}+q\Rightarrow q=E-\frac{p^2}{2}$ — a sideways parabola opening in the $-q$ direction. WHY a parabola: constant force means $p$ falls at a steady rate ($\dot{p}=-1$), while $q$ speeds along as $p$; that quadratic relation is a parabola. Answer: trajectories are parabolas $q=E-p^2/2$.

Fixed points need $\dot{p}=-V^{\prime }(q)=0$ and $p=0$. Here $V^{\prime }(q)=q=0\Rightarrow q=0$. So the only fixed point is $(0,0)$. Classify by the shape of $V$: $V^{\prime \prime }(0)=1>0$, so $q=0$ is a minimum of $V$ → a center (surrounded by closed loops, stable). See Stability and Fixed Points.

$V^{\prime }(q)=-q=0\Rightarrow q=0$, with $p=0$. $V^{\prime \prime }(0)=-1<0$ → a maximum of $V$ → a saddle (unstable). WHAT IT LOOKS LIKE — Figure s02. The two dashed yellow diagonals are the special "in" and "out" lines through the saddle (pink dot at the origin). Blue and pink curves are nearby trajectories: they swoop toward the saddle along one diagonal, then peel away along the other — a ball balanced on a hilltop that any nudge sends running off.

$V^{\prime }(q)=4q^3-2q=2q(2q^2-1)=0\Rightarrow q=0,q=\pm \frac{1}{\sqrt{2}}$. $V^{\prime \prime }(q)=12q^2-2$.

• $q=0:V^{\prime \prime }=-2<0$ → maximum → saddle.
• $q=\pm 1/\sqrt{2}:V^{\prime \prime }=12\cdot \frac{1}{2}-2=4>0$ → minima → centers.

The three energy regimes (all cases) — Figure s03. The saddle sits at height $V(0)=0$, so $E=0$ is the separatrix energy. The two wells have bottom energy $V(\pm 1/\sqrt{2})=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}$.

1. $-\frac{1}{4}<E<0$ (below separatrix): the particle lacks the energy to climb over the central bump at $q=0$. It is trapped in one well and traces a small closed loop (blue). Two families, one per well.
2. $E=0$ (on the separatrix): the yellow figure-eight. Each lobe just barely reaches the saddle with zero speed; the two lobes kiss at the origin.
3. $E>0$ (above separatrix): the particle now has enough energy to ride over the central hump, so it swings all the way from the left well to the right well and back. Its trajectory is one big loop (pink) that encircles both wells — a global oscillation crossing $q=0$ with nonzero speed. This is the important edge case: above the separatrix the two wells merge into a single large orbit.

See Energy Conservation.

Why $m{l}^2$ appears. For a bob of mass $m$ swinging on a rod of length $l$, its distance from the pivot is $l$, so its moment of inertia (rotational analogue of mass) is $I=m{l}^2$. The kinetic energy of a rotation with angular momentum $p$ is $p^2/(2I)=p^2/(2m{l}^2)$ — the exact rotational twin of $p^2/2m$ for straight-line motion. So $H=\frac{p^2}{2m{l}^2}+V(\theta )=E.$ Solve for $p$: $\frac{p^2}{2m{l}^2}=E+mgl\cos \theta \Rightarrow p(\theta )=\pm \sqrt{2m{l}^2(E+mgl\cos \theta )}.$ Separatrix condition: the orbit that just reaches the top $\theta =\pi$ with $p=0$. Set $p=0,\theta =\pi$: $0=E+mgl\cos \pi =E-mgl\Rightarrow E_{\text{sep}}=mgl.$ WHY this splits motions: below $E_{\text{sep}}$ the square root becomes imaginary before $\theta =\pi$ — the pendulum turns back (libration, closed loop). Above it, $p$ stays real for all $\theta$ — the bob whirls over the top (rotation, open wavy line). Answer: $p(\theta )=\pm \sqrt{2m{l}^2(E+mgl\cos \theta )},E_{\text{sep}}=mgl.$ See Pendulum.

Energy leaks. $\dot{E}=\dot{x}\ddot{x}+{\omega }_0^{2}x\dot{x}=\dot{x}(\ddot{x}+{\omega }_0^{2}x).$ From the equation, $\ddot{x}+{\omega }_0^{2}x=-2\gamma \dot{x}$, so $\dot{E}=\dot{x}(-2\gamma \dot{x})=-2\gamma {\dot{x}}^2\le 0.$ Energy falls whenever the object moves (since $\gamma >0$ and ${\dot{x}}^2\ge 0$), so the dot cannot stay on one contour — it drops onto smaller-energy ellipses.

WHAT "focus" means. A focus is a fixed point that trajectories reach by spiralling (going round and round while approaching), as opposed to a node where they come straight in without circling. "Stable" just means they come in, not out.

WHY the flow spirals (linearised picture). Look for solutions $x(t)=e^{\lambda t}$, where $\lambda$ is a (possibly complex) growth rate we must find. Substituting into $\ddot{x}+2\gamma \dot{x}+{\omega }_0^{2}x=0$ gives the characteristic equation ${\lambda }^2+2\gamma \lambda +{\omega }_0^2=0\Rightarrow \lambda =-\gamma \pm \sqrt{{\gamma }^2-{\omega }_0^2}.$ For light damping ($\gamma <{\omega }_0$) the quantity under the root is negative, so its square root is imaginary and $\lambda =-\gamma \pm i{\omega }_d,{\omega }_d=\sqrt{{\omega }_0^2-{\gamma }^2}.$ Here ${\omega }_d$ is the damped frequency — the actual (slightly slowed) rate of wiggling once friction is present, and $i=\sqrt{-1}$ is the imaginary unit. These roots are complex numbers. The imaginary part $\pm i{\omega }_d$ produces the rotation (sines and cosines, i.e. going round the loop); the real part $-\gamma$ produces shrinking (the factor $e^{-\gamma t}$ that pulls the radius toward zero). A rotation that continuously shrinks is an inward spiral — that is exactly a stable focus. The characteristic timescale of the shrink is $1/\gamma$: distance from the origin falls like $e^{-\gamma t}$.

Contrast with the conservative case — WHY undamped loops close. For a conservative Hamiltonian flow, phase-space area is preserved in time (this is Liouville's Theorem). Intuitively: Hamilton's equations make the flow "incompressible" — the divergence of the flow vector is $\frac{\partial \dot{q}}{\partial q}+\frac{\partial \dot{p}}{\partial p}=\frac{\partial }{\partial q}\left(\frac{\partial H}{\partial p}\right)+\frac{\partial }{\partial p}\left(-\frac{\partial H}{\partial q}\right)=\frac{{\partial }^{2}H}{\partial q\partial p}-\frac{{\partial }^{2}H}{\partial p\partial q}=0.$ Zero divergence means a little blob of states neither expands nor contracts, so trajectories cannot converge to a point — loops must close. Damping breaks this: it adds a term with negative divergence, the blob shrinks, and the flow is allowed to spiral into the origin. See Liouville's Theorem.

Ellipse area $=\pi ab$ with $a=\sqrt{2E/(m{\omega }^2)},b=\sqrt{2mE}$ (the semi-axes derived in L2.2). With $m=\omega =1$: $a=\sqrt{2E}$ and $b=\sqrt{2E}$, so $\text{area}=\pi ab=\pi \sqrt{2E}\sqrt{2E}=2\pi E.$ WHY the two axes are equal here: when $m=\omega =1$ the ellipse is actually a circle of radius $\sqrt{2E}$, so its area is $\pi (\sqrt{2E}{)}^2=2\pi E$. The ratio of the two areas is $\frac{{\text{area}}_2}{{\text{area}}_1}=\frac{2\pi E_2}{2\pi E_1}=\frac{E_2}{E_1}=\frac{8}{2}=4.$ Interpretation: phase-space area for a harmonic oscillator is directly proportional to energy ($2\pi E$), so quadrupling the energy quadruples the enclosed area. (This enclosed area is proportional to the action, the quantity conserved under slow, adiabatic changes.) Answer: the second ellipse's area is $4\times$ larger.

(a) $V^{\prime }(q)=3q^2-3=0\Rightarrow q=\pm 1$. $V^{\prime \prime }(q)=6q$.

• $q=+1:V^{\prime \prime }=6>0$ → minimum → center at $(1,0)$.
• $q=-1:V^{\prime \prime }=-6<0$ → maximum → saddle at $(-1,0)$.

(b) The separatrix goes through the saddle with $p=0$: $E_{\star }=V(-1)=(-1{)}^3-3(-1)=-1+3=2$. (c) Released from rest at $q=0$: its energy is $E=V(0)=0$, and $0<E_{\star }=2$, so it sits inside the well around $q=1$. It will oscillate back and forth in that well (a closed loop), never reaching the saddle. Answer: fixed points center $(1,0)$, saddle $(-1,0)$; $E_{\star }=2$; particle oscillates about $q=1$.

From Hamilton's equations $\dot{q}=p/m,\dot{p}=-m{\omega }^{2}q$. Differentiate the first and substitute the second: $\ddot{q}=\frac{\dot{p}}{m}=-{\omega }^{2}q.$ WHY this is the key step: it collapses the two first-order equations into a single equation whose solution is $q(t)=a\cos (\omega t+\varphi )$ — a pure cosine with frequency $\omega$ set by the equation, not by the amplitude $a$. Since $\omega$ has no $E$ in it, the period $T=\frac{2\pi }{\omega }$ is the same for every ellipse. Big loops just travel proportionally faster. This "amplitude-independent period" is exactly what makes the oscillator special — compare the pendulum, whose period grows near the separatrix.

Flow speed vanishes at fixed points. Near a fixed point the flow vector $(\dot{q},\dot{p})\to (0,0)$, so the dot moves ever more slowly.

• For a stable focus (damped oscillator), the linearised solution from L4.2 gives $x(t)=A{e}^{-\gamma t}\cos ({\omega }_{d}t+\varphi )$, where $\gamma$ is the damping coefficient and ${\omega }_d$ the damped frequency from L4.2. The envelope $e^{-\gamma t}$ is the radius shrinking, so the distance from the origin behaves like $r(t)\sim r_0{e}^{-\gamma t}$. WHERE this comes from: the real part of the eigenvalue $\lambda =-\gamma \pm i{\omega }_d$ is $-\gamma$, and any mode $\propto e^{\lambda t}$ decays with that rate. Since $e^{-\gamma t}>0$ for all finite $t$, the radius reaches $0$ only as $t\to \infty$ — infinite time, consistent with the dot never crossing itself (each loop is a different, smaller loop).
• For a separatrix into a saddle, the same $\dot{}\to 0$ slowdown means the incoming curve also takes infinite time to arrive. Resolution: trajectories touch fixed points only in the limit $t\to \pm \infty$, never at finite time, so the "one state → one future" rule is never violated. The apparent meeting of curves at a saddle is an asymptotic meeting, not a finite-time crossing.

Center ::: minimum of $V$, closed loops, stable oscillation. Saddle ::: maximum of $V$, unstable, separatrix through it. Separatrix energy (pendulum) ::: $E=mgl$ (just reaches the top with zero speed). Oscillator period ::: $T=2\pi /\omega$, independent of energy. Why damped loops don't close ::: $\dot{E}=-2\gamma {\dot{x}}^2<0$, trajectory cuts across contours inward. Why conservative loops close ::: Hamiltonian flow has zero divergence (Liouville), area is preserved.