The Phase Space

The central question of this lesson is short to ask: what is the state of a mechanical system, the smallest amount of information we must give, at one instant, to fix its entire future (and past) motion? Since Newton's law is second order, a position alone is not enough: we also need a velocity. Making that pair a single point in a space of states turns the laws of motion into a flow and, when energy is conserved, every possible motion (dynamics) can be read off as a level curve (geometry).

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1The state, the phase space, and the flow

The goal. We want the state of a mechanical system: the least information we must give, at one instant, to fix its whole future (and past). Newton's law is second order, so a position alone is not enough; we also need a velocity. We will first answer this question in the simplest case of one particle in a one-dimensional system; its generalization to many particles and higher spacial dimension is the subject of a later lesson.

From last time. The previous lesson built up conservative forces, which admit a potential \(F(q)=-dU/dq\) and the conservation of energy \(E=\half m\dot q^2+U(q)\), with two concrete examples we carry straight over: the simple pendulum, whose gravitational (height) potential is \(U(\theta)=mgL\,(1-\cos\theta)\), and the harmonic oscillator, \(U(q)=\half m\omega^2 q^2\). Both will return below in the discussion of phase portraits.

Take a single particle on a line. Pairing its position \(q\) with the momentum \(p=m\dot q\) puts position and velocity on the same footing, and the second-order law \(m\ddot q=F(q)\) splits into a pair of first-order equations,

\[ \dot q = \frac{p}{m}, \qquad \dot p = F(q). \]

Nothing has been solved: we have merely traded one second-order equation for two first-order ones. The gain is conceptual.

Definition: state & phase space The pair \(x=(q,p)\) is the state of the particle, and the set of all states \(\GG=\{(q,p)\}\) is its phase space, here an ordinary plane. A single point of \(\GG\) is a complete "identity card": it records where the particle is and where it is heading.

We keep the momentum \(p=m\dot q\) rather than the velocity because it is what is conserved in collisions and the natural partner of \(q\) in every later formulation of mechanics; here the two differ only by the factor \(m\).

A useful freedom. The coordinate \(q\) need not be a Cartesian position: any variable that fixes the configuration will do (an angle, a separation, a normal-mode amplitude), with \(p\) the momentum conjugate to that choice. The pendulum below is the first example, its natural coordinate being the angle \(\theta\). Choosing such coordinates well is the point of the Lagrangian formulation later in the course.

Examples. Already in one dimension the phase space need not be a flat plane; its shape is fixed by the configurations the system possesses. For the harmonic oscillator the position \(q\in\RR\) and \(p\in\RR\) are unrestricted, so \(\GG=\RR^2\); for the pendulum the angle \(\theta\) and \(\theta+2\pi\) are the same configuration, so the positions close into a circle and \(\GG=S^1\times\RR\), a cylinder.

Finally, write the equations of motion compactly as

\[ \dot x = \vb V(x), \qquad \vb V(x)=\Big(\frac{p}{m},\ F(q)\Big),\quad x=(q,p). \]

At every point of phase space the dynamics attaches a vector \(\vb V(x)\): a vector field, the phase velocity, like a stationary wind across \(\GG\). A solution is a curve \(x(t)\) everywhere tangent to this field, a phase curve (also called an orbit, or a trajectory). We now read this off in two examples; what makes the field so powerful, the existence–uniqueness theorem, we keep for the end.

2Harmonic oscillator

Last lesson we solved the oscillator outright: with \(U=\half m\omega^2 q^2\), Newton's law \(\ddot q=-\omega^2 q\) has the solution

\[ q(t)=q(0)\cos\omega t+\frac{\dot q(0)}{\omega}\sin\omega t. \]

Let us read that solution in the phase plane. Let us first take the simplest situation, release from rest at \(q(0)=q_0\): then \(\dot q(0)=0\), implying the initial momentum is \(p(0)=0\), and

\[ q(t)=q_0\cos\omega t,\qquad p(t)=m\dot q(t)=-m\omega q_0\,\sin\omega t. \]

As \(t\) advances, the phase point \(x(t)=(q(t),p(t))\) visits in turn

\[ \begin{array}{c|cccc} t & 0 & \pi/2\omega & \pi/\omega & 3\pi/2\omega\\ \hline x=(q,p) & (q_0,\,0) & (0,\,-m\omega q_0) & (-q_0,\,0) & (0,\,+m\omega q_0) \end{array} \]

right, bottom, left, top: the state runs clockwise around an ellipse and closes at \(t=2\pi/\omega\). This is no accident, and it holds for any initial state \(x_0=(q_0,p_0)\). In the generic case, the solution is \(q(t)=q_0\cos\omega t+\tfrac{p_0}{m\omega}\sin\omega t\) and \(p(t)=p_0\cos\omega t-m\omega q_0\sin\omega t\); substituting into the energy and using \(\cos^2(\omega t)+\sin^2(\omega t)=1\), the time cancels and

\[ E\big(q(t),p(t)\big)=\frac{p(t)^2}{2m}+\half m\omega^2 q(t)^2=\frac{p_0^{\,2}}{2m}+\half m\omega^2 q_0^{\,2}, \]

the same at every instant: the energy is conserved along every solution, for every \((q_0,p_0)\). Each trajectory is therefore confined to a level set \(E(q,p)=\text{const}\), the equation of an ellipse,

\[ \frac{p^2}{2m}+\half m\omega^2 q^2=E\quad\Longleftrightarrow\quad \frac{q^2}{2E/(m\omega^2)}+\frac{p^2}{2mE}=1. \]

And here is the point we keep: we did not really need the solution. Conservation of energy alone forces every trajectory onto a level curve \(E(q,p)=\text{const}\); the whole nested family of them, one ellipse per energy, run clockwise about the origin, is the phase portrait (i.e., the family of all possible trajectories). Its semi-axes are \(\sqrt{2E/(m\omega^2)}\) along \(q\) and \(\sqrt{2mE}\) along \(p\); where the ellipse meets the \(q\)-axis the particle is momentarily at rest (the turning points, \(U(q)=E\)), and at the bottom of the well the motion is all kinetic. The centre \((0,0)\) is the equilibrium: the particle at rest at the minimum.

Explore it. Pick a system and click in the lower (phase) plane to release a trajectory. The upper panel is the potential \(U(q)\) with the orbit's energy as a dashed line; the panel below the phase plane shows the actual physical motion of a spring following the harmonic motion. Two additional examples are considered: the pendulum and a particlein a double potential well potential.

separatrix equilibria sample orbits

click the plane to begin

potential energy U(q)

click to release a trajectory

the motion it represents

A phase portrait carries less information than an explicit solution (it has discarded the clock, so we no longer know when the particle is where) but in another sense far more, for it displays all motions at once, one curve per energy.

3Pendulum

The pendulum shows the method at its best. The plane pendulum of length \(\ell\) obeys the nonlinear equation

\[ \ddot\theta+\omega_0^2\sin\theta=0,\qquad \omega_0^2=\frac{g}{\ell}, \]

which has no solution in elementary functions. An exact solution does exist in terms of Jacobi elliptic functions, but it is the portrait, not the formula, that we are after. With \(U(\theta)=\omega_0^2(1-\cos\theta)\) (and \(p=\dot\theta\), the momentum conjugate to \(\theta\) in units where the moment of inertia \(m\ell^2=1\)) the conserved energy gives the level curves

\[ p(\theta)=\pm\sqrt{2\big(E-\omega_0^2(1-\cos\theta)\big)}, \]

read straight off the corrugated potential (select Pendulum in the explorer). Three kinds of motion appear:

Small oscillations. Near the bottom \(\sin\theta\approx\theta\) and the pendulum is a harmonic oscillator: the orbits there are the ellipses of §3. Every minimum hides an oscillator.
Libration \((E<2\omega_0^2)\): the bob swings between two turning points, a closed orbit.
Separatrix \((E=2\omega_0^2)\): the level curve \(p=\pm2\omega_0\cos(\theta/2)\) through the upright position \(\theta=\pi\), the boundary between the two kinds of motion.
Rotation \((E>2\omega_0^2)\): the momentum never vanishes and the bob swings over the top and circulates.

The saddle: an apparent crossing Watch the separatrix at the upright point \((\theta,p)=(\pi,0)\): it seems to cross itself there, but it does not. That point is a maximum of \(U\), an equilibrium where the motion halts, so the separatrix only approaches it, slowing without bound and taking infinite time to arrive. No two distinct trajectories ever share a point at finite time. That this holds for every curve, not just here, is the theorem of the next section.

Finally, \(\theta\) and \(\theta+2\pi\) are the same configuration, so the left and right edges of the portrait are identified: the phase plane becomes the cylinder \(\GG=S^1\times\RR\) of the Examples in §1. On it a rotation is a loop that wraps around and closes, a libration a small loop that does not, and the two saddles at \(\theta=\pm\pi\) are a single point on the back seam.

Exercise: the quartic double well Analyse the symmetric quartic \(U(q)=\tfrac14 q^4-\half q^2\): from \(U'(q)=q(q-1)(q+1)\) the critical points are \(q=0,\pm1\) with \(U''=3q^2-1\), so \(q=\pm1\) are centres and \(q=0\) a saddle, with separatrix energy \(E=U(0)=0\). Sketch the portrait, two wells joined by a figure-eight, and check it against the explorer's Double well tab.

4The phase flow: existence, uniqueness, no crossing

In both examples we quietly relied on something: distinct trajectories never meet, and the pendulum's separatrix only approaches the upright point without ever crossing. This is no accident; it is the content of one theorem, the foundation under every portrait we drew (see Appendix A for a proof).

Picard Theorem: existence and uniqueness Let the phase velocity \(\vb V\) be continuously differentiable (\(\vb V\in C^1\)) and autonomous (\(\vb V=\vb V(x)\), with no explicit time \(t\)). Then the initial-value problem \(\dot x=\vb V(x)\), \(x(t_0)=x_0\) has exactly one solution on some time interval around \(t_0\); equivalently, through every point of phase space passes a single trajectory.

The structural fact that makes a phase portrait legible is that orbits never cross:

Orbits never cross Through every point of phase space passes exactly one orbit; distinct orbits never meet, and none crosses itself except by closing into a loop. A crossing would be a single state with two different futures, which uniqueness forbids. This holds because the field is autonomous: \(\vb V(x)\) depends on the state, not on time \(t\), so each point carries a single velocity and hence a single future. (Strictly, it is the solutions that are unique, in the extended phase space below.)

Two further remarks round this out. Determinism: the present state fixes the entire future and, running the flow backward, the entire past. Equilibria: a point \(x^\star\) with \(\vb V(x^\star)=0\) is a motionless state, a trajectory by itself, which for the setup at hand means \(p=0\) and \(F(q)=0\), i.e. \(q\) at a critical point of the potential.

More precisely, the theorem gives a unique solution \(x(t)\). Adjoining the time axis to phase space forms the extended phase space \(\RR_t\times\GG\), in which the solution sits as its graph \(t\mapsto(t,x(t))\), an integral curve; distinct integral curves never meet. The orbit the portrait draws is this graph's projection onto \(\GG\), and orbits avoid crossing only because the field is autonomous; for a time-dependent \(\vb V(x,t)\) the projections may cross while the graphs stay disjoint, the clean picture being restored there by the autonomous extension \((\dot x,\dot t)=(\vb V,1)\).

Remark: determinism is not predictability Determinism does not mean practical predictability. Two very close initial states, separated only by a tiny distance \(\epsilon\), can evolve into trajectories that become exponentially separated at later times. This sensitivity to initial conditions is the seed of chaos: the future is fixed by the exact initial state, but a coarse-grained observer cannot distinguish nearby points with infinite precision. One might poetically view this gap as what an eighteenth-century theologian could have called a shadow of free will: not a violation of the celestial laws, but a limitation of our coarse-grained vision of their designs.

Outlook

Next lesson we will show that the same reading works for any one-dimensional conservative system, with no solving: every trajectory lies on a level curve of the energy,

\[ p(q)=\pm\sqrt{2m\,[\,E-U(q)\,]}, \]

so the whole portrait is read straight off the graph of \(U\): the motion confined to \(U(q)\le E\), every minimum of \(U\) a centre (expanded, a hidden harmonic oscillator of frequency \(\omega=\sqrt{U''/m}\)), every maximum a saddle with its separatrix. The double well already waiting in the explorer above is the first example, two centres and the saddle between them. Turning that qualitative picture into numbers, the period \(T(E)\) of a closed orbit, the energy method for an arbitrary \(U(q)\), and the way several wells partition the plane, is where we will pick up.

Looking ahead

We asked what the state of a system is, and we ended able to read every motion of a one-dimensional system off the geometry of a single picture, without solving a thing: equilibria, oscillations, and the boundary between swinging and rotating, all visible at a glance. That picture is not merely convenient; it is the natural home of mechanics and of much of physics beyond it. The phase space will lead later in the course to:

Hamiltonian mechanics. The flow \((\dot{\vb q},\dot{\vb p})\) is the seed of Hamilton's equations, and the enclosed area \(\oint \vb p\cdot\dd\vb q\) becomes their central invariant, the systematic theory we will build later on.
Chaos. Sensitive dependence, mixing, and strange attractors at the heart of chaos theory are all statements about the geometry of phase-space trajectories.

At the same time, more broadly, the phase space is the fundament of:

Statistical mechanics. A macroscopic system is a single point in a phase space of enormous dimension; an ensemble is a cloud of such points carried by the flow, and entropy counts the volume it occupies.
Quantum mechanics. The uncertainty principle forbids a sharp point, assigning each state a minimal phase-space area of order \(\hbar\); phase space is where the classical and quantum descriptions meet.

Appendix

A · Existence and uniqueness, by the contraction mapping theorem

We prove the existence–uniqueness theorem of §4 in any dimension, \(x\in\RR^{n}\). In fact, the hypothesis can be relaxed: it is necessary that \(\vb V\) is Lipschitz near \(x_0\), \(\|\vb V(x)-\vb V(y)\|\le L\|x-y\|\); this is automatic when \(\vb V\) is \(C^1\), since a bounded derivative supplies the constant \(L\).

When \(\GG\) is a manifold rather than \(\RR^n\), as for the pendulum's cylinder \(S^1\times\RR\), the statement still holds: it is local, every patch of the cylinder looks like an open piece of \(\RR^2\), the argument below runs there unchanged, and the identification \(\theta\sim\theta+2\pi\) only governs how trajectories close up, never their local existence or uniqueness.

Proof of the theorem. Throughout, \(x_0\in\RR^n\) denotes the fixed initial point, and curves are compared in the sup-norm \(\|x\|=\max_{t\in I}\|x(t)\|\). A continuous \(x(t)\) solves the IVP iff it solves the integral equation

\[ x(t)=x_0+\int_{t_0}^{t}\vb V(x(s))\,\dd s =: (Tx)(t). \]

For one direction, integrate \(\dot x=\vb V(x)\) from \(t_0\) to \(t\): the left side is \(x(t)-x(t_0)\) by the fundamental theorem of calculus, which gives the equation. For the other, differentiate the equation to recover \(\dot x=\vb V(x)\), and put \(t=t_0\) to make the integral vanish and get back \(x(t_0)=x_0\). The integral form is the convenient one, since it folds the initial condition into a single equation and makes sense for any merely continuous \(x\). A solution is therefore exactly a fixed point of the Picard operator \(T\). Fix a closed ball of radius \(r\) about \(x_0\) on which \(\vb V\) is bounded, \(\|\vb V\|\le\mu\), and Lipschitz, \(\|\vb V(x)-\vb V(y)\|\le L\|x-y\|\); the curves \(M=\{x\in C(I;\RR^n):\|x-x_0\|\le r\}\) on \(I=[t_0-h,t_0+h]\), measured in the sup-norm \(\|x\|=\max_{t\in I}\|x(t)\|\), form a complete space (because \(\RR^n\) is). Choosing \(h\le r/\mu\) makes \(T:M\to M\).

Contraction. Here, and only here, the Lipschitz hypothesis is used:

\[ \big\|(Tx)(t)-(Ty)(t)\big\|\le\int_{t_0}^{t} L\,\|x(s)-y(s)\|\,\dd s\le L\,h\,\|x-y\|, \]

so for \(h<1/L\), \(\kappa:=Lh<1\) and \(T\) is a contraction. The contraction-mapping lemma below (Banach–Caccioppoli) then gives the unique fixed point, the unique local solution.

A remark. The Picard iterates are explicit and in fact converge on the whole interval, not merely a short one: from the constant curve \(u_0(t)\equiv x_0\), \(\|u_{n+1}(t)-u_n(t)\|\le \mu L^n|t-t_0|^{n+1}/(n+1)!\), and the factorial (the series that builds \(e^{Lt}\)) converges for every \(t\). ∎

It remains to establish the tool we invoked.

Lemma (Banach–Caccioppoli, for Euclidean curves) Work on the continuous curves \(u:I\to\RR^n\) with the sup-norm \(\|u\|=\max_{t\in I}\|u(t)\|\) (the largest Euclidean size of \(x\) along \(I\)), so the distance between two curves is \(\|u-v\|\). Since \(\RR^n\) is complete, so is this space of curves. A map \(T\) that shrinks distances, \(\|Tu-Tv\|\le\kappa\,\|u-v\|\) with \(0\le\kappa<1\), then has exactly one fixed point, and the iterates \(u_{n+1}=Tu_n\) converge to it from any start.

This lemma has a more general form, valid for contractions on any complete metric space; here we use only the Euclidean case.

Proof of the lemma. Fix any curve \(u_0\) and iterate \(u_{n+1}=Tu_n\). Each step contracts, \(\|u_{n+1}-u_n\|=\|Tu_n-Tu_{n-1}\|\le\kappa\,\|u_n-u_{n-1}\|\le\kappa^{n}\,\|u_1-u_0\|\), so for \(m>n\) the triangle inequality and the geometric series give

\[ \|u_m-u_n\|\le\sum_{k=n}^{m-1}\|u_{k+1}-u_k\|\le\|u_1-u_0\|\sum_{k=n}^{m-1}\kappa^{k}\le\frac{\kappa^{n}}{1-\kappa}\,\|u_1-u_0\|, \]

which tends to \(0\) as \(n\to\infty\). The iterates therefore form a Cauchy sequence (their terms get arbitrarily close to one another), and by completeness of the curve space they converge to some curve \(u_\star\). A contraction is continuous, \(\|Tu_n-Tu_\star\|\le\kappa\,\|u_n-u_\star\|\to0\), so letting \(n\to\infty\) in \(u_{n+1}=Tu_n\) gives \(u_\star=Tu_\star\): a fixed point. It is the only one, since two fixed points would obey \(\|u_\star-v_\star\|=\|Tu_\star-Tv_\star\|\le\kappa\,\|u_\star-v_\star\|\), impossible for \(\kappa<1\) unless \(u_\star=v_\star\). ∎

?Questions you might ask

Why momentum \(p\), and not just the velocity \(\dot q\)?

At this level they differ only by the mass, \(p=m\dot q\), so either would draw the same picture. We keep \(p\) because it is what is conserved in collisions, and because it is the quantity every later formulation (Hamiltonian mechanics, statistical mechanics, quantum mechanics) selects as the natural partner of \(q\). Choosing it now costs nothing and saves a relabelling later.

Why is the phase space always even-dimensional?

Because a state needs a position and a momentum for each degree of freedom: \(\dim\GG=2f\). For \(N\) particles in three dimensions \(f=3N\), so \(\dim\GG=6N\). Oddness would mean some coordinate had no conjugate partner, impossible for a Newtonian system.

At the separatrix the curves seem to cross at the saddle. Doesn't that break uniqueness?

No. The saddle is an equilibrium, hence a complete trajectory by itself. The separatrix only approaches it, and the approach takes infinite time, so the two never actually meet at any finite time. No single state ever has two futures; uniqueness is intact, and the "crossing" is only apparent.

Why the hypothesis "continuously differentiable"? What if the force is rougher?

The proof (Appendix A) needs \(\vb V\) to be Lipschitz; \(C^1\) is a convenient sufficient condition. If it fails, uniqueness can genuinely fail: \(\dot x=\sqrt{|x|}\), \(x(0)=0\) is solved both by \(x\equiv0\) and by \(x=t^2/4\), a particle that may sit at the top of a cusp for an arbitrary time and then slide off (Norton's dome). Mere continuity still guarantees existence (Peano), but Newtonian determinism is a consequence of the smoothness of forces, not an automatic feature of the laws.

Why do you keep saying "every minimum hides a harmonic oscillator"?

Near a minimum \(q^\ast\), Taylor expansion gives \(U(q)\approx U(q^\ast)+\tfrac12 U''(q^\ast)(q-q^\ast)^2\), a parabola. So the small-amplitude orbits in any well are the ellipses of the oscillator, with frequency \(\omega=\sqrt{U''(q^\ast)/m}\). That is why the oscillator is the universal local model and why it is worth understanding completely.

How does this connect to what comes later in the course?

Directly. The first-order flow \((\dot{\vb q},\dot{\vb p})\) is the embryo of Hamilton's equations; the conserved area is the embryo of the symplectic invariants and of Liouville's theorem; and the same plane, with \([\hat q,\hat p]=i\hbar\), is where statistical and quantum mechanics are built. Today's picture is the foundation for all of them.