Phase space and Hamilton’s equations

1. A space of mechanical states

Configuration space records possible arrangements. A point $q$ tells us where the system is, but not where it is going. To specify a state of a regular mechanical system at an instant, supply the conjugate momenta as well. For $n$ degrees of freedom, the resulting phase space has $2n$ local coordinates:

(q_1,\ldots,q_n,p_1,\ldots,p_n).

(13.1)

A phase-space point represents the state of the entire system, not the position of one physical particle. A phase trajectory is the curve traced by that state as time passes. For a one-dimensional oscillator, physical motion lies on a line, configuration space is that line, and phase space is a plane with coordinates $(x,p)$ .

For a full pendulum, configuration space is a circle because angles separated by $2\pi$ represent the same orientation. Its phase space is a cylinder: angular position wraps around, while momentum ranges over the real line. Drawing a rectangular strip is a useful representation provided its two angular edges are identified.

2. Hamilton's equations

For a regular Hamiltonian obtained from $L$ , with no remaining generalized forces,

\boxed{\dot q_i=\frac{\partial H}{\partial p_i},\qquad \dot p_i=-\frac{\partial H}{\partial q_i}.}

(13.2)

All other phase-space coordinates are held fixed in each partial derivative. These are $2n$ first-order equations and require $2n$ initial values, just as the $n$ second-order Lagrange equations do. More equations do not mean more physical freedom.

The first equation relates momentum to velocity. The second describes momentum change, generalizing force. For $H=p^2/(2m)+V(x)$ they give $\dot x=p/m$ and $\dot p=-V'(x)$ , so differentiation of the first recovers $m\ddot x=-V'(x)$ .

The full derivation is optional deeper understanding, not required by the official syllabus. Here the equations are used to construct and interpret phase trajectories.

At each phase point, the pair $(\partial_pH,-\partial_qH)$ is a tangent direction to the motion. For an autonomous system with smooth, locally unique evolution, two distinct trajectories cannot cross at the same phase point: if they did, the same initial state would have two futures. Coordinate-space paths can cross because their momenta need not agree. In a time-dependent system, a projected phase-space path can pass through the same $(q,p)$ at different times; the evolution law also needs the time.

3. Energy contours and direction of motion

For an autonomous unforced Hamiltonian,

\frac{dH}{dt}=\sum_i\left(\frac{\partial H}{\partial q_i}\dot q_i+ \frac{\partial H}{\partial p_i}\dot p_i\right) =\sum_i\left(\frac{\partial H}{\partial q_i}\frac{\partial H}{\partial p_i} -\frac{\partial H}{\partial p_i}\frac{\partial H}{\partial q_i}\right)=0.

(13.3)

Thus motion stays on a constant- $H$ surface. For one degree of freedom these surfaces are curves. For several degrees of freedom, one energy value usually leaves a $(2n-1)$ -dimensional surface and does not determine a unique trajectory. A contour gives the possible path; Hamilton's equations give its direction and rate of traversal.

A fixed point satisfies both $\partial H/\partial p=0$ and $\partial H/\partial q=0$ . It is a state with no evolution. It must not be confused with a turning point where $\dot q=0$ but $\dot p\ne0$ .

Figure 13.1 · A harmonic oscillator traces a clockwise ellipse in physical phase coordinates. These scaled axes turn it into a circle; the physical particle moves along a line.

4. Applications

Example 1 · Oscillator ellipses

With $T=m\dot x^2/2$ , $V=kx^2/2$ , $p=m\dot x$ , the Hamiltonian is $H=p^2/(2m)+kx^2/2$ . On the energy curve $H=E>0$ ,

\frac{x^2}{2E/k}+\frac{p^2}{2mE}=1.

(13.4)

This is an ellipse with intercepts $x=\pm\sqrt{2E/k}$ and $p=\pm\sqrt{2mE}$ . At the right-hand intercept, $\dot x=0$ but $\dot p=-kx<0$ , so the trajectory goes downward. With $x$ horizontal and $p$ vertical, circulation is clockwise.

Using $\omega=\sqrt{k/m}$ , the solution $x=A\cos(\omega t+\phi)$ has $p=-m\omega A\sin(\omega t+\phi)$ . Substitution verifies both Hamilton equations and gives $E=kA^2/2$ . In dimensionless axes $x/A$ and $p/(m\omega A)$ , the ellipse becomes a circle; that rescaling does not make the physical quantities have the same dimensions.

Example 2 · Free-particle trajectories

For $H=p^2/(2m)$ , Hamilton's equations give $\dot p=0$ and $\dot x=p/m$ . Each nonzero-energy contour consists of two horizontal lines, $p=\pm\sqrt{2mE}$ , with opposite directions of motion. At $p=0$ , every position is a fixed point.

Energy alone does not decide whether the particle moves left or right. The initial momentum sign does. This simple example shows why a state needs momentum rather than speed alone.

Example 3 · Pendulum oscillation and rotation

For a fixed-length rod pendulum,

H=\frac{p_\theta^2}{2ma^2}+mga(1-\cos\theta),\qquad \dot\theta=\frac{p_\theta}{ma^2},\quad\dot p_\theta=-mga\sin\theta.

(13.5)

The stable bottom is $(\theta,p_\theta)=(0,0)$ modulo $2\pi$ . For $0<E<2mga$ , momentum reaches zero at two angles satisfying $\cos\theta=1-E/(mga)$ : the pendulum oscillates. For $E>2mga$ , momentum never reaches zero and the pendulum rotates with a fixed sign of angular velocity.

The curve $E=2mga$ separates these behaviors and is called a separatrix. The upper equilibrium $(\pi,0)$ lies on it. Near the top put $\theta=\pi+\eta$ . Since $\cos(\pi+\eta)\simeq-1+\eta^2/2$ , the separatrix energy equation gives $p_\theta^2/(2ma^2)\simeq mga\eta^2/2$ , hence $|\dot\eta|\simeq\sqrt{g/a}|\eta|$ . The time integral contains $\int d\eta/|\eta|$ , which diverges at zero: exact approach to the top takes infinite time.

A taut string can fail during some rotations, so this global phase portrait assumes a rod or another bilateral circular constraint.

Example 4 · An unstable equilibrium

Consider $V(x)=-kx^2/2$ with $k>0$ . Then $H=p^2/(2m)-kx^2/2$ , $\dot x=p/m$ , and $\dot p=kx$ . Eliminating $p$ gives $\ddot x=(k/m)x$ , with

x=Ae^{\omega t}+Be^{-\omega t},\qquad \omega=\sqrt{k/m}.

(13.6)

The energy contours are hyperbolas. At $E=0$ they are the two straight lines $p=\pm m\omega x$ . One direction approaches the origin and another departs from it. This saddle behavior explains instability: most arbitrarily small disturbances contain a growing exponential component. A picture resembling a cross at the origin does not violate uniqueness; the fixed point is reached only in infinite time along the approaching branch.

5. Reading a phase portrait

First identify the axes and their units. Then locate equilibria, turning points, and any separatrices. Use Hamilton's equations to put arrows on the curves; energy contours alone do not fix direction. Finally translate the curve back into motion: an oscillator ellipse represents back-and-forth motion on a line, not an elliptical path in physical space.

These distinctions will be reused for radial central-force motion, where $(r,p_r)$ describes the radial part of an orbit even though the particle simultaneously advances in angle.

Exercises

Hints and solutions are collected separately.

13.1 How many phase-space dimensions does a regular two-coordinate system have?
13.2 Why can configuration-space paths cross while autonomous phase trajectories cannot cross at a finite time?
13.3 Derive Newton’s equation from $H=p^2/(2m)+V(x)$ .
13.4 An oscillator has $m=2$ , $k=8$ , $E=4$ in SI units. Find its phase ellipse intercepts.
13.5 At the positive-position turning point of a spring oscillator, is the state a fixed point?
13.6 Prove that an autonomous unforced Hamiltonian is constant along motion.
13.7 What phase-space topology describes a full rod pendulum?
13.8 For a pendulum with $E=mga$ , find the turning angles.
13.9 Why does the pendulum separatrix take infinite time to reach the upper equilibrium?
13.10 For $V=-kx^2/2$ , find the two zero-energy lines in phase space.
13.11 Does one energy value specify a unique orbit for two degrees of freedom?
13.12 For a free particle with $p<0$ , describe its phase-space motion.

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