Phase space — trajectories, phase portraits: Question bank

Definition Symbols used on this page — read first

Everything below is built from a small dictionary. Nothing here is used before this box.

$q$ — the position (or angle) of the system. For a mass on a spring it is displacement $x$ ; for a Pendulum it is the angle $θ$ . So " $q$ " and " $x$ " mean the same kind of thing.
$p$ — the momentum, "how much motion" the system carries: $p = m \overset{q}{˙}$ , where $\overset{q}{˙}$ is the rate of change (velocity) of $q$ and $m$ is the mass. The dot always means "rate of change per unit time."
The state is the pair $(q, p)$ plotted as one point; the phase plane has $q$ across and $p$ up.
$H$ — the Hamiltonian, which for a conservative 1-D system is just the total energy written in terms of state: $H (q, p) = kinetic \frac{p ^{2}}{2 m} + potential V (q) .$
$E$ — a fixed number, the total energy of one orbit; each trajectory keeps its own $E$ .
$\partial_{q} H, \partial_{p} H$ — how $H$ changes when you nudge $q$ or $p$ alone (partial slopes of the energy landscape).
The flow rule (Hamiltonian Mechanics) that moves a state point is $\overset{q}{˙} = \partial_{p} H$ and $\overset{p}{˙} = - \partial_{q} H$ . These give the velocity arrow $(\overset{q}{˙}, \overset{p}{˙})$ at every point.

Figure — Phase space — trajectories, phase portraits

The figure above is the mental picture behind every question. Look at it now: the horizontal axis is position $q$ , the vertical axis is momentum $p$ . The cyan loop is one energy contour ( $H = E$ ). The amber dot is a state point $(q, p)$ sitting on it, and the white arrow is its flow velocity $(\overset{q}{˙}, \overset{p}{˙})$ — always tangent to the loop, sweeping clockwise. The white square at the origin is a center (fixed point). Refer back to it whenever a reveal mentions "the loop," "the arrow," or "the center."

Why the energy stays constant (do this once, then reuse it)

Several traps below hinge on the fact that $H$ never changes along a conservative orbit. Here is the cancellation spelled out, step by step, so you own it.

True or false — justify

The phase plane axes are position versus time.

False. The axes are position

q

versus momentum

p

(equivalently

x

\overset{x}{˙}

); time is the hidden parameter running along each curve, not an axis — look again at the figure, which has no time axis.

Knowing only the position

q

of a mass fixes its future motion.

False. Newton's law is second order, so you also need the velocity/momentum

p

— a spring released from rest and one whipped past the same point move differently. That is exactly why phase space needs two axes.

A closed loop in phase space means the motion repeats periodically.

True for a conservative 1-D system. The state returns to

(q, p)

, and since the flow arrow there is fixed, it must retrace the same loop forever — a genuine period. (In Harmonic Oscillator every loop is an ellipse.)

For a conservative system, every trajectory lies on a curve of constant energy

H = E

True. As shown above,

d H / d t = \partial_{q} H \overset{q}{˙} + \partial_{p} H \overset{p}{˙} = A B - A B = 0

once the flow rule is inserted, so

H (q, p) = E

stays fixed along the orbit. Trajectories are energy contours (see Energy Conservation).

Two phase trajectories can cross at an ordinary (non-fixed) point.

False. The flow arrow

(\overset{q}{˙}, \overset{p}{˙})

is single-valued, so a crossing would give one state two futures — forbidden by determinism. Curves meet only where the arrow vanishes (fixed points).

A bigger phase loop means the state point moves faster at every point of the loop.

False. Bigger loop = more energy, but the phase speed

\overset{q}{˙}^{2} + \overset{p}{˙}^{2}

varies around the loop — fast where

∣ p ∣

is large, nearly stopped near the turning points where

p \to 0

The area enclosed by a bundle of trajectories can shrink for a frictionless (Hamiltonian) system.

False. Liouville's Theorem says phase-space area (volume) is conserved by Hamiltonian flow — it can deform but never shrink. Shrinking areas require dissipation, as in a damped oscillator.

A damped oscillator has closed phase trajectories.

False. Energy falls,

\dot{E} = - 2 γ \overset{x}{˙}^{2} \leq 0

(the friction constant

γ > 0

), so each turn crosses a smaller ellipse — the orbit spirals inward to the origin (a stable focus), never closing.

Every fixed point in phase space is stable.

False. Minima of the potential

V (q)

give centers (stable, closed loops); maxima give saddles (unstable — states fly off along one direction). See Stability and Fixed Points.

The separatrix is a trajectory just like the loops around it.

False. Although it is a trajectory, it is a distinguished one: it carries exactly the critical energy, passes through a saddle, and divides qualitatively different motions (swinging vs whirling for the Pendulum) — ordinary loops do none of that.

At a fixed point the state has zero velocity, so trajectories can touch there.

True.

\overset{q}{˙} = \overset{p}{˙} = 0

means the flow arrow is zero, so the "no-crossing" rule (which forbids two nonzero arrows at one point) doesn't apply — separatrix branches genuinely meet at saddles.

Spot the error

\overset{p}{˙} = + \partial_{q} H

, by symmetry with

\overset{q}{˙} = \partial_{p} H

The minus sign is missing:

\overset{p}{˙} = - \partial_{q} H

. Without it the two terms in

d H / d t

would add instead of cancel, so energy wouldn't be conserved and oscillator loops would spiral out instead of closing.

"The harmonic oscillator traces a circle in the

(q, p)

plane."

Only if the axes are scaled so the two semi-axes match. Start from energy

\frac{p ^{2}}{2 m} + \frac{1}{2} m ω^{2} q^{2} = E

and divide both sides by

E

\frac{p ^{2}}{2 m E} + \frac{q ^{2}}{2 E / ( m ω ^{2} )} = 1

. The two denominators become the semi-axes-squared, so in raw

(q, p)

units it is an ellipse, generally not a circle.

"The oscillator loop runs counter-clockwise because time moves forward."

It runs clockwise: for

q > 0

the force gives

\overset{p}{˙} = - m ω^{2} q < 0

, so momentum decreases — the point sweeps downward on the right side, which is clockwise (matching the white arrow in the figure).

"For the pendulum,

θ = π

(bob straight up) is a stable resting point."

It is a maximum of

V = - m g l cos θ

(largest when

cos θ = - 1

), hence a saddle — unstable. The stable center is

θ = 0

(bob hanging down,

V

lowest).

"Since

d H / d t = 0

for the damped oscillator, its orbits are energy contours."

The damped system is not conservative:

\dot{E} = - 2 γ \overset{x}{˙}^{2} \neq = 0

, so no fixed

H

exists. Only for time-independent conservative

H

do trajectories coincide with energy contours.

"A fixed point sits wherever the velocity

\overset{q}{˙}

is zero."

You need both

\overset{q}{˙} = 0

and

\overset{p}{˙} = 0

. A turning point of an oscillation has

\overset{q}{˙} = 0

(momentum zero) yet

\overset{p}{˙} \neq = 0

, so it is not a fixed point — the state keeps moving.

Why questions

Why does the phase portrait of a conservative system reduce to a contour map?

Because

H

is constant along each orbit (

d H / d t = 0

, shown above), every trajectory is a level set

H (q, p) = E

— drawing the contours of the energy

H

is drawing the portrait.

Why must the flow arrow be tangent to the trajectory everywhere?

The trajectory is the path traced by the moving state, and

(\overset{q}{˙}, \overset{p}{˙})

is its instantaneous velocity — velocity is by definition tangent to the path it generates (see the white arrow on the loop in the figure).

Why does adding friction turn a loop into an inward spiral rather than a smaller loop?

Friction removes energy continuously in time (

\dot{E} = - 2 γ \overset{x}{˙}^{2} < 0

), so the state cannot return to its start — each cycle lands on a lower-energy curve, producing a spiral that converges to the equilibrium.

Why does the separatrix, not any random energy, mark the boundary between libration and rotation of a pendulum?

It is the exact energy that just reaches the unstable top (

θ = π

) with zero speed; below it the bob turns back (swings), above it it keeps going over the top (whirls).

Why can a smart reader "see" instability in a phase portrait without solving equations?

Instability shows as trajectories that diverge away from a point (saddle) rather than looping around it (center) — the geometry of the flow directly encodes the dynamics.

Edge cases

What does the phase trajectory of a mass at absolute rest at a stable equilibrium look like?

A single point — the state never moves, so the "curve" degenerates to the fixed point itself (zero-energy loop of zero size), like the white square at the origin of the figure.

What happens to the harmonic-oscillator ellipse as the energy

E \to 0

Both semi-axes

2 E / (m ω^{2})

and

2 m E

shrink to zero, so the ellipse collapses onto the center point at the origin — consistent with a mass sitting still at the bottom.

At a turning point of an oscillation, what is happening in phase space?

The trajectory crosses the

q

-axis where

p = 0

: momentum momentarily zero, but

\overset{p}{˙} = - \partial_{q} H \neq = 0

, so the point is at maximum displacement and about to reverse — it is not at rest.

For the pendulum spinning fast over the top, why is the trajectory an open wavy line, not a loop?

Because

θ

keeps increasing without bound — the bob just circulates. On a flat plane it runs off to the right forever; the wavy line reflects

p

rising and falling as the bob speeds through the bottom and slows over the top.

Why does the pendulum phase space really "want" to be a cylinder, not a flat plane?

The angle

θ

is periodic:

θ

and

θ + 2 π

are the same physical position. Rolling the

q

-axis into a loop turns the "open" whirling trajectories into closed circles going around the cylinder — the topological edge case that plane pictures hide.

What is the trajectory of a free particle (no potential,

V = 0

) in phase space?

A horizontal straight line

p = const

: with

H = p^{2} / (2 m)

the momentum is conserved so it never changes, while

q = q_{0} + (p / m) t

drifts steadily — the energy contours are just horizontal lines.

Recall One-line self-test

Cover everything: name the axes, the flow rule, the no-crossing reason, and the one thing friction changes. Axes ::: position $q$ and momentum $p$ (time hidden along the curve). Flow rule ::: $\overset{q}{˙} = + \partial_{p} H, \overset{p}{˙} = - \partial_{q} H$ (minus lives with $p$ ). No-crossing reason ::: single-valued flow arrow ⇒ one state, one future (determinism). What friction changes ::: energy drops, so closed loops become inward spirals.