Phase space — trajectories, phase portraits: Visual walkthrough

Step 1 — Two numbers, not one, describe "now"

WHAT. Take a block of mass $m$ on a spring. At one instant we write down two facts: where it is, and how fast it moves.

WHY. Newton's law $m \overset{x}{¨} = F$ is second order — it contains $\overset{x}{¨}$ , an acceleration. To predict the future of such an equation you must feed it two starting numbers: a position and a velocity. One is not enough. A spring released from rest at $x = 2$ moves differently from one flying through $x = 2$ at high speed. So the honest description of "now" is a pair.

We use position $x$ and momentum $p = m \overset{x}{˙}$ . Here:

$x$ = displacement from the rest point (metres), can be $+$ (stretched) or $-$ (compressed).
$\overset{x}{˙}$ = velocity (how fast $x$ changes), the dot means "rate of change in time".
$p = m \overset{x}{˙}$ = mass times velocity — a rescaled velocity that makes the later algebra clean.

PICTURE. Below, the block sits on its spring on the left. On the right we open a fresh blank plane: the horizontal axis is $x$ , the vertical axis is $p$ . The block's whole "now" collapses to one dot on that plane.

Figure — Phase space — trajectories, phase portraits

Step 2 — Give the dot a velocity: the flow arrows

WHAT. The dot is not frozen. As time ticks, both $x$ and $p$ change, so the dot drifts. We attach an arrow to it showing which way it moves and how fast.

WHY. To move a point you need its velocity. The dot lives in the $(x, p)$ plane, so its velocity has two parts: how fast $x$ changes ( $\overset{x}{˙}$ ) and how fast $p$ changes ( $\overset{p}{˙}$ ). We compute each from physics.

$\overset{x}{˙} = \frac{p}{m}$ . Read it term by term: $\overset{x}{˙}$ (rate of position) equals $p$ (momentum) divided by $m$ (mass). This is just $p = m \overset{x}{˙}$ turned around. It says: more momentum ⇒ faster sideways drift of the dot.
$\overset{p}{˙} = F = - k x$ . Here $\overset{p}{˙}$ (rate of momentum) equals the spring force $- k x$ . The $k$ is the spring's stiffness; the minus says the force always points back toward $x = 0$ . So a stretched spring ( $x > 0$ ) pushes $p$ down; a compressed spring ( $x < 0$ ) pushes $p$ up.

Together the arrow at each point is the little vector $(\overset{x}{˙}, \overset{p}{˙}) = (\frac{p}{m}, - k x)$ .

PICTURE. We drop this arrow at many points and get a whole field of arrows — the phase flow. Notice how they curl around the origin: rightward at the top, downward on the right, leftward at the bottom, upward on the left.

Step 3 — Follow one arrow to the next: a curve is born

WHAT. Start the dot somewhere, say stretched and at rest: $(x, p) = (a, 0)$ . Step it along its arrow a tiny bit, recompute the arrow, step again. Repeat.

WHY. A phase trajectory is exactly the curve you get by always moving along the local arrow. Formally it is the curve everywhere tangent to the flow field of Step 2. We trace it to see its shape before we prove that shape.

Tracking the signs tells the story of one lap:

Start at $(a, 0)$ : $\overset{p}{˙} = - k a < 0$ , so $p$ dives negative → dot moves down.
Now $p < 0$ : $\overset{x}{˙} = p / m < 0$ , so $x$ decreases → dot moves left.
Reach $(0, - p_{m a x})$ : fastest point, spring at rest length, all energy is motion.
Continue: $x < 0$ gives $\overset{p}{˙} = - k x > 0$ , so $p$ climbs back up → dot goes up and right.

PICTURE. The dot sweeps out a closed loop, running clockwise. It returns to $(a, 0)$ and repeats forever — this is the oscillation, drawn as geometry.

Step 4 — The hidden law that closes the loop: energy

WHAT. We now prove the loop closes — that the dot returns exactly, never spiralling. The tool is energy.

WHY a new tool? Tracing arrows suggests a loop but never proves it stays on one curve. We need a quantity that stays constant along the motion; its constant-value curve is then the trajectory. That quantity is the total energy.

Define the energy (the Hamiltonian) $H = kinetic: motion \frac{p ^{2}}{2 m} + potential: spring stretch \frac{1}{2} k x^{2} .$

$\frac{p ^{2}}{2 m}$ = energy of motion; grows with momentum, always $\geq 0$ .
$\frac{1}{2} k x^{2}$ = energy stored in the stretched/compressed spring; always $\geq 0$ .

Now watch how $H$ changes in time. We use the derivative $\frac{d H}{d t}$ because a derivative is precisely the tool that answers "is this quantity changing, and how fast?" $\frac{d H}{d t} = \frac{p}{m} \overset{p}{˙} + k x \overset{x}{˙} .$ Term by term: differentiate each piece and multiply by its rate ( $\overset{p}{˙}$ , $\overset{x}{˙}$ ) — the chain rule. Now substitute Step 2's flow ( $\overset{p}{˙} = - k x$ , $\overset{x}{˙} = p / m$ ): $\frac{d H}{d t} = \frac{p}{m} (- k x) + k x (\frac{p}{m}) = - \frac{k x p}{m} + \frac{k x p}{m} = 0.$ The two terms are equal and opposite — they cancel exactly. So $H$ never changes.

PICTURE. Energy is a bowl $H (x, p)$ sitting over the plane. Slicing the bowl at a fixed height $E$ leaves a closed rim; the shadow of that rim on the floor is our trajectory. Constant energy = constant height = a closed contour.

Step 5 — Why the contour is an ellipse (not a circle)

WHAT. Take the contour $H = E$ and read off its actual shape.

WHY. "Closed curve" is not enough — we want the precise geometry so we can label its width and height. We massage $H = E$ into the textbook ellipse equation.

Start from $\frac{p ^{2}}{2 m} + \frac{1}{2} k x^{2} = E .$ Divide every term by $E$ so the right side becomes $1$ : $\frac{x ^{2}}{2 E / k} + \frac{p ^{2}}{2 m E} = 1.$ Compare with the master ellipse form $\frac{x ^{2}}{a ^{2}} + \frac{p ^{2}}{b ^{2}} = 1$ :

$a = \frac{2 E}{k}$ — the half-width, the farthest $x$ reaches (the turning point where all energy is spring-stretch, $p = 0$ ).
$b = 2 m E$ — the half-height, the largest momentum (reached at $x = 0$ , all energy is motion).

Why an ellipse and not a circle? A circle needs $a = b$ . Here $a$ and $b$ carry different units and different physics — one is a distance set by stiffness $k$ , the other a momentum set by mass $m$ . They are equal only by fluke. In general the two axes differ, so the curve is a squashed circle: an ellipse.

PICTURE. The ellipse with its two half-axes marked, plus the special points: right/left tips are the turning points ( $p = 0$ , motion momentarily stops); top/bottom are the fastest points ( $x = 0$ ). Bigger $E$ ⇒ both $a$ and $b$ grow ⇒ a bigger nested ellipse.

Step 6 — Edge cases: what happens at zero and at the extremes

WHAT. We check the corner cases so no reader is ever surprised.

WHY. A derivation is only trustworthy once its boundaries behave. We test $E = 0$ , the tips, and what a stack of energies looks like.

$E = 0$ (degenerate). Then $a = b = 0$ : the ellipse shrinks to the single point $(0, 0)$ . Physically the block sits dead still at the spring's rest length forever. This is the fixed point — the flow arrow there is zero. (See Stability and Fixed Points.)
Right tip $x = a, p = 0$ . Momentum zero means momentarily not moving — but the arrow there points straight down ( $\overset{p}{˙} = - k a < 0$ ), so the dot is about to reverse. This is a turning point, not a stopping point.
Top $x = 0, p = b$ . Position zero, momentum maximal: flying through the middle at top speed. Arrow points left ( $\overset{x}{˙} = b / m > 0$ … wait sign: at top $p = b > 0$ so $\overset{x}{˙} > 0$ , moving right) — consistent with clockwise motion.
Many energies at once. Each $E$ gives its own nested ellipse; together they form the phase portrait, a set of nested rings around the center, like a contour map of a valley.

PICTURE. The degenerate point at the middle, then nested ellipses for growing $E$ , with the tip/top arrows drawn to show they are turning/fastest points — never true stops except the center.

The one-picture summary

Below, the whole derivation is compressed: the spring on the left, the energy bowl behind, one horizontal slice at height $E$ , and its shadow — a clockwise ellipse — on the $(x, p)$ floor, with turning points and fastest points labelled. Read it left to right and you have replayed Steps 1–6.

Recall Feynman retelling — the whole walkthrough in plain words

A block on a spring needs two numbers to describe it: where it is and how fast it's going. Draw those as a point on a map — left/right is position, up/down is momentum. That point isn't stuck; at every spot there's a little arrow saying which way it drifts, and the arrows curl around the middle. Follow the arrows and the point sweeps a loop, going clockwise. Why does the loop close instead of drifting away? Because the total energy — motion-energy plus spring-energy — never changes; and "energy stays the same" is a fence the point can't cross. That fence, when you write the algebra, turns out to be an ellipse: wide by an amount set by the spring's stiffness, tall by an amount set by the mass, so it's a squashed circle. The tips of the ellipse are the moments the block freezes and reverses; the top and bottom are the moments it whips through the middle at full speed. Give it more energy and you get a bigger ellipse around the same still center. That center — total energy zero — is the one place the block truly rests forever.

Recall

Why two axes, not one? ::: The equation of motion is second order, so you need both position and momentum to fix the future. What makes $\frac{d H}{d t} = 0$ ? ::: Substituting Hamilton's equations makes the two terms equal and opposite, so they cancel. Why an ellipse rather than a circle? ::: The half-axes $a = 2 E / k$ and $b = 2 m E$ differ (different physics/units), so the circle is squashed. What is the $E = 0$ case? ::: The ellipse collapses to the single fixed point at the origin — the block at permanent rest. Which way does the dot circulate and why? ::: Clockwise, because on the right ( $x > 0$ ) the restoring force drives $p$ downward.

Related: Pendulum (same method, closed loops become a separatrix) · Stability and Fixed Points · Liouville's Theorem.