Streamlines, pathlines, streaklines: Visual walkthrough

Step 1 — Draw the blank stage: space and the clock

WHAT. Before any curve exists, we need a place (the plane) and a time (a clock). Everything we draw lives on the plane; every rule we write refers to what the clock reads.

WHY. All three curves come from the same velocity field, but they ask time to behave differently. A streamline freezes the clock; a pathline runs it forward; a streakline remembers the whole history. So we must keep space and time visibly separate from line one.

PICTURE. Below: a horizontal axis $x$ (rightward position), a vertical axis $y$ (upward position), and a little clock face off to the side that we will spin or freeze depending on which curve we build.

Figure — Streamlines, pathlines, streaklines

Step 2 — Attach an arrow to every point: the velocity field

WHAT. At each point we plant an arrow whose direction and length are $v = (u, v) = (1, t)$ .

WHY. A velocity field is the raw material for all three constructions. The arrow says "if a speck of fluid sits here right now, this is how fast and which way it is being carried." Notice the arrow does not depend on where you are (no $x$ or $y$ inside), only on when — this is what makes the flow spatially uniform but temporally changing.

PICTURE. Two snapshots of the arrow-forest side by side:

$rightward speed u = 1, upward speed v = grows with clock t$

At $t = 0$ : every arrow is horizontal ( $v = 0$ ), pointing pure-right.
At $t = 2$ : every arrow tilts up steeply ( $v = 2$ ), because upward speed has caught up and overtaken rightward speed.

Step 3 — Streamline: freeze the clock, follow the arrows

WHAT. A streamline is a curve threaded so its tangent everywhere lines up with the arrows at one frozen instant.

WHY. We want the snapshot answer: "which way is everything pointing right now?" So we jam the clock at some value, say $t = 2$ , and draw the curve that never crosses an arrow but always runs along it.

HOW — the equation, term by term. "Tangent parallel to the arrow" means the tiny step along the curve $d r = (d x, d y)$ points the same way as $v = (u, v)$ . Parallel vectors have matching component-ratios:

$tiny rightward step \div rightward speed \frac{d x}{u} = tiny upward step \div upward speed \frac{d y}{v} (clock frozen at t)$

Plug in $u = 1$ , $v = t$ with $t$ held as a constant number:

$\frac{d x}{1} = \frac{d y}{t} ⟹ d y = t d x ⟹ a straight line of slope t y = t x + C$

PICTURE. At the frozen time $t = 2$ the streamlines are straight lines of slope $2$ . They are straight because at this one instant every arrow has the same tilt (the field is spatially uniform).

Step 4 — Pathline: release the clock, follow one particle

WHAT. A pathline is the actual trail of one speck of fluid as the clock genuinely runs.

WHY. Now we ask the tracking question: "where does THIS particle go?" The particle's velocity is the field evaluated at wherever-and-whenever the particle currently is. So instead of freezing $t$ , we let it flow and integrate in time.

HOW — the equations, term by term.

$how x changes per second \frac{d x}{d t} = u = 1, how y changes per second \frac{d y}{d t} = v = t$

Release the particle from the origin at $t_{0} = 0$ (so $x_{0} = y_{0} = 0$ ):

$x = \int_{0}^{t} 1 d t^{'} = drifts right at speed 1 t, y = \int_{0}^{t} t^{'} d t^{'} = climbs faster and faster \frac{1}{2} t^{2}$

Eliminate the clock: since $x = t$ , substitute $t = x$ :

$y = \frac{1}{2} x^{2} — a parabola opening upward$

PICTURE. One particle leaves the origin, drifts right steadily but climbs ever-faster (because $v = t$ keeps growing), carving a parabola.

Step 5 — Streakline: keep the clock running, remember every release

WHAT. A streakline is the dye plume: at one observation time, mark all particles that were ever released from one fixed source.

WHY. A smoke chimney or dye needle injects fluid continuously. Each puff was released at its own time $τ$ and has been drifting ever since. The visible streak is a photograph now of many different particles, each with a different amount of travel time.

HOW — the equations, term by term. A particle released from the origin at time $τ$ follows the same rules as Step 4, but its clock started at $τ$ :

$x = \int_{τ}^{t} 1 d t^{'} = time it has been drifting (t - τ), y = \int_{τ}^{t} t^{'} d t^{'} = its accumulated climb \frac{1}{2} (t^{2} - τ^{2})$

$t$ ::: the fixed observation instant (the shutter click) — same for every puff.
$τ$ ::: the release time — this is the knob we sweep, from long ago ( $τ \to - \infty$ ) up to now ( $τ = t$ ).

Fix the shutter at $t = 2$ and sweep $τ$ from $2$ down to negative values:

$x = 2 - τ, y = \frac{1}{2} (4 - τ^{2})$

Eliminate $τ$ using $τ = 2 - x$ :

$y = \frac{1}{2} (4 - (2 - x)^{2}) = \frac{1}{2} (4 x - x^{2}) = 2 x - \frac{1}{2} x^{2}$

$y = 2 x - \frac{1}{2} x^{2} — a *different* parabola$

PICTURE. At $t = 2$ , dots mark particles released at $τ = 2, 1, 0, - 1, \dots$ . Connect them: a downward-bending arc that is clearly not the pathline of Step 4.

Step 6 — Overlay: watch the three split apart

WHAT. Put all three on one grid at $t = 2$ , all through the origin.

WHY. Seeing them separate simultaneously cements that they are three genuinely different objects in unsteady flow.

PICTURE.

Streamline (straight, slope $2$ ): $y = 2 x$ .
Pathline (opens up): $y = \frac{1}{2} x^{2}$ .
Streakline (opens down): $y = 2 x - \frac{1}{2} x^{2}$ .

All three leave the origin, all three have the same starting slope there — check: streamline slope $= 2$ ; pathline slope $\frac{d y}{d x} = x = 0$ at origin... wait, look carefully — they agree only at the source and immediately fan out.

Step 7 — The degenerate case: make the flow steady and watch them fuse

WHAT. Replace the clock-dependent field $v = t$ with a steady one, $v = 1$ (constant). Redo all three.

WHY. The parent's "golden rule" says in steady flow all three coincide. We must see this, not just be told it. Covering this limiting case is what makes the derivation complete.

HOW. With $u = 1, v = 1$ :

Streamline (freeze $t$ ): $\frac{d x}{1} = \frac{d y}{1} \Rightarrow y = x + C$ .
Pathline (release from origin at $t_{0} = 0$ ): $x = t, y = t \Rightarrow y = x$ .
Streakline (release at $τ$ , observe at $t$ ): $x = t - τ, y = t - τ \Rightarrow y = x$ .

All three collapse to the same line $y = x$ . There is no clock-turning to make the arrows tilt, so no history effect, so no splitting.

PICTURE. The three overlaid curves lie exactly on top of each other — a single line.

The one-picture summary

This final panel stacks the whole story: the arrow-field turning upward over time (top), and the three curves fanning out from one source in the unsteady case beside the single fused line in the steady case (bottom). One glance = the entire derivation.

Recall Feynman retelling — the whole walkthrough in plain words

We laid down an empty grid and a clock. At each point we planted an arrow: it always pushes right at speed $1$ , but its upward push grows as the clock ticks. That growing upward push is the villain of the story.

Then we asked three questions:

"Which way does everything point RIGHT NOW?" — freeze the clock, thread a curve along the arrows: a straight line. That's the streamline.
"Where does ONE speck go?" — let the clock run and float one speck from the origin. Because the upward push keeps growing, it climbs faster and faster: a parabola opening up. That's the pathline.
"What does the smoke plume look like?" — keep puffing dye from the origin and take one photo. Old puffs have drifted far; new puffs are fresh. Connecting them gives a different parabola, opening down. That's the streakline.

They fanned apart because the arrows were turning. Finally we switched to a lazy flow where the arrows never turn — and instantly all three curves snapped onto the very same line. That's the golden rule: steady flow makes streamlines, pathlines, and streaklines identical.

Recall checkpoints

Streamline through origin at

t = 2

for

v = (1, t)

y = 2 x

(straight line, slope = frozen

t

Pathline from origin for

v = (1, t)

y = \frac{1}{2} x^{2}

Streakline from origin, observed at

t = 2

y = 2 x - \frac{1}{2} x^{2}

Why do the three split in this flow?

The upward speed

v = t

makes the arrows turn over time (unsteady) — that turning peels the curves apart.

What happens if we set

v = 1

(steady)?

All three fuse into

y = x

; no arrow-turning means no history effect.

In the streakline formulas, which symbol is fixed and which is swept?

t

(observation time) is fixed;

τ

(release time) is swept.

Connections

Velocity field and material derivative
Steady vs unsteady flow
Lagrangian vs Eulerian description
Stream function ψ
Continuity equation
Flow visualization techniques (dye, smoke, PIV)

Visual walkthrough — Streamlines, pathlines, streaklines