Visual walkthrough — Stream function, velocity potential

We will meet a river, count arrows, and never write a symbol we have not first drawn.

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Step 0 — What is a flow, in pictures?

Look at the figure: at each grid point sits one blue arrow. Its horizontal shadow is $u$, its vertical shadow is $v$. So describing the flow means knowing two numbers at every point — two whole functions $u(x,y)$ and $v(x,y)$. That is a lot of bookkeeping. The rest of this page is a hunt for a way to carry one number instead of two.

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Step 1 — The un-squishable rule (incompressibility)

WHAT the picture shows. A tiny square of side $\Delta x$ by $\Delta y$. Water enters the left face and leaves the right face (that's the $u$ traffic); it enters the bottom and leaves the top (the $v$ traffic). Count the net inflow and set it to zero.

• Left/right imbalance per unit area $=\frac{\partial u}{\partial x}$ — "how much the rightward speed changes as you walk right." If $u$ speeds up going right, more leaves than enters through those faces.
• Bottom/top imbalance $=\frac{\partial v}{\partial y}$ — "how much the upward speed changes as you walk up."

Setting the total change to zero gives the continuity equation (derived in Continuity Equation):

$$\underset{\text{net right-face outflow}}{\underbrace{\frac{\partial u}{\partial x}}}+\underset{\text{net top-face outflow}}{\underbrace{\frac{\partial v}{\partial y}}}=0.$$

This is one equation binding $u$ and $v$. It is the leash we will use in Step 2 to shrink two functions into one.

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Step 2 — Packing $(u,v)$ into one scalar $\psi$

We want a single function that automatically obeys Step 1's rule, so incompressibility comes for free. The trick: build $u$ and $v$ as slopes of one hill $\psi (x,y)$.

WHAT we do — the check. Drop these two definitions into the continuity equation:

$$\frac{\partial }{\partial x}\left(\underset{=u}{\underbrace{\frac{\partial \psi }{\partial y}}}\right)+\frac{\partial }{\partial y}\left(\underset{=v}{\underbrace{-\frac{\partial \psi }{\partial x}}}\right)={\psi }_{xy}-{\psi }_{yx}.$$

WHY it vanishes. The order of taking two partial derivatives does not matter for smooth fields: ${\psi }_{xy}={\psi }_{yx}$ (this is Clairaut's theorem, pictured as: differentiating a smooth hill east-then-north = north-then-east). So the two terms are equal and opposite — they cancel to $0$ no matter what $\psi$ is.

We have traded two functions for one, and the un-squishable rule now holds by construction.

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Step 3 — Why lines of constant $\psi$ are the streamlines

WHAT the tangent condition says. If a tiny step $(dx,dy)$ lies along a streamline, it must be parallel to $(u,v)$. Two vectors are parallel when their "cross" combination is zero:

$$udy-vdx=0.$$

Read it as: the sideways part of the step matches the flow's tilt.

WHAT the change in $\psi$ is over that same step. Adding up the two slope contributions:

$$d\psi =\underset{=-v}{\underbrace{\frac{\partial \psi }{\partial x}}}dx+\underset{=u}{\underbrace{\frac{\partial \psi }{\partial y}}}dy=-vdx+udy=udy-vdx.$$

WHY this is the punchline. The right side is exactly the tangent condition. So along a streamline $udy-vdx=0$, which means $d\psi =0$ — $\psi$ does not change as you walk along a streamline. Therefore:

$$\boxed{\psi =\text{const}⟺\text{a streamline}}$$

The green curves in the figure are level curves of the $\psi$-hill, and the blue arrows sit tangent to them — never crossing.

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Step 4 — $\psi$ counts the water (flux = difference of labels)

WHAT we compute. Draw any line joining a point on streamline ${\psi }_1$ to a point on streamline ${\psi }_2$. The water crossing that line per second is

$$Q=\int (udy-vdx)=\int d\psi ={\psi }_2-{\psi }_1.$$

WHY this is gorgeous. Because $udy-vdx$ is $d\psi$ (Step 3), the integral just adds up tiny changes in $\psi$ and telescopes to the endpoints — the shape of the crossing line does not matter, only which two streamlines it connects. So $\psi$ is a literal "how much water is to my left" meter. Two streamlines two units apart carry two units of flow between them, always.

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Step 5 — The non-spinning rule births a second scalar $\varphi$

Incompressibility gave us $\psi$. A different physical rule gives a different scalar.

WHAT this buys us. A theorem of calculus says: if a field has zero curl (in a simply-connected patch), it is the gradient of some scalar hill $\varphi$. So we may write

$$u=\frac{\partial \varphi }{\partial x},v=\frac{\partial \varphi }{\partial y}.$$

WHY this $\varphi$ automatically has no spin. The curl of any gradient is identically zero — same Clairaut cancellation as Step 2, now working for us: ${\partial }_x{\varphi }_y-{\partial }_y{\varphi }_x=0$. So writing velocity as a gradient makes irrotationality free, exactly as writing it via $\psi$ made incompressibility free.

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Step 6 — When BOTH exist: Cauchy–Riemann and Laplace

If the flow is both un-squishable and non-spinning ("ideal flow"), $\psi$ and $\varphi$ coexist. Now $u$ and $v$ each have two names — one from each scalar. Set the names equal:

These are the Cauchy-Riemann Equations — the very ones that define an analytic complex function.

WHY each scalar solves Laplace's equation (see Laplace Equation):

• $\varphi$: incompressibility says $u_x+v_y=0$; substitute $u={\varphi }_x,v={\varphi }_y$:

$$\underset{u_x}{\underbrace{{\varphi }_{xx}}}+\underset{v_y}{\underbrace{{\varphi }_{yy}}}=0\Rightarrow {\nabla }^2\varphi =0.$$

• $\psi$: irrotationality says $v_x-u_y=0$; substitute $v=-{\psi }_x,u={\psi }_y$:

$$\underset{v_x}{\underbrace{-{\psi }_{xx}}}-\underset{u_y}{\underbrace{{\psi }_{yy}}}=0\Rightarrow {\nabla }^2\psi =0.$$

Each rule feeds the other scalar into Laplace. Notice the beautiful crossover: incompressibility tames $\varphi$, irrotationality tames $\psi$. Every ideal 2D flow is therefore a pair of harmonic hills.

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Step 7 — Why the two maps cross at right angles

WHAT we compare. Two gradient arrows at any point:

$$\nabla \varphi =({\varphi }_x,{\varphi }_y)=(u,v),\nabla \psi =({\psi }_x,{\psi }_y)=(-v,u).$$

The first points across the $\varphi$-level-curves (the equipotentials); the second points across the $\psi$-level-curves (the streamlines).

WHY perpendicular. Take their dot product — the test for a right angle (zero means $90^{\circ }$):

$$\nabla \varphi \cdot \nabla \psi =(u)(-v)+(v)(u)=-uv+uv=0.$$

Zero, always. So the equipotentials and the streamlines meet everywhere at $90^{\circ }$: they weave the flow net, a curvy sheet of graph paper. This is the parent note's final claim, now seen.

Recall Edge cases you must not skip

Stagnation points ($u=v=0$): both gradients vanish, so the dot-product test reads $0=0$ trivially — the orthogonality statement is empty there and grid lines may fork (e.g. at the nose of a body). Nothing is broken; the net simply has a crossing point. Reveal ::: At a stagnation point $\nabla \varphi =\nabla \psi =\vec{0}$; the flow net can have an X-shaped junction. It is a degenerate, not a contradiction.

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Worked example threaded through the pictures — a point source

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The one-picture summary

Recall Feynman: the whole walkthrough in plain words

Picture a calm river seen from above. At every spot the water has an arrow — that's two numbers, sideways speed and upward speed, and juggling two numbers everywhere is tiring.

First trick. Water can't be squished, so as much flows out of any tiny box as flows in. That single rule lets us paint the river with "lane numbers" — call it $\psi$. Water never crosses lanes, so a line of equal lane number is a path the water truly follows (a streamline), and the gap between two lane numbers is exactly how much water runs in that channel. One number replaced two.

Second trick. If the water also has no little whirlpools, we can paint it with "heights" instead — call it $\varphi$ — and the water always slides toward higher ground. Again one number for two.

When both are true, each painting is smooth in a special way (Laplace's equation), and — the magic finale — the lane-lines and the height-lines cross everywhere at perfect right angles, like bent graph paper. Test it with a source: circles of equal height, spokes of equal lane, meeting at $90^{\circ }$, with the water that leaves the source adding up to exactly its strength.

Two scalars, born from two physical rules, woven into one orthogonal grid. That is the entire chapter in one breath.

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