Bernoulli's equation — derivation from F = ma along streamline: Visual walkthrough

Step 1 — What is a streamline, and what is a "parcel"?

WHAT. Water flowing steadily traces invisible highways called streamlines — a curve that is, at every point, pointing exactly the way the fluid is moving there. We pick a tiny blob of fluid sitting on one such highway: a little squat cylinder. Its flat faces point along the flow; its cross-section has area $A$ , and it is $d s$ long. The fluid at that spot moves along the streamline with a certain speed, which we call $v$ .

WHY. Newton's law, $F=ma$, talks about a definite chunk of matter. Fluid is a continuous smear, so we must first cut out one definite chunk — this cylinder — before we are allowed to say "force = mass × acceleration."

PICTURE.

Figure — Bernoulli's equation — derivation from F = ma along streamline

Step 2 — The push from pressure

WHAT. Pressure $P$ is how hard the surrounding fluid presses on each square metre of a face. It presses on the back face pushing the blob forward, and on the front face pushing it backward. If the pressure at the back is $P$ and at the front is $P + d P$ (a little more or less), we add the two pushes.

WHY. Only a difference in pressure across the blob leaves a net force. If both faces were squeezed equally, the pushes would cancel — like two people shoving a door equally from both sides. So we need $d P$ , the change in pressure over the length $d s$ .

PICTURE.

Step 3 — The pull of gravity along the slope

WHAT. Gravity pulls the blob straight down with force $d m g$ . But only the part of that pull pointing along the streamline can speed the blob up or slow it down. If the streamline climbs by height $d y$ while running a length $d s$ , it makes an angle $θ$ with the horizontal, and the streamline's steepness is captured by $sin θ = d y / d s$ .

WHY. A force sideways to the motion cannot change the speed — it only bends the path. So we keep only the slope-component of gravity. We use $sin θ$ because, on the little right-triangle formed by "rise $d y$ ", "run along streamline $d s$ ", the ratio (opposite over hypotenuse) is exactly what fraction of gravity acts along the flow.

PICTURE.

Step 4 — Why a "steady" flow still accelerates

WHAT. In steady flow the speed $v$ at each fixed spot never changes with time. Yet the blob speeds up — because it travels to a new spot where the fluid is naturally faster (e.g. into a narrower pipe). Its acceleration is $a = v \frac{d v}{d s}$ .

WHY THIS TOOL — the chain rule. We want $a = d v / d t$ , but we don't know how $v$ depends on time directly; we know how $v$ depends on position $s$ (fast in the narrow part, slow in the wide part). The chain rule lets us swap:

a = \frac{d v}{d t} = how v changes with place \frac{d v}{d s} \cdot = v, the speed \frac{d s}{d t} = v \frac{d v}{d s} .

This is called convective acceleration — the blob "carries itself" into a faster zone.

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Step 5 — Assemble $F = ma$

WHAT. Add the two forces (Steps 2, 3) and set them equal to mass × acceleration (Steps 1, 4).

WHY. This is Newton's law for our blob. Everything above was just measuring the ingredients honestly.

PICTURE.

Step 6 — Add up the tiny changes (integrate)

WHAT. We walk from point 1 to point 2 along the streamline, adding up every little $d P$ , $d y$ , and $v d v$ . "Adding up infinitely many tiny pieces" is exactly what the integral sign $\int$ means, so we write $\int_{1}^{2}$ for each term.

WHY. One tiny slice tells us the local rule; adding all slices gives the rule connecting two distant points. Because $ρ$ is constant, it slides outside the integral.

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Step 7 — Edge cases: check every corner

WHAT. A derivation you trust must survive the extreme inputs. Let's run them.

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The one-picture summary

Recall Feynman retelling — the whole walkthrough in plain words

We cut a tiny cylinder of water out of a flowing stream (Step 1). Two things can push it along its path: the water behind pressing harder than the water ahead (Step 2), and gravity, but only the slice of gravity that points along the slope (Step 3). Even though the stream looks unchanging, our blob still speeds up — because it slides into a spot where water naturally moves faster, and the chain rule turns that into $v d v / d s$ (Step 4). We write "push equals mass times how-fast-it-speeds-up," the area and length cancel, and out pops one tidy line of tiny changes (Step 5). We add up those tiny changes from start to finish, and the sum of $v d v$ becomes $\frac{1}{2} v^{2}$ (Step 6). The result: pressure, plus speed-energy, plus height-energy always add to the same total along one streamline. Test the corners — flat pipe gives venturi, still water gives hydrostatics, an open hole gives $2 g h$ , a blocked tube gives the pitot reading — and every corner behaves (Step 7).

Recall Quick self-check

Why does $A$ vanish from the final equation? ::: It multiplies every force term equally (pressure and gravity both scale with the face area), so dividing through cancels it — Bernoulli is independent of blob size. What does the $\frac{1}{2}$ in $\frac{1}{2} ρ v^{2}$ come from? ::: From integrating $v d v$ , exactly like $\int x d x = \frac{1}{2} x^{2}$ . On a flat pipe, if speed doubles what happens to dynamic pressure? ::: It quadruples (it goes as $v^{2}$ ), so the static-pressure drop is four times larger.