Area-velocity relation — dA - A = (M² − 1)(dV - V) — derivation (explains de Laval nozzle): Visual walkthrough

Step 0 — The picture we are all talking about

WHAT. A pipe (a "duct") whose cross-section changes along its length. Gas flows left to right. At one slice the pipe has area $A$ , the gas moves at speed $V$ , has density $ρ$ (how tightly packed the gas is), and pressure $p$ (how hard the gas pushes outward).

WHY. Before any equation, we must agree on what the letters point to in the picture. Every symbol below is a label on this drawing.

PICTURE.

Figure — Area-velocity relation — dA - A = (M² − 1)(dV - V) — derivation (explains de Laval nozzle)

The little letter $d$ in front of a symbol — as in $d V$ — means "a tiny change in that thing as we step a hair's-width downstream." So $d V$ is a small speed increase, $d A$ a small area change. We chase fractional changes like $d V / V$ ("what fraction did $V$ grow?") because those are dimensionless and add up cleanly.

Step 1 — Mass is conserved: $ρ A V = const$

WHAT. Count the mass crossing each slice per second. That amount, $\overset{m}{˙} = ρ A V$ , is identical at every slice.

WHY. In steady flow, gas cannot pile up in the middle of the pipe or leak out of it. Whatever mass enters the left of a chunk per second must leave the right per second — otherwise mass would appear or vanish. This is the Continuity Equation (Compressible Flow).

PICTURE. Two slices, thin and fat. In the thin slice the dots are dense and fast; in the fat slice they spread and slow — but the count per second through each is equal.

Now the clever move. A product $=$ constant is awkward to differentiate. So take the natural log first — logs turn multiplication into addition:

$ln ρ + ln A + ln V = const$

WHY the log? Because $d ( ln X ) = \frac{d X}{X}$ — the derivative of a log is automatically the fractional change. And fractional change is the language of this whole topic. Differentiating (a tiny step downstream, so the constant's change is $0$ ):

$\boxed{\ \frac{d\rho}{\rho} + \frac{dA}{A} + \frac{dV}{V} = 0\ }\tag{1}$

Step 2 — Newton's law along the flow: Euler's equation

WHAT. Look at a tiny slab of gas. The pressure behind it pushes forward; the pressure ahead pushes back. If the pressure ahead is lower, there is a net forward push, and the slab accelerates.

WHY. This is just $F = ma$ applied to a fluid slab with no friction and no gravity — the Euler's Equation for Steady Flow. Pressure is the only force in a horizontal inviscid gas.

PICTURE. A slab with a big red push from behind, a smaller red push from the front, and a green acceleration arrow.

Balancing force and acceleration for the slab gives:

$\underbrace{dp}_{\text{pressure change}} + \underbrace{\rho\, V\, dV}_{\text{mass}\times\text{accel term}} = 0 \quad\Longrightarrow\quad \frac{dp}{\rho} = -V\,dV \tag{2}$

Step 3 — What density does when pressure changes: the speed of sound

WHAT. Squeeze a gas a little (raise $p$ ) and it gets a little denser (raise $ρ$ ). The stiffness of that response — how much pressure jumps per unit of density change — turns out to be $a^{2}$ , the square of the sound speed.

WHY. A sound wave is a tiny travelling pressure pulse. How fast it runs depends on how stiffly the gas pushes back when compressed. Stiffer gas ⇒ faster sound. Formally this is the isentropic ("no heat lost, reversible") derivative — see Isentropic Flow Relations.

PICTURE. A little compression pulse riding along the pipe; a spring cartoon shows "stiffness = $a^{2}$ ".

Step 4 — Fold pressure out: density in terms of $M$

WHAT. Equations (2) and (3) both mention $d p$ . Eliminate $d p$ to get density directly in terms of speed.

WHY. Our final goal only wants $A$ and $V$ . Pressure and density are middlemen; we chase them out one at a time. First $d p$ , here; then $d ρ$ , in Step 5.

PICTURE. A flow diagram: $(2)$ and $(3)$ merge, $d p$ cancels, out drops a relation with $M^{2}$ in it.

Put (2) into (3):

$d ρ = \frac{d p}{a ^{2}} = \frac{- ρ V d V}{a ^{2}}$

Divide both sides by $ρ$ to get a fractional change, and group the speeds:

$\frac{d ρ}{ρ} = - \frac{V ^{2}}{a ^{2}} \frac{d V}{V}$

Now the magic substitution. Since $M = V / a$ , we have $\frac{V ^{2}}{a ^{2}} = M^{2}$ :

$\boxed{\ \frac{d\rho}{\rho} = -M^2\,\frac{dV}{V}\ }\tag{4}$

Step 5 — Combine into the area–velocity relation

WHAT. Substitute the density result (4) into the mass equation (1). Density vanishes; only $A$ and $V$ remain.

WHY. (1) linked three fractional changes. We now know $d ρ / ρ$ from (4), so we can trade it away and be left with the one link we wanted.

PICTURE. The $d ρ / ρ$ box in equation (1) gets replaced by $- M^{2} d V / V$ , then the $d V / V$ terms merge.

Start from (1) and drop in (4) for the density term:

$this is d ρ / ρ - M^{2} \frac{d V}{V} + \frac{d A}{A} + \frac{d V}{V} = 0$

Move the two $d V / V$ terms to the right and collect:

$\frac{d A}{A} = M^{2} \frac{d V}{V} - \frac{d V}{V} = (M^{2} - 1) \frac{d V}{V}$

Step 6 — Reading the switch: all three regimes

WHAT. Fix the goal "accelerate the gas," i.e. $d V > 0$ . Then the sign of $(M^{2} - 1)$ forces the sign of $d A$ . Walk all three cases.

WHY. A relation is only understood when you have covered every case. There are exactly three signs the switch can take, and each demands a different pipe shape.

PICTURE. Three ducts side by side — converging, throat, diverging — each labelled with its regime and sign.

Regime	$M^{2} - 1$	For $d V > 0$ we need	Shape
Subsonic $M < 1$	negative	$d A < 0$	converging
Sonic $M = 1$	zero	$d A = 0$	throat (min area)
Supersonic $M > 1$	positive	$d A > 0$	diverging

Step 7 — The degenerate case $M = 1$ : why sonic flow lives only at a throat

WHAT. At $M = 1$ the switch $(M^{2} - 1)$ is exactly $0$ , so the equation reads $d A / A = 0$ . Area is momentarily stationary — a minimum.

WHY. This is the linchpin of the De Laval Nozzle and Choked Flow. It says the gas can only become sonic where the pipe stops narrowing and starts widening — the throat.

PICTURE. Zoom on the throat: area curve flat ( $d A = 0$ ) at the neck, with $M$ passing smoothly through $1$ .

$At M = 1 : \frac{d A}{A} = = 0 (1 - 1) \frac{d V}{V} = 0 ⟹ d A = 0$

Worked examples (each a mini-picture in your head)

The one-picture summary

Recall Feynman retelling of the whole walkthrough

Picture gas in a pipe. Rule one: no gas is created or lost, so crowding × width × speed stays fixed all along (Step 1). Rule two: gas only speeds up when the pressure ahead is lower — it falls toward the low-pressure side (Step 2). Rule three: squeeze a gas and it gets denser, and the stiffness of that push-back is the sound speed squared (Step 3). Chain these: from rules two and three, when gas speeds up its crowding drops, and it drops harder the faster the gas already is compared to sound — that's the $M^{2}$ (Step 4). Feed that into rule one and the pressure and density middlemen cancel, leaving one clean sentence: $d A / A = (M^{2} - 1) d V / V$ (Step 5). The bracket $(M^{2} - 1)$ is a switch. Slow gas: squeeze to speed up. Fast-as-sound gas: the switch is zero, so the pipe must be at its narrowest — the throat (Step 7). Faster than sound: the switch flips positive, so you must open the pipe up to keep speeding up (Step 6). Squeeze, pinch to a neck, then flare wide — that hourglass is a de Laval nozzle, and now you know exactly why.