Visual walkthrough — Optimum expansion — P_e = P_a for maximum thrust

This is the visual walkthrough of the parent result: Optimum expansion — $P_e = P_a$. We build the whole argument from a bare picture of gas leaving a hole, and we do not use a single symbol until we can point to it in a figure.

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Step 1 — What "thrust" even is: a picture of stuff being thrown backward

WHAT. A rocket does one thing: it throws mass backward, and the recoil pushes it forward. Nothing more mysterious than jumping off a skateboard.

WHY start here. Every symbol we will use ($\dot{m}$, $v_e$, $A_e$, $P_e$, $P_a$) lives on this one picture. If we don't anchor them now, they float.

PICTURE. Look at the figure. The nozzle is the funnel. The flat disk at the mouth of the funnel — the exit plane — is where we watch the gas leave.

Read the labels off the picture:

• $\dot{m}$ (say "m-dot") $=$ the mass of gas leaving every second, in kilograms per second. The dot means "per second". Picture a certain number of gas-kilograms crossing the disk each tick of a clock.
• $v_e$ $=$ the speed of that gas as it crosses the exit plane, in metres per second. The little $e$ means "at the exit".
• $A_e$ $=$ the area of the exit disk, in square metres — literally the size of the hole.
• $P_e$ $=$ the pressure of the gas right at that disk (how hard the gas is pressing sideways on everything), in pascals.
• $P_a$ $=$ the ambient pressure — how hard the outside air presses back on that same disk.

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Step 2 — Counting the momentum push

WHAT. We put a number on effect 1: throwing mass backward.

WHY. Newton says force $=$ rate of change of momentum. Momentum is mass $\times$ velocity. If we launch $\dot{m}$ kilograms each second, each carrying speed $v_e$, the momentum leaving per second is $\dot{m}{v}_e$. By Newton's third law the rocket feels an equal forward push.

PICTURE. Each little gas parcel is an arrow; every second a fresh batch crosses the disk carrying its speed with it.

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Step 3 — Counting the pressure push (and why it can go either way)

WHAT. Now effect 2. The gas at the exit disk presses outward with pressure $P_e$; the atmosphere presses inward with $P_a$. Pressure $\times$ area $=$ force.

WHY. Pressure only makes a net force where it is unbalanced. On the exit disk, inside-gas pushes out over area $A_e$ giving $P_e{A}_e$; outside-air pushes in over the same area giving $P_a{A}_e$. Subtract.

PICTURE. Two sets of little arrows on the same disk — coral pushing out, lavender pushing in. Whichever set is longer wins.

Three cases, all on that same disk:

• $P_e>P_a$: coral wins → positive extra push.
• $P_e=P_a$: arrows equal → zero net pressure push.
• $P_e<P_a$: lavender wins → negative push (the air shoves the rocket back).

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Step 4 — The trap: $P_e$, $v_e$, $A_e$ are NOT free knobs

WHAT. The lazy move is to say "make the pressure term huge, done." It doesn't work, and the picture shows why.

WHY. The three quantities are chained together by the nozzle's shape. You have exactly one real design knob: the expansion ratio $\epsilon =A_e/A^{\ast }$, where $A^{\ast }$ is the narrowest point (the throat). Widen the bell (raise $\epsilon$) and all three respond at once: the gas keeps accelerating so $v_e$ rises, it thins out and cools so $P_e$ falls, and the mouth $A_e$ grows. You cannot move one without the others. This locking is exactly the Isentropic flow relations.

PICTURE. Two nozzles side by side: a short one (high $P_e$, low $v_e$, small $A_e$) and a long one (low $P_e$, high $v_e$, big $A_e$). One dial — three linked readouts.

So the honest question is: as I slowly open the bell (let $P_e$ slide down its allowed track), does thrust rise or fall? We answer with one derivative — done carefully.

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Step 5 — Two exact facts that live on the flow

Before differentiating we need two facts that are always true along this expanding flow.

Fact A — mass is conserved. The same $\dot{m}$ crosses every slice: $\dot{m}={\rho }_e{v}_e{A}_e$ where ${\rho }_e$ is the gas density at the exit (kg per cubic metre). Picture: a thin river must run faster where it's narrow, slower where wide, but the same water passes any bridge each second.

Fact B — Euler's momentum relation. For a steady stream, to speed the gas up you must spend pressure. In symbols: ${\rho }_e{v}_{e}d{v}_e=-d{P}_e$

WHY a minus sign? A tiny drop in pressure ($d{P}_e<0$) buys a tiny gain in speed ($d{v}_e>0$). Pressure energy converts into motion; the ledger balances with opposite signs.

PICTURE. A parcel sliding "downhill" in pressure while its speed-arrow lengthens.

Multiply Fact B by $A_e$ and use Fact A (${\rho }_e{v}_e{A}_e=\dot{m}$): $\dot{m}d{v}_e=-A_{e}d{P}_e⟹\dot{m}\frac{d{v}_e}{d{P}_e}=-A_e$

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Step 6 — The cancellation: watch the momentum gain eat half the pressure change

WHAT. Differentiate the whole thrust $F=\dot{m}{v}_e+(P_e-P_a)A_e$ with respect to $P_e$, letting everything ride.

WHY. We're looking for the flat top of the thrust hill.

The pressure term needs the product rule (two things vary: the gap $P_e-P_a$ and the area $A_e$): $\frac{d}{d{P}_e}[(P_e-P_a)A_e]=\underset{\text{gap grows}}{\underbrace{A_e}}+\underset{\text{area changes}}{\underbrace{(P_e-P_a)\frac{d{A}_e}{d{P}_e}}}$

Add the momentum term from Step 5 ($-A_e$): $\frac{dF}{d{P}_e}=\underset{\text{momentum}}{\underbrace{-A_e}}+\underset{\text{pressure, part 1}}{\underbrace{A_e}}+(P_e-P_a)\frac{d{A}_e}{d{P}_e}$

The magic: the first two terms are equal and opposite — they cancel exactly. The extra speed you gain from dropping the pressure precisely erases the "gap" part of the pressure term. What survives: $\boxed{\frac{dF}{d{P}_e}=(P_e-P_a)\frac{d{A}_e}{d{P}_e}}$

PICTURE. Two arrows, $-A_e$ and $+A_e$, annihilating; a lone survivor $(P_e-P_a)d{A}_e/d{P}_e$ left standing.

Opening the bell always changes the mouth size, so $\frac{d{A}_e}{d{P}_e}\ne 0$. Therefore the only way the derivative can be zero — the only flat top — is: $P_e-P_a=0⟹\boxed{P_e=P_a}$

At that point the pressure thrust term $(P_e-P_a)A_e$ vanishes and every newton of thrust is pure momentum, $F=\dot{m}{v}_e$ — the cleanest possible conversion.

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Step 7 — The thrust hill: under-, over-, and perfectly expanded

WHAT. Turn the derivative into the full landscape, covering every case.

WHY. The sign of $\frac{dF}{d{P}_e}=(P_e-P_a)\frac{d{A}_e}{d{P}_e}$ tells us which side of the peak we sit on. Note that opening the bell lowers $P_e$, so $\frac{d{A}_e}{d{P}_e}<0$ always (bigger area ↔ smaller pressure).

PICTURE. A single hump: thrust versus expansion. Left slope, summit, right slope — each labelled with its regime and its exhaust plume shape.

Regime	Sign of $P_e-P_a$	$\frac{dF}{d{P}_e}=(P_e-P_a)\frac{d{A}_e}{d{P}_e}$	Meaning
Under-expanded	$P_e>P_a$	$(+)(-)<0$	Lower $P_e$ (open the bell more) still raises $F$ → below peak, keep expanding. Plume bulges outward.
Optimum	$P_e=P_a$	$0$	Summit. $F=\dot{m}{v}_e$. Clean straight plume.
Over-expanded	$P_e<P_a$	$(-)(-)>0$	You went too far; you'd need $P_e$ up (shorter bell) to recover. Air pinches the plume, risking Supersonic flow separation.

Because $P_a$ falls as a rocket climbs but the fixed bell can't grow, no single nozzle stays at the summit for the whole flight — the reason for Altitude compensation and Rocket staging.

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The one-picture summary

Everything at once: the two pushes, the locking chain, the cancellation, and the resulting thrust hill with its peak sitting exactly at $P_e=P_a$.

Recall Feynman retelling — say it to a friend with no math

A rocket pushes forward two ways: by throwing gas out fast, and by the gas pressing on the mouth harder than the air presses back. You have exactly one thing to design — how far the exhaust bell opens up. Widen it and the gas comes out faster (more throwing-push) but thinner and softer (less pressing-push). So there's a tug-of-war. When I do the careful bookkeeping, the extra throwing-push I gain by opening the bell one notch exactly cancels part of the pressing-push I lose — every time. The only leftover is a term that says: keep going as long as the inside pressure still beats the outside pressure. The instant they're equal, there's nothing left to gain — that's the top of the hill. Open further and now the outside air actually presses back harder than the gas, shoving me backward and even tearing the flow off the walls. So the sweet spot, the maximum push, is precisely when the gas leaves at the same pressure as the air around it: $P_e=P_a$.

Recall Quick checks

What are the two contributions to thrust? ::: Momentum thrust $\dot{m}{v}_e$ and pressure thrust $(P_e-P_a)A_e$. Why can't we just maximize the pressure term alone? ::: Because $P_e$, $v_e$, $A_e$ are locked by the nozzle geometry (the expansion ratio); changing one changes all three. Which two terms cancel in $dF/d{P}_e$? ::: The momentum $-A_e$ and the first pressure-rule piece $+A_e$. What survives, and when is it zero? ::: $(P_e-P_a)d{A}_e/d{P}_e$; zero only when $P_e=P_a$ since $d{A}_e/d{P}_e\ne 0$. Under-expanded means which sign? ::: $P_e>P_a$; you are below the peak and should expand more.

See also: Specific impulse · Nozzle expansion ratio · Thrust equation derivation