This is the drill-yard for Kelvin's circulation theorem . The parent note told you what the theorem says and why it is true. Here we hit it from every angle: shrinking loops, stretching loops, zero-swirl loops, loops where the theorem fails , and the exam twists that trap people. Before every worked example you will be asked to forecast the answer — commit to a guess, then check yourself.
Before anything else we pin down every word and symbol the theorem uses, so you never have to hunt back to the parent.
Definition The words and symbols we build on
Material derivative D t D ::: the rate of change following a fluid particle as it moves , not a change measured at a fixed point in space. Written out, D t D ≡ ∂ t ∂ + u ⋅ ∇ : the first piece is how a quantity changes at a frozen location, the second is the extra change you feel because the particle is being carried through a varying field. When we say "D Γ/ D t " we mean "how the circulation of a loop changes while that loop drifts along with the fluid".
Line element d ℓ ::: a tiny straight arrow pointing along the loop C in the direction you walk around it; stitch all these little arrows head-to-tail and they trace the whole loop. Its length is the tiny distance you step, its direction is the local tangent.
Vorticity ω ::: the local "spin vector" of the fluid, ω = ∇ × u . Point your right thumb along ω and the fluid swirls the way your fingers curl; its length is twice the local angular speed. See Vorticity and the vorticity equation .
Scalar area A vs. vector element d A ::: scalar A is the total flat area of a small loop (used in the pocket formula Γ = ω A ). The bold d A in ∬ S ω ⋅ d A is a vector area element — an infinitesimal surface patch pointing along its normal n ^ . Add up all the d A patches over a small flat loop where ω is uniform and you recover scalar A . Don't mix the two.
Barotropic ::: density depends on pressure alone , ρ = ρ ( p ) — surfaces of constant pressure and constant density coincide. (Constant density is one special case.) When ρ also depends on temperature the flow is baroclinic and Kelvin can fail.
Conservative body force ::: a force expressible as f = − ∇Φ for some potential Φ (gravity is one). Such a force does zero net work around any closed loop, which is exactly why it drops out of Kelvin's proof.
Everything here rests on two facts from the parent:
Recall The two facts we reuse everywhere
Fact A — Circulation is conserved on a material loop. For an inviscid, barotropic flow with conservative body forces, using the material derivative D / D t ≡ ∂ / ∂ t + u ⋅ ∇ defined above,
D t D Γ = 0 , Γ = ∮ C u ⋅ d ℓ .
Here ∮ C u ⋅ d ℓ adds up, all the way around the loop, how much the velocity u points along each line element d ℓ .
Fact B — Circulation equals vorticity flux (Stokes' theorem ):
Γ = ∬ S ω ⋅ d A .
If the vorticity is roughly uniform over a small flat loop of scalar area A with perpendicular vorticity component ω , this collapses to the pocket formula Γ = ω A .
Every problem Kelvin's theorem can throw at you lands in one of these cells. The examples below are labelled by the cell they hit, and together they fill every cell.
#
Cell (case class)
What changes
Theorem's verdict
Example
C1
Area shrinks , positive ω
A ↓
ω ↑ (spin-up)
Ex 1
C2
Area grows
A ↑
ω ↓ (spin-down)
Ex 2
C3
Zero circulation start
Γ = 0 fixed
stays 0 forever
Ex 3
C4
Sign flip of enclosed swirl
loop turns inside-out
Γ keeps its signed value
Ex 4
C5
Degenerate loop (area → 0 )
A → 0
ω → ∞ , but is Γ finite?
Ex 5
C6
Real-world word problem (aerofoil)
starting vortex shed
Γ total = 0
Ex 6
C7
Theorem FAILS — baroclinic
∇ ρ × ∇ p = 0
D Γ/ D t = 0
Ex 7
C8
Theorem FAILS — viscous
friction diffuses ω
D Γ/ D t = 0
Ex 8
C9
Exam twist — fixed vs material loop
wrong loop chosen
trap!
Ex 9
We first define the sign convention we will use in every figure, because a wrong sign is the single most common mistake here.
Definition Sign convention (right-hand rule)
Choose a direction to walk around the loop C . Curl the fingers of your right hand that way; your thumb points along the positive normal n ^ . Then Γ > 0 means the fluid, on average, flows the same way you walk — a counter-clockwise swirl seen with n ^ pointing at you. Γ < 0 is clockwise. This is the signed convention: circulation carries a sign, not just a size.
Figure s01 draws this: the chalk loop with blue walk-arrows running counter-clockwise, the pale-yellow normal n ^ marked as coming out of the board at the centre, and the note that a clockwise flow flips the sign to Γ < 0 . Refer to the blue arrows whenever you need to fix which way is "positive".
Worked example Example 1 — Contracting vortex tube (cell C1)
A small material loop sits perpendicular to a vortex tube with area A 1 = 8 cm 2 and uniform perpendicular vorticity ω 1 = 5 s − 1 . The flow stretches the tube so the loop's area shrinks to A 2 = 2 cm 2 . Find the new vorticity ω 2 .
Forecast: area drops to a quarter — do you expect ω 2 to quarter, or quadruple?
Write the conserved quantity. Γ = ω A on this material loop.
Why this step? Fact A says Γ can't change; Fact B lets us write it as ω A for a small flat loop.
Set start equal to end. ω 1 A 1 = ω 2 A 2 .
Why this step? Same loop, same conserved Γ , two moments in time.
Solve. ω 2 = ω 1 A 2 A 1 = 5 ⋅ 2 8 = 20 s − 1 .
Why this step? Pure algebra — nothing physical left to decide.
Verify: Γ 1 = ω 1 A 1 = 5 ⋅ 8 × 1 0 − 4 = 4.0 × 1 0 − 3 m 2 / s , and Γ 2 = 20 ⋅ 2 × 1 0 − 4 = 4.0 × 1 0 − 3 m 2 / s . Equal ✓. Spin quadrupled because area quartered — the skater pulling in her arms.
Figure s02 shows the two states side by side: a big loop (light blue fill, few yellow spin-arrows, ω 1 = 5 ) shrinking to a small loop (pink fill, four times as many yellow spin-arrows, ω 2 = 20 ). The density of yellow arrows is the vorticity — denser arrows in the smaller loop is the spin-up made visible.
Worked example Example 2 — Loop dragged over a diverging region (cell C2)
Continue the very same material loop from Example 1. It has ended Example 1 at area 2 cm 2 carrying the conserved circulation Γ = ω 1 A 1 = 5 ⋅ 8 × 1 0 − 4 = 4.0 × 1 0 − 3 m 2 / s (that Γ value was computed in Ex 1's Verify line — nothing new is assumed). The tube now flows into a widening section and the loop expands to A 3 = 10 cm 2 . Find ω 3 .
Forecast: wider loop — faster or slower spin?
Reuse the conserved Γ . Γ = ω 3 A 3 , with Γ = 4.0 × 1 0 − 3 m 2 / s carried over unchanged from Ex 1.
Why this step? It is one continuous material loop; Kelvin locks its Γ for all time, so the value from Ex 1 is still valid here.
Solve for ω 3 . ω 3 = A 3 Γ = 10 × 1 0 − 4 4.0 × 1 0 − 3 = 4.0 s − 1 .
Why this step? ω ∝ 1/ A directly from Γ = ω A .
Verify: ω 3 A 3 = 4.0 ⋅ 10 × 1 0 − 4 = 4.0 × 1 0 − 3 ✓. Bigger loop ⇒ slower spin. Compare Ex 1 and Ex 2: same tube, same loop, ω went 5 → 20 → 4 , but Γ never moved.
Worked example Example 3 — Flow from rest (cell C3)
At t = 0 a fluid is completely at rest, u = 0 everywhere. It is then set into ideal (inviscid, barotropic) motion by conservative forces. Show that every material loop keeps Γ = 0 , and state what this implies about vorticity.
Forecast: can a smooth ideal flow ever "spontaneously" grow swirl from stillness?
Initial value. At rest, u = 0 ⇒ Γ ( 0 ) = ∮ 0 ⋅ d ℓ = 0 for every loop.
Why this step? No velocity means no line integral, on any loop whatsoever.
Apply Kelvin. D Γ/ D t = 0 , so Γ ( t ) = Γ ( 0 ) = 0 for all t .
Why this step? Fact A: the value is locked in for each material loop.
Convert to vorticity. If ∬ S ω ⋅ d A = 0 for every surface, then ω = 0 pointwise.
Why this step? A flux that vanishes through all surfaces forces the integrand to vanish everywhere.
Verify: The conclusion — an ideal flow started from rest stays irrotational forever — is a standard corollary of Kelvin's theorem (sometimes called the persistence of irrotationality). It justifies using a velocity potential. Sanity check against Ex 6: an aerofoil in air that was at rest must obey this — total Γ stays 0 , which is exactly what forces a starting vortex to appear.
Worked example Example 4 — Two counter-rotating vortices in one loop (cell C4)
A material loop encloses two vortex patches: patch + has Γ + = + 6 m 2 / s (counter-clockwise), patch − has Γ − = − 6 m 2 / s (clockwise). What is the loop's circulation, and what does Kelvin predict as they get stretched?
Forecast: two equal-and-opposite swirls inside one loop — total zero, or total twelve?
Add signed contributions. Γ = Γ + + Γ − = 6 + ( − 6 ) = 0 .
Why this step? Circulation is a signed flux (Fact B); opposite swirls cancel, they do not add in magnitude.
Apply Kelvin to the outer loop. Γ = 0 is conserved on that material loop.
Why this step? Fact A holds regardless of internal structure — only the total signed flux is locked.
What can still change? The two patches may stretch, migrate, and individually spin faster or slower — but their signed sum through the material loop stays 0 .
Why this step? Kelvin constrains the net , not the pieces; each sub-loop hugging one patch conserves its own Γ too.
Verify: Draw a tiny material loop around only patch + : it conserves + 6 . A tiny one around patch − : it conserves − 6 . Sum = 0 matches the big loop. Consistent ✓. This is why a vortex pair can travel far while the far-field loop reads zero circulation.
Figure s03 draws the big chalk loop with a blue counter-clockwise patch on the left (+ 6 ) and a pink clockwise patch on the right (− 6 ); the yellow banner across the top states the signed sum + 6 + ( − 6 ) = 0 . The arrow directions — not their count — carry the signs here.
Worked example Example 5 — Is a point vortex allowed? (cell C5)
Take the loop of Ex 1 and imagine squeezing its area toward zero, A → 0 , while keeping the same material fluid on it. What happens to ω ? Is anything conserved?
Forecast: as A → 0 , does Γ blow up, vanish, or stay put?
Γ is the invariant, not ω . Γ = ω A is fixed at 4.0 × 1 0 − 3 m 2 / s .
Why this step? Kelvin locks Γ ; ω and A are just its two factors.
Take the limit. As A → 0 with Γ fixed, ω = Γ/ A → ∞ .
Why this step? A finite numerator over a vanishing denominator diverges.
Interpret. The vorticity becomes a delta spike concentrated on a line — a point vortex (in 2D) — carrying finite circulation Γ but infinite local ω .
Why this step? Circulation is the physically finite, measurable, conserved quantity; local vorticity is idealized to infinity.
Verify: Evaluate ω at A = 2 × 1 0 − 4 , 2 × 1 0 − 6 , 2 × 1 0 − 8 m 2 : ω = 20 , 2000 , 2 × 1 0 5 s − 1 — growing without bound, while Γ = ω A = 4.0 × 1 0 − 3 every time ✓. Moral: Kelvin is a statement about Γ , which survives degeneration; ω need not.
Worked example Example 6 — Starting an aeroplane wing (cell C6)
A wing moving at U = 50 m/s through air of density ρ = 1.2 kg/m 3 must generate lift L = 1800 N per metre of span . Using Kutta–Joukowski L = ρ U Γ bound , find the bound circulation, then find the starting vortex circulation Kelvin demands.
Forecast: if the wing gains clockwise bound swirl, what must the wake do — nothing, or shed an opposite vortex?
Solve Kutta–Joukowski for bound circulation. Γ bound = ρ U L = 1.2 ⋅ 50 1800 = 30 m 2 / s .
Why this step? Lift is proportional to bound circulation; invert to get Γ bound .
Recall the pre-motion state. Before takeoff the air is at rest, so any material loop enclosing wing + wake has Γ = 0 (this is Ex 3's setup).
Why this step? Kelvin locks that 0 for all time on the material loop.
Balance the books. Γ bound + Γ starting = 0 ⇒ Γ starting = − 30 m 2 / s .
Why this step? The only way to keep the conserved total at 0 while the wing carries + 30 is to shed − 30 into the wake.
Verify: Total = 30 + ( − 30 ) = 0 ✓, matching the still-air initial condition. Units: [ Γ ] = m 2 / s , and ρ U Γ = 1.2 ⋅ 50 ⋅ 30 = 1800 kg m − 3 ⋅ m/s ⋅ m 2 / s = 1800 N/m ✓. Lift exists precisely because Kelvin allows a shed starting vortex.
Worked example Example 7 — A sea breeze spins up (cell C7)
Warm land air and cool sea air sit side by side. Pressure decreases with height, but density varies horizontally (warm=light on the left, cool=dense on the right). Explain, with the surviving term, why D Γ/ D t = 0 here.
Forecast: if ρ depends on temperature and not on p alone, does the pressure term still integrate to zero around a loop?
Recall the pressure term in the derivation. Kelvin needed ∮ ρ 1 ∇ p ⋅ d ℓ = 0 , which worked only because ρ 1 ∇ p = ∇ P was a pure gradient (barotropic).
Why this step? That gradient property was the whole reason the loop integral vanished.
Break barotropicity. When ρ = ρ ( p , T ) , ρ 1 ∇ p is not a gradient. Its curl is the baroclinic term:
D t D Γ = − ∮ C ρ ∇ p ⋅ d ℓ = − ∬ S ρ 2 ∇ ρ × ∇ p ⋅ d A = 0.
Why this step? Stokes' theorem turns the surviving loop integral into the flux of ∇ ρ × ∇ p — nonzero when density and pressure gradients are misaligned.
Read the geometry. A gradient points toward increasing values. Pressure is highest at the ground, so ∇ p points downward (toward higher pressure below). Density is higher toward the cool sea on the right, so ∇ ρ points to the right . Their cross product ∇ ρ × ∇ p (right × down) points out of the page ⇒ a counter-clockwise circulation is pumped in: air rises over warm land, sinks over cool sea — the sea breeze.
Why this step? The cross product's direction is the sign of the circulation being pumped in, and getting the gradient directions right is what fixes that sign.
Verify: If instead ∇ ρ ∥ ∇ p (barotropic), ∇ ρ × ∇ p = 0 and D Γ/ D t = 0 — Kelvin restored. So the failure is exactly measured by the misalignment angle between ∇ ρ and ∇ p ✓. This is the mechanism the parent note's third mistake warned about.
Figure s04 draws ∇ p as a blue arrow pointing down (toward higher pressure at the ground), ∇ ρ as a pink arrow pointing right (toward the dense sea air), and the resulting yellow counter-clockwise circulation loop; the caption states ∇ ρ × ∇ p points out of the page so Γ grows.
Worked example Example 8 — Vorticity leaking across a material loop (cell C8)
A real (viscous) fluid has kinematic viscosity ν . The vorticity equation gains a diffusion term ν ∇ 2 ω . Explain why a material loop's Γ now changes, and estimate over what time a boundary layer of thickness δ = 1 mm in water (ν = 1 0 − 6 m 2 / s ) lets vorticity diffuse across itself.
Forecast: friction — does it lock swirl in, or let it seep across your loop?
Add friction to Euler. The Navier–Stokes momentum balance replaces the clean Euler right-hand side − ρ 1 ∇ p − ∇Φ with an extra viscous force ν ∇ 2 u .
Why this step? Kelvin's Step 3 assumed D u / D t = − ρ 1 ∇ p − ∇Φ and nothing else; the new term breaks that assumption.
Show the extra term survives the loop integral. Feeding this into D Γ/ D t = ∮ C D t D u ⋅ d ℓ gives an extra piece ∮ C ν ∇ 2 u ⋅ d ℓ = ν ∮ C ∇ 2 u ⋅ d ℓ , which is not the loop integral of a pure gradient and generally does not vanish. Hence D Γ/ D t = 0 .
Why this step? The pressure and potential terms died in the proof only because each was a perfect differential; the viscous term is not, so it is the one term that can create or destroy circulation.
Estimate the diffusion time. Vorticity spreads a distance δ in time τ ∼ δ 2 / ν ; invert the diffusion length ν t ∼ δ to get τ ∼ δ 2 / ν = 1 0 − 6 ( 1 0 − 3 ) 2 = 1 s .
Why this step? This gives the concrete timescale over which viscosity ferries vorticity across a millimetre loop and visibly violates Kelvin.
Verify: Units: δ 2 / ν = m 2 / ( m 2 / s ) = s ✓. Numeric: τ = 1 s . So over ~1 second, viscosity carries vorticity through a 1 mm layer, so D Γ/ D t = 0 there. This is exactly the layer where lift-generating circulation is born, reconciling Ex 6 (needs a vortex shed) with Kelvin (needs viscosity to break it locally).
Worked example Example 9 — The loop that "seems" to violate Kelvin (cell C9)
A steady flow past a wing has, through a fixed rectangular loop in space (not moving with the fluid), a circulation that changes from 0 to 30 m 2 / s as the wing starts up. A student says "Kelvin is violated!" Where is the error?
Forecast: does Kelvin promise conservation on any loop, or only a special one?
Identify the loop type. The student used a loop fixed in space ; fluid particles pass through it.
Why this step? Kelvin's D / D t is a material derivative (defined at the top: ∂ / ∂ t + u ⋅ ∇ ) — it follows the fluid, not a stationary window in space.
State what Kelvin actually claims. Only for a material loop (always made of the same fluid particles) is D Γ/ D t = 0 . A fixed loop's circulation obeys a different equation and can freely change as vortices are advected across its boundary.
Why this step? Applying the theorem to the wrong kind of loop is the parent note's second listed mistake — the whole trap lives here.
Resolve the paradox. The starting vortex (Ex 6, circulation − 30 ) drifts downstream and eventually crosses out of the fixed loop; once it has left, the fixed loop reads + 30 (just the bound circulation). No contradiction — that fixed loop never obeyed Kelvin in the first place.
Why this step? Meanwhile the material loop that enclosed everything still reads 0 , so the physics is consistent; only the bookkeeping window was chosen wrongly.
Verify: Consistency chain: material loop total = Γ bound + Γ start = 30 + ( − 30 ) = 0 (Kelvin ✓), while the fixed loop reads + 30 after the − 30 vortex has drifted out (Kelvin not claimed for it ✓). Both statements are true at the same time — the "violation" was a category error.
Recall Which cell am I in? (decision check)
Loop shrinks, positive ω ::: C1 — spin-up, ω rises
Loop grows ::: C2 — spin-down, ω falls
Flow from rest, ideal ::: C3 — stays irrotational forever
Equal-and-opposite swirls in one loop ::: C4 — signed sum, total Γ = 0
Area → 0 ::: C5 — ω → ∞ but Γ finite (point vortex)
Wing generates lift ::: C6 — starting vortex − Γ bound shed
∇ ρ × ∇ p = 0 ::: C7 — baroclinic, Kelvin fails
Viscous boundary layer ::: C8 — diffusion, Kelvin fails
"Fixed loop violates Kelvin" ::: C9 — wrong loop; theorem needs a material loop
Mnemonic The one-line filter
"Same water, no stickiness, one-to-one density" — material loop, inviscid, barotropic. Miss any of the three and Γ can move (see Helmholtz vortex theorems and Bernoulli's principle for the neighbouring ideal-flow results that share these exact hypotheses).