Question bank — Kelvin's circulation theorem

False. Inviscid is not enough — you also need barotropic ($\rho =\rho (p)$) so that $\frac{1}{\rho }\nabla p$ becomes a pure gradient; if $\rho$ also depends on temperature the pressure term survives and generates circulation. See Baroclinic vorticity generation.

False. Only around a material loop that convects with the flow. A loop fixed in space generally sees $\Gamma$ change because fluid with different swirl passes through it.

True. Every material loop has $\Gamma =0$ initially, Kelvin keeps each at $0$ forever, so $\nabla \times \mathbf{u}=0$ persists — the Lagrange/Cauchy corollary (see Vorticity and the vorticity equation).

False. It conserves the flux of vorticity through a material loop ($\Gamma ={\iint }_S\mathbf{\omega }\cdot d\mathbf{A}$ via Stokes' theorem), not the local $\mathbf{\omega }$. Vorticity can intensify by stretching while $\Gamma$ stays fixed.

True. If $\rho =\text{const}$ then trivially $\rho =\rho (p)$, so the pressure term reduces to $\oint \nabla P\cdot d\mathbf{\ell }=0$. But barotropic is the more general (weaker) condition.

False. Away from thin boundary layers, viscous effects are tiny and the flow behaves nearly ideal, so $\Gamma$ is very nearly conserved — this is why airfoil theory (Kutta–Joukowski lift theorem) works so well.

False. Gravity is conservative ($\mathbf{f}=-\nabla \Phi$), and $\oint \nabla \Phi \cdot d\mathbf{\ell }=0$ around any closed loop. It does zero net work per unit mass on a loop, so it cannot alter $\Gamma$.

Broadly true as an analogy. Conserved $\Gamma$ with $\Gamma =\omega A$ forces $\omega$ up when area shrinks — the skater-pulling-in-arms effect. It is the fluid statement of rotational inertia being preserved along material tubes.

Circulation stays constant; it is the vorticity that quadruples ($\omega \propto 1/A$ since $\Gamma =\omega A$ is fixed). The speaker confused the conserved quantity with the intensified one.

Nothing is violated — the total $\Gamma$ of the original material loop stays $0$; the shed vortex $-{\Gamma }_{\text{bound}}$ exactly cancels the bound circulation $+{\Gamma }_{\text{bound}}$.

Note the notation carefully: the circulation itself is $\oint \mathbf{u}\cdot d\mathbf{\ell }$ (velocity dotted with a length step $d\mathbf{\ell }$), whereas this term is $\oint \mathbf{u}\cdot d\mathbf{u}$ (velocity dotted with a velocity change $d\mathbf{u}$) — a different object born from the stretching of the loop. It equals $\oint d(\frac{1}{2}|\mathbf{u}{|}^2)$, a perfect differential around a closed loop, which is exactly $0$; so the stretching contributes nothing.

Not always — only when $\frac{1}{\rho }\nabla p=\nabla P$ is a pure gradient, which needs barotropicity. In a baroclinic fluid this term is genuinely nonzero and creates circulation.

The element does not move rigidly; neighbouring particles have slightly different velocities, so $D(d\mathbf{\ell })/Dt=d\mathbf{u}\ne 0$. It stretches and rotates — that is the whole point of vortex stretching.

They are distinct. Bernoulli's principle relates pressure and speed along a streamline (an energy statement); Kelvin conserves circulation on a material loop (a rotational/angular-momentum statement). Both need ideal-flow assumptions but say different things.

False consequence. Conserved circulation along a tube (with $\nabla \cdot \mathbf{\omega }=0$) forces vortex lines to close or reach a boundary — they cannot terminate mid-fluid. See Helmholtz vortex theorems.

Because $D/Dt$ (the change following the moving fluid) is what makes the two pieces of the derivation — the stretching term and the "Euler substitution" (swapping $D\mathbf{u}/Dt$ for $-\frac{1}{\rho }\nabla p-\nabla \Phi$) — combine into perfect-differential loop integrals that cancel. A fixed loop uses $\partial /\partial t$ instead, letting fluid of differing swirl cross it, so its $\Gamma$ can change freely. See the deforming-loop figure at the top.

Only then can $\frac{1}{\rho }\nabla p$ be written as the gradient of a single-valued pressure function $P(p)=\int \frac{dp}{\rho (p)}$ — an antiderivative built purely from $p$. Because $\nabla P=\frac{dP}{dp}\nabla p=\frac{1}{\rho }\nabla p$, its loop integral $\oint \nabla P\cdot d\mathbf{\ell }=\oint dP=0$ returns to its start value. If $\rho$ also depends on temperature, no such single-valued $P$ exists and the term survives (baroclinic torque, from $\nabla \rho \times \nabla p$).

Viscosity adds a $\nu {\nabla }^2\mathbf{u}$ term to the Euler equation (Euler equation for ideal fluids is what Kelvin uses); this is not a gradient, so its loop integral need not vanish, and it diffuses vorticity across material loops, giving $D\Gamma /Dt\ne 0$. Vorticity leaks in or out through friction.

It forces the total circulation of a loop enclosing the wing to stay $0$, so the bound circulation needed for lift is only allowed if an equal-and-opposite starting vortex is shed. That shedding makes lift ($L=\rho U{\Gamma }_{\text{bound}}$, Kutta–Joukowski lift theorem) physically possible.

Because $\Gamma =\iint \mathbf{\omega }\cdot d\mathbf{A}$; when the tube's cross-section is stretched thinner (area shrinks), $\mathbf{\omega }$ must rise to keep the flux fixed. Conservation of $\Gamma$ causes the spin-up rather than preventing it.

Because a conservative force is $-\nabla \Phi$, and $\oint \nabla \Phi \cdot d\mathbf{\ell }=\oint d\Phi =0$ around a closed loop — the potential returns to its start value, so net work per unit mass is zero.

When surfaces of constant density and constant pressure are tilted relative to each other, $\nabla \rho$ and $\nabla p$ point in different directions; their cross product $\nabla \rho \times \nabla p$ is nonzero and acts like a torque that spins fluid up (see the tilted-surfaces figure below). Barotropic flow keeps these surfaces parallel, so the torque vanishes.

Its circulation approaches $0$, consistent with $\Gamma =\iint \mathbf{\omega }\cdot d\mathbf{A}\to 0$ as area $\to 0$; the theorem still holds trivially since a degenerate loop was $0$ all along if it started that way.

Exactly $0$, since $\oint \mathbf{u}\cdot d\mathbf{\ell }=\mathbf{u}\cdot \oint d\mathbf{\ell }=\mathbf{u}\cdot 0=0$ (the loop's total displacement closes). Kelvin then keeps it $0$ — no swirl can appear.

Yes — the stretching term and the conservative-force term still vanish; only the pressure term $\oint \frac{1}{\rho }\nabla p\cdot d\mathbf{\ell }$ survives, giving $D\Gamma /Dt=\oint \frac{1}{\rho }\nabla p\cdot d\mathbf{\ell }\ne 0$. That surviving term is the baroclinic source.

Not cleanly — the theorem assumes smooth, differentiable fields; across a sheet the velocity is discontinuous and $\mathbf{\omega }$ is singular, so the material-derivative manipulation is invalid there.

Nonzero. A paddle applies a non-conservative, non-pressure force that does net work around the loop, so it acts as an external circulation source — exactly the ingredient Kelvin's idealisations exclude.

Yes. At each instant $\oint \nabla \Phi \cdot d\mathbf{\ell }=\oint d\Phi =0$ regardless of explicit time dependence, because the spatial gradient still integrates to zero around a closed loop.

Yes. The domain is multiply-connected (the cylinder is a hole the loop cannot shrink through), so a loop wrapping it can hold a fixed nonzero $\Gamma$ that Kelvin conserves — this is exactly the bound circulation behind lift. See the multiply-connected figure below.

No. A loop not enclosing the obstacle can be shrunk to a point (giving $\Gamma =0$ in irrotational flow), but a loop that wraps the obstacle cannot — it may carry the trapped bound circulation. The two classes of loop are topologically different, and Kelvin conserves each class's value separately.