Exercises — Kelvin's circulation theorem

The three assumptions are: (1) inviscid (no friction), (2) conservative body force ($\mathbf{f}=-\nabla \Phi$), (3) barotropic ($\rho =\rho (p)$). (a) Satisfied. Incompressible $\Rightarrow \rho =$ const $\Rightarrow$ trivially barotropic; gravity is conservative ($\Phi =gz$); inviscid by assumption. ✓ (b) Fails (1) inviscid. Honey is highly viscous — friction diffuses vorticity, so $D\Gamma /Dt\ne 0$. (c) Fails (3) barotropic. When $\rho =\rho (p,T)$, the term $\oint \frac{1}{\rho }\nabla p\cdot d\mathbf{\ell }$ no longer vanishes — this is baroclinic generation.

Uniform $\omega$ over area $A$ gives $\Gamma =\omega A$ (from $\Gamma ={\iint }_S\mathbf{\omega }\cdot d\mathbf{A}$). So $\omega =\frac{\Gamma }{A}=\frac{6}{3}=2{\text{s}}^{-1}.$

Why Kelvin? The loop is material, so its circulation $\Gamma =\omega A$ is conserved. ${\omega }_1{A}_1={\omega }_2{A}_2\Rightarrow {\omega }_2={\omega }_1\frac{A_1}{A_2}=5\cdot \frac{8}{2}=20{\text{s}}^{-1}.$ Area dropped $4\times$, so spin rose $4\times$ — the skater-pulls-arms-in effect. See Vorticity and the vorticity equation for the stretching term $(\mathbf{\omega }\cdot \nabla )\mathbf{u}$ that does this.

(a) Before motion the air was still: the big material loop enclosing wing and wake had $\Gamma =0$. Kelvin keeps that total at $0$: ${\Gamma }_{\text{bound}}+{\Gamma }_{\text{starting}}=0\Rightarrow {\Gamma }_{\text{starting}}=-8{\text{m}}^2/\text{s}.$ (b) $L=\rho U{\Gamma }_{\text{bound}}=1.2\times 50\times 8=480\text{N/m}.$

Circulation adds algebraically (flux is additive): $\Gamma ={\Gamma }_a+{\Gamma }_b=3+(-1)=2{\text{m}}^2/\text{s}.$ By Kelvin's theorem, as long as both tubes stay threaded through the material loop, $\Gamma$ stays $2{\text{m}}^2/\text{s}$ — the tubes may move and deform but cannot cross the material loop (that's a Helmholtz consequence). $\Gamma$ does not change.

Notice $\mathbf{u}\cdot d\mathbf{u}=d\left(\frac{1}{2}\mathbf{u}\cdot \mathbf{u}\right)=d\left(\frac{1}{2}|\mathbf{u}{|}^2\right)$, because differentiating $\frac{1}{2}|\mathbf{u}{|}^2$ gives $\mathbf{u}\cdot d\mathbf{u}$ (product rule on a dot product). So ${\oint }_C\mathbf{u}\cdot d\mathbf{u}={\oint }_{C}d\left(\frac{1}{2}|\mathbf{u}{|}^2\right)={\left[\frac{1}{2}|\mathbf{u}{|}^2\right]}_{\text{start}}^{\text{end}}=0.$ Why zero: it is the integral of a perfect differential (a pure $d(\text{something})$). Around a closed loop you return to the start, so the value at the end equals the value at the start — their difference is $0$. The closedness of the loop is what does it.

With the kinetic and potential terms gone but the pressure term surviving, $\frac{D\Gamma }{Dt}=-{\oint }_C\frac{1}{\rho }\nabla p\cdot d\mathbf{\ell }.$ Converting to a surface integral with Stokes' theorem gives the baroclinic torque $\frac{D\Gamma }{Dt}={\iint }_S\frac{\nabla \rho \times \nabla p}{{\rho }^2}\cdot d\mathbf{A}.$ The mechanism is baroclinic vorticity generation: when surfaces of constant density and constant pressure are misaligned ($\nabla \rho \times \nabla p\ne 0$), circulation is created. This is how sea breezes and weather fronts spin up.

At $t=0$, every material loop $C$ has $\Gamma ={\iint }_S\mathbf{\omega }\cdot d\mathbf{A}=0$. By Kelvin, $\frac{D\Gamma }{Dt}=0$ on every material loop, so $\Gamma (t)=0$ for all $t$, for all loops. If ${\iint }_S\mathbf{\omega }\cdot d\mathbf{A}=0$ for every surface (shrink the loop to a point in any orientation), then $\mathbf{\omega }=0$ everywhere for all time. This is the Lagrange–Cauchy theorem. Shortcut justified: an irrotational velocity field can be written $\mathbf{u}=\nabla \varphi$ for a scalar velocity potential $\varphi$, collapsing the vector problem to one scalar equation (${\nabla }^2\varphi =0$ if incompressible).

(a) ${\omega }_1={\Gamma }_0/A_1=4/0.5=8{\text{s}}^{-1}.$ (b) Volume $V=A\ell$ fixed and $\ell \to 2\ell$, so $A_2=A_1/2=0.25{\text{m}}^2$. Kelvin keeps ${\Gamma }_0$ fixed on the material loop, so ${\omega }_2=\frac{{\Gamma }_0}{A_2}=\frac{4}{0.25}=16{\text{s}}^{-1}.$ Stretching to double length doubled the spin. (c) A tube of fluid narrowing while conserving $\Gamma$ is like a spinning mass pulling inward: less moment of inertia, faster spin, same angular momentum. Kelvin ⟷ conservation of angular momentum; vortex stretching ⟷ skater tucking arms.

Conserve circulation: $\Gamma =\omega A=10\times 0.2=2{\text{m}}^2/\text{s}$. ${\omega }^{\prime }=\frac{\Gamma }{A^{\prime }}=\frac{2}{0.05}=40{\text{s}}^{-1}.$ Ratio $\frac{{\omega }^{\prime 2}}{{\omega }^2}={\left(\frac{40}{10}\right)}^2=16.$ Because $\omega \propto 1/A$, squeezing area by $4\times$ raises $\omega$ by $4\times$ and the spin-energy proxy by $4^2=16\times$ — the concentrated swirl is far more energetic locally, consistent with Bernoulli's principle pressure drops in tight vortex cores.

With $\nabla p=(\partial p/\partial x)\hat{\mathbf{x}}=(-2)\hat{\mathbf{x}}$ and $\nabla \rho =(\partial \rho /\partial y)\hat{\mathbf{y}}=(3\times 10^{-4})\hat{\mathbf{y}}$, $\nabla \rho \times \nabla p=(3\times 10^{-4}\hat{\mathbf{y}})\times (-2\hat{\mathbf{x}})=(3\times 10^{-4})(-2)(\hat{\mathbf{y}}\times \hat{\mathbf{x}})=-6\times 10^{-4}(-\hat{\mathbf{z}})=6\times 10^{-4}\hat{\mathbf{z}}.$ Magnitude of integrand: $\left|\frac{\nabla \rho \times \nabla p}{{\rho }^2}\right|=\frac{6\times 10^{-4}}{(1.2{)}^2}=\frac{6\times 10^{-4}}{1.44}\approx 4.17\times 10^{-4}{\text{s}}^{-2}.$ Non-zero because the gradients are misaligned — this is a purely baroclinic spin-up, invisible to Kelvin's theorem in its barotropic form.

1. Before motion: still air, so any large material loop enclosing wing + wake has $\Gamma =0$.
2. On impulsive start: the potential (ideal) outer flow would slip around the sharp trailing edge with infinite velocity — physically impossible.
3. Viscosity acts in the thin boundary layer (Kelvin's inviscid assumption fails locally), generating vorticity that rolls up and is shed as a starting vortex of circulation $-{\Gamma }_{\text{bound}}$.
4. Kelvin still holds globally on the big loop: ${\Gamma }_{\text{bound}}+{\Gamma }_{\text{starting}}=0$, so the total stays $0$ while the wing keeps bound circulation ${\Gamma }_{\text{bound}}$.
5. Kutta–Joukowski then gives steady lift $L=\rho U{\Gamma }_{\text{bound}}$. Indispensable failure: the inviscid assumption's local breakdown in the boundary layer. Without viscosity to shed the starting vortex, the Kutta condition could never be met and no net bound circulation (hence no lift) could arise. Kelvin's theorem is what guarantees the shed vortex must be exactly $-{\Gamma }_{\text{bound}}$.

Recognise the three assumptions ⟶ apply $\Gamma =\omega A$ conservation ⟶ analyse the terms that vanish (and the one that need not) ⟶ synthesise stretching + energy ⟶ master real baroclinic and viscous exceptions.