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Use circulation to summarize how much a velocity field runs along an oriented closed curve $C$, such as a circle around a fluid vortex. Its definition is the line integralwhere $d\mathbf{l}$ is tangent to $C$ in the chosen direction. Reversing that direction reverses the sign of circulation. Its units are ${\mathrm{m}}^2{\mathrm{s}}^{-1}$, because velocity is multiplied by a length. Counterclockwise circulation in the horizontal plane is positive when the chosen unit normal vector is $+{\mathbf{e}}_z$, according to the right-hand rule.

For solid-body rotation $\mathbf{u}=(-\Omega y,\Omega x,0)$, a counterclockwise circle of radius $R$ has constant tangential velocity $\Omega R$. Multiplying by its circumference givesFor the potential line vortex $\mathbf{u}=K{\mathbf{e}}_{\theta }/r$, the same circle instead gives ${\Gamma }_C=(K/R)(2\pi R)=2\pi K$, independent of $R$. Stipulate $\Omega =1{\mathrm{s}}^{-1}$, $R=1\mathrm{m}$ and $K=1{\mathrm{m}}^2{\mathrm{s}}^{-1}$: both velocity fields have ${\Gamma }_C=2\pi {\mathrm{m}}^2{\mathrm{s}}^{-1}\approx 6.283185{\mathrm{m}}^2{\mathrm{s}}^{-1}$, although their vorticities in the fluid are $2{\mathbf{e}}_z{\mathrm{s}}^{-1}$ and zero respectively. On a circle of radius $2\mathrm{m}$, their circulations are $8\pi {\mathrm{m}}^2{\mathrm{s}}^{-1}$ and $2\pi {\mathrm{m}}^2{\mathrm{s}}^{-1}$. The comparison plot in vorticity shows this difference in radius dependence.

The bridge between circulation and vorticity is Stokes theorem:Here the velocity field must be continuously differentiable on a neighborhood of the entire spanning surface $S$ inside the fluid, and its boundary orientation must match the unit normal vector $\mathbf{n}$. For solid-body rotation, the disk is admissible and its surface integral is $2\Omega \pi R^2$. For the potential line vortex, the origin is singular and excluded from the fluid: a disk crossing it is not admissible. On an annulus, Stokes theorem instead gives ${\Gamma }_{R_{\mathrm{out}}}-{\Gamma }_{R_{\mathrm{in}}}=0$, with both circulations here measured counterclockwise; the induced inner boundary direction is clockwise. Nonzero circulation therefore does not imply nonzero local curl in the available fluid. ^[ref]

This is useful because Stokes theorem lets a local vorticity measurement determine a loop's circulation when a suitable surface exists, while Kelvin's circulation theorem tracks circulation along a moving material curve. The beauty is the precise link between a local derivative and a global line integral, with the domain deciding which conclusions are valid.

For smooth inviscid barotropic fluid with potential body forces, barotropic vorticity transport and $\nabla \cdot \mathbf{\omega }=0$ give $D\mathbf{\omega }/Dt=(\mathbf{\omega }\cdot \nabla )\mathbf{u}-\mathbf{\omega }\nabla \cdot \mathbf{u}$. Combine this with material surface element evolution to obtain $D(\mathbf{\omega }\cdot d\mathbf{S})/Dt=0$. Integrating over a material surface proves the flux law. Its boundary must move with the fluid; flux through a fixed open surface is generally not conserved. By Stokes theorem the same statement is Kelvin's circulation theorem around the material boundary.

In smooth barotropic inviscid flow with conservative body force, Kelvin's circulation theorem preserves circulation of material curves. Initially zero vorticity therefore gives zero circulation on contractible loops for all times. A loop winding around an excluded obstacle can have a nonzero constant circulation: the planar field $\mathbf{u}=\Gamma {\mathbf{e}}_{\theta }/(2\pi r)$ is irrotational outside the origin but has circulation $\Gamma$ on a positively oriented enclosing loop.

The surface must be a material surface $S(t)$, moving with the fluid; its boundary moves with it as well. The stated barotropic vorticity transport law, with $\mathbf{\omega }=\nabla \times \mathbf{v}$, gives $\nabla \cdot \mathbf{\omega }=0$. Expanding its curl therefore yieldsA material tangent vector $\mathbf{t}$ obeys $D\mathbf{t}/Dt=(\nabla \mathbf{v})\mathbf{t}$. Apply the product rule to the cross product of two such tangents to obtain material surface element evolution:Here $(\nabla \mathbf{v}{)}_{ij}={\partial }_j{v}_i$. The stretching and area-change terms cancel exactly in the transported flux element:Integrate over the fixed particle labels parameterizing the material surface. This proves the conservation of vorticity flux through a material surface:The conserved flux is also the circulation ${\oint }_{\partial S(t)}\mathbf{v}\cdot d\mathbf{\ell }$ by the Stokes theorem, so this agrees with Kelvin's circulation theorem. The assumptions are smooth inviscid barotropic flow and conservative body forces. A fixed open surface generally exchanges vorticity flux with its surroundings and does not satisfy this conservation statement; compressibility by itself does not invalidate the material-surface result.

Take positive circulation to mean counterclockwise in the $(x,z)$ diagram, with $x$ right and $z$ up. The upper surface is then traversed right to left. To first order in $a/b$,Thus the bound circulation is clockwise. Any exterior contour enclosing the hydrofoil has the same circulation in an irrotational flow; a globally completed two-dimensional flow must contain a circulation component decaying as $|\Gamma |/(2\pi r)$ at large distance. The simple local upper/lower velocity prescription cannot describe the entire exterior. For a start from rest, Kelvin's circulation theorem requires compensating shed vorticity, represented by a starting vortex in a physical wake.

Numerically,If the question's fluid force includes buoyancy, $B^{\prime }\simeq 1.25\times 10^3\mathrm{N}{\mathrm{m}}^{-1}$ and the total upward fluid force is about $2.05\times 10^3\mathrm{N}{\mathrm{m}}^{-1}$, using $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. Weight of the hydrofoil is not specified and is not part of this fluid force. The motion-induced lift also agrees with the Kutta–Joukowski theorem, $L^{\prime }=-\rho U_0\Gamma$ in this orientation.

The chord-based Reynolds number isAt this large Reynolds number, smooth laminar flow over the whole body is unlikely in ordinary conditions; wall layers may undergo transition and separate. A thin-layer inviscid flow approximation outside them can remain useful, but does not justify globally laminar flow.

For smooth inviscid flow of constant mass density under a conservative body force, Kelvin's circulation theorem states that the circulation around any closed material curve is conserved:The constant-mass density assumption is the usual homogeneous incompressible flow setting here. Merely requiring $\nabla \cdot \mathbf{u}=0$ in a fluid of varying mass density would not by itself make the pressure term below a gradient.

Parameterize the material curve by a periodic particle label $s\in [0,1]$, writing ${\mathbf{r}}_t(s,t)=\mathbf{u}(\mathbf{r}(s,t),t)$. Differentiate its line integral, keeping $s$ fixed:The second term vanishes by periodicity. If the conservative body force per unit mass is $-\nabla \Omega$, the Euler momentum equation givesIts line integral around a closed curve is zero, proving the claim. The velocity field must remain smooth on the moving curve, which cannot pass through a line vortex singularity during this argument. No assumption of irrotational flow was needed.

Each layer has constant mass density, initially negligible velocity and hence zero vorticity. Its Euler equations for an inviscid fluid have a conservative gravitational body force, so Kelvin's circulation theorem preserves zero bulk vorticity. In a simply connected layer, the irrotational flow therefore has ${\mathbf{v}}_i=\nabla {\varphi }_i$. The incompressible flow condition givesThis applies while the ideal layerwise solution is smooth. It allows an interfacial vortex sheet; it does not assert that the whole two-fluid flow has zero vorticity or that a smooth interface must exist for arbitrarily long times.

The rigid wall no-penetration boundary conditions arewith boundedness and ${\partial }_r{\varphi }_i=0$ at $r=0$ for regular axisymmetric fields. For a small interface displacement and small slope, linearization of the kinematic boundary condition about $z=0$ givesThe small-slope condition is an additional requirement for high-wavenumber modes: $|\alpha \eta |\ll 1$, not just $|\eta |\ll h,a$.

Choose the unperturbed hydrostatic pressures as ${\bar{p}}_i(z)=p_{\ast }-{\rho }_{i}gz$. The linearized Bernoulli equation gives the pressure perturbation $p_i^{\prime }=-{\rho }_i{\partial }_t{\varphi }_i$, after absorbing spatially constant gauges into ${\varphi }_i$. Since surface tension and viscous normal stress vanish, pressure continuity on $z=\eta$ givesorThe initial velocity is zero, so ${\eta }_t(r,0)=0$. Separate conservation of mass in each closed layer also requires ${\int }_0^a\eta (r,t)rdr=0$ relative to the initially horizontal interface.

An enclosing loop has no spanning disk wholly inside the fluid, so initial zero vorticity does not force its circulation to vanish. Kelvin's circulation theorem says only that the initial circulation is retained:For example $\mathbf{u}={\Gamma }_0{\mathbf{e}}_{\theta }/(2\pi r)$ has zero vorticity outside the cylinder but circulation ${\Gamma }_0$ around a once-enclosing positively oriented loop. Any two such loops have equal circulation, since their difference bounds an annulus in the fluid and Stokes theorem gives zero over it. More generally a loop with winding number $m$ has circulation $m{\Gamma }_0$. Thus irrotational flow need not have a single-valued velocity potential on the whole exterior domain.

A simple closed loop that does not surround the cylinder bounds a surface wholly in the fluid. At time zero, Stokes theorem gives zero circulation because its vorticity is zero. The smooth impermeable exterior flow transports this loop without changing its winding about the obstacle. Pulling a loop at time $t$ back to time zero and using Kelvin's circulation theorem therefore givesThis is the contractible-loop case of irrotational circulation in an exterior domain. Equivalently, the initially vanishing vorticity remains zero under the inviscid barotropic vorticity equation.

Kelvin's circulation theorem states that a smooth inviscid barotropic fluid subject to a conservative body force preserves circulation around every closed material curve. Write the pressure term as $\nabla w$ with $dw=dp/\rho$, and the body force per mass as $-\nabla \Phi$. Then the Euler equations give $D\mathbf{u}/Dt=-\nabla (w+\Phi )$. For a moving parametrization $\mathbf{x}(s,t)$ with ${\partial }_t\mathbf{x}=\mathbf{u}$,The last integral is the increment of a single-valued scalar around a closed curve. ConsequentlyConstant density is a special barotropic case with $w=p/\rho$. Viscosity, nonconservative forcing or baroclinic pressure gradients can invalidate this conservation law; the exterior-cylinder conclusions use the smooth inviscid hypotheses.

For constant density, divide the incompressible Euler equations by $\rho$. The vector identity $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times \mathbf{\omega }$ converts them into ${\partial }_t\mathbf{u}=\mathbf{u}\times \mathbf{\omega }-\nabla (p/\rho +|\mathbf{u}{|}^2/2)$. Taking the curl eliminates the gradient. Expand the curl of the cross product, use $\nabla \cdot \mathbf{u}=0$ and $\nabla \cdot \mathbf{\omega }=0$, and obtainThis is the vorticity equation, including vortex stretching. Without constant density there would in general be an additional baroclinic term; the displayed source equations use the constant-density interpretation.

For the specified velocity, direct differentiation gives $\mathbf{\omega }=2\Omega (t)\mathbf{k}$. It is spatially uniform, while $(\mathbf{\omega }\cdot \nabla )\mathbf{u}=2\beta \mathbf{\omega }$. Thus ${\Omega }^{\prime }=2\beta \Omega$ andIf the positive $z$ direction agrees with the initial vorticity, $2\Omega (0)={\omega }_0$ as written in the question. For opposite rotation the initial component is signed and the vector has the corresponding minus sign.

For a material particle, $d(x^2+y^2)/dt=2x(-\beta x-\Omega y)+2y(-\beta y+\Omega x)=-2\beta (x^2+y^2)$, while $dz/dt=2\beta z$. The initial material circle therefore evolves intoThe azimuthal rotation changes the labels of points on the circle but not the circle as a set. Its counterclockwise circulation, viewed from positive $z$, isfor the positive-orientation convention. The strain contributes no tangential velocity. Thus circulation stays constant although vorticity grows and the enclosed material area shrinks, illustrating Kelvin's circulation theorem for a material loop in constant-density inviscid flow with no nonconservative body force.

The vorticity of a fluid velocity field is $\mathbf{\omega }=\nabla \times \mathbf{v}$. A vortex line is an integral curve tangent to that vector, not generally a particle path or streamline. Since $\nabla \cdot \mathbf{\omega }=0$, vorticity flux through successive cross-sections of a vortex tube is equal; a regular line cannot simply end inside the fluid. By Stokes's theorem, the tube strength is the circulationFor an inviscid barotropic fluid with conservative body force, Kelvin's circulation theorem conserves circulation around a material circuit. Under those assumptions vortex tubes move with the fluid. In three-dimensional incompressible ideal flow,stretching a material tube raises its vorticity as its cross-section shrinks. In two-dimensional incompressible flow there is no such stretching, so scalar vorticity is materially conserved. Constant viscosity adds a diffusion term $\nu {\nabla }^2\mathbf{\omega }$; baroclinicity and boundaries can also generate vorticity.

An ideal line vortex has $v_{\varphi }=\Gamma /(2\pi r)$ and zero ordinary vorticity for $r>0$, with its circulation concentrated in the singular core. Circular trajectories therefore do not imply nonzero local vorticity everywhere. A finite physical core removes the velocity singularity.

Flow visualization and the idealization. A streamline is everywhere tangent to the instantaneous velocity field at one fixed time. A streakline is the locus, at one observation time, of particles that previously passed through a specified injection point. A marked particle's own trajectory is a pathline. These coincide geometrically in steady flow, but need not do so in an unsteady draining flow.

The intended ideal liquid is approximately incompressible and inviscid: viscosity and associated shear stresses are negligible on the scales and times being considered, the Mach number is small, and the density is effectively constant. This does not require zero vorticity; the initially rotating state is already a counterexample. It also does not justify ignoring viscosity in an arbitrarily thin boundary layer or at a real drain.

Spiralling flow and angular momentum. In an axisymmetric inviscid flow, pressure and gravity exert no torque about the vertical axis. A parcel therefore conserves axial specific angular momentum,The late, shallow remnant in the intended draining picture is supplied by parcels initially near the outer rim, so $r_0\simeq R$. The slow-drain angular-momentum amplification then givesThe assumption about the parcels' origin is necessary: torque conservation by itself does not assign $r_0=R$ to every parcel at every time.

A continuous dye feed near the outer free surface produces inward spirals that wind increasingly rapidly near the drain. The figure uses a late quasisteady model, $u_r=-D/r$ and $u_{\varphi }=C/r$ with positive constants $D,C$, giving $\varphi ={\varphi }_0-(C/D)\ln (r/r_0)$. Its streaklines and streamlines approximately coincide over an observation interval short compared with the overall draining time. Strictly, individually marked particles at the initial surface trace pathlines, not a continuously injected streakline.

The free-surface depression. Let $z=H(r)$ be the surface. Slow radial draining makes radial inertial corrections small compared with the azimuthal centripetal acceleration. The radial Euler equation and vertical hydrostatic equilibrium giveThus $g{H}^{\prime }(r)=u_{\varphi }^2/r=C^2/r^3$, with $C={\omega }_0{R}^2$. Integrating from $r$ to the rim givesFor $r\ll R$ this becomesThe surface rises outward because the inward pressure gradient force must sustain the circular motion. The dimensions are metres, and the $1/r^2$ profile is not valid inside the finite drain or vortex core.

The terrestrial estimate. Use the vertical component of Earth's rotation as the ideal initial angular speed, ${\omega }_0\simeq {\Omega }_E\sin \lambda$. At Cambridge, taking $\lambda \simeq 52.2^{\circ }$ and ${\Omega }_E=7.292\times 10^{-5}{\mathrm{s}}^{-1}$, this gives ${\omega }_0\simeq 5.76\times 10^{-5}{\mathrm{s}}^{-1}$. Hence, for $R=1\mathrm{m}$, $r=0.01\mathrm{m}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$,Taking the whole terrestrial angular speed rather than its local vertical projection would give about $2.7\mu \mathrm{m}$. The Cambridge estimate uses the projection appropriate to the vertical vortex. This very small head assumes terrestrial rotation supplies the only initial swirl; imposed residual rotation would change it quadratically.

Curl of the azimuthal velocity. For axisymmetric $u_{\varphi }=r\omega (r)$ with no vertical dependence, the rotational part of the vorticity is vertical:The ideal quasisteady radial sink has no curl outside its core. The source's symbol $\Omega$ denotes this vertical vorticity component, not the fluid's angular velocity. In general its magnitude is the absolute value of the bracket.

The two limiting states. For initial uniform rotation, $\omega ={\omega }_0$, whereas the late approximation has $\omega =C/r^2$. ThereforeThe late flow can have large angular velocity and nonzero circulation while being locally irrotational. The drain core is excluded from the domain of the $1/r$ velocity field.

What is actually conserved. Inviscid does not mean that each parcel's vorticity magnitude is fixed in three dimensions. For an incompressible barotropic liquid with conservative body forces, the vorticity equation isVertical compression of a material column changes its vorticity through vortex stretching, even while its volume is conserved by horizontal expansion. In the columnar shallow-layer limit, vorticity dilution in a shallow draining layer givesThus a remnant much shallower than its original column can have much smaller vertical vorticity. At finite depth there is generally a residual value; the exactly irrotational late profile is a leading limiting approximation, not an exact parcelwise conversion of nonzero vorticity to zero in finite time.

Kelvin's circulation theorem conserves circulation around a material loop, not around a fixed spatial circle. A loop whose original radius is near $R$ has circulation about $2\pi {\omega }_0{R}^2$, consistent with the late $u_{\varphi }=C/r$ field. Fluid also leaves through the drain, so the remaining tank contents are not a fixed material system. If the calculation is expressed explicitly in the Earth-fixed rotating frame, the corresponding shallow-water potential vorticity is $(\zeta +f)/H$, with $f=2{\Omega }_E\sin \lambda$; the conserved quantity involves absolute vorticity.

For inviscid nonrotating shallow water equations, columnar horizontal flow has material vorticity equation $D\zeta /Dt=-\zeta {\nabla }_h\cdot {\mathbf{u}}_h$ and depth equation $Dh/Dt=-h{\nabla }_h\cdot {\mathbf{u}}_h$. Thus $\zeta /h$ is conserved, not $\zeta$ itself. Squashing a column to much smaller depth decreases its vertical vorticity. This permits a late draining flow to approximate an irrotational $u_{\theta }=C/r$ profile outside the drain, despite nonzero initial solid-body vorticity. At finite nonzero depth the residual vorticity need not be exactly zero. In a rotating frame, replace $\zeta$ by absolute vorticity $\zeta +f$ in the conserved ratio. Kelvin's circulation theorem remains compatible with these changes because it follows a material loop, not a fixed spatial circle.