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In cylindrical coordinates, the transverse velocity is $A{\mathbf{e}}_{\theta }/r$, so its contribution to the circulation is $Ad\theta$. The vertical contribution is $U(z)dz$, the differential of an antiderivative of $U$, and therefore has zero integral around any closed curve. If $k$ is the oriented winding number of the curve's projection about the origin, thenIn particular, a single positive winding gives $2\pi A$; reversing orientation reverses the sign. The nonzero circulation is compatible with irrotational flow because the domain is not simply connected. For $A\ne 0$, an enclosing curve cannot bound a surface contained entirely in the domain to which Stokes theorem would apply.

Take a singly quantized vortex of positive circulation. The superfluid velocity is $\mathbf{u}=(\hslash /m)\nabla S$, where $S$ is the complex argument of the condensate. Since $S=\theta$, quantized circulation gives $u_{\theta }=\hslash /(mr)$. Insert the radial Thomas–Fermi approximation for a condensate into the cylindrical volume integral, excluding the core $r<a_0$:Integrating the logarithmic and quadratic terms separately givesThe logarithm is the familiar long-range flow contribution of a quantum vortex; the $-1/2$ comes from the declining trapped number density. If the bucket truncates the cloud, put $R=R_0<R_r$ and the same integration gives $E_{\mathrm{kin}}=\pi H_0{\hslash }^2{n}_0[\log (R/a_0)-(R^2-a_0^2)/(2R_r^2)]/m$, provided $a_0<R$. A vortex of integer winding number $q$ has $u_{\theta }=q\hslash /(mr)$ and multiplies this flow energy by $q^2$. The core radius acts as a cutoff of order the local healing length; its internal gradient and interaction energies are not computed by this shell estimate.

In a rotating reference frame at angular frequency $\Omega$, the relevant energy difference is approximately $E_{\mathrm{kin}}-\Omega L_z$. The vortex-free state has zero circulation and zero corresponding angular momentum. For $a_0/R_r\ll 1$, the preceding shell estimates reduce toTherefore the thermodynamic vortex-nucleation frequency for the first positive quantum vortex isThe common factors $H_0,n_0$ cancel explicitly, although the trap radius and core scale still depend on number density. In the strict leading-logarithm limit, ${\Omega }_c\sim 2\hslash \log (R_r/a_0)/(m{R}_r^2)$. The finite constant $-1/2$ is the one obtained from the stated Thomas–Fermi shell integral; a resolved vortex core and readjustment of the number density at fixed particle number can change the nonlogarithmic constant. This is an energetic threshold, not a dynamical guarantee of nucleation: a surface barrier can delay entry. Higher positive winding numbers have flow energy proportional to $q^2$ and angular momentum proportional to $q$, so their analogous threshold is larger in this approximation.

Work on a beta plane with $f=f_0+\beta y$ and constant depth $H$. Assume a steady, small-Rossby number interior, negligible friction there, small sea-surface displacements, and negligible depth gradients. Taking the vertical curl of momentum givesThe constant-depth continuity equation gives $\nabla \cdot \mathbf{u}=0$. Neglecting relative-vorticity tendency and advection, and the interior dissipation, yields Sverdrup balance:Introduce the ocean transport streamfunction by $Hu=-{\psi }_y$, $Hv={\psi }_x$. Then $\beta {\psi }_x=H(\nabla \times \mathbf{W}{)}_z$. Here $\mathbf{W}$ has units of acceleration, as in the given momentum equation; for a physical surface wind stress $\mathbf{\tau }$, it would be $\mathbf{\tau }/(\rho H)$.

To obtain the circulation law, use the vector identity $\mathbf{u}\cdot \nabla \mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)+\zeta \hat{\mathbf{k}}\times \mathbf{u}$. Its gradient term and the sea-surface pressure gradient integrate to zero around a closed contour. ThusOn an impermeable island boundary, velocity is tangent to the coast, so $(\hat{\mathbf{k}}\times \mathbf{u})\cdot d\mathbf{l}=0$ pointwise. Consequently its circulation obeys

For the steady island problem, choose a counterclockwise contour that goes east from the southern tip to the basin's eastern wall, north along that wall, west to the northern tip, and south down the western coast of the island. It avoids the eastern-island and western-basin dissipative layers. In the leading momentum balance, $\hat{\mathbf{k}}\times \mathbf{u}=-\nabla \psi /H$. Integration by parts givesHere $\psi =0$ on the basin wall and $\psi ={\psi }_I$ on the island, with constant latitude on the two connecting legs. Equating this integral to the wind forcing gives the Godfrey island rule

For the square island, $y_N=R/2$, $y_S=-R/2$, and its west coast is $x=-R/2$. Both horizontal contour legs have length $S+R/2$. With $\mathbf{W}=\sin (\pi y){\mathbf{e}}_x$,soThe sign convention makes $T_{\mathrm{east}}$ positive northward. For $0<R<2$ it is southward. The total transport includes the island's thin eastern boundary current: omitting that current would leave a latitude-dependent interior transport rather than this constant passage transport.

To quantify the island's effect on the western boundary current, compare interior streamfunctions west of the island. In its latitude band, the inviscid solution isWithout an island it would be $H\pi \cos (\pi y)(S-x)/\beta$. Since the western basin wall has zero streamfunction, the difference in its required northward boundary-current transport isThus the island changes the current's strength by redistributing transport across its latitude band. For a square contained in the central positive-cosine band, it weakens the northward current near $y=0$ and strengthens it near the island's northern and southern edges. The change averages to zero over $-R/2<y<R/2$; its local sign follows the formula for other island sizes.

A field is irrotational when $\nabla \times F=0$. In ${\mathbb{R}}^3$ definewhere $\gamma$ is any path from $x_0$ to $x$. The difference between two choices is the circulation around a closed loop. By the Stokes theorem it is the flux of $\nabla \times F$ through a spanning surface and hence zero. Thus $V$ is well-defined, $V(x_0)=0$, and differentiation gives $F=-\nabla V$.

If $\nabla \times A_0=B$, every solution isfor an arbitrary differentiable scalar $\chi$, since the difference is curl-free. A necessary condition is $\nabla \cdot B=0$, because the divergence of a curl vanishes.

Faraday's law isChoose the positive circulation to be anticlockwise as viewed from $+z$. Since ${\Phi }_B=\pi B{r}^2$,

If $\dot{r}>0$, the current is clockwise and its field points in the $-z$ direction, reducing the field inside the loop. Since $d\mathbf{F}=Id\mathbf{l}\times \mathbf{B}$, the force due to the imposed field is radially inward. If $\dot{r}<0$, the current is anticlockwise, its field points in the $+z$ direction and increases the field inside, while the force is radially outward. In both cases the magnetic force opposes the radial motion, in accordance with Lenz's law.

The term $-IS$ removes healthy foxes through mass-action contact with infectives. The Laplacian ${\nabla }^{2}I$ models random migration of infected foxes. The matching term $+IS$ creates newly infected animals, while $-\mu I$ removes infectives at per-capita rate $\mu$. The equation $R_t=\mu I$ records those removed from infectious circulation, for example foxes that have died from rabies; they neither migrate nor return to susceptibility in this model.

Single-valuedness of the condensate field gives total complex argument change $2\pi q$ on a closed loop avoiding zeros, where $q$ is an integer winding number. Integrating the superfluid velocity around that loop gives the displayed circulation. It is unchanged by smooth deformations of the loop that do not cross a quantum vortex.

A quantum vortex has integer winding number of a condensate's complex argument. For atomic mass $m$, its superfluid velocity gives circulation $2\pi \hslash q/m$ for integer $q$. A straight vortex has a density-depleted core and, outside it, tangential velocity $q\hslash /(mr)$. Its long-range flow energy is proportional to $q^2\log (R/a)$, while its angular momentum has the sign of $q$.