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For uniform flow $\mathbf{U}$ plus a line vortex of counterclockwise circulation $\Gamma$, the outward advective momentum flux around a large circle is $\rho \mathbf{\Gamma }\times \mathbf{U}/2$. The Bernoulli equation supplies an equal pressure contribution. Their sum is the force on the fluid; the force on an obstacle per unit span is its negative, $\rho \mathbf{U}\times \mathbf{\Gamma }$. Retaining only the advective term loses half the lift and obscures its sign.

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.

The method of images enforces impermeability at a plane wall by reflecting a line vortex across it and reversing its circulation. The two logarithmic stream functions cancel on the wall, which is consequently a streamline. For a vortex of circulation $\Gamma$ at $(0,b)$ above $y=0$, the image of circulation $-\Gamma$ at $(0,-b)$ induces velocity $(\Gamma /(4\pi b),0)$ at the real vortex. The signed drift parallel to the wall is therefore $\Gamma /(4\pi b)$.

Use irrotational flow to identify where a velocity potential can describe the velocity field locally. An irrotational flow has $\mathbf{\omega }=\nabla \times \mathbf{u}=0$ throughout the fluid region. For a continuously differentiable velocity field, the Poincaré lemma supplies a velocity potential on sufficiently small neighborhoods. A global single-valued velocity potential requires zero circulation around every closed curve; a simply connected domain is a sufficient domain assumption.

The potential line vortex $\mathbf{u}=K{\mathbf{e}}_{\theta }/r$ is an irrotational flow on the punctured plane, or on an annulus $a<r<b$ with $a>0$, even though its pathlines are circular. Every sufficiently small loop bounding a disk in that fluid has zero circulation, by Stokes theorem. A loop winding once counterclockwise around the excluded origin has circulation $2\pi K$. Thus local irrotational flow and nonzero circulation around a hole are compatible: the enclosing disk passes outside the domain of the velocity field.

An irrotational flow is a kinematic condition on a velocity field, independent of whether a particular model neglects viscosity or assumes incompressible flow. Adding incompressible flow makes the local velocity potential satisfy the Laplace equation. This reduces the number of unknowns and makes potential flow a powerful starting point for flow around bodies and for gravity waves. ^[ref]

An ideal straight line vortex in a two-dimensional incompressible flow has circulation $\Gamma$ and induces velocityIts stream function, with $\mathbf{u}=({\psi }_y,-{\psi }_x)$, is $\psi =-\Gamma \log |\mathbf{r}-{\mathbf{r}}_0|/(2\pi )$. Positive $\Gamma$ means counterclockwise circulation. In an ideal point-vortex description, the vortex position is advected by the regular velocity produced by the other vortices and the boundaries; its own singular velocity is excluded.

In the quadrant $x,y>0$, a line vortex of circulation $\Gamma$ at $(x,y)$ has three images: circulations $-\Gamma$ at $(-x,y)$ and $(x,-y)$, and $+\Gamma$ at $(-x,-y)$. Their induced velocity isIn polar coordinates, these equations give $\dot{r}=\Gamma \cot (2\theta )/(2\pi r)$ and $\dot{\theta }=-\Gamma /(4\pi r^2)$. Direct differentiation yields $\frac{d}{dt}(r\sin 2\theta )=0$, so the trajectory obeys $r\sin 2\theta =\text{constant}$.

Away from the axis, the cylindrical coordinates formula for curl gives $r^{-1}{\partial }_r(r{F}_{\theta })=2(1-n)r^{-2n}$ when $F_{\theta }=r^{1-2n}$. Its sign changes at $n=1$, even though the tangential velocity stays positive. Thus local angular velocity, one half of vorticity, can oppose orbital circulation. For $n=1$, the field is a line vortex with zero ordinary curl off the axis and circulation $2\pi$ around it. Stokes theorem applies on annuli excluding the axis with both boundary components included; it cannot be applied to a spanning disc containing the field singularity.

Define the line vortex strength $\kappa$ by $\mathbf{u}=\kappa {\mathbf{e}}_{\theta }/(2\pi r)$, with $\kappa >0$ meaning counterclockwise motion. On an enclosing circle of radius $R$,because the velocity is tangent to the circle. For any simple positively oriented enclosing curve, the same results follow by Stokes theorem and the divergence theorem on the region between that curve and a small inner circle. Both curl and divergence vanish throughout that intervening annulus. ConsequentlyEquivalently, $\Gamma =(\kappa /(2\pi ))\oint d\theta$ and the total angle change is $2\pi$. If strength is instead denoted by $K$ in $u_{\theta }=K/r$, the answer is $\Gamma =2\pi K$. An enclosing disk through the singular origin cannot be used to infer zero circulation from zero local vorticity.

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.

For planar incompressible flow and irrotational flow, combine a velocity potential and a streamfunction into the complex potential of a two-dimensional potential flow:The chosen conventions giveThese are the Cauchy-Riemann equations, so $w$ is a holomorphic function wherever the flow is regular and a consistent local potential is chosen. Its complex derivative is the complex velocityA spatially uniform flow has $w=(U-iV)z+$ a constant. The following contour integral identities also apply to nonuniform potential flow, including the specified line vortex. A complex potential of a two-dimensional potential flow may be multivalued around a source or vortex even though its derivative is single-valued.

For the same counterclockwise line vortex strength $\kappa$, its local velocity potential and streamfunction are $\varphi =\kappa \theta /(2\pi )$ and $\psi =-\kappa \log r/(2\pi )$. ConsequentlySubstitution givesin agreement with the direct circulation and volume flux calculation. The sign of the imaginary coefficient is essential for a positive counterclockwise circulation.

Use $w=\varphi +i\psi$ for the complex potential of a two-dimensional potential flow, so its complex velocity is $q=w^{\prime }=u-iv$. The incident uniform flow and line source haveThe coefficient of the complex logarithm is one because a line source of strength $Q$ has complex potential of a two-dimensional potential flow $(Q/(2\pi ))\log (z-b)$, and here $Q=2\pi$. Applying the Milne-Thomson circle theorem to the unit circle givesSince $1/z-b=-b(z-1/b)/z$, locally consistent branches of the complex logarithm giveChanges of branch only alter the constant in a local expression; $w^{\prime }$ is single valued away from its singularities. The added terms represent a two-dimensional potential doublet at the origin, an image line source at $1/b$, and a compensating line sink at the origin. The image line source and line sink have equal and opposite strengths, so no net volume flux enters the cylinder.

On $|z|=1$, the Milne-Thomson circle theorem makes $w=f(z)+\overline{f(z)}$ up to a constant. Consequently the streamfunction is constant there, establishing the no-penetration boundary condition. All line source strengths are real, and the two-dimensional potential doublet has zero circulation, so the circulation around the cylinder is zero. There is no additional line vortex term.

Take positive $\Gamma$ to mean counterclockwise circulation. A line vortex at ${\zeta }_0$ has complex potential of a two-dimensional potential flow $\Gamma \log (\zeta -{\zeta }_0)/(2\pi i)$. The method of images imposes the no-penetration boundary condition on the real axis by placing an image line vortex of strength $-\Gamma$ at ${\bar{\zeta }}_0$. ThusFor real $\zeta$, the numerator and denominator have equal modulus, so the streamfunction isThe real axis is therefore a streamline. The image line vortex lies outside the mapped fluid region, while the complex velocitydecays as $|\zeta |\to \infty$. This selects the flow due to the line vortex and the boundary, without superposing an imposed background velocity field.

The line vortex is mapped toIts imaginary part is positive because $c$ lies in the fluid region. A conformal map preserves the circulation of the local line vortex, so its strength in the mapped complex plane is still $\Gamma$.

A free point vortex is advected by the regular part of the velocity field at its position. Its own singular complex velocity $\Gamma /[2\pi i(z-c)]$ must first be subtracted. The image expression therefore givesEquivalently, $\dot{c}$ is the complex conjugate of the boxed expression. This is the instantaneous velocity, evaluated with the current value of $c$.

To see why mapping the image velocity alone would miss a term, expand the mapped physical singularity:Thus the same point-vortex drift under a conformal map can be writtenThe finite term involving $g^{\prime \prime }$ is part of the physical regular velocity field.

If $c=id$ with $d>1$, direct substitution into the image expression givesThe bracket is strictly positive, and the result is real. Hence a positive line vortex initially moves horizontally to the right, a negative line vortex moves to the left, and the initial vertical velocity is zero. As $d\to \infty$, the leading term $\Gamma /(4\pi d)$ is the familiar drift of a point vortex above a straight wall.

The physical complex potential of a two-dimensional potential flow is $w(z)=W(g(z))$. Applying the chain rule gives the complex velocityThe physical velocity is $(\mathrm{Re}{w}^{\prime },-\mathrm{Im}{w}^{\prime })$. A useful equivalent expression follows fromFactoring the image expression in the same way makes the $-1/z$ terms cancel, leavingThis represents the physical line vortex at $c$, an image line vortex of the same strength at $c^{-1}$, and image point vortices of opposite strength at $\bar{c}$ and ${\bar{c}}^{-1}$. All three image locations are outside the fluid region. Both boundary pieces inherit the constant streamfunction of the real axis in the mapped complex plane.

Take positive vortex strength $\kappa$ to mean anticlockwise circulation. A line vortex at $z_0$ has local velocity potential $\kappa \arg (z-z_0)/(2\pi )$. The image vortex at a plane wall must have strength $-\kappa$ to enforce zero normal velocity at $y=0$. At the initial instant, a potential is thereforeThis is a local, multivalued velocity potential: its gradient is single-valued away from the vortex, but the nonzero circulation precludes a globally single-valued potential around it. For a vortex centred at $(X,Y)$, the velocity isThe real vortex and its opposite image give cancelling $u_y$ at the wall. A point vortex moves with the regular velocity induced by the other vortex, not its singular self-field. Evaluating the image field at $(X,b)$ gives $\dot{X}=\kappa /(4\pi b)$ and $\dot{b}=0$. HenceThe time-dependent velocity potential is obtained by replacing $a$ with $X(t)$ above; the image remains at $(X(t),-b)$. Reversing the convention for positive circulation reverses the translation direction.

With polar unit vectors ${\mathbf{e}}_r=(\cos \theta ,\sin \theta ,0)$ and ${\mathbf{e}}_{\theta }=(-\sin \theta ,\cos \theta ,0)$, the second term is $\Gamma {\mathbf{e}}_{\theta }/(2\pi r)$. It is tangent to circles, independent of $z$, and has zero divergence and zero vorticity away from the axis. Its singular axis carries a line vortex; adding the constant velocity field gives a uniform flow plus that line vortex.

The counterclockwise circulation around a positively oriented loop is $\oint \mathbf{u}\cdot d\mathbf{r}$. For a radius-$R$ circle, the uniform flow contributes zero and the line vortex contributesAny loop winding once around the axis has the same circulation, by Stokes theorem in the nonsingular intervening region.

On $C_R$, $\mathbf{n}={\mathbf{e}}_r$, $\mathbf{u}\cdot \mathbf{n}=U\cos \theta$, and $dl=Rd\theta$. ThusIn fact this equality holds for every $R>0$ for the stated field, since the remaining vector integral is $(0,\pi ,0)$.

This is the advective part of the momentum flux. To obtain the force, include pressure. For steady irrotational flow, the Bernoulli equation givesThe constant terms integrate to zero against $\mathbf{n}$; the remaining pressure contribution is $\oint p\mathbf{n}dl=\rho \mathbf{\Gamma }\times \mathbf{U}/2$. The total outward momentum flux plus pressure term is therefore $\rho \mathbf{\Gamma }\times \mathbf{U}$, the force of the obstacle on the fluid. The opposite force, of fluid on the obstacle per unit span, isThis is the Kutta–Joukowski theorem, with positive $\Gamma$ defined counterclockwise: positive $U$ and $\Gamma$ give force in the negative $y$ direction. For an actual two-dimensional aerofoil, faster-decaying far-field terms do not change this limit. The displayed advective integral supplies half the lift; pressure supplies the other half.

For constant mass density and incompressible flow, use $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times \mathbf{\omega }$ in the Euler equations for an inviscid fluid. Taking the curl removes the pressure and kinetic energy gradients, giving ${\mathbf{\omega }}_t=\nabla \times (\mathbf{u}\times \mathbf{\omega })$. The supplied vector identity, with $\nabla \cdot \mathbf{u}=0$ and $\nabla \cdot \mathbf{\omega }=0$, yields the vorticity equationIn a two-dimensional flow $\mathbf{u}=(u,v,0)$ is independent of $z$, and $\mathbf{\omega }=\omega {\mathbf{e}}_z$. The right side vanishes, so $D\omega /Dt=0$: each material element retains its vorticity. The vortex lines are parallel to $z$ and are carried by the planar velocity. Also planar incompressibility preserves the area of a material cross-section; integrating its conserved vorticity over that area shows that its circulation, or vortex strength, stays constant. Thus the vortex lines move with the fluid without changing strength.

Place a line vortex of circulation $\Gamma$ at $(0,b)$ above the wall $y=0$. Its velocity field is $\Gamma {\mathbf{e}}_z\times (\mathbf{r}-{\mathbf{r}}_0)/(2\pi |\mathbf{r}-{\mathbf{r}}_0{|}^2)$. The method of images puts a vortex of circulation $-\Gamma$ at $(0,-b)$. With $\mathbf{u}=({\psi }_y,-{\psi }_x)$, the resulting stream function isIt is identically zero on the wall, so $u_y=-{\psi }_x=0$ there. In the fluid half-plane it has just the required vortex singularity, with no other singularities, and its velocity decays at infinity. This is the no-background-flow solution with the wall: the difference from another such solution is a regular irrotational, divergence-free decaying field with zero normal velocity. Reflect its tangential component evenly and its normal component oddly across the wall. The reflected components are bounded harmonic functions on the whole plane, so they are constant by harmonic Liouville theorem, and decay forces both constants to be zero. Thus replacing the wall by the image gives the same flow. The real vortex moves under the image velocity, excluding its own singular velocity, soIts signed velocity parallel to the wall is $\Gamma /(4\pi b)$, and its speed is $|\Gamma |/(4\pi b)$.

For the quarter-plane, place images of circulation $-\Gamma$ at $(-x,y)$ and $(x,-y)$, and $+\Gamma$ at $(-x,-y)$. The paired stream functions make both walls streamlines. Summing their velocities at the real vortex givesIn polar coordinates,It follows directly thatThus the line-vortex trajectory in a quarter-plane is $r\sin 2\theta =\text{constant}$.

Away from the axis, cylindrical coordinates put the field in the line vortex formThe only potentially nonzero component of its curl isThe other components vanish because there is no axial dependence and no radial or axial velocity. Thus the curl is zero for $\rho >0$; the field is singular on the axis, so a classical curl there is not defined.

For a spanning surface avoiding that axis, the smoothness hypotheses of Stokes theorem hold and give ${\oint }_C\mathbf{v}\cdot d\mathbf{r}=0$. If the enclosed planar region contains the axis, the singularity prevents applying that theorem directly. Excise a small disk about the origin. The resulting annulus has zero curl, so its outer circulation equals the counterclockwise circulation on the small circle:This sign assumes positive, counterclockwise orientation. Clockwise orientation gives $-\kappa$; more generally the value is $\kappa$ times the winding number about the origin. A curve through the origin is excluded because its integral is singular. In the distribution interpretation the missing vorticity is concentrated on the axis as $\kappa \delta (x)\delta (y){\mathbf{e}}_z$, explaining the nonzero circulation despite zero classical curl away from it.