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Start from the uniform-flow complex potential of a two-dimensional potential flow $Uz$. Milne-Thomson's circle theorem adds the image doublet $U{a}^2/z$, yielding constant stream function on $|z|=a$. Add a line vortex with positive counterclockwise circulation $\Gamma$:Indeed, on $z=a{e}^{i\theta }$ the uniform-flow and doublet terms sum to the real number $2Ua\cos \theta$, while the vortex has imaginary part $-\Gamma \log a/(2\pi )$, also constant. Thus the cylinder is a streamline and satisfies the impermeability condition. The vortex velocity is $\Gamma /(2\pi r)$ in the counterclockwise direction. Although the potential is multivalued when $\Gamma \ne 0$, its velocity is single-valued.

Differentiate and locate the coefficient of $z^{-1}$ in the square:For the counterclockwise contour in the printed Blasius force theorem, the residue theorem gives ${\oint }_C(w^{\prime }{)}^{2}dz=2U\Gamma$. ConsequentlyThese are force components per unit cylinder length; with the stated circulation convention the lift is downward for positive $U\Gamma$.

For the isolated line vortex, use polar coordinates $z=r{e}^{i\theta }$ to obtainThus for nonzero $\Gamma$ the streamlines are circles and their counterclockwise circulation is $2\pi r{u}_{\theta }=\Gamma$. If $\Gamma =0$ the fluid is stationary and the streamline description degenerates.

Place an opposite image vortex at $z=-1$. The method of images givesOn $z=iy$, the distances to the two vortex positions agree, so $\psi =-(\Gamma /2\pi )\log (|z-1|/|z+1|)=0$. Hence the wall is a streamline. Directly the wall velocity has $u=0$ andFor the stipulated steady flow, Bernoulli's equation in the irrotational fluid outside the vortex core givesThe force exerted by the fluid on this wall points in the $-x$ direction for positive pressure. Relative to uniform pressure $p_{\infty }$, however, its change is in the $+x$ direction. Using the supplied integral, the force change per unit span isThus the wall is attracted towards the vortex, with this magnitude if the other side remains at $p_{\infty }$. Subtracting the uniform pressure is essential for an infinite wall. A freely moving vortex would translate parallel to the wall under its image-induced velocity; the calculation here is for the stated steady vortex configuration, with its position held fixed.

A uniform flow has the same velocity at every position. A constant velocity field has zero divergence and zero vorticity; a velocity potential is $\varphi (\mathbf{x})=\mathbf{U}\cdot \mathbf{x}$. It describes the remote ambient flow in idealized obstacle and line vortex problems.

Use a velocity potential to reconstruct an irrotational flow from one scalar field. A velocity potential $\varphi (\mathbf{x},t)$ satisfiesIt has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$ and is unchanged as a description of the velocity field when any function of time alone is added. For twice continuously differentiable $\varphi$, the identity curl of a gradient proves zero vorticity. In incompressible flow, taking the divergence also gives ${\nabla }^2\varphi =0$, the Laplace equation. This turns a three-component velocity field into a single harmonic function with appropriate boundary conditions. ^[ref]

For uniform flow $\mathbf{u}=(U,0,0)$, take $\varphi =Ux$. With $U=2\mathrm{m}{\mathrm{s}}^{-1}$, the change from $x=0$ to $x=3\mathrm{m}$ is $\Delta \varphi =6{\mathrm{m}}^2{\mathrm{s}}^{-1}$. By the fundamental theorem for line integrals, the line integral of velocity between these points equals that change along any path in the domain; a closed curve gives zero circulation.

For the potential line vortex, choose a local angle branch in polar coordinates and set $\varphi =K\theta$. Then $u_r={\partial }_r\varphi =0$ and $u_{\theta }=r^{-1}{\partial }_{\theta }\varphi =K/r$. One counterclockwise turn changes the continued velocity potential by $2\pi K$, although the velocity field returns to its original value. With $K=1{\mathrm{m}}^2{\mathrm{s}}^{-1}$, this change is $2\pi {\mathrm{m}}^2{\mathrm{s}}^{-1}$, exactly the circulation. A full annulus therefore admits local angle branches and a multivalued velocity potential, but no global single-valued velocity potential when $K\ne 0$. Cutting the annulus along a radial line allows a single-valued angle branch on the remaining domain. The obstruction is path independence, not a failure of the local gradient formula.

For an ideal line vortex of circulation $\Gamma$ held stationary at distance $d$ from an impermeable plane wall, the opposite-sign image from the method of images gives wall-parallel velocityIn this stationary-flow model, subtracting uniform ambient pressure and applying the Bernoulli equation givesThe integral ${\int }_{-\infty }^{\infty }(d^2+y^2{)}^{-2}dy=\pi /(2d^3)$ yields force magnitude per unit vortex lengthdirected from the wall toward the fluid and vortex.

The held-stationary assumption matters. A free point vortex is advected parallel to the wall at $\Gamma /(4\pi d)$ by its image. That flow is time dependent in the wall frame, so its pressure also contains the time derivative of the velocity potential. The stationary pressure deficit cannot be transferred unchanged to that moving-vortex problem.

Use vorticity to distinguish the local spin of a fluid element from the motion of its center along a pathline. Circular streamlines can occur both with and without vorticity: the two velocity fields below even have the same circulation around the unit circle.

For a continuously differentiable velocity field $\mathbf{u}=(u_x,u_y,u_z)$, vorticity is the curl vectorEach partial derivative divides velocity by distance, so vorticity has units ${\mathrm{s}}^{-1}$. In a planar velocity field independent of $z$, only ${\omega }_z={\partial }_x{u}_y-{\partial }_y{u}_x$ survives. Positive ${\omega }_z$ points along $+{\mathbf{e}}_z$: the right-hand rule associates it with counterclockwise spin viewed from positive $z$ toward the plane. ^[ref]

The local rotational angular velocity is half the vorticity, ${\mathbf{\Omega }}_{\mathrm{spin}}=\mathbf{\omega }/2$. To see the factor, split the velocity gradient $L_{ij}={\partial }_j{u}_i$ into its symmetric rate-of-strain tensor $E=(L+L^T)/2$ and antisymmetric spin tensor $W=(L-L^T)/2$. A nearby displacement $\delta \mathbf{x}$ has relative velocity $L\delta \mathbf{x}$; its rotational part satisfies $W\delta \mathbf{x}=(\mathbf{\omega }/2)\times \delta \mathbf{x}$. This is a proof outline of the local spin interpretation. The rate-of-strain tensor can also change the angle of a particular material line, so that line's angular velocity need not equal $\mathbf{\omega }/2$.

For solid-body rotation, $\mathbf{u}=(-\Omega y,\Omega x,0)$ gives $\mathbf{\omega }=(0,0,2\Omega )$ and ${\mathbf{\Omega }}_{\mathrm{spin}}=(0,0,\Omega )$. Taking $\Omega =1{\mathrm{s}}^{-1}$, the vorticity is $2{\mathbf{e}}_z{\mathrm{s}}^{-1}$, although the angular velocity is only $1{\mathbf{e}}_z{\mathrm{s}}^{-1}$. For the potential line vortex, $\mathbf{u}=K{\mathbf{e}}_{\theta }/r$, the planar curl is $r^{-1}{\partial }_r(r{u}_{\theta })=r^{-1}{\partial }_{r}K=0$ wherever $r>0$. Its circular pathlines therefore coexist with zero local spin.

This distinction makes vorticity useful for recognizing shear flow as well as swirling motion; taking the curl of the Euler equations also removes a pressure gradient and leads to the vorticity equation. A historical landmark is Hermann von Helmholtz's 1858 paper on fluid vortex motion, an early study devoted to vortex dynamics. ^[ref]

A wingtip vortex is a concentrated rotating part of the wake behind a finite lifting wing. Spanwise variation of the wing’s circulation produces trailing vorticity, which can roll up into a strong counter-rotating pair. The two rotating wakes induce downward motion between them and upward motion farther outside. NASA’s account connects the tip-region pressure-driven flow with the pair and its induced change of local flow direction.^[ref]

A useful idealization treats each concentrated wake as a straight line vortex. Outside its core, circular symmetry and the definition of circulation giveFor stipulated $|\Gamma |=20{\mathrm{m}}^2/s$ and $r=0.50\mathrm{m}$, the ideal tangential speed is about $6.37m/s$. The singularity as $r\to 0$ is a defect of the line model: a real vortex has a finite core and a nonuniform vorticity distribution.

Consider an original cross-flow model with cores separated horizontally by $d$, strengths $-\Gamma$ and $+\Gamma$, and orientations chosen to give downwash between them. Each ideal vortex induces downward speed $|\Gamma |/(2\pi d)$ at the other core. At $d=10\mathrm{m}$ and $|\Gamma |=20{\mathrm{m}}^2/s$, this pair-interaction speed is about $0.318m/s$. At the midpoint, adding their two velocities gives $2|\Gamma |/(\pi d)\approx 1.273m/s$ downward. These are distinct locations and distinct speeds.

The wake is not generally two permanently straight tubes with fixed strength. A distributed trailing sheet rolls up, the cores move under induced velocities, and viscosity, turbulence and ambient motion alter the field. A wing’s actual span loading and operating state determine the wake, so the illustrative circulation and core size cannot be assigned to an aircraft merely from its visible span.

The connection to lift-induced drag is the complete induced-flow and load system. A short line sketched at each wingtip is not an extra independent friction force to add to an already complete drag calculation. The flow carries momentum and kinetic energy while changing the force direction experienced by the finite wing.

A proposed laboratory wake survey places a finite wing at a controlled load in a wind tunnel and measures transverse velocity at a downstream plane using calibrated directional probes or timed tracer images. Reverse the sign of lift to check the expected reversal of rotation, and compare measured circulation on several enclosing paths outside the cores. Tracers must follow the flow adequately, while finite sampling, tunnel walls and the probe’s own disturbance limit the inference. A visible tracer pattern alone does not supply the full three-dimensional velocity or pressure field.