Vorticity and circulation

Start with a moving liquid layer

A liquid layer is 20 mm high. Its lower boundary has zero horizontal speed and its upper boundary moves right at 0.100 m/s. Prescribe a horizontal speed that increases linearly with height, with no vertical velocity. The streamlines are straight. If we mark a tiny cross in the liquid, can it rotate locally? What changes if we instead add the velocity along an entire closed loop?

Choose x to the right, y upward and z out of the page in a fixed inertial frame. Write the velocity as v = (u, v_y, 0), where u is the horizontal component and v_y the vertical component. With coordinates in metres, the supplied field is u = ay, v_y = 0, with a = (0.100 m/s)/(0.020 m) = 5 s⁻¹. This profile is an assigned kinematic example; it supplies no viscosity, stress, pressure or wall-force prediction.

First follow the velocity samples, marked-line comparison, rectangular loop and circular-flow comparison. Vectors and vector spaces and Derivatives, Taylor series, and local approximation provide useful preparation. The curl and surface-integral notation is explained here; Gradient, divergence, curl, and vector calculus and Integrals and multivariable calculus are optional return paths for the final derivations.

Read local velocity change

A velocity gradient describes how neighbouring points have different velocities at the same instant. Here raising y by 1 mm raises u by 0.005 m/s. The ratio of these changes is ∂u/∂y = 5 s⁻¹. A spatial derivative of velocity has units (m/s)/m = s⁻¹; it is not an acceleration.

Vorticity is the vector curl of velocity: ω = ∇×v. For this planar field, only its z component can be nonzero: ω_z = ∂v_y/∂x − ∂u/∂y. Positive ω_z points out of the page and corresponds to counterclockwise local spin; negative ω_z points into the page and corresponds to clockwise local spin.

The supplied layer has ∂v_y/∂x = 0 and ∂u/∂y = 5 s⁻¹, so ω_z = −5 s⁻¹ everywhere. Straight streamlines therefore do not imply zero vorticity. Vorticity concerns neighbouring velocities and local rotation, rather than whether a parcel’s path happens to curve.

Separate spin from line rotation

The local spin vector is ω/2. In the layer it is −2.5 s⁻¹ along z. This is the rigid-rotation part of the infinitesimal motion. A material line follows the same marked fluid parcels. A deforming fluid element also stretches and shears, so a particular material line need not rotate at that rate.

At the initial instant, a horizontal material line in the shear field has zero angular rate, while a vertical line tips clockwise at 5 s⁻¹. Their signed rates are 0 and −5 s⁻¹. The mean for this initially orthogonal pair is −2.5 s⁻¹, matching the local spin. The cross also changes its right angle, so its motion is not a rigid rotation.

Compare the prescribed rigid-rotation field u = −Ωy, v_y = Ωx with Ω = −2.5 s⁻¹. Its derivatives give ω_z = 2Ω = −5 s⁻¹ and the same local spin. Both material lines now rotate clockwise at −2.5 s⁻¹, preserving their lengths and right angle. Equal vorticity does not imply equal deformation.

The first figure follows 4 mm half-arms for 0.040 s, relative to each moving centre. In shear, the horizontal line remains horizontal and the vertical line changes by −arctan(0.2) ≈ −11.309932°. In rigid rotation, both lines change by Ωt = −0.1 rad ≈ −5.729578°. These are exact finite-time changes for the assigned fields. Their mean is not a general finite-angle definition of spin; ω/2 describes the instantaneous local rate.

Walk an oriented closed loop

Circulation is the integral Γ_C = ∮_C v·dr around a closed, oriented contour C. Each short piece adds the velocity component in the direction of travel multiplied by that piece’s length. The contour is an integration path, not necessarily a streamline or a group of fluid parcels. Circulation has units (m/s)m = m²/s; it is not volume flow rate.

Use a rectangle from x = 0 to 0.040 m and y = 0 to 0.020 m. Choose counterclockwise traversal, viewed from +z: bottom to the right, right edge upward, top to the left and left edge downward. This orientation is compatible with surface normal +e_z.

The bottom contributes zero because u = 0 there. The vertical edges contribute zero because the velocity is horizontal and perpendicular to their travel directions. At the top, u = +0.100 m/s points right while the contour travels left, so v·t = −0.100 m/s. Multiplying by 0.040 m gives −0.004 m²/s. The total circulation is negative even though we chose a positive, counterclockwise contour.

Reversing the contour reverses Γ_C. It changes our signed measurement, rather than changing the fluid velocity. The second figure keeps the fluid arrows and contour arrows visually distinct so these two directions cannot be confused.

Connect circulation and vorticity flux

Stokes’ theorem connects the line integral to a surface integral: Γ_C = ∫_S ω·n dA. The contour must be the oriented boundary of an admitted surface S, with a compatible normal n, and the velocity must be sufficiently smooth on that surface and its neighbourhood. A missing point, excluded core or jump in the field can prevent use of the proposed spanning surface.

Our rectangle lies entirely within the smooth shear field. Its area is (0.040 m)(0.020 m) = 0.0008 m². Since ω_z = −5 s⁻¹ is uniform and n = +e_z, the flux is (−5 s⁻¹)(0.0008 m²) = −0.004 m²/s. This agrees with adding the four edge contributions.

For uniform planar vorticity, circulation divided by the admitted area gives ω_z. For nonuniform vorticity it gives the area average of ω·n, not the value at every point. Positive and negative patches can cancel, so zero circulation around one large loop does not prove zero vorticity everywhere inside.

Under the same smoothness conditions, shrinking an admitted planar loop to a point gives the local relation ω_z = lim Γ_C/A for normal +e_z. Circulation describes an entire loop; vorticity describes a local field. Both the area and the orientation matter when relating them.

Compare two circular velocity fields

Now compare two prescribed clockwise flows about the origin. In rigid rotation, the signed tangential velocity is v_θ = Ωr with Ω = −2.5 s⁻¹. A circular contour of radius R, traversed counterclockwise, has Γ_R = 2πR v_θ(R) = 2πΩR². The field is smooth at the origin, and its uniform ω_z = 2Ω gives the same result through the full disk.

For the second field, prescribe v_θ = C/r with C = −0.00225 m²/s only on the annulus 0.010 ≤ r ≤ 0.060 m. Its circular-contour integral is Γ_R = 2πR(C/R) = 2πC = −0.0045π m²/s ≈ −0.014137 m²/s. Every listed circular contour has the same circulation, although their tangential velocities differ.

At R = 0.030 m, both fields have v_θ = −0.075 m/s and the same circulation. Yet their local vorticities differ: −5 s⁻¹ for rigid rotation and zero within the smooth annular field. A circle’s velocity and circulation alone do not identify the spatial velocity gradient.

The third figure uses one physical radius scale and one circulation scale for both fields. The quadratic rigid-rotation curve and the constant annular curve meet at 0.030 m. Their streamline shapes are both circular; that resemblance does not make their local rotation or domain the same.

Respect the missing centre

Within the annulus, the Cartesian field is u = −Cy/(x²+y²), v_y = Cx/(x²+y²). Direct differentiation gives ∂v_y/∂x = ∂u/∂y = C(y²−x²)/(x²+y²)², so ω_z = 0 there. A field with zero vorticity is called irrotational. That word does not say that its fluid parcels travel along straight paths.

There is no contradiction with its nonzero circulation around the origin. A full disk spanning one of those circles would include the excluded inner region, where no velocity or vorticity has been assigned. Stokes’ smooth-surface relation cannot be used on that disk. We are not assigning a singular core, a delta-function vorticity or a physical core model behind the hole.

An admitted annular surface from r = 0.020 m to r = 0.040 m has two boundary components. With normal +e_z, the outer circle runs counterclockwise and the inner circle clockwise. Their contributions are −0.0045π and +0.0045π m²/s, summing to zero, exactly as the zero vorticity flux through that annular surface requires.

Any contour that bounds a smooth surface wholly inside this irrotational domain has zero circulation. An encircling contour cannot be shrunk through the missing centre while staying in the domain. Thus zero local curl does not guarantee zero circulation around every loop in a domain with a hole. The next subject, Potential flow, develops the related local and global velocity-potential questions.

Follow a material loop

The loops used so far were contours chosen at an instant. For Kelvin’s circulation theorem, C(t) is a material loop: the same identified fluid parcels form it while it moves and deforms. Conservation for that moving loop is different from asking how circulation changes around an arbitrarily fixed spatial contour.

A sufficient setting is a smooth inviscid fluid in an inertial frame, with conservative body-force acceleration and a pressure force per unit mass that is a gradient. Uniform density provides this pressure condition. A barotropic relation, with pressure determined by density, also provides it. Here inviscid means viscous stresses are omitted; conservative acceleration can be written as the gradient of a single-valued scalar potential, with either consistent sign convention.

Under those conditions, dΓ_C(t)/dt = 0. Circulation can remain nonzero and the loop can change shape. The theorem conserves circulation, not the speed of each parcel, the angular velocity of every line, or an arbitrary region’s angular momentum. It does not require a steady velocity field.

If viscous stresses influence the loop, density and pressure gradients are not aligned, or body-force acceleration is not conservative, the sufficient conditions fail. Circulation may then change. These mechanisms are not quantified by our three prescribed velocity fields. Viscosity, Reynolds number, and Navier–Stokes equations is the later subject for resolved viscous dynamics.

Understand vortex stretching

A vortex line is a curve tangent to the vorticity vector where that vector is nonzero. It differs from a streamline, which is tangent to velocity. A bundle of vortex lines forms a vortex tube whose side surface has no normal vorticity flux. Since the divergence of a smooth curl is zero, compatible vorticity flux through different tube cross-sections is the same at an instant.

In a smooth circulation-preserving motion, vortex lines are carried with the fluid. If a material cross-section shrinks while its circulation is conserved, its average axial vorticity increases in magnitude. Stretching can therefore strengthen local vorticity without generating additional circulation.

For a conditional straight-tube comparison, take initial Γ = −0.010 m²/s, uniform axial vorticity over area 0.002 m², and later area 0.001 m². Kelvin’s conditions are stipulated. The corresponding vorticities are Γ/A = −5 s⁻¹ and −10 s⁻¹. If material length increases from 0.020 m to 0.040 m at constant density, both volumes are 0.000040 m³. This illustrates incompressible three-dimensional stretching; it is not a full pressure-field or flow-trajectory solution.

In a strictly two-dimensional incompressible motion with no dependence on z, this stretching mechanism is absent. Under the same smooth inviscid and conservative conditions, a parcel’s scalar ω_z is conserved. Three-dimensional stretching, planar deformation and circulation generation should not be treated as interchangeable processes.

Optional return: gradient and transport laws

Return after the marked lines and loop comparisons for the matrix and evolution formulas. Define L_ij = ∂v_i/∂x_j. Split the velocity gradient as L = D + W, with D = (L+Lᵀ)/2 and W = (L−Lᵀ)/2. D gives local stretching and shearing; the skew part W acts like rigid rotation with axial vector ω/2. In the shear case, L = [[0,5],[0,0]] s⁻¹, D = [[0,2.5],[2.5,0]] s⁻¹ and W = [[0,2.5],[−2.5,0]] s⁻¹. In rigid rotation, D is zero while W is the same.

For a material loop, differentiation gives dΓ/dt = ∮_C(t) (Dv/Dt)·dr. Motion of the line elements adds ∮v·dv = ∮d(|v|²/2) = 0, because the loop is closed. The material derivative is D/Dt = ∂/∂t + v·∇: it follows the same moving matter rather than holding position fixed.

Euler’s momentum equation is Dv/Dt = −(1/ρ)∇p + b, where p is pressure, ρ density and b body-force acceleration. For constant density, −(1/ρ)∇p = −∇(p/ρ). For a smooth barotropic relation p = p(ρ) with ρ > 0, define H(ρ) = ∫[p′(s)/s] ds, where s is a density variable. Then ∇H = (1/ρ)∇p without needing to invert p(ρ). With b = −∇Φ and single-valued potentials on the loop, acceleration is −∇(H+Φ), or the constant-density counterpart. Its closed integral is zero, giving Kelvin’s theorem.

For smooth incompressible flow with uniform density and conservative body-force acceleration, taking the curl gives Dω/Dt = (ω·∇)v in the inviscid case. The right side represents stretching and tilting. If a constant kinematic viscosity ν is instead included in this same incompressible, uniform-density Newtonian model, a diffusion term ν∇²ω appears. Its units are (m²/s)(s⁻¹/m²) = s⁻². Variable material properties or compressible flow require their own terms and assumptions; this is not a general replacement for every fluid model.

One route for circulation generation is already visible without adopting a full viscous equation: ∇×[−(1/ρ)∇p] = (∇ρ×∇p)/ρ². Nonparallel density and pressure gradients can produce vorticity. A barotropic relation eliminates this source; it does not by itself eliminate every possible viscous or externally forced contribution.

Worked examples

Try it yourself

1. In a separate 0.030 m-high layer, lower horizontal speed is zero and upper horizontal speed is −0.120 m/s. Prescribe the same linear profile with no vertical velocity. Find ω_z and local spin along z.

   Show the answer and reasoning

   The gradient is −4 s⁻¹, so ω_z = +4 s⁻¹ and local spin is +2 s⁻¹, counterclockwise.

   Divide −0.120 m/s by 0.030 m to get ∂u/∂y = −4 s⁻¹. Since ∂v_y/∂x = 0, the curl subtracts this negative gradient. Changing the flow-profile sign changes the physical rotation sign.

2. In the original u = (5 s⁻¹)y field, use a counterclockwise rectangle from x = 0.010 to 0.070 m and y = 0.002 to 0.017 m. Find the bottom and top contributions and total circulation.

   Show the answer and reasoning

   Bottom contribution is +0.0006 m²/s, top contribution −0.0051 m²/s and total Γ = −0.0045 m²/s.

   Width is 0.060 m. Bottom and top speeds are 0.010 and 0.085 m/s; their traversal directions are +x and −x. Vertical contributions remain zero. Their sum also equals (−5 s⁻¹)(0.060 m)(0.015 m). Moving the loop changes each horizontal contribution, while uniform curl makes the net depend on its oriented area.

3. Traverse the original 0.040 m by 0.020 m shear rectangle clockwise. Which compatible surface normal should be used, and what are circulation and vorticity flux?

   Show the answer and reasoning

   Use normal −e_z. Both circulation and compatible vorticity flux are +0.004 m²/s.

   Contour reversal changes the line integral’s sign. Reversing the compatible surface normal also changes ω·n from −5 to +5 s⁻¹. The fluid itself still has ω_z = −5 s⁻¹ in the original fixed frame.

4. For the assigned inverse-radius annular field, use the surface from r = 0.020 to 0.040 m with normal +e_z. State both boundary directions and their circulations. Can one replace it with a filled disk through the missing centre?

   Show the answer and reasoning

   Outer traversal is counterclockwise with Γ = −0.0045π m²/s; inner traversal is clockwise with Γ = +0.0045π m²/s. Their total is zero. The filled disk is not admitted.

   The annulus stays within the smooth, zero-curl field, so Stokes gives zero flux. A filled disk would include the excluded inner region, where no velocity or vorticity is supplied. Ignoring the inner boundary would change the contour rather than evaluate the same surface.

5. Assume the stated smooth Kelvin conditions for a straight material vortex tube along z, with counterclockwise loop viewed from +z. Initial circulation is −0.010 m²/s. Uniform axial vorticity occupies initial area 0.002 m² and later area 0.001 m². Material lengths are 0.020 and 0.040 m; density stays 1000 kg/m³. Compare circulation, axial vorticity, volume and mass.

   Show the answer and reasoning

   Circulation remains −0.010 m²/s. Axial vorticity changes from −5 to −10 s⁻¹. Both volumes are 0.000040 m³ and both masses 0.040 kg.

   Kelvin conserves the material-loop circulation under the stipulated inviscid, barotropic and conservative-force conditions. For uniform axial vorticity, Γ = ω_z A, so halving area doubles its signed magnitude. Area times material length is unchanged, and mass is density times volume. The two supplied states illustrate incompressible three-dimensional stretching; no full dynamics or core profile has been solved.

What to carry forward

• Vorticity is local curl of velocity, with units s⁻¹; local spin is ω/2. A particular material line can rotate differently because it also deforms.
• Straight streamlines can have nonzero vorticity, and circular streamlines can be irrotational within their declared domain.
• Circulation is an oriented closed-loop velocity integral with units m²/s. Reversing the contour changes its sign.
• Stokes’ theorem requires a smooth admitted spanning surface and compatible orientation. Zero curl on a domain with a hole need not give zero circulation around that hole.
• Kelvin’s theorem follows a material loop and needs the stated smooth inviscid pressure and body-force conditions; it is not a general conservation law for any fixed contour.
• Conserved circulation can coexist with stronger three-dimensional vorticity when a material vortex tube stretches and its cross-section shrinks.