The circulation of a planar vortex
A planar vortex moves around a fixed axis rather than directly toward or away from it. In the ideal point-vortex model, the flow circles the axis, and its speed increases as the distance to the axis decreases. The vector field is undefined on the axis itself.
That missing point matters when calculating circulation. Direct parametrization gives a nonzero circulation around any circle enclosing the axis. Yet the classical curl of the field is zero at every point where the field is defined. These statements are consistent: Stokes’ theorem requires a sufficiently smooth field on a neighborhood of the entire spanning surface, and that requirement fails for a disk containing the singular axis.
Here \(\Gamma\) is a signed constant. Its sign determines the direction of circulation: positive \(\Gamma\) gives counterclockwise motion, and negative \(\Gamma\) gives clockwise motion. The speed at distance \(r\) from the axis is \(|\Gamma|/(2\pi r)\).
For a parametrized curve \(\mathbf{r}(t)=(x(t),y(t))\), the circulation of \(\mathbf{F}\) along the curve is the line integral \(\int \mathbf{F}\cdot d\mathbf{r}\). Its value depends on the direction in which the curve is traversed. We take a counterclockwise circle to have positive orientation when its surface normal points in the positive \(z\)-direction.
Direct calculation around a circle
Consider a circle of radius \(R>0\) centered on the vortex axis, traversed counterclockwise. A parametrization is \(\mathbf{r}(t)=(R\cos t,R\sin t)\), for \(0\leq t\leq 2\pi\). Along this curve the field is tangent to the circle, so the dot product with the tangent displacement can be evaluated directly.
Worked Example: Circulation around a circle of radius 3
Take \(\Gamma=12\) and let \(R=3\). Parametrize the counterclockwise circle by \(\mathbf{r}(t)=(3\cos t,3\sin t)\), for \(0\leq t\leq 2\pi\). Its derivative is \(\mathbf{r}'(t)=(-3\sin t,3\cos t)\).
On the circle, \(x^2+y^2=9\), so the field is \(\mathbf{F}(\mathbf{r}(t))=\left(-\frac{12(3\sin t)}{2\pi(9)},\frac{12(3\cos t)}{2\pi(9)}\right)=\left(-\frac{2\sin t}{\pi},\frac{2\cos t}{\pi}\right)\).
The dot product with the tangent displacement is \(\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)=(-2\sin t/\pi)(-3\sin t)+(2\cos t/\pi)(3\cos t)=\frac{6}{\pi}(\sin^2 t+\cos^2 t)=\frac{6}{\pi}\).
Therefore the circulation is \(\int_0^{2\pi}\frac{6}{\pi}\,dt=\frac{6}{\pi}(2\pi)=12\). As a check, the field’s tangential speed on the circle is \(12/(2\pi\cdot3)=2/\pi\), and the circumference is \(2\pi\cdot3=6\pi\). Their product is \((2/\pi)(6\pi)=12\), as obtained from the line integral.
The cancellation of the radius in this calculation is the key feature. A larger circle has a longer circumference, but the vortex speed along it is correspondingly smaller. The circulation depends on \(\Gamma\), not on the radius of a circle centered at the axis.
Why the classical curl vanishes away from the axis
Write the Cartesian components as \(P(x,y)=-\Gamma y/[2\pi(x^2+y^2)]\) and \(Q(x,y)=\Gamma x/[2\pi(x^2+y^2)]\). The planar curl is the \(z\)-component \(\partial Q/\partial x-\partial P/\partial y\). The derivatives exist wherever \(x^2+y^2>0\), but not at the origin.
For \(x^2+y^2>0\), differentiating gives \(\frac{\partial Q}{\partial x}=\frac{\Gamma}{2\pi}\frac{y^2-x^2}{(x^2+y^2)^2}\) and \(\frac{\partial P}{\partial y}=-\frac{\Gamma}{2\pi}\frac{x^2-y^2}{(x^2+y^2)^2}=\frac{\Gamma}{2\pi}\frac{y^2-x^2}{(x^2+y^2)^2}\). The two derivatives are equal, so their difference is zero.
Using curl integration on an annulus
Stokes’ theorem relates a circulation around an oriented boundary to the integral of curl over a spanning surface. It requires the field to be continuously differentiable on an open neighborhood of that surface. A disk centered at the origin contains the point where the vortex field is undefined, so that disk does not satisfy the required condition.
An annulus centered at the origin, with inner radius \(a>0\) and outer radius \(R>a\), does satisfy the condition. The origin lies outside the annulus, and the field is smooth on a neighborhood of it. The annulus has two boundary circles. With the upward normal, the outer circle is oriented counterclockwise and the inner circle clockwise.
Worked Example: Curl integration on an annulus
Let \(\Gamma=4\pi\), and take the annulus \(1\leq r\leq3\) with upward normal. Since the origin is excluded, the field is smooth on a neighborhood of the annulus. Its curl is zero throughout the region, so \(\displaystyle \iint_S (\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS=\iint_S 0\,dA=0\).
Stokes’ theorem therefore gives the sum of the oriented circulations on both boundary components as zero. On the outer circle, the direction is counterclockwise, so its circulation is \(4\pi\). On the inner circle, the direction is clockwise, so its circulation is \(-4\pi\). Their sum is \(4\pi+(-4\pi)=0\), matching the curl integral.
As a check on the inner-boundary sign, traversing that circle counterclockwise would give \(4\pi\), by the direct circle calculation. The required clockwise orientation reverses that value to \(-4\pi\).
The inner boundary is essential. Curl integration over the annulus gives the sum of both boundary circulations, not the outer circulation alone. The nonzero outer circulation is balanced by the oppositely oriented inner boundary.
Why a disk containing the axis is different
It is tempting to integrate the zero curl over the entire disk and conclude that the circulation around its boundary must be zero. That reasoning misuses Stokes’ theorem. The field fails to be defined at the disk’s center, so it is not continuously differentiable on a neighborhood of the whole disk. The hypotheses needed to equate the disk’s curl integral with its boundary circulation are absent.
A useful way to make the distinction precise is to remove a small disk of radius \(\varepsilon>0\) from the larger disk. The remaining region is an annulus, on which Stokes’ theorem does apply. Its curl integral is zero, and its boundary consists of the outer circle and the small inner circle, oriented in opposite senses. The inner circle does not disappear from the theorem simply because it is small: its circulation remains \(-\Gamma\) when oriented clockwise.
Worked Example: Shrinking the excluded circle
Let \(\Gamma=10\) and consider a counterclockwise outer circle of radius \(5\). Direct parametrization gives outer circulation \(10\). Now remove a disk of radius \(\varepsilon\), where \(0<\varepsilon<5\). The remaining annulus has zero curl, so Stokes’ theorem gives total boundary circulation zero.
The outer boundary contributes \(10\). The inner boundary is clockwise. If it were traversed counterclockwise, its circulation would be \(10\), regardless of \(\varepsilon\); with the required clockwise orientation, it contributes \(-10\). Thus the total is \(10+(-10)=0\) for every allowed \(\varepsilon\).
As \(\varepsilon\) decreases toward zero, the inner circle’s length \(2\pi\varepsilon\) decreases, but the speed on it is \(10/(2\pi\varepsilon)\). Their product remains \([10/(2\pi\varepsilon)](2\pi\varepsilon)=10\). The inner-boundary circulation therefore does not tend to zero, and its clockwise value remains \(-10\). This check explains why the limiting annular calculation cannot be replaced by applying Stokes’ theorem directly to the singular disk.
A complete comparison of the two methods
The direct line integral and the curl method answer related questions, but their permitted regions differ. Direct parametrization needs the field to be defined along the curve. Applying Stokes’ theorem also requires a smooth field on a neighborhood of the chosen spanning surface. For a circle enclosing the vortex axis, the boundary is well-defined even though the filled disk is not an admissible surface for the theorem.
The vortex field is defined for \(r>0\), but not on the axis \(r=0\).
Use a parametrization if the curve avoids the axis; account for its direction of traversal.
Stokes’ theorem applies only when the field is continuously differentiable on a neighborhood of the entire surface.
Use an annulus and include both oriented boundary circles in the circulation balance.
For a circle that encloses the origin, direct parametrization gives \(\Gamma\) in the counterclockwise direction. The classical curl is zero away from the origin, and its integral over an annular region is zero. There is no contradiction: the annulus has an inner boundary whose clockwise circulation cancels the outer one. The filled disk cannot be used in the classical theorem because it contains the singular point.
Check Your Understanding
Use the field \(\mathbf{F}(x,y)=(-\Gamma y/[2\pi(x^2+y^2)],\Gamma x/[2\pi(x^2+y^2)])\) where it is defined.
- For a counterclockwise circle centered at the origin, what is the circulation in terms of \(\Gamma\), and does it depend on the radius?
- What sign does the circulation have if the same circle is traversed clockwise?
- Where is the classical curl defined, and what is its value there?
- Why can Stokes’ theorem be applied to an annulus centered at the origin but not to the full disk?
- For an annulus with upward normal, which direction is assigned to its inner boundary, and why must that boundary be included in the circulation sum?