Vorticity and Circulation in Planar Flow

Vorticity and Circulation

A moving fluid can travel through a region while also tending to turn locally. The scalar curl of its velocity field describes this local rotational tendency. Circulation, in contrast, measures the accumulated tangential motion along a curve. These two quantities are connected: under suitable smoothness and orientation conditions, the total vorticity over a planar region equals the circulation around its boundary.

This connection is useful in both directions. If the vorticity is simple to integrate, it can make a boundary calculation easier. If the velocity field is simple along the boundary, direct circulation can check an area calculation. The sign of the answer matters: it records the direction of the circulation relative to the chosen orientation.

Scalar Curl and Local Rotation

Let a planar velocity field be \(\mathbf{v}(x,y)=(P(x,y),Q(x,y))\), where \(P\) is the horizontal velocity component and \(Q\) is the vertical component. Assume that both component functions have continuous first partial derivatives in the region under consideration.

Definition: The scalar curl, or planar vorticity, of \(\mathbf{v}\) is \(\omega(x,y)=\frac{\partial Q}{\partial x}(x,y)-\frac{\partial P}{\partial y}(x,y)\). It is the component perpendicular to the plane of the three-dimensional curl of \((P,Q,0)\).

The order of the terms is important: differentiate the vertical component \(Q\) with respect to \(x\), then subtract the derivative of the horizontal component \(P\) with respect to \(y\). Positive vorticity corresponds to a counterclockwise rotational component; negative vorticity corresponds to a clockwise one.

To see why vorticity represents local rotation, consider the derivative matrix of the field at a point:

D\mathbf{v} = \begin{pmatrix} P_x & P_y\\ Q_x & Q_y \end{pmatrix}.

Near that point, the field is approximated by its value at the point plus this linear transformation applied to a small displacement. The rotational part of the linear transformation is its antisymmetric part. Writing \(\omega=Q_x-P_y\), that part is

\frac{1}{2}\left(D\mathbf{v}-(D\mathbf{v})^{\top}\right) = \begin{pmatrix} 0 & -\omega/2\\ \omega/2 & 0 \end{pmatrix}.

This matrix rotates a displacement counterclockwise at angular velocity \(\omega/2\) in the linear approximation. Thus the scalar vorticity is twice the local angular velocity of the rotational component. This is an instantaneous, local description; it does not say that every nearby particle follows a circular path.

Circulation Around a Curve

Let \(C\) be a piecewise smooth, oriented curve in the plane. The circulation of \(\mathbf{v}\) along \(C\) is the line integral of the component of velocity tangent to the curve. If \(\mathbf{r}(t)=(x(t),y(t))\) parametrizes \(C\) in its chosen direction, then \(d\mathbf{r}=(dx,dy)\) and

\oint_C \mathbf{v}\cdot d\mathbf{r} = \oint_C P\,dx+Q\,dy.

The integral is signed. Reversing the direction of traversal reverses the sign of circulation. For a simple closed boundary, the positive orientation is counterclockwise: the region lies to the left as the boundary is traversed.

Interpretation: Circulation measures the net tangential component of the field along an oriented curve. Vorticity measures the local rotational tendency at a point. The planar Stokes theorem says that the area accumulation of this local quantity equals the circulation around the positively oriented boundary.

Why Small-Loop Circulation Measures Vorticity

The local relationship can be derived using a small rectangle centered at \((x_0,y_0)\), with horizontal side length \(h\) and vertical side length \(k\), traversed counterclockwise. Denote its left, right, bottom, and top coordinates by \(x_-,x_+,y_-,y_+\), respectively. The horizontal segments contribute \(\int_{x_-}^{x_+}P(x,y_-)\,dx-\int_{x_-}^{x_+}P(x,y_+)\,dx\). By the fundamental theorem of calculus in the \(y\)-variable, this is \(-\int_{x_-}^{x_+}\int_{y_-}^{y_+}P_y(x,y)\,dy\,dx\).

The upward right edge and downward left edge contribute \(\int_{y_-}^{y_+}Q(x_+,y)\,dy-\int_{y_-}^{y_+}Q(x_-,y)\,dy\). Applying the fundamental theorem of calculus in the \(x\)-variable gives \(\int_{y_-}^{y_+}\int_{x_-}^{x_+}Q_x(x,y)\,dx\,dy\). Adding the two contributions yields the exact identity

\oint_{\partial R}\mathbf{v}\cdot d\mathbf{r} = \int_{x_-}^{x_+}\int_{y_-}^{y_+} \left(Q_x-P_y\right)\,dy\,dx.

Divide both sides by the rectangle’s area \(hk\). The right side becomes the average value of \(\omega\) on the rectangle. As the rectangle shrinks to \((x_0,y_0)\), continuity of \(\omega\) implies that this average approaches \(\omega(x_0,y_0)\). Therefore,

\lim_{\substack{h,k\to 0\\h,k>0}} \frac{\text{counterclockwise circulation around the rectangle}}{hk} = \omega(x_0,y_0).

This explains the local meaning of scalar curl: it is the limiting circulation per unit area around a small positively oriented loop.

The Planar Stokes Theorem

The identity for a rectangle extends to suitable planar regions. In the plane, this result is also called Green’s theorem in circulation form.

Theorem: Let \(D\) be a bounded planar region whose boundary \(C=\partial D\) is a simple, piecewise smooth closed curve. Suppose \(P\) and \(Q\) have continuous first partial derivatives on an open region containing \(D\). Orient \(C\) counterclockwise. Then \(\displaystyle \oint_C P\,dx+Q\,dy=\iint_D(Q_x-P_y)\,dA=\iint_D\omega\,dA.\)

The assumptions ensure that the derivatives and integrals in the formula are defined and that the field has no singularity on or inside the region. The orientation is part of the theorem: if the boundary is traversed clockwise, the line integral has the opposite sign. The equality says that contributions along internal interfaces cancel when small pieces of the region are assembled, leaving the circulation around the outer boundary.

Find the scalar curl.
Compute \(\omega=Q_x-P_y\) for the given velocity field.

Check the region and orientation.
Confirm the required smoothness and use a counterclockwise boundary for the stated formula.

Choose the easier integral.
Evaluate \(\iint_D\omega\,dA\), or calculate \(\oint_{\partial D}P\,dx+Q\,dy\) directly if that is simpler.

Compare signed values.
The area integral and positively oriented boundary circulation must agree.

Worked Example: Circulation Around a Rectangle

Let \(\mathbf{v}(x,y)=(xy,x^2)\) and let \(D=[0,2]\times[0,1]\). Find the counterclockwise circulation around \(\partial D\) using the planar Stokes theorem, then check by direct boundary integration.

Here \(P=xy\) and \(Q=x^2\), so

Q_x=2x,\qquad P_y=x,\qquad \omega=Q_x-P_y=x.

The area integral is

\iint_D\omega\,dA=\int_0^2\int_0^1 x\,dy\,dx=\int_0^2 x(1)\,dx=\left[\frac{x^2}{2}\right]_0^2=2.

For a direct check, traverse the bottom from \((0,0)\) to \((2,0)\), the right edge upward, the top from right to left, and the left edge downward.

\begin{aligned} \text{Bottom: }&\int_0^2 P(x,0)\,dx=0,\\ \text{Right: }&\int_0^1 Q(2,y)\,dy=\int_0^1 4\,dy=4,\\ \text{Top: }&\int_2^0 P(x,1)\,dx=\int_2^0 x\,dx=-2,\\ \text{Left: }&\int_1^0 Q(0,y)\,dy=0. \end{aligned}

The total boundary circulation is \(0+4-2+0=2\), equal to the area integral. This verifies the theorem’s result and the counterclockwise sign.

Worked Example: A Different Rectangular Region

Take \(\mathbf{v}(x,y)=(y^2,3xy)\) on \(D=[-1,1]\times[0,2]\). Compute the counterclockwise circulation by integrating vorticity, and verify it along all four sides.

With \(P=y^2\) and \(Q=3xy\), the derivatives and scalar curl are

Q_x=3y,\qquad P_y=2y,\qquad \omega=3y-2y=y.

Thus the area integral is

\iint_D\omega\,dA=\int_{-1}^{1}\int_0^2 y\,dy\,dx=\int_{-1}^{1}\left[\frac{y^2}{2}\right]_0^2dx=\int_{-1}^{1}2\,dx=4.

Now calculate each boundary contribution. On the bottom, \(y=0\), so both \(P\) and \(Q\) vanish and the contribution is \(0\). On the right edge \(x=1\), \(y\) runs from \(0\) to \(2\), so the contribution is \(\int_0^2 3y\,dy=6\). On the top edge \(y=2\), \(x\) runs from \(1\) to \(-1\), so the contribution is \(\int_1^{-1}4\,dx=-8\). On the left edge \(x=-1\), \(y\) runs from \(2\) down to \(0\), so the contribution is \(\int_2^0(-3y)\,dy=6\).

The direct total is \(0+6-8+6=4\), which agrees with the vorticity integral. In particular, the top contribution is negative because that edge is traversed in the direction of decreasing \(x\).

Worked Example: Circulation Around a Circle

Let \(\mathbf{v}(x,y)=(0,x)\), and let \(D\) be the unit disk \(x^2+y^2\leq 1\). Find the circulation around its counterclockwise boundary in two ways.

Here \(P=0\) and \(Q=x\), so \(Q_x=1\), \(P_y=0\), and \(\omega=1\). The area integral is the area of the unit disk:

\iint_D\omega\,dA=\iint_D1\,dA=\pi.

For a direct calculation, parametrize the counterclockwise circle by \(x=\cos t,\ y=\sin t\), for \(0\leq t\leq 2\pi\). Then \(dx=-\sin t\,dt\) and \(dy=\cos t\,dt\). Since \(P=0\) and \(Q=x=\cos t\),

\oint_{\partial D}P\,dx+Q\,dy=\int_0^{2\pi}\cos t\cos t\,dt=\int_0^{2\pi}\cos^2 t\,dt=\pi.

The direct integral equals the area integral, so the circulation is \(\pi\). The result also matches the fact that this field has unit vorticity throughout a region of area \(\pi\).

A Sign Error to Avoid

A frequent mistake is reversing the order in the scalar curl. For the field in the previous rectangular example, an incorrect calculation might say

P_y-Q_x=2y-3y=-y, \qquad \iint_D(-y)\,dA=-4.

This gives the wrong sign because the planar scalar curl is \(Q_x-P_y\), not \(P_y-Q_x\). With the correct order, \(\omega=3y-2y=y\), and the area integral is \(4\). The direct counterclockwise boundary calculation also gives \(4\), exposing the sign error. If the curve were instead traversed clockwise, its circulation would be \(-4\), but that reversal would not change the definition of \(\omega\).

Using the Relationship Well

Vorticity and circulation describe different scales of the same rotational behavior. Vorticity is pointwise; circulation is accumulated along a curve. The planar Stokes theorem relates them only when the field is sufficiently smooth throughout the region and the curve is oriented consistently. It is therefore important to check both the hypotheses and the sign convention before choosing which integral to compute.

When applying the theorem, first calculate \(Q_x-P_y\) carefully. Then determine whether the region’s geometry makes its area integral manageable. If a boundary integral is used as a check, parameterize each segment in its actual traversal direction and retain the resulting signs. Agreement of the two methods is a strong check on both the derivatives and the orientation.

Check Your Understanding

Use the definitions and theorem above to answer the following questions.

For \(\mathbf{v}(x,y)=(P,Q)\), which partial derivatives form its scalar curl, and in what order?
What is the local angular velocity of the rotational component in terms of scalar vorticity?
State the planar Stokes theorem, including its boundary orientation and smoothness requirements.
If a counterclockwise circulation is \(7\), what is the circulation around the same curve traversed clockwise?
For \(\mathbf{v}(x,y)=(y,2x)\), calculate the scalar curl and determine whether it is positive, negative, or zero.

Vorticity and Circulation in Planar Flow

What You'll Learn