The second operation on a vector field that we examine is the curl, which measures the extent of rotation of the field about a point. Suppose that [latex]{\bf{F}}[/latex] represents the velocity field of a fluid. Then, the curl of [latex]{\bf{F}}[/latex] at point [latex]P[/latex] is a vector that measures the tendency of particles near [latex]P[/latex] to rotate about the axis that points in the direction of this vector. The magnitude of the curl vector at [latex]P[/latex] measures how quickly the particles rotate around this axis. In other words, the curl at a point is a measure of the vector field’s “spin” at that point. Visually, imagine placing a paddlewheel into a fluid at [latex]P[/latex], with the axis of the paddlewheel aligned with the curl vector (Figure 1). The curl measures the tendency of the paddlewheel to rotate.

Figure 1. To visualize curl at a point, imagine placing a small paddlewheel into the vector field at a point.

Consider the vector fields in Figure 1 from the Divergence learning objective. In part (a), the vector field is constant and there is no spin at any point. Therefore, we expect the curl of the field to be zero, and this is indeed the case. Part (b) shows a rotational field, so the field has spin. In particular, if you place a paddlewheel into a field at any point so that the axis of the wheel is perpendicular to a plane, the wheel rotates counterclockwise. Therefore, we expect the curl of the field to be nonzero, and this is indeed the case (the curl is [latex]2{\bf{k}}[/latex]).

To see what curl is measuring globally, imagine dropping a leaf into the fluid. As the leaf moves along with the fluid flow, the curl measures the tendency of the leaf to rotate. If the curl is zero, then the leaf doesn’t rotate as it moves through the fluid.

The definition of curl can be difficult to remember. To help with remembering, we use the notation [latex]\nabla\times{\bf{F}}[/latex] to stand for a “determinant” that gives the curl formula:

[latex]\large{\begin{vmatrix} {\bf{i}}&{\bf{j}}&{\bf{k}} \\ \frac{\partial}{\partial{x}}&\frac{\partial}{\partial{y}}&\frac{\partial}{\partial{z}} \\ P&Q&R \end{vmatrix}}[/latex].

The determinant of this matrix is

[latex]\large{(R_y-Q_z){\bf{i}}-(R_x-P_z){\bf{j}}+(Q_x-P_y){\bf{k}}=(R_y-Q_z){\bf{i}}+(P_z-R_x){\bf{j}}+(Q_x-P_y){\bf{k}}=\text{curl }{\bf{F}}}[/latex].

Thus, this matrix is a way to help remember the formula for curl. Keep in mind, though, that the word determinant is used very loosely. A determinant is not really defined on a matrix with entries that are three vectors, three operators, and three functions.

If [latex]{\bf{F}}=\langle{P},Q\rangle[/latex] is a vector field in [latex]\mathbb{R}^2[/latex], then the curl of [latex]{\bf{F}}[/latex], by definition, is

[latex]\large{\text{curl }{\bf{F}}=(Q_x-P_y){\bf{k}}=\left(\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}\right){\bf{k}}}[/latex].

Example: finding the curl of a three-dimensional vector field

Find the curl of [latex]{\bf{F}}(P,Q,R)=\langle{x}^2z,e^y+xz,xyz\rangle[/latex].

Show Solution

The curl is

[latex]\begin{aligned} \text{curl }{\bf{F}}&=\nabla\times{\bf{F}} \\ &=\begin{vmatrix} {\bf{i}}&{\bf{j}}&{\bf{k}} \\ \frac{\partial}{\partial{x}}&\frac{\partial}{\partial{y}}&\frac{\partial}{\partial{z}} \\ P&Q&R \end{vmatrix} \\ &=(R_y-Q)z){\bf{i}}+(P_z-R_x){\bf{j}}+(Q_x-P_y){\bf{k}} \\ &=(xz-z){\bf{i}}+(x^2-yz){\bf{j}}+z{\bf{k}} \end{aligned}[/latex].

try it

Find the curl of [latex]{\bf{F}}=\langle\sin{x}\cos{z},\sin{y}\sin{z},\cos{x}\cos{y}\rangle[/latex] at point [latex]\left(0,\frac{\pi}2,\frac{\pi}2\right)[/latex].

Show Solution

[latex]-{\bf{i}}[/latex].

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 6.43” here (opens in new window).

Try It

Example: finding the curl of a two-dimensional vector field

Find the curl of [latex]{\bf{F}}=\langle{P},Q\rangle=\langle{y},0\rangle[/latex].

Show Solution

Notice that this vector field consists of vectors that are all parallel. In fact, each vector in the field is parallel to the x-axis. This fact might lead us to the conclusion that the field has no spin and that the curl is zero. To test this theory, note that

[latex]\large{\text{curl }{\bf{F}}=(Q_x-P_t){\bf{k}}=-{\bf{k}}\ne0}[/latex].

Therefore, this vector field does have spin. To see why, imagine placing a paddlewheel at any point in the first quadrant (Figure 2). The larger magnitudes of the vectors at the top of the wheel cause the wheel to rotate. The wheel rotates in the clockwise (negative) direction, causing the coefficient of the curl to be negative.

Figure 2. Vector field [latex]{\bf{F}}(x,y)=\langle{y},0\rangle[/latex] consists of vectors that are all parallel.

Note that if [latex]{\bf{F}}=\langle{P},Q\rangle[/latex] is a vector field in a plane, then [latex]\text{curl }{\bf{F}}\cdot{\bf{k}}=(Q_x-P_y){\bf{k}}\cdot{\bf{k}}=Q_x-P_y[/latex]. Therefore, the circulation form of Green’s theorem is sometimes written as

[latex]\large{\displaystyle\oint_C{\bf{F}}\cdot{d}{\bf{r}}=\displaystyle\iint_D\text{curl }{\bf{}}\cdot{\bf{k}}dA}[/latex],

where [latex]C[/latex] is a simple closed curve and [latex]D[/latex] is the region enclosed by [latex]C[/latex]. Therefore, the circulation form of Green’s theorem can be written in terms of the curl. If we think of curl as a derivative of sorts, then Green’s theorem says that the “derivative” of [latex]{\bf{F}}[/latex] on a region can be translated into a line integral of [latex]{\bf{F}}[/latex] along the boundary of the region. This is analogous to the Fundamental Theorem of Calculus, in which the derivative of a function [latex]f[/latex] on line segment [latex][a,b][/latex] can be translated into a statement about [latex]f[/latex] on the boundary of [latex][a,b][/latex]. Using curl, we can see the circulation form of Green’s theorem is a higher-dimensional analog of the Fundamental Theorem of Calculus.

We can now use what we have learned about curl to show that gravitational fields have no “spin.” Suppose there is an object at the origin with mass [latex]m_1[/latex] at the origin and an object with mass [latex]m_2[/latex]. Recall that the gravitational force that object 1 exerts on object 2 is given by field

[latex]\large{{\bf{F}}(x,y,z)=-Gm_1m_2\left\langle\frac{x}{(x^2+y^2+z^2)^{3/2}},\frac{y}{(x^2+y^2+z^2)^{3/2}},\frac{z}{(x^2+y^2+z^2)^{3/2}}\right\rangle}[/latex].

Example: determining the spin of a gravitational field

Show that a gravitational field has no spin.

Show Solution

To show that [latex]{\bf{F}}[/latex] has no spin, we calculate its curl. Let [latex]P(x,y,z)=\frac{x}{(x^2+y^2+z^2)^{3/2}}[/latex], [latex]Q(x,y,z)=\frac{y}{(x^2+y^2+z^2)^{3/2}}[/latex], and [latex]R(x,y,z)=\frac{z}{(x^2+y^2+z^2)^{3/2}}[/latex]. Then,

[latex]\begin{aligned} \text{curl }{\bf{F}}&=-Gm_1m_2[(R_y-Q_z){\bf{i}}+(P_z-R_x){\bf{j}}+(Q_x-P_y){\bf{k}}] \\ &=-Gm_1m_2 \left[\begin{aligned} &\left(\frac{-3yz}{(x^2+y^2+z^2)^{5/2}}-\left(\frac{-3yz}{(x^2+y^2+z^2)^{5/2}}\right)\right){\bf{i}} \\ &+\left(\frac{-3xz}{(x^2+y^2+z^2)^{5/2}}-\left(\frac{-3xz}{(x^2+y^2+z^2)^{5/2}}\right)\right){\bf{j}} \\ &+\left(\frac{-3xy}{(x^2+y^2+z^2)^{5/2}}-\left(\frac{-3xy}{(x^2+y^2+z^2)^{5/2}}\right)\right){\bf{k}} \end{aligned}\right]\\ &=0\ \end{aligned}[/latex].

Since the curl of the gravitational field is zero, the field has no spin.

try it

Field [latex]{\bf{v}}(x,y)=\left\langle-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2}\right\rangle[/latex] models the flow of a fluid. Show that if you drop a leaf into this fluid, as the leaf moves over time, the leaf does not rotate.

Show Solution

curl [latex]{\bf{v}}=0[/latex].