Multivariable calculus

Divergence of the Rotation Field F = (-y, x)

Q: How can one field have divergence 0 and curl 2?

They read different partials. Divergence uses \frac{\partial P}{\partial x} and \frac{\partial Q}{\partial y}, which are both 0 here. Curl uses \frac{\partial Q}{\partial x} and \frac{\partial P}{\partial y}, which are 1 and -1.

For the rotation field F(x, y) = (-y, x) the divergence is 0 at every point. The first component contains no x and the second contains no y, so both diagonal partials are derivatives of a constant. The field spins but never spreads: it is incompressible. Its curl, by contrast, is 2.

\mathbf{F}(x,y) = \langle -y,\ x \rangle

\frac{\partial P}{\partial x} = \frac{\partial}{\partial x}\left(-y\right) = 0

\frac{\partial Q}{\partial y} = \frac{\partial}{\partial y}\left(x\right) = 0

\nabla \cdot \mathbf{F} = 0 + 0 = 0

Both diagonal partials are derivatives of a constant

Divergence pairs each component with its own variable, so the work is $\frac{\partial P}{\partial x}$ and $\frac{\partial Q}{\partial y}$ and nothing else.

When you differentiate $P = -y$ with respect to $x$ , the variable $y$ is held fixed. A held-fixed variable is a number, and the derivative of a number is zero. The same argument kills the second term.

\frac{\partial}{\partial x}\left(-y\right) = 0, \qquad \frac{\partial}{\partial y}\left(x\right) = 0

So $\nabla \cdot \mathbf{F} = 0$ everywhere. Notice what made this happen: each component is a function of the other variable only, so the diagonal partials had nothing to bite on.

The mistake: reading zero divergence as no motion

A field with zero divergence is not a field with zero arrows. Every arrow here has length $\sqrt{x^2 + y^2}$ , and at $(3,0)$ the arrow $\langle 0, 3 \rangle$ is pointing hard upward. Plenty is happening.

Zero divergence says something narrower: no net expansion. Drop a small disc into this flow and it will be carried around the origin and rotated, but its area will not change. That is what incompressible means.

The second half of the mistake is concluding that a divergence-free field is boring. Compute the other two partials and the motion shows up immediately.

\text{scalar curl} = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 - (-1) = 2

What the pair of numbers tells you

Divergence and curl answer different questions about the same field, which is why this example appears in every course.

Divergence $0$ : no source, no sink, area is preserved.
Curl $2$ : constant counterclockwise spin, the same at every point.
Taken together: rigid rotation, like a turntable, at angular speed $1$ .

Compare with $\langle x, y \rangle$ , which has divergence $2$ and curl $0$ . Between the two examples you can see that the diagonal partials and the off-diagonal partials really are measuring separate things.

Checked on every build: each component is differentiated numerically with respect to its own variable, and the sum is compared against the divergence above at six sample points.

Frequently asked questions

Does zero divergence mean the vector field is zero?

No. It means no net outflow from any small region. Here the arrows are long far from the origin, but the flow only circles: what enters a small disc on one side leaves on the other.

How can one field have divergence 0 and curl 2?

They read different partials. Divergence uses $\frac{\partial P}{\partial x}$ and $\frac{\partial Q}{\partial y}$ , which are both $0$ here. Curl uses $\frac{\partial Q}{\partial x}$ and $\frac{\partial P}{\partial y}$ , which are $1$ and $-1$ .