Multivariable calculus

Curl of the Shear Flow (z^2, 0, 0)

Q: Is this field conservative?

No. Its curl is nonzero wherever z \neq 0, and a conservative field needs zero curl on the whole region. You can also see the failure directly: a potential would need \frac{\partial f}{\partial x} = z^2, giving f = xz^2 + g(y,z), and then \frac{\partial f}{\partial z} = 2xz + \frac{\partial g}{\partial z} cannot be zero, since g has no x.

Q: What is the divergence of this field?

Zero. P = z^2 has no x, and Q and R are identically zero, so all three matched partials vanish. The field is therefore incompressible and rotational, the exact opposite pairing to \langle x^2, y^2, z^2 \rangle, which is compressible and irrotational.

The curl of the field with components z squared, 0 and 0 is the vector (0, 2z, 0). Only the middle component survives, from the z-partial of the first component, which is 2z. Every flow line is straight and parallel to the x-axis, so this is the standard example that curl measures local spin, not curved motion.

P(x,y,z) = z^2

Q(x,y,z) = 0

R(x,y,z) = 0

\nabla \times \mathbf{F} = \langle 0,\ 2z,\ 0 \rangle

Only one partial is not zero

With $Q = R = 0$ , four of the six partials in the curl formula are gone before you start. $P = z^2$ has no $y$ in it either, so the only survivor is $\frac{\partial P}{\partial z} = 2z$ .

\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} = 0 - 0 = 0

\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} = 2z - 0 = 2z

\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0 - 0 = 0

So the curl points along $\mathbf{j}$ , perpendicular to the flow direction $\mathbf{i}$ and perpendicular to the direction $\mathbf{k}$ in which the speed changes. That is the general rule for a shear, and it is worth remembering as a geometric fact rather than a formula.

Why straight flow lines can still spin

Every particle here moves in the $+x$ direction and nothing ever turns a corner. What varies is speed: a particle at height $z$ moves at $z^2$ , so at $z = 1$ the speed is $1$ and at $z = 2$ it is $4$ .

Drop a small paddle wheel into the flow with its axle along the $y$ -axis, so its spokes reach up and down. The upper spoke sits in faster fluid than the lower one, so the wheel turns, carrying the $+z$ direction toward $+x$ . That is a rotation about the $+y$ axis, matching $2z\,\mathbf{j}$ for $z > 0$ .

Below the plane $z = 0$ the speed increases as $z$ decreases, so the wheel turns the other way and the curl flips sign, which is what $2z$ says.
On the plane $z = 0$ the speed profile is flat to first order and the curl is the zero vector.
The curl grows without bound as you move away from that plane, because the shear itself does.

The mistake: judging curl from the shape of the picture

This page exists because of one sentence students say: the flow lines are straight, so there is no rotation. This field is the counterexample, and it is not exotic; it is what a river does near its bed and what air does near a wing.

The converse error is just as common. Circular flow lines look like rotation, but the field $\left\langle \frac{-y}{x^2+y^2},\ \frac{x}{x^2+y^2},\ 0 \right\rangle$ has circular flow lines and zero curl everywhere it is defined.

Curl is a derivative, not a shape. It compares the flow at a point with the flow just beside it, so the only reliable way to find it is to differentiate.

The middle component is recomputed numerically on every build from P, Q and R, so the factor of two above is measured rather than asserted.

Frequently asked questions

Is this field conservative?

No. Its curl is nonzero wherever $z \neq 0$ , and a conservative field needs zero curl on the whole region. You can also see the failure directly: a potential would need $\frac{\partial f}{\partial x} = z^2$ , giving $f = xz^2 + g(y,z)$ , and then $\frac{\partial f}{\partial z} = 2xz + \frac{\partial g}{\partial z}$ cannot be zero, since $g$ has no $x$ .

What is the divergence of this field?

Zero. $P = z^2$ has no $x$ , and $Q$ and $R$ are identically zero, so all three matched partials vanish. The field is therefore incompressible and rotational, the exact opposite pairing to $\langle x^2, y^2, z^2 \rangle$ , which is compressible and irrotational.