Drop a tiny paddle wheel into this flow. Which way does it spin - and why?

The Other Six Derivatives

In §3.4 we stared at the $3 \times 3$ Jacobian matrix of a vector field and grabbed the diagonal - those three terms gave us divergence, a single number measuring net outflow. But we deliberately left six entries untouched:

$\dfrac{\partial v_x}{\partial x}$ $\dfrac{\partial v_x}{\partial y}$ $\dfrac{\partial v_x}{\partial z}$ $\dfrac{\partial v_y}{\partial x}$ $\dfrac{\partial v_y}{\partial y}$ $\dfrac{\partial v_y}{\partial z}$ $\dfrac{\partial v_z}{\partial x}$ $\dfrac{\partial v_z}{\partial y}$ $\dfrac{\partial v_z}{\partial z}$

The diagonal (muted) told us about stretching. Now the off-diagonal entries (orange) tell us about shearing: how flow in one direction changes as you move in a perpendicular direction. $\partial v_y / \partial x$ answers: "as I step in the $x$-direction, does the $y$-component of velocity change?" If the $y$-velocity increases as we move right but decreases as we move left, the field is twisting around us.

These six entries naturally pair up: $(\partial v_y / \partial x, \partial v_x / \partial y)$, $(\partial v_z / \partial y, \partial v_y / \partial z)$, $(\partial v_x / \partial z, \partial v_z / \partial x)$. Within each pair, the two terms pull in opposite rotational directions. Their differences are what matter - and those differences are precisely the three components of the curl.

The Curl Formula

Where divergence was a dot product ($\nabla \cdot \mathbf{v}$, producing a scalar), the curl is a cross product ($\nabla \times \mathbf{v}$, producing a vector). We can express it as a symbolic $3 \times 3$ determinant:

$$\nabla \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ v_x & v_y & v_z \end{vmatrix}$$
$$= \mathbf{i}\!\left(\frac{\partial v_z}{\partial y} - \frac{\partial v_y}{\partial z}\right)$$
$$\color{#dc2626}{-\;\mathbf{j}}\!\left(\frac{\partial v_z}{\partial x} - \frac{\partial v_x}{\partial z}\right)$$
$$+ \mathbf{k}\!\left(\frac{\partial v_y}{\partial x} - \frac{\partial v_x}{\partial y}\right)$$

Worked example. Take $\mathbf{v} = xz\,\mathbf{i} + xyz\,\mathbf{j} + y^2\,\mathbf{k}$. Then:

$\nabla \times \mathbf{v} = (2y - xy)\,\mathbf{i} - (0 - x)\,\mathbf{j} + (yz - 0)\,\mathbf{k} = (2y - xy)\,\mathbf{i} + x\,\mathbf{j} + yz\,\mathbf{k}$

Notice: curl eats a vector field and spits out a vector field. At every point in space, the curl gives us a vector whose direction is the axis of rotation and whose magnitude tells us how fast things spin around that axis.

What Curl Measures

Here's the physical picture. Imagine dropping a tiny paddle wheel into a fluid flow. If the flow pushes harder on one side of the wheel than the other, it spins. The curl tells us exactly how fast and which way:

Drag the paddle wheel through the field below. Watch it spin based on the local curl value. Try each preset to build intuition.

Curl (z-component): 0 Paddle wheel: stationary

Drag the paddle wheel through the field. Watch it spin based on the local curl.

For pure rotation $(-y, x)$: the paddle spins everywhere. We can verify: $(\nabla \times \mathbf{v})_z = \partial x / \partial x - \partial(-y) / \partial y = 1 - (-1) = 2$. Constant positive curl - uniform rotation.

For the source field $(x, y)$: the paddle wheel sits still. $(\nabla \times \mathbf{v})_z = \partial y / \partial x - \partial x / \partial y = 0 - 0 = 0$. The field pushes radially outward with no twist.

For shear $(y, 0)$: the paddle spins! $(\nabla \times \mathbf{v})_z = \partial(0) / \partial x - \partial y / \partial y = 0 - 1 = -1$. Layers slide past each other, and the imbalance in shearing makes the wheel rotate clockwise. This surprises many students - a flow that "looks" non-rotational actually has curl.

Playground - your own field

Type any two functions of $x$ and $y$. The page computes the curl numerically and colors each arrow. Purple = paddle spins counter-clockwise ($(\nabla \times \mathbf{v})_z > 0$), green = clockwise, gray = irrotational. Drag the paddle wheel to feel the local spin.

Supported: + - * / ^, parentheses, and sin cos tan exp log sqrt abs. Constants: pi, e. Examples: x*y, sin(x)*cos(y), -y/(x^2+y^2+0.5).

Curl at paddle: 0.00 Paddle wheel: stationary

Edit either field and the visualization updates instantly. Drag the paddle wheel to move it.

Curl in Three Dimensions

So far we've been watching 2D slices. But curl is really a 3D creature - at every point it produces a vector that can point in any direction, not just straight up or down. The ball below sits in a 3D vector field. The arrows show the flow, and the orange arrow through the ball shows the curl vector - its direction is the axis of rotation, and its length tells you how fast the fluid spins around that axis.

Orbit the scene by dragging. Try each preset and watch how the curl arrow changes direction.

$\nabla \times \mathbf{v}$ = (0, 0, 0)

Properties of the Curl

Curl behaves well with addition and scalar multiplication, just like divergence - but with a twist (pun intended). The two key rules:

$$\nabla \times (\mathbf{u} + \mathbf{v}) = \nabla \times \mathbf{u} + \nabla \times \mathbf{v}$$

Linearity - curl distributes over sums. No surprises here, since partial derivatives are linear.

$$\nabla \times (f\,\mathbf{u}) = f\,(\nabla \times \mathbf{u}) + (\nabla f) \times \mathbf{u}$$

This one is more interesting - and compare it carefully to the divergence product rule. For divergence we had $\nabla \cdot (f\mathbf{u}) = f(\nabla \cdot \mathbf{u}) + (\nabla f) \cdot \mathbf{u}$, with a dot product. For curl the correction term is a cross product: $(\nabla f) \times \mathbf{u}$.

This makes sense dimensionally: divergence produces a scalar (dot products give scalars), while curl produces a vector (cross products give vectors). The correction term inherits the type of the operation.

Common mistake: writing $\nabla \times (f\mathbf{u}) = f\,\nabla \times \mathbf{u}$, pulling the scalar out. But $f$ varies in space, so the product rule kicks in and we get the extra $(\nabla f) \times \mathbf{u}$ term. Another common mistake: using a dot product instead of a cross product in the correction term.

Practice Problems - §3.5

From Kaplan, problems after §3.6

3 Prove $\nabla \times (\mathbf{u} + \mathbf{v}) = \nabla \times \mathbf{u} + \nabla \times \mathbf{v}$ and $\nabla \times (f\,\mathbf{u}) = f\,\nabla \times \mathbf{u} + (\nabla f) \times \mathbf{u}$

Prove the two basic properties of curl: linearity over vector sums, and the product rule with a scalar.

Step 1: Linearity - the $\mathbf{k}$-component

Write $\mathbf{u} = (u_1, u_2, u_3)$ and $\mathbf{v} = (v_1, v_2, v_3)$. The $k$-component of curl is:

$[\nabla \times (\mathbf{u} + \mathbf{v})]_z = \dfrac{\partial(u_2 + v_2)}{\partial x} - \dfrac{\partial(u_1 + v_1)}{\partial y}$

$= \left(\dfrac{\partial u_2}{\partial x} - \dfrac{\partial u_1}{\partial y}\right) + \left(\dfrac{\partial v_2}{\partial x} - \dfrac{\partial v_1}{\partial y}\right) = [\nabla \times \mathbf{u}]_z + [\nabla \times \mathbf{v}]_z$

The $\mathbf{i}$ and $\mathbf{j}$ components split identically. Linearity of partial derivatives does all the work.

Step 2: Product rule - the $\mathbf{k}$-component

Now consider $f\mathbf{u} = (fu_1, fu_2, fu_3)$. The $k$-component of $\nabla \times (f\mathbf{u})$ is:

$[\nabla \times (f\mathbf{u})]_z = \dfrac{\partial(fu_2)}{\partial x} - \dfrac{\partial(fu_1)}{\partial y}$

Apply the single-variable product rule to each term:

$= \left(f\dfrac{\partial u_2}{\partial x} + u_2\dfrac{\partial f}{\partial x}\right) - \left(f\dfrac{\partial u_1}{\partial y} + u_1\dfrac{\partial f}{\partial y}\right)$

Group the $f$ terms and the $\nabla f$ terms:

$= f\left(\dfrac{\partial u_2}{\partial x} - \dfrac{\partial u_1}{\partial y}\right) + \left(\dfrac{\partial f}{\partial x} u_2 - \dfrac{\partial f}{\partial y} u_1\right)$

$= f\,[\nabla \times \mathbf{u}]_z + [(\nabla f) \times \mathbf{u}]_z$ ✓

The second group is the $z$-component of the cross product $(\nabla f) \times \mathbf{u} = (f_y u_3 - f_z u_2,\; f_z u_1 - f_x u_3,\; f_x u_2 - f_y u_1)$. The other two components work the same way.

16 Rigid body rotation: show $\mathbf{v} = \boldsymbol{\omega} \times \overrightarrow{OP}$ and compute div and curl

A rigid body rotates about the $z$-axis with angular velocity $\omega$. Show the velocity field is $\mathbf{v} = \boldsymbol{\omega} \times \overrightarrow{OP}$ where $\boldsymbol{\omega} = \omega\,\mathbf{k}$. Evaluate $\operatorname{div}\mathbf{v}$ and $\operatorname{curl}\mathbf{v}$.

Step 1: Write the position and differentiate

A point $P$ at distance $r$ from the $z$-axis traces a circle. Its position is:

$\overrightarrow{OP}(t) = r\cos(\omega t + \alpha)\,\mathbf{i} + r\sin(\omega t + \alpha)\,\mathbf{j} + z_0\,\mathbf{k}$

Differentiating:

$\mathbf{v} = -r\omega\sin(\omega t + \alpha)\,\mathbf{i} + r\omega\cos(\omega t + \alpha)\,\mathbf{j} = -\omega y\,\mathbf{i} + \omega x\,\mathbf{j}$

where $x = r\cos(\omega t + \alpha)$ and $y = r\sin(\omega t + \alpha)$.

Step 2: Verify $\mathbf{v} = \boldsymbol{\omega} \times \overrightarrow{OP}$

Compute the cross product with $\boldsymbol{\omega} = \omega\,\mathbf{k} = (0, 0, \omega)$ and $\overrightarrow{OP} = (x, y, z_0)$:

$\boldsymbol{\omega} \times \overrightarrow{OP} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 0 & \omega \\ x & y & z_0 \end{vmatrix} = (0 \cdot z_0 - \omega y)\,\mathbf{i} - (0 \cdot z_0 - \omega x)\,\mathbf{j} + (0 \cdot y - 0 \cdot x)\,\mathbf{k}$

$= -\omega y\,\mathbf{i} + \omega x\,\mathbf{j} = \mathbf{v}$ ✓

Step 3: Compute $\operatorname{div}\mathbf{v}$

With $\mathbf{v} = (-\omega y, \;\omega x, \;0)$:

$\nabla \cdot \mathbf{v} = \dfrac{\partial(-\omega y)}{\partial x} + \dfrac{\partial(\omega x)}{\partial y} + \dfrac{\partial(0)}{\partial z} = 0 + 0 + 0 = 0$

The rotation is incompressible - spinning fluid neither expands nor compresses.

Step 4: Compute $\operatorname{curl}\mathbf{v}$

$\nabla \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \partial_x & \partial_y & \partial_z \\ -\omega y & \omega x & 0 \end{vmatrix}$

$\mathbf{i}$: $\;\partial(0)/\partial y - \partial(\omega x)/\partial z = 0 - 0 = 0$
$\mathbf{j}$: $\;-[\partial(0)/\partial x - \partial(-\omega y)/\partial z] = -[0 - 0] = 0$
$\mathbf{k}$: $\;\partial(\omega x)/\partial x - \partial(-\omega y)/\partial y = \omega - (-\omega) = 2\omega$

$$\nabla \times \mathbf{v} = 2\omega\,\mathbf{k}$$

A beautiful result: the curl of a rigid rotation is twice the angular velocity vector. This is why the factor of $\frac{1}{2}$ appears in the vorticity-to-angular-velocity conversion in fluid mechanics.