Review / Chapter 2 — Vector Calculus

Curl

Chapter 2 — Vector Calculus

The curl is a second type of vector derivative — divergence's rotational sibling. While divergence measures spreading (flux per unit volume), curl measures twisting (circulation per unit area). The result is a vector field whose direction and magnitude encode how much, and about which axis, the field \(\vec{F}\) locally swirls.

From circulation to curl

The circulation of \(\vec{F}\) around a closed curve \(C\) is

\[\Gamma = \oint_C \vec{F}\cdot d\vec{r}.\]

If \(C\) bounds a tiny flat patch of area \(A\) with unit normal \(\hat{n}\) (oriented by the right-hand rule), the curl at that point, projected onto \(\hat{n}\), is

\[(\vec{\nabla}\times\vec{F})\cdot\hat{n} = \lim_{A\to 0}\frac{1}{A}\oint_C \vec{F}\cdot d\vec{r}.\]

Doing this for loops perpendicular to each coordinate axis picks off the three components of \(\vec{\nabla}\times\vec{F}\).

Cartesian formula

The usual "determinant" mnemonic:

\[\vec{\nabla}\times\vec{F} = \begin{vmatrix}\hat{x} & \hat{y} & \hat{z} \\ \partial_x & \partial_y & \partial_z \\ F_x & F_y & F_z\end{vmatrix}\]

which expands to

\[\vec{\nabla}\times\vec{F} = \left(\partial_y F_z - \partial_z F_y,\ \partial_z F_x - \partial_x F_z,\ \partial_x F_y - \partial_y F_x\right).\]

Each component is "cyclic": the \(z\)-component involves \(\partial_x F_y - \partial_y F_x\), and the others follow by cyclic permutation \(x\to y\to z\to x\).

Physical picture

Imagine dropping a tiny paddle wheel into the field \(\vec{F}\) (thinking of it as fluid velocity). Where \(\vec{\nabla}\times\vec{F}\neq 0\), the wheel spins. The axis of spin is parallel to \(\vec{\nabla}\times\vec{F}\); the rate of spin is proportional to its magnitude. A "shearing" field like \(\vec{F}=(y,0,0)\) has a nonzero curl even though none of the arrows appear to rotate individually.

A swirling vector field with a small test loop: arrows circulating around the loop indicate a nonzero curl, whose direction is given by the right-hand rule perpendicular to the loop.
Circulation around a small loop gives the component of \(\vec{\nabla}\times\vec{F}\) perpendicular to the loop, oriented by the right-hand rule.

A crucial identity

For any smooth scalar field \(\phi\),

\[\vec{\nabla}\times(\vec{\nabla}\phi) = 0.\]

The curl of any gradient is zero. This is a consequence of the equality of mixed partials: each component of \(\vec{\nabla}\times\vec{\nabla}\phi\) is of the form \(\partial_x\partial_y\phi - \partial_y\partial_x\phi = 0\). Since \(\vec{E}=-\vec{\nabla} V\), it follows immediately that electrostatic fields have zero curl everywhere.

Worked examples

(1) Let \(\vec{F}=(cy,cx,0)\). Then \(\partial_x F_y - \partial_y F_x = c - c = 0\), and the other components involve \(\partial_z\) of zeros. So \(\vec{\nabla}\times\vec{F} = 0\). This field is a valid electrostatic field, and indeed comes from the potential \(V = -cxy\).

(2) Now try \(\vec{F}=(cy,-cx,0)\). Same recipe: \(\partial_x(-cx) - \partial_y(cy) = -c - c = -2c \neq 0\). So \(\vec{\nabla}\times\vec{F} = -2c\,\hat{z}\). This one is not electrostatic — no potential exists for it.

Practice Problems

Problem 1easy
Compute \(\vec{\nabla}\times\vec{F}\) for \(\vec{F} = (0,\ x,\ 0)\).
Hint
Only the \(z\)-component is nonzero — \(\partial_x F_y - \partial_y F_x\).
Solution

\(\partial_x(x) - \partial_y(0) = 1\). Other components are zero.

Answer: \(\vec{\nabla}\times\vec{F} = \hat{z}\).

Problem 2easy
Compute \(\vec{\nabla}\times\vec{F}\) for \(\vec{F} = (y,-x,0)\).
Hint
This is a rigid rotation about the \(z\)-axis, so expect a \(z\)-component.
Solution

\(z\)-component: \(\partial_x(-x) - \partial_y(y) = -1-1 = -2\). Others zero.

Answer: \(\vec{\nabla}\times\vec{F} = -2\hat{z}\).

Problem 3easy
Does \(\vec{F} = (x,y,z)\) have zero curl?
Hint
It's a gradient of what scalar?
Solution

\(\vec{F} = \vec{\nabla}(\tfrac12(x^2+y^2+z^2))\), so it's a gradient. Any gradient has zero curl. Direct: each component involves \(\partial_i x_j\) with \(i\neq j\), all zero.

Answer: yes, \(\vec{\nabla}\times\vec{F} = 0\).

Problem 4medium
Compute \(\vec{\nabla}\times\vec{F}\) for \(\vec{F} = (y z,\ z x,\ x y)\).
Hint
Use the determinant expansion carefully.
Solution

\(x\)-comp: \(\partial_y(xy) - \partial_z(zx) = x - x = 0\). \(y\)-comp: \(\partial_z(yz) - \partial_x(xy) = y - y = 0\). \(z\)-comp: \(\partial_x(zx) - \partial_y(yz) = z - z = 0\).

Answer: \(\vec{\nabla}\times\vec{F} = \vec{0}\). (It's a gradient: \(\vec{F} = \vec{\nabla}(xyz)\).)

Problem 5medium
Is \(\vec{F} = (-y,\ x,\ 0)/(x^2+y^2)\) a possible electrostatic field on the punctured plane \((x,y)\neq(0,0)\)?
Hint
Compute \(\vec{\nabla}\times\vec{F}\); this is the classic "winding around the origin" field.
Solution

Direct computation gives \(\vec{\nabla}\times\vec{F} = 0\) everywhere \(\vec{F}\) is defined — so locally it looks fine. But it has nonzero circulation \(2\pi\) around any loop enclosing the origin, so no single-valued potential exists globally. In physics terms, it's not a valid electrostatic field because the origin isn't part of the simply-connected domain.

Answer: curl is zero on its domain, but circulation around the \(z\)-axis is nonzero — no global potential exists, so not a valid electrostatic field in vacuum.

Problem 6hard
Prove that \(\vec{\nabla}\times(\vec{\nabla}\phi) = \vec{0}\) in Cartesian coordinates for any smooth \(\phi\).
Hint
Write out one component and use \(\partial_x\partial_y = \partial_y\partial_x\).
Solution

The \(z\)-component of \(\vec{\nabla}\times(\vec{\nabla}\phi)\) is \(\partial_x(\partial_y\phi) - \partial_y(\partial_x\phi) = \partial_x\partial_y\phi - \partial_y\partial_x\phi = 0\) by Clairaut's theorem. The same argument (with cyclic swaps) kills the other components.

Answer: equality of mixed partials makes every component vanish.