alternate characterization of curl

Let $\mathbf{F}$ be a smooth vector field on (an open subset of) $\mathbb{R}^{3}$ .

We show that $\operatorname{curl}\mathbf{F}$ defined using the coordinate-free definition given on the parent entry (http://planetmath.org/curl) is the same as the curl defined by $\nabla\times\mathbf{F}$ in Cartesian coordinates.

The case for spherical surfaces

This will be done by directly computing the limit $\mathbf{L}$ of surface integrals defining $\operatorname{curl}\mathbf{F}(\mathbf{p})$ , using spheres $S^{2}(r,\mathbf{p})$ centered at $\mathbf{p}$ of radius $r$ . The formula is:

	$\displaystyle\operatorname{curl}\mathbf{F}(\mathbf{p})=\mathbf{L}$	$\displaystyle=\lim_{r\to 0}\frac{3}{4\pi r^{3}}\iint_{S^{2}(r,\mathbf{p})}% \mathbf{n}\times\mathbf{F}\,dA$
		$\displaystyle=\lim_{r\to 0}\frac{3r^{2}}{4\pi r^{3}}\iint_{S^{2}}\mathbf{n}% \times\mathbf{F}(r\mathbf{n}+\mathbf{p})\,dA\,,$

where $\mathbf{n}$ is the outward unit normal to the surface (at each point of the surface), and $S^{2}$ is the unit sphere at the origin.

We simplify the last integral. Expanding $\mathbf{F}(r\mathbf{n}+\mathbf{p})$ in a first-degree Taylor polynomial about $\mathbf{p}$ , we have

	$\displaystyle\iint_{S^{2}}\mathbf{n}\times\mathbf{F}(r\mathbf{n}+\mathbf{p})\,dA$	$\displaystyle=\iint_{S^{2}}\mathbf{n}\times\mathbf{F}(\mathbf{p})\,dA$
		$\displaystyle\quad+\iint_{S^{2}}\mathbf{n}\times\operatorname{D}\mathbf{F}(% \mathbf{p})r\mathbf{n}\,dA+\iint_{S^{2}}\mathbf{n}\times o(\lVert r\mathbf{n}% \rVert)\,dA\,.$

The integral $\iint_{S^{2}}\mathbf{n}\times\mathbf{F}(\mathbf{p})\,dA$ vanishes by symmetry of the sphere, while

\displaystyle\left\lVert\iint_{S^{2}}\mathbf{n}\times o(\lVert r\mathbf{n}% \rVert)\,dA\right\rVert\leq\iint_{S^{2}}\lVert\mathbf{n}\rVert\,o(r)\,dA=o(r)\,.

Combining these facts, we obtain

	$\displaystyle\mathbf{L}$	$\displaystyle=\lim_{r\to 0}\left[0+\frac{3}{4\pi r}\iint_{S^{2}}\mathbf{n}% \times\operatorname{D}\mathbf{F}(\mathbf{p})r\mathbf{n}\,dA+o(1)\right]$
		$\displaystyle=\frac{3}{4\pi}\iint_{S^{2}}\mathbf{n}\times\operatorname{D}% \mathbf{F}(\mathbf{p})\mathbf{n}\,dA\,.$

Notice that $\mathbf{L}$ depends only on the derivative of $\mathbf{F}$ at $\mathbf{p}$ .

We want to evaluate the last integral in Cartesian coordinates. Let $\mathbf{e}_{k}$ be an orthonormal basis of $\mathbb{R}^{3}$ oriented positively, and let $B$ be the matrix of the derivative $\operatorname{D}\mathbf{F}(p)$ in this basis. Then the $k$ th coordinate of $\mathbf{L}$ with respect to the same basis is

\left(\iint_{S^{2}}\mathbf{n}\times B\mathbf{n}\,dA\right)\cdot\mathbf{e}_{k}=% \iint_{S^{2}}(\mathbf{n}\times B\mathbf{n})\cdot\mathbf{e}_{k}\,dA

The $k$ th coordinate of the integrand is

(\mathbf{n}\times B\mathbf{n})\cdot\mathbf{e}_{k}=n^{i}\,(B\mathbf{n})^{j}\,% \epsilon_{ijk}=n^{i}\,B^{j}_{l}n^{l}\,\epsilon_{ijk}\,,

where to lessen the writing, we employ the Einstein summation convention, along with the Levi-Civita permutation symbol $\epsilon_{ijk}$ , and $B^{j}_{l}$ denotes the entry at the $j$ th row, $l$ th column of $B$ .

In the summation above, if a summmand has $i\neq l$ , then the integral of that summand over the sphere is zero, by symmetry. This means that in the summation the index $l$ may be set to $i$ , and thus

\left(\iint_{S^{2}}\mathbf{n}\times B\mathbf{n}\,dA\right)\cdot\mathbf{e}_{k}=% \iint_{S^{2}}n^{i}B_{i}^{j}n^{i}\epsilon_{ijk}\,dA=B_{i}^{j}\epsilon_{ijk}% \iint_{S^{2}}(n^{i})^{2}\,dA\,.

Now there is a formula for the evaluation of integrals of polynomials over $S^{m-1}\subset\mathbb{R}^{m}$ , in terms of the gamma function; in our case ( $m=3$ ) the formula reads:

\iint_{S^{2}}(n^{i})^{2}\,dA=\frac{2\Gamma(\frac{3}{2})\,\Gamma(\frac{1}{2})\,% \Gamma(\frac{1}{2})}{\Gamma(\frac{3}{2}+\frac{1}{2}+\frac{1}{2})}=\frac{2% \Gamma(\frac{3}{2})\,\sqrt{\pi}\,\sqrt{\pi}}{\frac{3}{2}\Gamma(\frac{3}{2})}=% \frac{4\pi}{3}\,.

(If you do not know this formula, the integral in our case can be computed directly using spherical coordinates.) Therefore the $k$ th component of $\mathbf{L}$ is

\mathbf{L}\cdot\mathbf{e}_{k}=\frac{3}{4\pi}\,B_{i}^{j}\epsilon_{ijk}\iint_{S^% {2}}(n^{i})^{2}\,dA=B_{i}^{j}\epsilon_{ijk}=\left.\frac{\partial F^{j}}{% \partial x^{i}}\right|_{\mathbf{p}}\,\epsilon_{ijk}\,.

But this is just $(\nabla\times\mathbf{F}(\mathbf{p}))\cdot\mathbf{e}_{k}$ .

The case for arbitrary surfaces

Although we have only computed

\mathbf{L}=\operatorname{curl}\mathbf{F}(\mathbf{p})=\lim_{V\to 0}\frac{1}{V}% \iint_{S}\mathbf{n}\times\mathbf{F}\,dA

only for spheres $S=S^{2}(r,\mathbf{p})$ , this formula holds for arbitrary closed surfaces $S$ that shrink nicely to $\mathbf{p}$ . It is hardly obvious, especially since our computation before depended on the symmetry of the sphere extensively.

To show the general result, consider the triple scalar product $(\mathbf{v}\times\mathbf{F})\cdot\mathbf{e}_{k}$ . This is a linear functional in the vector $\mathbf{v}$ , so there exists a unique vector function $\mathbf{g}_{k}$ such that $(\mathbf{v}\times\mathbf{F})\cdot\mathbf{e}_{k}=\mathbf{g}_{k}\cdot\mathbf{v}$ for all $\mathbf{v}\in\mathbb{R}^{3}$ . We can find the components of this $\mathbf{g}_{k}$ by evaluating the functional at $\mathbf{v}=\mathbf{e}_{i}$ :

g_{k}^{i}=\mathbf{g}_{k}\cdot\mathbf{e}_{i}=(\mathbf{e}_{i}\times\mathbf{F})% \cdot\mathbf{e}_{k}=\det(\mathbf{e}_{i},\mathbf{F},\mathbf{e}_{k})=F^{j}% \epsilon_{ijk}\,.

The reason for considering such expressions is that, putting $\mathbf{v}=\mathbf{n}$ , we have

\iint_{S}(\mathbf{n}\times\mathbf{F})\cdot\mathbf{e}_{k}\,dA=\iint_{S}\mathbf{% g}_{k}\cdot\mathbf{n}\,dA=\iint_{S}\mathbf{g}_{k}\cdot d\mathbf{A}\,.

So we have converted the original integral into an ordinary surface integral. And this surface integral can be changed into a volume integral, by using the divergence theorem:

\iint_{S}\mathbf{g}_{k}\cdot d\mathbf{A}=\iiint_{M}\operatorname{div}\mathbf{g% }_{k}\,dV=\iiint_{M}\frac{\partial F^{j}}{\partial x^{i}}\,\epsilon_{ijk}\,dV\,,

where $M$ is the volume whose boundary is $S$ . Hence

	$\displaystyle\mathbf{L}\cdot\mathbf{e}_{k}$	$\displaystyle=\lim_{V\to 0}\frac{1}{V}\iint_{S}(\mathbf{n}\times\mathbf{F})% \cdot\mathbf{e}_{k}\,dA$
		$\displaystyle=\lim_{V\to 0}\frac{1}{V}\iiint_{M}\frac{\partial F^{j}}{\partial x% ^{i}}\,\epsilon_{ijk}\,dV$
		$\displaystyle=\left.\frac{\partial F^{j}}{\partial x^{i}}\right\|_{\mathbf{p}}% \,\epsilon_{ijk}=(\nabla\times\mathbf{F}(\mathbf{p}))\cdot\mathbf{e}_{k}\,.$

Definition in terms of differential forms

We mention, in passing, a computational, yet coordinate-free, alternative to the definition of the curl, using differential forms. If $\omega$ is a 1-form on $\mathbb{R}^{3}$ such that $\omega(\mathbf{v})=\langle\mathbf{F},\mathbf{v}\rangle$ , then the curl of $\mathbf{F}$ is defined as the vector function $\mathbf{g}=g^{k}\,e_{k}$ such that

d\omega(\mathbf{u},\mathbf{v})=\langle\mathbf{g},\mathbf{u}\times\mathbf{v}% \rangle\,.

In Cartesian coordinates, we have

	$\displaystyle\omega$	$\displaystyle=F^{1}\,dx^{1}+F^{2}\,dx^{2}+F^{3}\,dx^{3}$
	$\displaystyle d\omega$	$\displaystyle=g^{1}\,dx^{2}\wedge dx^{3}+g^{2}\,dx^{3}\wedge\,dx^{1}+g^{3}\,dx% ^{1}\wedge dx^{2}\,,$

If we take the exterior derivative of the first equation for $\omega$ , and then equate components with the second equation for $d\omega$ , we find that $g^{k}$ = $(\nabla\times\mathbf{F})\cdot\mathbf{e}_{k}$ , so our new definition is equivalent to the others.

Title	alternate characterization of curl
Canonical name	AlternateCharacterizationOfCurl
Date of creation	2013-03-22 15:29:10
Last modified on	2013-03-22 15:29:10
Owner	stevecheng (10074)
Last modified by	stevecheng (10074)
Numerical id	9
Author	stevecheng (10074)
Entry type	Derivation
Classification	msc 53-01
Related topic	curl
Related topic	nabla
Related topic	FirstOrderOperatorsInRiemannianGeometry
Related topic	Curl
Related topic	NablaNabla