# 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 AlternateCharacterizationOfCurl 2013-03-22 15:29:10 2013-03-22 15:29:10 stevecheng (10074) stevecheng (10074) 9 stevecheng (10074) Derivation msc 53-01 curl nabla FirstOrderOperatorsInRiemannianGeometry Curl NablaNabla