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 definedby $\abla\\times\\mathbf{F}$ in Cartesian coordinates.

The case for spherical surfaces

This will be done by directly computing the limit $\\mathbf{L}$ of surfaceintegrals 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 thesame basisis

\\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 employthe Einstein summation convention, alongwith 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 $i\eq l$ ,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 integralsof 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 casecan 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 $(\abla\\times\\mathbf{F}(\\mathbf{p}))\\cdot\\mathbf{e}_{k}$ .

The case for arbitrary surfaces

Although we have onlycomputed

\\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 dependedon 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 existsa 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 functionalat $\\mathbf{v}=\\mathbf{e}_{i}$ :

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

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

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 componentswith the second equation for $d\\omega$ ,we find that $g^{k}$ = $(\abla\\times\\mathbf{F})\\cdot\\mathbf{e}_{k}$ ,so our new definition is equivalent to the others.