classical Stokes' theoremClassicalStokesTheorem2005-08-12 14:53:422005-08-14 17:28:48Theoremcurl\usepackage{amssymb}
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\DeclareMathOperator{\curl}{curl}Let $M$ be a compact, oriented two-dimensional differentiable manifold (surface) with boundary in $\real^3$,
and $\vF$ be a $C^2$-smooth vector field defined on an open set in $\real^3$ containing $M$.
Then
\[
\iint_M (\nabla \times \vF) \cdot d\vA = \int_{\partial M} \vF \cdot d\vs\,.
\]
Here, the boundary of $M$, $\partial M$ (which is a curve)
is given the induced orientation from $M$. The symbol $\nabla \times \vF$
denotes the curl of $\vF$.
The symbol $d \vs$ denotes the line element $ds$ with a direction
parallel to the unit tangent vector $\vt$ to $\partial M$, while $d \vA$ denotes
the area element $dA$ of the surface $M$ with a direction parallel to the unit outward normal $\vn$
to $M$. In precise terms:
\[
d\vA = \vn \, dA\,, \quad d\vs = \vt \, ds\,.
\]
The classical Stokes' theorem reduces to Green's theorem on the plane if the surface $M$
is taken to lie in the xy-plane.
The classical Stokes' theorem, and
the other ``Stokes' type'' theorems
are special cases of the general Stokes' theorem involving
differential forms.
In fact, in the proof we present below, we appeal to the general Stokes' theorem.
\section*{Physical interpretation}
(To be written.)
\section*{Proof using differential forms}
The proof becomes a triviality once we express
$(\nabla \times \vF) \cdot d\vA$
and $\vF \cdot d\vs$ in terms of differential forms.
\begin{proof}
Define the differential forms $\eta$ and $\omega$ by
\begin{align*}
\eta_p(\vu, \vv) &= \langle \curl \vF(p), \vu \times \vv \rangle\,, \\
\omega_p(\vv) &= \langle \vF(p), \vv \rangle\,.
\end{align*}
for points $p \in \real^3$, and tangent vectors $\vu, \vv \in \real^3$.
The symbol $\langle, \rangle$ denotes the dot product in $\real^3$.
Clearly, the functions $\eta_p$ and $\omega_p$ are linear and alternating in
$\vu$ and $\vv$.
We claim
\begin{align}
\eta &= \nabla \times \vF \cdot d\vA & \text{ on $M$. } \\
\omega &= \vF \cdot d\vs & \text{ on $\partial M$.}
\end{align}
To prove (1), it suffices to check
it holds true when we evaluate the left- and right-hand sides
on an orthonormal basis $\vu, \vv$ for the tangent space of $M$
corresponding to the orientation of $M$,
given by the unit outward normal $\vn$.
We calculate
\begin{align*}
\nabla \times \vF \cdot d\vA(\vu,\vv) &=
\langle \curl \vF, \vn \rangle \,
dA(\vu, \vv)
& \text{definition of $d\vA = \vn \, dA$}
\\
&=
\langle \curl \vF, \vn \rangle
& \text{definition of volume form $dA$}
\\
&=
\langle \curl \vF, \vu \times \vv \rangle
& \text{since $\vu \times \vv = \vn$} \\
&=
\eta(\vu, \vv)\,.
\end{align*}
For equation (2), similarly, we only have to check that it holds
when both sides are evaluated at $\vv = \vt$,
the unit tangent vector of $\partial M$
with the induced orientation of $\partial M$.
We calculate again,
\begin{align*}
\vF \cdot d\vs(\vt) &= \langle \vF, \vt \rangle \, ds(\vt)
& \text{definition of $d\vs = \vt \, ds$} \\
&= \langle \vF, \vt \rangle
& \text{definition of volume form $ds$} \\
&= \omega(\vt)\,.
\end{align*}
Furthermore, $d \omega$ = $\eta$.
(This can be checked by a calculation
in Cartesian coordinates, but in fact this equation
is one of the coordinate-free \emph{definitions} of the curl.)
The classical Stokes' Theorem now follows
from the general Stokes' Theorem,
\[
\int_M \eta = \int_M d\omega = \int_{\partial M} \omega\,. \qedhere
\]
\end{proof}
\begin{thebibliography}{3}
\bibitem{Spivak} Michael Spivak. {\it Calculus on Manifolds}. Perseus Books, 1998.
\end{thebibliography}