Implications of Stokes' Theorem

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Theorem

Stokes' Theorem implies all of the following results:


Proof

Classical Stokes' Theorem

We note that given $f_1, f_2, f_3: \R^3 \to \R$, $f_1 \, \mathrm d x + f_2 \, \mathrm d y + f_3 \, \mathrm d z$ is a $1$-form defined on a surface.

Since we are integrating over the boundary of a surface, which is a $1$-manifold, we may use the general Stokes' theorem:

\(\displaystyle \oint_{\partial S} f_1 \, \mathrm d x + f_2 \, \mathrm d y + f_3 \, \mathrm d z\) \(=\) \(\displaystyle \iint_S \, \mathrm d \left({f_1 \, \mathrm d x + f_2 \, \mathrm d y + f_3 \, \mathrm d z}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \iint_S \frac {\partial f_1} {\partial x} \, \mathrm d x \wedge \mathrm d x + \frac {\partial f_1} {\partial y} \, \mathrm d y \wedge \mathrm d x + \frac {\partial f_1} {\partial z} \, \mathrm d z \wedge \mathrm d x\) $\quad$ $\quad$
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \frac {\partial f_2} {\partial x} \, \mathrm d x \wedge \mathrm d y + \frac {\partial f_2} {\partial y} \, \mathrm d y \wedge \mathrm d y + \frac {\partial f_2} {\partial z} \, \mathrm d z \wedge \mathrm d y\) $\quad$ $\quad$
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \frac {\partial f_3} {\partial x} \, \mathrm d x \wedge \mathrm d z + \frac {\partial f_3} {\partial y} \, \mathrm d z \wedge \mathrm d y + \frac {\partial f_3} {\partial z} \, \mathrm d z \wedge \mathrm d z\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \iint \left({\frac {\partial f_2} {\partial x} - \frac{\partial f_1} {\partial y} }\right) \, \mathrm d x \wedge \mathrm d y + \left({\frac{\partial f_3}{\partial y} - \frac{\partial f_2}{\partial z} }\right) \, \mathrm d y \wedge \mathrm d z + \left({\frac{\partial f_1}{\partial z} - \frac{\partial f_3}{\partial x} }\right) \, \mathrm d z \wedge \mathrm d x\) $\quad$ $\quad$


If we define $\mathbf F: \R^3 \to \R^3$ as:

$\mathbf F = f_1 \mathbf e_1 + f_2 \mathbf e_2 + \mathbf e_3$

we recognize this expression as:

$\nabla \times \mathbf F$

with:

$\mathrm d y \wedge \mathrm d z$ replacing $\mathbf e_1$
$\mathrm d z \wedge \mathrm d x$ replacing $\mathbf e_2$

and:

$\mathrm d x \wedge \mathrm d y$ replacing $\mathbf e_3$.



Green's Theorem

We note that given $A, B: \R^2 \to \R$, $A \, \mathrm d x + B \, \mathrm d y$ is a $1$-form defined on some open set $U \subset \R^2$.

Since we are integrating a $1$-form on $\partial U$, a $1$-manifold, we may use the general Stokes' theorem:

\(\displaystyle \oint_{\partial U} \left({A \, \mathrm d x + B \, \mathrm d y}\right)\) \(=\) \(\displaystyle \iint_U \, \mathrm d \left({A \, \mathrm d x + B \, \mathrm d y}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \iint_U \left({\frac {\partial A} {\partial x} \, \mathrm d x \wedge \mathrm d x + \frac {\partial A} {\partial y} \, \mathrm d y \wedge \mathrm d x + \frac {\partial B} {\partial x} \, \mathrm d x \wedge \mathrm d y + \frac {\partial B} {\partial y} \, \mathrm d y \wedge \mathrm d y}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \iint_U \left({0 - \frac {\partial A} {\partial y} \, \mathrm d y \wedge \mathrm d x + \frac {\partial B} {\partial x} \, \mathrm d x \wedge \mathrm d y + 0}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \iint_U \left({\frac {\partial B} {\partial x} - \frac{ \partial A} {\partial y} }\right) \, \mathrm d x \, \mathrm d y\) $\quad$ $\quad$

$\Box$


Divergence Theorem

We note that the expression $\mathbf F \cdot \mathbf n \, \mathrm d S$ is a $2$-form, since it takes two vectors ($F$ and $n$) and outputs a scalar.

Less formally:

$\mathbf F \cdot \mathbf n \, \mathrm d S = $ (scalar)(area $2$-form) $=$ $2$-form

Since we are integrating this expression over the boundary of a region in three space, which is a $2$-manifold, we may use the general Stokes' theorem:

\(\displaystyle \iint \limits_{\partial U} \mathbf F \cdot \mathbf n \, \mathrm d S\) \(=\) \(\displaystyle \iiint_U \, \mathrm d \left({\mathbf F \cdot \mathbf n \, \mathrm d S}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \iiint_U \, \mathrm d \left({\mathbf F \cdot \mathbf n}\right) \, \mathrm d S + \mathbf F \cdot \mathbf n \, \mathrm d^2 S\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \iiint_U \, \mathrm d \left({\mathbf F \cdot \mathbf n}\right) \, \mathrm d S\) $\quad$ $\quad$



Fundamental Theorem of Calculus

Given a function $f: \R \to \R$, integrable on some open set $I$, we can recognize such a function a $0$-form -- that is, defined at a point and taking no vectors as inputs.

Hence the exterior derivative is just the ordinary differential:

$f' \left({x}\right) \, \mathrm d x$

Since the ordinary differential is a $1$-form, and we are integrating on a $1$-manifold, we may use the general Stokes' theorem:

$\int_I f \left({x}\right) \, \mathrm d x = \int_{\partial I} f \left({x}\right)$

By Induced Orientations on Boundaries, $\partial I = \left\{ {x_0', x_1}\right\}$ for some points $x_0, x_1$ where $y'$ indicates a point with "negative" orientation.

We then have

$\int_{\partial I} f \left({x}\right) = -f \left({x_0}\right) + f \left({x_1}\right) = f \left({x_1}\right) - f \left({x_0}\right)$

$\Box$


Cauchy's Residue Theorem

Let $z = x + iy$ and $f \left({z}\right) = u \left({x, y}\right) + i v\left({x, y}\right)$ for two real functions $u, v$.

Let $D$ be a disk and its boundary $\partial D = C$ a circle.

Then by Stokes' Theorem:

$\displaystyle \int_C f \left({z}\right) \, \mathrm d z = \int_C u \left({z}\right) \, \mathrm d x + v \left({z}\right) \, \mathrm d y = \int_D \frac {\partial u} {\partial y} - \frac {\partial v} {\partial x}$

which is zero if $f$ satisfies the Cauchy-Riemman conditions.

$\blacksquare$