**The (general) Stokes Theorem:**this unique theorem puts

Green’s/Gauss/Stokes Theorems (from intermediate calculus)

under one umbrella. Let \(C\) be a polyhedron with boundary \(\partial C\). Then

$$ {\int_{\partial C} \omega = \int_C d\omega }$$

**Example:**integrate the form \(\omega = y dx \) on the boundary

of the polygon \(C = \text{conv}\{(0,0), (1,0), (3,2), (0,3)\} \subset \mathbb{R}^2\).

Indeed, by the Stokes theorem we have

$$ \int_{\partial C} \omega = \int_{C} d\omega $$

The differential of \(\omega\) is

\(d\omega = d(ydx) = dy \wedge dx + y(d(dx)) = dy\wedge dx = – dx\wedge dy \)

which is merely the area form. Therefore

$$ \int_{C} d\omega = -\int_{C} dx\wedge dy = -|\text{area}| =-6.5 $$

We can verify that the line integral over the boundary has

$$ \int_0^3 2(x-2) dx + \int_3^0 (3-x/3) dx + 1 = (3x-x^2)\Big |_{3}^0 +1 = -7.5+1 = -6.5 $$

**Example (sum of area normals):**consider a polyhedra \(P\subset R^n\).

A common result asserts that the sum of all outward pointing

face area normals (normals whose length is the area of the fact) vanish, i.e

$$ \sum_{\text{faces}} \hat n_F \cdot \text{area}(F) = 0 $$

This is a direct result of the Stokes theorem: the form \(\omega = dx_2\wedge \ldots \wedge dx_n\)

(which is an area form) has

\( \int_{\partial P} \omega = \int_P d\omega \)

but \(d\omega =0\).

**Example (Archimedes’ buoyancy principle):**this well

known physical law states that the buoyant force acting on a body

equals to the weight it displaces in the fluid. Stokes theorem

provides an elegant proof for this principle.

Assume a coordinate system \(x_1,x_2,x_3\) such that

gravity (known our vertical reference) acts along \(x_1\).

The local vertical force acting on a small boundary patch is

$$ f_1 = x_1 \rho g dx_2\wedge dx_3 $$

The overall vertical component of this force is the integral over the boundary,

$$ \begin{align}

F_1 &= \int_{\partial C} f_1 =

– \rho g \int_{\partial C} x_1 dx_2\wedge dx_3 \\

& = -\rho g \int_C d(x_1 dx_2\wedge dx_3\\

&= -\rho g \int_C dx_1\wedge dx_2 \wedge dx_3) \\

&= -g(\rho \text{vol}(C)) = -g \cdot \text{mass}

\end{align} $$

**Flux Integrals:**consider the form flux form

$$ \omega(x) = \sum (-1)^i a_i(x)dx_{J_i}, $$

here

\( J_i = (1,\ldots, i-1,i+1,\ldots, n) \)

are all the indices between \(1,\ldots,n\) with \(i\) omitted.

In particular, the form in \(\mathbb{R}^3\) looks like

\( \omega(x) = a_1 dx_2 \wedge dx_3 + a_2 dx_3 \wedge dx_1 + a_3 dx_1 dx_2 \),

which is an area form (\(2\)-form)). Integral of this form over a surface \(S\)

can be interpreted the flux:

$$ \int_S \omega = \int (\bar a \cdot \bar n) dS $$

here \(\bar n\) is the unit normal pointing outwards.

Flux can be used to determine areas over surfaces: if we set, for example, \(|a|=|n|\) then

the integral above amounts to the area of \(S\):

$$ |A| = \int_A \underbrace{x_1 dx_2\wedge dx_3 + x_2 dx_3\wedge dx_1 + x_3 dx_1\wedge dx_2}_{\omega} $$

We can use this idea to compute the area of patches on the unit sphere \( x_1^2+x_2^2 +x_3^2 =1 \).

The form must admit (by differentiating a constant) \( 2x_1dx_1 + 2x_2dx_2 + 2x_3dx_3 = 0 \),

therefore

$$ dx_3 = – \frac{x_1dx_2 +x_2 dx_2 }{x_3}. $$

Substitute that in \(\omega\) above

$$

\begin{align}

\omega &= x_3 dx_1\wedge dx_2

-x_1 dx_2\wedge \frac{x_1dx_2 +x_2 dx_2 }{x_3}

-x_2\frac{x_1dx_2 +x_2 dx_2 }{x_3} \wedge dx_1 \\

& = \frac{x_1^2+x_2^2+x_3^2}{x_3} dx_1\wedge dx_2 = \frac{rdr \wedge d\phi}{x_3} \end{align}

$$

Set \(x_3 = \sin(\theta)\) \( r = \cos(\theta) \)

$$ \frac{\cos(\theta)(-\sin(\theta))}{\sin(\theta)}d\theta \wedge d\phi = d(\overbrace{\sin(\theta)}^{x_3})d\phi, $$

as though we projected the sphere to a cylinder about the \(x_3\) axis, and measure the area

on the cylinder surface.

The Lambert projection maps the surface of earth to a Rectangular map,

maintaining areas on the expense of geographical distortion.

This gives a very easy way to compute the area of a cap of a sphere. Define

the cap by the * height * of the cap (from the pole down to the base),

and apparently the area is proportional to the height; the area is

$$ |A| = \int_{A} dx_3 \wedge d\phi $$

and note that the differential of the form \( \alpha = x_3 d\phi \)

is precisely the area form being integrated \(d\alpha = dx_3\wedge d\phi \).

This yields, in fact yet another application of the Stokes integral: we can compute

the area of a patch (say, a country’s area) bound by the points \(p_k = (lat_k,lon_n)\)

by the approximation

$$ \int_{A} dx_3 \wedge d\phi = \int_{\partial A} x_3 d\phi \approx \sum_k \sin(lat_k)\cdot [lon_{k+1}-lon_k] $$

**Flux of Electric field:**the attenuation of the electrostatic

(Coulomb force) is reciprocal squared, i.e \( |F| \sim |r|^{-2} \).

Gauss’s law of flux asserts that the overall flux through a surface

is proportional to the amount of electric charge in it. Thus, if if no

charge is present inside a surfaace, the flux should vanish.

The integral of this force through a surface \(S\) is in fact the Flux integral

$$ \int_S \omega = \int_{S} \frac{\bar n\cdot \bar r}{|r|^3} dS . $$

described by the differential \(2\)-form:

$$ \omega = \frac{\overbrace{x_1 dx_2 \wedge dx_3 + x_2 dx_3\wedge dx_1 + x_3 dx_1\wedge dx_2}^{\tilde \omega}}{|r|^3} $$

here \(|r|^2 = x_1^2+x_2^2+x_3^2 \), where

the total Flux is \( \int_S \omega \). Differentiate \(\tilde \omega\):

$$ d(x_1 dx_2\wedge dx_3) = \underbrace{dx_1\wedge dx_2\wedge dx_3}_{\lambda} + x_1\underbrace{d(dx_2\wedge dx_2)}_{=0}, $$

similarly \( d(x_2 dx_3\wedge dx_1) = dx_2\wedge dx_3 \wedge dx_1 \), and same for

the third term, resulting with \(d\tilde \omega = 3\lambda\), the Lebesgue volume form.

For the denominator, we have

$$ d|r|^{-3} = -\frac{3}{2}(|r|^2)^{-5/2} d|r|^2 = -\frac{3}{2} \frac{2(\bar r\cdot d\bar r)}{|r|^{-5}}

=-3\frac{\bar r \cdot d\bar r}{|r|^5} $$

since \(d|r|^2 = 2(x_1dx_2 + x_2dx_2 + x_3dx_3)\).

Going back to \(\omega\)

$$ d\omega = d(|r|^{-3}\tilde \omega) = d|r|^{-3}\wedge \tilde \omega + |r|^{-3} d\tilde \omega

= -3 \frac{\bar r d\bar r}{|r|^5}\wedge \tilde \omega + 3 |r|^{-3}\lambda

= -3 \frac{|r|^2\lambda }{|r|^5} + 3 |r|^{-3}\lambda = 0 $$

so that the electrostatic flux is a * closed form * (definition follows).

**Closed forms**: if \(d\omega = 0\) then \(\omega\) is called

*closed*.

Integrating over closed forms does not depend on the actual surface of integration,

a property we call

*robustness*(as Gauss’s Law of flux we have seen earlier).

Another example, the form in \(\mathbb{R}^2\)

\( \omega = \frac{x_1 dx_2 – x_2 dx_1}{x_1^2 +x_2^2} \)

is closed, i.e \( d\omega = 0 \).

** Exact vs closed: ** recall that a differential form \(\omega\) is exact if

\(\omega = d\alpha \). For example, ** sometimes ** the

Coulomb force field can be differentiated from a scalar potential field (a \(1\)-form).

* Every exact form is closed, but not the other way around. *

For example, the form above (that has singularity at \(x_1=x_2=0\)) is closed but not

exact: an integral over the circle equals to \(2\pi\), whereas for exact forms

this integral (over closed path) should vanish.

**Applications in Vector Calculus**: let \(A\) be a vector field in \(\mathbb{R}^3\), i.e

$$ A(x) = (A_1(x),A_2(x),A_3(x)). $$

Classically, the

*rotor/curl*is defined as

$$ \text{curl}(A) = \left|\begin{matrix} e_1 & e_2 & e_3 \\

\frac{\partial}{\partial x_1} & \frac{\partial}{\partial x_2} & \frac{\partial}{\partial x_3}\\

A_x & A_y & A_z \end{matrix}\right| $$

and the divergence as

\( \text{div}(A) = \frac{\partial A_1}{\partial x_1} + \frac{\partial A_2}{\partial x_2} +

\frac{\partial A_3}{\partial x_3} \), and similarly the gradient for a scalar field.

In the language of differential forms, we have

$$ \alpha_A = A_1 dx_1 + A_2 dx_2 + A_3 dx_3,~~~~

\beta_A = A_1 dx_2\wedge dx_3 + A_2 dx_3\wedge dx_1 + A_3 dx_1\wedge dx_2 $$

For a scalar field \(u\) the gradient field is the form

\( du = \alpha_{\text{grad}(u)} \).

Also, for example, we have

$$ \alpha_A = \beta_{\text{curl}(A)} $$

$$ d\beta_A = \text{div}(A)\cdot \lambda $$

A well known identity in vector analysis is that the

curl of a gradient vanish, and indeed

$$ \beta_{curl(grad(f))} = d \alpha_{grad(f)} = d(df) = 0 $$

Another standard formula in vector analysis is

the curl of a vector field multiplied by a scalar field:

$$ \beta_{curl(fA)} = d\alpha_{fA} = d(f\alpha_A)=df\wedge \alpha_A+fd\alpha_A = df\times A +f\text{curl}(A) $$

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