The Generalized Stokes Formula

In calculus we learn a whole family of major integral theorems, for example:

  • Stokes’ theorem

∮∂DF→⋅dr→=∬D(▽×F→)⋅dA→\oint_{\partial D}\overrightarrow{F}\cdot d \overrightarrow{r} = \iint_{D}(\triangledown\times \overrightarrow{F})\cdot d \overrightarrow{A}

  • Gauss’ theorem

∬∂DF→⋅dA→=∭D(▽⋅F→)dV\iint_{\partial D}\overrightarrow{F}\cdot d \overrightarrow{A}=\iiint_{D}(\triangledown\cdot \overrightarrow{F})dV

  • Green’s formula

∮∂DPdx+Qdy=∬D(∂Q∂x−∂P∂y)dA\oint_{\partial D}Pdx + Qdy=\iint_{D}\left( \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right)dA

  • Even the Newton-Leibniz formula

f(a)−f(b)=∫baf′(x) dxf(a)-f(b) = \int _{b}^a f'(x) \, dx

At least for Green’s formula, most people have probably heard the intuitive explanation: many tiny line integrals add up to the integral over the whole boundary, and then basic calculus lets you rewrite that as a double integral over the region.

But if you look closely at these formulas, they all really have the same shape. Or rather, if you think about it, integrating over the boundary of a region is exactly the same kind of object as integrating all the infinitesimal variation inside it. That is the generalized Stokes formula:

∫D dω=∫∂D ω\int _{D} \, d \omega=\int _{\partial D} \, \omega

Compare that with the formulas above:

f(b)−f(a)⏟∫∂Dω=∫abf′(x)dx⏟∫Ddω\underbrace{f(b) - f(a)}_{\int_{\partial D} \omega} = \underbrace{\int_a^b f'(x) dx}_{\int_D d\omega}

∮∂DPdx+Qdy⏟∫∂Dω=∬D(∂Q∂x−∂P∂y)dA⏟∫Ddω\underbrace{\oint_{\partial D} P dx + Q dy}_{\int_{\partial D} \omega} = \underbrace{\iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA}_{\int_D d\omega}

∬∂DF⃗⋅dA⃗⏟∫∂Dω=∭D(∇⋅F⃗) dV⏟∫Ddω\underbrace{\iint_{\partial D} \vec{F} \cdot d\vec{A}}_{\int_{\partial D} \omega} = \underbrace{\iiint_D (\nabla \cdot \vec{F}) \,dV}_{\int_D d\omega}

∮∂DF⃗⋅dr⃗⏟∫∂Dω=∬D(∇×F⃗)⋅dA⃗⏟∫Ddω\underbrace{\oint_{\partial D} \vec{F} \cdot d\vec{r}}_{\int_{\partial D} \omega} = \underbrace{\iint_D (\nabla \times \vec{F}) \cdot d\vec{A}}_{\int_D d\omega}

The visual idea is simple: integrating a function along the boundary is the same as integrating its rate of change throughout the interior.

So what are \omega and d\omega here? The symbol d is the exterior derivative, and to talk about it properly we need another operation first: the wedge product \wedge.

In \mathbb{R}^3, its basic rules look like this:

dx∧dx=0, dy∧dy=0, dz∧dz=0dx∧dy=−dy∧dx, dx∧dy=−dy∧dzdy∧dz=−dz∧dy\begin{align*} dx \wedge dx &= 0,\ dy \wedge dy = 0,\ dz \wedge dz = 0 \\ dx \wedge dy &= -dy \wedge dx ,\ dx \wedge dy = -dy \wedge dz \\ dy \wedge dz &= -dz \wedge dy \end{align*}

This naturally reminds people of the cross product: swapping the order flips the sign, and wedging something with itself gives zero. In three dimensions the wedge product is indeed closely related to the cross product, but the point is that the wedge product generalizes to other dimensions as well.

Illustration of orientation

The wedge product represents an oriented geometric element. A one-dimensional element like dx is a tiny directed line segment. Wedge it with dy, and what you get is an oriented area.

That is where the jump to higher dimensions comes from. A volume element in general is whatever fills the underlying space: dx fills a one-dimensional direction, an area element fills a surface, and higher-dimensional analogues do the same in higher dimensions. In that sense, the wedge product is really a way of increasing dimension.

There is another subtle point here. In formulas like

∮∂DPdx+Qdy=∬D(∂Q∂x−∂P∂y)dA\oint_{\partial D}Pdx + Qdy=\iint_{D}\left( \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right)dA

it may look as if orientation does not matter and everything is just dx, dy \to dxdy. But the orientation is hidden in the fact that the reference frame does not change. Once the directions of dx and dy are fixed, the orientation of dA = dxdy is fixed too. Differential elements are treated as objective geometric objects, not something that rotates just because the observer turns around.

Now go back to the exterior derivative d. What is the relation between a function-like object \omega and its differential d\omega? It is simply variation.

To explain why that matters, it helps to rethink what a function is. The key idea is intrinsic dimension. In an integral like \int f(x)\,dx, the differential element is dx, so what the function really does is assign an intensity to each tiny one-dimensional element. Likewise, in a two-dimensional case the function assigns values to oriented area elements. In that sense, the dimensionality of the object is determined by the differential form that carries it.

Then what is a derivative? It is the variation of that intensity from one tiny element to the next. So in

∫D dω=∫∂D ω\int _{D} \, d \omega=\int _{\partial D} \, \omega

the “disturbance” or infinitesimal change inside the region is precisely the differential of the original object.

Once the concepts are in place, the computation itself becomes very straightforward. For example, let us derive Stokes’ theorem:

∮∂DF→⋅dr→=∮Pdx+Qdy+Rdz=(∂P∂xdx+∂P∂ydy+∂P∂zdz)∧dx+(∂Q∂xdx+∂Q∂ydy+∂Q∂zdz)∧dy+(∂R∂xdx+∂R∂ydy+∂R∂zdz)∧dz=(∂R∂y−∂Q∂z)dydz+(∂P∂z−∂R∂x)dzdx+(∂Q∂x−∂P∂y)dxdy=∬D(▽×F→)⋅dA→\begin{align*} \oint_{\partial D}\overrightarrow{F}\cdot d \overrightarrow{r} &=\oint Pdx+Qdy+Rdz \\ &=\left( \frac{\partial P}{\partial x}dx+\frac{\partial P}{\partial y}dy+\frac{\partial P}{\partial z}dz \right)\wedge dx+\left( \frac{\partial Q}{\partial x}dx+\frac{\partial Q}{\partial y}dy+\frac{\partial Q}{\partial z}dz \right)\wedge dy+\left( \frac{\partial R}{\partial x}dx+\frac{\partial R}{\partial y}dy+\frac{\partial R}{\partial z}dz \right)\wedge dz \\ &=\left( \frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z} \right)dydz+\left( \frac{\partial P}{\partial z}-\frac{\partial R}{\partial x} \right)dzdx+\left( \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right)dxdy\\ &= \iint_{D}(\triangledown\times \overrightarrow{F})\cdot d \overrightarrow{A} \end{align*}

That is really all there is to it. The other classical formulas can be derived in the same spirit.

Afterword

I originally thought about breaking this into more numbered sections, but every idea depends on the previous one, so it was hard to find a clean stopping point. In the end this is mostly a personal note anyway.

To be honest, if someone does not care about the underlying structure, memorizing the standard formulas is enough for ordinary problems. And if someone already understands all of this, then they probably do not need my explanation.

Still, I care a lot about definitions. To me, a good definition is like driving a nail into a tower from high up: even when it is not anchored at ground level, it can still hold other things in place.

The funny part is that the real nature of many things often turns out to be completely different from what you first imagined. People always say science is flexible, but the truly flexible things are often not inside science at all.

References