Stokes' theorem

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An illustration of Stokes' theorem, with surface Template:Math, its boundary Template:Math and the normal vector Template:Mvar. The direction of positive circulation of the bounding contour Template:Math, and the direction Template:Mvar of positive flux through the surface Template:Math, are related by a right-hand-rule (i.e., the right hand the fingers circulate along Template:Math and the thumb is directed along Template:Mvar).

Stokes' theorem, also known as the Kelvin–Stokes theorem after Lord Kelvin and George Stokes, the fundamental theorem for curls, or simply the curl theorem, is a theorem in vector calculus on ${ℝ}^3$. Given a vector field, the theorem relates the integral of the curl of the vector field over some surface, to the line integral of the vector field around the boundary of the surface. The classical theorem of Stokes can be stated in one sentence:

The line integral of a vector field over a loop is equal to the surface integral of its curl over the enclosed surface.

Stokes' theorem is a special case of the generalized Stokes theorem. In particular, a vector field on ${ℝ}^3$ can be considered as a 1-form in which case its curl is its exterior derivative, a 2-form.

Theorem

Let $\mathrm{\Sigma }$ be a smooth oriented surface in ${ℝ}^3$ with boundary $\partial \mathrm{\Sigma }\equiv \mathrm{\Gamma }$. If a vector field $𝐅(x,y,z)=(F_x(x,y,z),F_y(x,y,z),F_z(x,y,z))$ is defined and has continuous first order partial derivatives in a region containing $\mathrm{\Sigma }$, then $$\underset{\mathrm{\Sigma }}{\iint }(\mathrm{\nabla }\times 𝐅)\cdot \mathrm{d}\mathrm{\Sigma }={\oint }_{\partial \mathrm{\Sigma }}𝐅\cdot \mathrm{d}\mathrm{\Gamma }.$$ More explicitly, the equality says that $$\begin{array}{cc}& \underset{\mathrm{\Sigma }}{\iint }((\frac{\partial F_z}{\partial y}-\frac{\partial F_y}{\partial z})\mathrm{d}y\mathrm{d}z+(\frac{\partial F_x}{\partial z}-\frac{\partial F_z}{\partial x})\mathrm{d}z\mathrm{d}x+(\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y})\mathrm{d}x\mathrm{d}y)\\ & ={\oint }_{\partial \mathrm{\Sigma }}(F_x\mathrm{d}x+F_y\mathrm{d}y+F_z\mathrm{d}z).\end{array}$$

The main challenge in a precise statement of Stokes' theorem is in defining the notion of a boundary. Surfaces such as the Koch snowflake, for example, are well-known not to exhibit a Riemann-integrable boundary, and the notion of surface measure in Lebesgue theory cannot be defined for a non-Lipschitz surface. One (advanced) technique is to pass to a weak formulation and then apply the machinery of geometric measure theory; for that approach see the coarea formula. In this article, we instead use a more elementary definition, based on the fact that a boundary can be discerned for full-dimensional subsets of ${ℝ}^2$.

A more detailed statement will be given for subsequent discussions. Let $\gamma :[a,b]\to {ℝ}^2$ be a piecewise smooth Jordan plane curve. The Jordan curve theorem implies that $\gamma$ divides ${ℝ}^2$ into two components, a compact one and another that is non-compact. Let $D$ denote the compact part; then $D$ is bounded by $\gamma$. It now suffices to transfer this notion of boundary along a continuous map to our surface in ${ℝ}^3$. But we already have such a map: the parametrization of $\mathrm{\Sigma }$.

Suppose $\psi :D\to {ℝ}^3$ is piecewise smooth at the neighborhood of $D$, with $\mathrm{\Sigma }=\psi (D)$. If $\mathrm{\Gamma }$ is the space curve defined by $\mathrm{\Gamma }(t)=\psi (\gamma (t))$ then we call $\mathrm{\Gamma }$ the boundary of $\mathrm{\Sigma }$, written $\partial \mathrm{\Sigma }$.

With the above notation, if $𝐅$ is any smooth vector field on ${ℝ}^3$, then$${\oint }_{\partial \mathrm{\Sigma }}𝐅\cdot \mathrm{d}\mathrm{\Gamma }=\underset{\mathrm{\Sigma }}{\iint }\mathrm{\nabla }\times 𝐅\cdot \mathrm{d}\mathrm{\Sigma }.$$

Here, the "$\cdot$" represents the dot product in ${ℝ}^3$.

Special case of a more general theorem

Stokes' theorem can be viewed as a special case of the following identity: $${\oint }_{\partial \mathrm{\Sigma }}(𝐅\cdot \mathrm{d}\mathrm{\Gamma })𝐠=\underset{\mathrm{\Sigma }}{\iint }[\mathrm{d}\mathrm{\Sigma }\cdot (\mathrm{\nabla }\times 𝐅-𝐅\times \mathrm{\nabla })]𝐠,$$ where $𝐠$ is any smooth vector or scalar field in ${ℝ}^3$. When $𝐠$ is a uniform scalar field, the standard Stokes' theorem is recovered.

Proof

The proof of the theorem consists of 4 steps. We assume Green's theorem, so what is of concern is how to boil down the three-dimensional complicated problem (Stokes' theorem) to a two-dimensional rudimentary problem (Green's theorem). When proving this theorem, mathematicians normally deduce it as a special case of a more general result, which is stated in terms of differential forms, and proved using more sophisticated machinery. While powerful, these techniques require substantial background, so the proof below avoids them, and does not presuppose any knowledge beyond a familiarity with basic vector calculus and linear algebra. At the end of this section, a short alternative proof of Stokes' theorem is given, as a corollary of the generalized Stokes' theorem.

Elementary proof

First step of the elementary proof (parametrization of integral)

As in Template:Slink, we reduce the dimension by using the natural parametrization of the surface. Let Template:Math and Template:Mvar be as in that section, and note that by change of variables $${\oint }_{\partial \mathrm{\Sigma }}𝐅(𝐱)\cdot \mathrm{d}\mathrm{\Gamma }={\oint }_{\gamma }𝐅(\mathit{\psi }(\mathit{\gamma }))\cdot \mathrm{d}\mathit{\psi }(\mathit{\gamma })={\oint }_{\gamma }𝐅(\mathit{\psi }(𝐲))\cdot J_{𝐲}(\mathit{\psi })\mathrm{d}\gamma$$ where Template:Mvar stands for the Jacobian matrix of Template:Mvar at Template:Math.

Now let Template:Math be an orthonormal basis in the coordinate directions of Template:Math.

Recognizing that the columns of Template:Math are precisely the partial derivatives of Template:Math at Template:Math, we can expand the previous equation in coordinates as $$\begin{array}{cc}{\oint }_{\partial \mathrm{\Sigma }}𝐅(𝐱)\cdot \mathrm{d}\mathrm{\Gamma }& ={\oint }_{\gamma }𝐅(\mathit{\psi }(𝐲))J_{𝐲}(\mathit{\psi }){𝐞}_u({𝐞}_u\cdot \mathrm{d}𝐲)+𝐅(\mathit{\psi }(𝐲))J_{𝐲}(\mathit{\psi }){𝐞}_v({𝐞}_v\cdot \mathrm{d}𝐲)\\ & ={\oint }_{\gamma }((𝐅(\mathit{\psi }(𝐲))\cdot \frac{\partial \mathit{\psi }}{\partial u}(𝐲)){𝐞}_u+(𝐅(\mathit{\psi }(𝐲))\cdot \frac{\partial \mathit{\psi }}{\partial v}(𝐲)){𝐞}_v)\cdot \mathrm{d}𝐲\end{array}$$

Second step in the elementary proof (defining the pullback)

The previous step suggests we define the function $$𝐏(u,v)=(𝐅(\mathit{\psi }(u,v))\cdot \frac{\partial \mathit{\psi }}{\partial u}(u,v)){𝐞}_u+(𝐅(\mathit{\psi }(u,v))\cdot \frac{\partial \mathit{\psi }}{\partial v}(u,v)){𝐞}_v$$

Now, if the scalar value functions $P_u$ and $P_v$ are defined as follows, $$P_u(u,v)=(𝐅(\mathit{\psi }(u,v))\cdot \frac{\partial \mathit{\psi }}{\partial u}(u,v))$$ $$P_v(u,v)=(𝐅(\mathit{\psi }(u,v))\cdot \frac{\partial \mathit{\psi }}{\partial v}(u,v))$$ then, $$𝐏(u,v)=P_u(u,v){𝐞}_u+P_v(u,v){𝐞}_v.$$

This is the pullback of Template:Math along Template:Math, and, by the above, it satisfies $${\oint }_{\partial \mathrm{\Sigma }}𝐅(𝐱)\cdot \mathrm{d}𝐥={\oint }_{\gamma }𝐏(𝐲)\cdot \mathrm{d}𝐥={\oint }_{\gamma }(P_u(u,v){𝐞}_u+P_v(u,v){𝐞}_v)\cdot \mathrm{d}𝐥$$

We have successfully reduced one side of Stokes' theorem to a 2-dimensional formula; we now turn to the other side.

Third step of the elementary proof (second equation)

First, calculate the partial derivatives appearing in Green's theorem, via the product rule: $$\begin{array}{cc}\frac{\partial P_u}{\partial v}& =\frac{\partial (𝐅\circ \mathit{\psi })}{\partial v}\cdot \frac{\partial \mathit{\psi }}{\partial u}+(𝐅\circ \mathit{\psi })\cdot \frac{{\partial }^2\mathit{\psi }}{\partial v\partial u}\\ \frac{\partial P_v}{\partial u}& =\frac{\partial (𝐅\circ \mathit{\psi })}{\partial u}\cdot \frac{\partial \mathit{\psi }}{\partial v}+(𝐅\circ \mathit{\psi })\cdot \frac{{\partial }^2\mathit{\psi }}{\partial u\partial v}\end{array}$$

Conveniently, the second term vanishes in the difference, by equality of mixed partials. So, $$\begin{array}{cc}\frac{\partial P_v}{\partial u}-\frac{\partial P_u}{\partial v}& =\frac{\partial (𝐅\circ \mathit{\psi })}{\partial u}\cdot \frac{\partial \mathit{\psi }}{\partial v}-\frac{\partial (𝐅\circ \mathit{\psi })}{\partial v}\cdot \frac{\partial \mathit{\psi }}{\partial u}\\ & =\frac{\partial \mathit{\psi }}{\partial v}\cdot (J_{\mathit{\psi }(u,v)}𝐅)\frac{\partial \mathit{\psi }}{\partial u}-\frac{\partial \mathit{\psi }}{\partial u}\cdot (J_{\mathit{\psi }(u,v)}𝐅)\frac{\partial \mathit{\psi }}{\partial v}& & \text{(chain rule)}\\ & =\frac{\partial \mathit{\psi }}{\partial v}\cdot (J_{\mathit{\psi }(u,v)}𝐅-{(J_{\mathit{\psi }(u,v)}𝐅)}^{𝖳})\frac{\partial \mathit{\psi }}{\partial u}\end{array}$$

But now consider the matrix in that quadratic form—that is, $J_{\mathit{\psi }(u,v)}𝐅-(J_{\mathit{\psi }(u,v)}𝐅{)}^{𝖳}$. We claim this matrix in fact describes a cross product. Here the superscript "${}^{𝖳}$" represents the transposition of matrices.

To be precise, let $A=(A_{ij}{)}_{ij}$ be an arbitrary Template:Math matrix and let $$𝐚=\left[\begin{array}{c}{a}_1\\ a_2\\ a_3\end{array}\right]=\left[\begin{array}{c}{A}_{32}-A_{23}\\ A_{13}-A_{31}\\ A_{21}-A_{12}\end{array}\right]$$

Note that Template:Math is linear, so it is determined by its action on basis elements. But by direct calculation $$\begin{array}{cc}(A-A^{𝖳}){𝐞}_1& =\left[\begin{array}{c}0\\ a_3\\ -a_2\end{array}\right]=𝐚\times {𝐞}_1\\ (A-A^{𝖳}){𝐞}_2& =\left[\begin{array}{c}-a_3\\ 0\\ a_1\end{array}\right]=𝐚\times {𝐞}_2\\ (A-A^{𝖳}){𝐞}_3& =\left[\begin{array}{c}{a}_2\\ -a_1\\ 0\end{array}\right]=𝐚\times {𝐞}_3\end{array}$$ Here, Template:Math represents an orthonormal basis in the coordinate directions of ${ℝ}^3$.

Thus Template:Math for any Template:Math.

Substituting $(J_{\mathit{\psi }(u,v)}𝐅)$ for Template:Mvar, we obtain $$((J_{\mathit{\psi }(u,v)}𝐅)-{(J_{\mathit{\psi }(u,v)}𝐅)}^{𝖳})𝐱=(\mathrm{\nabla }\times 𝐅)\times 𝐱,\text{for all}𝐱\in {ℝ}^3$$

We can now recognize the difference of partials as a (scalar) triple product: $$\begin{array}{cc}\frac{\partial P_v}{\partial u}-\frac{\partial P_u}{\partial v}& =\frac{\partial \mathit{\psi }}{\partial v}\cdot (\mathrm{\nabla }\times 𝐅)\times \frac{\partial \mathit{\psi }}{\partial u}=(\mathrm{\nabla }\times 𝐅)\cdot \frac{\partial \mathit{\psi }}{\partial u}\times \frac{\partial \mathit{\psi }}{\partial v}\end{array}$$

On the other hand, the definition of a surface integral also includes a triple product—the very same one! $$\begin{array}{cc}\underset{\mathrm{\Sigma }}{\iint }(\mathrm{\nabla }\times 𝐅)\cdot d\mathrm{\Sigma }& =\underset{D}{\iint }(\mathrm{\nabla }\times 𝐅)(\mathit{\psi }(u,v))\cdot \frac{\partial \mathit{\psi }}{\partial u}(u,v)\times \frac{\partial \mathit{\psi }}{\partial v}(u,v)\mathrm{d}u\mathrm{d}v\end{array}$$

So, we obtain $$\underset{\mathrm{\Sigma }}{\iint }(\mathrm{\nabla }\times 𝐅)\cdot \mathrm{d}\mathrm{\Sigma }=\underset{D}{\iint }(\frac{\partial P_v}{\partial u}-\frac{\partial P_u}{\partial v})\mathrm{d}u\mathrm{d}v$$

Fourth step of the elementary proof (reduction to Green's theorem)

Combining the second and third steps and then applying Green's theorem completes the proof. Green's theorem asserts the following: for any region D bounded by the Jordans closed curve γ and two scalar-valued smooth functions $P_u(u,v),P_v(u,v)$ defined on D;

$${\oint }_{\gamma }(P_u(u,v){𝐞}_u+P_v(u,v){𝐞}_v)\cdot \mathrm{d}𝐥=\underset{D}{\iint }(\frac{\partial P_v}{\partial u}-\frac{\partial P_u}{\partial v})\mathrm{d}u\mathrm{d}v$$

We can substitute the conclusion of STEP2 into the left-hand side of Green's theorem above, and substitute the conclusion of STEP3 into the right-hand side. Q.E.D.

Proof via differential forms

The functions ${ℝ}^3\to {ℝ}^3$ can be identified with the differential 1-forms on ${ℝ}^3$ via the map $$F_x{𝐞}_1+F_y{𝐞}_2+F_z{𝐞}_3\mapsto F_x\mathrm{d}x+F_y\mathrm{d}y+F_z\mathrm{d}z.$$

Write the differential 1-form associated to a function Template:Math as Template:Math. Then one can calculate that $$\star {\omega }_{\mathrm{\nabla }\times 𝐅}=\mathrm{d}{\omega }_{𝐅},$$ where Template:Math is the Hodge star and $\mathrm{d}$ is the exterior derivative. Thus, by generalized Stokes' theorem, $${\oint }_{\partial \mathrm{\Sigma }}𝐅\cdot \mathrm{d}\mathit{\gamma }={\oint }_{\partial \mathrm{\Sigma }}{\omega }_{𝐅}=\underset{\mathrm{\Sigma }}{\int }\mathrm{d}{\omega }_{𝐅}=\underset{\mathrm{\Sigma }}{\int }\star {\omega }_{\mathrm{\nabla }\times 𝐅}=\underset{\mathrm{\Sigma }}{\iint }\mathrm{\nabla }\times 𝐅\cdot \mathrm{d}\mathrm{\Sigma }$$

Applications

Irrotational fields

In this section, we will discuss the irrotational field (lamellar vector field) based on Stokes' theorem.

Definition 2-1 (irrotational field). A smooth vector field Template:Math on an open $U\subseteq {ℝ}^3$ is irrotational (lamellar vector field) if Template:Math.

This concept is very fundamental in mechanics; as we'll prove later, if Template:Math is irrotational and the domain of Template:Math is simply connected, then Template:Math is a conservative vector field.

Helmholtz's theorem

In this section, we will introduce a theorem that is derived from Stokes' theorem and characterizes vortex-free vector fields. In classical mechanics and fluid dynamics it is called Helmholtz's theorem.

Theorem 2-1 (Helmholtz's theorem in fluid dynamics).Template:Rp Let $U\subseteq {ℝ}^3$ be an open subset with a lamellar vector field Template:Math and let Template:Math be piecewise smooth loops. If there is a function Template:Math such that

• [TLH0] Template:Mvar is piecewise smooth,
• [TLH1] Template:Math for all Template:Math,
• [TLH2] Template:Math for all Template:Math,
• [TLH3] Template:Math for all Template:Math.

Then, $$\underset{c_0}{\int }𝐅\mathrm{d}{c}_0=\underset{c_1}{\int }𝐅\mathrm{d}{c}_1$$

Some textbooks such as Lawrence call the relationship between Template:Math and Template:Math stated in theorem 2-1 as "homotopic" and the function Template:Math as "homotopy between Template:Math and Template:Math". However, "homotopic" or "homotopy" in above-mentioned sense are different (stronger than) typical definitions of "homotopic" or "homotopy"; the latter omit condition [TLH3]. So from now on we refer to homotopy (homotope) in the sense of theorem 2-1 as a tubular homotopy (resp. tubular-homotopic).Template:Refn

Proof of Helmholtz's theorem

The definitions of Template:Math

In what follows, we abuse notation and use "$\oplus$" for concatenation of paths in the fundamental groupoid and "$\ominus$" for reversing the orientation of a path.

Let Template:Math, and split Template:Math into four line segments Template:Math. $$\begin{array}{cc}{\gamma }_1:[0,1]\to D;& {\gamma }_1(t)=(t,0)\\ {\gamma }_2:[0,1]\to D;& {\gamma }_2(s)=(1,s)\\ {\gamma }_3:[0,1]\to D;& {\gamma }_3(t)=(1-t,1)\\ {\gamma }_4:[0,1]\to D;& {\gamma }_4(s)=(0,1-s)\end{array}$$ so that $$\partial D={\gamma }_1\oplus {\gamma }_2\oplus {\gamma }_3\oplus {\gamma }_4$$

By our assumption that Template:Math and Template:Math are piecewise smooth homotopic, there is a piecewise smooth homotopy Template:Math $$\begin{array}{cccc}{\mathrm{\Gamma }}_i(t)& =H({\gamma }_i(t))& & i=1,2,3,4\\ \mathrm{\Gamma }(t)& =H(\gamma (t))=({\mathrm{\Gamma }}_1\oplus {\mathrm{\Gamma }}_2\oplus {\mathrm{\Gamma }}_3\oplus {\mathrm{\Gamma }}_4)(t)\end{array}$$

Let Template:Mvar be the image of Template:Mvar under Template:Mvar. That $$\underset{S}{\iint }\mathrm{\nabla }\times 𝐅\mathrm{d}S={\oint }_{\mathrm{\Gamma }}𝐅\mathrm{d}\mathrm{\Gamma }$$ follows immediately from Stokes' theorem. Template:Math is lamellar, so the left side vanishes, i.e. $$0={\oint }_{\mathrm{\Gamma }}𝐅\mathrm{d}\mathrm{\Gamma }=\sum _{i=1}^4{\oint }_{{\mathrm{\Gamma }}_i}𝐅\mathrm{d}\mathrm{\Gamma }$$

As Template:Mvar is tubular(satisfying [TLH3]),${\mathrm{\Gamma }}_2=\ominus {\mathrm{\Gamma }}_4$ and ${\mathrm{\Gamma }}_2=\ominus {\mathrm{\Gamma }}_4$. Thus the line integrals along Template:Math and Template:Math cancel, leaving $$0={\oint }_{{\mathrm{\Gamma }}_1}𝐅\mathrm{d}\mathrm{\Gamma }+{\oint }_{{\mathrm{\Gamma }}_3}𝐅\mathrm{d}\mathrm{\Gamma }$$

On the other hand, Template:Math, $c_3=\ominus {\mathrm{\Gamma }}_3$, so that the desired equality follows almost immediately.

Conservative forces

Above Helmholtz's theorem gives an explanation as to why the work done by a conservative force in changing an object's position is path independent. First, we introduce the Lemma 2-2, which is a corollary of and a special case of Helmholtz's theorem.

Lemma 2-2. Let $U\subseteq {ℝ}^3$ be an open subset, with a Lamellar vector field Template:Math and a piecewise smooth loop Template:Math. Fix a point Template:Math, if there is a homotopy Template:Math such that

• [SC0] Template:Mvar is piecewise smooth,
• [SC1] Template:Math for all Template:Math,
• [SC2] Template:Math for all Template:Math,
• [SC3] Template:Math for all Template:Math.

Then, $$\underset{c_0}{\int }𝐅\mathrm{d}{c}_0=0$$

Above Lemma 2-2 follows from theorem 2–1. In Lemma 2-2, the existence of Template:Mvar satisfying [SC0] to [SC3] is crucial;the question is whether such a homotopy can be taken for arbitrary loops. If Template:Mvar is simply connected, such Template:Mvar exists. The definition of simply connected space follows:

Definition 2-2 (simply connected space). Let $M\subseteq {ℝ}^n$ be non-empty and path-connected. Template:Mvar is called simply connected if and only if for any continuous loop, Template:Math there exists a continuous tubular homotopy Template:Math from Template:Mvar to a fixed point Template:Math; that is,

• [SC0'] Template:Mvar is continuous,
• [SC1] Template:Math for all Template:Math,
• [SC2] Template:Math for all Template:Math,
• [SC3] Template:Math for all Template:Math.

The claim that "for a conservative force, the work done in changing an object's position is path independent" might seem to follow immediately if the M is simply connected.　However, recall that simple-connection only guarantees the existence of a continuous homotopy satisfying [SC1-3]; we seek a piecewise smooth homotopy satisfying those conditions instead.

Fortunately, the gap in regularity is resolved by the Whitney's approximation theorem.Template:Rp In other words, the possibility of finding a continuous homotopy, but not being able to integrate over it, is actually eliminated with the benefit of higher mathematics.　We thus obtain the following theorem.

Theorem 2-2. Let $U\subseteq {ℝ}^3$ be open and simply connected with an irrotational vector field Template:Math. For all piecewise smooth loops Template:Math $$\underset{c_0}{\int }𝐅\mathrm{d}{c}_0=0$$

Maxwell's equations

Template:See also In the physics of electromagnetism, Stokes' theorem provides the justification for the equivalence of the differential form of the Maxwell–Faraday equation and the Maxwell–Ampère equation and the integral form of these equations. For Faraday's law, Stokes theorem is applied to the electric field, $𝐄$: $${\oint }_{\partial \mathrm{\Sigma }}𝐄\cdot \mathrm{d}𝒍=\underset{\mathrm{\Sigma }}{\iint }\mathrm{\nabla }\times 𝐄\cdot \mathrm{d}𝐒.$$

For Ampère's law, Stokes' theorem is applied to the magnetic field, $𝐁$: $${\oint }_{\partial \mathrm{\Sigma }}𝐁\cdot \mathrm{d}𝒍=\underset{\mathrm{\Sigma }}{\iint }\mathrm{\nabla }\times 𝐁\cdot \mathrm{d}𝐒.$$