# Kelvin–Stokes theorem An illustration of the Kelvin–Stokes theorem, with surface Σ, its boundary ∂Σ and the "normal" vector n.

The Kelvin–Stokes theorem, named after Lord Kelvin and George Stokes, also known as the Stokes' theorem, the fundamental theorem for curls or simply the curl theorem, is a theorem in vector calculus on $\mathbb {R} ^{3}$ . 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.

If a vector field $\mathbf {A} =(P(x,y,z),Q(x,y,z),R(x,y,z))$ is defined in a region with smooth oriented surface $\Sigma$ and has first order continuous partial derivatives then:

{\begin{aligned}\iint \limits _{\Sigma }(\nabla \times \mathbf {A} )d\mathbf {a} &=\iint \limits _{\Sigma }{\Bigg (}\left({\frac {\partial R}{\partial y}}-{\frac {\partial Q}{\partial z}}\right)\,dy\,dz+\left({\frac {\partial P}{\partial z}}-{\frac {\partial R}{\partial x}}\right)\,dz\,dx+\left({\frac {\partial Q}{\partial x}}-{\frac {\partial P}{\partial y}}\right)\,dx\,dy{\Bigg )}\\&=\oint \limits _{\partial \Sigma }{\Big (}P\,dx+Q\,dy+R\,dz{\Big )}=\oint \mathbf {A} \cdot d\mathbf {l} ,\end{aligned}} where $\partial \Sigma$ is boundary of region with smooth surface $\Sigma$ .

The above classical Kelvin-Stokes theorem can be stated in one sentence: The integral of a vector field over a loop is equal to the flux of its curl through the enclosed surface.

The Kelvin–Stokes theorem is a special case of the “generalized Stokes' theorem.” In particular, a vector field on $\mathbb {R} ^{3}$ can be considered as a 1-form in which case curl is the exterior derivative.

## Theorem

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 2.

Let γ: [a, b] → R2 be a piecewise smooth Jordan plane curve. The Jordan curve theorem implies that γ divides R2 into two components, a compact one and another that is non-compact. Let D denote the compact part; then D is bounded by γ. It now suffices to transfer this notion of boundary along a continuous map to our surface in 3. But we already have such a map: the parametrization of Σ.

Suppose ψ: DR3 is smooth, with Σ = ψ(D). If Γ is the space curve defined by Γ(t) = ψ(γ(t))[note 1], then we call Γ the boundary of Σ, written Σ.

With the above notation, if F is any smooth vector field on R3, then

$\oint _{\partial \Sigma }\mathbf {F} \,\cdot \,d{\mathbf {\Gamma } }=\iint _{\Sigma }\nabla \times \mathbf {F} \,\cdot \,d\mathbf {S} .$ ## 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 (Kelvin–Stokes theorem) to a two-dimensional rudimentary problem (Green's theorem). When proving this theorem, mathematicians normally use the differential form. The "pull-back of a differential form" is a very powerful tool for this situation, but learning differential forms requires substantial background knowledge. So, the proof below does not require knowledge of differential forms, and may be helpful for understanding the notion of differential forms.

### Elementary proof

#### First step of the proof (parametrization of integral)

As in Theorem § Notes, we reduce the dimension by using the natural parametrization of the surface. Let ψ and γ be as in that section, and note that by change of variables

$\oint _{\partial \Sigma }{\mathbf {F} (\mathbf {x} )\cdot \,d\mathbf {l} }=\oint _{\gamma }{\mathbf {F} (\mathbf {\psi } (\mathbf {y} ))\cdot \,d\mathbf {\psi } (\mathbf {y} )}=\oint _{\gamma }{\mathbf {F} (\mathbf {\psi } (\mathbf {y} ))J_{\mathbf {y} }(\mathbf {\psi } )\,d\mathbf {y} }$ where stands for the Jacobian matrix of ψ.

Now let {eu,ev} be an orthonormal basis in the coordinate directions of 2. Recognizing that the columns of Jyψ are precisely the partial derivatives of ψ at y , we can expand the previous equation in coordinates as

{\begin{aligned}\oint _{\partial \Sigma }{\mathbf {F} (\mathbf {x} )\cdot \,d\mathbf {l} }&=\oint _{\gamma }{\mathbf {F} (\mathbf {\psi } (\mathbf {y} ))J_{\mathbf {y} }(\mathbf {\psi } )\mathbf {e} _{u}(\mathbf {e} _{u}\cdot \,d\mathbf {y} )+\mathbf {F} (\mathbf {\psi } (\mathbf {y} ))J_{\mathbf {y} }(\mathbf {\psi } )\mathbf {e} _{v}(\mathbf {e} _{v}\cdot \,d\mathbf {y} )}\\&=\oint _{\gamma }{\left(\left(\mathbf {F} (\mathbf {\psi } (\mathbf {y} ))\cdot {\frac {\partial \mathbf {\psi } }{\partial u}}(\mathbf {y} )\right)\mathbf {e} _{u}+\left(\mathbf {F} (\mathbf {\psi } (\mathbf {y} ))\cdot {\frac {\partial \mathbf {\psi } }{\partial v}}(\mathbf {y} )\right)\mathbf {e} _{v}\right)\cdot \,d\mathbf {y} }\end{aligned}} #### Second step in the proof (defining the pullback)

The previous step suggests we define the function

$\mathbf {P} (u,v)=\left(\mathbf {F} (\mathbf {\psi } (u,v))\cdot {\frac {\partial \mathbf {\psi } }{\partial u}}(u,v)\right)\mathbf {e} _{u}+\left(\mathbf {F} (\mathbf {\psi } (u,v))\cdot {\frac {\partial \mathbf {\psi } }{\partial v}}\right)\mathbf {e} _{v}$ This is the pullback of F along ψ , and, by the above, it satisfies

$\oint _{\partial \Sigma }{\mathbf {F} (\mathbf {x} )\cdot \,d\mathbf {l} }=\oint _{\gamma }{\mathbf {P} (\mathbf {y} )\cdot \,d\mathbf {l} }$ 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 proof (second equation)

First, calculate the partial derivatives appearing in Green's theorem, via the product rule:

{\begin{aligned}{\frac {\partial P_{1}}{\partial v}}&={\frac {\partial (\mathbf {F} \circ \psi )}{\partial v}}\cdot {\frac {\partial \psi }{\partial u}}+(\mathbf {F} \circ \psi )\cdot {\frac {\partial ^{2}\psi }{\partial v\partial u}}\\{\frac {\partial P_{2}}{\partial u}}&={\frac {\partial (\mathbf {F} \circ \psi )}{\partial u}}\cdot {\frac {\partial \psi }{\partial v}}+(\mathbf {F} \circ \psi )\cdot {\frac {\partial ^{2}\psi }{\partial u\partial v}}\end{aligned}} Conveniently, the second term vanishes in the difference, by equality of mixed partials. So,

{\begin{aligned}{\frac {\partial P_{1}}{\partial v}}-{\frac {\partial P_{2}}{\partial u}}&={\frac {\partial (\mathbf {F} \circ \psi )}{\partial v}}\cdot {\frac {\partial \psi }{\partial u}}-{\frac {\partial (\mathbf {F} \circ \psi )}{\partial u}}\cdot {\frac {\partial \psi }{\partial v}}\\&={\frac {\partial \psi }{\partial u}}(J_{\psi (u,v)}\mathbf {F} ){\frac {\partial \psi }{\partial v}}-{\frac {\partial \psi }{\partial v}}(J_{\psi (u,v)}\mathbf {F} ){\frac {\partial \psi }{\partial u}}&&{\text{ (chain rule)}}\\&={\frac {\partial \psi }{\partial u}}(J_{\psi (u,v)}\mathbf {F} -(J_{\psi (u,v)}\mathbf {F} )^{\mathsf {T}}){\frac {\partial \psi }{\partial v}}\end{aligned}} But now consider the matrix in that quadratic form—that is, $J_{\psi (u,v)}\mathbf {F} -(J_{\psi (u,v)}\mathbf {F} )^{\mathsf {T}}$ . We claim this matrix in fact describes a cross product.

To be precise, let $A=(A_{ij})_{ij}$ be an arbitrary 3 × 3 matrix and let

$\mathbf {a} ={\begin{pmatrix}A_{32}-A_{23}\\A_{13}-A_{31}\\A_{21}-A_{12}\end{pmatrix}}$ Note that x a × x is linear, so it is determined by its action on basis elements. But by direct calculation

{\begin{aligned}(A-A^{\mathsf {T}})\mathbf {e} _{1}&={\begin{pmatrix}0\\a_{3}\\-a_{2}\end{pmatrix}}=\mathbf {a} \times \mathbf {e} _{1}\\(A-A^{\mathsf {T}})\mathbf {e} _{2}&={\begin{pmatrix}-a_{3}\\0\\a_{1}\end{pmatrix}}=\mathbf {a} \times \mathbf {e} _{2}\\(A-A^{\mathsf {T}})\mathbf {e} _{3}&={\begin{pmatrix}a_{2}\\-a_{1}\\0\end{pmatrix}}=\mathbf {a} \times \mathbf {e} _{3}\end{aligned}} Thus (A-AT) x = a × x for any x . Substituting J F for A, we obtain

$({(J_{\psi (u,v)}\mathbf {F} )}_{\psi (u,v)}-{(J_{\psi (u,v)}\mathbf {F} )}^{\mathsf {T}})\mathbf {x} =(\nabla \times \mathbf {F} )\times \mathbf {x} ,\quad {\text{for all}}\,\mathbf {x} \in \mathbf {R} ^{3}$ We can now recognize the difference of partials as a (scalar) triple product:

{\begin{aligned}{\frac {\partial P_{1}}{\partial v}}-{\frac {\partial P_{2}}{\partial u}}&={\frac {\partial \psi }{\partial u}}\cdot (\nabla \times \mathbf {F} )\times {\frac {\partial \psi }{\partial v}}\\&=\det \left[(\nabla \times \mathbf {F} )(\psi (u,v))\quad {\frac {\partial \psi }{\partial u}}(u,v)\quad {\frac {\partial \psi }{\partial v}}(u,v)\right]\end{aligned}} On the other hand, the definition of a surface integral also includes a triple product — the very same one!

{\begin{aligned}\iint _{S}(\nabla \times \mathbf {F} )\cdot \,d^{2}\mathbf {S} &=\iint _{D}{(\nabla \times \mathbf {F} )(\psi (u,v))\cdot \left({\frac {\partial \psi }{\partial u}}(u,v)\times {\frac {\partial \psi }{\partial v}}(u,v)\,du\,dv\right)}\\&=\iint _{D}\det \left[(\nabla \times \mathbf {F} )(\psi (u,v))\quad {\frac {\partial \psi }{\partial u}}(u,v)\quad {\frac {\partial \psi }{\partial v}}(u,v)\right]\,du\,dv\end{aligned}} So, we obtain

$\iint _{S}(\nabla \times \mathbf {F} )\cdot \,d^{2}\mathbf {S} =\iint _{D}\left({\frac {\partial P_{2}}{\partial u}}-{\frac {\partial P_{1}}{\partial v}}\right)\,du\,dv$ #### Fourth step of the proof (reduction to Green's theorem)

Combining the second and third steps, and then applying Green's theorem completes the proof.

### Proof via differential forms

ℝ→ℝ3 can be identified with the degree-1 one differential forms on 3 via the map

$F_{1}\mathbf {e} _{1}+F_{2}\mathbf {e} _{2}+F_{3}\mathbf {e} _{3}\mapsto F_{1}\,dx+F_{2}\,dy+F_{3}dy$ .

Write the differential 1-form associated to a function F as ωF. Then one can calculate that

$\star \omega _{\nabla \times \mathbf {F} }=d\omega _{\mathbf {F} }$ where is the Hodge star. Thus, by generalized Stokes' theorem, 

$\oint _{\partial \Sigma }{\mathbf {F} \cdot \,d\mathbf {l} }=\oint _{\partial \Sigma }{\omega _{\mathbf {F} }}=\int _{\Sigma }{d\omega _{\mathbf {F} }}=\int _{\Sigma }{\star \omega _{\nabla \times \mathbf {F} }}=\iint _{\Sigma }{\nabla \times \mathbf {F} \cdot \,d^{2}\mathbf {S} }$ ## Applications

### In fluid dynamics

In this section, we will discuss the lamellar vector field based on Kelvin–Stokes theorem.

### Irrotational fields

If the domain of F is simply connected, then F is a conservative vector field.

#### Helmholtz's theorems

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

Some textbooks such as Lawrence call the relationship between c0 and c1 stated in Theorem 2-1 as “homotopic” and the function H: [0, 1] × [0, 1] → U as “homotopy between c0 and c1”. 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).[note 2]

##### Proof of the theorem

In what follows, we abuse notation and use "+" for concatenation of paths in the fundamental groupoid and "-" for reversing the orientation of a path.

Let D = [0, 1] × [0, 1], and split D into 4 line segments γj.

{\begin{aligned}\gamma _{1}:[0,1]\to D;\quad &\gamma _{1}(t)=(t,0)\\\gamma _{2}:[0,1]\to D;\quad &\gamma _{2}(s)=(1,s)\\\gamma _{3}:[0,1]\to D;\quad &\gamma _{3}(t)=(1-t,1)\\\gamma _{4}:[0,1]\to D;\quad &\gamma _{4}(s)=(0,1-s)\partial D=\gamma _{1}+\gamma _{2}+\gamma _{3}+\gamma _{4}\end{aligned}} By our assumption that c1 and c2 are piecewise smooth homotopic, there is a piecewise smooth homotopy H: DM

{\begin{aligned}\Gamma _{i}(t)&=H(\gamma _{i}(t))&&i=1,2,3,4\\\Gamma (t)&=H(\gamma (t))=(\Gamma _{1}\oplus \Gamma _{2}\oplus \Gamma _{3}\oplus \Gamma _{4})(t)\end{aligned}} Let S be the image of D under H. That

$\iint _{S}\nabla \times \mathbf {F} \,dS=\oint _{\Gamma }\mathbf {F} \,d\Gamma$ follows immediately from the Kelvin–Stokes theorem. F is lamellar, so the left side vanishes, i.e.

$0=\oint _{\Gamma }\mathbf {F} \,d\Gamma =\sum _{i=1}^{4}\oint _{\Gamma _{i}}\mathbf {F} \,d\Gamma$ As H is tubular, Γ2=-Γ4. Thus the line integrals along Γ2(s) and Γ4(s) cancel, leaving

$0=\oint _{\Gamma _{1}}\mathbf {F} \,d\Gamma +\oint _{\Gamma _{3}}\mathbf {F} d\Gamma$ On the other hand, c1=Γ1 and c3=-Γ3, so that the desired equality follows almost immediately.

### Conservative forces

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 follows from Theorem 2-1. In Lemma 2-2, the existence of H satisfying [SC0] to [SC3] is crucial. If U is simply connected, such H exists. The definition of Simply connected space follows:

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

However, the gap in regularity is resolved by the Whitney approximation theorem.:136,421 We thus obtain the following theorem.