# Kelvin–Stokes theorem

The Kelvin–Stokes theorem, also known as the curl theorem, is a theorem in vector calculus on R3. 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 Kelvin–Stokes theorem is a special case of the “generalized Stokes' theorem.” In particular, the vector field on R3 can be considered as a 1-form in which case curl is the exterior derivative.

## Theorem

Let γ : [a, b] → R2 be a Piecewise smooth Jordan plane curve, that bounds the domain DR2.[note 1] Suppose ψ : DR3 is smooth, with S := ψ[D] [note 2] , and Γ is the space curve defined by Γ(t) = ψ(γ(t)).[note 3] If F a smooth vector field on R3, then

$\oint _{\Gamma }\mathbf {F} \,d\Gamma =\iint _{S}\nabla \times \mathbf {F} \,dS$ ## Proof

The proof of the Theorem consists of 4 steps.[note 4] We assume that the Green's theorem is known, 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)". In an ordinary, mathematicians use the differential form, especially "pull-back[note 4] of differential form" is very powerful tool for this situation, however, learning differential form needs too much background knowledge. So, the proof below does not require background information on differential form, and may be helpful for understanding the notion of differential form.

### First step of the proof (defining the pullback)

Define

$\mathbf {P} (u,v)=(P_{1}(u,v),P_{2}(u,v))$ so that P is the pull-back[note 4] of F, and that P(u, v) is R2-valued function, depends on two parameter u, v. In order to do so we define P1 and P2 as follows.

$P_{1}(u,v)=\left\langle \mathbf {F} (\psi (u,v)){\bigg |}{\frac {\partial \psi }{\partial u}}\right\rangle ,\qquad P_{2}(u,v)=\left\langle \mathbf {F} (\psi (u,v)){\bigg |}{\frac {\partial \psi }{\partial v}}\right\rangle$ ### Second step of the proof (first equation)

According to the definition of line integral,

{\begin{aligned}\oint _{\Gamma }\mathbf {F} d\Gamma &=\int _{a}^{b}\left\langle (\mathbf {F} \circ \psi (t)){\bigg |}{\frac {d\Gamma }{dt}}(t)\right\rangle \,dt\\&=\int _{a}^{b}\left\langle (\mathbf {F} \circ \psi (t)){\bigg |}{\frac {d(\psi \circ \gamma )}{dt}}(t)\right\rangle \,dt\\&=\int _{a}^{b}\left\langle (\mathbf {F} \circ \psi (t)){\bigg |}(J\psi )_{\gamma (t)}\cdot {\frac {d\gamma }{dt}}(t)\right\rangle \,dt\end{aligned}} where, Jψ stands for the Jacobian matrix of ψ. Hence,[note 5][note 6]

{\begin{aligned}\left\langle (\mathbf {F} \circ \Gamma (t)){\bigg |}(J\psi )_{\gamma (t)}{\frac {d\gamma }{dt}}(t)\right\rangle &=\left\langle (\mathbf {F} \circ \Gamma (t)){\bigg |}(J\psi )_{\gamma (t)}{\bigg |}{\frac {d\gamma }{dt}}(t)\right\rangle \\&=\left\langle ({}^{t}\mathbf {F} \circ \Gamma (t))\cdot (J\psi )_{\gamma (t)}\ {\bigg |}\ {\frac {d\gamma }{dt}}(t)\right\rangle \\&=\left\langle \left(\left\langle (\mathbf {F} (\psi (\gamma (t)))){\bigg |}{\frac {\partial \psi }{\partial u}}(\gamma (t))\right\rangle ,\left\langle (\mathbf {F} (\psi (\gamma (t)))){\bigg |}{\frac {\partial \psi }{\partial v}}(\gamma (t))\right\rangle \right){\bigg |}{\frac {d\gamma }{dt}}(t)\right\rangle \\&=\left\langle (P_{1}(u,v),P_{2}(u,v)){\bigg |}{\frac {d\gamma }{dt}}(t)\right\rangle \\&=\left\langle \mathbf {P} (u,v)\ {\bigg |}{\frac {d\gamma }{dt}}(t)\right\rangle \end{aligned}} So, we obtain following equation

$\oint _{\Gamma }\mathbf {F} d\Gamma =\oint _{\gamma }\mathbf {P} d\gamma$ ### Third step of the proof (second equation)

First, calculate the partial derivatives, using Leibniz rule of inner product

{\begin{aligned}{\frac {\partial P_{1}}{\partial v}}&=\left\langle {\frac {\partial (\mathbf {F} \circ \psi )}{\partial v}}{\bigg |}{\frac {\partial \psi }{\partial u}}\right\rangle +\left\langle \mathbf {F} \circ \psi {\bigg |}{\frac {\partial ^{2}\psi }{\partial v\partial u}}\right\rangle \\{\frac {\partial P_{2}}{\partial u}}&=\left\langle {\frac {\partial (\mathbf {F} \circ \psi )}{\partial u}}{\bigg |}{\frac {\partial \psi }{\partial v}}\right\rangle +\left\langle \mathbf {F} \circ \psi {\bigg |}{\frac {\partial ^{2}\psi }{\partial u\partial v}}\right\rangle \end{aligned}} {\begin{aligned}{\frac {\partial P_{1}}{\partial v}}-{\frac {\partial P_{2}}{\partial u}}&=\left\langle {\frac {\partial (\mathbf {F} \circ \psi )}{\partial v}}{\bigg |}{\frac {\partial \psi }{\partial u}}\right\rangle -\left\langle {\frac {\partial (\mathbf {F} \circ \psi )}{\partial u}}{\bigg |}{\frac {\partial \psi }{\partial v}}\right\rangle \\&=\left\langle (J\mathbf {F} )_{\psi (u,v)}\cdot {\frac {\partial \psi }{\partial v}}{\bigg |}{\frac {\partial \psi }{\partial u}}\right\rangle -\left\langle (J\mathbf {F} )_{\psi (u,v)}\cdot {\frac {\partial \psi }{\partial u}}{\bigg |}{\frac {\partial \psi }{\partial v}}\right\rangle &&{\text{ chain rule}}\\&=\left\langle {\frac {\partial \psi }{\partial u}}{\bigg |}(J\mathbf {F} )_{\psi (u,v)}{\bigg |}{\frac {\partial \psi }{\partial v}}\right\rangle -\left\langle {\frac {\partial \psi }{\partial u}}{\bigg |}{}^{t}(J\mathbf {F} )_{\psi (u,v)}{\bigg |}{\frac {\partial \psi }{\partial v}}\right\rangle \\&=\left\langle {\frac {\partial \psi }{\partial u}}{\bigg |}(J\mathbf {F} )_{\psi (u,v)}-{}^{t}{(J\mathbf {F} )}_{\psi (u,v)}{\bigg |}{\frac {\partial \psi }{\partial v}}\right\rangle \\&=\left\langle {\frac {\partial \psi }{\partial u}}{\bigg |}\left((J\mathbf {F} )_{\psi (u,v)}-{}^{t}(J\mathbf {F} )_{\psi (u,v)}\right)\cdot {\frac {\partial \psi }{\partial v}}\right\rangle \\&=\left\langle {\frac {\partial \psi }{\partial u}}{\bigg |}(\nabla \times \mathbf {F} )\times {\frac {\partial \psi }{\partial v}}\right\rangle &&\left((J\mathbf {F} )_{\psi (u,v)}-{}^{t}(J\mathbf {F} )_{\psi (u,v)}\right)\cdot \mathbf {x} =(\nabla \times \mathbf {F} )\times \mathbf {x} \\&=\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]&&{\text{ Scalar Triple Product}}\end{aligned}} On the other hand, according to the definition of surface integral,

{\begin{aligned}\iint _{S}(\nabla \times \mathbf {F} )\,dS&=\iint _{D}\left\langle (\nabla \times \mathbf {F} )(\psi (u,v)){\bigg |}{\frac {\partial \psi }{\partial u}}(u,v)\times {\frac {\partial \psi }{\partial v}}(u,v)\right\rangle \,du\,dv\\&=\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&&{\text{ scalar triple product}}\end{aligned}} So, we obtain

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

According to the result of the second step, and according to the result of Third step, and further considering the Green's theorem, subjected equation is proved.

## Application for conservative force and scalar potential

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

First, we define the notarization map, $\theta _{[a,b]}:[0,1]\to [a,b]$ as follows.

$\theta _{[a,b]}=s(b-a)+a$ Above-mentioned $\theta _{[a,b]}$ is a strongly increasing function that, for all piece wise smooth paths c:[a,b] → R3, for all smooth vector field F, domain of which includes $c[[a,b]]$ (image of [a,b] under c.), following equation is satisfied.

$\int _{c}\mathbf {F} \ dc\ =\int _{c\circ \theta _{[a,b]}}\ \mathbf {F} \ d(c\circ \theta _{[a,b]})$ So, we can unify the domain of the curve from the beginning to [0,1].

### The Lamellar vector field

In mechanics a lamellar vector field is called a conservative force; in fluid dynamics, it is called a Vortex-free vector field. So, lamellar vector field, conservative force, and vortex-free vector field are the same notion.

### 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,.[note 8]

That theorem is also important in the area of Homotopy theorem.

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

However, “homotope” or “homotopy” in above-mentioned sense are different toward (stronger than) typical definitions of “homotope” or “homotopy”.[note 9]

So there are no appropriate terminology which can discriminate between homotopy in typical sense and sense of Theorem 2-1. So, in this article, to discriminate between them, we say “Theorem 2-1 sense homotopy as tube-like-homotopy and, we say “Theorem 2-1 sense homotope” as tube-like homotope.[note 10]

### Proof of the theorem

Hereinafter, the ⊕ stands for joining paths [note 11] the $\ominus$ stands for backwards of curve [note 12]

Let D = [0, 1] × [0, 1]. By our assumption, c1 and c2 are piecewise smooth homotopic, there are the piecewise smooth homogony H : DM

{\begin{aligned}{\begin{cases}\gamma _{1}:[0,1]\to D\\\gamma _{1}(t):=(t,0)\end{cases}},\qquad &{\begin{cases}\gamma _{2}:[0,1]\to D\\\gamma _{2}(s):=(1,s)\end{cases}}\\{\begin{cases}\gamma _{3}:[0,1]\to D\\\gamma _{3}(t):=(-t+0+1,1)\end{cases}},\qquad &{\begin{cases}\gamma _{4}:[0,1]\to D\\\gamma _{4}(s):=(0,1-s)\end{cases}}\end{aligned}} $\gamma (t):=(\gamma _{1}\oplus \gamma _{2}\oplus \gamma _{3}\oplus \gamma _{4})(t)$ $\Gamma _{i}(t):=H(\gamma _{i}(t)),\qquad i=1,2,3,4$ $\Gamma (t):=H(\gamma (t))=(\Gamma _{1}\oplus \Gamma _{2}\oplus \Gamma _{3}\oplus \Gamma _{4})(t)$ And, let S be the image of D under H. Then,

$\oint _{\Gamma }\mathbf {F} \,d\Gamma =\iint _{S}\nabla \times \mathbf {F} \,dS$ will be obvious according to the Theorem 1 and, F is Lamellar vector field that, right side of that equation is zero, so,

$\oint _{\Gamma }\mathbf {F} \,d\Gamma =0$ Here,

$\oint _{\Gamma }\mathbf {F} \,d\Gamma =\sum _{i=1}^{4}\oint _{\Gamma _{i}}\mathbf {F} d\Gamma$ [note 11]

and, H is Tubeler-Homotopy that,

$\Gamma _{2}(s)={\Gamma }_{4}(1-s)=\ominus {\Gamma }_{4}(s)$ $\oint _{{\Gamma }_{1}}\mathbf {F} d\Gamma +\oint _{\Gamma _{3}}\mathbf {F} d\Gamma =0$ On the other hand,

$c_{1}(t)=H(t,0)=H({\gamma }_{1}(t))={\Gamma }_{1}(t)$ $c_{2}(t)=H(t,1)=H(\ominus {\gamma }_{3}(t))=\ominus {\Gamma }_{3}(t)$ that, subjected equation is proved.

### Application for conservative force

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, obviously follows from Theorem 2-1. In Lemma 2-2, the existence of H satisfying [SC0] to [SC3]" is crucial. It is a well-known fact that, if U is simply connected, such H exists. The definition of Simply connected space follows:

You will find that, the [SC1] to [SC3] of both Lemma 2-2 and Definition 2-2 is same.

So, someone may think that, the issue, "when the Conservative Force, the work done in changing an object's position is path independent" is elucidated. However there are very large gap between following two.

• There are continuous H such that it satisfies [SC1] to [SC3]
• There are piecewise smooth H such that it satisfies [SC1] to [SC3]

To fill that gap, the deep knowledge of Homotopy Theorem is required. For example, to fill the gap, following resources may be helpful for you.

Considering above-mentioned fact and Lemma 2-2, we will obtain following theorem. That theorem is anser for subjecting issue.

## Kelvin–Stokes theorem on singular 2-cube and cube subdivisionable sphere

### Singular 2-cube and boundary

${\theta }_{D}({u}_{1},{u}_{2})=\left({\begin{array}{c}{u}_{1}({b}_{1}-{a}_{1})+{a}_{1}\\{u}_{2}({b}_{2}-{a}_{2})+{a}_{2}\end{array}}\right)$ Above-mentioned lemma is obverse that, we neglects the proof. Acceding to the above-mentioned lemma, hereinafter, we consider that, domain of all singular 2-cube are notarized (that means, hereinafter, we consider that domain of all singular 2-cube are from the beginning, I2.

In order to facilitate the discussion of boundary, we define $\delta _{[k,j,c]}:\mathbb {R} ^{k}\to \mathbb {R} ^{k+1}$ by

$\delta _{[k,j,c]}(t_{1},\cdots ,t_{k}):=(t_{1},\cdots ,t_{j-1},c,t_{j+1},\cdots ,t_{k})$ γ1, ..., γ4 are the one-dimensional edges of the image of I2.Hereinafter, the ⊕ stands for joining paths[note 11] and,　 the $\ominus$ stands for backwards of curve .[note 12]

{\begin{aligned}{\begin{cases}\gamma _{1}:[0,1]\to {I}^{2}\\\gamma _{1}(t):={\delta }_{[1,2,0]}(t)=(t,0)\end{cases}},\qquad &{\begin{cases}\gamma _{2}:[0,1]\to {I}^{2}\\\gamma _{2}(t):={\delta }_{[1,1,1]}(t)=(1,t)\end{cases}}\\{\begin{cases}\gamma _{3}:[0,1]\to {I}^{2}\\\gamma _{3}(t):=\ominus {\delta }_{[1,2,1]}(t)=(-t+0+1,1)\end{cases}},\qquad &{\begin{cases}\gamma _{4}:[0,1]\to {I}^{2}\\\gamma _{4}(t):=\ominus {\delta }_{[1,1,0]}(t)=(0,1-t)\end{cases}}\end{aligned}} $\gamma (t):=(\gamma _{1}\oplus \gamma _{2}\oplus \gamma _{3}\oplus \gamma _{4})(t)$ $\Gamma _{i}(t):=\varphi (\gamma _{i}(t)),\qquad i=1,2,3,4$ $\Gamma (t):=\varphi (\gamma (t))=(\Gamma _{1}\oplus \Gamma _{2}\oplus \Gamma _{3}\oplus \Gamma _{4})(t)$ ### Cube subdivision

The definition of the boundary of the Definitions 3-3 is apparently depends on the cube subdevision. However, considering the following fact, the boundary is not depends on the cube subdevision.

So, considering the above-mentioned fact, following "Definition3-4" is well-defined.