Kelvin–Stokes theorem

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The Kelvin–Stokes theorem,[1][2][3][4][5] also known as the curl theorem,[6] 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.”[7][8] In particular, the vector field on R3 can be considered as a 1-form in which case curl is the exterior derivative.

Proof

The proof of the Theorem consists of 4 steps [2][3][note 4] The proof below does not require background information on differential form, and may be helpful for understanding the notion of differential form, especially pull-back of differential form.

First Step of 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.

$P_1(u,v)=\left\langle \mathbf{F}(\psi(u,v)) \bigg| \frac{\partial \psi}{\partial u} \right\rangle, \qquad P_2(u,v)=\left\langle \mathbf{F}(\psi(u,v)) \bigg| \frac{\partial \psi}{\partial v} \right\rangle$

Where, $\langle \ |\ \rangle$ is the normal inner product of R3 and hereinafter, $\langle \ |A|\ \rangle$ stands for bilinear form according to matrix A [note 5] .[note 6]

Second Step of Proof (First Equation)

According to the definition of line integral,

\begin{align} \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{align}

where, Jψ stands for the Jacobian matrix of ψ. Hence,[note 5][note 6]

\begin{align} \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{align}

So, we obtain following equation

$\oint_{\Gamma} \mathbf{F} d\Gamma = \oint_{\gamma} \mathbf{P} d\gamma$

Third Step of Proof (Second Equation)

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

\begin{align} \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{align}
\begin{align} \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{align}

On the other hand, according to the definition of surface integral,

\begin{align} \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 dudv\\ &= \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{align}

So, we obtain

$\iint_S (\nabla\times\mathbf{F}) dS =\iint_{D} \left( \frac{\partial P_2}{\partial u} - \frac{\partial P_1}{\partial v} \right) dudv$

Fourth Step of Proof (Reduction to Green's Theorem)

According to the result of 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 strongly increase function that, for all piece wise sooth path 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 7]

That theorem is also important in the area of Homoropy thorem.[7]

Some textbooks such as Lawrence[7] 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 8]

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 9]

Proof of the Theorem

The definitions of γ1, ..., γ4

Hereinafter, the ⊕ stands for joining paths [note 10] the $\ominus$ stands for backwards of curve [note 11]

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{align} \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{align}
$\gamma(t):= (\gamma_{1} \oplus \gamma_{2} \oplus \gamma_{3} \oplus \gamma_{4})(t)$
$\Gamma_{i}(t):= H(\gamma_{i}(t)), \qquad i=1, 2, 3, 4$
$\Gamma(t):= H(\gamma(t)) =(\Gamma_{1} \oplus \Gamma_{2} \oplus \Gamma_{3} \oplus \Gamma_{4})(t)$

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 10]

and, H is Tubeler-Homotopy that,

$\Gamma_{2}(s)= {\Gamma}_{4}(1-s)=\ominus{\Gamma}_{4}(s)$

that, line integral along $\Gamma_{2}(s)$ and line integral along $\Gamma_{4}(s)$ are compensated each other[note 11] so,

$\oint_{{\Gamma}_{1}} \mathbf{F} d\Gamma +\oint_{\Gamma_3} \mathbf{F} d\Gamma =0$

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

Given $D:=[{a}_{1},{b}_{1}]\times[{a}_{2},{b}_{2}]$, we define the notarization map of sngler two cube ${\theta}_{D}:{I}^{2}\to D$

${\theta}_{D}({u}_{1},{u}_{2})= \left( \begin{array}{c} {u}_{1}({b}_{1}-{a}_{1}) +{a}_{1}\\ {u}_{2}({b}_{2}-{a}_{2}) +{a}_{2} \end{array} \right)$

here, the I:=[0,1] and I2 stands for ${I}^{2}=I\times I$.

Above mentioned is strongly increase function (that means $det(J(\theta_{D})_{({u}_{1},{u}_{2})})>0$ (for all $({u}_{1},{u}_{2})\in\mathbb{R}^{3}$) that, following lemma is satisfied.

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 10] and,　 the $\ominus$ stands for backwards of curve .[note 11]

\begin{align} \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{align}
$\gamma(t):= (\gamma_{1} \oplus \gamma_{2} \oplus \gamma_{3} \oplus \gamma_{4})(t)$
$\Gamma_{i}(t):= \varphi(\gamma_{i}(t)), \qquad i=1, 2, 3, 4$
$\Gamma(t):= \varphi(\gamma(t)) =(\Gamma_{1} \oplus \Gamma_{2} \oplus \Gamma_{3} \oplus \Gamma_{4})(t)$

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, considerting above mentiond fact, following "Definition3-4" is well-defined.

Notes and References

Notes

1. ^ The Jordan curve theorem implies that the Jordan curve divides R2 into two components, a compact one (the bounded area) and another is non-compact.
2. ^ When ψ is a mapping and D is a subset of the domain of ψ, ψ[D] stands for the image of D under ψ.
3. ^ γ and Γ are both loops, however, Γ is not necessarily a Jordan curve
4. ^ a b . If you know the differential form, when we considering following identification of the vector field A = (a1, a2, a3),
$\mathbf{A} = \omega_{\mathbf{A}} = a_1 dx_1+{a}_{2}d{x}_{2}+a_3dx_{3}$
$\mathbf{A} = {}^{*} \omega_{\mathbf{A}}= {a}_{1}d{x}_{2} \wedge d{x}_{3}+{a}_{2}d{x}_{3} \wedge d{x}_{1}+{a}_{3}d{x}_{1} \wedge d{x}_{2}$,
the proof here is similar to the proof using the pull-back of ωF. In actual, under the identification of ωF = F following equations are satisfied.
\begin{align} \nabla\times\mathbf{F} &= d\omega_{\mathbf{F}} \\ \psi^{*}\omega_{\mathbf{F}} &= P_1du +P_2dv \\ \psi^{*}(d \omega_{\mathbf{F}}) &= \left (\frac{\partial P_2}{\partial u} - \frac{\partial P_1}{\partial v} \right ) du\wedge dv \end{align}
Here, d stands for Exterior derivative of Differential form, ψ* stands for pull back by ψ and, P1 and P2 of above mentuoned are same as P1 and P2 of the body text of this article respectively.
5. ^ a b c Given a n × m matrix A we define a bilinear form:
$\mathbf{x} \in \mathbf{R}^m, \mathbf{y} \in \mathbf{R}^n \ : \qquad \left\langle \mathbf{x} |A| \mathbf{y} \right\rangle ={}^{t}\mathbf{x}A\mathbf{y}$
which also satisfies:
$\left \langle \mathbf{x} |A| \mathbf{y} \right \rangle= \left \langle \mathbf{y} |{}^{t}A| \mathbf{x} \right \rangle.$
$\left \langle \mathbf{x} |A| \mathbf{y} \right \rangle + \left \langle \mathbf{x} |B| \mathbf{y}\right \rangle = \left \langle \mathbf{x} |A+B|\mathbf{y}\right \rangle$
6. ^ a b c Given a n × m matrix A , tA stands for transposed matrix of A.
7. ^ There are a number of theorems with the same name, however they are not necessarily the same.
8. ^ Typical definition of homotopy and homotope are as follows.
9. ^ In Some textbooks such as Lawrence Conlon;"Differentiable Manifolds (Modern Birkhauser Classics)" Birkhaeuser Boston (2008/1/11)[1] use the term of homotopy and homotope in Theorem 2-1 sense. homotopy and homotope in Theorem 2-1 sense Indeed, it is convenience to adopt such sense to discuss conservative force. However, homotopy in Theorem 2-1 sense and homotope in Theorem 2-1 sense are different from and stronger than homotopy in typical sense and homotope in typical sence. So there are no appropriate terminology which can discriminate between homotopy in typical sense and sense of Theorem 2-1. In this article, to avoid ambiguity and to discriminate between them, we will define two “just-in-time term”, Tube-like-Homotopy and Tube-like-Homotope as follows.
10. ^ a b c Given two curves α: [a1, b1] → M , β: [a2, b2, ] → M, if α and β satisfy α(b1) = β(a2) then, we can define new curve α ⊕ β so that, for all smooth vector field F (if domain of which includes image of α ⊕ β )
${\int}_{\alpha\oplus \beta} \mathbf{F} d(\alpha\oplus \beta)= {\int}_{\alpha}\mathbf{F} d\alpha +{\int}_{\beta} \mathbf{F} d\beta$
,which is also used when we define Fundamental group. To do so, accurate definition of the “Joint of paths” is as follows.
11. ^ a b c Given curve on M, α: [a1, b1] → M , we can define new curve $\ominus$α so that, for all smooth vector field F (if domain of which includes image of α )
${\int}_{\ominus\alpha} \mathbf{F} d(\ominus\alpha)= -{\int}_{\alpha}\mathbf{F} d\alpha$
,which is also used when we define Fundamental group. To do so, accurate definition of the “Backwards of curve” is as follows.

And, given two curves on M, α: [a1, b1] → M β: [a2, b2] → M which satisfies α(b1)=β(b2) (that means α(b1)=$\ominus$β(a2), we can define $\alpha\ominus\beta$ as following manner.

$\alpha\ominus\beta:=\alpha\oplus(\ominus\beta)$

References

1. ^ a b James Stewart;"Essential Calculus: Early Transcendentals" Cole Pub Co (2010)[2]
2. ^ a b c This proof is based on the Lecture Notes given by Prof. Robert Scheichl (University of Bath, U.K) [3], please refer the [4]
3. ^ a b c This proof is also same to the proof shown in
4. Nagayoshi Iwahori, et.al:"Bi-Bun-Seki-Bun-Gaku" Sho-Ka-Bou(jp) 1983/12 ISBN978-4-7853-1039-4 [5](Written in Japanese)
5. ^ a b Atsuo Fujimoto;"Vector-Kai-Seki Gendai su-gaku rekucha zu. C(1)" Bai-Fu-Kan(jp)(1979/01) ISBN 978-4563004415 [6](Written in Japanese)
6. ^ http://mathworld.wolfram.com/CurlTheorem.html
7. Lawrence Conlon; "Differentiable Manifolds (Modern Birkhauser Classics)" Birkhaeuser Boston (2008/1/11) [7]
8. John M. Lee; "Introduction to Smooth Manifolds (Graduate Texts in Mathematics, 218) "Springer (2002/9/23) [8] [9]
9. ^ L. S. Pontryagin, Smooth manifolds and their applications in homotopy theory, American Mathematical Society Translations, Ser. 2, Vol. 11, American Mathematical Society, Providence, R.I., 1959, pp. 1–114. MR 0115178 (22 #5980 [10])[11]
10. ^ Michael Spivak:"Calculus On Manifolds: A Modern Approach To Classical Theorems Of Advanced Calculus" Westview Press, 1971 [12]