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Stokes' theorem

Stokes' theorem,[1] also known as the Kelvin–Stokes theorem[2][3] after Lord Kelvin and George Stokes, the fundamental theorem for curls or simply the curl theorem,[4] is a theorem in vector calculus on . 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 flux of its curl through the enclosed surface. It is illustrated in the figure, where the direction of positive circulation of the bounding contour ∂Σ, and the direction n of positive flux through the surface Σ, are related by a right-hand-rule. For the right hand the fingers circulate along ∂Σ and the thumb is directed along n.

An illustration of Stokes' theorem, with surface Σ, its boundary ∂Σ and the normal vector n.

Stokes' theorem is a special case of the generalized Stokes theorem.[5][6] In particular, a vector field on can be considered as a 1-form in which case its curl is its exterior derivative, a 2-form.

Theorem Edit

Let   be a smooth oriented surface in   with boundary  . If a vector field   is defined and has continuous first order partial derivatives in a region containing  , then

 
More explicitly, the equality says that
 

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  .

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

Suppose   is piecewise smooth at the neighborhood of  , with  .[note 1] If   is the space curve defined by  [note 2] then we call   the boundary of  , written  .

With the above notation, if   is any smooth vector field on  , then[7][8]

 

Here, the " " represents the dot product in  .

Proof Edit

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).[9] 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.[8] 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 Edit

First step of the elementary proof (parametrization of integral) Edit

As in § Theorem, 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

 
where Jyψ stands for the Jacobian matrix of ψ at y = γ(t).

Now let {eu, ev} be an orthonormal basis in the coordinate directions of R2.[note 3]

Recognizing that the columns of Jyψ are precisely the partial derivatives of ψ at y, we can expand the previous equation in coordinates as

 

Second step in the elementary proof (defining the pullback) Edit

The previous step suggests we define the function

 

Now, if the scalar value functions   and   are defined as follows,

 
 
then,
 

This is the pullback of F along ψ, and, by the above, it satisfies

 

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) Edit

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

 

Conveniently, the second term vanishes in the difference, by equality of mixed partials. So,[note 4]

 

But now consider the matrix in that quadratic form—that is,  . We claim this matrix in fact describes a cross product. Here the superscript " " represents the transposition of matrices.

To be precise, let   be an arbitrary 3 × 3 matrix and let

 

Note that xa × x is linear, so it is determined by its action on basis elements. But by direct calculation

 
Here, {e1, e2, e3} represents an orthonormal basis in the coordinate directions of  .[note 5]

Thus (AAT)x = a × x for any x.

Substituting   for A, we obtain

 

We can now recognize the difference of partials as a (scalar) triple product:

 

On the other hand, the definition of a surface integral also includes a triple product—the very same one!

 

So, we obtain

 

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

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   defined on D;

 

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 Edit

The functions   can be identified with the differential 1-forms on   via the map

 

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

 

where is the Hodge star and   is the exterior derivative. Thus, by generalized Stokes' theorem,[10]

 

Applications Edit

Irrotational fields Edit

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 F on an open   is irrotational (lamellar vector field) if ∇ × F = 0.

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

The Helmholtz's theorems Edit

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

Theorem 2-1 (Helmholtz's theorem in fluid dynamics).[5][3]: 142  Let   be an open subset with a lamellar vector field F and let c0, c1: [0, 1] → U be piecewise smooth loops. If there is a function H: [0, 1] × [0, 1] → U such that

  • [TLH0] H is piecewise smooth,
  • [TLH1] H(t, 0) = c0(t) for all t ∈ [0, 1],
  • [TLH2] H(t, 1) = c1(t) for all t ∈ [0, 1],
  • [TLH3] H(0, s) = H(1, s) for all s ∈ [0, 1].

Then,

 

Some textbooks such as Lawrence[5] 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 6]

Proof of the Helmholtz's theorems Edit
 
The definitions of γ1, ..., γ4

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 four line segments γj.

 
so that
 

By our assumption that c0 and c1 are piecewise smooth homotopic, there is a piecewise smooth homotopy H: DM

 

Let S be the image of D under H. That

 

follows immediately from Stokes' theorem. F is lamellar, so the left side vanishes, i.e.

 

As H is tubular(satisfying [TLH3]),  and  . Thus the line integrals along Γ2(s) and Γ4(s) cancel, leaving

 

On the other hand, c1 = Γ1,  , so that the desired equality follows almost immediately.

Conservative forces Edit

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.[5][6] Let   be an open subset, with a Lamellar vector field F and a piecewise smooth loop c0: [0, 1] → U. Fix a point pU, if there is a homotopy H: [0, 1] × [0, 1] → U such that

  • [SC0] H is piecewise smooth,
  • [SC1] H(t, 0) = c0(t) for all t ∈ [0, 1],
  • [SC2] H(t, 1) = p for all t ∈ [0, 1],
  • [SC3] H(0, s) = H(1, s) = p for all s ∈ [0, 1].

Then,

 

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

Definition 2-2 (simply connected space).[5][6] Let   be non-empty and path-connected. M is called simply connected if and only if for any continuous loop, c: [0, 1] → M there exists a continuous tubular homotopy H: [0, 1] × [0, 1] → M from c to a fixed point pc; that is,

  • [SC0'] H is continuous,
  • [SC1] H(t, 0) = c(t) for all t ∈ [0, 1],
  • [SC2] H(t, 1) = p for all t ∈ [0, 1],
  • [SC3] H(0, s) = H(1, s) = p for all s ∈ [0, 1].

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.[6]: 136, 421 [11] 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.[5][6] Let   be open and simply connected with an irrotational vector field F. For all piecewise smooth loops c: [0, 1] → U

 

Maxwell's equations Edit

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,  :

 

For Ampère's law, Stokes' theorem is applied to the magnetic field,  :

 

Notes Edit

  1. ^   represents the image set of   by  
  2. ^   may not be a Jordan curve if the loop   interacts poorly with  . Nonetheless,   is always a loop, and topologically a connected sum of countably many Jordan curves, so that the integrals are well-defined.
  3. ^ In this article,
     
    Note that, in some textbooks on vector analysis, these are assigned to different things. For example, in some text book's notation, {eu, ev} can mean the following {tu, tv} respectively. In this article, however, these are two completely different things.
     
    Here,
     
    and the " " represents Euclidean norm.
  4. ^ For all  , for all   square matrix,   and therefore  .
  5. ^ In this article,
     
    Note that, in some textbooks on vector analysis, these are assigned to different things.
  6. ^ There do exist textbooks that use the terms "homotopy" and "homotopic" in the sense of Theorem 2-1.[5] Indeed, this is very convenient for the specific problem of conservative forces. However, both uses of homotopy appear sufficiently frequently that some sort of terminology is necessary to disambiguate, and the term "tubular homotopy" adopted here serves well enough for that end.

References Edit

  1. ^ Stewart, James (2012). Calculus – Early Transcendentals (7th ed.). Brooks/Cole Cengage Learning. p. 1122. ISBN 978-0-538-49790-9.
  2. ^ Nagayoshi Iwahori, et al.:"Bi-Bun-Seki-Bun-Gaku" Sho-Ka-Bou(jp) 1983/12 ISBN 978-4-7853-1039-4 [1](Written in Japanese)
  3. ^ a b Atsuo Fujimoto;"Vector-Kai-Seki Gendai su-gaku rekucha zu. C(1)" Bai-Fu-Kan(jp)(1979/01) ISBN 978-4563004415 [2] (Written in Japanese)
  4. ^ Griffiths, David (2013). Introduction to Electrodynamics. Pearson. p. 34. ISBN 978-0-321-85656-2.
  5. ^ a b c d e f g Conlon, Lawrence (2008). Differentiable Manifolds. Modern Birkhauser Classics. Boston: Birkhaeuser.
  6. ^ a b c d e Lee, John M. (2002). Introduction to Smooth Manifolds. Graduate Texts in Mathematics. Vol. 218. Springer.
  7. ^ Stewart, James (2010). Essential Calculus: Early Transcendentals. Cole.
  8. ^ a b Robert Scheichl, lecture notes for University of Bath mathematics course [3]
  9. ^ Colley, Susan Jane (2002). Vector Calculus (4th ed.). Boston: Pearson. pp. 500–3.
  10. ^ Edwards, Harold M. (1994). Advanced Calculus: A Differential Forms Approach. Birkhäuser. ISBN 0-8176-3707-9.
  11. ^ 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. MR0115178 (22 #5980 [4]). See theorems 7 & 8.

stokes, theorem, also, known, kelvin, stokes, theorem, after, lord, kelvin, george, stokes, fundamental, theorem, curls, simply, curl, theorem, theorem, vector, calculus, displaystyle, mathbb, given, vector, field, theorem, relates, integral, curl, vector, fie. Stokes theorem 1 also known as the Kelvin Stokes theorem 2 3 after Lord Kelvin and George Stokes the fundamental theorem for curls or simply the curl theorem 4 is a theorem in vector calculus on R 3 displaystyle 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 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 flux of its curl through the enclosed surface It is illustrated in the figure where the direction of positive circulation of the bounding contour S and the direction n of positive flux through the surface S are related by a right hand rule For the right hand the fingers circulate along S and the thumb is directed along n An illustration of Stokes theorem with surface S its boundary S and the normal vector n Stokes theorem is a special case of the generalized Stokes theorem 5 6 In particular a vector field on R 3 displaystyle mathbb R 3 can be considered as a 1 form in which case its curl is its exterior derivative a 2 form Contents 1 Theorem 2 Proof 2 1 Elementary proof 2 1 1 First step of the elementary proof parametrization of integral 2 1 2 Second step in the elementary proof defining the pullback 2 1 3 Third step of the elementary proof second equation 2 1 4 Fourth step of the elementary proof reduction to Green s theorem 2 2 Proof via differential forms 3 Applications 3 1 Irrotational fields 3 1 1 The Helmholtz s theorems 3 1 1 1 Proof of the Helmholtz s theorems 3 2 Conservative forces 3 3 Maxwell s equations 4 Notes 5 ReferencesTheorem EditLet S displaystyle Sigma nbsp be a smooth oriented surface in R 3 displaystyle mathbb R 3 nbsp with boundary S G displaystyle partial Sigma equiv Gamma nbsp If a vector field F x y z F x x y z F y x y z F z x y z displaystyle mathbf F x y z F x x y z F y x y z F z x y z nbsp is defined and has continuous first order partial derivatives in a region containing S displaystyle Sigma nbsp then S F d S S F d G displaystyle iint Sigma nabla times mathbf F cdot mathrm d mathbf Sigma oint partial Sigma mathbf F cdot mathrm d mathbf Gamma nbsp More explicitly the equality says that S F z y F y z d y d z F x z F z x d z d x F y x F x y d x d y S F x d x F y d y F z d z displaystyle begin aligned amp iint Sigma left left frac partial F z partial y frac partial F y partial z right mathrm d y mathrm d z left frac partial F x partial z frac partial F z partial x right mathrm d z mathrm d x left frac partial F y partial x frac partial F x partial y right mathrm d x mathrm d y right amp oint partial Sigma Bigl F x mathrm d x F y mathrm d y F z mathrm d z Bigr end aligned nbsp 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 displaystyle mathbb R 2 nbsp A more detailed statement will be given for subsequent discussions Let g a b R 2 displaystyle gamma a b to mathbb R 2 nbsp be a piecewise smooth Jordan plane curve The Jordan curve theorem implies that g displaystyle gamma nbsp divides R 2 displaystyle mathbb R 2 nbsp into two components a compact one and another that is non compact Let D displaystyle D nbsp denote the compact part then D displaystyle D nbsp is bounded by g displaystyle gamma nbsp It now suffices to transfer this notion of boundary along a continuous map to our surface in R 3 displaystyle mathbb R 3 nbsp But we already have such a map the parametrization of S displaystyle Sigma nbsp Suppose ps D R 3 displaystyle psi D to mathbb R 3 nbsp is piecewise smooth at the neighborhood of D displaystyle D nbsp with S ps D displaystyle Sigma psi D nbsp note 1 If G displaystyle Gamma nbsp is the space curve defined by G t ps g t displaystyle Gamma t psi gamma t nbsp note 2 then we call G displaystyle Gamma nbsp the boundary of S displaystyle Sigma nbsp written S displaystyle partial Sigma nbsp With the above notation if F displaystyle mathbf F nbsp is any smooth vector field on R 3 displaystyle mathbb R 3 nbsp then 7 8 S F d G S F d S displaystyle oint partial Sigma mathbf F cdot mathrm d mathbf Gamma iint Sigma nabla times mathbf F cdot mathrm d mathbf Sigma nbsp Here the displaystyle cdot nbsp represents the dot product in R 3 displaystyle mathbb R 3 nbsp Proof EditThe 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 9 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 8 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 Edit First step of the elementary proof parametrization of integral Edit As in Theorem we reduce the dimension by using the natural parametrization of the surface Let ps and g be as in that section and note that by change of variables S F x d G g F ps g d ps g g F ps y J y ps d g displaystyle oint partial Sigma mathbf F mathbf x cdot mathrm d mathbf Gamma oint gamma mathbf F boldsymbol psi mathbf gamma cdot mathrm d boldsymbol psi mathbf gamma oint gamma mathbf F boldsymbol psi mathbf y cdot J mathbf y boldsymbol psi mathrm d gamma nbsp where Jyps stands for the Jacobian matrix of ps at y g t Now let eu ev be an orthonormal basis in the coordinate directions of R2 note 3 Recognizing that the columns of Jyps are precisely the partial derivatives of ps at y we can expand the previous equation in coordinates as S F x d G g F ps y J y ps e u e u d y F ps y J y ps e v e v d y g F ps y ps u y e u F ps y ps v y e v d y displaystyle begin aligned oint partial Sigma mathbf F mathbf x cdot mathrm d mathbf Gamma amp oint gamma mathbf F boldsymbol psi mathbf y J mathbf y boldsymbol psi mathbf e u mathbf e u cdot mathrm d mathbf y mathbf F boldsymbol psi mathbf y J mathbf y boldsymbol psi mathbf e v mathbf e v cdot mathrm d mathbf y amp oint gamma left left mathbf F boldsymbol psi mathbf y cdot frac partial boldsymbol psi partial u mathbf y right mathbf e u left mathbf F boldsymbol psi mathbf y cdot frac partial boldsymbol psi partial v mathbf y right mathbf e v right cdot mathrm d mathbf y end aligned nbsp Second step in the elementary proof defining the pullback Edit The previous step suggests we define the functionP u v F ps u v ps u u v e u F ps u v ps v u v e v displaystyle mathbf P u v left mathbf F boldsymbol psi u v cdot frac partial boldsymbol psi partial u u v right mathbf e u left mathbf F boldsymbol psi u v cdot frac partial boldsymbol psi partial v u v right mathbf e v nbsp Now if the scalar value functions P u displaystyle P u nbsp and P v displaystyle P v nbsp are defined as follows P u u v F ps u v ps u u v displaystyle P u u v left mathbf F boldsymbol psi u v cdot frac partial boldsymbol psi partial u u v right nbsp P v u v F ps u v ps v u v displaystyle P v u v left mathbf F boldsymbol psi u v cdot frac partial boldsymbol psi partial v u v right nbsp then P u v P u u v e u P v u v e v displaystyle mathbf P u v P u u v mathbf e u P v u v mathbf e v nbsp This is the pullback of F along ps and by the above it satisfies S F x d l g P y d l g P u u v e u P v u v e v d l displaystyle oint partial Sigma mathbf F mathbf x cdot mathrm d mathbf l oint gamma mathbf P mathbf y cdot mathrm d mathbf l oint gamma P u u v mathbf e u P v u v mathbf e v cdot mathrm d mathbf l nbsp 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 Edit First calculate the partial derivatives appearing in Green s theorem via the product rule P u v F ps v ps u F ps 2 ps v u P v u F ps u ps v F ps 2 ps u v displaystyle begin aligned frac partial P u partial v amp frac partial mathbf F circ boldsymbol psi partial v cdot frac partial boldsymbol psi partial u mathbf F circ boldsymbol psi cdot frac partial 2 boldsymbol psi partial v partial u 5pt frac partial P v partial u amp frac partial mathbf F circ boldsymbol psi partial u cdot frac partial boldsymbol psi partial v mathbf F circ boldsymbol psi cdot frac partial 2 boldsymbol psi partial u partial v end aligned nbsp Conveniently the second term vanishes in the difference by equality of mixed partials So note 4 P v u P u v F ps u ps v F ps v ps u ps v J ps u v F ps u ps u J ps u v F ps v chain rule ps v J ps u v F J ps u v F T ps u displaystyle begin aligned frac partial P v partial u frac partial P u partial v amp frac partial mathbf F circ boldsymbol psi partial u cdot frac partial boldsymbol psi partial v frac partial mathbf F circ boldsymbol psi partial v cdot frac partial boldsymbol psi partial u 5pt amp frac partial boldsymbol psi partial v cdot J boldsymbol psi u v mathbf F frac partial boldsymbol psi partial u frac partial boldsymbol psi partial u cdot J boldsymbol psi u v mathbf F frac partial boldsymbol psi partial v amp amp text chain rule 5pt amp frac partial boldsymbol psi partial v cdot left J boldsymbol psi u v mathbf F J boldsymbol psi u v mathbf F mathsf T right frac partial boldsymbol psi partial u end aligned nbsp But now consider the matrix in that quadratic form that is J ps u v F J ps u v F T displaystyle J boldsymbol psi u v mathbf F J boldsymbol psi u v mathbf F mathsf T nbsp We claim this matrix in fact describes a cross product Here the superscript T displaystyle mathsf T nbsp represents the transposition of matrices To be precise let A A i j i j displaystyle A A ij ij nbsp be an arbitrary 3 3 matrix and leta a 1 a 2 a 3 A 32 A 23 A 13 A 31 A 21 A 12 displaystyle mathbf a begin bmatrix a 1 a 2 a 3 end bmatrix begin bmatrix A 32 A 23 A 13 A 31 A 21 A 12 end bmatrix nbsp Note that x a x is linear so it is determined by its action on basis elements But by direct calculation A A T e 1 0 a 3 a 2 a e 1 A A T e 2 a 3 0 a 1 a e 2 A A T e 3 a 2 a 1 0 a e 3 displaystyle begin aligned left A A mathsf T right mathbf e 1 amp begin bmatrix 0 a 3 a 2 end bmatrix mathbf a times mathbf e 1 left A A mathsf T right mathbf e 2 amp begin bmatrix a 3 0 a 1 end bmatrix mathbf a times mathbf e 2 left A A mathsf T right mathbf e 3 amp begin bmatrix a 2 a 1 0 end bmatrix mathbf a times mathbf e 3 end aligned nbsp Here e1 e2 e3 represents an orthonormal basis in the coordinate directions of R 3 displaystyle mathbb R 3 nbsp note 5 Thus A AT x a x for any x Substituting J ps u v F displaystyle J boldsymbol psi u v mathbf F nbsp for A we obtain J ps u v F J ps u v F T x F x for all x R 3 displaystyle left J boldsymbol psi u v mathbf F J boldsymbol psi u v mathbf F mathsf T right mathbf x nabla times mathbf F times mathbf x quad text for all mathbf x in mathbb R 3 nbsp We can now recognize the difference of partials as a scalar triple product P v u P u v ps v F ps u F ps u ps v displaystyle begin aligned frac partial P v partial u frac partial P u partial v amp frac partial boldsymbol psi partial v cdot nabla times mathbf F times frac partial boldsymbol psi partial u nabla times mathbf F cdot frac partial boldsymbol psi partial u times frac partial boldsymbol psi partial v end aligned nbsp On the other hand the definition of a surface integral also includes a triple product the very same one S F d S D F ps u v ps u u v ps v u v d u d v displaystyle begin aligned iint Sigma nabla times mathbf F cdot d mathbf Sigma amp iint D nabla times mathbf F boldsymbol psi u v cdot frac partial boldsymbol psi partial u u v times frac partial boldsymbol psi partial v u v mathrm d u mathrm d v end aligned nbsp So we obtain S F d S D P v u P u v d u d v displaystyle iint Sigma nabla times mathbf F cdot mathrm d mathbf Sigma iint D left frac partial P v partial u frac partial P u partial v right mathrm d u mathrm d v nbsp Fourth step of the elementary proof reduction to Green s theorem Edit 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 g and two scalar valued smooth functions P u u v P v u v displaystyle P u u v P v u v nbsp defined on D g P u u v e u P v u v e v d l D P v u P u v d u d v displaystyle oint gamma P u u v mathbf e u P v u v mathbf e v cdot mathrm d mathbf l iint D left frac partial P v partial u frac partial P u partial v right mathrm d u mathrm d v nbsp 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 Edit The functions R 3 R 3 displaystyle mathbb R 3 to mathbb R 3 nbsp can be identified with the differential 1 forms on R 3 displaystyle mathbb R 3 nbsp via the mapF x e 1 F y e 2 F z e 3 F x d x F y d y F z d z displaystyle F x mathbf e 1 F y mathbf e 2 F z mathbf e 3 mapsto F x mathrm d x F y mathrm d y F z mathrm d z nbsp Write the differential 1 form associated to a function F as wF Then one can calculate that w F d w F displaystyle star omega nabla times mathbf F mathrm d omega mathbf F nbsp where is the Hodge star and d displaystyle mathrm d nbsp is the exterior derivative Thus by generalized Stokes theorem 10 S F d g S w F S d w F S w F S F d S displaystyle oint partial Sigma mathbf F cdot mathrm d mathbf gamma oint partial Sigma omega mathbf F int Sigma mathrm d omega mathbf F int Sigma star omega nabla times mathbf F iint Sigma nabla times mathbf F cdot mathrm d mathbf Sigma nbsp Applications EditIrrotational fields Edit 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 F on an open U R 3 displaystyle U subseteq mathbb R 3 nbsp is irrotational lamellar vector field if F 0 This concept is very fundamental in mechanics as we ll prove later if F is irrotational and the domain of F is simply connected then F is a conservative vector field The Helmholtz s theorems Edit In this section we will introduce a theorem that is derived from Stokes theorem and characterizes vortex free vector fields In fluid dynamics it is called Helmholtz s theorems Theorem 2 1 Helmholtz s theorem in fluid dynamics 5 3 142 Let U R 3 displaystyle U subseteq mathbb R 3 nbsp be an open subset with a lamellar vector field F and let c0 c1 0 1 U be piecewise smooth loops If there is a function H 0 1 0 1 U such that TLH0 H is piecewise smooth TLH1 H t 0 c0 t for all t 0 1 TLH2 H t 1 c1 t for all t 0 1 TLH3 H 0 s H 1 s for all s 0 1 Then c 0 F d c 0 c 1 F d c 1 displaystyle int c 0 mathbf F mathrm d c 0 int c 1 mathbf F mathrm d c 1 nbsp Some textbooks such as Lawrence 5 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 6 Proof of the Helmholtz s theorems Edit nbsp The definitions of g1 g4In what follows we abuse notation and use displaystyle oplus nbsp for concatenation of paths in the fundamental groupoid and displaystyle ominus nbsp for reversing the orientation of a path Let D 0 1 0 1 and split D into four line segments gj g 1 0 1 D g 1 t t 0 g 2 0 1 D g 2 s 1 s g 3 0 1 D g 3 t 1 t 1 g 4 0 1 D g 4 s 0 1 s displaystyle begin aligned gamma 1 0 1 to D quad amp gamma 1 t t 0 gamma 2 0 1 to D quad amp gamma 2 s 1 s gamma 3 0 1 to D quad amp gamma 3 t 1 t 1 gamma 4 0 1 to D quad amp gamma 4 s 0 1 s end aligned nbsp so that D g 1 g 2 g 3 g 4 displaystyle partial D gamma 1 oplus gamma 2 oplus gamma 3 oplus gamma 4 nbsp By our assumption that c0 and c1 are piecewise smooth homotopic there is a piecewise smooth homotopy H D MG i t H g i t i 1 2 3 4 G t H g t G 1 G 2 G 3 G 4 t displaystyle begin aligned Gamma i t amp H gamma i t amp amp i 1 2 3 4 Gamma t amp H gamma t Gamma 1 oplus Gamma 2 oplus Gamma 3 oplus Gamma 4 t end aligned nbsp Let S be the image of D under H That S F d S G F d G displaystyle iint S nabla times mathbf F mathrm d S oint Gamma mathbf F mathrm d Gamma nbsp follows immediately from Stokes theorem F is lamellar so the left side vanishes i e 0 G F d G i 1 4 G i F d G displaystyle 0 oint Gamma mathbf F mathrm d Gamma sum i 1 4 oint Gamma i mathbf F mathrm d Gamma nbsp As H is tubular satisfying TLH3 G 2 G 4 displaystyle Gamma 2 ominus Gamma 4 nbsp and G 2 G 4 displaystyle Gamma 2 ominus Gamma 4 nbsp Thus the line integrals along G2 s and G4 s cancel leaving0 G 1 F d G G 3 F d G displaystyle 0 oint Gamma 1 mathbf F mathrm d Gamma oint Gamma 3 mathbf F mathrm d Gamma nbsp On the other hand c1 G1 c 3 G 3 displaystyle c 3 ominus Gamma 3 nbsp so that the desired equality follows almost immediately Conservative forces Edit 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 5 6 Let U R 3 displaystyle U subseteq mathbb R 3 nbsp be an open subset with a Lamellar vector field F and a piecewise smooth loop c0 0 1 U Fix a point p U if there is a homotopy H 0 1 0 1 U such that SC0 H is piecewise smooth SC1 H t 0 c0 t for all t 0 1 SC2 H t 1 p for all t 0 1 SC3 H 0 s H 1 s p for all s 0 1 Then c 0 F d c 0 0 displaystyle int c 0 mathbf F mathrm d c 0 0 nbsp Above Lemma 2 2 follows from theorem 2 1 In Lemma 2 2 the existence of H satisfying SC0 to SC3 is crucial the question is whether such a homotopy can be taken for arbitrary loops If U is simply connected such H exists The definition of simply connected space follows Definition 2 2 simply connected space 5 6 Let M R n displaystyle M subseteq mathbb R n nbsp be non empty and path connected M is called simply connected if and only if for any continuous loop c 0 1 M there exists a continuous tubular homotopy H 0 1 0 1 M from c to a fixed point p c that is SC0 H is continuous SC1 H t 0 c t for all t 0 1 SC2 H t 1 p for all t 0 1 SC3 H 0 s H 1 s p for all s 0 1 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 6 136 421 11 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 5 6 Let U R 3 displaystyle U subseteq mathbb R 3 nbsp be open and simply connected with an irrotational vector field F For all piecewise smooth loops c 0 1 U c 0 F d c 0 0 displaystyle int c 0 mathbf F mathrm d c 0 0 nbsp Maxwell s equations Edit See also Maxwell s equations Circulation and curl 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 Ampere equation and the integral form of these equations For Faraday s law Stokes theorem is applied to the electric field E displaystyle mathbf E nbsp S E d l S E d S displaystyle oint partial Sigma mathbf E cdot mathrm d boldsymbol l iint Sigma mathbf nabla times mathbf E cdot mathrm d mathbf S nbsp For Ampere s law Stokes theorem is applied to the magnetic field B displaystyle mathbf B nbsp S B d l S B d S displaystyle oint partial Sigma mathbf B cdot mathrm d boldsymbol l iint Sigma mathbf nabla times mathbf B cdot mathrm d mathbf S nbsp Notes Edit S ps D displaystyle Sigma psi D nbsp represents the image set of D displaystyle D nbsp by ps displaystyle psi nbsp G displaystyle Gamma nbsp may not be a Jordan curve if the loop g displaystyle gamma nbsp interacts poorly with ps displaystyle psi nbsp Nonetheless G displaystyle Gamma nbsp is always a loop and topologically a connected sum of countably many Jordan curves so that the integrals are well defined In this article e u 1 0 e v 0 1 displaystyle mathbf e u begin bmatrix 1 0 end bmatrix mathbf e v begin bmatrix 0 1 end bmatrix nbsp Note that in some textbooks on vector analysis these are assigned to different things For example in some text book s notation eu ev can mean the following tu tv respectively In this article however these are two completely different things t u 1 h u f u t v 1 h v f v displaystyle mathbf t u frac 1 h u frac partial varphi partial u mathbf t v frac 1 h v frac partial varphi partial v nbsp Here h u f u h v f v displaystyle h u left frac partial varphi partial u right h v left frac partial varphi partial v right nbsp and the displaystyle cdot nbsp represents Euclidean norm For all a b R n displaystyle textbf a textbf b in mathbb R n nbsp for all A n n displaystyle A n times n nbsp square matrix a A b a T A b displaystyle textbf a cdot A textbf b textbf a mathsf T A textbf b nbsp and therefore a A b b A T a displaystyle textbf a cdot A textbf b textbf b cdot A mathsf T textbf a nbsp In this article e 1 1 0 0 e 2 0 1 0 e 3 0 0 1 displaystyle mathbf e 1 begin bmatrix 1 0 0 end bmatrix mathbf e 2 begin bmatrix 0 1 0 end bmatrix mathbf e 3 begin bmatrix 0 0 1 end bmatrix nbsp Note that in some textbooks on vector analysis these are assigned to different things There do exist textbooks that use the terms homotopy and homotopic in the sense of Theorem 2 1 5 Indeed this is very convenient for the specific problem of conservative forces However both uses of homotopy appear sufficiently frequently that some sort of terminology is necessary to disambiguate and the term tubular homotopy adopted here serves well enough for that end References Edit Stewart James 2012 Calculus Early Transcendentals 7th ed Brooks Cole Cengage Learning p 1122 ISBN 978 0 538 49790 9 Nagayoshi Iwahori et al Bi Bun Seki Bun Gaku Sho Ka Bou jp 1983 12 ISBN 978 4 7853 1039 4 1 Written in Japanese a b Atsuo Fujimoto Vector Kai Seki Gendai su gaku rekucha zu C 1 Bai Fu Kan jp 1979 01 ISBN 978 4563004415 2 Written in Japanese Griffiths David 2013 Introduction to Electrodynamics Pearson p 34 ISBN 978 0 321 85656 2 a b c d e f g Conlon Lawrence 2008 Differentiable Manifolds Modern Birkhauser Classics Boston Birkhaeuser a b c d e Lee John M 2002 Introduction to Smooth Manifolds Graduate Texts in Mathematics Vol 218 Springer Stewart James 2010 Essential Calculus Early Transcendentals Cole a b Robert Scheichl lecture notes for University of Bath mathematics course 3 Colley Susan Jane 2002 Vector Calculus 4th ed Boston Pearson pp 500 3 Edwards Harold M 1994 Advanced Calculus A Differential Forms Approach Birkhauser ISBN 0 8176 3707 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 MR0115178 22 5980 4 See theorems 7 amp 8 Retrieved from https en wikipedia org w index php title Stokes 27 theorem amp oldid 1173024535, wikipedia, wiki, book, books, library,

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