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Stokes' theorem

From Wikipedia, the free encyclopedia

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

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 surface integral of its curl over 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.

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.

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Transcription

So I've drawn multiple versions of the exact same surface S, five copies of that exact same surface. And what I want to do is think about the value of the line integral-- let me write this down-- the value of the line integral of F dot dr, where F is the vector field that I've drawn in magenta in each of these diagrams. And obviously, it's different in each of these diagrams. And the only part of the vector field that I've drawn is a part that's along the surface. I could have drawn the part of the vector field that's off the surface, but we're only going to be concerned with what's going on on the surface. So the vector field could be defined in this entire three-dimensional space in here as well. And these are obviously different vector fields, and we can see that visually based on how we drew it. And the contour that we care about-- remember, we're going to take a line integral, so the path matters. The path that we care about is the counterclockwise boundary of our surface, so it's going to be this right over here. The counterclockwise boundary of our surface is what we're going to be taking F dot dr along. So this right over here, and let me draw the orientation. It's going to be counterclockwise. And we're going to do it in every one of these situations, for every one of these surfaces and every one of these F's. And what I really want to think about is how the value of F dot dr over that contour, how it might change from example to example. And obviously, the only difference between each of these is what the vector field F is doing. So first let's think about this example right over here. At this part of the contour, this bottom part right over here, our vector field is going in the exact same direction as our line, as our contour. So we're going to get positive values of F dot dr down here, and we're going to keep summing them up. We're taking an integral. Then as we go up the curve or as we go kind of uphill right over here, we see that our vector field is going essentially orthogonal. It's going perpendicular. It's going perpendicular to our contour. So our F dot dr, we're not going to get any value from that. F dot dr and all of these, this part of the contour, it's all going to be 0. So we're not going to get anything right over there. Actually, let me just write nothing. So we get nothing right over there. Maybe I'll write 0. I'll write we'll get 0 right over there. And then up here, when we're at this part of the contour, our vector field is going in the exact opposite direction as our path. Up here our path is going, I guess, from the right to the left, while our vector field is going from the left to the right. And so we're actually going to get negative values. We are going to get negative values up here, and they're going to sum up to a reasonable negative value. And if the vector field is constant, and I kind of drew it like it is, and if this length is equal to this length, then these two values are going to cancel out. When you add this positive sum to this negative sum, they're going to be 0. And then once again, when you go downhill again, the vector field is perpendicular to our actual path, so we're going to get 0. So based on the way I've described it, your line integral of F dot dr for this version of F in this example right over here, it might all cancel out. You will get something, if we make the assumptions that I made, this might be equal to 0. So in this example, F dot dr could be equal to 0. Now let's think about what's going on in this situation. In this one, just like the last one, as we go along the bottom, the vector field is going in the exact same direction as our contour, so we're going to get positive values. And once we go up the hill, the vector field is going perpendicular to our path, so it's not going to add anything really to it. So we're just going to get 0 along this part. But then up here, our vector field has switched directions. And once again, it's going in the exact same direction as our path, so we're going to get more positive values right over there. And then as we go down here, it's not going to add anything because our vector field is perpendicular to our path. So we're going to get 0. But notice, now these two ends don't cancel out with each other. We're going to get a positive value. We are going to get a positive value. And what was the difference between this version of F, this vector field, and this vector field right over here? Well, this vector field switched directions so that the top part didn't cancel out with the bottom part. Or another way to think about it is, it had some curl. There's a little bit of spinning going on. If this was describing the velocity of a fluid, and if you were to put a stick right over there on the surface, the stick would spin. It has some spin or some curl, however you want to describe it. This right over here has no curl. If you put a stick right over here, it would just flow with the fluid, but the stick itself would not spin. So we got a positive value for the line integral in this situation. And we also have, it looks like, a positive curl. Now let's think about this one. In this situation, as we go along this part of our contour, our vector field F is going in the exact same direction, so we're going to get positive values. Now as we go uphill, our vector field F, it's also kind of turned in that direction, and so we'll get more positive values. And now as we go in that direction, our vector field F is still going in the direction of our contour, we're going to get more positive values. And as we go down, once again the vector field F is going in the direction of our contour, so we'll get even more positive values. So in this situation, the value of our line integral of F dot dr is even more positive. And we see that the actual vector field along the surface-- and remember, the vector field might be doing all sorts of crazy things off the surface. Actually, let me draw that in that same magenta color. It might be doing all sorts of crazy things off the surface, but what we really care about is what's happening on the surface. And because this vector field is, I guess you could say curling, or it's spinning along the surface, it allows it to go with the boundary along all the points, and we get a very positive value for this line integral. So we have a higher curl. So more curl, it looks like, is leading to a more positive line integral. Now let's think about what's happening in this situation right over here. This situation down here, our vector field is going in the same direction as our path, so we're going to get positive values. Just like the first situation, as we go up the hill like this or up the surface like that, our vector field is perpendicular to our surface, so it's really not going to add anything to our line integral. And then as we go along this top part, this first part of the top part right over here, the vector field is going against us. So it's negative right over here. We're going in the exact opposite direction of our path. And then, right as we get to the end, the vector field flips direction, and we get a little bit of positive right over here because it's a little bit of it going in the same direction. And then we go back downhill. When we go back downhill, it adds nothing because our vector field is going perpendicular to our path. So the big difference between this case and the case up here is the case up here-- well, actually, I could compare between these two or these two. But the difference between this one and this one is that at least this part of the vector field has switched directions. So we get a little bit of positive value. And one way to think about it, this one is going to be less positive than that, if we take the line integral, but more positive than that. And one other way to think about it is, we have a little bit of curl going on right over here. Our vector field switched directions right around there, or I guess you could say that it's spinning right around there. So if you put a stick, if that was in the water, it would start spinning. But everywhere else, there isn't a lot of curl. So you have some curl, but it's over a little small region of the surface. While over here, you had curl going on over a larger portion of our surface. And so up here, you had a more positive curl, more positive line integral. Here you have a curl over less of the surface, and you're going to have a less positive line integral. Now let's think about this one over here. This vector field, along the surface there is some curl. There's some curl going on right over there. If you put a stick in the water, if you view that as the velocity of water, the stick would spin. So you have some curl. But then it switches direction again, so you have curl there as well, and it's actually curl in the opposite direction. So to some degree, if you were to sum all of this up, maybe it would cancel out. And it makes sense. It makes sense that it would cancel out because when you take the line integral around the whole thing, just like this first situation, it looks like it will add up to 0. Because even though you have some curl, the curls cancel out each other. And so when you go to this top part of the surface, the vector field is going in the exact same direction as this bottom part of the surface. So if you were to take your line integral, the same one that we care about, just like the first situation it would be positive down here, 0 as we go up the curve. And then as we go down here, the vector field has switched directions twice, so it's still going against the path of our contour, just like this first situation. So it would be negative up there. And then as we go down, it would be 0. So this thing right over here also looks just like the first one because the curls essentially cancel out. We switched direction twice. So over here, our line integral might also be 0. Now, the whole reason why I went through this exercise is to give you an intuition of why it might make sense that if we have more curl happening over more of this surface, why that might make the value of this line integral be larger. And so, hopefully, it starts to give you an intuition that maybe, just maybe, the value of this line integral, the value of F dot dr over that contour, over this contour that's going in the counterclockwise direction-- we'll talk more about orientation in future videos-- maybe this is equal to the sum of the curls over the surface. And so let's think about this. It could be a surface integral. So we're going to go over the surface, and what we care about is the curl of F. We care about the curl of F. But we don't just care about the curl of F generally because F might be spinning in a direction, in a way, that's off the surface. We care about how much it's curling on the surface. So what we would want to do is we'd want to take the curl of F and dot it with the normal vector at any point to the surface and then multiply that times the surface itself. And this is just to say that the more surface where we have more curling going on, the more that the line integral, the value of the line integral, might be. And we saw that when we compared these three examples. And another way of writing all of this is the surface integral-- let me write the surface in that same brown color-- the surface integral of the curl of F, which would just be another vector that tells us how much we are spinning generally, but we want to care how much we're spinning along the surface. So we're dotting it with the normal vector. Or another way to write this whole thing is to say dot ds. So if you take essentially the sum across the entire surface of how much we're curling, how much we're spinning along that surface, then maybe, just maybe, this will be equal to the value of the line integral as we go around the boundary of the surface. And it actually turns out that this is the case. And obviously, I haven't proven it to you here, but hopefully you have some intuition why this makes sense. And this idea that this is equal to this is called Stokes' theorem, and we'll explore it more in the next few videos.

Theorem

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

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

First step of the elementary proof (parametrization of integral)

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)

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)

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)

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

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

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 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

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
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

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

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

  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

  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.
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