Sarahisme Posted August 24, 2006 Share Posted August 24, 2006 Hello all, I am completely new to this stokes theorem bussiness..what i have got so far is the nabla x F part, but i am unsure of how to find N (the unit normal field i think its called). any suggestions people? i get that nabla x F or curl(F) = i + j + k cheers -sarah Link to comment Share on other sites More sharing options...

woelen Posted August 24, 2006 Share Posted August 24, 2006 Are you referring to the general Stokes theorem http://mathworld.wolfram.com/StokesTheorem.html ? Or to the special case of 3D fields, which are integrated over a 2D surface and also can be expressed as a closed line integral along the border of the 2D surface? This question gives me the impression that you did not really learn the Stokes theorem, but only the specialized version of it for a vector field in 3D space. For the general Stokes theorem nothing more specific can be said, if you don't specify your differential form dω, over which the integral is computed. For the 3D variation of a vector field F, one can write [∫S] (nabla × F)•dS = [∫∂S] F•dl Here with [∫S] I mean the surface integral over surface S, with [∫∂S] I mean the closed line integral along the border of S. Here • is the interior product of two 3D vectors, and × is the exterior product of two 3D vectors. Now suppose F is a field, with F = [FX, FY, FZ]. You want to compute dS, being the unit normal vector to S. Suppose you are working in cartesian coordinates, then this "vector" can be written as [dy^dz , dz^dx , dx^dy] The total expression (nabla × F)•dS can be written as (∂FZ/∂y - ∂FY/dz)dy^dz + (∂FX/∂z - ∂FZ/∂x)dz^dx + (∂FY/∂x - ∂FX/∂y)dx^dy Now you integrate over the surface S, expressing all in cartesian coordinates. For F•dl the expression is simpler. FX dx + FY dy + Fz dz and now you integrate along the rim of the surface S, one time around in positive (counter clock wise) direction. I hope this helps. Deriving these things can be done, using the generalized Stokes theorem for N+1/N dimensions, and using so-called forms. Do you have knowledge of forms? If you know these, and you know the concept of the wedge-product, then all this kind of formula's can be easily derived. http://mathworld.wolfram.com/WedgeProduct.html I really would suggest you to try to understand the theory of forms. That will make life much easier . Link to comment Share on other sites More sharing options...

woelen Posted August 24, 2006 Share Posted August 24, 2006 Ah... now I see more of this question. While I posted the other reply, i did not see the white block with the parameterization of S, so hence my general reply. You have to transform your euclidian coordinates to u,v coordinates. Now I need the wedge product ^: dx = cos(v) du - u sin(v) dv dy = sin(v) du + u cos(v) dv dx^dy = cos(v)sin(v) du^du + u cos²(v) du^dv - u sin²(v) dv^du + u² sin(v) cos(v) dv^dv du^du and dv^dv are equal to 0, dv^du = - du^dv, so dx^dy = u*(cos²(v) + sin²(v)) du^dv = u du^dv Using dv^dv = 0, and du^dv = -dv^du, one can easily derive dz^dx: dz^dx = -cos(v) du^dv dy^dz = sin(v) du^dv So, the unit normal field is [sin(v) du^dv , -cos(v) du^dv, u du^dv]. ---------------------------------------------------------- Please carefully check my computation, a little minus sign error is introduced easily, but at least, you should be able to grasp how I did this computation. Link to comment Share on other sites More sharing options...

Sarahisme Posted August 24, 2006 Author Share Posted August 24, 2006 i don't understand wedge stuff. but i got the unit normal field as (sin(v), -cos(v), u) by taking the cross product of the paramaterization of r. hmm but once we have this unit normal field, how do use it in the formula? that is how can we integrate with u's and v's in there? Link to comment Share on other sites More sharing options...

Sarahisme Posted August 24, 2006 Author Share Posted August 24, 2006 i thought it was a cylinder... i guess not, lol this is what we want to do yeah? [math] \int \int_{S} \nabla \times \vec{F} \dot \vec{\hat{n}} dS [/math] and you can work out that [math] \nabla \times \vec{F} =[/math] i + j + k is this ok so far? but then i don't see how to get the n part, or the dS part.... :S or well i don't know how to use the n part once i get it.... Link to comment Share on other sites More sharing options...

woelen Posted August 24, 2006 Share Posted August 24, 2006 Well, things are quite easy. F = [z x y] You already determined nabla × F and state it is equal to [1 1 1]. No need to transform F for that to u, v coordinates, simply compute partial derivatives to x,y,z. I agree with your answer. Your ndS is my "vector" [dy^dz, dz^dx, dx^dy]. We both already derived that beast in completely different ways, but we agree on its value , being [sin(v) du^dv , -cos(v) du^dv, u du^dv]. Now, simply compute the interior product [1 1 1] • [sin(v) du^dv , -cos(v) du^dv, u du^dv]. That yields the simple 2-form (sin(v) - cos(v) + u) du^dv Integration now is easy. Your integral is ∫∫(sin(v) - cos(v) + u) du^dv, with u going from 0 to 1 and v going from 0 to pi/2. Let's first integrate over v. Then we get as result: ∫(1 - 1 + u*pi/2) du for u going from 0 to 1. Now when you compute that integral, you end up with pi/4 as the final answer. I hope this explanation is clear to you. Link to comment Share on other sites More sharing options...

woelen Posted August 24, 2006 Share Posted August 24, 2006 What I did here is mixing up a little the forms notation and your notation with interior and exterior products. That is not really clean to do, but I did this in order to demonstrate what I did, otherwise you might not understand all of it. Once you have the concept of forms, you can very easily perform coordinate transformations (e.g. from dxdydz to some weird set of coordinates u,v,w). You can easily do that for any number of dimensions (here in this example we had only two of them). Also, all this hassling with interior and exterior products is not necessary anymore, in fact, the elements of the exterior product simply are the elements of the wedge product of two 1-forms Adx+Bdy+Cdz: (Adx+Bdy+Cdz) ^ (Pdx+Qdy+Rdz) = AP dx^dx + AQ dx^dy + AR dx^dz + BP dy^dx + BQ dy^dy + BR dy^dz + CP dz^dx + CQ dz^dy + CR dz^dz. Applying the rule that dw^dw equals 0 and dv^dw = -dw^dv one finds: (BR - CQ) dy^dz + (CP - AR) dz^dx + (AQ - BP) dx^dy I'm sure you will recognize the elements of the classical exterior product or cross product. Link to comment Share on other sites More sharing options...

Sarahisme Posted August 25, 2006 Author Share Posted August 25, 2006 how do i work out what dS is? that is i get the same normal as you except for the du^dv parts. what does the ^ stand for? or does du^dv = [math]\frac{du}{dv}[/math] ? how does this look for the integral we need to evaluate? = i used the unit normal here, instead of the normal that you used and so i get a different answer...(as you can see above) is it possible to do this without wedge product stuff? i.e. just by using cross and dot products Link to comment Share on other sites More sharing options...

woelen Posted August 25, 2006 Share Posted August 25, 2006 You did not find another normal than I did, they are the same. I just included the differentials in the vector and you didn't. What I derived is ndS. This contains a differential (better: 2-form). In your notation, my normal can be written as (sin(v)*i - cos(v)*j + u*k) * dudv You should not normalize this vector anymore. I see you divided by its length in order to obtain a vector of unit length. Important thing to mention here is that normally you do not compute n and dS separately, you at once compute ndS, without caring for n and dS. That is why in many formulations one simply writes dS, with this object being a vector, normal to the surface. The Stokes equations are in cartesian coordinates. In those coordinates the object ndS can be written as [dydz dzdx dxdy] and what I merely did is converting this vector to your (u, v) coordinates. A vector, which has unit length in one coordinate system need not have unit length in another coordinate system! Now, deriving dl and dS from my original formula [∫S] (nabla × F)•dS = [∫∂S] F•dl in cartesian coordinates. First the easy thing, deriving dl. Suppose we move along the rim of the surface, a distance dl. This means that we move a small portion dx, a small portion dy and a small portion dz. The length |dl| simply is the length of the vector [dx dy dz]. Now the more difficult part. We integrate over the surface. Suppose along the surface we move a portion dx, dy, dz. We cannot simply say how much surface area we covered, but we also do not need that, we only need the product of the unit normal vector and the surface area. You can imagine a surface, tilted in some complicated way, but what we are merely interested in is in the following: Sx*i + Sy*j + Sz*k Sx is the projection of the surface along the x-axis, this has surface area dy*dz. Sy is the projection of the surface along the y-axis, which equals dx*dz and Sz is the projection of the surface along the z-axis, which equals dx*dy. (normal products, no wedge products, hence the *). So, what is dS in my formula (and ndS in yours). It simply is dydz*i + dxdz*j + dxdy*k You see, one can derive such things without wedge products. Now we need to transform this to your u,v coordinates. Unfortunately, that is not easy to describe without wedge products, but it can be done of course. You have a parameterization [x y z] = G([u v]) Now you need to convert each of the three products dydz, dxdz and dxdy to dudv form. This can be done by deriving three vector functions, from [u v] to [y z], from [u v] to [x z] and from [u v] to [x y] Let's call these functions G1, G2, G3. You have to compute the determinants of the Jacobians of G1, G2 and G3, in terms of u and v. And then you can compute dydz = det(Jac(G1))*dudv dxdz = det(Jac(G2))*dudv dxdy = det(Jac(G3))*dudv One example I will give: G1([u v]) = [u*sin(v) v] Jac(G1) = [sin(v) u*cos(v)] [0 1] The determinant of this matrix is sin(v). So, dydz = sin(v)*dudv Similarly you can do this for the other two determinants of the Jacobian and these are -cos(v)*dudv and u*dudv. You mentioned something about computing the normal in terms of u and v by taking a cross product, but I see no way how that should be done. Either do it in terms of forms and wedge products, and if you are not familiar with that, then you need to use the determinant of the Jacobian of the transforms from one coordinate system to another coordinate system. I prefer the wedge products, these are much less cumbersome, especially if the dimension is increasing. Finally, what does dx^dy mean? It does not mean dx/dy. It is an abstraction, very useful in performing differential computations on so-called manifolds. A manifold can be seen as a curved space of N dimensions (e.g. a sphere is a 2D manifold, our 3D space can be seen as a 3D manifold). Locally, manifolds are like N-dimensional euclidian space, but globally, they can have all kinds of weird shapes. It goes to far to explain all details over here. If you really want to know, try to understand the concepts of "global analysis". For this issue of Stokes theorem, it is interesting to know that for computation of so-called volume elements dV (in case of 2D manifolds S one usually writes dS, because the "volume" then is a surface), one can use the wedge product. The following link may be helpful to you: http://mathworld.wolfram.com/VolumeElement.html In that link you also see the determinant of the Jacobian mentioned, but things are much easier expressed in terms of wedge products. The use of the notation dxdy, dxdydz etc, is somewhat informal, as that page states. Indeed, it is, because these are not real products, multiplication of differentials is very tricky, because one may not simply multiply the expressions of dx and dy in order to get dxdy. Try it for your example, dxdy = u dudv, but if you simply multiply the expressions in u and v for dx and dy, then you'll not find u dudv, but some complicated expression, which also has du*du and dv*dv in it. Hopefully this lengthy post makes things clear for you. Yes, the math behind Stokes' theorem is quite involved. It is a beautiful piece of theory, but unfortunately many people are REALLY struggling with it. I am inclined to blame the teachers of this stuff, because it frequently is tought while other very useful concepts are not covered. This makes the subject very complicated and also quite ad-hoc and scattered, while with understanding of some basic concepts, all pieces nicely fall together and life becomes much easier. Ask your teacher about a reformulation in terms of wedge products. 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Sarahisme Posted August 26, 2006 Author Share Posted August 26, 2006 yep that helps a lot thanks! i was told this aswell, that [math]dS=\left| \frac{\partial \vec{r}}{\partial u}\times\frac{\partial \vec{r}}{\partial v}\right|\,du\,dv[/math] so we want to evaluate this integral... and that gives [math] \frac{\pi}{4} [/math] Link to comment Share on other sites More sharing options...

woelen Posted August 27, 2006 Share Posted August 27, 2006 Hmmm, interesting to see that formula you give. I've never seen this before. Indeed, it works out to be the same as what I found, but I do not see the general underlying principle behind this formula. This formula only is valid for 2D surfaces in 3D space. The formalism, I outlined in the previous post, is general and works for any manifold. Also the mechanism with the determinant of the Jacobian is general, but it is somewhat more cumbersome. But it is good to see that our answers now agree and that this issue is settled for you. Link to comment Share on other sites More sharing options...

matt grime Posted August 27, 2006 Share Posted August 27, 2006 is it possible to do this without wedge product stuff? i.e. just by using cross and dot products yes, and you should avoid going into these things for a while. there is no need to be at all concerned with manifolds, wedge products and the like, since you're only doing this with surfaces in 3 space. The motivation behind that is because you're doing stuff that is 'real life'. What woelen is talking about is the generalization to arbitrary dimensions, and not relevant to you. Link to comment Share on other sites More sharing options...

Sarahisme Posted August 29, 2006 Author Share Posted August 29, 2006 yes, and you should avoid going into these things for a while. there is no need to be at all concerned with manifolds, wedge products and the like, since you're only doing this with surfaces in 3 space. The motivation behind that is because you're doing stuff that is 'real life'. What woelen is talking about is the generalization to arbitrary dimensions, and not relevant to you. ah ok, that would explain why i have never heard my lecturer mention them Link to comment Share on other sites More sharing options...

Sarahisme Posted August 29, 2006 Author Share Posted August 29, 2006 Hmmm' date=' interesting to see that formula you give. I've never seen this before. Indeed, it works out to be the same as what I found, but I do not see the general underlying principle behind this formula. This formula only is valid for 2D surfaces in 3D space. The formalism, I outlined in the previous post, is general and works for any manifold. Also the mechanism with the determinant of the Jacobian is general, but it is somewhat more cumbersome. But it is good to see that our answers now agree and that this issue is settled for you.[/quote'] yep its all good now thanks for all your help and especially your time -Sarah Link to comment Share on other sites More sharing options...

woelen Posted August 29, 2006 Share Posted August 29, 2006 I have been thinking further on that formula you supplied with the cross-product. I now understand how it can be derived. At a certain point along the surface, you take a tangent along that service along one coordinate, and you take a tangent along that service along the other coordinate. Both vectors are tangent to the surface. The cross-product is perpendicular to both of them, and hence it is normal to the surface. If you want just the normal, then you have to divide by its length, if you want ndS, then you should not divide by its length, that would throw away the information on S (change of surface, relative to change of coordinate), so ndS simply is the cross-product of these two tangent vectors. But remember, this only works for surfaces, embedded in 3D space. Link to comment Share on other sites More sharing options...

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