20071026, 15:11  #1 
May 2005
Argentina
2×3×31 Posts 
I generalized the fundamental theorem of calculus
I don't know if this is already known, so I am interested in knowing references to this issue.
It is known from Vector Analisis the theorems of Green (Stokes in the plane), Stokes (rotor), and Gauss (divergence). The Stokes (rotor) theorem relates an integral over an open surface, with a line integral over the curve of its boundary. What I've done is to apply the rotor to closed surfaces. For a general closed surface I want to know how "electrons" circle over the surface, so the answer to this question is a "vector" that points in the direction of the rotation axis. This vector should be: where But any of these 3 integrals can be related to a surface integral, so take for example Where is the versor , is any vector field, is a versor tangent to the surface, proyected over the xy plane (so it doesnt have z component), and is directed in the "positive" direction. For any point on any surface on there is only one versor with these characteristics. And finally is the versor normal to the surface. is the dot product between and is the norm of the vector product between and It is interesting to note that carefully choosing the vector field so that , one can calculate the volume, by integrating over the surface, I checked this for a parallelepiped, a cylinder and sphere. Does anybody know all this? I have many books on vector analysis and none mentions this subject. 
20071026, 17:31  #2 
∂^{2}ω=0
Sep 2002
República de California
3×5^{3}×31 Posts 
I'll have to go back and consult my copy of Spivak's Calculus on Manifolds [sheesh, Amazon's price seems outrageous for such a slender paperback], but IIRC Green's theorem is simply the 2D special case of Stokes' moregeneral theorem, which applies to [suitably defined] manifolds in R^{n}. After all, even through one generally only calculates the explicit curl operator in 2D and 3D in most college calculus classes, these operators are fully generalizable to nD.
Also, some of your terminology is unfamiliar to me: by rot(V) do your mean curl of the vector field V? And what is as "versor"? 
20071026, 17:45  #3 
May 2005
Argentina
186_{10} Posts 
By I meant and by "versor" I meant unit vector.
Sorry for the unfamiliar terminology, I based on spanish terminology. By the way, I'm not sure if the integral over a surface always works, I only verified for a cube, a cylinder and a sphere, but may not work for any surface, so If anyone verifies it for other surface I would be interested on hearing about that. 
20071026, 18:50  #4 
∂^{2}ω=0
Sep 2002
República de California
3×5^{3}×31 Posts 
Suggest you have a look at the Wikipedia page on Stokes' Theorem  I'll leave it to you to figure out whether your generalization is indeed equivalent, either in general or under some restricted circumstances.
[Hint: it is at best the latter.] Last fiddled with by ewmayer on 20071026 at 18:55 
20071026, 19:32  #5 
May 2005
Argentina
272_{8} Posts 
Yes, I'd read Wikipedia on stokes theorem before and I agree that stocks theorem is more general that my formula, as it can be generalized to R^n.
But, what surprises me is that being my formula on R^3, it isn't taught on vector calculus courses, and it isn't even mentioned in vector calculus texts. Anyway, there is something different between my formula and the general stocks theorem: my formula returns a vector, not a number. And where on the general stokes theorem do you see the factor  n x k  ? 
20071026, 19:49  #6 
May 2005
Argentina
2·3·31 Posts 
By the way, I've a doubt: In R^3 the curl of a vector (P, Q, R) is (dR/dy  dQ/dz, dP/dZ  dR/dx, dQ/dx  dP/dy).
How do you calculate the curl in R^4 of a vector (P, Q, R, S) ? 
20071026, 22:20  #7 
Aug 2002
Ann Arbor, MI
433 Posts 
There is no good equivalent notion of curl or cross product in higher dimension. The reason it works in three dimensional space is because 1=3(1+1).
Hopefully what I can remember from my class two years is ago is a good enough combination of accurate and simple. In general dimensions, you work with things called differential forms. They have an abstract definition, but you can think of them as being vectors whose coefficients are functions and whose basis is kfold products of differentials (a differential kform). So in R3, w=f(x,y,z)dx+g(x,y,z)dy+h(x,y,z)dz would be a 1form, w=f(x,y,z)dxdy+g(x,y,z)dydz+h(x,y,z)dxdz would be a two form. Like with vectors, they're some abstract thing, and really only make sense for a choice of basis (in R3, you choose x y and z). They have some rules like dxdx=0 and dxdy=dydx, so you can only put your differentials in a canonical increasing order. The two important operators on different forms are the exterior derivative and the hodge star operator. An exterior derivative is very much like a total derivative. For w=f(x,y)dy, d(w) would be partial f / partial x dydx + partial f / partial y dydy. Using the reduction rules, dydy=0 and dydx=dxdy you get d(w)=partial f / partial x dxdy. It's more or less the chain rule, except you have some extra rules with the dx's. The important thing is, if you put in a kform, you get out a k+1 form. The Hodgestar operator basically takes a differential form, and in each term you replace the differential with the complementary set of differentials (assuming your basis is orthogonal). In R3, this would boil down something like *(dx)=dydz, and one of them would have a negative sign because of the dxdy=dydx rule. But the important thing is, if you put in a kform, you get an nk form (where n is the dimension you're working in). Finally, the div/grad/curl operators are defined in terms of these things. In R3, a function is just a 0 form, and a vector is a 1form. You can see that if you take the exterior derivative of a 0form, you get the gradient of that function. The div function is defined as curl(w)=*d(w) on 1forms. You put in a 1form, exterior derivative of it is a 1+1form, hodge star of it is a 3(1+1) form, and so you get the result is again a 1form, or a vector. Similarly, div(w)=*d*(w) on 0forms. You can check that if you put in you input a vector or function into those things in the appropiate form, you get the same result you learned in Calculus. The formulas for grad and curl work in any dimension, but the fact that you can put a 1 form into the curl operator and get back out a 1 form only works when you have n=3. 
20071026, 22:25  #8  
Feb 2007
2^{4}·3^{3} Posts 
Quote:
in 3D you have the completely antisymmetric tensor E_ijk that allows to see the 2form dv (where v = Pdx+Qdy+Rdz) again as a vector. in other dimensions this is not possible. in general you integrate the (n1) form on a hypersurface (again of dim. n1 of course) to get the same than integrating its exterior derivative over the enclosed volume (or the like  did I mix up something?) 

20071031, 16:02  #9 
May 2005
Argentina
10111010_{2} Posts 
Thank you for your answers.
I've been reading about differential forms and I have a doubt. On a website I read this: "In general, it is true that in R^3 the operation of d on a differential 0form gives the gradient of that differential 0form, that on a differential 1form give the curl of that differential 1form, and that on a differential 2form gives its divergence." My question is: in R^2 how can I obtain de 2d version of the divergence by differentiating a form? Because if I differentiate a 0form it gives me the 2d version of the gradient, and if I differentiate a 1form it gives me the 2d version of the curl. 
20071031, 16:09  #10  
Nov 2003
2^{2}×5×373 Posts 
Quote:
The reason it works in 3D space is also related to the structure of the quaternions. Curl also works in 7D space. (and is related to the Octonions). For higher dimensional spaces, one can find an analogue of the curl operator based on Clifford algebras. But I don't know very much about this. I only had a very brief intro to Clifford algebras many years ago, and have forgotten the little that I learned. 

20071031, 21:51  #11 
Aug 2002
Ann Arbor, MI
433 Posts 
I only learned about this stuff as a small sidetangent in a differential geometry class I took two years when we were going over differential forms and generalized Stoke's theorem, so I don't know much about generalizations. But even if you can define an analagous concept for dimensions 2^n 1, the point still remains that what you have in dimension three doesn't clearly generalize to higher dimensions.

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