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 2007-10-26, 15:11 #1 Damian     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: $R = (xr, yr, zr)$ where $xr = \int \int \int_V rot(f) * (1, 0, 0) dV$ $yr = \int \int \int_V rot(f) * (0, 1, 0) dV$ $zr = \int \int \int_V rot(f) * (0, 0, 1) dV$ But any of these 3 integrals can be related to a surface integral, so take for example $zr$ $\int \int \int_V rot(f) * k dV = \int \int_S f \mathbf{.} t_{xy} \left\|n \mathscr{\^} k \right\| dS$ Where $k$ is the versor $(0,0,1)$, $f$ is any vector field, $t_{xy}$ 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 $R^3$ there is only one versor with these characteristics. And finally $n$ is the versor normal to the surface. $f \mathbf{.} t_{xy}$ is the dot product between $f$ and $t_{xy}$ $\left\|n \mathscr{\^} k \right\|$ is the norm of the vector product between $n$ and $k$ It is interesting to note that carefully choosing the vector field $f$ so that $rot(f) * k = 1$, 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.
 2007-10-26, 17:31 #2 ewmayer ∂2ω=0     Sep 2002 República de California 3×53×31 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 2-D special case of Stokes' more-general theorem, which applies to [suitably defined] manifolds in Rn. After all, even through one generally only calculates the explicit curl operator in 2-D and 3-D in most college calculus classes, these operators are fully generalizable to n-D. 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"?
 2007-10-26, 17:45 #3 Damian     May 2005 Argentina 18610 Posts By $rot(f)$ I meant $curl(f)$ 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.
 2007-10-26, 18:50 #4 ewmayer ∂2ω=0     Sep 2002 República de California 3×53×31 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 2007-10-26 at 18:55
 2007-10-26, 19:32 #5 Damian     May 2005 Argentina 2728 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 | ?
 2007-10-26, 19:49 #6 Damian     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) ?
 2007-10-26, 22:20 #7 Kevin     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 k-fold products of differentials (a differential k-form). So in R3, w=f(x,y,z)dx+g(x,y,z)dy+h(x,y,z)dz would be a 1-form, 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 k-form, you get out a k+1 form. The Hodge-star 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 k-form, you get an n-k 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 1-form. You can see that if you take the exterior derivative of a 0-form, you get the gradient of that function. The div function is defined as curl(w)=*d(w) on 1-forms. You put in a 1-form, exterior derivative of it is a 1+1-form, hodge star of it is a 3-(1+1) form, and so you get the result is again a 1-form, or a vector. Similarly, div(w)=*d*(w) on 0-forms. 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.
2007-10-26, 22:25   #8
m_f_h

Feb 2007

24·33 Posts

Quote:
 Originally Posted by Damian 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) ?
there is no (vector) curl in other than 3 dim's

in 3D you have the completely antisymmetric tensor E_ijk
that allows to see the 2-form dv (where v = Pdx+Qdy+Rdz)
again as a vector. in other dimensions this is not possible.

in general you integrate the (n-1) form on a hypersurface (again of dim. n-1 of course) to get the same than integrating its exterior derivative over the enclosed volume
(or the like - did I mix up something?)

 2007-10-31, 16:02 #9 Damian     May 2005 Argentina 101110102 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 0-form gives the gradient of that differential 0-form, that on a differential 1-form give the curl of that differential 1-form, and that on a differential 2-form gives its divergence." My question is: in R^2 how can I obtain de 2-d version of the divergence by differentiating a form? Because if I differentiate a 0-form it gives me the 2-d version of the gradient, and if I differentiate a 1-form it gives me the 2-d version of the curl.
2007-10-31, 16:09   #10
R.D. Silverman

Nov 2003

22×5×373 Posts

Quote:
 Originally Posted by Kevin 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).

The reason it works in 3-D space is also related to the structure of the
quaternions. Curl also works in 7-D space. (and is related to the Octonions).

For higher dimensional spaces, one can find an analogue of the curl
I only had a very brief intro to Clifford algebras many years ago, and have
forgotten the little that I learned.

 2007-10-31, 21:51 #11 Kevin     Aug 2002 Ann Arbor, MI 433 Posts I only learned about this stuff as a small side-tangent 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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