Differential Forms, Vector Calculus and Vector analysis in Higher Dimensions

[Rule]

Recently I've been studying differential forms and am amazed by their power: I can prove almost any result of calculus (including all the fundamental theorems, theorems of green, gauss, stokes', etc).

I do this through the generalized stokes' theorem -
the integral over the boundary of a manifold of omega equals the integral over the manifold of d(Omega).

Most of you have probably been trained to think of a differential, e.g. "dx" as an infinitesimal "distance", that approximates Delta(X) as Delta(X) approaches zero.  Sometimes it is even taught as being a formality, e.g. to be translated to "with respect to x".   

What you might not know is that it is infact an operator, that takes in vectors and outputs a scalar.  For example, if I have the vector (98, 144) multiplied by "dx" the output is "98", the "x" component of the vector.

If I multiply two differentials, we must consider the exterior or the "wedge" product.  dx^dy takes two vectors as input and again outputs a scalar.  If you had v1 and v2 multiplied by dx^dy, we would have the determinant of a matrix, the first row being the "x" components of v1 and v2 , the second row being the "y" components of v1 and v2 .  From this, think about what dx^dy represents geometrically... (hint, you could use vector projections)


It follows from this that dx^dy = -dy^dx  (anti-commutative),
and that dx^dx = 0.

This is where the reasoning of an "infinitesimal piece of length" doesn't hold: if you are say trying to find the moment of inertia of a rectangular object (say around an x axis), you might try the double integral of density times y^2 dxdy, thinking of dx and dy as "infinitesimal pieces of length" being summed up.  By this logic it seems that "dxdy = dydx".

With the (very brief) background information I have presented, I will briefly outline some steps to prove a couple important results of calculus.

I return to the generalized Stokes' theorem -
the integral over the boundary of a manifold of w equals the integral over the manifold of dw.

1)

Fundamental theorem of Calculus for Scalar valued functions in R1:
w = f(x)
dw = df/dx * dx     
(recall that if f = f(x, y, z, r, q, ....), df = df/dx*dx + df/dy*dy... where the "d"s in this case are partials.

So the boundary of the manifold in this case has orientation -- it will be "+B" and "-A". 
Using generalized stokes' theorem, the integral of df/dx*dx from A to B = f(B) - f(A).

2)
Follow the same logic and let w = f(x,y) to prove the fundamental theorem for line integrals.

3) Green's Theorem
w = M(x,y) dx  + N(x,y) dy.
dw = (dM/dx*dx + dM/dy *dy) dx + (dN/dx*dx + dN/dy*dy) dy
dw = dM/dy (dy^dx) + dN/dx (dx^dy)
dw = (dN/dx - dM/dy) dx^dy

The manifold in this case will be a surface in R2, e.g. x^2+y^2 <= 9419
and its boundary will be the surrounding curve.

Using generalized stokes' theorem, the contour integral of M(x,y) dx + N(x,y) dy over the boundary curve equals the double integral of (dN/dx - dM/dy) dx^dy over the region of integration.  This is Green's theorem.

3) Divergence theorem in R2
Let w = M(x,y)dy - N(x,y)dx,  proceed in a similar manner as before.

4) Stokes' Theorem
(Vector line integral over a closed curve equals the double integral of CurlF (dot) dS)
Let w = F1(x,y,z) + F2(x,y,z)  + F3(x,y,z)
Calculate dw in a manner similar to before. 
F1, F2, and F3, represent the "x", "y", and "z" components of the vector field, as "N" represented the "y" component, and "M" represented the "x" component previously.

To show that (dy^dz, -dx^dz, dx^dy) leads to dS try parametrizing x, y, and z as functions of q and p.  You'll find that you'll get something like
Tq dq (cross) Tp dp. 


5) Gauss' Theorem (Divergence Theorem in R3)
(Double Integral of F (dot) dS) over a surface equals the triple integral of Div(F) dV over the volume)

Let w = F1(x,y,z)dy^dz - F2(x,y,z) dx^dz + F3(x,y,z)dx^dy
Calculate dw in a manner similar to before. 

6) Extending this to higher dimensions

Want Stokes' theorem for a four dimensional object?
w = F1(x, y, z, r) + F2 + F3 + F4
etc.

To get the flux through a 4Dimensional object's surface (volume), let
w = F1 dy^dz^dr - F2 dx^dz^dr + F3 dx^dy^dr - F4 dx^dy^dz

Let me know if you need clarification.  Note ->  Div(F) = del (dot) F, and Curl(F) = del (cross) F.

Aside -

Some other interesting results....  It is Poincare's Lemma, that d^(2)w = 0.
This follows from the generalized stokes' theorem.
Integral of d^(2)w over M = Integral of dw of the boundary of M = Integral of w of the boundary of the boundary of M.  The boundary of the boundary of something is just the empty set, so this integral must be zero. 

You can prove by somewhat lengthy but straightforward calculation that
Curl(Grad(F)) = 0, and that Div(Curl(F)) = 0.  Or you can use differential forms.

Let w = F(x,y), calculate d^(2)w and you'll find that it's zero, and that it's curl(GradF).

Let w = F1(x,y,z)dx + F2(x,y,z)dy + F3(x,y,z)dz
Calculate d^(2)w and you'll find it's Div(CurlF) = 0.

d^(2)w = d(dw)