6760, Math 21a, Fall 2009
Exhibits page Math 21a 2009, Multivariable Calculus
Stokes becomes Green
Course head: Oliver Knill
Office: SciCtr 434

About Green and Stokes

Here is how the three dimensional Stokes Theorem can be reduced to the two dimensional Green theorem. This is surprising because Stokes theorem is in a completely different space. It deals with vector fields which are non planar, even so if the surface is flat. Reducing the three dimensional theorem to two dimensions needs some work. Andrew Cotton-Clay managed to write it down in an elementary way:

Stokes theorem states that
intS curl F . dS = intC F . dr
If we rewrite the left hand side in terms of a parametrization r(u,v) from a region D in the u-v plane, we get intD curl(F) . (ru x rv) du dv If we rewrite the right hand side in terms of this parametrization, we get, letting B be the boundary of D
    intB (F . ru, F . rv) . dr 
(Check this! it amounts to the chain rule). By Green's theorem (applied in the u-v plane) this is
intD d/du (F . rv) - d/dv (F . ru) du dv
Thus if we can show that
curl(F) . (ru x rv) = d/du (F . rv) - d/dv (F . ru) 
we'll have shown that Stokes theorem follows from Green's theorem. This can be done using vector triple product identities as follows: Lets pretend first the vectors ru and rv are constant in u and v, so that only derivatives of F matter: by identities which can be found in 9.4 of the book and neglecting further derivatives of ru because ru and rv are constant:
(nabla x F) . (ru x rv) = ((nabla x F) x ru) . rv  
           = ((ru . nabla) F - nabla (F . ru)) . rv
           = (ru . nabla)(F . rv) - (rv . nabla)(F . ru)
           =  d/du (F . rv) - d/dv (F . ru)   (*) 
           =  dF/du . rv - dF/dv . ru         (**)
Now lets stop pretending ru and rv are constant. We'd really like the original expression (*), but with ru and rv not constant, so let's try expanding (*) via the chain rule. We get:
dF/du . rv + F . rvu - dF/dv . ru - F . ruv.
Thankfully, two terms cancel by Clairaut's theorem, and the two expressions (*) and (**) are equal, even if ru and rv are allow to vary. If P(u,v) = F . ru and Q(u,v) = F . rv,, we see that the line integral around the boundary, using the parametrization, is precisely the line integral with respect to the vector field (P,Q). Stokes Theorem is reduced to Green's Theorem.

Andy adds the following remark for mathematicians: we showed using vector identities that the curl commutes with the "pull back" to the parameterizing region in the uv-plane. The majority of this is proving that (curl F) . u, for u a unit vector, is what you'd expect by taking coordinates on the plane perpendicular to u with total area 1 and computing the 2-dimensional curl of F with respect to those coordinates.

Questions and comments to knill@math.harvard.edu
Math21b | Math 21a | Fall 2009 | Department of Mathematics | Faculty of Art and Sciences | Harvard University