Stokes becomes Green

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

int_S curl F . dS = int_C 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 _{int_D curl(F) . (r_u x r_v) du dv} If we rewrite the right hand side in terms of this parametrization, we get, letting B be the boundary of D

    int_B (F . r_u, F . r_v) . dr

(Check this! it amounts to the chain rule). By Green's theorem (applied in the u-v plane) this is

int_D d/du (F . r_v) - d/dv (F . r_u) du dv

Thus if we can show that

curl(F) . (r_u x r_v) = d/du (F . r_v) - d/dv (F . r_u)

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 r_u and r_v 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 r_u because r_u and r_v are constant:

(nabla x F) . (r_u x r_v) = ((nabla x F) x r_u) . r_v  
           = ((r_u . nabla) F - nabla (F . r_u)) . r_v
           = (r_u . nabla)(F . r_v) - (r_v . nabla)(F . r_u)
           =  d/du (F . r_v) - d/dv (F . r_u)   (*) 
           =  dF/du . r_v - dF/dv . r_u         (**)

Now lets stop pretending r_u and r_v are constant. We'd really like the original expression (*), but with r_u and r_v not constant, so let's try expanding (*) via the chain rule. We get:

dF/du . r_v + F . r_vu - dF/dv . r_u - F . r_uv.

Thankfully, two terms cancel by Clairaut's theorem, and the two expressions (*) and (**) are equal, even if r_u and r_v are allow to vary. If P(u,v) = F . r_u and Q(u,v) = F . r_v,, 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.