bivector form of Stokes theorem
Posted by peeterjoot on July 18, 2009
Obsolete with potential errors.
This post may be in error. I wrote this before understanding that the gradient used in Stokes Theorem must be projected onto the tangent space of the parameterized surface, as detailed in Alan MacDonald’s Vector and Geometric Calculus.
A parallelepiped volume element is depicted in the figure below. Three parameters , , generate a set of differential vector displacements spanning the three dimensional subspace
Writing the displacements
We have for the front, right and top face area elements
These are the surfaces of constant parameterization, respectively, , , and . For a bivector, the flux through the surface is therefore
Written out in full this is a bit of a mess
It should equal, at least up to a sign, . Expanding the latter is probably easier than regrouping the mess, and doing so we have
Expanding just that trivector-vector dot product
So we have
Noting that an , interchange in the first term inverts the sign, we have an exact match with (5), thus fixing the sign for the bivector form of Stokes theorem for the orientation picked in this diagram
Like the vector case, there is a requirement to be very specific about the meaning given to the oriented surfaces, and the corresponding oriented volume element (which could be a volume subspace of a greater than three dimensional space).