## 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.

See the post ‘stokes theorem in geometric algebra‘ [PDF], where this topic has been revisited with this in mind.

# Original Post:

[Click here for a PDF of this post with nicer formatting]

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).

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