Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

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 \alpha, \beta, \sigma generate a set of differential vector displacements spanning the three dimensional subspace

volume element

volume element

Writing the displacements

\begin{aligned}dx_\alpha &= \frac{\partial {x}}{\partial {\alpha}} d\alpha \\ dx_\beta &= \frac{\partial {x}}{\partial {\beta}} d\beta \\ dx_\sigma &= \frac{\partial {x}}{\partial {\sigma}} d\sigma \end{aligned}

We have for the front, right and top face area elements

\begin{aligned}dA_F &= dx_\alpha \wedge dx_\beta \\ dA_R &= dx_\beta \wedge dx_\sigma \\ dA_T &= dx_\sigma \wedge dx_\alpha \\ \end{aligned}

These are the surfaces of constant parameterization, respectively, \sigma = \sigma_1, \alpha = \alpha_1, and \beta = \beta_1. For a bivector, the flux through the surface is therefore

\begin{aligned}\int B \cdot dA &= (B_{\sigma_1} \cdot dA_F - B_{\sigma_0} \cdot dA_P ) + (B_{\alpha_1} \cdot dA_R - B_{\alpha_0} \cdot dA_L) + (B_{\beta_1} \cdot dA_T - B_{\beta_0} \cdot dA_B) \\ &= d \sigma \frac{\partial {B}}{\partial {\sigma}} \cdot (dx_\alpha \wedge dx_\beta ) + d \alpha \frac{\partial {B}}{\partial {\alpha}} \cdot (dx_\beta \wedge dx_\sigma) + d \beta \frac{\partial {B}}{\partial {\beta}} \cdot (dx_\sigma \wedge dx_\alpha ) \\ \end{aligned}

Written out in full this is a bit of a mess

\begin{aligned}\int B \cdot dA &= d \alpha d\beta d\sigma \partial_\mu B \cdot \left( \left( - \frac{\partial {x^\mu}}{\partial {\sigma}} \frac{\partial {x^\nu}}{\partial {\beta}} \frac{\partial {x^\epsilon}}{\partial {\alpha}} + \frac{\partial {x^\mu}}{\partial {\alpha}} \frac{\partial {x^\nu}}{\partial {\beta}} \frac{\partial {x^\epsilon}}{\partial {\sigma}} + \frac{\partial {x^\mu}}{\partial {\beta}} \frac{\partial {x^\nu}}{\partial {\sigma}} \frac{\partial {x^\epsilon}}{\partial {\alpha}} \right) (\gamma_\nu \wedge \gamma_\epsilon ) \right) \end{aligned} \quad\quad\quad(5)

It should equal, at least up to a sign, \int (\nabla \wedge B) \cdot d^3 x. Expanding the latter is probably easier than regrouping the mess, and doing so we have

\begin{aligned}(\nabla \wedge B) \cdot d^3 x &= d\alpha d\beta d\sigma ( \gamma^\mu \wedge \partial_\mu B) \cdot \left( \frac{\partial {x}}{\partial {\alpha}} \wedge \frac{\partial {x}}{\partial {\beta}} \wedge \frac{\partial {x}}{\partial {\sigma}} \right) \\ &= d\alpha d\beta d\sigma \frac{1}{{2}} ( \gamma^\mu \partial_\mu B + \partial_\mu B \gamma^\mu ) \cdot \left( \frac{\partial {x}}{\partial {\alpha}} \wedge \frac{\partial {x}}{\partial {\beta}} \wedge \frac{\partial {x}}{\partial {\sigma}} \right) \\ &= d\alpha d\beta d\sigma \frac{1}{{2}} \left\langle{{ ( \gamma^\mu \partial_\mu B + \partial_\mu B \gamma^\mu ) \left( \frac{\partial {x}}{\partial {\alpha}} \wedge \frac{\partial {x}}{\partial {\beta}} \wedge \frac{\partial {x}}{\partial {\sigma}} \right) }}\right\rangle \\ &= d\alpha d\beta d\sigma \frac{1}{{2}} \partial_\mu B \cdot {\left\langle{{ \left( \frac{\partial {x}}{\partial {\alpha}} \wedge \frac{\partial {x}}{\partial {\beta}} \wedge \frac{\partial {x}}{\partial {\sigma}} \right) \gamma^\mu + \gamma^\mu \left( \frac{\partial {x}}{\partial {\alpha}} \wedge \frac{\partial {x}}{\partial {\beta}} \wedge \frac{\partial {x}}{\partial {\sigma}} \right) }}\right\rangle}_{2} \\ &= d\alpha d\beta d\sigma \partial_\mu B \cdot \left( \left( \frac{\partial {x}}{\partial {\alpha}} \wedge \frac{\partial {x}}{\partial {\beta}} \wedge \frac{\partial {x}}{\partial {\sigma}} \right) \cdot \gamma^\mu \right) \\ \end{aligned}

Expanding just that trivector-vector dot product

\begin{aligned}\left( \frac{\partial {x}}{\partial {\alpha}} \wedge \frac{\partial {x}}{\partial {\beta}} \wedge \frac{\partial {x}}{\partial {\sigma}} \right) \cdot \gamma^\mu &= \frac{\partial {x^\lambda}}{\partial {\alpha}} \frac{\partial {x^\nu}}{\partial {\beta}} \frac{\partial {x^\epsilon}}{\partial {\sigma}} \left( \gamma_\lambda \wedge \gamma_\nu \wedge \gamma_\epsilon \right) \cdot \gamma^\mu \\ &= \frac{\partial {x^\lambda}}{\partial {\alpha}} \frac{\partial {x^\nu}}{\partial {\beta}} \frac{\partial {x^\epsilon}}{\partial {\sigma}} \left( \gamma_\lambda \wedge \gamma_\nu {\delta_\epsilon}^\mu -\gamma_\lambda \wedge \gamma_\epsilon {\delta_\nu}^\mu +\gamma_\nu \wedge \gamma_\epsilon {\delta_\lambda}^\mu \right) \end{aligned}

So we have

\begin{aligned}(\nabla \wedge B) \cdot d^3 x &= d\alpha d\beta d\sigma \frac{\partial {x^\lambda}}{\partial {\alpha}} \frac{\partial {x^\nu}}{\partial {\beta}} \frac{\partial {x^\epsilon}}{\partial {\sigma}} \partial_\mu B \cdot \left( \gamma_\lambda \wedge \gamma_\nu {\delta_\epsilon}^\mu -\gamma_\lambda \wedge \gamma_\epsilon {\delta_\nu}^\mu +\gamma_\nu \wedge \gamma_\epsilon {\delta_\lambda}^\mu \right) \\ &= d\alpha d\beta d\sigma \partial_\mu B \cdot \left( \frac{\partial {x^\lambda}}{\partial {\alpha}} \frac{\partial {x^\nu}}{\partial {\beta}} \frac{\partial {x^\mu}}{\partial {\sigma}} \gamma_\lambda \wedge \gamma_\nu + \frac{\partial {x^\lambda}}{\partial {\alpha}} \frac{\partial {x^\mu}}{\partial {\beta}} \frac{\partial {x^\epsilon}}{\partial {\sigma}} \gamma_\epsilon \wedge \gamma_\lambda + \frac{\partial {x^\mu}}{\partial {\alpha}} \frac{\partial {x^\nu}}{\partial {\beta}} \frac{\partial {x^\epsilon}}{\partial {\sigma}} \gamma_\nu \wedge \gamma_\epsilon \right) \\ &= d\alpha d\beta d\sigma \partial_\mu B \cdot \left( \left( \frac{\partial {x^\nu}}{\partial {\alpha}} \frac{\partial {x^\epsilon}}{\partial {\beta}} \frac{\partial {x^\mu}}{\partial {\sigma}} + \frac{\partial {x^\epsilon}}{\partial {\alpha}} \frac{\partial {x^\mu}}{\partial {\beta}} \frac{\partial {x^\nu}}{\partial {\sigma}} + \frac{\partial {x^\mu}}{\partial {\alpha}} \frac{\partial {x^\nu}}{\partial {\beta}} \frac{\partial {x^\epsilon}}{\partial {\sigma}} \right) \gamma_\nu \wedge \gamma_\epsilon \right) \\ \end{aligned}

Noting that an \epsilon, \nu 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

\begin{aligned}\int (\nabla \wedge B) \cdot d^3 x &= \int B \cdot d^2 x \end{aligned}

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).


One Response to “bivector form of Stokes theorem”

  1. lidiodu said

    Fundamental Theorem for Line Integrals
    The following result for line integrals is analogous to the Fundamental Theorem of Calculus.
    Let C be a curve in the xyz space parameterized by the vector function r(t)= for a<=t<=b. Suppose that f(x,y,z) is a differentiable function whose gradient grad f= is continuous on C. Then

    The above result states that the line integral of a vector field derived from a gradient depends only on the function f(x,y,z) and on the initial point (x(a),y(a),z(a)) and final point (x(b),y(b),z(b)) and not on the particular curve C. Hence, the integral is path independent. (Compare this with an example of a path dependent line integral.) as prof dr mircea orasanu
    We have given the result for a function f(x,y,z) of three variables and a curve in 3 dimensional space. The above result is valid for functions of any number of variables.

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