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## Stokes theorem applied to a vector field.

Posted by peeterjoot on July 17, 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.

# Motivation

I found my self forgetting stokes theorem once again. Redo this for the simplest case of a parallelogram area element.

What I recall is that we have on one side the curl dotted into the plane of the surface area element

\begin{aligned} \int ( \nabla \wedge A ) \cdot d^2 x \end{aligned}

and on the other side a loop integral (implying here a counterclockwise orientation: any idea how to do ointctrclockwise in wordpress?)

\begin{aligned} \int A \cdot dx \end{aligned}

Comparing the two we should end up with the same form and thus determine the form of the grade two Stokes equation (i.e. for curl of a vector).

# Bivector product part.

\begin{aligned} ( \nabla \wedge A ) \cdot d^2 x &= ( \nabla \wedge A ) \cdot \left(\frac{\partial x}{\partial \alpha} \wedge \frac{\partial x}{\partial \beta}\right) d\alpha d\beta \\ &= \partial_\mu A_\nu \frac{\partial x^\sigma}{\partial \alpha} \frac{\partial x^\epsilon}{\partial \beta} (\gamma^\mu \wedge \gamma^\nu) \cdot (\gamma_\sigma \wedge \gamma_\epsilon) d\alpha d\beta \\ &= \partial_\mu A_\nu \frac{\partial x^\sigma}{\partial \alpha} \frac{\partial x^\epsilon}{\partial \beta} ( {\delta^\mu}_\epsilon {\delta^\nu}_\sigma - {\delta^\mu}_\sigma {\delta^\nu}_\epsilon ) d\alpha d\beta \\ &= \partial_\mu A_\nu \left( \frac{\partial x^\nu}{\partial \alpha} \frac{\partial x^\mu}{\partial \beta} - \frac{\partial x^\mu}{\partial \alpha} \frac{\partial x^\nu}{\partial \beta} \right) d\alpha d\beta \\ \end{aligned}

So we have

\begin{aligned} ( \nabla \wedge A ) \cdot d^2 x &= -\partial_\mu A_\nu \frac{\partial (x^\mu, x^\nu)}{\partial (\alpha, \beta)} d\alpha d\beta \end{aligned} \quad\quad\quad(3)

# Loop integral part.

Integrating around a parallelogram spacetime area element with sides $d\alpha \partial x/\partial \alpha$ and $d\beta \partial x/\partial \beta$ we have

surface area element

\begin{aligned} \int A \cdot dx &= \int {\left. A \right\vert}_{\beta=\beta_0} \cdot \frac{\partial x}{\partial \alpha} d\alpha + {\left. A \right\vert}_{\alpha=\alpha_1} \cdot \frac{\partial x}{\partial \beta} d\beta + {\left. A \right\vert}_{\beta=\beta_1} \cdot \left( -\frac{\partial x}{\partial \alpha} d\alpha \right) + {\left. A \right\vert}_{\alpha=\alpha_0} \cdot \left( -\frac{\partial x}{\partial \beta} d\beta \right) \\ &= \int \left( {\left. A \right\vert}_{\alpha=\alpha_1} - {\left. A \right\vert}_{\alpha=\alpha_0} \right) \cdot \frac{\partial x}{\partial \beta} d\beta -\left( {\left. A \right\vert}_{\beta=\beta_1} - {\left. A \right\vert}_{\beta=\beta_0} \right) \cdot \frac{\partial x}{\partial \alpha} d\alpha \\ &= \int \frac{\partial A}{\partial \alpha} \cdot \frac{\partial x}{\partial \beta} d\alpha d\beta -\frac{\partial A}{\partial \beta} \cdot \frac{\partial x}{\partial \alpha} d\beta d\alpha \end{aligned}

Expanding the derivatives in terms of coordinates we have

\begin{aligned} \frac{\partial A}{\partial \sigma} &= \frac{\partial A_mu}{\partial \sigma} \gamma^\mu \\ &= \frac{\partial A_mu}{\partial x^\nu}\frac{\partial x^\nu}{\partial \sigma} \gamma^\mu \\ &= \partial_\nu A_\mu \frac{\partial x^\nu}{\partial \sigma} \gamma^\mu \\ \end{aligned}

and

\begin{aligned} \frac{\partial x}{\partial \sigma} &= \frac{\partial x^\nu}{\partial \sigma} \gamma_\nu \end{aligned}

Assembling we have

\begin{aligned} \int A \cdot dx &= \int \partial_\nu A_\mu \left( \frac{\partial x^\nu}{\partial \alpha} \frac{\partial x^\mu}{\partial \beta} - \frac{\partial x^\nu}{\partial \beta} \frac{\partial x^\mu}{\partial \alpha} \right) d\alpha d\beta \end{aligned}

In terms of the Jacobian used in (3) we have

\begin{aligned} \int A \cdot dx &= \int \partial_\mu A_\nu \frac{\partial (x^\mu, x^\nu)}{\partial (\alpha, \beta)} d\alpha d\beta \end{aligned}

Comparing the two we have only a sign difference so the conclusion is that Stokes for a vector field (considering only a flat parallelogram area element) is

\begin{aligned} \int ( \nabla \wedge A ) \cdot d^2 x &= -\int A \cdot dx \end{aligned}

Observe that there’s an implied orientation of the area element on the LHS, required to match up with the (reversed) counterclockwise orientation of the RHS integral.