# Peeter Joot's Blog.

• ## Recent Comments

 ivor on Just Energy Canada nasty busin… A final pre-exam upd… on An updated compilation of note… Anon on About peeterjoot on About Anon on About
• ## People not reading this blog: 6,973,738,433 minus:

• 132,259 hits

# Posts Tagged ‘stokes theorem’

## PHY450H1S. Relativistic Electrodynamics Lecture 18 (Taught by Prof. Erich Poppitz). Green’s function solution to Maxwell’s equation.

Posted by peeterjoot on March 12, 2011

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

Covering chapter 8 material from the text [1].

Covering lecture notes pp. 136-146: continued reminder of electrostatic Greens function (136); the retarded Greens function of the d’Alembert operator: derivation and properties (137-140); the solution of the d’Alembert equation with a source: retarded potentials (141-142)

# Solving the forced wave equation.

See the notes for a complex variables and Fourier transform method of deriving the Green’s function. In class, we’ll just pull it out of a magic hat. We wish to solve

\begin{aligned}\square A^k = \partial_i \partial^i A^k = \frac{4 \pi}{c} j^k\end{aligned} \hspace{\stretch{1}}(2.1)

(with a $\partial_i A^i = 0$ gauge choice).

Our Green’s method utilizes

\begin{aligned}\square_{(\mathbf{x}, t)} G(\mathbf{x} - \mathbf{x}', t - t') = \delta^3( \mathbf{x} - \mathbf{x}') \delta( t - t')\end{aligned} \hspace{\stretch{1}}(2.2)

If we know such a function, our solution is simple to obtain

\begin{aligned}A^k(\mathbf{x}, t)= \int d^3 \mathbf{x}' dt' \frac{4 \pi}{c} j^k(\mathbf{x}', t') G(\mathbf{x} - \mathbf{x}', t - t')\end{aligned} \hspace{\stretch{1}}(2.3)

Proof:

\begin{aligned}\square_{(\mathbf{x}, t)} A^k(\mathbf{x}, t)&=\int d^3 \mathbf{x}' dt' \frac{4 \pi}{c} j^k(\mathbf{x}', t')\square_{(\mathbf{x}, t)}G(\mathbf{x} - \mathbf{x}', t - t') \\ &=\int d^3 \mathbf{x}' dt' \frac{4 \pi}{c} j^k(\mathbf{x}', t')\delta^3( \mathbf{x} - \mathbf{x}') \delta( t - t') \\ &=\frac{4 \pi}{c} j^k(\mathbf{x}, t)\end{aligned}

Claim:

\begin{aligned}G(\mathbf{x}, t) = \frac{\delta(t - {\left\lvert{\mathbf{x}}\right\rvert}/c)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }\end{aligned} \hspace{\stretch{1}}(2.4)

This is the retarded Green’s function of the operator $\square$, where

\begin{aligned}\square G(\mathbf{x}, t) = \delta^3(\mathbf{x}) \delta(t)\end{aligned} \hspace{\stretch{1}}(2.5)

## Proof of the d’Alembertian Green’s function

Our Prof is excellent at motivating any results that he pulls out of magic hats. He’s said that he’s included a derivation using Fourier transforms and tricky contour integration arguments in the class notes for anybody who is interested (and for those who also know how to do contour integration). For those who don’t know contour integration yet (some people are taking it concurrently), one can actually prove this by simply applying the wave equation operator to this function. This treats the delta function as a normal function that one can take the derivatives of, something that can be well defined in the context of generalized functions. Chugging ahead with this approach we have

\begin{aligned}\square G(\mathbf{x}, t)=\left(\frac{1}{{c^2}} \frac{\partial^2 {{}}}{\partial {{t}}^2} - \Delta\right)\frac{\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }=\frac{\delta''\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi c^2 {\left\lvert{\mathbf{x}}\right\rvert} }- \Delta \frac{\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }.\end{aligned} \hspace{\stretch{1}}(2.6)

This starts things off and now things get a bit hairy. It’s helpful to consider a chain rule expansion of the Laplacian

\begin{aligned}\Delta (u v)&=\partial_{\alpha\alpha} (u v) \\ &=\partial_{\alpha} (v \partial_\alpha u+ u\partial_\alpha v) \\ &=(\partial_\alpha v) (\partial_\alpha u ) + v \partial_{\alpha\alpha} u+(\partial_\alpha u) (\partial_\alpha v ) + u \partial_{\alpha\alpha} v).\end{aligned}

In vector form this is

\begin{aligned}\Delta (u v) = u \Delta v + 2 (\boldsymbol{\nabla} u) \cdot (\boldsymbol{\nabla} v) + v \Delta u.\end{aligned} \hspace{\stretch{1}}(2.7)

Applying this to the Laplacian portion of 2.6 we have

\begin{aligned}\Delta \frac{\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }=\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)\Delta\frac{1}{{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }}+\left(\boldsymbol{\nabla} \frac{1}{{2 \pi {\left\lvert{\mathbf{x}}\right\rvert} }}\right)\cdot\left(\boldsymbol{\nabla}\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \right)+\frac{1}{{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }}\Delta\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right).\end{aligned} \hspace{\stretch{1}}(2.8)

Here we make the identification

\begin{aligned}\Delta \frac{1}{{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }} = - \delta^3(\mathbf{x}).\end{aligned} \hspace{\stretch{1}}(2.9)

This could be considered a given from our knowledge of electrostatics, but it’s not too much work to just do so.

### An aside. Proving the Laplacian Green’s function.

If $-1/{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }$ is a Green’s function for the Laplacian, then the Laplacian of the convolution of this with a test function should recover that test function

\begin{aligned}\Delta \int d^3 \mathbf{x}' \left(-\frac{1}{{4 \pi {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert} }} \right) f(\mathbf{x}') = f(\mathbf{x}).\end{aligned} \hspace{\stretch{1}}(2.10)

We can directly evaluate the LHS of this equation, following the approach in [2]. First note that the Laplacian can be pulled into the integral and operates only on the presumed Green’s function. For that operation we have

\begin{aligned}\Delta \left(-\frac{1}{{4 \pi {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert} }} \right)=-\frac{1}{{4 \pi}} \boldsymbol{\nabla} \cdot \boldsymbol{\nabla} {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}.\end{aligned} \hspace{\stretch{1}}(2.11)

It will be helpful to compute the gradient of various powers of ${\left\lvert{\mathbf{x}}\right\rvert}$

\begin{aligned}\boldsymbol{\nabla} {\left\lvert{\mathbf{x}}\right\rvert}^a&=e_\alpha \partial_\alpha (x^\beta x^\beta)^{a/2} \\ &=e_\alpha \left(\frac{a}{2}\right) 2 x^\beta {\delta_\beta}^\alpha {\left\lvert{\mathbf{x}}\right\rvert}^{a - 2}.\end{aligned}

In particular we have, when $\mathbf{x} \ne 0$, this gives us

\begin{aligned}\boldsymbol{\nabla} {\left\lvert{\mathbf{x}}\right\rvert} &= \frac{\mathbf{x}}{{\left\lvert{\mathbf{x}}\right\rvert}} \\ \boldsymbol{\nabla} \frac{1}{{{\left\lvert{\mathbf{x}}\right\rvert}}} &= -\frac{\mathbf{x}}{{\left\lvert{\mathbf{x}}\right\rvert}^3} \\ \boldsymbol{\nabla} \frac{1}{{{\left\lvert{\mathbf{x}}\right\rvert}^3}} &= -3 \frac{\mathbf{x}}{{\left\lvert{\mathbf{x}}\right\rvert}^5}.\end{aligned} \hspace{\stretch{1}}(2.12)

For the Laplacian of $1/{\left\lvert{\mathbf{x}}\right\rvert}$, at the points $\mathbf{e} \ne 0$ where this is well defined we have

\begin{aligned}\Delta \frac{1}{{{\left\lvert{\mathbf{x}}\right\rvert}}} &=\boldsymbol{\nabla} \cdot \boldsymbol{\nabla} \frac{1}{{{\left\lvert{\mathbf{x}}\right\rvert}}} \\ &= -\partial_\alpha \frac{x^\alpha}{{\left\lvert{\mathbf{x}}\right\rvert}^3} \\ &= -\frac{3}{{\left\lvert{\mathbf{x}}\right\rvert}^3} - x^\alpha \partial_\alpha \frac{1}{{\left\lvert{\mathbf{x}}\right\rvert}^3} \\ &= -\frac{3}{{\left\lvert{\mathbf{x}}\right\rvert}^3} - \mathbf{x} \cdot \boldsymbol{\nabla} \frac{1}{{\left\lvert{\mathbf{x}}\right\rvert}^3} \\ &= -\frac{3}{{\left\lvert{\mathbf{x}}\right\rvert}^3} + 3 \frac{\mathbf{x}^2}{{\left\lvert{\mathbf{x}}\right\rvert}^5}\end{aligned}

So we have a zero. This means that the Laplacian operation

\begin{aligned}\Delta \int d^3 \mathbf{x}' \frac{1}{{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert} }} f(\mathbf{x}') =\lim_{\epsilon = {\left\lvert{\mathbf{x} -\mathbf{x}'}\right\rvert} \rightarrow 0}f(\mathbf{x}) \int d^3 \mathbf{x}' \Delta \frac{1}{{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}},\end{aligned} \hspace{\stretch{1}}(2.15)

can only have a value in a neighborhood of point $\mathbf{x}$. Writing $\Delta = \boldsymbol{\nabla} \cdot \boldsymbol{\nabla}$ we have

\begin{aligned}\Delta \int d^3 \mathbf{x}' \frac{1}{{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert} }} f(\mathbf{x}') =\lim_{\epsilon = {\left\lvert{\mathbf{x} -\mathbf{x}'}\right\rvert} \rightarrow 0}f(\mathbf{x}) \int d^3 \mathbf{x}' \boldsymbol{\nabla} \cdot -\frac{\mathbf{x} - \mathbf{x}'}{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}.\end{aligned} \hspace{\stretch{1}}(2.16)

Observing that $\boldsymbol{\nabla} \cdot f(\mathbf{x} -\mathbf{x}') = -\boldsymbol{\nabla}' f(\mathbf{x} - \mathbf{x}')$ we can put this in a form that allows for use of Stokes theorem so that we can convert this to a surface integral

\begin{aligned}\Delta \int d^3 \mathbf{x}' \frac{1}{{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert} }} f(\mathbf{x}') &=\lim_{\epsilon = {\left\lvert{\mathbf{x} -\mathbf{x}'}\right\rvert} \rightarrow 0}f(\mathbf{x}) \int d^3 \mathbf{x}' \boldsymbol{\nabla}' \cdot \frac{\mathbf{x} - \mathbf{x}'}{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}^3} \\ &=\lim_{\epsilon = {\left\lvert{\mathbf{x} -\mathbf{x}'}\right\rvert} \rightarrow 0}f(\mathbf{x}) \int d^2 \mathbf{x}' \mathbf{n} \cdot \frac{\mathbf{x} - \mathbf{x}'}{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}^3} \\ &= \int_{\phi=0}^{2\pi} \int_{\theta = 0}^\pi \epsilon^2 \sin\theta d\theta d\phi \frac{\mathbf{x}' - \mathbf{x}}{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}} \cdot \frac{\mathbf{x} - \mathbf{x}'}{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}^3} \\ &= -\int_{\phi=0}^{2\pi} \int_{\theta = 0}^\pi \epsilon^2 \sin\theta d\theta d\phi \frac{\epsilon^2}{\epsilon^4}\end{aligned}

where we use $(\mathbf{x}' - \mathbf{x})/{\left\lvert{\mathbf{x}' - \mathbf{x}}\right\rvert}$ as the outwards normal for a sphere centered at $\mathbf{x}$ of radius $\epsilon$. This integral is just $-4 \pi$, so we have

\begin{aligned}\Delta \int d^3 \mathbf{x}' \frac{1}{{-4 \pi {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert} }} f(\mathbf{x}') =f(\mathbf{x}).\end{aligned} \hspace{\stretch{1}}(2.17)

The convolution of $f(\mathbf{x})$ with $-\Delta/4 \pi {\left\lvert{\mathbf{x}}\right\rvert}$ produces $f(\mathbf{x})$, allowing an identification of this function with a delta function, since the two have the same operational effect

\begin{aligned}\int d^3 \mathbf{x}' \delta(\mathbf{x} - \mathbf{x}') f(\mathbf{x}') =f(\mathbf{x}).\end{aligned} \hspace{\stretch{1}}(2.18)

### Returning to the d’Alembertian Green’s function.

We need two additional computations to finish the job. The first is the gradient of the delta function

\begin{aligned}\boldsymbol{\nabla} \delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) &= ? \\ \Delta \delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) &= ?\end{aligned}

Consider $\boldsymbol{\nabla} f(g(\mathbf{x}))$. This is

\begin{aligned}\boldsymbol{\nabla} f(g(\mathbf{x}))&=e_\alpha \frac{\partial {f(g(\mathbf{x}))}}{\partial {x^\alpha}} \\ &=e_\alpha \frac{\partial {f}}{\partial {g}} \frac{\partial {g}}{\partial {x^\alpha}},\end{aligned}

so we have

\begin{aligned}\boldsymbol{\nabla} f(g(\mathbf{x}))=\frac{\partial {f}}{\partial {g}} \boldsymbol{\nabla} g.\end{aligned} \hspace{\stretch{1}}(2.19)

The Laplacian is similar

\begin{aligned}\Delta f(g)&= \boldsymbol{\nabla} \cdot \left(\frac{\partial {f}}{\partial {g}} \boldsymbol{\nabla} g \right) \\ &= \partial_\alpha \left(\frac{\partial {f}}{\partial {g}} \partial_\alpha g \right) \\ &= \left( \partial_\alpha \frac{\partial {f}}{\partial {g}} \right) \partial_\alpha g +\frac{\partial {f}}{\partial {g}} \partial_{\alpha\alpha} g \\ &= \frac{\partial^2 {{f}}}{\partial {{g}}^2} \left( \partial_\alpha g \right) (\partial_\alpha g)+\frac{\partial {f}}{\partial {g}} \Delta g,\end{aligned}

so we have

\begin{aligned}\Delta f(g)= \frac{\partial^2 {{f}}}{\partial {{g}}^2} (\boldsymbol{\nabla} g)^2 +\frac{\partial {f}}{\partial {g}} \Delta g\end{aligned} \hspace{\stretch{1}}(2.20)

With $g(\mathbf{x}) = {\left\lvert{\mathbf{x}}\right\rvert}$, we’ll need the Laplacian of this vector magnitude

\begin{aligned}\Delta {\left\lvert{\mathbf{x}}\right\rvert}&=\partial_\alpha \frac{x_\alpha}{{\left\lvert{\mathbf{x}}\right\rvert}} \\ &=\frac{3}{{\left\lvert{\mathbf{x}}\right\rvert}} + x_\alpha \partial_\alpha (x^\beta x^\beta)^{-1/2} \\ &=\frac{3}{{\left\lvert{\mathbf{x}}\right\rvert}} - \frac{x_\alpha x_\alpha}{{\left\lvert{\mathbf{x}}\right\rvert}^3} \\ &= \frac{2}{{\left\lvert{\mathbf{x}}\right\rvert}} \end{aligned}

So that we have

\begin{aligned}\boldsymbol{\nabla} \delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) &= -\frac{1}{{c}} \delta'\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \frac{\mathbf{x}}{{\left\lvert{\mathbf{x}}\right\rvert}} \\ \Delta \delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) &=\frac{1}{{c^2}} \delta''\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) -\frac{1}{{c}} \delta'\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \frac{2}{{\left\lvert{\mathbf{x}}\right\rvert}} \end{aligned} \hspace{\stretch{1}}(2.21)

Now we have all the bits and pieces of 2.8 ready to assemble

\begin{aligned}\Delta \frac{\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }&=-\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \delta^3(\mathbf{x}) \\ &\quad +\frac{1}{{2\pi}} \left( - \frac{\mathbf{x}}{{\left\lvert{\mathbf{x}}\right\rvert}^3} \right)\cdot-\frac{1}{{c}} \delta'\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \frac{\mathbf{x}}{{\left\lvert{\mathbf{x}}\right\rvert}} \\ &\quad +\frac{1}{{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }}\left(\frac{1}{{c^2}} \delta''\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) -\frac{1}{{c}} \delta'\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \frac{2}{{\left\lvert{\mathbf{x}}\right\rvert}} \right) \\ &=-\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \delta^3(\mathbf{x}) +\frac{1}{{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} c^2 }}\delta''\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \end{aligned}

Since we also have

\begin{aligned}\frac{1}{{c^2}} \partial_{tt}\frac{\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }=\frac{\delta''\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} c^2}\end{aligned} \hspace{\stretch{1}}(2.23)

The $\delta''$ terms cancel out in the d’Alembertian, leaving just

\begin{aligned}\square \frac{\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }=\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \delta^3(\mathbf{x}) \end{aligned} \hspace{\stretch{1}}(2.24)

Noting that the spatial delta function is non-zero only when $\mathbf{x} = 0$, which means $\delta(t - {\left\lvert{\mathbf{x}}\right\rvert}/c) = \delta(t)$ in this product, and we finally have

\begin{aligned}\square \frac{\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }=\delta(t) \delta^3(\mathbf{x}) \end{aligned} \hspace{\stretch{1}}(2.25)

We write

\begin{aligned}G(\mathbf{x}, t) = \frac{\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} },\end{aligned} \hspace{\stretch{1}}(2.26)

# Elaborating on the wave equation Green’s function

The Green’s function 2.26 is a distribution that is non-zero only on the future lightcone. Observe that for $t < 0$ we have

\begin{aligned}\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)&=\delta\left(-{\left\lvert{t}\right\rvert} - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \\ &= 0.\end{aligned}

We say that $G$ is supported only on the future light cone. At $\mathbf{x} = 0$, only the contributions for $t > 0$ matter. Note that in the “old days”, Green’s functions used to be called influence functions, a name that works particularly well in this case. We have other Green’s functions for the d’Alembertian. The one above is called the retarded Green’s functions and we also have an advanced Green’s function. Writing $+$ for advanced and $-$ for retarded these are

\begin{aligned}G_{\pm} = \frac{\delta\left(t \pm \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert}}\end{aligned} \hspace{\stretch{1}}(3.27)

There are also causal and non-causal variations that won’t be of interest for this course.

This arms us now to solve any problem in the Lorentz gauge

\begin{aligned}A^k(\mathbf{x}, t) = \frac{1}{{c}} \int d^3 \mathbf{x}' dt' \frac{\delta\left(t - t' - \frac{{\left\lvert{\mathbf{x} -\mathbf{x}'}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}j^k(\mathbf{x}', t')+\text{An arbitrary collection of EM waves.}\end{aligned} \hspace{\stretch{1}}(3.28)

The additional EM waves are the possible contributions from the homogeneous equation.

Since $\delta(t - t' - {\left\lvert{\mathbf{x} -\mathbf{x}'}\right\rvert}/c)$ is non-zero only when $t' = t - {\left\lvert{\mathbf{x} -\mathbf{x}'}\right\rvert}/c)$, the non-homogeneous parts of 3.28 reduce to

\begin{aligned}A^k(\mathbf{x}, t) = \frac{1}{{c}} \int d^3 \mathbf{x}' \frac{j^k(\mathbf{x}', t - {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}/c)}{4 \pi {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}.\end{aligned} \hspace{\stretch{1}}(3.29)

Our potentials at time $t$ and spatial position $\mathbf{x}$ are completely specified in terms of the sums of the currents acting at the retarded time $t - {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}/c$. The field can only depend on the charge and current distribution in the past. Specifically, it can only depend on the charge and current distribution on the past light cone of the spacetime point at which we measure the field.

# Example of the Green’s function. Consider a charged particle moving on a worldline

\begin{aligned}(c t, \mathbf{x}_c(t))\end{aligned} \hspace{\stretch{1}}(4.30)

($c$ for classical)

For this particle

\begin{aligned}\rho(\mathbf{x}, t) &= e \delta^3(\mathbf{x} - \mathbf{x}_c(t)) \\ \mathbf{j}(\mathbf{x}, t) &= e \dot{\mathbf{x}}_c(t) \delta^3(\mathbf{x} - \mathbf{x}_c(t))\end{aligned} \hspace{\stretch{1}}(4.31)

\begin{aligned}\begin{bmatrix}A^0(\mathbf{x}, t)\mathbf{A}(\mathbf{x}, t)\end{bmatrix}&=\frac{1}{{c}}\int d^3 \mathbf{x}' dt'\frac{ \delta( t - t' - {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}/c }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}\begin{bmatrix}c e \\ e \dot{\mathbf{x}}_c(t)\end{bmatrix}\delta^3(\mathbf{x} - \mathbf{x}_c(t)) \\ &=\int_{-\infty}^\infty\frac{ \delta( t - t' - {\left\lvert{\mathbf{x} - \mathbf{x}_c(t')}\right\rvert}/c }{{\left\lvert{\mathbf{x}_c(t') - \mathbf{x}}\right\rvert}}\begin{bmatrix}e \\ e \frac{\dot{\mathbf{x}}_c(t)}{c}\end{bmatrix}\end{aligned}

PICTURE: light cones, and curved worldline. Pick an arbitrary point $(\mathbf{x}_0, t_0)$, and draw the past light cone, looking at where this intersects with the trajectory

For the arbitrary point $(\mathbf{x}_0, t_0)$ we see that this point and the retarded time $(\mathbf{x}_c(t_r), t_r)$ obey the relation

\begin{aligned}c (t_0 - t_r) = {\left\lvert{\mathbf{x}_0 - \mathbf{x}_c(t_r)}\right\rvert}\end{aligned} \hspace{\stretch{1}}(4.33)

This retarded time is unique. There is only one such intersection.

Our job is to calculate

\begin{aligned}\int_{-\infty}^\infty \delta(f(x)) g(x) = \frac{g(x_{*})}{f'(x_{*})}\end{aligned} \hspace{\stretch{1}}(4.34)

where $f(x_{*}) = 0$.

\begin{aligned}f(t') = t - t' - {\left\lvert{\mathbf{x} - \mathbf{x}_c(t')}\right\rvert}/c\end{aligned} \hspace{\stretch{1}}(4.35)

\begin{aligned}\frac{\partial {f}}{\partial {t'}}&= -1 - \frac{1}{{c}} \frac{\partial {}}{\partial {t'}} \sqrt{ (\mathbf{x} - \mathbf{x}_c(t')) \cdot (\mathbf{x} - \mathbf{x}_c(t')) } \\ &= -1 + \frac{1}{{c}} \frac{\partial {}}{\partial {t'}} \frac{(\mathbf{x} - \mathbf{x}_c(t')) \cdot \mathbf{v}_c(t_r)}{{\left\lvert{\mathbf{x} - \mathbf{x}_c(t_r)}\right\rvert}}\end{aligned}

# References

[1] L.D. Landau and E.M. Lifshitz. The classical theory of fields. Butterworth-Heinemann, 1980.

[2] M. Schwartz. Principles of Electrodynamics. Dover Publications, 1987.

## PHY450H1S. Relativistic Electrodynamics Lecture 16 (Taught by Prof. Erich Poppitz). Monochromatic EM fields. Poynting vector and energy density conservation

Posted by peeterjoot on March 3, 2011

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

Covering chapter 6 material from the text [1].

Covering lecture notes pp. 115-127: properties of monochromatic plane EM waves (122-124); energy and energy flux of the EM field and energy conservation from the equations of motion (125-127) [Wednesday, Mar. 2]

# Review. Solution to the wave equation.

Recall that in the Coulomb gauge

\begin{aligned}A^0 &= 0 \\ \boldsymbol{\nabla} \cdot \mathbf{A} &= 0\end{aligned} \hspace{\stretch{1}}(2.1)

our equation to solve is

\begin{aligned}\left( \frac{1}{{c^2}} \frac{\partial^2 {{}}}{\partial {{t}}^2} - \Delta \right) \mathbf{A} = 0.\end{aligned} \hspace{\stretch{1}}(2.3)

We found that the general solution was

\begin{aligned}\mathbf{A}(\mathbf{x}, t) = \int \frac{d^3\mathbf{k}}{(2 \pi)^3} \left(e^{i (\mathbf{k} \cdot \mathbf{x} + \omega_k t)} \boldsymbol{\beta}^{*}(-\mathbf{k})+e^{i (\mathbf{k} \cdot \mathbf{x} - \omega_k t)} \boldsymbol{\beta}(\mathbf{k})\right)\end{aligned} \hspace{\stretch{1}}(2.4)

where

\begin{aligned}\mathbf{k} \cdot \boldsymbol{\beta}(\mathbf{k}) = 0\end{aligned} \hspace{\stretch{1}}(2.5)

It is clear that this is a solution since

\begin{aligned}\left( \frac{1}{{c^2}} \frac{\partial^2 {{}}}{\partial {{t}}^2} - \Delta \right) e^{i (\mathbf{k} \cdot \mathbf{x} \pm \omega_k t)} = 0\end{aligned} \hspace{\stretch{1}}(2.6)

# Moving to physically relevant results.

Since the most general solution is a sum over $\mathbf{k}$, it is enough to consider only a single $\mathbf{k}$, or equivalently, take

\begin{aligned}\boldsymbol{\beta}(\mathbf{k}) &= \boldsymbol{\beta} ( 2\pi)^3 \delta^3(\mathbf{k} - \mathbf{p}) \\ \boldsymbol{\beta}^{*}(-\mathbf{k}) &= \boldsymbol{\beta}^{*} ( 2\pi)^3 \delta^3(-\mathbf{k} - \mathbf{p})\end{aligned} \hspace{\stretch{1}}(3.7)

but we have the freedom to pick a real and constant $\boldsymbol{\beta}$. Now our solution is

\begin{aligned}\mathbf{A}(\mathbf{x}, t) = \boldsymbol{\beta} \left(e^{-i (\mathbf{p} \cdot \mathbf{x} + \omega_k t)}+e^{i (\mathbf{p} \cdot \mathbf{x} - \omega_k t)}\right)= \boldsymbol{\beta} \cos( \omega t - \mathbf{p} \cdot \mathbf{x})\end{aligned} \hspace{\stretch{1}}(3.9)

where

\begin{aligned}\boldsymbol{\beta} \cdot \mathbf{p} = 0\end{aligned} \hspace{\stretch{1}}(3.10)

FIXME:DIY: show that also using $\boldsymbol{\beta}$ complex also works.

Let’s choose

\begin{aligned}\mathbf{p} = (p, 0, 0)\end{aligned} \hspace{\stretch{1}}(3.11)

Since

\begin{aligned}\mathbf{p} \cdot \boldsymbol{\beta} = p_x \beta_x\end{aligned} \hspace{\stretch{1}}(3.12)

we must have

\begin{aligned}\boldsymbol{\beta}_x = 0\end{aligned} \hspace{\stretch{1}}(3.13)

so

\begin{aligned}\boldsymbol{\beta} = (0, \beta_y, \beta_z)\end{aligned} \hspace{\stretch{1}}(3.14)

\paragraph{Claim:} The Coulomb gauge $0 = \boldsymbol{\nabla} \cdot \mathbf{A} = (\boldsymbol{\beta} \cdot \mathbf{p})\sin(\omega t - \mathbf{p} \cdot \mathbf{x})$ implies that there are two linearly independent choices of $\boldsymbol{\beta}$ and $\mathbf{p}$.

FIXME: missing exactly how this is?

PICTURE:

$\boldsymbol{\beta}_1$, $\boldsymbol{\beta}_2$, $\mathbf{p}$ all mutually perpendicular.

\begin{aligned}\mathbf{E}&= -\frac{\partial {\mathbf{A}}}{\partial {ct}} \\ &= -\frac{\boldsymbol{\beta}}{c} \frac{\partial {}}{\partial {t}} \cos(\omega t - \mathbf{p} \cdot \mathbf{x}) \\ &= -\frac{1}{{c}} \boldsymbol{\beta} \omega_p\sin(\omega t - \mathbf{p} \cdot \mathbf{x})\end{aligned}

(recall: $\omega_p = c{\left\lvert{\mathbf{p}}\right\rvert}$)

\begin{aligned}\boxed{\mathbf{E} = \boldsymbol{\beta} {\left\lvert{\mathbf{p}}\right\rvert} \sin(\omega t - \mathbf{p} \cdot \mathbf{x})}\end{aligned} \hspace{\stretch{1}}(3.15)

\begin{aligned}\mathbf{B}&= \boldsymbol{\nabla} \times \mathbf{A} \\ &= \boldsymbol{\nabla} \times ( \boldsymbol{\beta} \cos(\omega t - \mathbf{p} \cdot \mathbf{x}) \\ &= (\boldsymbol{\nabla} \cos(\omega t - \mathbf{p} \cdot \mathbf{x})) \times \boldsymbol{\beta} \\ &= \sin(\omega t - \mathbf{p} \cdot \mathbf{x}) \mathbf{p} \times \boldsymbol{\beta}\end{aligned}

\begin{aligned}\boxed{\mathbf{B} = (\mathbf{p} \times \boldsymbol{\beta}) \sin(\omega t - \mathbf{p} \cdot \mathbf{x})}\end{aligned} \hspace{\stretch{1}}(3.16)

\paragraph{Example:} $\mathbf{p} \parallel \mathbf{e}_x$, $\mathbf{B} \parallel \mathbf{e}_y$ or $\mathbf{e}_z$

(since we have two linearly independent choices)

\paragraph{Example:} take $\boldsymbol{\beta} \parallel \mathbf{e}_y$

\begin{aligned}\mathbf{E} &= \boldsymbol{\beta} p \sin(c p t - p x) \\ \mathbf{B} &= (\mathbf{p} \times \boldsymbol{\beta}) \sin(c p t - p x)\end{aligned} \hspace{\stretch{1}}(3.17)

At $t = 0$

\begin{aligned}\mathbf{E} &= -\boldsymbol{\beta} p \sin( p x) \\ B_z &= - {\left\lvert{\boldsymbol{\beta}}\right\rvert} \mathbf{e}_z c p \sin(p x)\end{aligned} \hspace{\stretch{1}}(3.19)

PICTURE: two oscillating mutually perpendicular sinusoids.

So physically, we see that $\mathbf{p}$ is the direction of propagation. We have always

\begin{aligned}\mathbf{p} \perp \mathbf{E}\end{aligned} \hspace{\stretch{1}}(3.21)

and we have two possible polarizations.

Convention is usually to take the direction of oscillation of $\mathbf{E}$ the polarization of the wave.

This is the starting point for the field of optics, because the polarization of the incident wave, is strongly tied to how much of the wave will reflect off of a surface with a given index of refraction $n$.

# EM waves carrying energy and momentum

Maxwell field in vacuum is the sum of plane monochromatic waves, two per wave vector.

PICTURE:

\begin{aligned}\mathbf{E} &\parallel \mathbf{e}_3 \\ \mathbf{B} &\parallel \mathbf{e}_1 \\ \mathbf{k} &\parallel \mathbf{e}_2\end{aligned}

PICTURE:

\begin{aligned}\mathbf{B} &\parallel -\mathbf{e}_3 \\ \mathbf{E} &\parallel \mathbf{e}_1 \\ \mathbf{k} &\parallel \mathbf{e}_2\end{aligned}

(two linearly independent polarizations)

Our wave frequency is

\begin{aligned}\omega_{\mathbf{k}} = c {\left\lvert{\mathbf{k}}\right\rvert}\end{aligned} \hspace{\stretch{1}}(4.22)

The wavelength, the value such that $x \rightarrow x + \frac{2 \pi}{k}$

FIXME:DIY: see:

\begin{aligned}\sin(k c t - k x)\end{aligned} \hspace{\stretch{1}}(4.23)

\begin{aligned}\lambda_{\mathbf{k}} = \frac{2 \pi}{k}\end{aligned} \hspace{\stretch{1}}(4.24)

period

\begin{aligned}T = \frac{ 2 \pi} {k c} = \frac{\lambda_\mathbf{k}}{c}\end{aligned} \hspace{\stretch{1}}(4.25)

# Energy and momentum of EM waves.

## Classical mechanics motivation.

To motivate our approach, let’s recall one route from our equations of motion in classical mechanics, to the energy conservation relation. Our EOM in one dimension is

\begin{aligned}m \frac{d}{dt} \dot{x} = - \mathcal{U}'(x).\end{aligned} \hspace{\stretch{1}}(5.26)

We can multiply both sides by what we take the time derivative of

\begin{aligned}m \dot{x} \frac{d{{\dot{x}}}}{dt} = - \dot{x} \mathcal{U}'(x),\end{aligned} \hspace{\stretch{1}}(5.27)

and then manipulate it a bit so that we have time derivatives on both sides

\begin{aligned}\frac{d{{}}}{dt} \frac{m \dot{x}^2}{2} = - \frac{d{{ \mathcal{U}(x) }}}{dt}.\end{aligned} \hspace{\stretch{1}}(5.28)

Taking differences, we have

\begin{aligned}\frac{d{{}}}{dt} \left( \frac{m \dot{x}^2}{2} + \mathcal{U}(x) \right) = 0,\end{aligned} \hspace{\stretch{1}}(5.29)

which allows us to find a conservation relationship that we label energy conservation ($\mathcal{E} = K + \mathcal{U}$).

## Doing the same thing for Maxwell’s equations.

Poppitz claims we have very little tricks in physics, and we really just do the same thing for our EM case. Our equations are a bit messier to start with, and for the vacuum, our non-divergence equations are

\begin{aligned}\boldsymbol{\nabla} \times \mathbf{B} -\frac{1}{{c}} \frac{\partial {\mathbf{E}}}{\partial {t}} &= \frac{4 \pi}{c} \mathbf{j} \\ \boldsymbol{\nabla} \times \mathbf{E} +\frac{1}{{c}} \frac{\partial {\mathbf{B}}}{\partial {t}} &= 0\end{aligned} \hspace{\stretch{1}}(5.30)

We can dot these with $\mathbf{E}$ and $\mathbf{B}$ respectively, repeating the trick of “multiplying” by what we take the time derivative of

\begin{aligned}\mathbf{E} \cdot (\boldsymbol{\nabla} \times \mathbf{B}) -\frac{1}{{c}} \mathbf{E} \cdot \frac{\partial {\mathbf{E}}}{\partial {t}} &= \frac{4 \pi}{c} \mathbf{E} \cdot \mathbf{j} \\ \mathbf{B} \cdot (\boldsymbol{\nabla} \times \mathbf{E}) +\frac{1}{{c}} \mathbf{B} \cdot \frac{\partial {\mathbf{B}}}{\partial {t}} &= 0,\end{aligned} \hspace{\stretch{1}}(5.32)

and then take differences

\begin{aligned}\frac{1}{{c}} \left(\mathbf{B} \cdot \frac{\partial {\mathbf{B}}}{\partial {t}}+ \mathbf{E} \cdot \frac{\partial {\mathbf{E}}}{\partial {t}} \right) + \mathbf{B} \cdot (\boldsymbol{\nabla} \times \mathbf{E}) -\mathbf{E} \cdot (\boldsymbol{\nabla} \times \mathbf{B}) =-\frac{4 \pi}{c} \mathbf{E} \cdot \mathbf{j}.\end{aligned} \hspace{\stretch{1}}(5.34)

\paragraph{Claim:}

\begin{aligned}-\mathbf{B} \cdot (\boldsymbol{\nabla} \times \mathbf{E}) +\mathbf{E} \cdot (\boldsymbol{\nabla} \times \mathbf{B}) = \boldsymbol{\nabla} \cdot ( \mathbf{B} \times \mathbf{E} ).\end{aligned} \hspace{\stretch{1}}(5.35)

This is almost trivial with an expansion of the RHS in tensor notation

\begin{aligned}\boldsymbol{\nabla} \cdot ( \mathbf{B} \times \mathbf{E} )&=\partial_\alpha e^{\alpha \beta \sigma} B^\beta E^\sigma \\ &=e^{\alpha \beta \sigma} (\partial_\alpha B^\beta) E^\sigma+e^{\alpha \beta \sigma} B^\beta (\partial_\alpha E^\sigma) \\ &=\mathbf{E} \cdot (\boldsymbol{\nabla} \times \mathbf{B})-\mathbf{B} \cdot (\boldsymbol{\nabla} \times \mathbf{E})\qquad \square\end{aligned}

Regrouping we have

\begin{aligned}\frac{1}{{2 c}} \frac{\partial {}}{\partial {t}} \left(\mathbf{B}^2 + \mathbf{E}^2 \right) - \boldsymbol{\nabla} \cdot ( \mathbf{B} \times \mathbf{E} )=-\frac{4 \pi}{c} \mathbf{E} \cdot \mathbf{j}.\end{aligned} \hspace{\stretch{1}}(5.36)

A final rescaling makes the units natural

\begin{aligned}\frac{\partial {}}{\partial {t}} \frac{ \mathbf{E}^2 + \mathbf{B}^2 }{8 \pi} - \boldsymbol{\nabla} \cdot \left( \frac{c}{4 \pi} \mathbf{B} \times \mathbf{E} \right) = - \mathbf{E} \cdot \mathbf{j}.\end{aligned} \hspace{\stretch{1}}(5.37)

We define the cross product term as the Poynting vector

\begin{aligned}\mathbf{S} &= \frac{c}{4 \pi} \mathbf{B} \times \mathbf{E}.\end{aligned} \hspace{\stretch{1}}(5.38)

Suppose we integrate over a spatial volume. This gives us

\begin{aligned}\frac{\partial {}}{\partial {t}}\int_V d^3 \mathbf{x} \frac{ \mathbf{E}^2 + \mathbf{B}^2 }{8 \pi} - \int_V d^3 \mathbf{x} \boldsymbol{\nabla} \cdot \mathbf{S} = - \int_V d^3 \mathbf{x} \mathbf{E} \cdot \mathbf{j}.\end{aligned} \hspace{\stretch{1}}(5.39)

Our Poynting integral can be converted to a surface integral utilizing Stokes theorem

\begin{aligned}\int_V d^3 \mathbf{x} \boldsymbol{\nabla} \cdot \mathbf{S} = \int_{\partial V} d^2 \sigma \mathbf{n} \cdot \mathbf{S} =\int_{\partial V} d^2 \boldsymbol{\sigma} \cdot \mathbf{S}\end{aligned} \hspace{\stretch{1}}(5.40)

We make the interpretations

\begin{aligned}\int_V d^3 \mathbf{x} \frac{ \mathbf{E}^2 + \mathbf{B}^2 }{8 \pi} &= \mbox{energy} \\ \int_V d^3 \mathbf{x} \boldsymbol{\nabla} \cdot \mathbf{S} &= \mbox{momentum change through surface per unit time} \\ - \int_V d^3 \mathbf{x} \mathbf{E} \cdot \mathbf{j} &= \mbox{work done}\end{aligned}

\paragraph{Justifying the sign, and clarifying work done by what, above.}

Recall that the energy term of the Lorentz force equation was

\begin{aligned}\frac{d{{\mathcal{E}_{\text{kinetic}}}}}{dt} = e \mathbf{E} \cdot \mathbf{v}\end{aligned} \hspace{\stretch{1}}(5.41)

and

\begin{aligned}\mathbf{j} = e \rho \mathbf{v}\end{aligned} \hspace{\stretch{1}}(5.42)

so

\begin{aligned}\int_V d^3 \mathbf{x} \mathbf{E} \cdot \mathbf{j}\end{aligned} \hspace{\stretch{1}}(5.43)

represents the rate of change of kinetic energy of the charged particles as they move through through a field. If this is positive, then the charge distribution has gained energy. The negation of this quantity would represent energy transfer to the field from the charge distribution, the work done \underline{on the field} by the charge distribution.

## Aside: As a four vector relationship.

In tutorial today (after this lecture, but before typing up these lecture notes in full), we used $\mathcal{U}$ for the energy density term above

\begin{aligned}\mathcal{U} = \frac{ \mathbf{E}^2 + \mathbf{B}^2 }{8 \pi} .\end{aligned} \hspace{\stretch{1}}(5.44)

This allows us to group the quantities in our conservation relationship above nicely

\begin{aligned}\frac{\partial {\mathcal{U}}}{\partial {t}} - \boldsymbol{\nabla} \cdot \mathbf{S} = - \mathbf{E} \cdot \mathbf{j}.\end{aligned} \hspace{\stretch{1}}(5.45)

It appears natural to write 5.45 in the form of a four divergence. Suppose we define

\begin{aligned}P^i = (\mathcal{U}, \mathbf{S}/c^2)\end{aligned} \hspace{\stretch{1}}(5.46)

then we have

\begin{aligned}\partial_i P^i = - c \mathbf{E} \cdot \mathbf{j}.\end{aligned} \hspace{\stretch{1}}(5.47)

Since the LHS has the appearance of a four scalar, this seems to imply that $\mathbf{E} \cdot \mathbf{j}$ is a Lorentz invariant. It is curious that we have only the four scalar that comes from the energy term of the Lorentz force on the RHS of the conservation relationship. Peeking ahead at the text, this appears to be why a rank two energy tensor $T^{ij}$ is introduced. For a relativistically natural quantity, we ought to have a conservation relationship also associated with each of the momentum change components of the four vector Lorentz force equation too.

# References

[1] L.D. Landau and E.M. Lifshitz. The classical theory of fields. Butterworth-Heinemann, 1980.

## Exploring Stokes Theorem in tensor form.

Posted by peeterjoot on February 22, 2011

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

# Motivation.

I’ve worked through Stokes theorem concepts a couple times on my own now. One of the first times, I was trying to formulate this in a Geometric Algebra context. I had to resort to a tensor decomposition, and pictures, before ending back in the Geometric Algebra description. Later I figured out how to do it entirely with a Geometric Algebra description, and was able to eliminate reliance on the pictures that made the path to generalization to higher dimensional spaces unclear.

It’s my expectation that if one started with a tensor description, the proof entirely in tensor form would not be difficult. This is what I’d like to try this time. To start off, I’ll temporarily use the Geometric Algebra curl expression so I know what my tensor equation starting point will be, but once that starting point is found, we can work entirely in coordinate representation. For somebody who already knows that this is the starting point, all of this initial motivation can be skipped.

# Translating the exterior derivative to a coordinate representation.

Our starting point is a curl, dotted with a volume element of the same grade, so that the result is a scalar

\begin{aligned}\int d^n x \cdot (\nabla \wedge A).\end{aligned} \hspace{\stretch{1}}(2.1)

Here $A$ is a blade of grade $n-1$, and we wedge this with the gradient for the space

\begin{aligned}\nabla \equiv e^i \partial_i = e_i \partial^i,\end{aligned} \hspace{\stretch{1}}(2.2)

where we with with a basis (not necessarily orthonormal) $\{e_i\}$, and the reciprocal frame for that basis $\{e^i\}$ defined by the relation

\begin{aligned}e^i \cdot e_j = {\delta^i}_j.\end{aligned} \hspace{\stretch{1}}(2.3)

Our coordinates in these basis sets are

\begin{aligned}x \cdot e^i & \equiv x^i \\ x \cdot e_i & \equiv x_i\end{aligned} \hspace{\stretch{1}}(2.4)

so that

\begin{aligned}x = x^i e_i = x_i e^i.\end{aligned} \hspace{\stretch{1}}(2.6)

The operator coordinates of the gradient are defined in the usual fashion

\begin{aligned}\partial_i & \equiv \frac{\partial }{\partial {x^i}} \\ \partial^i & \equiv \frac{\partial}{\partial {x_i}}\end{aligned} \hspace{\stretch{1}}(2.7)

The volume element for the subspace that we are integrating over we will define in terms of an arbitrary parametrization

\begin{aligned}x = x(\alpha_1, \alpha_2, \cdots, \alpha_n)\end{aligned} \hspace{\stretch{1}}(2.9)

The subspace can be considered spanned by the differential elements in each of the respective curves where all but the $i$th parameter are held constant.

\begin{aligned}dx_{\alpha_i}= d\alpha_i \frac{\partial x}{\partial {\alpha_i}}= d\alpha_i \frac{\partial {x^j}}{\partial {\alpha_i}} e_j.\end{aligned} \hspace{\stretch{1}}(2.10)

We assume that the integral is being performed in a subspace for which none of these differential elements in that region are linearly dependent (i.e. our Jacobean determinant must be non-zero).

The magnitude of the wedge product of all such differential elements provides the volume of the parallelogram, or parallelepiped (or higher dimensional analogue), and is

\begin{aligned}d^n x=d\alpha_1 d\alpha_2\cdots d\alpha_n\frac{\partial x}{\partial {\alpha_n}} \wedge\cdots \wedge\frac{\partial x}{\partial {\alpha_2}}\wedge\frac{\partial x}{\partial {\alpha_1}}.\end{aligned} \hspace{\stretch{1}}(2.11)

The volume element is a oriented quantity, and may be adjusted with an arbitrary sign (or equivalently an arbitrary permutation of the differential elements in the wedge product), and we’ll see that it is convenient for the translation to tensor form, to express these in reversed order.

Let’s write

\begin{aligned}d^n \alpha = d\alpha_1 d\alpha_2 \cdots d\alpha_n,\end{aligned} \hspace{\stretch{1}}(2.12)

so that our volume element in coordinate form is

\begin{aligned}d^n x = d^n \alpha\frac{\partial {x^i}}{\partial {\alpha_1}}\frac{\partial {x^j}}{\partial {\alpha_2}}\cdots \frac{\partial {x^k}}{\partial {\alpha_{n-1}}}\frac{\partial {x^l}}{\partial {\alpha_n}}( e_l \wedge e_k \wedge \cdots \wedge e_j \wedge e_i ).\end{aligned} \hspace{\stretch{1}}(2.13)

Our curl will also also be a grade $n$ blade. We write for the $n-1$ grade blade

\begin{aligned}A = A_{b c \cdots d} (e^b \wedge e^c \wedge \cdots e^d),\end{aligned} \hspace{\stretch{1}}(2.14)

where $A_{b c \cdots d}$ is antisymmetric (i.e. $A = a_1 \wedge a_2 \wedge \cdots a_{n-1}$ for a some set of vectors $a_i, i \in 1 .. n-1$).

With our gradient in coordinate form

\begin{aligned}\nabla = e^a \partial_a,\end{aligned} \hspace{\stretch{1}}(2.15)

the curl is then

\begin{aligned}\nabla \wedge A = \partial_a A_{b c \cdots d} (e^a \wedge e^b \wedge e^c \wedge \cdots e^d).\end{aligned} \hspace{\stretch{1}}(2.16)

The differential form for our integral can now be computed by expanding out the dot product. We want

\begin{aligned}( e_l \wedge e_k \wedge \cdots \wedge e_j \wedge e_i )\cdot(e^a \wedge e^b \wedge e^c \wedge \cdots e^d)=((((( e_l \wedge e_k \wedge \cdots \wedge e_j \wedge e_i ) \cdot e^a ) \cdot e^b ) \cdot e^c ) \cdot \cdots ) \cdot e^d.\end{aligned} \hspace{\stretch{1}}(2.17)

Evaluation of the interior dot products introduces the intrinsic antisymmetry required for Stokes theorem. For example, with

\begin{aligned}( e_n \wedge e_{n-1} \wedge \cdots \wedge e_2 \wedge e_1 ) \cdot e^a a & =( e_n \wedge e_{n-1} \wedge \cdots \wedge e_3 \wedge e_2 ) (e_1 \cdot e^a) \\ & -( e_n \wedge e_{n-1} \wedge \cdots \wedge e_3 \wedge e_1 ) (e_2 \cdot e^a) \\ & +( e_n \wedge e_{n-1} \wedge \cdots \wedge e_2 \wedge e_1 ) (e_3 \cdot e^a) \\ & \cdots \\ & (-1)^{n-1}( e_{n-1} \wedge e_{n-2} \wedge \cdots \wedge e_2 \wedge e_1 ) (e_n \cdot e^a)\end{aligned}

Since $e_i \cdot e^a = {\delta_i}^a$ our end result is a completely antisymmetric set of permutations of all the deltas

\begin{aligned}( e_l \wedge e_k \wedge \cdots \wedge e_j \wedge e_i )\cdot(e^a \wedge e^b \wedge e^c \wedge \cdots e^d)={\delta^{[a}}_i{\delta^b}_j\cdots {\delta^{d]}}_l,\end{aligned} \hspace{\stretch{1}}(2.18)

and the curl integral takes it’s coordinate form

\begin{aligned}\int d^n x \cdot ( \nabla \wedge A ) =\int d^n \alpha\frac{\partial {x^i}}{\partial {\alpha_1}}\frac{\partial {x^j}}{\partial {\alpha_2}}\cdots \frac{\partial {x^k}}{\partial {\alpha_{n-1}}}\frac{\partial {x^l}}{\partial {\alpha_n}}\partial_a A_{b c \cdots d}{\delta^{[a}}_i{\delta^b}_j\cdots {\delta^{d]}}_l.\end{aligned} \hspace{\stretch{1}}(2.19)

One final contraction of the paired indexes gives us our Stokes integral in its coordinate representation

\begin{aligned}\boxed{\int d^n x \cdot ( \nabla \wedge A ) =\int d^n \alpha\frac{\partial {x^{[a}}}{\partial {\alpha_1}}\frac{\partial {x^b}}{\partial {\alpha_2}}\cdots \frac{\partial {x^c}}{\partial {\alpha_{n-1}}}\frac{\partial {x^{d]}}}{\partial {\alpha_n}}\partial_a A_{b c \cdots d}}\end{aligned} \hspace{\stretch{1}}(2.20)

We now have a starting point that is free of any of the abstraction of Geometric Algebra or differential forms. We can identify the products of partials here as components of a scalar hypervolume element (possibly signed depending on the orientation of the parametrization)

\begin{aligned}d\alpha_1 d\alpha_2\cdots d\alpha_n\frac{\partial {x^{[a}}}{\partial {\alpha_1}}\frac{\partial {x^b}}{\partial {\alpha_2}}\cdots \frac{\partial {x^c}}{\partial {\alpha_{n-1}}}\frac{\partial {x^{d]}}}{\partial {\alpha_n}}\end{aligned} \hspace{\stretch{1}}(2.21)

This is also a specific computation recipe for these hypervolume components, something that may not be obvious when we allow for general metrics for the space. We are also allowing for non-orthonormal coordinate representations, and arbitrary parametrization of the subspace that we are integrating over (our integral need not have the same dimension as the underlying vector space).

Observe that when the number of parameters equals the dimension of the space, we can write out the antisymmetric term utilizing the determinant of the Jacobian matrix

\begin{aligned}\frac{\partial {x^{[a}}}{\partial {\alpha_1}}\frac{\partial {x^b}}{\partial {\alpha_2}}\cdots \frac{\partial {x^c}}{\partial {\alpha_{n-1}}}\frac{\partial {x^{d]}}}{\partial {\alpha_n}}= \epsilon^{a b \cdots d} {\left\lvert{ \frac{\partial(x^1, x^2, \cdots x^n)}{\partial(\alpha_1, \alpha_2, \cdots \alpha_n)} }\right\rvert}\end{aligned} \hspace{\stretch{1}}(2.22)

When the dimension of the space $n$ is greater than the number of parameters for the integration hypervolume in question, the antisymmetric sum of partials is still the determinant of a Jacobian matrix

\begin{aligned}\frac{\partial {x^{[a_1}}}{\partial {\alpha_1}}\frac{\partial {x^{a_2}}}{\partial {\alpha_2}}\cdots \frac{\partial {x^{a_{n-1}}}}{\partial {\alpha_{n-1}}}\frac{\partial {x^{a_n]}}}{\partial {\alpha_n}}= {\left\lvert{ \frac{\partial(x^{a_1}, x^{a_2}, \cdots x^{a_n})}{\partial(\alpha_1, \alpha_2, \cdots \alpha_n)} }\right\rvert},\end{aligned} \hspace{\stretch{1}}(2.23)

however, we will have one such Jacobian for each unique choice of indexes.

# The Stokes work starts here.

The task is to relate our integral to the boundary of this volume, coming up with an explicit recipe for the description of that bounding surface, and determining the exact form of the reduced rank integral. This job is essentially to reduce the ranks of the tensors that are being contracted in our Stokes integral. With the derivative applied to our rank $n-1$ antisymmetric tensor $A_{b c \cdots d}$, we can apply the chain rule and examine the permutations so that this can be rewritten as a contraction of $A$ itself with a set of rank $n-1$ surface area elements.

\begin{aligned}\int d^n \alpha\frac{\partial {x^{[a}}}{\partial {\alpha_1}}\frac{\partial {x^b}}{\partial {\alpha_2}}\cdots \frac{\partial {x^c}}{\partial {\alpha_{n-1}}}\frac{\partial {x^{d]}}}{\partial {\alpha_n}}\partial_a A_{b c \cdots d} = ?\end{aligned} \hspace{\stretch{1}}(3.24)

Now, while the setup here has been completely general, this task is motivated by study of special relativity, where there is a requirement to work in a four dimensional space. Because of that explicit goal, I’m not going to attempt to formulate this in a completely abstract fashion. That task is really one of introducing sufficiently general notation. Instead, I’m going to proceed with a simpleton approach, and do this explicitly, and repeatedly for each of the rank 1, rank 2, and rank 3 tensor cases. It will be clear how this all generalizes by doing so, should one wish to work in still higher dimensional spaces.

## The rank 1 tensor case.

The equation we are working with for this vector case is

\begin{aligned}\int d^2 x \cdot (\nabla \wedge A) =\int d{\alpha_1} d{\alpha_2}\frac{\partial {x^{[a}}}{\partial {\alpha_1}}\frac{\partial {x^{b]}}}{\partial {\alpha_2}}\partial_a A_{b}(\alpha_1, \alpha_2)\end{aligned} \hspace{\stretch{1}}(3.25)

Expanding out the antisymmetric partials we have

\begin{aligned}\frac{\partial {x^{[a}}}{\partial {\alpha_1}}\frac{\partial {x^{b]}}}{\partial {\alpha_2}} & =\frac{\partial {x^{a}}}{\partial {\alpha_1}}\frac{\partial {x^{b}}}{\partial {\alpha_2}}-\frac{\partial {x^{b}}}{\partial {\alpha_1}}\frac{\partial {x^{a}}}{\partial {\alpha_2}},\end{aligned}

with which we can reduce the integral to

\begin{aligned}\int d^2 x \cdot (\nabla \wedge A) & =\int \left( d{\alpha_1}\frac{\partial {x^{a}}}{\partial {\alpha_1}}\frac{\partial {A_{b}}}{\partial {x^a}} \right)\frac{\partial {x^{b}}}{\partial {\alpha_2}} d{\alpha_2}-\left( d{\alpha_2}\frac{\partial {x^{a}}}{\partial {\alpha_2}}\frac{\partial {A_{b}}}{\partial {x^a}} \right)\frac{\partial {x^{b}}}{\partial {\alpha_1}} d{\alpha_1} \\ & =\int \left( d\alpha_1 \frac{\partial {A_b}}{\partial {\alpha_1}} \right)\frac{\partial {x^{b}}}{\partial {\alpha_2}} d{\alpha_2}-\left( d\alpha_2 \frac{\partial {A_b}}{\partial {\alpha_2}} \right)\frac{\partial {x^{b}}}{\partial {\alpha_1}} d{\alpha_1} \\ \end{aligned}

Now, if it happens that

\begin{aligned}\frac{\partial}{\partial {\alpha_1}}\frac{\partial {x^{a}}}{\partial {\alpha_2}} = \frac{\partial}{\partial {\alpha_2}}\frac{\partial {x^{a}}}{\partial {\alpha_1}} = 0\end{aligned} \hspace{\stretch{1}}(3.26)

then each of the individual integrals in $d\alpha_1$ and $d\alpha_2$ can be carried out. In that case, without any real loss of generality we can designate the integration bounds over the unit parametrization space square $\alpha_i \in [0,1]$, allowing this integral to be expressed as

\begin{aligned}\begin{aligned} & \int d{\alpha_1} d{\alpha_2}\frac{\partial {x^{[a}}}{\partial {\alpha_1}}\frac{\partial {x^{b]}}}{\partial {\alpha_2}}\partial_a A_{b}(\alpha_1, \alpha_2) \\ & =\int \left( A_b(1, \alpha_2) - A_b(0, \alpha_2) \right)\frac{\partial {x^{b}}}{\partial {\alpha_2}} d{\alpha_2}-\left( A_b(\alpha_1, 1) - A_b(\alpha_1, 0) \right)\frac{\partial {x^{b}}}{\partial {\alpha_1}} d{\alpha_1}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.27)

It’s also fairly common to see ${\left.{{A}}\right\vert}_{{\partial \alpha_i}}$ used to designate evaluation of this first integral on the boundary, and using this we write

\begin{aligned}\int d{\alpha_1} d{\alpha_2}\frac{\partial {x^{[a}}}{\partial {\alpha_1}}\frac{\partial {x^{b]}}}{\partial {\alpha_2}}\partial_a A_{b}(\alpha_1, \alpha_2)=\int {\left.{{A_b}}\right\vert}_{{\partial \alpha_1}}\frac{\partial {x^{b}}}{\partial {\alpha_2}} d{\alpha_2}-{\left.{{A_b}}\right\vert}_{{\partial \alpha_2}}\frac{\partial {x^{b}}}{\partial {\alpha_1}} d{\alpha_1}.\end{aligned} \hspace{\stretch{1}}(3.28)

Also note that since we are summing over all $a,b$, and have

\begin{aligned}\frac{\partial {x^{[a}}}{\partial {\alpha_1}}\frac{\partial {x^{b]}}}{\partial {\alpha_2}}=-\frac{\partial {x^{[b}}}{\partial {\alpha_1}}\frac{\partial {x^{a]}}}{\partial {\alpha_2}},\end{aligned} \hspace{\stretch{1}}(3.29)

we can write this summing over all unique pairs of $a,b$ instead, which eliminates a small bit of redundancy (especially once the dimension of the vector space gets higher)

\begin{aligned}\boxed{\sum_{a < b}\int d{\alpha_1} d{\alpha_2}\frac{\partial {x^{[a}}}{\partial {\alpha_1}}\frac{\partial {x^{b]}}}{\partial {\alpha_2}}\left( \partial_a A_{b}-\partial_b A_{a} \right)=\int {\left.{{A_b}}\right\vert}_{{\partial \alpha_1}}\frac{\partial {x^{b}}}{\partial {\alpha_2}} d{\alpha_2}-{\left.{{A_b}}\right\vert}_{{\partial \alpha_2}}\frac{\partial {x^{b}}}{\partial {\alpha_1}} d{\alpha_1}.}\end{aligned} \hspace{\stretch{1}}(3.30)

In this form we have recovered the original geometric structure, with components of the curl multiplied by the component of the area element that shares the orientation and direction of that portion of the curl bivector.

This form of the result with evaluation at the boundaries in this form, assumed that ${\partial {x^a}}/{\partial {\alpha_1}}$ was not a function of $\alpha_2$ and ${\partial {x^a}}/{\partial {\alpha_2}}$ was not a function of $\alpha_1$. When that is not the case, we appear to have a less pretty result

\begin{aligned}\boxed{\sum_{a < b}\int d{\alpha_1} d{\alpha_2}\frac{\partial {x^{[a}}}{\partial {\alpha_1}}\frac{\partial {x^{b]}}}{\partial {\alpha_2}}\left( \partial_a A_{b}-\partial_b A_{a} \right)=\int d\alpha_2\int d\alpha_1\frac{\partial {A_b}}{\partial {\alpha_1}}\frac{\partial {x^{b}}}{\partial {\alpha_2}}-\int d\alpha_2\int d\alpha_1\frac{\partial {A_b}}{\partial {\alpha_2}}\frac{\partial {x^{b}}}{\partial {\alpha_1}}}\end{aligned} \hspace{\stretch{1}}(3.31)

Can this be reduced any further in the general case? Having seen the statements of Stokes theorem in it’s differential forms formulation, I initially expected the answer was yes, and only when I got to evaluating my $\mathbb{R}^{4}$ spacetime example below did I realize that the differentials displacements for the parallelogram that constituted the area element were functions of both parameters. Perhaps this detail is there in the differential forms version of the general Stokes theorem too, but is just hidden in a tricky fashion by the compact notation.

### Sanity check: $\mathbb{R}^{2}$ case in rectangular coordinates.

For $x^1 = x, x^2 = y$, and $\alpha_1 = x, \alpha_2 = y$, we have for the LHS

\begin{aligned} & \int_{x=x_0}^{x_1}\int_{y=y_0}^{y_1}dx dy\left(\frac{\partial {x^{1}}}{\partial {\alpha_1}}\frac{\partial {x^{2}}}{\partial {\alpha_2}}-\frac{\partial {x^{2}}}{\partial {\alpha_1}}\frac{\partial {x^{1}}}{\partial {\alpha_2}}\right)\partial_1 A_{2}+\left(\frac{\partial {x^{2}}}{\partial {\alpha_1}}\frac{\partial {x^{1}}}{\partial {\alpha_2}}-\frac{\partial {x^{1}}}{\partial {\alpha_1}}\frac{\partial {x^{2}}}{\partial {\alpha_2}}\right)\partial_2 A_{1} \\ & =\int_{x=x_0}^{x_1}\int_{y=y_0}^{y_1}dx dy\left( \frac{\partial {A_y}}{\partial x} - \frac{\partial {A_x}}{\partial y} \right)\end{aligned}

Our RHS expands to

\begin{aligned} & \int_{y=y_0}^{y_1} dy\left(\left( A_1(x_1, y) - A_1(x_0, y) \right)\frac{\partial {x^{1}}}{\partial y}+\left( A_2(x_1, y) - A_2(x_0, y) \right)\frac{\partial {x^{2}}}{\partial y}\right) \\ & \qquad-\int_{x=x_0}^{x_1} dx\left(\left( A_1(x, y_1) - A_1(x, y_0) \right)\frac{\partial {x^{1}}}{\partial x}+\left( A_2(x, y_1) - A_2(x, y_0) \right)\frac{\partial {x^{2}}}{\partial x}\right) \\ & =\int_{y=y_0}^{y_1} dy\left( A_y(x_1, y) - A_y(x_0, y) \right)-\int_{x=x_0}^{x_1} dx\left( A_x(x, y_1) - A_x(x, y_0) \right)\end{aligned}

We have

\begin{aligned}\begin{aligned} & \int_{x=x_0}^{x_1}\int_{y=y_0}^{y_1}dx dy\left( \frac{\partial {A_y}}{\partial x} - \frac{\partial {A_x}}{\partial y} \right) \\ & =\int_{y=y_0}^{y_1} dy\left( A_y(x_1, y) - A_y(x_0, y) \right)-\int_{x=x_0}^{x_1} dx\left( A_x(x, y_1) - A_x(x, y_0) \right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.32)

The RHS is just a positively oriented line integral around the rectangle of integration

\begin{aligned}\int A_x(x, y_0) \hat{\mathbf{x}} \cdot ( \hat{\mathbf{x}} dx )+ A_y(x_1, y) \hat{\mathbf{y}} \cdot ( \hat{\mathbf{y}} dy )+ A_x(x, y_1) \hat{\mathbf{x}} \cdot ( -\hat{\mathbf{x}} dx )+ A_y(x_0, y) \hat{\mathbf{y}} \cdot ( -\hat{\mathbf{y}} dy )= \oint \mathbf{A} \cdot d\mathbf{r}.\end{aligned} \hspace{\stretch{1}}(3.33)

This special case is also recognizable as Green’s theorem, evident with the substitution $A_x = P$, $A_y = Q$, which gives us

\begin{aligned}\int_A dx dy \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)=\oint_C P dx + Q dy.\end{aligned} \hspace{\stretch{1}}(3.34)

Strictly speaking, Green’s theorem is more general, since it applies to integration regions more general than rectangles, but that generalization can be arrived at easily enough, once the region is broken down into adjoining elementary regions.

### Sanity check: $\mathbb{R}^{3}$ case in rectangular coordinates.

It is expected that we can recover the classical Kelvin-Stokes theorem if we use rectangular coordinates in $\mathbb{R}^{3}$. However, we see that we have to consider three different parametrizations. If one picks rectangular parametrizations $(\alpha_1, \alpha_2) = \{ (x,y), (y,z), (z,x) \}$ in sequence, in each case holding the value of the additional coordinate fixed, we get three different independent Green’s function like relations

\begin{aligned}\int_A dx dy \left( \frac{\partial {A_y}}{\partial x} - \frac{\partial {A_x}}{\partial y} \right) & = \oint_C A_x dx + A_y dy \\ \int_A dy dz \left( \frac{\partial {A_z}}{\partial y} - \frac{\partial {A_y}}{\partial z} \right) & = \oint_C A_y dy + A_z dz \\ \int_A dz dx \left( \frac{\partial {A_x}}{\partial z} - \frac{\partial {A_z}}{\partial x} \right) & = \oint_C A_z dz + A_x dx.\end{aligned} \hspace{\stretch{1}}(3.35)

Note that we cannot just add these to form a complete integral $\oint \mathbf{A} \cdot d\mathbf{r}$ since the curves are all have different orientations. To recover the $\mathbb{R}^{3}$ Stokes theorem in rectangular coordinates, it appears that we’d have to consider a Riemann sum of triangular surface elements, and relate that to the loops over each of the surface elements. In that limiting argument, only the boundary of the complete surface would contribute to the RHS of the relation.

All that said, we shouldn’t actually have to go to all this work. Instead we can stick to a two variable parametrization of the surface, and use 3.30 directly.

### An illustration for a $\mathbb{R}^{4}$ spacetime surface.

Suppose we have a particle trajectory defined by an active Lorentz transformation from an initial spacetime point

\begin{aligned}x^i = O^{ij} x_j(0) = O^{ij} g_{jk} x^k = {O^{i}}_k x^k(0)\end{aligned} \hspace{\stretch{1}}(3.38)

Let the Lorentz transformation be formed by a composition of boost and rotation

\begin{aligned}{O^i}_j & = {L^i}_k {R^k}_j \\ {L^i}_j & =\begin{bmatrix}\cosh_\alpha & -\sinh\alpha & 0 & 0 \\ -\sinh_\alpha & \cosh\alpha & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \\ {R^i}_j & =\begin{bmatrix}1 & 0 & 0 & 0 \\ \cos_\alpha & \sin\alpha & 0 & 0 \\ -\sin_\alpha & \cos\alpha & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.39)

Different rates of evolution of $\alpha$ and $\theta$ define different trajectories, and taken together we have a surface described by the two parameters

\begin{aligned}x^i(\alpha, \theta) = {L^i}_k {R^k}_j x^j(0, 0).\end{aligned} \hspace{\stretch{1}}(3.42)

We can compute displacements along the trajectories formed by keeping either $\alpha$ or $\theta$ fixed and varying the other. Those are

\begin{aligned}\frac{\partial {x^i}}{\partial {\alpha}} d\alpha & = \frac{d{L^i}_k}{d\alpha} {R^k}_j x^j(0, 0) \\ \frac{\partial {x^i}}{\partial {\theta}} d\theta & = {L^i}_k \frac{d{R^k}_j}{d\theta} x^j(0, 0) .\end{aligned} \hspace{\stretch{1}}(3.43)

Writing $y^i = x^i(0,0)$ the computation of the partials above yields

\begin{aligned}\frac{\partial {x^i}}{\partial {\alpha}} & =\begin{bmatrix}\sinh\alpha y^0 -\cosh\alpha (\cos\theta y^1 + \sin\theta y^2) \\ -\cosh\alpha y^0 +\sinh\alpha (\cos\theta y^1 + \sin\theta y^2) \\ 0 \\ 0\end{bmatrix} \\ \frac{\partial {x^i}}{\partial {\theta}} & =\begin{bmatrix}-\sinh\alpha (-\sin\theta y^1 + \cos\theta y^2 ) \\ \cosh\alpha (-\sin\theta y^1 + \cos\theta y^2 ) \\ -(\cos\theta y^1 + \sin\theta y^2 ) \\ 0\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.45)

Different choices of the initial point $y^i$ yield different surfaces, but we can get the idea by picking a simple starting point $y^i = (0, 1, 0, 0)$ leaving

\begin{aligned}\frac{\partial {x^i}}{\partial {\alpha}} & =\begin{bmatrix}-\cosh\alpha \cos\theta \\ \sinh\alpha \cos\theta \\ 0 \\ 0\end{bmatrix} \\ \frac{\partial {x^i}}{\partial {\theta}} & =\begin{bmatrix}\sinh\alpha \sin\theta \\ -\cosh\alpha \sin\theta \\ -\cos\theta \\ 0\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(3.47)

We can now compute our Jacobian determinants

\begin{aligned}\frac{\partial {x^{[a}}}{\partial {\alpha}} \frac{\partial {x^{b]}}}{\partial {\theta}}={\left\lvert{\frac{\partial(x^a, x^b)}{\partial(\alpha, \theta)}}\right\rvert}.\end{aligned} \hspace{\stretch{1}}(3.49)

Those are

\begin{aligned}{\left\lvert{\frac{\partial(x^0, x^1)}{\partial(\alpha, \theta)}}\right\rvert} & = \cos\theta \sin\theta \\ {\left\lvert{\frac{\partial(x^0, x^2)}{\partial(\alpha, \theta)}}\right\rvert} & = \cosh\alpha \cos^2\theta \\ {\left\lvert{\frac{\partial(x^0, x^3)}{\partial(\alpha, \theta)}}\right\rvert} & = 0 \\ {\left\lvert{\frac{\partial(x^1, x^2)}{\partial(\alpha, \theta)}}\right\rvert} & = -\sinh\alpha \cos^2\theta \\ {\left\lvert{\frac{\partial(x^1, x^3)}{\partial(\alpha, \theta)}}\right\rvert} & = 0 \\ {\left\lvert{\frac{\partial(x^2, x^3)}{\partial(\alpha, \theta)}}\right\rvert} & = 0\end{aligned} \hspace{\stretch{1}}(3.50)

Using this, let’s see a specific 4D example in spacetime for the integral of the curl of some four vector $A^i$, enumerating all the non-zero components of 3.31 for this particular spacetime surface

\begin{aligned}\sum_{a < b}\int d{\alpha} d{\theta}{\left\lvert{\frac{\partial(x^a, x^b)}{\partial(\alpha, \theta)}}\right\rvert}\left( \partial_a A_{b}-\partial_b A_{a} \right)=\int d\theta\int d\alpha\frac{\partial {A_b}}{\partial {\alpha}}\frac{\partial {x^{b}}}{\partial {\theta}}-\int d\theta\int d\alpha\frac{\partial {A_b}}{\partial {\theta}}\frac{\partial {x^{b}}}{\partial {\alpha}}\end{aligned} \hspace{\stretch{1}}(3.56)

The LHS is thus found to be

\begin{aligned} & \int d{\alpha} d{\theta}\left({\left\lvert{\frac{\partial(x^0, x^1)}{\partial(\alpha, \theta)}}\right\rvert} \left( \partial_0 A_{1} -\partial_1 A_{0} \right)+{\left\lvert{\frac{\partial(x^0, x^2)}{\partial(\alpha, \theta)}}\right\rvert} \left( \partial_0 A_{2} -\partial_2 A_{0} \right)+{\left\lvert{\frac{\partial(x^1, x^2)}{\partial(\alpha, \theta)}}\right\rvert} \left( \partial_1 A_{2} -\partial_2 A_{1} \right)\right) \\ & =\int d{\alpha} d{\theta}\left(\cos\theta \sin\theta \left( \partial_0 A_{1} -\partial_1 A_{0} \right)+\cosh\alpha \cos^2\theta \left( \partial_0 A_{2} -\partial_2 A_{0} \right)-\sinh\alpha \cos^2\theta \left( \partial_1 A_{2} -\partial_2 A_{1} \right)\right)\end{aligned}

On the RHS we have

\begin{aligned}\int d\theta\int d\alpha & \frac{\partial {A_b}}{\partial {\alpha}}\frac{\partial {x^{b}}}{\partial {\theta}}-\int d\theta\int d\alpha\frac{\partial {A_b}}{\partial {\theta}}\frac{\partial {x^{b}}}{\partial {\alpha}} \\ & =\int d\theta\int d\alpha\begin{bmatrix}\sinh\alpha \sin\theta & -\cosh\alpha \sin\theta & -\cos\theta & 0\end{bmatrix}\frac{\partial}{\partial {\alpha}}\begin{bmatrix}A_0 \\ A_1 \\ A_2 \\ A_3 \\ \end{bmatrix} \\ & -\int d\theta\int d\alpha\begin{bmatrix}-\cosh\alpha \cos\theta & \sinh\alpha \cos\theta & 0 & 0\end{bmatrix}\frac{\partial}{\partial {\theta}}\begin{bmatrix}A_0 \\ A_1 \\ A_2 \\ A_3 \\ \end{bmatrix} \\ \end{aligned}

\begin{aligned}\begin{aligned} & \int d{\alpha} d{\theta}\cos\theta \sin\theta \left( \partial_0 A_{1} -\partial_1 A_{0} \right) \\ & \qquad+\int d{\alpha} d{\theta}\cosh\alpha \cos^2\theta \left( \partial_0 A_{2} -\partial_2 A_{0} \right) \\ & \qquad-\int d{\alpha} d{\theta}\sinh\alpha \cos^2\theta \left( \partial_1 A_{2} -\partial_2 A_{1} \right) \\ & =\int d\theta \sin\theta \int d\alpha \left( \sinh\alpha \frac{\partial {A_0}}{\partial {\alpha}} - \cosh\alpha \frac{\partial {A_1}}{\partial {\alpha}} \right) \\ & \qquad-\int d\theta \cos\theta \int d\alpha \frac{\partial {A_2}}{\partial {\alpha}} \\ & \qquad+\int d\alpha \cosh\alpha \int d\theta \cos\theta \frac{\partial {A_0}}{\partial {\theta}} \\ & \qquad-\int d\alpha \sinh\alpha \int d\theta \cos\theta \frac{\partial {A_1}}{\partial {\theta}}\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.57)

Because of the complexity of the surface, only the second term on the RHS has the “evaluate on the boundary” characteristic that may have been expected from a Green’s theorem like line integral.

It is also worthwhile to point out that we have had to be very careful with upper and lower indexes all along (and have done so with the expectation that our application would include the special relativity case where our metric determinant is minus one.) Because we worked with upper indexes for the area element, we had to work with lower indexes for the four vector and the components of the gradient that we included in our curl evaluation.

## The rank 2 tensor case.

Let’s consider briefly the terms in the contraction sum

\begin{aligned}{\left\lvert{ \frac{\partial(x^a, x^b, x^c)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert} \partial_a A_{bc}\end{aligned} \hspace{\stretch{1}}(3.58)

For any choice of a set of three distinct indexes $(a, b, c) \in (0, 1, 2), (0, 1, 3), (0, 2, 3), (1, 2, 3)$), we have $6 = 3!$ ways of permuting those indexes in this sum

\begin{aligned}{\left\lvert{ \frac{\partial(x^a, x^b, x^c)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert} \partial_a A_{bc} & =\sum_{a < b < c} {\left\lvert{ \frac{\partial(x^a, x^b, x^c)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert} \partial_a A_{bc} + {\left\lvert{ \frac{\partial(x^a, x^c, x^b)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert} \partial_a A_{cb} + {\left\lvert{ \frac{\partial(x^b, x^c, x^a)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert} \partial_b A_{ca} \\ & \qquad + {\left\lvert{ \frac{\partial(x^b, x^a, x^c)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert} \partial_b A_{ac} + {\left\lvert{ \frac{\partial(x^c, x^a, x^b)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert} \partial_c A_{ab} + {\left\lvert{ \frac{\partial(x^c, x^b, x^a)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert} \partial_c A_{ba} \\ & =2!\sum_{a < b < c}{\left\lvert{ \frac{\partial(x^a, x^b, x^c)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert}\left( \partial_a A_{bc} + \partial_b A_{c a} + \partial_c A_{a b} \right)\end{aligned}

Observe that we have no sign alternation like we had in the vector (rank 1 tensor) case. That sign alternation in this summation expansion appears to occur only for odd grade tensors.

Returning to the problem, we wish to expand the determinant in order to apply a chain rule contraction as done in the rank-1 case. This can be done along any of rows or columns of the determinant, and we can write any of

\begin{aligned}{\left\lvert{ \frac{\partial(x^a, x^b, x^c)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert} & =\frac{\partial {x^a}}{\partial {\alpha_1}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_2, \alpha_3)} }\right\rvert}-\frac{\partial {x^a}}{\partial {\alpha_2}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_1, \alpha_3)} }\right\rvert}+\frac{\partial {x^a}}{\partial {\alpha_3}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_1, \alpha_2)} }\right\rvert} \\ & =\frac{\partial {x^b}}{\partial {\alpha_1}} {\left\lvert{ \frac{\partial(x^c, x^a)}{\partial(\alpha_2, \alpha_3)} }\right\rvert}-\frac{\partial {x^b}}{\partial {\alpha_2}} {\left\lvert{ \frac{\partial(x^c, x^a)}{\partial(\alpha_1, \alpha_3)} }\right\rvert}+\frac{\partial {x^b}}{\partial {\alpha_3}} {\left\lvert{ \frac{\partial(x^c, x^a)}{\partial(\alpha_1, \alpha_2)} }\right\rvert} \\ & =\frac{\partial {x^c}}{\partial {\alpha_1}} {\left\lvert{ \frac{\partial(x^a, x^b)}{\partial(\alpha_2, \alpha_3)} }\right\rvert}-\frac{\partial {x^c}}{\partial {\alpha_2}} {\left\lvert{ \frac{\partial(x^a, x^b)}{\partial(\alpha_1, \alpha_3)} }\right\rvert}+\frac{\partial {x^c}}{\partial {\alpha_3}} {\left\lvert{ \frac{\partial(x^a, x^b)}{\partial(\alpha_1, \alpha_2)} }\right\rvert} \\ \end{aligned}

This allows the contraction of the index $a$, eliminating it from the result

\begin{aligned}{\left\lvert{ \frac{\partial(x^a, x^b, x^c)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert} \partial_a A_{bc} & =\left( \frac{\partial {x^a}}{\partial {\alpha_1}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_2, \alpha_3)} }\right\rvert}-\frac{\partial {x^a}}{\partial {\alpha_2}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_1, \alpha_3)} }\right\rvert}+\frac{\partial {x^a}}{\partial {\alpha_3}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_1, \alpha_2)} }\right\rvert} \right) \frac{\partial {A_{bc}}}{\partial {x^a}} \\ & =\frac{\partial {A_{bc}}}{\partial {\alpha_1}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_2, \alpha_3)} }\right\rvert}-\frac{\partial {A_{bc}}}{\partial {\alpha_2}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_1, \alpha_3)} }\right\rvert}+\frac{\partial {A_{bc}}}{\partial {\alpha_3}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_1, \alpha_2)} }\right\rvert} \\ & =2!\sum_{b < c}\frac{\partial {A_{bc}}}{\partial {\alpha_1}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_2, \alpha_3)} }\right\rvert}-\frac{\partial {A_{bc}}}{\partial {\alpha_2}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_1, \alpha_3)} }\right\rvert}+\frac{\partial {A_{bc}}}{\partial {\alpha_3}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_1, \alpha_2)} }\right\rvert} \\ \end{aligned}

Dividing out the common $2!$ terms, we can summarize this result as

\begin{aligned}\boxed{\begin{aligned}\sum_{a < b < c} & \int d\alpha_1 d\alpha_2 d\alpha_3 {\left\lvert{ \frac{\partial(x^a, x^b, x^c)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert}\left( \partial_a A_{bc} + \partial_b A_{c a} + \partial_c A_{a b} \right) \\ & =\sum_{b < c}\int d\alpha_2 d\alpha_3 \int d\alpha_1\frac{\partial {A_{bc}}}{\partial {\alpha_1}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_2, \alpha_3)} }\right\rvert} \\ & -\sum_{b < c}\int d\alpha_1 d\alpha_3 \int d\alpha_2\frac{\partial {A_{bc}}}{\partial {\alpha_2}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_1, \alpha_3)} }\right\rvert} \\ & +\sum_{b < c}\int d\alpha_1 d\alpha_2 \int d\alpha_3\frac{\partial {A_{bc}}}{\partial {\alpha_3}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_1, \alpha_2)} }\right\rvert}\end{aligned}}\end{aligned} \hspace{\stretch{1}}(3.59)

In general, as observed in the spacetime surface example above, the two index Jacobians can be functions of the integration variable first being eliminated. In the special cases where this is not the case (such as the $\mathbb{R}^{3}$ case with rectangular coordinates), then we are left with just the evaluation of the tensor element $A_{bc}$ on the boundaries of the respective integrals.

## The rank 3 tensor case.

The key step is once again just a determinant expansion

\begin{aligned} {\partial(\alpha_1, \alpha_2, \alpha_3, \alpha_4)} }\right\rvert} \\ & =\frac{\partial {x^a}}{\partial {\alpha_1}} {\left\lvert{ \frac{\partial(x^b, x^c, x^d)}{\partial(\alpha_2, \alpha_3, \alpha_4)} }\right\rvert}-\frac{\partial {x^a}}{\partial {\alpha_2}} {\left\lvert{ \frac{\partial(x^b, x^c, x^d)}{\partial(\alpha_1, \alpha_3, \alpha_4)} }\right\rvert}+\frac{\partial {x^a}}{\partial {\alpha_3}} {\left\lvert{ \frac{\partial(x^b, x^c, x^d)}{\partial(\alpha_1, \alpha_2, \alpha_4)} }\right\rvert}+\frac{\partial {x^a}}{\partial {\alpha_4}} {\left\lvert{ \frac{\partial(x^b, x^c, x^d)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert}\\ \end{aligned}

so that the sum can be reduced from a four index contraction to a 3 index contraction

\begin{aligned} {\partial(\alpha_1, \alpha_2, \alpha_3, \alpha_4)} }\right\rvert} \partial_a A_{bcd} \\ & =\frac{\partial {A_{bcd}}}{\partial {\alpha_1}} {\left\lvert{ \frac{\partial(x^b, x^c, x^d)}{\partial(\alpha_2, \alpha_3, \alpha_4)} }\right\rvert}-\frac{\partial {A_{bcd}}}{\partial {\alpha_2}} {\left\lvert{ \frac{\partial(x^b, x^c, x^d)}{\partial(\alpha_1, \alpha_3, \alpha_4)} }\right\rvert}+\frac{\partial {A_{bcd}}}{\partial {\alpha_3}} {\left\lvert{ \frac{\partial(x^b, x^c, x^d)}{\partial(\alpha_1, \alpha_2, \alpha_4)} }\right\rvert}+\frac{\partial {A_{bcd}}}{\partial {\alpha_4}} {\left\lvert{ \frac{\partial(x^b, x^c, x^d)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert}\end{aligned}

That’s the essence of the theorem, but we can play the same combinatorial reduction games to reduce the built in redundancy in the result

\begin{aligned}\boxed{\begin{aligned}\frac{1}{{3!}} & \int d^4 \alpha {\left\lvert{ \frac{\partial(x^a, x^b, x^c, x^d)}{\partial(\alpha_1, \alpha_2, \alpha_3, \alpha_4)} }\right\rvert} \partial_a A_{bcd} \\ & =\sum_{a < b < c < d}\int d^4 \alpha {\left\lvert{ \frac{\partial(x^a, x^b, x^c, x^d)}{\partial(\alpha_1, \alpha_2, \alpha_3, \alpha_4)} }\right\rvert} \left( \partial_a A_{bcd} -\partial_b A_{cda} +\partial_c A_{dab} -\partial_d A_{abc} \right) \\ & =\qquad \sum_{b < c < d}\int d\alpha_2 d\alpha_3 d\alpha_4 \int d\alpha_1\frac{\partial {A_{bcd}}}{\partial {\alpha_1}} {\left\lvert{ \frac{\partial(x^b, x^c, x^d)}{\partial(\alpha_2, \alpha_3, \alpha_4)} }\right\rvert} \\ & \qquad -\sum_{b < c < d}\int d\alpha_1 d\alpha_3 d\alpha_4 \int d\alpha_2\frac{\partial {A_{bcd}}}{\partial {\alpha_2}} {\left\lvert{ \frac{\partial(x^b, x^c, x^d)}{\partial(\alpha_1, \alpha_3, \alpha_4)} }\right\rvert} \\ & \qquad +\sum_{b < c < d}\int d\alpha_1 d\alpha_2 d\alpha_4 \int d\alpha_3\frac{\partial {A_{bcd}}}{\partial {\alpha_3}} {\left\lvert{ \frac{\partial(x^b, x^c, x^d)}{\partial(\alpha_1, \alpha_2, \alpha_4)} }\right\rvert} \\ & \qquad +\sum_{b < c < d}\int d\alpha_1 d\alpha_2 d\alpha_3 \int d\alpha_4\frac{\partial {A_{bcd}}}{\partial {\alpha_4}} {\left\lvert{ \frac{\partial(x^b, x^c, x^d)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert} \\ \end{aligned}}\end{aligned} \hspace{\stretch{1}}(3.60)

## A note on Four diverence.

Our four divergence integral has the following form

\begin{aligned}\int d^4 \alpha {\left\lvert{ \frac{\partial(x^1, x^2, x^2, x^4)}{\partial(\alpha_1, \alpha_2, \alpha_3, \alpha_4)} }\right\rvert} \partial_a A^a\end{aligned} \hspace{\stretch{1}}(3.61)

We can relate this to the rank 3 Stokes theorem with a duality transformation, multiplying with a pseudoscalar

\begin{aligned}A^a = \epsilon^{abcd} T_{bcd},\end{aligned} \hspace{\stretch{1}}(3.62)

where $T_{bcd}$ can also be related back to the vector by the same sort of duality transformation

\begin{aligned}A^a \epsilon_{a b c d} = \epsilon^{abcd} \epsilon_{a b c d} T_{bcd} = 4! T_{bcd}.\end{aligned} \hspace{\stretch{1}}(3.63)

The divergence integral in terms of the rank 3 tensor is

\begin{aligned}\int d^4 \alpha {\left\lvert{ \frac{\partial(x^1, x^2, x^2, x^4)}{\partial(\alpha_1, \alpha_2, \alpha_3, \alpha_4)} }\right\rvert} \partial_a \epsilon^{abcd} T_{bcd}=\int d^4 \alpha {\left\lvert{ \frac{\partial(x^a, x^b, x^c, x^d)}{\partial(\alpha_1, \alpha_2, \alpha_3, \alpha_4)} }\right\rvert} \partial_a T_{bcd},\end{aligned} \hspace{\stretch{1}}(3.64)

and we are free to perform the same Stokes reduction of the integral. Of course, this is particularly simple in rectangular coordinates. I still have to think though one sublty that I feel may be important. We could have started off with an integral of the following form

\begin{aligned}\int dx^1 dx^2 dx^3 dx^4 \partial_a A^a,\end{aligned} \hspace{\stretch{1}}(3.65)

and I think this differs from our starting point slightly because this has none of the antisymmetric structure of the signed 4 volume element that we have used. We do not take the absolute value of our Jacobians anywhere.

## Stokes Theorem for antisymmetric tensors.

Posted by peeterjoot on January 18, 2011

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

In [3] I worked through the Geometric Algebra expression for Stokes Theorem. For a $k-1$ grade blade, the final result of that work was

\begin{aligned}\int( \nabla \wedge F ) \cdot d^k x =\frac{1}{{(k-1)!}} \epsilon^{ r s \cdots t u } \int da_u \frac{\partial {F}}{\partial {a_{u}}} \cdot (dx_r \wedge dx_s \wedge \cdots \wedge dx_t)\end{aligned} \hspace{\stretch{1}}(7.44)

Let’s expand this in coordinates to attempt to get the equivalent expression for an antisymmetric tensor of rank $k-1$.

Starting with the RHS of 7.44 we have

\begin{aligned}F &= \frac{1}{{(k-1)!}}F_{\mu_1 \mu_2 \cdots \mu_{k-1} }\gamma^{\mu_1} \wedge \gamma^{ \mu_2 } \wedge \cdots \wedge \gamma^{\mu_{k-1}} \\ dx_r \wedge dx_s \wedge \cdots \wedge dx_t &=\frac{\partial {x^{\nu_1}}}{\partial {a_r}}\frac{\partial {x^{\nu_2}}}{\partial {a_s}}\cdots\frac{\partial {x^{\nu_{k-1}}}}{\partial {a_t}}\gamma_{\nu_1} \wedge \gamma_{ \nu_2 } \wedge \cdots \wedge \gamma_{\nu_{k-1}}da_r da_s \cdots da_t\end{aligned} \hspace{\stretch{1}}(7.45)

We need to expand the dot product of the wedges, for which we have

\begin{aligned}\left( \gamma^{\mu_1} \wedge \gamma^{ \mu_2 } \wedge \cdots \wedge \gamma^{\mu_{k-1}} \right) \cdot\left( \gamma_{\nu_1} \wedge \gamma_{ \nu_2 } \wedge \cdots \wedge \gamma_{\nu_{k-1}}\right) ={\delta^{\mu_{k-1}}}_{\nu_1} {\delta^{ \mu_{k-2} }}_{\nu_2} \cdots {\delta^{\mu_{1}} }_{\nu_{k-1}}\epsilon^{\nu_1 \nu_2 \cdots \nu_{k-1}}\end{aligned} \hspace{\stretch{1}}(7.47)

Putting all the LHS bits together we have

\begin{aligned}&\frac{1}{{((k-1)!)^2}} \epsilon^{ r s \cdots t u } \int da_u \frac{\partial {}}{\partial {a_{u}}} F_{\mu_1 \mu_2 \cdots \mu_{k-1} }{\delta^{\mu_{k-1}}}_{\nu_1} {\delta^{ \mu_{k-2} }}_{\nu_2} \cdots {\delta^{\mu_{1}} }_{\nu_{k-1}}\epsilon^{\nu_1 \nu_2 \cdots \nu_{k-1}}\frac{\partial {x^{\nu_1}}}{\partial {a_r}}\frac{\partial {x^{\nu_2}}}{\partial {a_s}}\cdots\frac{\partial {x^{\nu_{k-1}}}}{\partial {a_t}}da_r da_s \cdots da_t \\ &=\frac{1}{{((k-1)!)^2}} \epsilon^{ r s \cdots t u } \int da_u \frac{\partial {}}{\partial {a_{u}}} F_{\mu_1 \mu_2 \cdots \mu_{k-1} }\epsilon^{\mu_{k-1} \mu_{k-2} \cdots \mu_{1}}\frac{\partial {x^{\mu_{k-1}}}}{\partial {a_r}}\frac{\partial {x^{\mu_{k-2}}}}{\partial {a_s}}\cdots\frac{\partial {x^{\mu_1}}}{\partial {a_t}}da_r da_s \cdots da_t \\ &=\frac{1}{{((k-1)!)^2}} \epsilon^{ r s \cdots t u } \int da_u \frac{\partial {}}{\partial {a_{u}}} F_{\mu_1 \mu_2 \cdots \mu_{k-1} }{\left\lvert{\frac{\partial(x^{\mu_{k-1}},x^{\mu_{k-2}},\cdots,x^{\mu_1})}{\partial(a_r, a_s, \cdots, a_t)}}\right\rvert}da_r da_s \cdots da_t \\ \end{aligned}

Now, for the LHS of 7.44 we have

\begin{aligned}\nabla \wedge F &=\gamma^\mu \wedge \partial_\mu F \\ &=\frac{1}{{(k-1)!}}\frac{\partial {}}{\partial {x^{\mu_k}}} F_{\mu_1 \mu_2 \cdots \mu_{k-1}}\gamma^{\mu_k} \wedge\gamma^{\mu_1} \wedge \gamma^{ \mu_2 } \wedge \cdots \wedge \gamma^{\mu_{k-1}} \end{aligned}

and the volume element of

\begin{aligned}d^k x &=\frac{\partial {x^{\nu_1}}}{\partial {a_1}}\frac{\partial {x^{\nu_2}}}{\partial {a_2}}\cdots\frac{\partial {x^{\nu_{k}}}}{\partial {a_k}}\gamma_{\nu_1} \wedge \gamma_{ \nu_2 } \wedge \cdots \wedge \gamma_{\nu_k}da_1 da_2 \cdots da_k\end{aligned}

Our dot product is

\begin{aligned}\left(\gamma^{\mu_k} \wedge\gamma^{\mu_1} \wedge \gamma^{ \mu_2 } \wedge \cdots \wedge \gamma^{\mu_{k-1}} \right) \cdot\left( \gamma_{\nu_1} \wedge \gamma_{ \nu_2 } \wedge \cdots \wedge \gamma_{\nu_k} \right)={\delta^{\mu_{k-1}}}_{\nu_1} {\delta^{ \mu_{k-2} }}_{\nu_2} \cdots {\delta^{\mu_{1}} }_{\nu_{k-1}}{\delta^{\mu_{k}} }_{\nu_{k}}\epsilon^{\nu_1 \nu_2 \cdots \nu_{k}}\end{aligned} \hspace{\stretch{1}}(7.48)

The LHS of our k-form now evaluates to

\begin{aligned}(\gamma^\mu \wedge \partial_\mu F) \cdot d^k x &= \frac{1}{(k-1)!}\frac{\partial }{\partial {x^{\mu_k}}} F_{\mu_1 \mu_2 \cdots \mu_{k-1}}{\delta^{\mu_{k-1}}}_{\nu_1} {\delta^{ \mu_{k-2} }}_{\nu_2} \cdots {\delta^{\mu_1} }_{\nu_{k-1}}{\delta^{\mu_k} }_{\nu_k}\epsilon^{\nu_1 \nu_2 \cdots \nu_k}\frac{\partial {x^{\nu_1}}}{\partial {a_1}}\frac{\partial {x^{\nu_2}}}{\partial {a_2}} \cdots \frac{\partial {x^{\nu_k}}}{\partial {a_k}} da_1 da_2 \cdots da_k \\ &= \frac{1}{(k-1)!}\frac{\partial }{\partial {x^{\mu_k}}} F_{\mu_1 \mu_2 \cdots \mu_{k-1}}\epsilon^{\mu_{k-1} \mu_{k-2} \cdots \mu_1 \mu_k}\frac{\partial {x^{\mu_{k-1}}}}{\partial {a_1}}\frac{\partial {x^{\mu_{k-2}}}}{\partial {a_2}} \cdots \frac{\partial {x^{\mu_1}}}{\partial {a_{k-1}}}\frac{\partial {x^{\mu_k}}}{\partial {a_k}} da_1 da_2 \cdots da_k \\ &= \frac{1}{(k-1)!}\frac{\partial }{\partial {x^{\mu_k}}} F_{\mu_1 \mu_2 \cdots \mu_{k-1}}{\left\lvert{\frac{\partial(x^{\mu_{k-1}},x^{\mu_{k-2}},\cdots x^{\mu_1},x^{\mu_k})}{\partial(a_1, a_2, \cdots, a_{k-1}, a_k)}}\right\rvert} da_1 da_2 \cdots da_k \\ \end{aligned}

Presuming no mistakes were made anywhere along the way (including in the original Geometric Algebra expression), we have arrived at Stokes Theorem for rank $k-1$ antisymmetric tensors $F$

\boxed{ \begin{aligned}&\int\frac{\partial }{\partial {x^{\mu_k}}} F_{\mu_1 \mu_2 \cdots \mu_{k-1}}{\left\lvert{\frac{\partial(x^{\mu_{k-1}},x^{\mu_{k-2}},\cdots x^{\mu_1},x^{\mu_k})}{\partial(a_1, a_2, \cdots, a_{k-1}, a_k)}}\right\rvert} da_1 da_2 \cdots da_k \\ &= \frac{1}{(k-1)!} \epsilon^{ r s \cdots t u } \int da_u \frac{\partial }{\partial {a_u}} F_{\nu_1 \nu_2 \cdots \nu_{k-1} }{\left\lvert{\frac{\partial(x^{\nu_{k-1}},x^{\nu_{k-2}}, \cdots ,x^{\nu_1})}{\partial(a_r, a_s, \cdots, a_t)}}\right\rvert} da_r da_s \cdots da_t \end{aligned} } \hspace{\stretch{1}}(7.49)

The next task is to validate this, expanding it out for some specific ranks and hypervolume element types, and to compare the results with the familiar 3d expressions.

# References

[1] L.D. Landau and E.M. Lifshits. The classical theory of fields. Butterworth-Heinemann, 1980.

[2] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

[3] Peeter Joot. Stokes theorem derivation without tensor expansion of the blade [online]. http://sites.google.com/site/peeterjoot/math2009/stokesNoTensor.pdf.

## 4D divergence theorem, continued.

Posted by peeterjoot on July 23, 2009

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

The basic idea of using duality to express the 4D divergence integral as a stokes boundary surface integral has been explored. Lets consider this in more detail picking a specific parametrization, namely rectangular four vector coordinates. For the volume element write

\begin{aligned}d^4 x &= ( \gamma_0 dx^0 ) \wedge ( \gamma_1 dx^1 ) \wedge ( \gamma_2 dx^2 ) \wedge ( \gamma_3 dx^3 ) \\ &= \gamma_0 \gamma_1 \gamma_2 \gamma_3 dx^0 dx^1 dx^2 dx^3 \\ &= i dx^0 dx^1 dx^2 dx^3 \\ \end{aligned}

As seen previously (but not separately), the divergence can be expressed as the dual of the curl

\begin{aligned}\nabla \cdot f&=\left\langle{{ \nabla f }}\right\rangle \\ &=-\left\langle{{ \nabla i (\underbrace{i f}_{\text{grade 3}}) }}\right\rangle \\ &=\left\langle{{ i \nabla (i f) }}\right\rangle \\ &=\left\langle{{ i ( \underbrace{\nabla \cdot (i f)}_{\text{grade 2}} + \underbrace{\nabla \wedge (i f)}_{\text{grade 4}} ) }}\right\rangle \\ &=i (\nabla \wedge (i f)) \\ \end{aligned}

So we have $\nabla \wedge (i f) = -i (\nabla \cdot f)$. Putting things together, and writing $i f = -f i$ we have

\begin{aligned}\int (\nabla \wedge (i f)) \cdot d^4 x&= \int (\nabla \cdot f) dx^0 dx^1 dx^2 dx^3 \\ &=\int dx^0 \partial_0 (f i) \cdot \gamma_{123} dx^1 dx^2 dx^3 \\ &-\int dx^1 \partial_1 (f i) \cdot \gamma_{023} dx^0 dx^2 dx^3 \\ &+\int dx^2 \partial_2 (f i) \cdot \gamma_{013} dx^0 dx^1 dx^3 \\ &-\int dx^3 \partial_3 (f i) \cdot \gamma_{012} dx^0 dx^1 dx^2 \\ \end{aligned}

It is straightforward to reduce each of these dot products. For example

\begin{aligned}\partial_2 (f i) \cdot \gamma_{013}&=\left\langle{{ \partial_2 f \gamma_{0123013} }}\right\rangle \\ &=-\left\langle{{ \partial_2 f \gamma_{2} }}\right\rangle \\ &=- \gamma_2 \partial_2 \cdot f \\ &=\gamma^2 \partial_2 \cdot f \end{aligned}

The rest proceed the same and rather anticlimactically we end up coming full circle

\begin{aligned}\int (\nabla \cdot f) dx^0 dx^1 dx^2 dx^3 &=\int dx^0 \gamma^0 \partial_0 \cdot f dx^1 dx^2 dx^3 \\ &+\int dx^1 \gamma^1 \partial_1 \cdot f dx^0 dx^2 dx^3 \\ &+\int dx^2 \gamma^2 \partial_2 \cdot f dx^0 dx^1 dx^3 \\ &+\int dx^3 \gamma^3 \partial_3 \cdot f dx^0 dx^1 dx^2 \\ \end{aligned}

This is however nothing more than the definition of the divergence itself and no need to resort to Stokes theorem is required. However, if we are integrating over a rectangle and perform each of the four integrals, we have (with c=1) from the dual Stokes equation the perhaps less obvious result

\begin{aligned}\int \partial_\mu f^\mu dt dx dy dz&=\int (f^0(t_1) - f^0(t_0)) dx dy dz \\ &+\int (f^1(x_1) - f^1(x_0)) dt dy dz \\ &+\int (f^2(y_1) - f^2(y_0)) dt dx dz \\ &+\int (f^3(z_1) - f^3(z_0)) dt dx dy \\ \end{aligned}

When stated this way one sees that this could have just as easily have followed directly from the left hand side. What’s the point then of the divergence theorem or Stokes theorem? I think that the value must really be the fact that the Stokes formulation naturally builds the volume element in a fashion independent of any specific parametrization. Here in rectangular coordinates the result seems obvious, but would the equivalent result seem obvious if non-rectangular spacetime coordinates were employed? Probably not.

## Stokes theorem in Geometric Algebra formalism.

Posted by peeterjoot on July 22, 2009

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

# Motivation

Relying on pictorial means and a brute force ugly comparison of left and right hand sides, a verification of Stokes theorem for the vector and bivector cases was performed ([1]). This was more of a confirmation than a derivation, and the technique fails the transition to the trivector case. The trivector case is of particular interest in electromagnetism since that and a duality transformation provides a four-vector divergence theorem.

The fact that the pictorial means of defining the boundary surface doesn’t work well in four vector space is not the only unsatisfactory aspect of the previous treatment. The fact that a coordinate expansion of the hypervolume element and hypersurface element was performed in the LHS and RHS comparisons was required is particularly ugly. It is a lot of work and essentially has to be undone on the opposing side of the equation. Comparing to previous attempts to come to terms with Stokes theorem in ([2]) and ([3]) this more recent attempt at least avoids the requirement for a tensor expansion of the vector or bivector. It should be possible to build on this and minimize the amount of coordinate expansion required and go directly from the volume integral to the expression of the boundary surface.

# Do it.

## Notation and Setup.

The desire is to relate the curl hypervolume integral to a hypersurface integral on the boundary

\begin{aligned}\int (\nabla \wedge F) \cdot d^k x = \int F \cdot d^{k-1} x\end{aligned} \hspace{\stretch{1}}(2.1)

In order to put meaning to this statement the volume and surface elements need to be properly defined. In order that this be a scalar equation, the object $F$ in the integral is required to be of grade $k-1$, and $k \le n$ where $n$ is the dimension of the vector space that generates the object $F$.

## Reciprocal frames.

As evident in equation (2.1) a metric is required to define the dot product. If an affine non-metric formulation
of Stokes theorem is possible it will not be attempted here. A reciprocal basis pair will be utilized, defined by

\begin{aligned}\gamma^\mu \cdot \gamma_\nu = {\delta^\mu}_\nu\end{aligned} \hspace{\stretch{1}}(2.2)

Both of the sets $\{\gamma_\mu\}$ and $\{\gamma^\mu\}$ are taken to span the space, but are not required to be orthogonal. The notation is consistent with the Dirac reciprocal basis, and there will not be anything in this treatment that prohibits the Minkowski metric signature required for such a relativistic space.

Vector decomposition in terms of coordinates follows by taking dot products. We write

\begin{aligned}x = x^\mu \gamma_\mu = x_\nu \gamma^\nu\end{aligned} \hspace{\stretch{1}}(2.3)

When working with a non-orthonormal basis, use of the reciprocal frame can be utilized to express the gradient.

\begin{aligned}\nabla \equiv \gamma^\mu \partial_\mu \equiv \sum_\mu \gamma^\mu \frac{\partial {}}{\partial {x^\mu}}\end{aligned} \hspace{\stretch{1}}(2.4)

This contains what may perhaps seem like an odd seeming mix of upper and lower indexes in this definition. This is how the gradient is defined in [4]. Although it is possible to accept this definition and work with it, this form can be justified by require of the gradient consistency with the the definition of directional derivative. A definition of the directional derivative that works for single and multivector functions, in $\mathbb{R}^{3}$ and other more general spaces is

\begin{aligned}a \cdot \nabla F \equiv \lim_{\lambda \rightarrow 0} \frac{F(x + a\lambda) - F(x)}{\lambda} = {\left.\frac{\partial {F(x + a\lambda)}}{\partial {\lambda}} \right\vert}_{\lambda=0}\end{aligned} \hspace{\stretch{1}}(2.5)

Taylor expanding about $\lambda=0$ in terms of coordinates we have

\begin{aligned}{\left.\frac{\partial {F(x + a\lambda)}}{\partial {\lambda}} \right\vert}_{\lambda=0}&= a^\mu \frac{\partial {F}}{\partial {x^\mu}} \\ &= (a^\nu \gamma_\nu) \cdot (\gamma^\mu \partial_\mu) F \\ &= a \cdot \nabla F \quad\quad\quad\square\end{aligned}

The lower index representation of the vector coordinates could also have been used, so using the directional derivative to imply a definition of the gradient, we have an additional alternate representation of the gradient

\begin{aligned}\nabla \equiv \gamma_\mu \partial^\mu \equiv \sum_\mu \gamma_\mu \frac{\partial {}}{\partial {x_\mu}}\end{aligned} \hspace{\stretch{1}}(2.6)

## Volume element

We define the hypervolume in terms of parametrized vector displacements $x = x(a_1, a_2, ... a_k)$. For the vector x we can form a pseudoscalar for the subspace spanned by this parametrization by wedging the displacements in each of the directions defined by variation of the parameters. For $m \in [1,k]$ let

\begin{aligned}dx_i = \frac{\partial {x}}{\partial {a_i}} da_i = \gamma_\mu \frac{\partial {x^\mu}}{\partial {a_i}} da_i,\end{aligned} \hspace{\stretch{1}}(2.7)

so the hypervolume element for the subspace in question is

\begin{aligned}d^k x \equiv dx_1 \wedge dx_2 \cdots dx_k\end{aligned} \hspace{\stretch{1}}(2.8)

This can be expanded explicitly in coordinates

\begin{aligned}d^k x &= da_1 da_2 \cdots da_k \left(\frac{\partial {x^{\mu_1}}}{\partial {a_1}} \frac{\partial {x^{\mu_2}}}{\partial {a_2}} \cdots\frac{\partial {x^{\mu_k}}}{\partial {a_k}} \right)( \gamma_{\mu_1} \wedge \gamma_{\mu_2} \wedge \cdots \wedge \gamma_{\mu_k} ) \\ \end{aligned}

Observe that when $k$ is also the dimension of the space, we can employ a pseudoscalar $I = \gamma_0 \gamma_1 \cdots \gamma_k$ and can specify our volume element in terms of the Jacobian determinant.

This is

\begin{aligned}d^k x =I da_1 da_2 \cdots da_k {\left\lvert{\frac{\partial {(x^1, x^2, \cdots, x^k)}}{\partial {(a_1, a_2, \cdots, a_k)}}}\right\rvert}\end{aligned} \hspace{\stretch{1}}(2.9)

However, we won’t have a requirement to express the Stokes result in terms of such Jacobians.

## Expansion of the curl and volume element product

We are now prepared to go on to the meat of the issue. The first order of business is the expansion of the curl and volume element product

\begin{aligned}( \nabla \wedge F ) \cdot d^k x&=( \gamma^\mu \wedge \partial_\mu F ) \cdot d^k x \\ &=\left\langle{{ ( \gamma^\mu \wedge \partial_\mu F ) d^k x }}\right\rangle \\ \end{aligned}

The wedge product within the scalar grade selection operator can be expanded in symmetric or antisymmetric sums, but this is a grade dependent operation. For odd grade blades $A$ (vector, trivector, …), and vector $a$ we have for the dot and wedge product respectively

\begin{aligned}a \wedge A = \frac{1}{{2}} (a A - A a) \\ a \cdot A = \frac{1}{{2}} (a A + A a)\end{aligned}

Similarly for even grade blades we have

\begin{aligned}a \wedge A = \frac{1}{{2}} (a A + A a) \\ a \cdot A = \frac{1}{{2}} (a A - A a)\end{aligned}

First treating the odd grade case for $F$ we have

\begin{aligned}( \nabla \wedge F ) \cdot d^k x&=\frac{1}{{2}} \left\langle{{ \gamma^\mu \partial_\mu F d^k x }}\right\rangle - \frac{1}{{2}} \left\langle{{ \partial_\mu F \gamma^\mu d^k x }}\right\rangle \\ \end{aligned}

Employing cyclic scalar reordering within the scalar product for the first term

\begin{aligned}\left\langle{{a b c}}\right\rangle = \left\langle{{b c a}}\right\rangle\end{aligned} \hspace{\stretch{1}}(2.10)

we have

\begin{aligned}( \nabla \wedge F ) \cdot d^k x&=\frac{1}{{2}} \left\langle{{ \partial_\mu F (d^k x \gamma^\mu - \gamma^\mu d^k x)}}\right\rangle \\ &=\frac{1}{{2}} \left\langle{{ \partial_\mu F (d^k x \cdot \gamma^\mu - \gamma^\mu d^k x)}}\right\rangle \\ &=\left\langle{{ \partial_\mu F (d^k x \cdot \gamma^\mu)}}\right\rangle \\ \end{aligned}

The end result is

\begin{aligned}( \nabla \wedge F ) \cdot d^k x &= \partial_\mu F \cdot (d^k x \cdot \gamma^\mu) \end{aligned} \hspace{\stretch{1}}(2.11)

For even grade $F$ (and thus odd grade $d^k x$) it is straightforward to show that (2.11) also holds.

## Expanding the volume dot product

We want to expand the volume integral dot product

\begin{aligned}d^k x \cdot \gamma^\mu\end{aligned} \hspace{\stretch{1}}(2.12)

Picking $k = 4$ will serve to illustrate the pattern, and the generalization (or degeneralization to lower grades) will be clear. We have

\begin{aligned}d^4 x \cdot \gamma^\mu&=( dx_1 \wedge dx_2 \wedge dx_3 \wedge dx_4 ) \cdot \gamma^\mu \\ &= ( dx_1 \wedge dx_2 \wedge dx_3 ) dx_4 \cdot \gamma^\mu \\ &-( dx_1 \wedge dx_2 \wedge dx_4 ) dx_3 \cdot \gamma^\mu \\ &+( dx_1 \wedge dx_3 \wedge dx_4 ) dx_2 \cdot \gamma^\mu \\ &-( dx_2 \wedge dx_3 \wedge dx_4 ) dx_1 \cdot \gamma^\mu \\ \end{aligned}

This avoids the requirement to do the entire Jacobian expansion of (2.9). The dot product of the differential displacement $dx_m$ with $\gamma^\mu$ can now be made explicit without as much mess.

\begin{aligned}dx_m \cdot \gamma^\mu &=da_m \frac{\partial {x^\nu}}{\partial {a_m}} \gamma_\nu \cdot \gamma^\mu \\ &=da_m \frac{\partial {x^\mu}}{\partial {a_m}} \\ \end{aligned}

We now have products of the form

\begin{aligned}\partial_\mu F da_m \frac{\partial {x^\mu}}{\partial {a_m}} &=da_m \frac{\partial {x^\mu}}{\partial {a_m}} \frac{\partial {F}}{\partial {x^\mu}} \\ &=da_m \frac{\partial {F}}{\partial {a_m}} \\ \end{aligned}

Now we see that the differential form of (2.11) for this $k=4$ example is reduced to

\begin{aligned}( \nabla \wedge F ) \cdot d^4 x &= da_4 \frac{\partial {F}}{\partial {a_4}} \cdot ( dx_1 \wedge dx_2 \wedge dx_3 ) \\ &- da_3 \frac{\partial {F}}{\partial {a_3}} \cdot ( dx_1 \wedge dx_2 \wedge dx_4 ) \\ &+ da_2 \frac{\partial {F}}{\partial {a_2}} \cdot ( dx_1 \wedge dx_3 \wedge dx_4 ) \\ &- da_1 \frac{\partial {F}}{\partial {a_1}} \cdot ( dx_2 \wedge dx_3 \wedge dx_4 ) \\ \end{aligned}

While 2.11 was a statement of Stokes theorem in this Geometric Algebra formulation, it was really incomplete without this explicit expansion of $(\partial_\mu F) \cdot (d^k x \cdot \gamma^\mu)$. This expansion for the $k=4$ case serves to illustrate that we would write Stokes theorem as

\begin{aligned}\boxed{\int( \nabla \wedge F ) \cdot d^k x =\frac{1}{{(k-1)!}} \epsilon^{ r s \cdots t u } \int da_u \frac{\partial {F}}{\partial {a_{u}}} \cdot (dx_r \wedge dx_s \wedge \cdots \wedge dx_t)}\end{aligned} \hspace{\stretch{1}}(2.13)

Here the indexes have the range $\{r, s, \cdots, t, u\} \in \{1, 2, \cdots k\}$. This with the definitions 2.7, and 2.8 is really Stokes theorem in its full glory.

Observe that in this Geometric algebra form, the one forms $dx_i = da_i {\partial {x}}/{\partial {a_i}}, i \in [1,k]$ are nothing more abstract that plain old vector differential elements. In the formalism of differential forms, this would be vectors, and $(\nabla \wedge F) \cdot d^k x$ would be a $k$ form. In a context where we are working with vectors, or blades already, the Geometric Algebra statement of the theorem avoids a requirement to translate to the language of forms.

With a statement of the general theorem complete, let’s return to our $k=4$ case where we can now integrate over each of the $a_1, a_2, \cdots, a_k$ parameters. That is

\begin{aligned}\int ( \nabla \wedge F ) \cdot d^4 x &= \int (F(a_4(1)) - F(a_4(0))) \cdot ( dx_1 \wedge dx_2 \wedge dx_3 ) \\ &- \int (F(a_3(1)) - F(a_3(0))) \cdot ( dx_1 \wedge dx_2 \wedge dx_4 ) \\ &+ \int (F(a_2(1)) - F(a_2(0))) \cdot ( dx_1 \wedge dx_3 \wedge dx_4 ) \\ &- \int (F(a_1(1)) - F(a_1(0))) \cdot ( dx_2 \wedge dx_3 \wedge dx_4 ) \\ \end{aligned}

This is precisely Stokes theorem for the trivector case and makes the enumeration of the boundary surfaces explicit. As derived there was no requirement for an orthonormal basis, nor a Euclidean metric, nor a parametrization along the basis directions. The only requirement of the parametrization is that the associated volume element is non-trivial (i.e. none of $dx_q \wedge dx_r = 0$).

For completeness, note that our boundary surface and associated Stokes statement for the bivector and vector cases is, by inspection respectively

\begin{aligned}\int ( \nabla \wedge F ) \cdot d^3 x &= \int (F(a_3(1)) - F(a_3(0))) \cdot ( dx_1 \wedge dx_2 ) \\ &- \int (F(a_2(1)) - F(a_2(0))) \cdot ( dx_1 \wedge dx_3 ) \\ &+ \int (F(a_1(1)) - F(a_1(0))) \cdot ( dx_2 \wedge dx_3 ) \\ \end{aligned}

and

\begin{aligned}\int ( \nabla \wedge F ) \cdot d^2 x &= \int (F(a_2(1)) - F(a_2(0))) \cdot dx_1 \\ &- \int (F(a_1(1)) - F(a_1(0))) \cdot dx_2 \\ \end{aligned}

These three expansions can be summarized by the original single statement of (2.1), which repeating for reference, is

\begin{aligned}\int ( \nabla \wedge F ) \cdot d^k x = \int F \cdot d^{k-1} x \end{aligned}

Where it is implied that the blade $F$ is evaluated on the boundaries and dotted with the associated hypersurface boundary element. However, having expanded this we now have an explicit statement of exactly what that surface element is now for any desired parametrization.

# Duality relations and special cases.

Some special (and more recognizable) cases of (2.1) are possible considering specific grades of $F$, and in some cases employing duality relations.

## curl surface integral

One important case is the $\mathbb{R}^{3}$ vector result, which can be expressed in terms of the cross product.

Write $\hat{\mathbf{n}} d^2 x = -i dA$. Then we have

\begin{aligned}( \boldsymbol{\nabla} \wedge \mathbf{f} ) \cdot d^2 x&=\left\langle{{ i (\boldsymbol{\nabla} \times \mathbf{f}) (- \hat{\mathbf{n}} i dA) }}\right\rangle \\ &=(\boldsymbol{\nabla} \times \mathbf{f}) \cdot \hat{\mathbf{n}} dA\end{aligned}

This recovers the familiar cross product form of Stokes law.

\begin{aligned}\int (\boldsymbol{\nabla} \times \mathbf{f}) \cdot \hat{\mathbf{n}} dA = \oint \mathbf{f} \cdot d\mathbf{x}\end{aligned} \hspace{\stretch{1}}(3.14)

## 3D divergence theorem

Duality applied to the bivector Stokes result provides the divergence theorem in $\mathbb{R}^{3}$. For bivector $B$, let $iB = \mathbf{f}$, $d^3 x = i dV$, and $d^2 x = i \hat{\mathbf{n}} dA$. We then have

\begin{aligned}( \boldsymbol{\nabla} \wedge B ) \cdot d^3 x&=\left\langle{{ ( \boldsymbol{\nabla} \wedge B ) \cdot d^3 x }}\right\rangle \\ &=\frac{1}{{2}} \left\langle{{ ( \boldsymbol{\nabla} B + B \boldsymbol{\nabla} ) i dV }}\right\rangle \\ &=\boldsymbol{\nabla} \cdot \mathbf{f} dV \\ \end{aligned}

Similarly

\begin{aligned}B \cdot d^2 x&=\left\langle{{ -i\mathbf{f} i \hat{\mathbf{n}} dA}}\right\rangle \\ &=(\mathbf{f} \cdot \hat{\mathbf{n}}) dA \\ \end{aligned}

This recovers the $\mathbb{R}^{3}$ divergence equation

\begin{aligned}\int \boldsymbol{\nabla} \cdot \mathbf{f} dV = \int (\mathbf{f} \cdot \hat{\mathbf{n}}) dA\end{aligned} \hspace{\stretch{1}}(3.15)

## 4D divergence theorem

How about the four dimensional spacetime divergence? Write, express a trivector as a dual four-vector $T = if$, and the four volume element $d^4 x = i dQ$. This gives

\begin{aligned}(\nabla \wedge T) \cdot d^4 x&=\frac{1}{{2}} \left\langle{{ (\nabla T - T \nabla) i }}\right\rangle dQ \\ &=\frac{1}{{2}} \left\langle{{ (\nabla i f - if \nabla) i }}\right\rangle dQ \\ &=\frac{1}{2} \left\langle{{ (\nabla f + f \nabla) }}\right\rangle dQ \\ &=(\nabla \cdot f) dQ\end{aligned}

For the boundary volume integral write $d^3 x = n i dV$, for

\begin{aligned}T \cdot d^3 x &= \left\langle{{ (if) ( n i ) }}\right\rangle dV \\ &= \left\langle{{ f n }}\right\rangle dV \\ &= (f \cdot n) dV\end{aligned}

So we have

\begin{aligned}\int \partial_\mu f^\mu dQ = \int f^\nu n_\nu dV\end{aligned}

the orientation of the fourspace volume element and the boundary normal is defined in terms of the parametrization, the duality relations and our explicit expansion of the 4D stokes boundary integral above.

## 4D divergence theorem, continued.

The basic idea of using duality to express the 4D divergence integral as a stokes boundary surface integral has been explored. Lets consider this in more detail picking a specific parametrization, namely rectangular four vector coordinates. For the volume element write

\begin{aligned}d^4 x &= ( \gamma_0 dx^0 ) \wedge ( \gamma_1 dx^1 ) \wedge ( \gamma_2 dx^2 ) \wedge ( \gamma_3 dx^3 ) \\ &= \gamma_0 \gamma_1 \gamma_2 \gamma_3 dx^0 dx^1 dx^2 dx^3 \\ &= i dx^0 dx^1 dx^2 dx^3 \\ \end{aligned}

As seen previously (but not separately), the divergence can be expressed as the dual of the curl

\begin{aligned}\nabla \cdot f&=\left\langle{{ \nabla f }}\right\rangle \\ &=-\left\langle{{ \nabla i (\underbrace{i f}_{\text{grade 3}}) }}\right\rangle \\ &=\left\langle{{ i \nabla (i f) }}\right\rangle \\ &=\left\langle{{ i ( \underbrace{\nabla \cdot (i f)}_{\text{grade 2}} + \underbrace{\nabla \wedge (i f)}_{\text{grade 4}} ) }}\right\rangle \\ &=i (\nabla \wedge (i f)) \\ \end{aligned}

So we have $\nabla \wedge (i f) = -i (\nabla \cdot f)$. Putting things together, and writing $i f = -f i$ we have

\begin{aligned}\int (\nabla \wedge (i f)) \cdot d^4 x&= \int (\nabla \cdot f) dx^0 dx^1 dx^2 dx^3 \\ &=\int dx^0 \partial_0 (f i) \cdot \gamma_{123} dx^1 dx^2 dx^3 \\ &-\int dx^1 \partial_1 (f i) \cdot \gamma_{023} dx^0 dx^2 dx^3 \\ &+\int dx^2 \partial_2 (f i) \cdot \gamma_{013} dx^0 dx^1 dx^3 \\ &-\int dx^3 \partial_3 (f i) \cdot \gamma_{012} dx^0 dx^1 dx^2 \\ \end{aligned}

It is straightforward to reduce each of these dot products. For example

\begin{aligned}\partial_2 (f i) \cdot \gamma_{013}&=\left\langle{{ \partial_2 f \gamma_{0123013} }}\right\rangle \\ &=-\left\langle{{ \partial_2 f \gamma_{2} }}\right\rangle \\ &=- \gamma_2 \partial_2 \cdot f \\ &=\gamma^2 \partial_2 \cdot f \end{aligned}

The rest proceed the same and rather anticlimactically we end up coming full circle

\begin{aligned}\int (\nabla \cdot f) dx^0 dx^1 dx^2 dx^3 &=\int dx^0 \gamma^0 \partial_0 \cdot f dx^1 dx^2 dx^3 \\ &+\int dx^1 \gamma^1 \partial_1 \cdot f dx^0 dx^2 dx^3 \\ &+\int dx^2 \gamma^2 \partial_2 \cdot f dx^0 dx^1 dx^3 \\ &+\int dx^3 \gamma^3 \partial_3 \cdot f dx^0 dx^1 dx^2 \\ \end{aligned}

This is however nothing more than the definition of the divergence itself and no need to resort to Stokes theorem is required. However, if we are integrating over a rectangle and perform each of the four integrals, we have (with $c=1$) from the dual Stokes equation the perhaps less obvious result

\begin{aligned}\int \partial_\mu f^\mu dt dx dy dz&=\int (f^0(t_1) - f^0(t_0)) dx dy dz \\ &+\int (f^1(x_1) - f^1(x_0)) dt dy dz \\ &+\int (f^2(y_1) - f^2(y_0)) dt dx dz \\ &+\int (f^3(z_1) - f^3(z_0)) dt dx dy \\ \end{aligned}

When stated this way one sees that this could have just as easily have followed directly from the left hand side. What’s the point then of the divergence theorem or Stokes theorem? I think that the value must really be the fact that the Stokes formulation naturally builds the volume element in a fashion independent of any specific parametrization. Here in rectangular coordinates the result seems obvious, but would the equivalent result seem obvious if non-rectangular spacetime coordinates were employed? Probably not.

# References

[1] Peeter Joot. Stokes theorem applied to vector and bivector fields [online]. http://sites.google.com/site/peeterjoot/math2009/stokesGradeTwo.pdf.

[2] Peeter Joot. Stokes law in wedge product form [online]. http://sites.google.com/site/peeterjoot/geometric-algebra/vector_integral_relations.pdf.

[3] Peeter Joot. Stokes Law revisited with algebraic enumeration of boundary [online]. http://sites.google.com/site/peeterjoot/geometric-algebra/stokes_revisited.pdf.

[4] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

## bivector form of Stokes theorem

Posted by peeterjoot on July 18, 2009

[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

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).

## Stokes theorem applied to a vector field.

Posted by peeterjoot on July 17, 2009

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

# 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.