## Spherical gradient, divergence, curl and Laplacian

### Unit vectors

Two of the spherical unit vectors we can immediately write by inspection.

\label{eqn:sphericalLaplacian:20}
\begin{aligned}
\rcap &= \Be_1 \sin\theta \cos\phi + \Be_2 \sin\theta \sin\phi + \Be_3 \cos\theta \\
\phicap &= -\Be_1 \sin\theta + \Be_2 \cos\phi
\end{aligned}

We can compute $$\thetacap$$ by utilizing the right hand triplet property

\label{eqn:sphericalLaplacian:40}
\begin{aligned}
\thetacap
&=
\phicap \cross \rcap \\
&=
\begin{vmatrix}
\Be_1 & \Be_2 & \Be_3 \\
-S_\phi & C_\phi & 0 \\
S_\theta C_\phi & S_\theta S_\phi & C_\theta \\
\end{vmatrix} \\
&=
\Be_1 \lr{ C_\theta C_\phi }
+\Be_2 \lr{ C_\theta S_\phi }
+\Be_3 \lr{ -S_\theta \lr{ S_\phi^2 + C_\phi^2 } } \\
&=
\Be_1 \cos\theta \cos\phi
+\Be_2 \cos\theta \sin\phi
-\Be_3 \sin\theta.
\end{aligned}

Here I’ve used $$C_\theta = \cos\theta, S_\phi = \sin\phi, \cdots$$ as a convenient shorthand. Observe that with $$i = \Be_1 \Be_2$$, these unit vectors admit a small factorization that makes further manipulation easier

\label{eqn:sphericalLaplacian:80}
\boxed{
\begin{aligned}
\rcap &= \Be_1 e^{i\phi} \sin\theta + \Be_3 \cos\theta \\
\thetacap &= \cos\theta \Be_1 e^{i\phi} – \sin\theta \Be_3 \\
\phicap &= \Be_2 e^{i\phi}
\end{aligned}
}

It should also be the case that $$\rcap \thetacap \phicap = I$$, where $$I = \Be_1 \Be_2 \Be_3 = \Be_{123}$$ is the \R{3} pseudoscalar, which is straightforward to check

\label{eqn:sphericalLaplacian:60}
\begin{aligned}
\rcap \thetacap \phicap
&=
\lr{ \Be_1 e^{i\phi} \sin\theta + \Be_3 \cos\theta }
\lr{ \cos\theta \Be_1 e^{i\phi} – \sin\theta \Be_3 }
\Be_2 e^{i\phi} \\
&=
\lr{ \sin\theta \cos\theta – \cos\theta \sin\theta + \Be_{31} e^{i\phi} \lr{ \cos^2\theta + \sin^2\theta } }
\Be_2 e^{i\phi} \\
&=
\Be_{31} \Be_2 e^{-i\phi} e^{i\phi} \\
&=
\Be_{123}.
\end{aligned}

This property could also have been used to compute $$\thetacap$$.

To compute the gradient, note that the coordinate vectors for the spherical parameterization are
\label{eqn:sphericalLaplacian:120}
\begin{aligned}
\Bx_r
&= \PD{r}{\Br} \\
&= \PD{r}{\lr{r \rcap}} \\
&= \rcap + r \PD{r}{\rcap} \\
&= \rcap,
\end{aligned}

\label{eqn:sphericalLaplacian:140}
\begin{aligned}
\Bx_\theta
&= \PD{\theta}{\lr{r \rcap} } \\
&= r \PD{\theta}{} \lr{ S_\theta \Be_1 e^{i\phi} + C_\theta \Be_3 } \\
&= r \PD{\theta}{} \lr{ C_\theta \Be_1 e^{i\phi} – S_\theta \Be_3 } \\
&= r \thetacap,
\end{aligned}

\label{eqn:sphericalLaplacian:160}
\begin{aligned}
\Bx_\phi
&= \PD{\phi}{\lr{r \rcap} } \\
&= r \PD{\phi}{} \lr{ S_\theta \Be_1 e^{i\phi} + C_\theta \Be_3 } \\
&= r S_\theta \Be_2 e^{i\phi} \\
&= r \sin\theta \phicap.
\end{aligned}

Since these are all normal, the dual vectors defined by $$\Bx^j \cdot \Bx_k = \delta^j_k$$, can be obtained by inspection
\label{eqn:sphericalLaplacian:180}
\begin{aligned}
\Bx^r &= \rcap \\
\Bx^\theta &= \inv{r} \thetacap \\
\Bx^\phi &= \inv{r \sin\theta} \phicap.
\end{aligned}

\label{eqn:sphericalLaplacian:200}
\Bx^r \PD{r}{} +
\Bx^\theta \PD{\theta}{} +
\Bx^\phi \PD{\phicap}{},

or
\label{eqn:sphericalLaplacian:240}
\boxed{
=
\rcap \PD{r}{} +
\frac{\thetacap}{r} \PD{\theta}{} +
\frac{\phicap}{r\sin\theta} \PD{\phicap}{}.
}

More information on this general dual-vector technique of computing the gradient in curvilinear coordinate systems can be found in
[2].

### Partials

To compute the divergence, curl and Laplacian, we’ll need the partials of each of the unit vectors $$\PDi{\theta}{\rcap}, \PDi{\phi}{\rcap}, \PDi{\theta}{\thetacap}, \PDi{\phi}{\thetacap}, \PDi{\phi}{\phicap}$$.

The $$\thetacap$$ partials are

\label{eqn:sphericalLaplacian:260}
\begin{aligned}
\PD{\theta}{\thetacap}
&=
\PD{\theta}{} \lr{
C_\theta \Be_1 e^{i\phi} – S_\theta \Be_3
} \\
&=
-S_\theta \Be_1 e^{i\phi} – C_\theta \Be_3 \\
&=
-\rcap,
\end{aligned}

\label{eqn:sphericalLaplacian:280}
\begin{aligned}
\PD{\phi}{\thetacap}
&=
\PD{\phi}{} \lr{
C_\theta \Be_1 e^{i\phi} – S_\theta \Be_3
} \\
&=
C_\theta \Be_2 e^{i\phi} \\
&=
C_\theta \phicap.
\end{aligned}

The $$\phicap$$ partials are

\label{eqn:sphericalLaplacian:300}
\begin{aligned}
\PD{\theta}{\phicap}
&=
\PD{\theta}{} \Be_2 e^{i\phi} \\
&=
0.
\end{aligned}

\label{eqn:sphericalLaplacian:320}
\begin{aligned}
\PD{\phi}{\phicap}
&=
\PD{\phi}{} \Be_2 e^{i \phi} \\
&=
-\Be_1 e^{i \phi} \\
&=
-\rcap \gpgradezero{ \rcap \Be_1 e^{i \phi} }
– \thetacap \gpgradezero{ \thetacap \Be_1 e^{i \phi} }
– \phicap \gpgradezero{ \phicap \Be_1 e^{i \phi} } \\
&=
\Be_1 e^{i\phi} S_\theta + \Be_3 C_\theta
} \Be_1 e^{i \phi} }
C_\theta \Be_1 e^{i\phi} – S_\theta \Be_3
} \Be_1 e^{i \phi} } \\
&=
-\rcap \gpgradezero{ e^{-i\phi} S_\theta e^{i \phi} }
– \thetacap \gpgradezero{ C_\theta e^{-i\phi} e^{i \phi} } \\
&=
-\rcap S_\theta
– \thetacap C_\theta.
\end{aligned}

The $$\rcap$$ partials are were computed as a side effect of evaluating $$\Bx_\theta$$, and $$\Bx_\phi$$, and are

\label{eqn:sphericalLaplacian:340}
\PD{\theta}{\rcap}
=
\thetacap,

\label{eqn:sphericalLaplacian:360}
\PD{\phi}{\rcap}
=
S_\theta \phicap.

In summary
\label{eqn:sphericalLaplacian:380}
\boxed{
\begin{aligned}
\partial_{\theta}{\rcap} &= \thetacap \\
\partial_{\phi}{\rcap} &= S_\theta \phicap \\
\partial_{\theta}{\thetacap} &= -\rcap \\
\partial_{\phi}{\thetacap} &= C_\theta \phicap \\
\partial_{\theta}{\phicap} &= 0 \\
\partial_{\phi}{\phicap} &= -\rcap S_\theta – \thetacap C_\theta.
\end{aligned}
}

### Divergence and curl.

The divergence and curl can be computed from the vector product of the spherical coordinate gradient and the spherical representation of a vector. That is

\label{eqn:sphericalLaplacian:400}

\label{eqn:sphericalLaplacian:420}
\begin{aligned}
&=
\lr{
\rcap \partial_{r}
+ \frac{\thetacap}{r} \partial_{\theta}
+ \frac{\phicap}{rS_\theta} \partial_{\phi}
}
\lr{ \rcap A_r + \thetacap A_\theta + \phicap A_\phi} \\
&=
\rcap \partial_{r}
\lr{ \rcap A_r + \thetacap A_\theta + \phicap A_\phi} \\
&+ \frac{\thetacap}{r} \partial_{\theta}
\lr{ \rcap A_r + \thetacap A_\theta + \phicap A_\phi} \\
&+ \frac{\phicap}{rS_\theta} \partial_{\phicap}
\lr{ \rcap A_r + \thetacap A_\theta + \phicap A_\phi} \\
&=
\lr{ \partial_r A_r + \rcap \thetacap \partial_r A_\theta + \rcap \phicap \partial_r A_\phi} \\
&+ \frac{1}{r}
\lr{
\thetacap (\partial_\theta \rcap) A_r + \thetacap (\partial_\theta \thetacap) A_\theta + \thetacap (\partial_\theta \phicap) A_\phi
+\thetacap \rcap \partial_\theta A_r + \partial_\theta A_\theta + \thetacap \phicap \partial_\theta A_\phi
} \\
&+ \frac{1}{rS_\theta}
\lr{
\phicap (\partial_\phi \rcap) A_r + \phicap (\partial_\phi \thetacap) A_\theta + \phicap (\partial_\phi \phicap) A_\phi
+\phicap \rcap \partial_\phi A_r + \phicap \thetacap \partial_\phi A_\theta + \partial_\phi A_\phi
} \\
&=
\lr{ \partial_r A_r + \rcap \thetacap \partial_r A_\theta + \rcap \phicap \partial_r A_\phi} \\
&+ \frac{1}{r}
\lr{
\thetacap (\thetacap) A_r + \thetacap (-\rcap) A_\theta + \thetacap (0) A_\phi
+\thetacap \rcap \partial_\theta A_r + \partial_\theta A_\theta + \thetacap \phicap \partial_\theta A_\phi
} \\
&+ \frac{1}{r S_\theta}
\lr{
\phicap (S_\theta \phicap) A_r + \phicap (C_\theta \phicap) A_\theta – \phicap (\rcap S_\theta + \thetacap C_\theta) A_\phi
+\phicap \rcap \partial_\phi A_r + \phicap \thetacap \partial_\phi A_\theta + \partial_\phi A_\phi
}.
\end{aligned}

The scalar component of this is the divergence
\label{eqn:sphericalLaplacian:440}
\begin{aligned}
&=
\partial_r A_r
+ \frac{A_r}{r}
+ \inv{r} \partial_\theta A_\theta
+ \frac{1}{r S_\theta}
\lr{ S_\theta A_r + C_\theta A_\theta + \partial_\phi A_\phi
} \\
&=
\partial_r A_r
+ 2 \frac{A_r}{r}
+ \inv{r} \partial_\theta A_\theta
+ \frac{1}{r S_\theta}
C_\theta A_\theta
+ \frac{1}{r S_\theta} \partial_\phi A_\phi \\
&=
\partial_r A_r
+ 2 \frac{A_r}{r}
+ \inv{r} \partial_\theta A_\theta
+ \frac{1}{r S_\theta}
C_\theta A_\theta
+ \frac{1}{r S_\theta} \partial_\phi A_\phi,
\end{aligned}

which can be factored as
\label{eqn:sphericalLaplacian:460}
\boxed{
=
\inv{r^2} \partial_r (r^2 A_r)
+ \inv{r S_\theta} \partial_\theta (S_\theta A_\theta)
+ \frac{1}{r S_\theta} \partial_\phi A_\phi.
}

The bivector grade of $$\spacegrad \BA$$ is the bivector curl
\label{eqn:sphericalLaplacian:480}
\begin{aligned}
&=
\lr{
\rcap \thetacap \partial_r A_\theta + \rcap \phicap \partial_r A_\phi
} \\
\lr{
\thetacap (-\rcap) A_\theta
+\thetacap \rcap \partial_\theta A_r + \thetacap \phicap \partial_\theta A_\phi
} \\
\frac{1}{r S_\theta}
\lr{
-\phicap (\rcap S_\theta + \thetacap C_\theta) A_\phi
+\phicap \rcap \partial_\phi A_r + \phicap \thetacap \partial_\phi A_\theta
} \\
&=
\lr{
\rcap \thetacap \partial_r A_\theta – \phicap \rcap \partial_r A_\phi
} \\
\lr{
\rcap \thetacap A_\theta
-\rcap \thetacap \partial_\theta A_r + \thetacap \phicap \partial_\theta A_\phi
} \\
\frac{1}{r S_\theta}
\lr{
-\phicap \rcap S_\theta A_\phi + \thetacap \phicap C_\theta A_\phi
+\phicap \rcap \partial_\phi A_r – \thetacap \phicap \partial_\phi A_\theta
} \\
&=
\thetacap \phicap \lr{
\inv{r S_\theta} C_\theta A_\phi
+\frac{1}{r} \partial_\theta A_\phi
-\frac{1}{r S_\theta} \partial_\phi A_\theta
} \\
-\partial_r A_\phi
+
\frac{1}{r S_\theta}
\lr{
-S_\theta A_\phi
+ \partial_\phi A_r
}
} \\
\partial_r A_\theta
+ \frac{1}{r} A_\theta
– \inv{r} \partial_\theta A_r
} \\
&=
I
\rcap \lr{
\inv{r S_\theta} \partial_\theta (S_\theta A_\phi)
-\frac{1}{r S_\theta} \partial_\phi A_\theta
}
+ I \thetacap \lr{
\frac{1}{r S_\theta} \partial_\phi A_r
-\inv{r} \partial_r (r A_\phi)
}
+ I \phicap \lr{
\inv{r} \partial_r (r A_\theta)
– \inv{r} \partial_\theta A_r
}
\end{aligned}

This gives
\label{eqn:sphericalLaplacian:500}
\boxed{
=
\rcap \lr{
\inv{r S_\theta} \partial_\theta (S_\theta A_\phi)
-\frac{1}{r S_\theta} \partial_\phi A_\theta
}
+ \thetacap \lr{
\frac{1}{r S_\theta} \partial_\phi A_r
-\inv{r} \partial_r (r A_\phi)
}
+ \phicap \lr{
\inv{r} \partial_r (r A_\theta)
– \inv{r} \partial_\theta A_r
}.
}

This and the divergence result above both check against the back cover of [1].

### Laplacian

Using the divergence and curl it’s possible to compute the Laplacian from those, but we saw in cylindrical coordinates that it was much harder to do it that way than to do it directly.

\label{eqn:sphericalLaplacian:540}
\begin{aligned}
&=
\lr{
\rcap \partial_{r} +
\frac{\thetacap}{r} \partial_{\theta} +
\frac{\phicap}{r S_\theta} \partial_{\phi}
}
\lr{
\rcap \partial_{r} \psi
+ \frac{\thetacap}{r} \partial_{\theta} \psi
+ \frac{\phicap}{r S_\theta} \partial_{\phi} \psi
} \\
&=
\partial_{rr} \psi
+ \rcap \thetacap \partial_r \lr{ \inv{r} \partial_\theta \psi}
+ \rcap \phicap \inv{S_\theta} \partial_r \lr{ \inv{r} \partial_\phi \psi } \\
&
\quad + \frac{\thetacap}{r} \partial_{\theta} \lr{ \rcap \partial_{r} \psi }
+ \frac{\thetacap}{r^2} \partial_{\theta} \lr{ \thetacap \partial_{\theta} \psi }
+ \frac{\thetacap}{r^2} \partial_{\theta} \lr{ \frac{\phicap}{S_\theta} \partial_{\phi} \psi } \\
&
\quad + \frac{\phicap}{r S_\theta} \partial_{\phi} \lr{ \rcap \partial_{r} \psi }
+ \frac{\phicap}{r^2 S_\theta} \partial_{\phi} \lr{ \thetacap \partial_{\theta} \psi }
+ \frac{\phicap}{r^2 S_\theta^2} \partial_{\phi} \lr{ \phicap \partial_{\phi} \psi } \\
&=
\partial_{rr} \psi
+ \rcap \thetacap \partial_r \lr{ \inv{r} \partial_\theta \psi}
+ \rcap \phicap \inv{S_\theta} \partial_r \lr{ \inv{r} \partial_\phi \psi } \\
&
\quad + \frac{\thetacap\rcap}{r} \partial_{\theta} \lr{ \partial_{r} \psi }
+ \frac{1}{r^2} \partial_{\theta \theta} \psi
+ \frac{\thetacap \phicap}{r^2} \partial_{\theta} \lr{ \frac{1}{S_\theta} \partial_{\phi} \psi } \\
&
\quad + \frac{\phicap \rcap}{r S_\theta} \partial_{\phi r} \psi
+ \frac{\phicap\thetacap}{r^2 S_\theta} \partial_{\phi\theta} \psi
+ \frac{1}{r^2 S_\theta^2} \partial_{\phi \phi} \psi \\
&
\quad + \frac{\thetacap}{r} (\partial_\theta \rcap) \partial_{r} \psi
+ \frac{\thetacap}{r^2} (\partial_\theta \thetacap) \partial_{\theta} \psi
+ \frac{\thetacap}{r^2} (\partial_\theta \phicap) \frac{\phicap}{S_\theta} \partial_{\phi} \psi \\
&
\quad + \frac{\phicap}{r S_\theta} (\partial_\phi \rcap) \partial_{r} \psi
+ \frac{\phicap}{r^2 S_\theta} (\partial_\phi \thetacap) \partial_{\theta} \psi
+ \frac{\phicap}{r^2 S_\theta^2} (\partial_\phi \phicap) \partial_{\phi} \psi \\
&=
\partial_{rr} \psi
+ \rcap \thetacap \partial_r \lr{ \inv{r} \partial_\theta \psi}
+ \rcap \phicap \inv{S_\theta} \partial_r \lr{ \inv{r} \partial_\phi \psi } \\
&
\quad + \frac{\thetacap\rcap}{r} \partial_{\theta} \lr{ \partial_{r} \psi }
+ \frac{1}{r^2} \partial_{\theta \theta} \psi
+ \frac{\thetacap \phicap}{r^2} \partial_{\theta} \lr{ \frac{1}{S_\theta} \partial_{\phi} \psi } \\
&
\quad + \frac{\phicap \rcap}{r S_\theta} \partial_{\phi r} \psi
+ \frac{\phicap\thetacap}{r^2 S_\theta} \partial_{\phi\theta} \psi
+ \frac{1}{r^2 S_\theta^2} \partial_{\phi \phi} \psi \\
&
\quad + \frac{\thetacap}{r} (\thetacap) \partial_{r} \psi
+ \frac{\thetacap}{r^2} (-\rcap) \partial_{\theta} \psi
+ \frac{\thetacap}{r^2} (0) \frac{\phicap}{S_\theta} \partial_{\phi} \psi \\
&
\quad + \frac{\phicap}{r S_\theta} (S_\theta \phicap) \partial_{r} \psi
+ \frac{\phicap}{r^2 S_\theta} (C_\theta \phicap) \partial_{\theta} \psi
+ \frac{\phicap}{r^2 S_\theta^2} (-\rcap S_\theta – \thetacap C_\theta) \partial_{\phi} \psi
\end{aligned}

All the bivector factors are expected to cancel out, but this should be checked. Those with an $$\rcap \thetacap$$ factor are

\label{eqn:sphericalLaplacian:560}
\partial_r \lr{ \inv{r} \partial_\theta \psi}
– \frac{1}{r} \partial_{\theta r} \psi
+ \frac{1}{r^2} \partial_{\theta} \psi
=
-\inv{r^2} \partial_\theta \psi
+\inv{r} \partial_{r \theta} \psi
– \frac{1}{r} \partial_{\theta r} \psi
+ \frac{1}{r^2} \partial_{\theta} \psi
= 0,

and those with a $$\thetacap \phicap$$ factor are
\label{eqn:sphericalLaplacian:580}
\frac{1}{r^2} \partial_{\theta} \lr{ \frac{1}{S_\theta} \partial_{\phi} \psi }
– \frac{1}{r^2 S_\theta} \partial_{\phi\theta} \psi
+ \frac{1}{r^2 S_\theta^2} C_\theta \partial_{\phi} \psi
=
– \frac{1}{r^2} \frac{C_\theta}{S_\theta^2} \partial_{\phi} \psi
+ \frac{1}{r^2 S_\theta} \partial_{\theta \phi} \psi
– \frac{1}{r^2 S_\theta} \partial_{\phi\theta} \psi
+ \frac{1}{r^2 S_\theta^2} C_\theta \partial_{\phi} \psi
= 0,

and those with a $$\phicap \rcap$$ factor are
\label{eqn:sphericalLaplacian:600}
– \inv{S_\theta} \partial_r \lr{ \inv{r} \partial_\phi \psi }
+ \frac{1}{r S_\theta} \partial_{\phi r} \psi
– \frac{1}{r^2 S_\theta^2} S_\theta \partial_{\phi} \psi
=
\inv{S_\theta} \frac{1}{r^2} \partial_\phi \psi
– \inv{r S_\theta} \partial_{r \phi} \psi
+ \frac{1}{r S_\theta} \partial_{\phi r} \psi
– \frac{1}{r^2 S_\theta} \partial_{\phi} \psi
= 0.

This leaves
\label{eqn:sphericalLaplacian:620}
=
\partial_{rr} \psi
+ \frac{2}{r} \partial_{r} \psi
+ \frac{1}{r^2} \partial_{\theta \theta} \psi
+ \frac{1}{r^2 S_\theta} C_\theta \partial_{\theta} \psi
+ \frac{1}{r^2 S_\theta^2} \partial_{\phi \phi} \psi.

This factors nicely as

\label{eqn:sphericalLaplacian:640}
\boxed{
=
\inv{r^2} \PD{r}{} \lr{ r^2 \PD{r}{ \psi} }
+ \frac{1}{r^2 \sin\theta} \PD{\theta}{} \lr{ \sin\theta \PD{\theta}{ \psi } }
+ \frac{1}{r^2 \sin\theta^2} \PDSq{\phi}{ \psi}
,
}

which checks against the back cover of Jackson. Here it has been demonstrated explicitly that this operator expression is valid for multivector fields $$\psi$$ as well as scalar fields $$\psi$$.

# References

[1] JD Jackson. Classical Electrodynamics. John Wiley and Sons, 2nd edition, 1975.

[2] A. Macdonald. Vector and Geometric Calculus. CreateSpace Independent Publishing Platform, 2012.

## Corollaries to Stokes and Divergence theorems

In [1] a few problems are set to prove some variations of Stokes theorem. He gives some cool tricks to prove each one using just the classic 3D Stokes and divergence theorems. We can also do them directly from the more general Stokes theorem $$\int d^k \Bx \cdot (\spacegrad \wedge F) = \oint d^{k-1} \Bx \cdot F$$.

## Question: Stokes theorem on scalar function. ([1] pr. 1.60a)

Prove
\label{eqn:stokesCorollariesGriffiths:20}
\int \spacegrad T dV = \oint T d\Ba.

The direct way to prove this is to apply Stokes theorem

\label{eqn:stokesCorollariesGriffiths:80}
\int d^3 \Bx \cdot (\spacegrad \wedge T) = \oint d^2 \Bx \cdot T

Here $$d^3 \Bx = d\Bx_1 \wedge d\Bx_2 \wedge d\Bx_3$$, a pseudoscalar (trivector) volume element, and the wedge and dot products take their most general meanings. For $$k$$-blade $$F$$, and $$k’$$-blade $$F’$$, that is

\label{eqn:stokesCorollariesGriffiths:100}
\begin{aligned}
F \wedge F’ &= \gpgrade{F F’}{k+k’} \\
F \cdot F’ &= \gpgrade{F F’}{\Abs{k-k’}}
\end{aligned}

With $$d^3\Bx = I dV$$, and $$d^2 \Bx = I \ncap dA = I d\Ba$$, we have

\label{eqn:stokesCorollariesGriffiths:120}
\int I dV \spacegrad T = \oint I d\Ba T.

Cancelling the factors of $$I$$ proves the result.

Griffith’s trick to do this was to let $$\Bv = \Bc T$$, where $$\Bc$$ is a constant. For this, the divergence theorem integral is

\label{eqn:stokesCorollariesGriffiths:160}
\begin{aligned}
\int dV \spacegrad \cdot (\Bc T)
&=
\int dV \Bc \cdot \spacegrad T \\
&=
\Bc \cdot \int dV \spacegrad T \\
&=
\oint d\Ba \cdot (\Bc T) \\
&=
\Bc \cdot \oint d\Ba T.
\end{aligned}

This is true for any constant $$\Bc$$, so is also true for the unit vectors. This allows for summing projections in each of the unit directions

\label{eqn:stokesCorollariesGriffiths:180}
\begin{aligned}
&=
\sum \Be_k \lr{ \Be_k \cdot \int dV \spacegrad T } \\
&=
\sum \Be_k \lr{ \Be_k \cdot \oint d\Ba T } \\
&=
\oint d\Ba T.
\end{aligned}

## Question: ([1] pr. 1.60b)

Prove
\label{eqn:stokesCorollariesGriffiths:40}
\int \spacegrad \cross \Bv dV = -\oint \Bv \cross d\Ba.

This also follows directly from the general Stokes theorem

\label{eqn:stokesCorollariesGriffiths:200}
\int d^3 \Bx \cdot \lr{ \spacegrad \wedge \Bv } = \oint d^2 \Bx \cdot \Bv

The volume integrand is

\label{eqn:stokesCorollariesGriffiths:220}
\begin{aligned}
d^3 \Bx \cdot \lr{ \spacegrad \wedge \Bv }
&=
&=
\end{aligned}

and the surface integrand is
\label{eqn:stokesCorollariesGriffiths:240}
\begin{aligned}
d^2 \Bx \cdot \Bv
&=
\gpgradeone{ I d\Ba \Bv } \\
&=
\gpgradeone{ I (d\Ba \wedge \Bv) } \\
&=
I^2 (d\Ba \cross \Bv) \\
&=
-d\Ba \cross \Bv \\
&=
\Bv \cross d\Ba.
\end{aligned}

Plugging these into \ref{eqn:stokesCorollariesGriffiths:200} proves the result.

Griffiths trick for the same is to apply the divergence theorem to $$\Bv \cross \Bc$$. Such a volume integral is

\label{eqn:stokesCorollariesGriffiths:260}
\begin{aligned}
\int dV \spacegrad \cdot (\Bv \cross \Bc)
&=
\int dV \Bc \cdot (\spacegrad \cross \Bv) \\
&=
\Bc \cdot \int dV \spacegrad \cross \Bv.
\end{aligned}

This must equal
\label{eqn:stokesCorollariesGriffiths:280}
\begin{aligned}
\oint d\Ba \cdot (\Bv \cross \Bc)
&=
\Bc \cdot \oint d\Ba \cross \Bv \\
&=
-\Bc \cdot \oint \Bv \cross d\Ba
\end{aligned}

Again, assembling projections, we have
\label{eqn:stokesCorollariesGriffiths:300}
\begin{aligned}
&=
\sum \Be_k \lr{ \Be_k \cdot \int dV \spacegrad \cross \Bv } \\
&=
-\sum \Be_k \lr{ \Be_k \cdot \oint \Bv \cross d\Ba } \\
&=
-\oint \Bv \cross d\Ba.
\end{aligned}

## Question: ([1] pr. 1.60e)

Prove
\label{eqn:stokesCorollariesGriffiths:60}
\int \spacegrad T \cross d\Ba = -\oint T d\Bl.

This one follows from
\label{eqn:stokesCorollariesGriffiths:320}
\int d^2 \Bx \cdot \lr{ \spacegrad \wedge T } = \oint d^1 \Bx \cdot T.

The surface integrand can be written
\label{eqn:stokesCorollariesGriffiths:340}
\begin{aligned}
d^2 \Bx \cdot \lr{ \spacegrad \wedge T }
&=
&=
I (d\Ba \wedge \spacegrad T ) \\
&=
I^2 ( d\Ba \cross \spacegrad T ) \\
&=
\end{aligned}

The line integrand is

\label{eqn:stokesCorollariesGriffiths:360}
d^1 \Bx \cdot T = d^1 \Bx T.

Given a two parameter representation of the surface area element $$d^2 \Bx = d\Bx_1 \wedge d\Bx_2$$, the line element representation is
\label{eqn:stokesCorollariesGriffiths:380}
\begin{aligned}
d^1 \Bx
&= (\Bx_1 \wedge d\Bx_2) \cdot \Bx^1 + (d\Bx_1 \wedge \Bx_2) \cdot \Bx^2 \\
&= -d\Bx_2 + d\Bx_1,
\end{aligned}

giving

\label{eqn:stokesCorollariesGriffiths:400}
\begin{aligned}
&=
\int
-\evalbar{\lr{ \PD{u_2}{\Bx} T }}{\Delta u_1} du_2
+\evalbar{\lr{ \PD{u_1}{\Bx} T }}{\Delta u_2} du_1 \\
&=
-\oint d\Bl T,
\end{aligned}

or
\label{eqn:stokesCorollariesGriffiths:420}
=
-\oint d\Bl T.

Griffiths trick for the same is to use $$\Bv = \Bc T$$ for constant $$\Bc$$ in (the usual 3D) Stokes’ theorem. That is

\label{eqn:stokesCorollariesGriffiths:440}
\begin{aligned}
\int d\Ba \cdot (\spacegrad \cross (\Bc T))
&=
\Bc \cdot \int d\Ba \cross \spacegrad T \\
&=
-\Bc \cdot \int \spacegrad T \cross d\Ba \\
&=
\oint d\Bl \cdot (\Bc T) \\
&=
\Bc \cdot \oint d\Bl T.
\end{aligned}

Again assembling projections we have
\label{eqn:stokesCorollariesGriffiths:460}
\begin{aligned}
&=
\sum \Be_k \lr{ \Be_k \cdot \int \spacegrad T \cross d\Ba} \\
&=
-\sum \Be_k \lr{ \Be_k \cdot \oint d\Bl T } \\
&=
-\oint d\Bl T.
\end{aligned}

# References

[1] David Jeffrey Griffiths and Reed College. Introduction to electrodynamics. Prentice hall Upper Saddle River, NJ, 3rd edition, 1999.

## Helmholtz theorem

This is a problem from ece1228. I attempted solutions in a number of ways. One using Geometric Algebra, one devoid of that algebra, and then this method, which combined aspects of both. Of the three methods I tried to obtain this result, this is the most compact and elegant. It does however, require a fair bit of Geometric Algebra knowledge, including the Fundamental Theorem of Geometric Calculus, as detailed in [1], [3] and [2].

## Question: Helmholtz theorem

Prove the first Helmholtz’s theorem, i.e. if vector $$\BM$$ is defined by its divergence

\label{eqn:helmholtzDerviationMultivector:20}

and its curl
\label{eqn:helmholtzDerviationMultivector:40}

within a region and its normal component $$\BM_{\textrm{n}}$$ over the boundary, then $$\BM$$ is
uniquely specified.

The gradient of the vector $$\BM$$ can be written as a single even grade multivector

\label{eqn:helmholtzDerviationMultivector:60}
= s + I \BC.

We will use this to attempt to discover the relation between the vector $$\BM$$ and its divergence and curl. We can express $$\BM$$ at the point of interest as a convolution with the delta function at all other points in space

\label{eqn:helmholtzDerviationMultivector:80}
\BM(\Bx) = \int_V dV’ \delta(\Bx – \Bx’) \BM(\Bx’).

The Laplacian representation of the delta function in \R{3} is

\label{eqn:helmholtzDerviationMultivector:100}
\delta(\Bx – \Bx’) = -\inv{4\pi} \spacegrad^2 \inv{\Abs{\Bx – \Bx’}},

so $$\BM$$ can be represented as the following convolution

\label{eqn:helmholtzDerviationMultivector:120}
\BM(\Bx) = -\inv{4\pi} \int_V dV’ \spacegrad^2 \inv{\Abs{\Bx – \Bx’}} \BM(\Bx’).

Using this relation and proceeding with a few applications of the chain rule, plus the fact that $$\spacegrad 1/\Abs{\Bx – \Bx’} = -\spacegrad’ 1/\Abs{\Bx – \Bx’}$$, we find

\label{eqn:helmholtzDerviationMultivector:720}
\begin{aligned}
-4 \pi \BM(\Bx)
&= \int_V dV’ \spacegrad^2 \inv{\Abs{\Bx – \Bx’}} \BM(\Bx’) \\
} } \\
&=
\ncap \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}}
}
\frac{s(\Bx’) + I\BC(\Bx’)}{\Abs{\Bx – \Bx’}}
} \\
&=
\ncap \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}}
}
\frac{s(\Bx’)}{\Abs{\Bx – \Bx’}}
\frac{I\BC(\Bx’)}{\Abs{\Bx – \Bx’}}.
\end{aligned}

By inserting a no-op grade selection operation in the second step, the trivector terms that would show up in subsequent steps are automatically filtered out. This leaves us with a boundary term dependent on the surface and the normal and tangential components of $$\BM$$. Added to that is a pair of volume integrals that provide the unique dependence of $$\BM$$ on its divergence and curl. When the surface is taken to infinity, which requires $$\Abs{\BM}/\Abs{\Bx – \Bx’} \rightarrow 0$$, then the dependence of $$\BM$$ on its divergence and curl is unique.

In order to express final result in traditional vector algebra form, a couple transformations are required. The first is that

\label{eqn:helmholtzDerviationMultivector:800}
\gpgradeone{ \Ba I \Bb } = I^2 \Ba \cross \Bb = -\Ba \cross \Bb.

For the grade selection in the boundary integral, note that

\label{eqn:helmholtzDerviationMultivector:740}
\begin{aligned}
&=
+
&=
+
&=

\end{aligned}

These give

\label{eqn:helmholtzDerviationMultivector:721}
\boxed{
\begin{aligned}
\BM(\Bx)
&=
\spacegrad \inv{4\pi} \int_{\partial V} dA’ \ncap \cdot \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}}

\spacegrad \cross \inv{4\pi} \int_{\partial V} dA’ \ncap \cross \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}} \\
\frac{s(\Bx’)}{\Abs{\Bx – \Bx’}}
\frac{\BC(\Bx’)}{\Abs{\Bx – \Bx’}}.
\end{aligned}
}

# References

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

[2] A. Macdonald. Vector and Geometric Calculus. CreateSpace Independent Publishing Platform, 2012.

[3] Garret Sobczyk and Omar Le’on S’anchez. Fundamental theorem of calculus. Advances in Applied Clifford Algebras, 21:221–231, 2011. URL http://arxiv.org/abs/0809.4526.

## Does the divergence and curl uniquely determine the vector?

A problem posed in the ece1228 problem set was the following

### Helmholtz theorem.

Prove the first Helmholtz’s theorem, i.e. if vector $$\BM$$ is defined by its divergence

\label{eqn:emtProblemSet1Problem5:20}

and its curl
\label{eqn:emtProblemSet1Problem5:40}

within a region and its normal component $$\BM_{\textrm{n}}$$ over the boundary, then $$\BM$$ is uniquely specified.

### Solution.

This problem screams for an attempt using Geometric Algebra techniques, since
the gradient of this vector can be written as a single even grade multivector

\label{eqn:emtProblemSet1Problem5AppendixGA:60}
\begin{aligned}
&= s + I \BC.
\end{aligned}

Observe that the Laplacian of $$\BM$$ is vector valued

\label{eqn:emtProblemSet1Problem5AppendixGA:400}

This means that $$\spacegrad \BC$$ must be a bivector $$\spacegrad \BC = \spacegrad \wedge \BC$$, or that $$\BC$$ has zero divergence

\label{eqn:emtProblemSet1Problem5AppendixGA:420}

This required constraint on $$\BC$$ will show up in subsequent analysis. An equivalent problem to the one posed
is to show that the even grade multivector equation $$\spacegrad \BM = s + I \BC$$ has an inverse given the constraint
specified by \ref{eqn:emtProblemSet1Problem5AppendixGA:420}.

The Green’s function for the gradient can be found in [1], where it is used to generalize the Cauchy integral equations to higher dimensions.

\label{eqn:emtProblemSet1Problem5AppendixGA:80}
\begin{aligned}
G(\Bx ; \Bx’) &= \inv{4 \pi} \frac{ \Bx – \Bx’ }{\Abs{\Bx – \Bx’}^3} \\
\end{aligned}

The inversion equation is an application of the Fundamental Theorem of (Geometric) Calculus, with the gradient operating bidirectionally on the Green’s function and the vector function

\label{eqn:emtProblemSet1Problem5AppendixGA:100}
\begin{aligned}
\oint_{\partial V} G(\Bx, \Bx’) d^2 \Bx’ \BM(\Bx’)
&=
\int_V G(\Bx, \Bx’) d^3 \Bx \lrspacegrad’ \BM(\Bx’) \\
&=
\int_V d^3 \Bx (G(\Bx, \Bx’) \lspacegrad’) \BM(\Bx’)
+
\int_V d^3 \Bx G(\Bx, \Bx’) (\spacegrad’ \BM(\Bx’)) \\
&=
-\int_V d^3 \Bx \delta(\Bx – \By) \BM(\Bx’)
+
\int_V d^3 \Bx G(\Bx, \Bx’) \lr{ s(\Bx’) + I \BC(\Bx’) } \\
&=
-I \BM(\Bx)
+
\inv{4 \pi} \int_V d^3 \Bx \frac{ \Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 } \lr{ s(\Bx’) + I \BC(\Bx’) }.
\end{aligned}

The integrals are in terms of the primed coordinates so that the end result is a function of $$\Bx$$. To rearrange for $$\BM$$, let $$d^3 \Bx’ = I dV’$$, and $$d^2 \Bx’ \ncap(\Bx’) = I dA’$$, then right multiply with the pseudoscalar $$I$$, noting that in \R{3} the pseudoscalar commutes with any grades

\label{eqn:emtProblemSet1Problem5AppendixGA:440}
\begin{aligned}
\BM(\Bx)
&=
I \oint_{\partial V} G(\Bx, \Bx’) I dA’ \ncap \BM(\Bx’)

I \inv{4 \pi} \int_V I dV’ \frac{ \Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 } \lr{ s(\Bx’) + I \BC(\Bx’) } \\
&=
-\oint_{\partial V} dA’ G(\Bx, \Bx’) \ncap \BM(\Bx’)
+
\inv{4 \pi} \int_V dV’ \frac{ \Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 } \lr{ s(\Bx’) + I \BC(\Bx’) }.
\end{aligned}

This can be decomposed into a vector and a trivector equation. Let $$\Br = \Bx – \Bx’ = r \rcap$$, and note that

\label{eqn:emtProblemSet1Problem5AppendixGA:500}
\begin{aligned}
&=
\gpgradeone{ I \rcap \BC } \\
&=
I \rcap \wedge \BC \\
&=
-\rcap \cross \BC,
\end{aligned}

so this pair of equations can be written as

\label{eqn:emtProblemSet1Problem5AppendixGA:520}
\begin{aligned}
\BM(\Bx)
&=
-\inv{4 \pi} \oint_{\partial V} dA’ \frac{\gpgradeone{ \rcap \ncap \BM(\Bx’) }}{r^2}
+
\inv{4 \pi} \int_V dV’ \lr{
\frac{\rcap}{r^2} s(\Bx’) –
\frac{\rcap}{r^2} \cross \BC(\Bx’) } \\
0
&=
-\inv{4 \pi} \oint_{\partial V} dA’ \frac{\rcap}{r^2} \wedge \ncap \wedge \BM(\Bx’)
+
\frac{I}{4 \pi} \int_V dV’ \frac{ \rcap \cdot \BC(\Bx’) }{r^2}.
\end{aligned}

Consider the last integral in the pseudoscalar equation above. Since we expect no pseudoscalar components, this must be zero, or cancel perfectly. It’s not obvious that this is the case, but a transformation to a surface integral shows the constraints required for that to be the case. To do so note

\label{eqn:emtProblemSet1Problem5AppendixGA:540}
\begin{aligned}
&= -\spacegrad’ \inv{\Bx – \Bx’} \\
&=
-\frac{\Bx – \Bx’}{\Abs{\Bx – \Bx’}^3} \\
&= -\frac{\rcap}{r^2}.
\end{aligned}

Using this and the chain rule we have

\label{eqn:emtProblemSet1Problem5AppendixGA:560}
\begin{aligned}
\frac{I}{4 \pi} \int_V dV’ \frac{ \rcap \cdot \BC(\Bx’) }{r^2}
&=
\frac{I}{4 \pi} \int_V dV’ \lr{ \spacegrad’ \inv{ r } } \cdot \BC(\Bx’) \\
&=
\frac{I}{4 \pi} \int_V dV’ \spacegrad’ \cdot \frac{\BC(\Bx’)}{r}

\frac{I}{4 \pi} \int_V dV’ \frac{ \spacegrad’ \cdot \BC(\Bx’) }{r} \\
&=
\frac{I}{4 \pi} \int_V dV’ \spacegrad’ \cdot \frac{\BC(\Bx’)}{r} \\
&=
\frac{I}{4 \pi} \int_{\partial V} dA’ \ncap(\Bx’) \cdot \frac{\BC(\Bx’)}{r}.
\end{aligned}

The divergence of $$\BC$$ above was killed by recalling the constraint \ref{eqn:emtProblemSet1Problem5AppendixGA:420}. This means that we can rewrite entirely as surface integral and eventually reduced to a single triple product

\label{eqn:emtProblemSet1Problem5AppendixGA:580}
\begin{aligned}
0
&=
-\frac{I}{4 \pi} \oint_{\partial V} dA’ \lr{
\frac{\rcap}{r^2} \cdot (\ncap \cross \BM(\Bx’))
-\ncap \cdot \frac{\BC(\Bx’)}{r}
} \\
&=
\frac{I}{4 \pi} \oint_{\partial V} dA’ \ncap \cdot \lr{
\frac{\rcap}{r^2} \cross \BM(\Bx’)
+ \frac{\BC(\Bx’)}{r}
} \\
&=
\frac{I}{4 \pi} \oint_{\partial V} dA’ \ncap \cdot \lr{
+ \frac{\BC(\Bx’)}{r}
} \\
&=
\frac{I}{4 \pi} \oint_{\partial V} dA’ \ncap \cdot \lr{
} \\
&=
\frac{I}{4 \pi} \oint_{\partial V} dA’
\frac{\BM(\Bx’) \cross \ncap}{r}
&=
\frac{I}{4 \pi} \oint_{\partial V} dA’
\frac{\BM(\Bx’) \cross \ncap}{r}.
\end{aligned}

### Final results.

Assembling things back into a single multivector equation, the complete inversion integral for $$\BM$$ is

\label{eqn:emtProblemSet1Problem5AppendixGA:600}
\BM(\Bx)
=
\inv{4 \pi} \oint_{\partial V} dA’
\lr{
\frac{\BM(\Bx’) \wedge \ncap}{r}
}
+
\inv{4 \pi} \int_V dV’ \lr{
\frac{\rcap}{r^2} s(\Bx’) –
\frac{\rcap}{r^2} \cross \BC(\Bx’) }.

This shows that vector $$\BM$$ can be recovered uniquely from $$s, \BC$$ when $$\Abs{\BM}/r^2$$ vanishes on an infinite surface. If we restrict attention to a finite surface, we have to add to the fixed solution a specific solution that depends on the value of $$\BM$$ on that surface. The vector portion of that surface integrand contains

\label{eqn:emtProblemSet1Problem5AppendixGA:640}
\begin{aligned}
&=
\rcap (\ncap \cdot \BM )
+
\rcap \cdot (\ncap \wedge \BM ) \\
&=
\rcap (\ncap \cdot \BM )
+
(\rcap \cdot \ncap) \BM

(\rcap \cdot \BM ) \ncap.
\end{aligned}

The constraints required by a zero triple product $$\spacegrad’ \cdot (\BM(\Bx’) \cross \ncap(\Bx’))$$ are complicated on a such a general finite surface. Consider instead, for simplicity, the case of a spherical surface, which can be analyzed more easily. In that case the outward normal of the surface centred on the test charge point $$\Bx$$ is $$\ncap = -\rcap$$. The pseudoscalar integrand is not generally killed unless the divergence of its tangential component on this surface is zero. One way that this can occur is for $$\BM \cross \ncap = 0$$, so that $$-\gpgradeone{ \rcap \ncap \BM } = \BM = (\BM \cdot \ncap) \ncap = \BM_{\textrm{n}}$$.

This gives

\label{eqn:emtProblemSet1Problem5AppendixGA:620}
\BM(\Bx)
=
\inv{4 \pi} \oint_{\Abs{\Bx – \Bx’} = r} dA’ \frac{\BM_{\textrm{n}}(\Bx’)}{r^2}
+
\inv{4 \pi} \int_V dV’ \lr{
\frac{\rcap}{r^2} s(\Bx’) +
\BC(\Bx’) \cross \frac{\rcap}{r^2} },

or, in terms of potential functions, which is arguably tidier

\label{eqn:emtProblemSet1Problem5AppendixGA:300}
\boxed{
\BM(\Bx)
=
\inv{4 \pi} \oint_{\Abs{\Bx – \Bx’} = r} dA’ \frac{\BM_{\textrm{n}}(\Bx’)}{r^2}
-\spacegrad \int_V dV’ \frac{ s(\Bx’)}{ 4 \pi r }
+\spacegrad \cross \int_V dV’ \frac{ \BC(\Bx’) }{ 4 \pi r }.
}

### Commentary

I attempted this problem in three different ways. My first approach (above) assembled the divergence and curl relations above into a single (Geometric Algebra) multivector gradient equation and applied the vector valued Green’s function for the gradient to invert that equation. That approach logically led from the differential equation for $$\BM$$ to the solution for $$\BM$$ in terms of $$s$$ and $$\BC$$. However, this strategy introduced some complexities that make me doubt the correctness of the associated boundary analysis.

Even if the details of the boundary handling in my multivector approach is not correct, I thought that approach was interesting enough to share.

# References

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

## Stokes Theorem

The Fundamental Theorem of (Geometric) Calculus is a generalization of Stokes theorem to multivector integrals. Notationally, it looks like Stokes theorem with all the dot and wedge products removed. It is worth restating Stokes theorem and all the definitions associated with it for reference

## Stokes’ Theorem

For blades $$F \in \bigwedge^{s}$$, and $$m$$ volume element $$d^k \Bx, s < k$$, \begin{equation*} \int_V d^k \Bx \cdot (\boldpartial \wedge F) = \oint_{\partial V} d^{k-1} \Bx \cdot F. \end{equation*} This is a loaded and abstract statement, and requires many definitions to make it useful

• The volume integral is over a $$m$$ dimensional surface (manifold).
• Integration over the boundary of the manifold $$V$$ is indicated by $$\partial V$$.
• This manifold is assumed to be spanned by a parameterized vector $$\Bx(u^1, u^2, \cdots, u^k)$$.
• A curvilinear coordinate basis $$\setlr{ \Bx_i }$$ can be defined on the manifold by
\label{eqn:fundamentalTheoremOfCalculus:40}
\Bx_i \equiv \PD{u^i}{\Bx} \equiv \partial_i \Bx.

• A dual basis $$\setlr{\Bx^i}$$ reciprocal to the tangent vector basis $$\Bx_i$$ can be calculated subject to the requirement $$\Bx_i \cdot \Bx^j = \delta_i^j$$.
• The vector derivative $$\boldpartial$$, the projection of the gradient onto the tangent space of the manifold, is defined by
\label{eqn:fundamentalTheoremOfCalculus:100}
\boldpartial = \Bx^i \partial_i = \sum_{i=1}^k \Bx_i \PD{u^i}{}.

• The volume element is defined by
\label{eqn:fundamentalTheoremOfCalculus:60}
d^k \Bx = d\Bx_1 \wedge d\Bx_2 \cdots \wedge d\Bx_k,

where

\label{eqn:fundamentalTheoremOfCalculus:80}
d\Bx_k = \Bx_k du^k,\qquad \text{(no sum)}.

• The volume element is non-zero on the manifold, or $$\Bx_1 \wedge \cdots \wedge \Bx_k \ne 0$$.
• The surface area element $$d^{k-1} \Bx$$, is defined by
\label{eqn:fundamentalTheoremOfCalculus:120}
d^{k-1} \Bx = \sum_{i = 1}^k (-1)^{k-i} d\Bx_1 \wedge d\Bx_2 \cdots \widehat{d\Bx_i} \cdots \wedge d\Bx_k,

where $$\widehat{d\Bx_i}$$ indicates the omission of $$d\Bx_i$$.

• My proof for this theorem was restricted to a simple “rectangular” volume parameterized by the ranges
$$[u^1(0), u^1(1) ] \otimes [u^2(0), u^2(1) ] \otimes \cdots \otimes [u^k(0), u^k(1) ]$$

• The precise meaning that should be given to oriented area integral is
\label{eqn:fundamentalTheoremOfCalculus:140}
\oint_{\partial V} d^{k-1} \Bx \cdot F
=
\sum_{i = 1}^k (-1)^{k-i} \int \evalrange{
\lr{ \lr{ d\Bx_1 \wedge d\Bx_2 \cdots \widehat{d\Bx_i} \cdots \wedge d\Bx_k } \cdot F }
}{u^i = u^i(0)}{u^i(1)},

where both the a area form and the blade $$F$$ are evaluated at the end points of the parameterization range.

After the work of stating exactly what is meant by this theorem, most of the proof follows from the fact that for $$s < k$$ the volume curl dot product can be expanded as $$\label{eqn:fundamentalTheoremOfCalculus:160} \int_V d^k \Bx \cdot (\boldpartial \wedge F) = \int_V d^k \Bx \cdot (\Bx^i \wedge \partial_i F) = \int_V \lr{ d^k \Bx \cdot \Bx^i } \cdot \partial_i F.$$ Each of the $$du^i$$ integrals can be evaluated directly, since each of the remaining $$d\Bx_j = du^j \PDi{u^j}{}, i \ne j$$ is calculated with $$u^i$$ held fixed. This allows for the integration over a rectangular'' parameterization region, proving the theorem for such a volume parameterization. A more general proof requires a triangulation of the volume and surface, but the basic principle of the theorem is evident, without that additional work.

## Fundamental Theorem of Calculus

There is a Geometric Algebra generalization of Stokes theorem that does not have the blade grade restriction of Stokes theorem. In [2] this is stated as

\label{eqn:fundamentalTheoremOfCalculus:180}
\int_V d^k \Bx \boldpartial F = \oint_{\partial V} d^{k-1} \Bx F.

A similar expression is used in [1] where it is also pointed out there is a variant with the vector derivative acting to the left

\label{eqn:fundamentalTheoremOfCalculus:200}
\int_V F d^k \Bx \boldpartial = \oint_{\partial V} F d^{k-1} \Bx.

In [3] it is pointed out that a bidirectional formulation is possible, providing the most general expression of the Fundamental Theorem of (Geometric) Calculus

\label{eqn:fundamentalTheoremOfCalculus:220}
\boxed{
\int_V F d^k \Bx \boldpartial G = \oint_{\partial V} F d^{k-1} \Bx G.
}

Here the vector derivative acts both to the left and right on $$F$$ and $$G$$. The specific action of this operator is
\label{eqn:fundamentalTheoremOfCalculus:240}
\begin{aligned}
F \boldpartial G
&=
(F \boldpartial) G
+
F (\boldpartial G) \\
&=
(\partial_i F) \Bx^i G
+
F \Bx^i (\partial_i G).
\end{aligned}

The fundamental theorem can be demonstrated by direct expansion. With the vector derivative $$\boldpartial$$ and its partials $$\partial_i$$ acting bidirectionally, that is

\label{eqn:fundamentalTheoremOfCalculus:260}
\begin{aligned}
\int_V F d^k \Bx \boldpartial G
&=
\int_V F d^k \Bx \Bx^i \partial_i G \\
&=
\int_V F \lr{ d^k \Bx \cdot \Bx^i + d^k \Bx \wedge \Bx^i } \partial_i G.
\end{aligned}

Both the reciprocal frame vectors and the curvilinear basis span the tangent space of the manifold, since we can write any reciprocal frame vector as a set of projections in the curvilinear basis

\label{eqn:fundamentalTheoremOfCalculus:280}
\Bx^i = \sum_j \lr{ \Bx^i \cdot \Bx^j } \Bx_j,

so $$\Bx^i \in sectionpan \setlr{ \Bx_j, j \in [1,k] }$$.
This means that $$d^k \Bx \wedge \Bx^i = 0$$, and

\label{eqn:fundamentalTheoremOfCalculus:300}
\begin{aligned}
\int_V F d^k \Bx \boldpartial G
&=
\int_V F \lr{ d^k \Bx \cdot \Bx^i } \partial_i G \\
&=
\sum_{i = 1}^{k}
\int_V
du^1 du^2 \cdots \widehat{ du^i} \cdots du^k
F \lr{
(-1)^{k-i}
\Bx_1 \wedge \Bx_2 \cdots \widehat{\Bx_i} \cdots \wedge \Bx_k } \partial_i G du^i \\
&=
\sum_{i = 1}^{k}
(-1)^{k-i}
\int_{u^1}
\int_{u^2}
\cdots
\int_{u^{i-1}}
\int_{u^{i+1}}
\cdots
\int_{u^k}
\evalrange{ \lr{
F d\Bx_1 \wedge d\Bx_2 \cdots \widehat{d\Bx_i} \cdots \wedge d\Bx_k G
}
}{u^i = u^i(0)}{u^i(1)}.
\end{aligned}

Adding in the same notational sugar that we used in Stokes theorem, this proves the Fundamental theorem \ref{eqn:fundamentalTheoremOfCalculus:220} for “rectangular” parameterizations. Note that such a parameterization need not actually be rectangular.

## Example: Application to Maxwell’s equation

{example:fundamentalTheoremOfCalculus:1}

Maxwell’s equation is an example of a first order gradient equation

\label{eqn:fundamentalTheoremOfCalculus:320}
\grad F = \inv{\epsilon_0 c} J.

Integrating over a four-volume (where the vector derivative equals the gradient), and applying the Fundamental theorem, we have

\label{eqn:fundamentalTheoremOfCalculus:340}
\inv{\epsilon_0 c} \int d^4 x J = \oint d^3 x F.

Observe that the surface area element product with $$F$$ has both vector and trivector terms. This can be demonstrated by considering some examples

\label{eqn:fundamentalTheoremOfCalculus:360}
\begin{aligned}
\gamma_{012} \gamma_{01} &\propto \gamma_2 \\
\gamma_{012} \gamma_{23} &\propto \gamma_{023}.
\end{aligned}

On the other hand, the four volume integral of $$J$$ has only trivector parts. This means that the integral can be split into a pair of same-grade equations

\label{eqn:fundamentalTheoremOfCalculus:380}
\begin{aligned}
\inv{\epsilon_0 c} \int d^4 x \cdot J &=
\oint \gpgradethree{ d^3 x F} \\
0 &=
\oint d^3 x \cdot F.
\end{aligned}

The first can be put into a slightly tidier form using a duality transformation
\label{eqn:fundamentalTheoremOfCalculus:400}
\begin{aligned}
&=
-\gpgradethree{ d^3 x I^2 F} \\
&=
\gpgradethree{ I d^3 x I F} \\
&=
(I d^3 x) \wedge (I F).
\end{aligned}

Letting $$n \Abs{d^3 x} = I d^3 x$$, this gives

\label{eqn:fundamentalTheoremOfCalculus:420}
\oint \Abs{d^3 x} n \wedge (I F) = \inv{\epsilon_0 c} \int d^4 x \cdot J.

Note that this normal is normal to a three-volume subspace of the spacetime volume. For example, if one component of that spacetime surface area element is $$\gamma_{012} c dt dx dy$$, then the normal to that area component is $$\gamma_3$$.

A second set of duality transformations

\label{eqn:fundamentalTheoremOfCalculus:440}
\begin{aligned}
n \wedge (IF)
&=
&=
&=
-\gpgradethree{ I (n \cdot F)} \\
&=
-I (n \cdot F),
\end{aligned}

and
\label{eqn:fundamentalTheoremOfCalculus:460}
\begin{aligned}
I d^4 x \cdot J
&=
\gpgradeone{ I d^4 x \cdot J } \\
&=
\gpgradeone{ I d^4 x J } \\
&=
\gpgradeone{ (I d^4 x) J } \\
&=
(I d^4 x) J,
\end{aligned}

can further tidy things up, leaving us with

\label{eqn:fundamentalTheoremOfCalculus:500}
\boxed{
\begin{aligned}
\oint \Abs{d^3 x} n \cdot F &= \inv{\epsilon_0 c} \int (I d^4 x) J \\
\oint d^3 x \cdot F &= 0.
\end{aligned}
}

The Fundamental theorem of calculus immediately provides relations between the Faraday bivector $$F$$ and the four-current $$J$$.

# References

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

[2] A. Macdonald. Vector and Geometric Calculus. CreateSpace Independent Publishing Platform, 2012.

[3] Garret Sobczyk and Omar Le\’on S\’anchez. Fundamental theorem of calculus. Advances in Applied Clifford Algebras, 21\penalty0 (1):\penalty0 221–231, 2011. URL http://arxiv.org/abs/0809.4526.

## Motivation

fig 1. Two surfaces normal to the interface.

Most electrodynamics textbooks either start with or contain a treatment of boundary value conditions. These typically involve evaluating Maxwell’s equations over areas or volumes of decreasing height, such as those illustrated in fig. 1, and fig. 2. These represent surfaces and volumes where the height is allowed to decrease to infinitesimal levels, and are traditionally used to find the boundary value constraints of the normal and tangential components of the electric and magnetic fields.

fig 2. A pillbox volume encompassing the interface.

More advanced topics, such as evaluation of the Fresnel reflection and transmission equations, also rely on similar consideration of boundary value constraints. I’ve wondered for a long time how the Fresnel equations could be attacked by looking at the boundary conditions for the combined field $$F = \BE + I c \BB$$, instead of the considering them separately.

## A unified approach.

The Geometric Algebra (and relativistic tensor) formulations of Maxwell’s equations put the electric and magnetic fields on equal footings. It is in fact possible to specify the boundary value constraints on the fields without first separating Maxwell’s equations into their traditional forms. The starting point in Geometric Algebra is Maxwell’s equation, premultiplied by a stationary observer’s timelike basis vector

\label{eqn:maxwellBoundaryConditions:20}
\gamma_0 \grad F = \inv{\epsilon_0 c} \gamma_0 J,

or

\label{eqn:maxwellBoundaryConditions:40}
\lr{ \partial_0 + \spacegrad} F = \frac{\rho}{\epsilon_0} – \frac{\BJ}{\epsilon_0}.

The electrodynamic field $$F = \BE + I c \BB$$ is a multivector in this spatial domain (whereas it is a bivector in the spacetime algebra domain), and has vector and bivector components. The product of the spatial gradient and the field can still be split into dot and curl components $$\spacegrad M = \spacegrad \cdot M + \spacegrad \wedge M$$. If $$M = \sum M_i$$, where $$M_i$$ is an grade $$i$$ blade, then we give this the Hestenes’ [1] definitions

\label{eqn:maxwellBoundaryConditions:60}
\begin{aligned}
\end{aligned}

With that said, Maxwell’s equation can be rearranged into a pair of multivector equations

\label{eqn:maxwellBoundaryConditions:80}
\begin{aligned}
\spacegrad \cdot F &= \gpgrade{-\partial_0 F + \frac{\rho}{\epsilon_0} – \frac{\BJ}{\epsilon_0 c}}{0,1} \\
\end{aligned}

The latter equation can be integrated with Stokes theorem, but we need to apply a duality transformation to the latter in order to apply Stokes to it

\label{eqn:maxwellBoundaryConditions:120}
\begin{aligned}
&=
&=
&=
&=
\end{aligned}

so

\label{eqn:maxwellBoundaryConditions:100}
\begin{aligned}
\spacegrad \wedge (I F) &= I \lr{ -\inv{c} \partial_t \BE + \frac{\rho}{\epsilon_0} – \frac{\BJ}{\epsilon_0 c} } \\
\spacegrad \wedge F &= -I \partial_t \BB.
\end{aligned}

Integrating each of these over the pillbox volume gives

\label{eqn:maxwellBoundaryConditions:140}
\begin{aligned}
\oint_{\partial V} d^2 \Bx \cdot (I F)
&=
\int_{V} d^3 \Bx \cdot \lr{ I \lr{ -\inv{c} \partial_t \BE + \frac{\rho}{\epsilon_0} – \frac{\BJ}{\epsilon_0 c} } } \\
\oint_{\partial V} d^2 \Bx \cdot F
&=
– \partial_t \int_{V} d^3 \Bx \cdot \lr{ I \BB }.
\end{aligned}

In the absence of charges and currents on the surface, and if the height of the volume is reduced to zero, the volume integrals vanish, and only the upper surfaces of the pillbox contribute to the surface integrals.

\label{eqn:maxwellBoundaryConditions:200}
\begin{aligned}
\oint_{\partial V} d^2 \Bx \cdot (I F) &= 0 \\
\oint_{\partial V} d^2 \Bx \cdot F &= 0.
\end{aligned}

With a multivector $$F$$ in the mix, the geometric meaning of these integrals is not terribly clear. They do describe the boundary conditions, but to see exactly what those are, we can now resort to the split of $$F$$ into its electric and magnetic fields. Let’s look at the non-dual integral to start with

\label{eqn:maxwellBoundaryConditions:160}
\begin{aligned}
\oint_{\partial V} d^2 \Bx \cdot F
&=
\oint_{\partial V} d^2 \Bx \cdot \lr{ \BE + I c \BB } \\
&=
\oint_{\partial V} d^2 \Bx \cdot \BE + I c d^2 \Bx \wedge \BB \\
&=
0.
\end{aligned}

No component of $$\BE$$ that is normal to the surface contributes to $$d^2 \Bx \cdot \BE$$, whereas only components of $$\BB$$ that are normal contribute to $$d^2 \Bx \wedge \BB$$. That means that we must have tangential components of $$\BE$$ and the normal components of $$\BB$$ matching on the surfaces

\label{eqn:maxwellBoundaryConditions:180}
\begin{aligned}
\lr{\BE_2 \wedge \ncap} \ncap – \lr{\BE_1 \wedge (-\ncap)} (-\ncap) &= 0 \\
\lr{\BB_2 \cdot \ncap} \ncap – \lr{\BB_1 \cdot (-\ncap)} (-\ncap) &= 0 .
\end{aligned}

Similarly, for the dot product of the dual field, this is

\label{eqn:maxwellBoundaryConditions:220}
\begin{aligned}
\oint_{\partial V} d^2 \Bx \cdot (I F)
&=
\oint_{\partial V} d^2 \Bx \cdot (I \BE – c \BB) \\
&=
\oint_{\partial V} I d^2 \Bx \wedge \BE – c d^2 \Bx \cdot \BB.
\end{aligned}

For this integral, only the normal components of $$\BE$$ contribute, and only the tangential components of $$\BB$$ contribute. This means that

\label{eqn:maxwellBoundaryConditions:240}
\begin{aligned}
\lr{\BE_2 \cdot \ncap} \ncap – \lr{\BE_1 \cdot (-\ncap)} (-\ncap) &= 0 \\
\lr{\BB_2 \wedge \ncap} \ncap – \lr{\BB_1 \wedge (-\ncap)} (-\ncap) &= 0.
\end{aligned}

This is why we end up with a seemingly strange mix of tangential and normal components of the electric and magnetic fields. These constraints can be summarized as

\label{eqn:maxwellBoundaryConditions:260}
\begin{aligned}
( \BE_2 – \BE_1 ) \cdot \ncap &= 0 \\
( \BE_2 – \BE_1 ) \wedge \ncap &= 0 \\
( \BB_2 – \BB_1 ) \cdot \ncap &= 0 \\
( \BB_2 – \BB_1 ) \wedge \ncap &= 0
\end{aligned}

These relationships are usually expressed in terms of all of $$\BE, \BD, \BB$$ and $$\BH$$. Because I’d started with Maxwell’s equations for free space, I don’t have the $$\epsilon$$ and $$\mu$$ factors that produce those more general relationships. Those more general boundary value relationships are usually the starting point for the Fresnel interface analysis. It is also possible to further generalize these relationships to include charges and currents on the surface.

# References

[1] D. Hestenes. New Foundations for Classical Mechanics. Kluwer Academic Publishers, 1999.