## Potential solutions to the static Maxwell’s equation using geometric algebra

When neither the electromagnetic field strength $$F = \BE + I \eta \BH$$, nor current $$J = \eta (c \rho – \BJ) + I(c\rho_m – \BM)$$ is a function of time, then the geometric algebra form of Maxwell’s equations is the first order multivector (gradient) equation
\label{eqn:staticPotentials:20}

While direct solutions to this equations are possible with the multivector Green’s function for the gradient
\label{eqn:staticPotentials:40}
G(\Bx, \Bx’) = \inv{4\pi} \frac{\Bx – \Bx’}{\Norm{\Bx – \Bx’}^3 },

the aim in this post is to explore second order (potential) solutions in a geometric algebra context. Can we assume that it is possible to find a multivector potential $$A$$ for which
\label{eqn:staticPotentials:60}

is a solution to the Maxwell statics equation? If such a solution exists, then Maxwell’s equation is simply
\label{eqn:staticPotentials:80}

which can be easily solved using the scalar Green’s function for the Laplacian
\label{eqn:staticPotentials:240}
G(\Bx, \Bx’) = -\inv{\Norm{\Bx – \Bx’} },

a beastie that may be easier to convolve than the vector valued Green’s function for the gradient.

It is immediately clear that some restrictions must be imposed on the multivector potential $$A$$. In particular, since the field $$F$$ has only vector and bivector grades, this gradient must have no scalar, nor pseudoscalar grades. That is
\label{eqn:staticPotentials:100}

This constraint on the potential can be avoided if a grade selection operation is built directly into the assumed potential solution, requiring that the field is given by
\label{eqn:staticPotentials:120}

However, after imposing such a constraint, Maxwell’s equation has a much less friendly form
\label{eqn:staticPotentials:140}

Luckily, it is possible to introduce a transformation of potentials, called a gauge transformation, that eliminates the ugly grade selection term, and allows the potential equation to be expressed as a plain old Laplacian. We do so by assuming first that it is possible to find a solution of the Laplacian equation that has the desired grade restrictions. That is
\label{eqn:staticPotentials:160}
\begin{aligned}
\end{aligned}

for which $$F = \spacegrad A’$$ is a grade 1,2 solution to $$\spacegrad F = J$$. Suppose that $$A$$ is any formal solution, free of any grade restrictions, to $$\spacegrad^2 A = J$$, and $$F = \gpgrade{\spacegrad A}{1,2}$$. Can we find a function $$\tilde{A}$$ for which $$A = A’ + \tilde{A}$$?

Maxwell’s equation in terms of $$A$$ is
\label{eqn:staticPotentials:180}
\begin{aligned}
J
\end{aligned}

or
\label{eqn:staticPotentials:200}

This non-homogeneous Laplacian equation that can be solved as is for $$\tilde{A}$$ using the Green’s function for the Laplacian. Alternatively, we may also solve the equivalent first order system using the Green’s function for the gradient.
\label{eqn:staticPotentials:220}

Clearly $$\tilde{A}$$ is not unique, as we can add any function $$\psi$$ satisfying the homogeneous Laplacian equation $$\spacegrad^2 \psi = 0$$.

In summary, if $$A$$ is any multivector solution to $$\spacegrad^2 A = J$$, that is
\label{eqn:staticPotentials:260}
A(\Bx)
= \int dV’ G(\Bx, \Bx’) J(\Bx’)
= -\int dV’ \frac{J(\Bx’)}{\Norm{\Bx – \Bx’} },

then $$F = \spacegrad A’$$ is a solution to Maxwell’s equation, where $$A’ = A – \tilde{A}$$, and $$\tilde{A}$$ is a solution to the non-homogeneous Laplacian equation or the non-homogeneous gradient equation above.

### Integral form of the gauge transformation.

Additional insight is possible by considering the gauge transformation in integral form. Suppose that
\label{eqn:staticPotentials:280}
A(\Bx) = -\int_V dV’ \frac{J(\Bx’)}{\Norm{\Bx – \Bx’} } – \tilde{A}(\Bx),

is a solution of $$\spacegrad^2 A = J$$, where $$\tilde{A}$$ is a multivector solution to the homogeneous Laplacian equation $$\spacegrad^2 \tilde{A} = 0$$. Let’s look at the constraints on $$\tilde{A}$$ that must be imposed for $$F = \spacegrad A$$ to be a valid (i.e. grade 1,2) solution of Maxwell’s equation.
\label{eqn:staticPotentials:300}
\begin{aligned}
F
&=
-\int_V dV’ \lr{ \spacegrad \inv{\Norm{\Bx – \Bx’} } } J(\Bx’)
&=
\int_V dV’ \lr{ \spacegrad’ \inv{\Norm{\Bx – \Bx’} } } J(\Bx’)
&=
\int_V dV’ \spacegrad’ \frac{J(\Bx’)}{\Norm{\Bx – \Bx’} } – \int_V dV’ \frac{\spacegrad’ J(\Bx’)}{\Norm{\Bx – \Bx’} }
&=
\int_{\partial V} dA’ \ncap’ \frac{J(\Bx’)}{\Norm{\Bx – \Bx’} } – \int_V \frac{\spacegrad’ J(\Bx’)}{\Norm{\Bx – \Bx’} }
\end{aligned}

Where $$\ncap’ = (\Bx’ – \Bx)/\Norm{\Bx’ – \Bx}$$, and the fundamental theorem of geometric calculus has been used to transform the gradient volume integral into an integral over the bounding surface. Operating on Maxwell’s equation with the gradient gives $$\spacegrad^2 F = \spacegrad J$$, which has only grades 1,2 on the left hand side, meaning that $$J$$ is constrained in a way that requires $$\spacegrad J$$ to have only grades 1,2. This means that $$F$$ has grades 1,2 if
\label{eqn:staticPotentials:320}
= \int_{\partial V} dA’ \frac{ \gpgrade{\ncap’ J(\Bx’)}{0,3} }{\Norm{\Bx – \Bx’} }.

The product $$\ncap J$$ expands to
\label{eqn:staticPotentials:340}
\begin{aligned}
\ncap J
&=
&=
\ncap \cdot (-\eta \BJ) + \gpgradethree{\ncap (-I \BM)} \\
&=- \eta \ncap \cdot \BJ -I \ncap \cdot \BM,
\end{aligned}

so
\label{eqn:staticPotentials:360}
=
-\int_{\partial V} dA’ \frac{ \eta \ncap’ \cdot \BJ(\Bx’) + I \ncap’ \cdot \BM(\Bx’)}{\Norm{\Bx – \Bx’} }.

Observe that if there is no flux of current density $$\BJ$$ and (fictitious) magnetic current density $$\BM$$ through the surface, then $$F = \spacegrad A$$ is a solution to Maxwell’s equation without any gauge transformation. Alternatively $$F = \spacegrad A$$ is also a solution if $$\lim_{\Bx’ \rightarrow \infty} \BJ(\Bx’)/\Norm{\Bx – \Bx’} = \lim_{\Bx’ \rightarrow \infty} \BM(\Bx’)/\Norm{\Bx – \Bx’} = 0$$ and the bounding volume is taken to infinity.

# References

## Motivation

Geometric algebra (GA) allows for a compact description of Maxwell’s equations in either an explicit 3D representation or a STA (SpaceTime Algebra [2]) representation. The 3D GA and STA representations Maxwell’s equation both the form

\label{eqn:potentialMethods:1280}
L \boldsymbol{\mathcal{F}} = J,

where $$J$$ represents the sources, $$L$$ is a multivector gradient operator that includes partial derivative operator components for each of the space and time coordinates, and

\label{eqn:potentialMethods:1020}
\boldsymbol{\mathcal{F}} = \boldsymbol{\mathcal{E}} + \eta I \boldsymbol{\mathcal{H}},

is an electromagnetic field multivector, $$I = \Be_1 \Be_2 \Be_3$$ is the \R{3} pseudoscalar, and $$\eta = \sqrt{\mu/\epsilon}$$ is the impedance of the media.

When Maxwell’s equations are extended to include magnetic sources in addition to conventional electric sources (as used in antenna-theory [1] and microwave engineering [3]), they take the form

\label{eqn:chapter3Notes:20}
\spacegrad \cross \boldsymbol{\mathcal{E}} = – \boldsymbol{\mathcal{M}} – \PD{t}{\boldsymbol{\mathcal{B}}}

\label{eqn:chapter3Notes:40}
\spacegrad \cross \boldsymbol{\mathcal{H}} = \boldsymbol{\mathcal{J}} + \PD{t}{\boldsymbol{\mathcal{D}}}

\label{eqn:chapter3Notes:60}

\label{eqn:chapter3Notes:80}

The corresponding GA Maxwell equations in their respective 3D and STA forms are

\label{eqn:potentialMethods:300}
\lr{ \spacegrad + \inv{v} \PD{t}{} } \boldsymbol{\mathcal{F}}
=
\eta
\lr{ v q_{\textrm{e}} – \boldsymbol{\mathcal{J}} }
+ I \lr{ v q_{\textrm{m}} – \boldsymbol{\mathcal{M}} }

\label{eqn:potentialMethods:320}
\grad \boldsymbol{\mathcal{F}} = \eta J – I M,

where the wave group velocity in the medium is $$v = 1/\sqrt{\epsilon\mu}$$, and the medium is isotropic with
$$\boldsymbol{\mathcal{B}} = \mu \boldsymbol{\mathcal{H}}$$, and $$\boldsymbol{\mathcal{D}} = \epsilon \boldsymbol{\mathcal{E}}$$. In the STA representation, $$\grad, J, M$$ are all four-vectors, the specific meanings of which will be spelled out below.

How to determine the potential equations and the field representation using the conventional distinct Maxwell’s \ref{eqn:chapter3Notes:20}, … is well known. The basic procedure is to consider the electric and magnetic sources in turn, and observe that in each case one of the electric or magnetic fields must have a curl representation. The STA approach is similar, except that it can be observed that the field must have a four-curl representation for each type of source. In the explicit 3D GA formalism
\ref{eqn:potentialMethods:300} how to formulate a natural potential representation is not as obvious. There is no longer an reason to set any component of the field equal to a curl, and the representation of the four curl from the STA approach is awkward. Additionally, it is not obvious what form gauge invariance takes in the 3D GA representation.

### Ideas explored in these notes

• GA representation of Maxwell’s equations including magnetic sources.
• STA GA formalism for Maxwell’s equations including magnetic sources.
• Explicit form of the GA potential representation including both electric and magnetic sources.
• Demonstration of exactly how the 3D and STA potentials are related.
• Explore the structure of gauge transformations when magnetic sources are included.
• Explore the structure of gauge transformations in the 3D GA formalism.
• Specify the form of the Lorentz gauge in the 3D GA formalism.

### No magnetic sources

When magnetic sources are omitted, it follows from \ref{eqn:chapter3Notes:80} that there is some $$\boldsymbol{\mathcal{A}}^{\mathrm{e}}$$ for which

\label{eqn:potentialMethods:20}
\boxed{
}

Substitution into Faraday’s law \ref{eqn:chapter3Notes:20} gives

\label{eqn:potentialMethods:40}

or
\label{eqn:potentialMethods:60}
\spacegrad \cross \lr{ \boldsymbol{\mathcal{E}} + \PD{t}{ \boldsymbol{\mathcal{A}}^{\mathrm{e}} } } = 0.

A gradient representation of this curled quantity, say $$-\spacegrad \phi$$, will provide the required zero

\label{eqn:potentialMethods:80}
\boxed{
\boldsymbol{\mathcal{E}} = -\spacegrad \phi -\PD{t}{ \boldsymbol{\mathcal{A}}^{\mathrm{e}} }.
}

The final two Maxwell equations yield

\label{eqn:potentialMethods:100}
\begin{aligned}
-\spacegrad^2 \boldsymbol{\mathcal{A}}^{\mathrm{e}} + \spacegrad \lr{ \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{e}} } &= \mu \lr{ \boldsymbol{\mathcal{J}} + \epsilon \PD{t}{} \lr{ -\spacegrad \phi -\PD{t}{ \boldsymbol{\mathcal{A}}^{\mathrm{e}} } } } \\
\end{aligned}

or
\label{eqn:potentialMethods:120}
\boxed{
\begin{aligned}
\spacegrad^2 \boldsymbol{\mathcal{A}}^{\mathrm{e}} – \inv{v^2} \PDSq{t}{ \boldsymbol{\mathcal{A}}^{\mathrm{e}} }
\inv{v^2} \PD{t}{\phi}
}
&= -\mu \boldsymbol{\mathcal{J}} \\
\end{aligned}
}

Note that the Lorentz condition $$\PDi{t}{(\phi/v^2)} + \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{e}} = 0$$ can be imposed to decouple these, leaving non-homogeneous wave equations for the vector and scalar potentials respectively.

### No electric sources

Without electric sources, a curl representation of the electric field can be assumed, satisfying Gauss’s law

\label{eqn:potentialMethods:140}
\boxed{
\boldsymbol{\mathcal{D}} = – \spacegrad \cross \boldsymbol{\mathcal{A}}^{\mathrm{m}}.
}

Substitution into the Maxwell-Faraday law gives
\label{eqn:potentialMethods:160}
\spacegrad \cross \lr{ \boldsymbol{\mathcal{H}} + \PD{t}{\boldsymbol{\mathcal{A}}^{\mathrm{m}}} } = 0.

This is satisfied with any gradient, say, $$-\spacegrad \phi_m$$, providing a potential representation for the magnetic field

\label{eqn:potentialMethods:180}
\boxed{
\boldsymbol{\mathcal{H}} = -\spacegrad \phi_m – \PD{t}{\boldsymbol{\mathcal{A}}^{\mathrm{m}}}.
}

The remaining Maxwell equations provide the required constraints on the potentials

\label{eqn:potentialMethods:220}
\lr{
-\boldsymbol{\mathcal{M}} – \mu \PD{t}{}
\lr{
}
}

\label{eqn:potentialMethods:240}
\lr{
}
= \inv{\mu} q_m,

or
\label{eqn:potentialMethods:260}
\boxed{
\begin{aligned}
\spacegrad^2 \boldsymbol{\mathcal{A}}^{\mathrm{m}} – \inv{v^2} \PDSq{t}{\boldsymbol{\mathcal{A}}^{\mathrm{m}}} – \spacegrad \lr{ \inv{v^2} \PD{t}{\phi_m} + \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{m}} } &= -\epsilon \boldsymbol{\mathcal{M}} \\
\end{aligned}
}

The general solution to Maxwell’s equations is therefore
\label{eqn:potentialMethods:280}
\begin{aligned}
\boldsymbol{\mathcal{E}} &=
– \inv{\epsilon} \spacegrad \cross \boldsymbol{\mathcal{A}}^{\mathrm{m}} \\
\boldsymbol{\mathcal{H}} &=
\end{aligned}

subject to the constraints \ref{eqn:potentialMethods:120} and \ref{eqn:potentialMethods:260}.

### Potential operator structure

Knowing that there is a simple underlying structure to the potential representation of the electromagnetic field in the STA formalism inspires the question of whether that structure can be found directly using the scalar and vector potentials determined above.

Specifically, what is the multivector representation \ref{eqn:potentialMethods:1020} of the electromagnetic field in terms of all the individual potential variables, and can an underlying structure for that field representation be found? The composite field is

\label{eqn:potentialMethods:280b}
\boldsymbol{\mathcal{F}}
=
– \inv{\epsilon} \spacegrad \cross \boldsymbol{\mathcal{A}}^{\mathrm{m}} \\
+ I \eta
\lr{
}.

Can this be factored into into multivector operator and multivector potentials? Expanding the cross products provides some direction

\label{eqn:potentialMethods:1040}
\begin{aligned}
\boldsymbol{\mathcal{F}}
&=
– \PD{t}{ \boldsymbol{\mathcal{A}}^{\mathrm{e}} }
– \eta \PD{t}{I \boldsymbol{\mathcal{A}}^{\mathrm{m}}}
– \spacegrad \lr{ \phi – \eta I \phi_m } \\
+ \frac{1}{2 \epsilon} \lr{ \rspacegrad I \boldsymbol{\mathcal{A}}^{\mathrm{m}} – I \boldsymbol{\mathcal{A}}^{\mathrm{m}} \lspacegrad }.
\end{aligned}

Observe that the
gradient and the time partials can be grouped together

\label{eqn:potentialMethods:1060}
\begin{aligned}
\boldsymbol{\mathcal{F}}
&=
– \PD{t}{ } \lr{\boldsymbol{\mathcal{A}}^{\mathrm{e}} + \eta I \boldsymbol{\mathcal{A}}^{\mathrm{m}}}
– \spacegrad \lr{ \phi + \eta I \phi_m }
+ \frac{v}{2} \lr{ \rspacegrad (\boldsymbol{\mathcal{A}}^{\mathrm{e}} + I \eta \boldsymbol{\mathcal{A}}^{\mathrm{m}}) – (\boldsymbol{\mathcal{A}}^{\mathrm{e}} + I \eta \boldsymbol{\mathcal{A}}^{\mathrm{m}}) \lspacegrad } \\
&=
\inv{2} \lr{
\lr{ \rspacegrad – \inv{v} {\stackrel{ \rightarrow }{\partial_t}} } \lr{ v \boldsymbol{\mathcal{A}}^{\mathrm{e}} + \eta v I \boldsymbol{\mathcal{A}}^{\mathrm{m}} }

\lr{ v \boldsymbol{\mathcal{A}}^{\mathrm{e}} + \eta v I \boldsymbol{\mathcal{A}}^{\mathrm{m}}} \lr{ \lspacegrad + \inv{v} {\stackrel{ \leftarrow }{\partial_t}} }
} \\
\lr{ \rspacegrad – \inv{v} {\stackrel{ \rightarrow }{\partial_t}} } \lr{ -\phi – \eta I \phi_m }
– \lr{ \phi + \eta I \phi_m } \lr{ \lspacegrad + \inv{v} {\stackrel{ \leftarrow }{\partial_t}} }
}
,
\end{aligned}

or

\label{eqn:potentialMethods:1080}
\boxed{
\boldsymbol{\mathcal{F}}
=
\inv{2} \Biglr{
\lr{ \rspacegrad – \inv{v} {\stackrel{ \rightarrow }{\partial_t}} }
\lr{
– \phi
+ v \boldsymbol{\mathcal{A}}^{\mathrm{e}}
+ \eta I v \boldsymbol{\mathcal{A}}^{\mathrm{m}}
– \eta I \phi_m
}

\lr{
\phi
+ v \boldsymbol{\mathcal{A}}^{\mathrm{e}}
+ \eta I v \boldsymbol{\mathcal{A}}^{\mathrm{m}}
+ \eta I \phi_m
}
\lr{ \lspacegrad + \inv{v} {\stackrel{ \leftarrow }{\partial_t}} }
}
.
}

There’s a conjugate structure to the potential on each side of the curl operation where we see a sign change for the scalar and pseudoscalar elements only. The reason for this becomes more clear in the STA formalism.

## Potentials in the STA formalism.

Maxwell’s equation in its explicit 3D form \ref{eqn:potentialMethods:300} can be
converted to STA form, by introducing a four-vector basis $$\setlr{ \gamma_\mu }$$, where the spatial basis
$$\setlr{ \Be_k = \gamma_k \gamma_0 }$$
is expressed in terms of the Dirac basis $$\setlr{ \gamma_\mu }$$.
By multiplying from the left with $$\gamma_0$$ a STA form of Maxwell’s equation
\ref{eqn:potentialMethods:320}
is obtained,
where
\label{eqn:potentialMethods:340}
\begin{aligned}
J &= \gamma^\mu J_\mu = ( v q_e, \boldsymbol{\mathcal{J}} ) \\
M &= \gamma^\mu M_\mu = ( v q_m, \boldsymbol{\mathcal{M}} ) \\
I &= \gamma_0 \gamma_1 \gamma_2 \gamma_3,
\end{aligned}

Here the metric choice is $$\gamma_0^2 = 1 = -\gamma_k^2$$. Note that in this representation the electromagnetic field $$\boldsymbol{\mathcal{F}} = \boldsymbol{\mathcal{E}} + \eta I \boldsymbol{\mathcal{H}}$$ is a bivector, not a multivector as it is explicit (frame dependent) 3D representation of \ref{eqn:potentialMethods:300}.

A potential representation can be obtained as before by considering electric and magnetic sources in sequence and using superposition to assemble a complete potential.

### No magnetic sources

Without magnetic sources, Maxwell’s equation splits into vector and trivector terms of the form

\label{eqn:potentialMethods:380}
\grad \cdot \boldsymbol{\mathcal{F}} = \eta J

\label{eqn:potentialMethods:400}

A four-vector curl representation of the field will satisfy \ref{eqn:potentialMethods:400} allowing an immediate potential solution

\label{eqn:potentialMethods:560}
\boxed{
\begin{aligned}
&\boldsymbol{\mathcal{F}} = \grad \wedge {A^{\mathrm{e}}} \\
\end{aligned}
}

This can be put into correspondence with \ref{eqn:potentialMethods:120} by noting that

\label{eqn:potentialMethods:460}
\begin{aligned}
\grad^2 &= (\gamma^\mu \partial_\mu) \cdot (\gamma^\nu \partial_\nu) = \inv{v^2} \partial_{tt} – \spacegrad^2 \\
\gamma_0 {A^{\mathrm{e}}} &= \gamma_0 \gamma^\mu {A^{\mathrm{e}}}_\mu = {A^{\mathrm{e}}}_0 + \Be_k {A^{\mathrm{e}}}_k = {A^{\mathrm{e}}}_0 + \BA^{\mathrm{e}} \\
\gamma_0 \grad &= \gamma_0 \gamma^\mu \partial_\mu = \inv{v} \partial_t + \spacegrad \\
\grad \cdot {A^{\mathrm{e}}} &= \partial_\mu {A^{\mathrm{e}}}^\mu = \inv{v} \partial_t {A^{\mathrm{e}}}_0 – \spacegrad \cdot \BA^{\mathrm{e}},
\end{aligned}

so multiplying from the left with $$\gamma_0$$ gives

\label{eqn:potentialMethods:480}
\lr{ \inv{v^2} \partial_{tt} – \spacegrad^2 } \lr{ {A^{\mathrm{e}}}_0 + \BA^{\mathrm{e}} } – \lr{ \inv{v} \partial_t + \spacegrad }\lr{ \inv{v} \partial_t {A^{\mathrm{e}}}_0 – \spacegrad \cdot \BA^{\mathrm{e}} } = \eta( v q_e – \boldsymbol{\mathcal{J}} ),

or

\label{eqn:potentialMethods:520}
\lr{ \inv{v^2} \partial_{tt} – \spacegrad^2 } \BA^{\mathrm{e}} – \spacegrad \lr{ \inv{v} \partial_t {A^{\mathrm{e}}}_0 – \spacegrad \cdot \BA^{\mathrm{e}} } = -\eta \boldsymbol{\mathcal{J}}

\label{eqn:potentialMethods:540}
\spacegrad^2 {A^{\mathrm{e}}}_0 – \inv{v} \partial_t \lr{ \spacegrad \cdot \BA^{\mathrm{e}} } = -q_e/\epsilon.

So $${A^{\mathrm{e}}}_0 = \phi$$ and $$-\ifrac{\BA^{\mathrm{e}}}{v} = \boldsymbol{\mathcal{A}}^{\mathrm{e}}$$, or

\label{eqn:potentialMethods:600}
\boxed{
{A^{\mathrm{e}}} = \gamma_0\lr{ \phi – v \boldsymbol{\mathcal{A}}^{\mathrm{e}} }.
}

### No electric sources

Without electric sources, Maxwell’s equation now splits into

\label{eqn:potentialMethods:640}

\label{eqn:potentialMethods:660}
\grad \wedge \boldsymbol{\mathcal{F}} = -I M.

Here the dual of an STA curl yields a solution

\label{eqn:potentialMethods:680}
\boxed{
\boldsymbol{\mathcal{F}} = I ( \grad \wedge {A^{\mathrm{m}}} ).
}

Substituting this gives

\label{eqn:potentialMethods:720}
\begin{aligned}
0
&=
&=
&=
\end{aligned}

\label{eqn:potentialMethods:740}
\begin{aligned}
-I M
&=
&=
&=
\end{aligned}

The $$\grad \cdot \boldsymbol{\mathcal{F}}$$ relation \ref{eqn:potentialMethods:720} is identically zero as desired, leaving

\label{eqn:potentialMethods:760}
\boxed{
=
M.
}

So the general solution with both electric and magnetic sources is

\label{eqn:potentialMethods:800}
\boxed{
}

subject to the constraints of \ref{eqn:potentialMethods:560} and \ref{eqn:potentialMethods:760}. As before the four-potential $${A^{\mathrm{m}}}$$ can be put into correspondence with the conventional scalar and vector potentials by left multiplying with $$\gamma_0$$, which gives

\label{eqn:potentialMethods:820}
\lr{ \inv{v^2} \partial_{tt} – \spacegrad^2 } \lr{ {A^{\mathrm{m}}}_0 + \BA^{\mathrm{m}} } – \lr{ \inv{v} \partial_t + \spacegrad }\lr{ \inv{v} \partial_t {A^{\mathrm{m}}}_0 – \spacegrad \cdot \BA^{\mathrm{m}} } = v q_m – \boldsymbol{\mathcal{M}},

or
\label{eqn:potentialMethods:860}
\lr{ \inv{v^2} \partial_{tt} – \spacegrad^2 } \BA^{\mathrm{m}} – \spacegrad \lr{ \inv{v} \partial_t {A^{\mathrm{m}}}_0 – \spacegrad \cdot \BA^{\mathrm{m}} } = – \boldsymbol{\mathcal{M}}

\label{eqn:potentialMethods:880}

Comparing with \ref{eqn:potentialMethods:260} shows that $${A^{\mathrm{m}}}_0/v = \mu \phi_m$$ and $$-\ifrac{\BA^{\mathrm{m}}}{v^2} = \mu \boldsymbol{\mathcal{A}}^{\mathrm{m}}$$, or

\label{eqn:potentialMethods:900}
\boxed{
{A^{\mathrm{m}}} = \gamma_0 \eta \lr{ \phi_m – v \boldsymbol{\mathcal{A}}^{\mathrm{m}} }.
}

### Potential operator structure

Observe that there is an underlying uniform structure of the differential operator that acts on the potential to produce the electromagnetic field. Expressed as a linear operator of the
gradient and the potentials, that is

$$\boldsymbol{\mathcal{F}} = L(\lrgrad, {A^{\mathrm{e}}}, {A^{\mathrm{m}}})$$

\label{eqn:potentialMethods:980}
\begin{aligned}
\boldsymbol{\mathcal{F}}
&=
&=
&=
+ \frac{1}{2} \lr{ -\rgrad I {A^{\mathrm{m}}} – I {A^{\mathrm{m}}} \lgrad } \\
&=
\inv{2} \lr{ \rgrad ({A^{\mathrm{e}}} -I {A^{\mathrm{m}}}) – ({A^{\mathrm{e}}} + I {A^{\mathrm{m}}}) \lgrad }
,
\end{aligned}

or
\label{eqn:potentialMethods:1000}
\boxed{
\boldsymbol{\mathcal{F}}
=
\inv{2} \lr{ \rgrad ({A^{\mathrm{e}}} -I {A^{\mathrm{m}}}) – ({A^{\mathrm{e}}} – I {A^{\mathrm{m}}})^\dagger \lgrad }
.
}

Observe that \ref{eqn:potentialMethods:1000} can be
put into correspondence with \ref{eqn:potentialMethods:1080} using a factoring of unity $$1 = \gamma_0 \gamma_0$$

\label{eqn:potentialMethods:1100}
\boldsymbol{\mathcal{F}}
=
\inv{2} \lr{ (-\rgrad \gamma_0) (-\gamma_0 ({A^{\mathrm{e}}} -I {A^{\mathrm{m}}})) – (({A^{\mathrm{e}}} + I {A^{\mathrm{m}}}) \gamma_0)(\gamma_0 \lgrad) },

where

\label{eqn:potentialMethods:1140}
\begin{aligned}
&=
-(\gamma^0 \partial_0 + \gamma^k \partial_k) \gamma_0 \\
&=
-\partial_0 – \gamma^k \gamma_0 \partial_k \\
&=
-\inv{v} \partial_t
,
\end{aligned}

\label{eqn:potentialMethods:1160}
\begin{aligned}
&=
\gamma_0 (\gamma^0 \partial_0 + \gamma^k \partial_k) \\
&=
\partial_0 – \gamma^k \gamma_0 \partial_k \\
&=
+ \inv{v} \partial_t
,
\end{aligned}

and
\label{eqn:potentialMethods:1200}
\begin{aligned}
-\gamma_0 ( {A^{\mathrm{e}}} – I {A^{\mathrm{m}}} )
&=
-\gamma_0 \gamma_0 \lr{ \phi -v \boldsymbol{\mathcal{A}}^{\mathrm{e}} + \eta I \lr{ \phi_m – v \boldsymbol{\mathcal{A}}^{\mathrm{m}} } } \\
&=
-\lr{ \phi -v \boldsymbol{\mathcal{A}}^{\mathrm{e}} + \eta I \phi_m – \eta v I \boldsymbol{\mathcal{A}}^{\mathrm{m}} } \\
&=
– \phi
+ v \boldsymbol{\mathcal{A}}^{\mathrm{e}}
+ \eta v I \boldsymbol{\mathcal{A}}^{\mathrm{m}}
– \eta I \phi_m
\end{aligned}

\label{eqn:potentialMethods:1220}
\begin{aligned}
( {A^{\mathrm{e}}} + I {A^{\mathrm{m}}} )\gamma_0
&=
\lr{ \gamma_0 \lr{ \phi -v \boldsymbol{\mathcal{A}}^{\mathrm{e}} } + I \gamma_0 \eta \lr{ \phi_m – v \boldsymbol{\mathcal{A}}^{\mathrm{m}} } } \gamma_0 \\
&=
\phi + v \boldsymbol{\mathcal{A}}^{\mathrm{e}} + I \eta \phi_m + I \eta v \boldsymbol{\mathcal{A}}^{\mathrm{m}} \\
&=
\phi
+ v \boldsymbol{\mathcal{A}}^{\mathrm{e}}
+ \eta v I \boldsymbol{\mathcal{A}}^{\mathrm{m}}
+ \eta I \phi_m
,
\end{aligned}

This recovers \ref{eqn:potentialMethods:1080} as desired.

## Potentials in the 3D Euclidean formalism

In the conventional scalar plus vector differential representation of Maxwell’s equations \ref{eqn:chapter3Notes:20}…, given electric(magnetic) sources the structure of the electric(magnetic) potential follows from first setting the magnetic(electric) field equal to the curl of a vector potential. The procedure for the STA GA form of Maxwell’s equation was similar, where it was immediately evident that the field could be set to the four-curl of a four-vector potential (or the dual of such a curl for magnetic sources).

In the 3D GA representation, there is no immediate rationale for introducing a curl or the equivalent to a four-curl representation of the field. Reconciliation of this is possible by recognizing that the fact that the field (or a component of it) may be represented by a curl is not actually fundamental. Instead, observe that the two sided gradient action on a potential to generate the electromagnetic field in the STA representation of \ref{eqn:potentialMethods:1000} serves to select the grade two component product of the gradient and the multivector potential $${A^{\mathrm{e}}} – I {A^{\mathrm{m}}}$$, and that this can in fact be written as
a single sided gradient operation on a potential, provided the multivector product is filtered with a four-bivector grade selection operation

\label{eqn:potentialMethods:1240}
\boxed{
}

Similarly, it can be observed that the
specific function of the conjugate structure in the two sided potential representation of
\ref{eqn:potentialMethods:1080}
is to discard all the scalar and pseudoscalar grades in the multivector product. This means that a single sided potential can also be used, provided it is wrapped in a grade selection operation

\label{eqn:potentialMethods:1260}
\boxed{
\boldsymbol{\mathcal{F}} =
\lr{
– \phi
+ v \boldsymbol{\mathcal{A}}^{\mathrm{e}}
+ \eta I v \boldsymbol{\mathcal{A}}^{\mathrm{m}}
– \eta I \phi_m
} }{1,2}.
}

It is this grade selection operation that is really the fundamental defining action in the potential of the STA and conventional 3D representations of Maxwell’s equations. So, given Maxwell’s equation in the 3D GA representation, defining a potential representation for the field is really just a demand that the field have the structure

\label{eqn:potentialMethods:1320}
\boldsymbol{\mathcal{F}} = \gpgrade{ (\alpha \spacegrad + \beta \partial_t)( A_0 + A_1 + I( A_0′ + A_1′ ) }{1,2}.

This is a mandate that the electromagnetic field is the grades 1 and 2 components of the vector product of space and time derivative operators on a multivector field $$A = \sum_{k=0}^3 A_k = A_0 + A_1 + I( A_0′ + A_1′ )$$ that can potentially have any grade components. There are more degrees of freedom in this specification than required, since the multivector can absorb one of the $$\alpha$$ or $$\beta$$ coefficients, so without loss of generality, one of these (say $$\alpha$$) can be set to 1.

Expanding \ref{eqn:potentialMethods:1320} gives

\label{eqn:potentialMethods:1340}
\begin{aligned}
\boldsymbol{\mathcal{F}}
&=
+ \beta \partial_t A_1
+ \beta \partial_t A_1′
&=
\boldsymbol{\mathcal{E}} + I \eta \boldsymbol{\mathcal{H}}.
\end{aligned}

This naturally has all the right mixes of curls, gradients and time derivatives, all following as direct consequences of applying a grade selection operation to the action of a “spacetime gradient” on a general multivector potential.

The conclusion is that the potential representation of the field is

\label{eqn:potentialMethods:1360}
\boldsymbol{\mathcal{F}} =

where $$A$$ is a multivector potentially containing all grades, where grades 0,1 are required for electric sources, and grades 2,3 are required for magnetic sources. When it is desirable to refer back to the conventional scalar and vector potentials this multivector potential can be written as $$A = -\phi + v \boldsymbol{\mathcal{A}}^{\mathrm{e}} + \eta I \lr{ -\phi_m + v \boldsymbol{\mathcal{A}}^{\mathrm{m}} }$$.

## Gauge transformations

Recall that for electric sources the magnetic field is of the form

\label{eqn:potentialMethods:1380}

so adding the gradient of any scalar field to the potential $$\boldsymbol{\mathcal{A}}’ = \boldsymbol{\mathcal{A}} + \spacegrad \psi$$
does not change the magnetic field

\label{eqn:potentialMethods:1400}
\begin{aligned}
\boldsymbol{\mathcal{B}}’
&= \boldsymbol{\mathcal{B}}.
\end{aligned}

The electric field with this changed potential is

\label{eqn:potentialMethods:1420}
\begin{aligned}
\boldsymbol{\mathcal{E}}’
&= -\spacegrad \lr{ \phi + \partial_t \psi } – \partial_t \BA,
\end{aligned}

so if
\label{eqn:potentialMethods:1440}
\phi = \phi’ – \partial_t \psi,

the electric field will also be unaltered by this transformation.

In the STA representation, the field can similarly be altered by adding any (four)gradient to the potential. For example with only electric sources

\label{eqn:potentialMethods:1460}

and for electric or magnetic sources

\label{eqn:potentialMethods:1480}

In the 3D GA representation, where the field is given by \ref{eqn:potentialMethods:1360}, there is no field that is being curled to add a gradient to. However, if the scalar and vector potentials transform as

\label{eqn:potentialMethods:1500}
\begin{aligned}
\boldsymbol{\mathcal{A}} &\rightarrow \boldsymbol{\mathcal{A}} + \spacegrad \psi \\
\phi &\rightarrow \phi – \partial_t \psi,
\end{aligned}

then the multivector potential transforms as
\label{eqn:potentialMethods:1520}
-\phi + v \boldsymbol{\mathcal{A}}
\rightarrow -\phi + v \boldsymbol{\mathcal{A}} + \partial_t \psi + v \spacegrad \psi,

so the electromagnetic field is unchanged when the multivector potential is transformed as

\label{eqn:potentialMethods:1540}
A \rightarrow A + \lr{ \spacegrad + \inv{v} \partial_t } \psi,

where $$\psi$$ is any field that has scalar or pseudoscalar grades. Viewed in terms of grade selection, this makes perfect sense, since the transformed field is

\label{eqn:potentialMethods:1560}
\begin{aligned}
\boldsymbol{\mathcal{F}}
&\rightarrow
\gpgrade{ \lr{ \spacegrad – \inv{v} \PD{t}{} } \lr{ A + \lr{ \spacegrad + \inv{v} \partial_t } \psi } }{1,2} \\
&=
\gpgrade{ \lr{ \spacegrad – \inv{v} \PD{t}{} } A + \lr{ \spacegrad^2 – \inv{v^2} \partial_{tt} } \psi }{1,2} \\
&=
\end{aligned}

The $$\psi$$ contribution to the grade selection operator is killed because it has scalar or pseudoscalar grades.

## Lorenz gauge

Maxwell’s equations are completely decoupled if the potential can be found such that

\label{eqn:potentialMethods:1580}
\begin{aligned}
\boldsymbol{\mathcal{F}}
&=
&=
\lr{ \spacegrad – \inv{v} \PD{t}{} } A.
\end{aligned}

When this is the case, Maxwell’s equations are reduced to four non-homogeneous potential wave equations

\label{eqn:potentialMethods:1620}
\lr{ \spacegrad^2 – \inv{v^2} \PDSq{t}{} } A = J,

that is

\label{eqn:potentialMethods:1600}
\begin{aligned}
\lr{ \spacegrad^2 – \inv{v^2} \PDSq{t}{} } \phi &= – \inv{\epsilon} q_e \\
\lr{ \spacegrad^2 – \inv{v^2} \PDSq{t}{} } \boldsymbol{\mathcal{A}}^{\mathrm{e}} &= – \mu \boldsymbol{\mathcal{J}} \\
\lr{ \spacegrad^2 – \inv{v^2} \PDSq{t}{} } \phi_m &= – \frac{I}{\mu} q_m \\
\lr{ \spacegrad^2 – \inv{v^2} \PDSq{t}{} } \boldsymbol{\mathcal{A}}^{\mathrm{m}} &= – I \epsilon \boldsymbol{\mathcal{M}}.
\end{aligned}

There should be no a-priori assumption that such a field representation has no scalar, nor no pseudoscalar components. That explicit expansion in grades is

\label{eqn:potentialMethods:1640}
\begin{aligned}
\lr{ \spacegrad – \inv{v} \PD{t}{} } A
&=
\lr{ \spacegrad – \inv{v} \PD{t}{} } \lr{ -\phi + v \boldsymbol{\mathcal{A}}^{\mathrm{e}} + \eta I \lr{ -\phi_m + v \boldsymbol{\mathcal{A}}^{\mathrm{m}} } } \\
&=
\inv{v} \partial_t \phi
+ v \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{e}} \\
+ I \eta v \spacegrad \wedge \boldsymbol{\mathcal{A}}^{\mathrm{m}}
– \partial_t \boldsymbol{\mathcal{A}}^{\mathrm{e}} \\
– I \eta \partial_t \boldsymbol{\mathcal{A}}^{\mathrm{m}} \\
&+ \eta I \inv{v} \partial_t \phi_m
+ I \eta v \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{m}},
\end{aligned}

so if this potential representation has only vector and bivector grades, it must be true that

\label{eqn:potentialMethods:1660}
\begin{aligned}
\inv{v} \partial_t \phi + v \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{e}} &= 0 \\
\inv{v} \partial_t \phi_m + v \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{m}} &= 0.
\end{aligned}

The first is the well known Lorenz gauge condition, whereas the second is the dual of that condition for magnetic sources.

Should one of these conditions, say the Lorenz condition for the electric source potentials, be non-zero, then it is possible to make a potential transformation for which this condition is zero

\label{eqn:potentialMethods:1680}
\begin{aligned}
0
&\ne
\inv{v} \partial_t \phi + v \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{e}} \\
&=
\inv{v} \partial_t (\phi’ – \partial_t \psi) + v \spacegrad \cdot (\boldsymbol{\mathcal{A}}’ + \spacegrad \psi) \\
&=
\inv{v} \partial_t \phi’ + v \spacegrad \boldsymbol{\mathcal{A}}’
+ v \lr{ \spacegrad^2 – \inv{v^2} \partial_{tt} } \psi,
\end{aligned}

so if $$\inv{v} \partial_t \phi’ + v \spacegrad \boldsymbol{\mathcal{A}}’$$ is zero, $$\psi$$ must be found such that
\label{eqn:potentialMethods:1700}
\inv{v} \partial_t \phi + v \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{e}}
= v \lr{ \spacegrad^2 – \inv{v^2} \partial_{tt} } \psi.

# References

[1] Constantine A Balanis. Antenna theory: analysis and design. John Wiley \& Sons, 3rd edition, 2005.

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

[3] David M Pozar. Microwave engineering. John Wiley \& Sons, 2009.

## Vector wave equation in spherical coordinates

For a vector $$\BA$$ in spherical coordinates, let’s compute the Laplacian

\label{eqn:vectorWaveEquationSpherical:20}

to see the form of the wave equation. The spherical vector representation has a curvilinear basis
\label{eqn:vectorWaveEquationSpherical:40}
\BA = \rcap A_r + \thetacap A_\theta + \phicap A_\phi,

and the spherical Laplacian has been found to have the representation

\label{eqn:vectorWaveEquationSpherical:60}
=
\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^2\theta} \PDSq{\phi}{ \psi}.

Evaluating the Laplacian will require the following curvilinear basis derivatives

\label{eqn:vectorWaveEquationSpherical:80}
\begin{aligned}
\partial_\theta \rcap &= \thetacap \\
\partial_\theta \thetacap &= -\rcap \\
\partial_\theta \phicap &= 0 \\
\partial_\phi \rcap &= S_\theta \phicap \\
\partial_\phi \thetacap &= C_\theta \phicap \\
\partial_\phi \phicap &= -\rcap S_\theta – \thetacap C_\theta.
\end{aligned}

We’ll need to evaluate a number of derivatives. Starting with the $$\rcap$$ components

\label{eqn:vectorWaveEquationSpherical:120}
\partial_r \lr{ r^2 \partial_r \lr{ \rcap \psi} }
=
\rcap \partial_r \lr{ r^2 \partial_r \psi }

\label{eqn:vectorWaveEquationSpherical:140}
\begin{aligned}
\partial_\theta \lr{ S_\theta \partial_\theta \lr{ \rcap \psi } }
&=
\partial_\theta \lr{ S_\theta (\thetacap \psi + \rcap \partial_\theta \psi ) } \\
&=
C_\theta (\thetacap \psi + \rcap \partial_\theta \psi )
+ S_\theta \partial_\theta (\thetacap \psi + \rcap \partial_\theta \psi ) \\
&=
C_\theta (\thetacap \psi + \rcap \partial_\theta \psi )
+ S_\theta \partial_\theta ((\partial_\theta \thetacap) \psi + (\partial_\theta \rcap) \partial_\theta \psi )
+ S_\theta \partial_\theta (\thetacap \partial_\theta \psi + \rcap \partial_{\theta \theta} \psi ) \\
&=
C_\theta (\thetacap \psi + \rcap \partial_\theta \psi )
+ S_\theta ((-\rcap) \psi + (\thetacap) \partial_\theta \psi )
+ S_\theta (\thetacap \partial_\theta \psi + \rcap \partial_{\theta \theta} \psi ) \\
&=
\rcap \lr{
C_\theta \partial_\theta \psi
– S_\theta \psi
+ S_\theta \partial_{\theta \theta} \psi
}
+\thetacap \lr{
C_\theta \psi
+ 2 S_\theta \partial_\theta \psi
}
\end{aligned}

\label{eqn:vectorWaveEquationSpherical:160}
\begin{aligned}
\partial_{\phi \phi} \lr{ \rcap \psi}
&=
\partial_\phi \lr{ (\partial_\phi \rcap) \psi + \rcap \partial_\phi \psi } \\
&=
\partial_\phi \lr{ (S_\theta \phicap) \psi + \rcap \partial_\phi \psi } \\
&=
S_\theta \partial_\phi (\phicap \psi)
+ \partial_\phi \lr{ \rcap \partial_\phi \psi } \\
&=
S_\theta (\partial_\phi \phicap) \psi
+ S_\theta \phicap \partial_\phi \psi
+ (\partial_\phi \rcap) \partial_\phi \psi
+ \rcap \partial_{\phi\phi} \psi \\
&=
S_\theta (-S_\theta \rcap – C_\theta \thetacap) \psi
+ S_\theta \phicap \partial_\phi \psi
+ (S_\theta \phicap) \partial_\phi \psi
+ \rcap \partial_{\phi\phi} \psi \\
&=
\rcap \lr{
– S_\theta^2 \psi
+ \partial_{\phi\phi} \psi
}
+
\thetacap \lr{
– S_\theta C_\theta \psi
}
+
\phicap \lr{
2 S_\theta \phicap \partial_\phi \psi
}
\end{aligned}

This gives

\label{eqn:vectorWaveEquationSpherical:180}
\begin{aligned}
&=
\rcap \lr{
\inv{r^2}
\partial_r \lr{ r^2 \partial_r A_r }
+
\inv{r^2 S_\theta}
\lr{
C_\theta \partial_\theta A_r
– S_\theta A_r
+ S_\theta \partial_{\theta \theta} A_r
}
+ \inv{r^2 S_\theta^2}
\lr{
– S_\theta^2 A_r
+ \partial_{\phi\phi} A_r
}
} \\
\thetacap
\lr{
\inv{r^2 S_\theta}
\lr{
C_\theta A_r
+ 2 S_\theta \partial_\theta A_r
}

\inv{r^2 S_\theta}
S_\theta C_\theta A_r
} \\
\phicap
\lr{
\inv{r^2 S_\theta^2}
2 S_\theta \partial_\phi A_r
} \\
&=
\rcap \lr{
-\frac{2}{r^2 } A_r
}
+
\frac{\thetacap}{r^2}
\lr{
\frac{C_\theta}{S_\theta} A_r
+ 2 \partial_\theta A_r
– C_\theta A_r
}
+
\phicap
\frac{2}{r^2 S_\theta} \partial_\phi A_r.
\end{aligned}

Next, let’s compute the derivatives of the $$\thetacap$$ projection.

\label{eqn:vectorWaveEquationSpherical:220}
\partial_r \lr{ r^2 \partial_r \lr{ \thetacap \psi} }
=
\thetacap \partial_r \lr{ r^2 \partial_r \psi }

\label{eqn:vectorWaveEquationSpherical:240}
\begin{aligned}
\partial_\theta \lr{ S_\theta \partial_\theta \lr{ \thetacap \psi } }
&=
\partial_\theta \lr{ S_\theta
\lr{
(\partial_\theta \thetacap ) \psi
+\thetacap \partial_\theta \psi
}
} \\
&=
\partial_\theta
\lr{ S_\theta
\lr{
(-\rcap ) \psi
+\thetacap \partial_\theta \psi
}
} \\
&=
C_\theta \lr{
-\rcap \psi
+\thetacap \partial_\theta \psi
}
+
S_\theta
\lr{
-(\partial_\theta \rcap) \psi
-\rcap \partial_\theta \psi
+(\partial_\theta \thetacap) \partial_\theta \psi
+\thetacap \partial_{\theta \theta} \psi
} \\
&=
C_\theta \lr{
-\rcap \psi
+\thetacap \partial_\theta \psi
}
+
S_\theta
\lr{
-(\thetacap) \psi
-\rcap \partial_\theta \psi
+(-\rcap) \partial_\theta \psi
+\thetacap \partial_{\theta \theta} \psi
} \\
&=
\rcap \lr{
-C_\theta \psi
-2 S_\theta \partial_\theta \psi
}
+
\thetacap \lr{
+C_\theta \partial_\theta \psi
-S_\theta \psi
+S_\theta \partial_{\theta \theta} \psi
} \\
&=
\rcap \lr{
-C_\theta \psi
-2 S_\theta \partial_\theta \psi
}
+
\thetacap \lr{
+\partial_\theta (S_\theta \partial_\theta \psi)
-S_\theta \psi
}
\end{aligned}

\label{eqn:vectorWaveEquationSpherical:260}
\begin{aligned}
\partial_{\phi \phi} \lr{ \thetacap \psi}
&=
\partial_{\phi} \lr{
(\partial_\phi \thetacap) \psi
+\thetacap \partial_\phi \psi
} \\
&=
\partial_{\phi} \lr{
(C_\theta \phicap) \psi
+\thetacap \partial_\phi \psi
} \\
&=
C_\theta \partial_{\phi} (\phicap \psi)
+
\partial_{\phi} ( \thetacap \partial_\phi \psi ) \\
&=
C_\theta (\partial_\phi \phicap) \psi
+C_\theta \phicap \partial_{\phi} \psi
+ (\partial_\phi \thetacap) \partial_\phi \psi
+\thetacap \partial_{\phi\phi} \psi \\
&=
C_\theta (-\rcap S_\theta – \thetacap C_\theta) \psi
+C_\theta \phicap \partial_{\phi} \psi
+ (C_\theta \phicap) \partial_\phi \psi
+\thetacap \partial_{\phi\phi} \psi \\
&=
-\rcap C_\theta S_\theta \psi
+\thetacap \lr{
-C_\theta C_\theta \psi
+\partial_{\phi\phi} \psi
}
+2 \phicap C_\theta \partial_\phi \psi,
\end{aligned}

which gives
\label{eqn:vectorWaveEquationSpherical:360}
\begin{aligned}
&=
\rcap
\lr{
\inv{r^2 S_\theta}
\lr{
-C_\theta A_\theta
-2 S_\theta \partial_\theta A_\theta
}

\inv{r^2 S_\theta^2}
C_\theta S_\theta A_\theta
} \\
\thetacap \lr{
\inv{r^2} \partial_r \lr{ r^2 \partial_r A_\theta }
+
\inv{r^2 S_\theta}
\lr{
+\partial_\theta (S_\theta \partial_\theta A_\theta)
-S_\theta A_\theta
}
+\inv{r^2 S_\theta^2}
\lr{
-C_\theta C_\theta A_\theta
+\partial_{\phi\phi} A_\theta
}
} \\
\phicap \lr{
\inv{r^2 S_\theta^2}
2 C_\theta \partial_\phi A_\theta
} \\
&=
-2 \rcap
\inv{r^2 S_\theta}
\partial_\theta (S_\theta A_\theta)
+
\thetacap \lr{
-\inv{r^2}
A_\theta
-\inv{r^2 S_\theta^2} C_\theta^2 A_\theta
}
+
2 \phicap \lr{
\inv{r^2 S_\theta^2}
C_\theta \partial_\phi A_\theta
}.
\end{aligned}

Finally, we can compute the derivatives of the $$\phicap$$ projection.

\label{eqn:vectorWaveEquationSpherical:300}
\partial_r \lr{ r^2 \partial_r \lr{ \phicap \psi} }
=
\phicap \partial_r \lr{ r^2 \partial_r \psi }

\label{eqn:vectorWaveEquationSpherical:320}
\partial_\theta \lr{ S_\theta \partial_\theta \lr{ \phicap \psi } }
=
\phicap \partial_\theta \lr{ S_\theta \partial_\theta \psi }

\label{eqn:vectorWaveEquationSpherical:340}
\begin{aligned}
\partial_{\phi \phi} \lr{ \phicap \psi}
&=
\partial_{\phi} \lr{
(\partial_\phi \phicap) \psi
+\phicap \partial_\phi \psi
} \\
&=
\partial_{\phi} \lr{
(-\rcap S_\theta – \thetacap C_\theta) \psi
+\phicap \partial_\phi \psi
} \\
&=
-((\partial_\phi \rcap) S_\theta + (\partial_\phi \thetacap) C_\theta) \psi
-(\rcap S_\theta + \thetacap C_\theta) \partial_\phi \psi
+(\partial_\phi \phicap \partial_\phi \psi
+\phicap \partial_{\phi \phi} \psi \\
&=
-((S_\theta \phicap) S_\theta + (C_\theta \phicap) C_\theta) \psi
-(\rcap S_\theta + \thetacap C_\theta) \partial_\phi \psi
+(-\rcap S_\theta – \thetacap C_\theta) \partial_\phi \psi
+\phicap \partial_{\phi \phi} \psi \\
&=
– 2 \rcap S_\theta \partial_\phi \psi
– 2 \thetacap C_\theta \partial_\phi \psi
+ \phicap \lr{
\partial_{\phi \phi} \psi
-\psi
},
\end{aligned}

which gives
\label{eqn:vectorWaveEquationSpherical:380}
\begin{aligned}
&=
-2 \rcap \inv{r^2 S_\theta} \partial_\phi A_\phi
-2 \thetacap \inv{r^2 S_\theta^2} C_\theta \partial_\phi A_\phi \\
\phicap \lr{
\inv{r^2}
\partial_r \lr{ r^2 \partial_r A_\phi }
+
\inv{r^2 S_\theta}
\partial_\theta \lr{ S_\theta \partial_\theta A_\phi }
+
\inv{r^2 S_\theta^2}
\lr{
\partial_{\phi \phi} A_\phi -A_\phi
}
} \\
&=
-2 \rcap \inv{r^2 S_\theta} \partial_\phi A_\phi
-2 \thetacap \inv{r^2 S_\theta^2} C_\theta \partial_\phi A_\phi
+
\phicap \lr{
}.
\end{aligned}

The vector Laplacian resolves into three augmented scalar wave equations, all highly coupled

\label{eqn:vectorWaveEquationSpherical:420}
\boxed{
\begin{aligned}
\rcap \cdot \lr{ \spacegrad^2 \BA }
&=
-\frac{2}{r^2 } A_r
– \frac{2}{r^2 S_\theta} \partial_\theta (S_\theta A_\theta)
– \frac{2}{r^2 S_\theta} \partial_\phi A_\phi \\
\thetacap \cdot \lr{ \spacegrad^2 \BA }
&=
\frac{1}{r^2} \frac{C_\theta}{S_\theta} A_r
+ \frac{2}{r^2} \partial_\theta A_r
– \frac{1}{r^2} C_\theta A_r
– \inv{r^2} A_\theta
– \inv{r^2 S_\theta^2} C_\theta^2 A_\theta
-2 \inv{r^2 S_\theta^2} C_\theta \partial_\phi A_\phi \\
\phicap \cdot \lr{ \spacegrad^2 \BA }
&=
\frac{2}{r^2 S_\theta} \partial_\phi A_r
+ \frac{2}{r^2 S_\theta^2} C_\theta \partial_\phi A_\theta
+ \spacegrad^2 A_\phi – \inv{r^2} A_\phi.
\end{aligned}
}

I’d guess one way to decouple these equations would be to impose a constraint that allows all the non-wave equation terms in one of the component equations to be killed, and then substitute that constraint into the remaining equations. Let’s try one such constraint

\label{eqn:vectorWaveEquationSpherical:480}
A_r
=
– \inv{S_\theta} \partial_\theta (S_\theta A_\theta)
– \inv{S_\theta} \partial_\phi A_\phi.

This gives

\label{eqn:vectorWaveEquationSpherical:520}
\begin{aligned}
\rcap \cdot \lr{ \spacegrad^2 \BA }
&=
\thetacap \cdot \lr{ \spacegrad^2 \BA }
&=
\lr{
\frac{1}{r^2} \frac{C_\theta}{S_\theta}
+ \frac{2}{r^2} \partial_\theta
– \frac{1}{r^2} C_\theta
}
\lr{
– \inv{S_\theta} \partial_\theta (S_\theta A_\theta)
– \inv{S_\theta} \partial_\phi A_\phi
} \\
– \inv{r^2} A_\theta
– \inv{r^2 S_\theta^2} C_\theta^2 A_\theta
-\frac{2}{r^2 S_\theta^2} C_\theta \partial_\phi A_\phi \\
\phicap \cdot \lr{ \spacegrad^2 \BA }
&=
– \frac{2}{r^2 S_\theta} \partial_\phi
\lr{
\inv{S_\theta} \partial_\theta (S_\theta A_\theta)
+ \inv{S_\theta} \partial_\phi A_\phi
}
+ \frac{2}{r^2 S_\theta^2} C_\theta \partial_\phi A_\theta
+ \spacegrad^2 A_\phi – \inv{r^2} A_\phi \\
&=
-\frac{2}{r^2 S_\theta} \partial_\theta A_\theta
-\frac{2}{r^2 S_\theta^2} \partial_{\phi\phi} A_\theta
+ \spacegrad^2 A_\phi – \inv{r^2} A_\phi
\end{aligned}

It looks like some additional cancellations may be had in the $$\thetacap$$ projection of this constrained vector Laplacian. I’m not inclined to try to take this reduction any further without a thorough check of all the algebra (using Mathematica to do so would make sense).

I also guessing that such a solution might be how the $$\textrm{TE}^r$$ and $$\textrm{TM}^r$$ modes were defined, but that doesn’t appear to be the case according to [1]. There the wave equation is formulated in terms of the vector potentials (picking one to be zero and the other to be radial only). The solution obtained from such a potential wave equation then directly defines the $$\textrm{TE}^r$$ and $$\textrm{TM}^r$$ modes. It would be interesting to see how the modes derived in that analysis transform with application of the vector Laplacian derived above.

# References

[1] Constantine A Balanis. Advanced engineering electromagnetics. Wiley New York, 1989.

## 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.

## Gradient, divergence, curl and Laplacian in cylindrical coordinates

In class it was suggested that the identity

\label{eqn:laplacianCylindrical:20}

can be used to compute the Laplacian in non-rectangular coordinates. Is that the easiest way to do this?

\label{eqn:laplacianCylindrical:80}
\spacegrad = \rhocap \partial_\rho + \frac{\phicap}{\rho} \partial_\phi + \zcap \partial_z,

where
\label{eqn:laplacianCylindrical:100}
\begin{aligned}
\rhocap &= \Be_1 e^{\Be_1 \Be_2 \phi} \\
\phicap &= \Be_2 e^{\Be_1 \Be_2 \phi} \\
\end{aligned}

Taking $$\phi$$ derivatives of \ref{eqn:laplacianCylindrical:100}, we have

\label{eqn:laplacianCylindrical:120}
\begin{aligned}
\partial_\phi \rhocap &= \Be_1 \Be_1 \Be_2 e^{\Be_1 \Be_2 \phi} = \Be_2 e^{\Be_1 \Be_2 \phi} = \phicap \\
\partial_\phi \phicap &= \Be_2 \Be_1 \Be_2 e^{\Be_1 \Be_2 \phi} = -\Be_1 e^{\Be_1 \Be_2 \phi} = -\rhocap.
\end{aligned}

The gradient of a vector $$\BA = \rhocap A_\rho + \phicap A_\phi + \zcap A_z$$ is

\label{eqn:laplacianCylindrical:60}
\begin{aligned}
&=
\lr{ \rhocap \partial_\rho + \frac{\phicap}{\rho} \partial_\phi + \zcap \partial_z }
\lr{ \rhocap A_\rho + \phicap A_\phi + \zcap A_z } \\
&=
\quad \rhocap \partial_\rho \lr{ \rhocap A_\rho + \phicap A_\phi + \zcap A_z } \\
&\quad + \frac{\phicap}{\rho} \partial_\phi \lr{ \rhocap A_\rho + \phicap A_\phi + \zcap A_z } \\
&\quad + \zcap \partial_z \lr{ \rhocap A_\rho + \phicap A_\phi + \zcap A_z } \\
&=
\quad \rhocap \lr{ \rhocap \partial_\rho A_\rho + \phicap \partial_\rho A_\phi + \zcap \partial_\rho A_z } \\
&\quad + \frac{\phicap}{\rho} \lr{ \partial_\phi(\rhocap A_\rho) + \partial_\phi(\phicap A_\phi) + \zcap \partial_\phi A_z } \\
&\quad + \zcap \lr{ \rhocap \partial_z A_\rho + \phicap \partial_z A_\phi + \zcap \partial_z A_z } \\
&=
\quad \partial_\rho A_\rho + \rhocap \phicap \partial_\rho A_\phi + \rhocap \zcap \partial_\rho A_z \\
&\quad +\frac{1}{\rho} \lr{ A_\rho + \phicap \rhocap \partial_\phi A_\rho – \phicap \rhocap A_\phi + \partial_\phi A_\phi + \phicap \zcap \partial_\phi A_z } \\
&\quad + \zcap \rhocap \partial_z A_\rho + \zcap \phicap \partial_z A_\phi + \partial_z A_z \\
&=
\quad \partial_\rho A_\rho + \frac{1}{\rho} \lr{ A_\rho + \partial_\phi A_\phi } + \partial_z A_z \\
\zcap \rhocap \lr{
\partial_z A_\rho
-\partial_\rho A_z
} \\
\phicap \zcap \lr{
\inv{\rho} \partial_\phi A_z
– \partial_z A_\phi
} \\
\rhocap \phicap \lr{
\partial_\rho A_\phi
– \inv{\rho} \lr{ \partial_\phi A_\rho – A_\phi }
},
\end{aligned}

As expected, we see that the gradient splits nicely into a dot and curl

\label{eqn:laplacianCylindrical:160}
\begin{aligned}
&= \spacegrad \cdot \BA + \rhocap \phicap \zcap (\spacegrad \cross \BA ),
\end{aligned}

where the cylindrical representation of the divergence is seen to be

\label{eqn:laplacianCylindrical:140}
=
\inv{\rho} \partial_\rho (\rho A_\rho) + \frac{1}{\rho} \partial_\phi A_\phi + \partial_z A_z,

and the cylindrical representation of the curl is

\label{eqn:laplacianCylindrical:180}
=
\rhocap
\lr{
\inv{\rho} \partial_\phi A_z
– \partial_z A_\phi
}
+
\phicap
\lr{
\partial_z A_\rho
-\partial_\rho A_z
}
+
\inv{\rho} \zcap \lr{
\partial_\rho ( \rho A_\phi )
– \partial_\phi A_\rho
}.

Should we want to, it is now possible to evaluate the Laplacian of $$\BA$$ using
\ref{eqn:laplacianCylindrical:20}
, which will have the following components

\label{eqn:laplacianCylindrical:220}
\begin{aligned}
\rhocap \cdot \lr{ \spacegrad^2 \BA }
&=
\partial_\rho
\lr{
\inv{\rho} \partial_\rho (\rho A_\rho) + \frac{1}{\rho} \partial_\phi A_\phi + \partial_z A_z
}

\lr{
\inv{\rho} \partial_\phi \lr{
\inv{\rho} \lr{
\partial_\rho ( \rho A_\phi ) – \partial_\phi A_\rho
}
}
– \partial_z \lr{
\partial_z A_\rho -\partial_\rho A_z
}
} \\
&=
\partial_\rho \lr{ \inv{\rho} \partial_\rho (\rho A_\rho)}
+ \partial_\rho \lr{ \frac{1}{\rho} \partial_\phi A_\phi}
+ \partial_{\rho z} A_z
– \inv{\rho^2}\partial_{\phi \rho} ( \rho A_\phi )
+ \inv{\rho^2}\partial_{\phi\phi} A_\rho
+ \partial_{zz} A_\rho
– \partial_{z\rho} A_z \\
&=
\partial_\rho \lr{ \inv{\rho} \partial_\rho (\rho A_\rho)}
+ \inv{\rho^2}\partial_{\phi\phi} A_\rho
+ \partial_{zz} A_\rho
– \frac{1}{\rho^2} \partial_\phi A_\phi
+ \frac{1}{\rho} \partial_{\rho\phi} A_\phi
– \inv{\rho^2}\partial_{\phi} A_\phi
– \inv{\rho}\partial_{\phi\rho} A_\phi \\
&=
\partial_\rho \lr{ \inv{\rho} \partial_\rho (\rho A_\rho)}
+ \inv{\rho^2}\partial_{\phi\phi} A_\rho
+ \partial_{zz} A_\rho
– \frac{2}{\rho^2} \partial_\phi A_\phi \\
&=
\inv{\rho} \partial_\rho \lr{ \rho \partial_\rho A_\rho}
+ \inv{\rho^2}\partial_{\phi\phi} A_\rho
+ \partial_{zz} A_\rho
– \frac{A_\rho}{\rho^2}
– \frac{2}{\rho^2} \partial_\phi A_\phi,
\end{aligned}

\label{eqn:laplacianCylindrical:240}
\begin{aligned}
\phicap \cdot \lr{ \spacegrad^2 \BA }
&=
\inv{\rho} \partial_\phi
\lr{
\inv{\rho} \partial_\rho (\rho A_\rho) + \frac{1}{\rho} \partial_\phi A_\phi + \partial_z A_z
}

\lr{
\lr{
\partial_z \lr{
\inv{\rho} \partial_\phi A_z – \partial_z A_\phi
}
-\partial_\rho \lr{
\inv{\rho} \lr{ \partial_\rho ( \rho A_\phi ) – \partial_\phi A_\rho}
}
}
} \\
&=
\inv{\rho^2} \partial_{\phi\rho} (\rho A_\rho)
+ \frac{1}{\rho^2} \partial_{\phi\phi} A_\phi
+ \inv{\rho}\partial_{\phi z} A_z
– \inv{\rho} \partial_{z\phi} A_z
+ \partial_{z z} A_\phi
+\partial_\rho \lr{ \inv{\rho} \partial_\rho ( \rho A_\phi ) }
– \partial_\rho \lr{ \inv{\rho} \partial_\phi A_\rho} \\
&=
\partial_\rho \lr{ \inv{\rho} \partial_\rho ( \rho A_\phi ) }
+ \frac{1}{\rho^2} \partial_{\phi\phi} A_\phi
+ \partial_{z z} A_\phi
+ \inv{\rho^2} \partial_{\phi\rho} (\rho A_\rho)
+ \inv{\rho}\partial_{\phi z} A_z
– \inv{\rho} \partial_{z\phi} A_z
– \partial_\rho \lr{ \inv{\rho} \partial_\phi A_\rho} \\
&=
\partial_\rho \lr{ \inv{\rho} \partial_\rho ( \rho A_\phi ) }
+ \frac{1}{\rho^2} \partial_{\phi\phi} A_\phi
+ \partial_{z z} A_\phi
+ \inv{\rho^2} \partial_{\phi} A_\rho
+ \inv{\rho} \partial_{\phi\rho} A_\rho
+ \inv{\rho^2} \partial_\phi A_\rho
– \inv{\rho} \partial_{\rho\phi} A_\rho \\
&=
\partial_\rho \lr{ \inv{\rho} \partial_\rho ( \rho A_\phi ) }
+ \frac{1}{\rho^2} \partial_{\phi\phi} A_\phi
+ \partial_{z z} A_\phi
+ \frac{2}{\rho^2} \partial_{\phi} A_\rho \\
&=
\inv{\rho} \partial_\rho \lr{ \rho \partial_\rho A_\phi }
+ \frac{1}{\rho^2} \partial_{\phi\phi} A_\phi
+ \partial_{z z} A_\phi
+ \frac{2}{\rho^2} \partial_{\phi} A_\rho
– \frac{A_\phi}{\rho^2},
\end{aligned}

\label{eqn:laplacianCylindrical:260}
\begin{aligned}
\zcap \cdot \lr{ \spacegrad^2 \BA }
&=
\partial_z
\lr{
\inv{\rho} \partial_\rho (\rho A_\rho) + \frac{1}{\rho} \partial_\phi A_\phi + \partial_z A_z
}

\inv{\rho} \lr{
\partial_\rho \lr{ \rho \lr{
\partial_z A_\rho -\partial_\rho A_z
}
}
– \partial_\phi \lr{
\inv{\rho} \partial_\phi A_z – \partial_z A_\phi
}
} \\
&=
\inv{\rho} \partial_{z\rho} (\rho A_\rho)
+ \frac{1}{\rho} \partial_{z\phi} A_\phi
+ \partial_{zz} A_z
– \inv{\rho}\partial_\rho \lr{ \rho \partial_z A_\rho }
+ \inv{\rho}\partial_\rho \lr{ \rho \partial_\rho A_z }
+ \inv{\rho^2} \partial_{\phi\phi} A_z
– \inv{\rho} \partial_{\phi z} A_\phi \\
&=
\inv{\rho}\partial_\rho \lr{ \rho \partial_\rho A_z }
+ \inv{\rho^2} \partial_{\phi\phi} A_z
+ \partial_{zz} A_z
+ \inv{\rho} \partial_{z} A_\rho
+\partial_{z\rho} A_\rho
+ \frac{1}{\rho} \partial_{z\phi} A_\phi
– \inv{\rho}\partial_z A_\rho
– \partial_{\rho z} A_\rho
– \inv{\rho} \partial_{\phi z} A_\phi \\
&=
\inv{\rho}\partial_\rho \lr{ \rho \partial_\rho A_z }
+ \inv{\rho^2} \partial_{\phi\phi} A_z
+ \partial_{zz} A_z
\end{aligned}

Evaluating these was a fairly tedious and mechanical job, and would have been better suited to a computer algebra system than by hand as done here.

### Explicit cylindrical Laplacian

Let’s try this a different way. The most obvious potential strategy is to just apply the Laplacian to the vector itself, but we need to include the unit vectors in such an operation

\label{eqn:laplacianCylindrical:280}
\spacegrad^2 \lr{ \rhocap A_\rho + \phicap A_\phi + \zcap A_z }.

First we need to know the explicit form of the cylindrical Laplacian. From the painful expansion, we can guess that it is

\label{eqn:laplacianCylindrical:300}
=
\inv{\rho}\partial_\rho \lr{ \rho \partial_\rho \psi }
+ \inv{\rho^2} \partial_{\phi\phi} \psi
+ \partial_{zz} \psi.

Let’s check that explicitly. Here I use the vector product where $$\rhocap^2 = \phicap^2 = \zcap^2 = 1$$, and these vectors anticommute when different

\label{eqn:laplacianCylindrical:320}
\begin{aligned}
&=
\lr{ \rhocap \partial_\rho + \frac{\phicap}{\rho} \partial_\phi + \zcap \partial_z }
\lr{ \rhocap \partial_\rho \psi + \frac{\phicap}{\rho} \partial_\phi \psi + \zcap \partial_z \psi } \\
&=
\rhocap \partial_\rho
\lr{ \rhocap \partial_\rho \psi + \frac{\phicap}{\rho} \partial_\phi \psi + \zcap \partial_z \psi }
+ \frac{\phicap}{\rho} \partial_\phi
\lr{ \rhocap \partial_\rho \psi + \frac{\phicap}{\rho} \partial_\phi \psi + \zcap \partial_z \psi }
+ \zcap \partial_z
\lr{ \rhocap \partial_\rho \psi + \frac{\phicap}{\rho} \partial_\phi \psi + \zcap \partial_z \psi } \\
&=
\partial_{\rho\rho} \psi
+ \rhocap \phicap \partial_\rho \lr{ \frac{1}{\rho} \partial_\phi \psi}
+ \rhocap \zcap \partial_{\rho z} \psi
+ \frac{\phicap}{\rho} \partial_\phi \lr{ \rhocap \partial_\rho \psi }
+ \frac{\phicap}{\rho} \partial_\phi \lr{ \frac{\phicap}{\rho} \partial_\phi \psi }
+ \frac{\phicap \zcap }{\rho} \partial_{\phi z} \psi
+ \zcap \rhocap \partial_{z\rho} \psi
+ \frac{\zcap \phicap}{\rho} \partial_{z\phi} \psi
+ \partial_{zz} \psi \\
&=
\partial_{\rho\rho} \psi
+ \inv{\rho} \partial_\rho \psi
+ \frac{1}{\rho^2} \partial_{\phi \phi} \psi
+ \partial_{zz} \psi
+ \rhocap \phicap
\lr{
-\frac{1}{\rho^2} \partial_\phi \psi
+\frac{1}{\rho} \partial_{\rho \phi} \psi
-\inv{\rho} \partial_{\phi \rho} \psi
+ \frac{1}{\rho^2} \partial_\phi \psi
}
+ \zcap \rhocap \lr{
-\partial_{\rho z} \psi
+ \partial_{z\rho} \psi
}
+ \phicap \zcap \lr{
\inv{\rho} \partial_{\phi z} \psi
– \inv{\rho} \partial_{z\phi} \psi
} \\
&=
\partial_{\rho\rho} \psi
+ \inv{\rho} \partial_\rho \psi
+ \frac{1}{\rho^2} \partial_{\phi \phi} \psi
+ \partial_{zz} \psi,
\end{aligned}

so the Laplacian operator is

\label{eqn:laplacianCylindrical:340}
\boxed{
=
\inv{\rho} \PD{\rho}{} \lr{ \rho \PD{\rho}{} }
+ \frac{1}{\rho^2} \PDSq{\phi}{}
+ \PDSq{z}{}.
}

All the bivector grades of the Laplacian operator are seen to explicitly cancel, regardless of the grade of $$\psi$$, just as if we had expanded the scalar Laplacian as a dot product
$$\spacegrad^2 \psi = \spacegrad \cdot \lr{ \spacegrad \psi}$$.
Unlike such a scalar expansion, this derivation is seen to be valid for any grade $$\psi$$. We know now that we can trust this result when $$\psi$$ is a scalar, a vector, a bivector, a trivector, or even a multivector.

### Vector Laplacian

Now that we trust that the typical scalar form of the Laplacian applies equally well to multivectors as it does to scalars, that cylindrical coordinate operator can now be applied to a
vector. Consider the projections onto each of the directions in turn

\label{eqn:laplacianCylindrical:360}
=
\rhocap \inv{\rho} \partial_\rho \lr{ \rho \partial_\rho A_\rho }
+ \frac{1}{\rho^2} \partial_{\phi\phi} \lr{\rhocap A_\rho}
+ \rhocap \partial_{zz} A_\rho

\label{eqn:laplacianCylindrical:380}
\begin{aligned}
\partial_{\phi\phi} \lr{\rhocap A_\rho}
&=
\partial_\phi \lr{ \phicap A_\rho + \rhocap \partial_\phi A_\rho } \\
&=
-\rhocap A_\rho
+\phicap \partial_\phi A_\rho
+ \phicap \partial_\phi A_\rho
+ \rhocap \partial_{\phi\phi} A_\rho \\
&=
\rhocap \lr{ \partial_{\phi\phi} A_\rho -A_\rho }
+ 2 \phicap \partial_\phi A_\rho
\end{aligned}

so this component of the vector Laplacian is

\label{eqn:laplacianCylindrical:400}
\begin{aligned}
&=
\rhocap
\lr{
\inv{\rho} \partial_\rho \lr{ \rho \partial_\rho A_\rho }
+ \inv{\rho^2} \partial_{\phi\phi} A_\rho
– \inv{\rho^2} A_\rho
+ \partial_{zz} A_\rho
}
+
\phicap
\lr{
2 \inv{\rho^2} \partial_\phi A_\rho
} \\
&=
\rhocap \lr{
– \inv{\rho^2} A_\rho
}
+
\phicap
\frac{2}{\rho^2} \partial_\phi A_\rho
.
\end{aligned}

The Laplacian for the projection of the vector onto the $$\phicap$$ direction is

\label{eqn:laplacianCylindrical:420}
=
\phicap \inv{\rho} \partial_\rho \lr{ \rho \partial_\rho A_\phi }
+ \frac{1}{\rho^2} \partial_{\phi\phi} \lr{\phicap A_\phi}
+ \phicap \partial_{zz} A_\phi,

Again, since the unit vectors are $$\phi$$ dependent, the $$\phi$$ derivatives have to be treated carefully

\label{eqn:laplacianCylindrical:440}
\begin{aligned}
\partial_{\phi\phi} \lr{\phicap A_\phi}
&=
\partial_{\phi} \lr{-\rhocap A_\phi + \phicap \partial_\phi A_\phi} \\
&=
-\phicap A_\phi
-\rhocap \partial_\phi A_\phi
– \rhocap \partial_\phi A_\phi
+ \phicap \partial_{\phi \phi} A_\phi \\
&=
– 2 \rhocap \partial_\phi A_\phi
+
\phicap
\lr{
\partial_{\phi \phi} A_\phi
– A_\phi
},
\end{aligned}

so the Laplacian of this projection is
\label{eqn:laplacianCylindrical:460}
\begin{aligned}
&=
\phicap
\lr{
\inv{\rho} \partial_\rho \lr{ \rho \partial_\rho A_\phi }
+ \phicap \partial_{zz} A_\phi,
\inv{\rho^2} \partial_{\phi \phi} A_\phi
– \frac{A_\phi }{\rho^2}
}
– \rhocap \frac{2}{\rho^2} \partial_\phi A_\phi \\
&=
\phicap \lr{
– \frac{A_\phi}{\rho^2}
}
– \rhocap \frac{2}{\rho^2} \partial_\phi A_\phi.
\end{aligned}

Since $$\zcap$$ is fixed we have

\label{eqn:laplacianCylindrical:480}
=

Putting all the pieces together we have
\label{eqn:laplacianCylindrical:500}
\boxed{
=
\rhocap \lr{
– \inv{\rho^2} A_\rho
– \frac{2}{\rho^2} \partial_\phi A_\phi
}
+\phicap \lr{
– \frac{A_\phi}{\rho^2}
+ \frac{2}{\rho^2} \partial_\phi A_\rho
}
+
}

This matches the results of \ref{eqn:laplacianCylindrical:220}, …, from the painful expansion of
$$\spacegrad \lr{ \spacegrad \cdot \BA } – \spacegrad \cross \lr{ \spacegrad \cross \BA }$$.

## Line charge field and potential.

When computing the most general solution of the electrostatic potential in a plane, Jackson [1] mentions that $$-2 \lambda_0 \ln \rho$$ is the well known potential for an infinite line charge (up to the unit specific factor). Checking that statement, since I didn’t recall what that potential was offhand, I encountered some inconsistencies and non-convergent integrals, and thought it was worthwhile to explore those a bit more carefully. This will be done here.

### Using Gauss’s law.

For an infinite length line charge, we can find the radial field contribution using Gauss’s law, imagining a cylinder of length $$\Delta l$$ of radius $$\rho$$ surrounding this charge with the midpoint at the origin. Ignoring any non-radial field contribution, we have

\label{eqn:lineCharge:20}
\int_{-\Delta l/2}^{\Delta l/2} \ncap \cdot \BE (2 \pi \rho) dl = \frac{\lambda_0}{\epsilon_0} \Delta l,

or

\label{eqn:lineCharge:40}
\BE = \frac{\lambda_0}{2 \pi \epsilon_0} \frac{\rhocap}{\rho}.

Since

\label{eqn:lineCharge:60}

this means that the potential is

\label{eqn:lineCharge:80}
\phi = -\frac{2 \lambda_0}{4 \pi \epsilon_0} \ln \rho.

### Finite line charge potential.

Let’s try both these calculations for a finite charge distribution. Gauss’s law looses its usefulness, but we can evaluate the integrals directly. For the electric field

\label{eqn:lineCharge:100}
\BE
= \frac{\lambda_0}{4 \pi \epsilon_0} \int \frac{(\Bx – \Bx’)}{\Abs{\Bx – \Bx’}^3} dl’.

Using cylindrical coordinates with the field point $$\Bx = \rho \rhocap$$ for convience, the charge point $$\Bx’ = z’ \zcap$$, and a the charge distributed over $$[a,b]$$ this is

\label{eqn:lineCharge:120}
\BE
= \frac{\lambda_0}{4 \pi \epsilon_0} \int_a^b \frac{(\rho \rhocap – z’ \zcap)}{\lr{\rho^2 + (z’)^2}^{3/2}} dz’.

When the charge is uniformly distributed around the origin $$[a,b] = b[-1,1]$$ the $$\zcap$$ component of this field is killed because the integrand is odd. This justifies ignoring such contributions in the Gaussing cylinder analysis above. The general solution to this integral is found to be

\label{eqn:lineCharge:140}
\BE
=
\frac{\lambda_0}{4 \pi \epsilon_0}
\evalrange{
\lr{
\frac{z’ \rhocap }{\rho \sqrt{ \rho^2 + (z’)^2 } }
+\frac{\zcap}{ \sqrt{ \rho^2 + (z’)^2 } }
}
}{a}{b},

or
\label{eqn:lineCharge:240}
\boxed{
\BE
=
\frac{\lambda_0}{4 \pi \epsilon_0}
\lr{
\frac{\rhocap }{\rho}
\lr{
\frac{b}{\sqrt{ \rho^2 + b^2 } }
-\frac{a}{\sqrt{ \rho^2 + a^2 } }
}
+ \zcap
\lr{
\frac{1}{ \sqrt{ \rho^2 + b^2 } }
-\frac{1}{ \sqrt{ \rho^2 + a^2 } }
}
}.
}

When $$b = -a = \Delta l/2$$, this reduces to

\label{eqn:lineCharge:160}
\BE
=
\frac{\lambda_0}{4 \pi \epsilon_0}
\frac{\rhocap }{\rho}
\frac{\Delta l}{\sqrt{ \rho^2 + (\Delta l/2)^2 } },

which further reduces to \ref{eqn:lineCharge:40} when $$\Delta l \gg \rho$$.

### Finite line charge potential. Wrong but illuminating.

Again, putting the field point at $$z’ = 0$$, we have

\label{eqn:lineCharge:180}
\phi(\rho)
= \frac{\lambda_0}{4 \pi \epsilon_0} \int_a^b \frac{dz’}{\lr{\rho^2 + (z’)^2}^{1/2}},

which integrates to
\label{eqn:lineCharge:260}
\phi(\rho)
= \frac{\lambda_0}{4 \pi \epsilon_0 }
\ln \frac{ b + \sqrt{ \rho^2 + b^2 }}{ a + \sqrt{\rho^2 + a^2}}.

With $$b = -a = \Delta l/2$$, this approaches

\label{eqn:lineCharge:200}
\phi
\approx
\frac{\lambda_0}{4 \pi \epsilon_0 }
\ln \frac{ (\Delta l/2) }{ \rho^2/2\Abs{\Delta l/2}}
=
\frac{-2 \lambda_0}{4 \pi \epsilon_0 } \ln \rho
+
\frac{\lambda_0}{4 \pi \epsilon_0 }
\ln \lr{ (\Delta l)^2/2 }.

Before $$\Delta l$$ is allowed to tend to infinity, this is identical (up to a difference in the reference potential) to \ref{eqn:lineCharge:80} found using Gauss’s law. It is, strictly speaking, singular when $$\Delta l \rightarrow \infty$$, so it does not seem right to infinity as a reference point for the potential.

There’s another weird thing about this result. Since this has no $$z$$ dependence, it is not obvious how we would recover the non-radial portion of the electric field from this potential using $$\BE = -\spacegrad \phi$$? Let’s calculate the elecric field from \ref{eqn:lineCharge:180} explicitly

\label{eqn:lineCharge:220}
\begin{aligned}
\BE
&=
-\frac{\lambda_0}{4 \pi \epsilon_0}
\ln \frac{ b + \sqrt{ \rho^2 + b^2 }}{ a + \sqrt{\rho^2 + a^2}} \\
&=
-\frac{\lambda_0 \rhocap}{4 \pi \epsilon_0 }
\PD{\rho}{}
\ln \frac{ b + \sqrt{ \rho^2 + b^2 }}{ a + \sqrt{\rho^2 + a^2}} \\
&=
-\frac{\lambda_0 \rhocap}{4 \pi \epsilon_0}
\lr{
\inv{ b + \sqrt{ \rho^2 + b^2 }} \frac{ \rho }{\sqrt{ \rho^2 + b^2 }}
-\inv{ a + \sqrt{ \rho^2 + a^2 }} \frac{ \rho }{\sqrt{ \rho^2 + a^2 }}
} \\
&=
-\frac{\lambda_0 \rhocap}{4 \pi \epsilon_0 \rho}
\lr{
\frac{ -b + \sqrt{ \rho^2 + b^2 }}{\sqrt{ \rho^2 + b^2 }}
-\frac{ -a + \sqrt{ \rho^2 + a^2 }}{\sqrt{ \rho^2 + a^2 }}
} \\
&=
\frac{\lambda_0 \rhocap}{4 \pi \epsilon_0 \rho}
\lr{
\frac{ b }{\sqrt{ \rho^2 + b^2 }}
-\frac{ a }{\sqrt{ \rho^2 + a^2 }}
}.
\end{aligned}

This recovers the radial component of the field from \ref{eqn:lineCharge:240}, but where did the $$\zcap$$ component go? The required potential appears to be

\label{eqn:lineCharge:340}
\phi(\rho, z)
=
\frac{\lambda_0}{4 \pi \epsilon_0 }
\ln \frac{ b + \sqrt{ \rho^2 + b^2 }}{ a + \sqrt{\rho^2 + a^2}}

\frac{z \lambda_0}{4 \pi \epsilon_0 }
\lr{ \frac{1}{\sqrt{\rho^2 + b^2}}
-\frac{1}{\sqrt{\rho^2 + a^2}}
}.

When computing the electric field $$\BE(\rho, \theta, z)$$, it was convienent to pick the coordinate system so that $$z = 0$$. Doing this with the potential gives the wrong answers. The reason for this appears to be that this kills the potential term that is linear in $$z$$ before taking its gradient, and we need that term to have the $$\zcap$$ field component that is expected for a charge distribution that is non-symmetric about the origin on the z-axis!

### Finite line charge potential. Take II.

Let the point at which the potential is evaluated be

\label{eqn:lineCharge:360}
\Bx = \rho \rhocap + z \zcap,

and the charge point be
\label{eqn:lineCharge:380}
\Bx’ = z’ \zcap.

This gives

\label{eqn:lineCharge:400}
\begin{aligned}
\phi(\rho, z)
&= \frac{\lambda_0}{4\pi \epsilon_0} \int_a^b \frac{dz’}{\Abs{\rho^2 + (z – z’)^2 }} \\
&= \frac{\lambda_0}{4\pi \epsilon_0} \int_{a-z}^{b-z} \frac{du}{ \Abs{\rho^2 + u^2} } \\
&= \frac{\lambda_0}{4\pi \epsilon_0}
\evalrange{\ln \lr{ u + \sqrt{ \rho^2 + u^2 }}}{b-z}{a-z} \\
&=
\frac{\lambda_0}{4\pi \epsilon_0}
\ln \frac
{ b-z + \sqrt{ \rho^2 + (b-z)^2 }}
{ a-z + \sqrt{ \rho^2 + (a-z)^2 }}.
\end{aligned}

The limit of this potential $$a = -\Delta/2 \rightarrow -\infty, b = \Delta/2 \rightarrow \infty$$ doesn’t exist in any strict sense. If we are cavilier about the limits, as in \ref{eqn:lineCharge:200}, this can be evaluated as

\label{eqn:lineCharge:n}
\phi \approx
\frac{\lambda_0}{4\pi \epsilon_0} \lr{ -2 \ln \rho + \textrm{constant} }.

however, the constant ($$\ln \Delta^2/2$$) is infinite, so there isn’t really a good justification for using that constant as the potential reference point directly.

It seems that the “right” way to calculate the potential for the infinite distribution, is to

• Calculate the field from the potential.
• Take the PV limit of that field with the charge distribution extending to infinity.
• Compute the corresponding potential from this limiting value of the field.

Doing that doesn’t blow up. That field calculation, for the finite case, should include a $$\zcap$$ component. To verify, let’s take the respective derivatives

\label{eqn:lineCharge:420}
\begin{aligned}
-\PD{z}{} \phi
&=
-\frac{\lambda_0}{4\pi \epsilon_0}
\lr{
\frac{ -1 + \frac{z – b}{\sqrt{ \rho^2 + (b-z)^2 }} }{
b-z + \sqrt{ \rho^2 + (b-z)^2 }
}

\frac{ -1 + \frac{z – a}{\sqrt{ \rho^2 + (a-z)^2 }} }{
a-z + \sqrt{ \rho^2 + (a-z)^2 }
}
} \\
&=
\frac{\lambda_0}{4\pi \epsilon_0}
\lr{
\frac{ 1 + \frac{b – z}{\sqrt{ \rho^2 + (b-z)^2 }} }{
b-z + \sqrt{ \rho^2 + (b-z)^2 }
}

\frac{ 1 + \frac{a – z}{\sqrt{ \rho^2 + (a-z)^2 }} }{
a-z + \sqrt{ \rho^2 + (a-z)^2 }
}
} \\
&=
\frac{\lambda_0}{4\pi \epsilon_0}
\lr{
\inv{\sqrt{ \rho^2 + (b-z)^2 }}
-\inv{\sqrt{ \rho^2 + (a-z)^2 }}
},
\end{aligned}

and

\label{eqn:lineCharge:440}
\begin{aligned}
-\PD{\rho}{} \phi
&=
-\frac{\lambda_0}{4\pi \epsilon_0}
\lr{
\frac{ \frac{\rho}{\sqrt{ \rho^2 + (b-z)^2 }} }{
b-z + \sqrt{ \rho^2 + (b-z)^2 }
}

\frac{ \frac{\rho}{\sqrt{ \rho^2 + (a-z)^2 }} }{
a-z + \sqrt{ \rho^2 + (a-z)^2 }
}
} \\
&=
-\frac{\lambda_0}{4\pi \epsilon_0}
\lr{
\frac{\rho \lr{
-(b-z) + \sqrt{ \rho^2 + (b-z)^2 }
}}{ \rho^2 \sqrt{ \rho^2 + (b-z)^2 } }

\frac{\rho \lr{
-(a-z) + \sqrt{ \rho^2 + (a-z)^2 }
}}{ \rho^2 \sqrt{ \rho^2 + (a-z)^2 } }
} \\
&=
\frac{\lambda_0}{4\pi \epsilon_0 \rho}
\lr{
\frac{b-z}{\sqrt{ \rho^2 + (b-z)^2 }}
-\frac{a-z}{\sqrt{ \rho^2 + (a-z)^2 }}
}
.
\end{aligned}

Putting the pieces together, the electric field is
\label{eqn:lineCharge:460}
\BE =
\frac{\lambda_0}{4\pi \epsilon_0}
\lr{
\frac{\rhocap}{\rho} \lr{
\frac{b-z}{\sqrt{ \rho^2 + (b-z)^2 }}
-\frac{a-z}{\sqrt{ \rho^2 + (a-z)^2 }}
}
+
\zcap \lr{
\inv{\sqrt{ \rho^2 + (b-z)^2 }}
-\inv{\sqrt{ \rho^2 + (a-z)^2 }}
}
}.

For has a PV limit of \ref{eqn:lineCharge:40} at $$z = 0$$, and also for the finite case, has the $$\zcap$$ field component that was obtained when the field was obtained by direct integration.

### Conclusions

• We have to evaluate the potential at all points in space, not just on the axis that we evaluate the field on (should we choose to do so).
• In this case, we found that it was not directly meaningful to take the limit of a potential distribution. We can, however, compute the field from a potential for a finite charge distribution,
take the limit of that field, and then calculate the corresponding potential for the infinite distribution.

Is there a more robust mechanism that can be used to directly calculate the potential for an infinite charge distribution, instead of calculating the potential from the field of such an infinite distribution?

I think that were things go wrong is that the integral of \ref{eqn:lineCharge:180} does not apply to charge distributions that are not finite on the infinite range $$z \in [-\infty, \infty]$$. That solution was obtained by utilizing an all-space Green’s function, and the boundary term in that Green’s analysis was assumed to tend to zero. That isn’t the case when the charge distribution is $$\lambda_0 \delta( z )$$.

# References

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

## Jackson’s electrostatic self energy analysis

### Motivation

I was reading my Jackson [1], which characteristically had the statement “the […] integral can easily be shown to have the value $$4 \pi$$”, in a discussion of electrostatic energy and self energy. After a few attempts and a couple of pages of calculations, I figured out how this can be easily shown.

### Context

Let me walk through the context that leads to the “easy” integral, and then the evaluation of that integral. Unlike my older copy of Jackson, I’ll do this in SI units.

The starting point is a statement that the work done (potential energy) of one charge $$q_i$$ in a set of $$n$$ charges, where that charge is brought to its position $$\Bx_i$$ from infinity, is

\label{eqn:electrostaticJacksonSelfEnergy:20}
W_i = q_i \Phi(\Bx_i),

where the potential energy due to the rest of the charge configuration is

\label{eqn:electrostaticJacksonSelfEnergy:40}
\Phi(\Bx_i) = \inv{4 \pi \epsilon} \sum_{i \ne j} \frac{q_j}{\Abs{\Bx_i – \Bx_j}}.

This means that the total potential energy, making sure not to double count, to move all the charges in from infinity is

\label{eqn:electrostaticJacksonSelfEnergy:60}
W = \inv{4 \pi \epsilon} \sum_{1 \le i < j \le n} \frac{q_i q_j}{\Abs{\Bx_i - \Bx_j}}. This sum over all unique pairs is somewhat unwieldy, so it can be adjusted by explicitly double counting with a corresponding divide by two $$\label{eqn:electrostaticJacksonSelfEnergy:80} W = \inv{2} \inv{4 \pi \epsilon} \sum_{1 \le i \ne j \le n} \frac{q_i q_j}{\Abs{\Bx_i - \Bx_j}}.$$ The point that causes the trouble later is the continuum equivalent to this relationship, which is $$\label{eqn:electrostaticJacksonSelfEnergy:100} W = \inv{8 \pi \epsilon} \int \frac{\rho(\Bx) \rho(\Bx')}{\Abs{\Bx - \Bx'}} d^3 \Bx d^3 \Bx',$$ or $$\label{eqn:electrostaticJacksonSelfEnergy:120} W = \inv{2} \int \rho(\Bx) \Phi(\Bx) d^3 \Bx.$$ There's a subtlety here that is often passed over. When the charge densities represent point charges $$\rho(\Bx) = q \delta^3(\Bx - \Bx')$$ are located at, notice that this integral equivalent is evaluated over all space, including the spaces that the charges that the charges are located at. Ignoring that subtlety, this potential energy can be expressed in terms of the electric field, and then integrated by parts \label{eqn:electrostaticJacksonSelfEnergy:140} \begin{aligned} W &= \inv{2 } \int (\spacegrad \cdot (\epsilon \BE)) \Phi(\Bx) d^3 \Bx \\ &= \frac{\epsilon}{2 } \int \lr{ \spacegrad \cdot (\BE \Phi) - (\spacegrad \Phi) \cdot \BE } d^3 \Bx \\ &= \frac{\epsilon}{2 } \oint dA \ncap \cdot (\BE \Phi) + \frac{\epsilon}{2 } \int \BE \cdot \BE d^3 \Bx. \end{aligned} The presumption is that $$\BE \Phi$$ falls off as the bounds of the integration volume tends to infinity. That leaves us with an energy density proportional to the square of the field $$\label{eqn:electrostaticJacksonSelfEnergy:160} w = \frac{\epsilon}{2 } \BE^2.$$

### Inconsistency

It’s here that Jackson points out the inconsistency between \ref{eqn:electrostaticJacksonSelfEnergy:160} and the original
discrete analogue \ref{eqn:electrostaticJacksonSelfEnergy:80} that this was based on. The energy density is positive definite, whereas the discrete potential energy can be negative if there is a difference in the sign of the charges.

Here Jackson uses a two particle charge distribution to help resolve this conundrum. For a superposition $$\BE = \BE_1 + \BE_2$$, we have

\label{eqn:electrostaticJacksonSelfEnergy:180}
\BE
=
\inv{4 \pi \epsilon} \frac{q_1 (\Bx – \Bx_1)}{\Abs{\Bx – \Bx_1}^3}
+ \inv{4 \pi \epsilon} \frac{q_2 (\Bx – \Bx_2)}{\Abs{\Bx – \Bx_2}^3},

so the energy density is
\label{eqn:electrostaticJacksonSelfEnergy:200}
w =
\frac{1}{32 \pi^2 \epsilon} \frac{q_1^2}{\Abs{\Bx – \Bx_1}^4 }
+
\frac{1}{32 \pi^2 \epsilon} \frac{q_2^2}{\Abs{\Bx – \Bx_2}^4 }
+
2 \frac{q_1 q_2}{32 \pi^2 \epsilon}
\frac{(\Bx – \Bx_1)}{\Abs{\Bx – \Bx_1}^3} \cdot
\frac{(\Bx – \Bx_2)}{\Abs{\Bx – \Bx_2}^3}.

The discrete potential had only an interaction energy, whereas the potential from this squared field has an interaction energy plus two self energy terms. Those two strictly positive self energy terms are what forces this field energy positive, independent of the sign of the interaction energy density. Jackson makes a change of variables of the form

\label{eqn:electrostaticJacksonSelfEnergy:220}
\begin{aligned}
\Brho &= (\Bx – \Bx_1)/R \\
R &= \Abs{\Bx_1 – \Bx_2} \\
\ncap &= (\Bx_1 – \Bx_2)/R,
\end{aligned}

for which we find

\label{eqn:electrostaticJacksonSelfEnergy:240}
\Bx = \Bx_1 + R \Brho,

so
\label{eqn:electrostaticJacksonSelfEnergy:260}
\Bx – \Bx_2 =
\Bx_1 – \Bx_2 + R \Brho
R (\ncap + \Brho),

and
\label{eqn:electrostaticJacksonSelfEnergy:280}
d^3 \Bx = R^3 d^3 \Brho,

so the total interaction energy is
\label{eqn:electrostaticJacksonSelfEnergy:300}
\begin{aligned}
W_{\textrm{int}}
&=
\frac{q_1 q_2}{16 \pi^2 \epsilon}
\int d^3 \Bx
\frac{(\Bx – \Bx_1)}{\Abs{\Bx – \Bx_1}^3} \cdot
\frac{(\Bx – \Bx_2)}{\Abs{\Bx – \Bx_2}^3} \\
&=
\frac{q_1 q_2}{16 \pi^2 \epsilon}
\int R^3 d^3 \Brho
\frac{ R \Brho }{ R^3 \Abs{\Brho}^3 } \cdot
\frac{R (\ncap + \Brho)}{R^3 \Abs{\ncap + \Brho}^3} \\
&=
\frac{q_1 q_2}{16 \pi^2 \epsilon R}
\int d^3 \Brho
\frac{ \Brho }{ \Abs{\Brho}^3 } \cdot
\frac{(\ncap + \Brho)}{ \Abs{\ncap + \Brho}^3}.
\end{aligned}

Evaluating this integral is what Jackson calls easy. The technique required is to express the integrand in terms of gradients in the $$\Brho$$ coordinate system

\label{eqn:electrostaticJacksonSelfEnergy:320}
\begin{aligned}
\int d^3 \Brho
\frac{ \Brho }{ \Abs{\Brho}^3 } \cdot
\frac{(\ncap + \Brho)}{ \Abs{\ncap + \Brho}^3}
&=
\int d^3 \Brho
\cdot
\lr{ – \spacegrad_\Brho \inv{\Abs{\ncap + \Brho}} } \\
&=
\int d^3 \Brho
\cdot
\lr{ \spacegrad_\Brho \inv{\Abs{\ncap + \Brho}} }.
\end{aligned}

I found it somewhat non-trivial to find the exact form of the chain rule that is required to simplify this integral, but after some trial and error, figured it out by working backwards from
\label{eqn:electrostaticJacksonSelfEnergy:340}
\spacegrad_\Brho^2 \inv{ \Abs{\Brho} \Abs{\ncap + \Brho}}
=
+

In integral form this is
\label{eqn:electrostaticJacksonSelfEnergy:360}
\begin{aligned}
\oint dA’ \ncap’ \cdot \spacegrad_\Brho \inv{ \Abs{\Brho} \Abs{\ncap + \Brho}}
&=
\int d^3 \Brho’
+
\int d^3 \Brho
\spacegrad_\Brho \cdot \lr{ \inv{\Abs{\ncap + \Brho}} \spacegrad_\Brho \inv{ \Abs{\Brho} } } \\
&=
\int d^3 \Brho’
\lr{ \spacegrad_{\Brho’} \inv{\Abs{\Brho’ – \ncap} } \cdot \spacegrad_{\Brho’} \inv{ \Abs{\Brho’} } }
+
\int d^3 \Brho’
\inv{\Abs{\Brho’ – \ncap}} \spacegrad_{\Brho’}^2 \inv{ \Abs{\Brho’} } \\
&+
\int d^3 \Brho
+
\int d^3 \Brho
\inv{\Abs{\ncap + \Brho}} \spacegrad_\Brho^2 \inv{ \Abs{\Brho} } \\
&=
2 \int d^3 \Brho
&- 4 \pi
\int d^3 \Brho’
\inv{\Abs{\Brho’ – \ncap}} \delta^3(\Brho’)
– 4 \pi
\int d^3 \Brho
\inv{\Abs{\Brho + \ncap}} \delta^3(\Brho) \\
&=
2 \int d^3 \Brho
– 8 \pi.
\end{aligned}

This used the Laplacian representation of the delta function $$\delta^3(\Bx) = -(1/4\pi) \spacegrad^2 (1/\Abs{\Bx})$$. Back-substitution gives

\label{eqn:electrostaticJacksonSelfEnergy:380}
\int d^3 \Brho
\frac{ \Brho }{ \Abs{\Brho}^3 } \cdot
\frac{(\ncap + \Brho)}{ \Abs{\ncap + \Brho}^3}
=
4 \pi
+
\oint dA’ \ncap’ \cdot \spacegrad_\Brho \inv{ \Abs{\Brho} \Abs{\ncap + \Brho}}.

We can argue that this last integral tends to zero, since

\label{eqn:electrostaticJacksonSelfEnergy:400}
\begin{aligned}
\oint dA’ \ncap’ \cdot \spacegrad_\Brho \inv{ \Abs{\Brho} \Abs{\ncap + \Brho}}
&=
\oint dA’ \ncap’ \cdot \lr{
\lr{ \spacegrad_\Brho \inv{ \Abs{\Brho}} } \inv{\Abs{\ncap + \Brho}}
+
\inv{ \Abs{\Brho}} \lr{ \spacegrad_\Brho \inv{\Abs{\ncap + \Brho}} }
} \\
&=
-\oint dA’ \ncap’ \cdot \lr{
\frac{ \Brho } {\inv{ \Abs{\Brho}}^3 } \inv{\Abs{\ncap + \Brho}}
+
\inv{ \Abs{\Brho}} \frac{ (\Brho + \ncap) }{ \Abs{\ncap + \Brho}^3 }
} \\
&=
-\oint dA’ \inv{\Abs{\Brho} \Abs{\Brho + \ncap}}
\lr{
\frac{ \ncap’ \cdot \Brho }{
{\Abs{\Brho}}^2 }
+\frac{ \ncap’ \cdot (\Brho + \ncap) }{
{\Abs{\Brho + \ncap}}^2 }
}.
\end{aligned}

The integrand in this surface integral is of $$O(1/\rho^3)$$ so tends to zero on an infinite surface in the $$\Brho$$ coordinate system. This completes the “easy” integral, leaving

\label{eqn:electrostaticJacksonSelfEnergy:420}
\int d^3 \Brho
\frac{ \Brho }{ \Abs{\Brho}^3 } \cdot
\frac{(\ncap + \Brho)}{ \Abs{\ncap + \Brho}^3}
=
4 \pi.

The total field energy can now be expressed as a sum of the self energies and the interaction energy
\label{eqn:electrostaticJacksonSelfEnergy:440}
W =
\frac{1}{32 \pi^2 \epsilon} \int d^3 \Bx \frac{q_1^2}{\Abs{\Bx – \Bx_1}^4 }
+
\frac{1}{32 \pi^2 \epsilon} \int d^3 \Bx \frac{q_2^2}{\Abs{\Bx – \Bx_2}^4 }
+ \inv{ 4 \pi \epsilon}
\frac{q_1 q_2}{\Abs{\Bx_1 – \Bx_2} }.

The interaction energy is exactly the potential energies for the two particles, the this total energy in the field is biased in the positive direction by the pair of self energies. It is interesting that the energy obtained from integrating the field energy density contains such self energy terms, but I don’t know exactly what to make of them at this point in time.

# References

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

## 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 https://arxiv.org/abs/0809.4526.

## Motivation

I initially thought that I might submit a problem set solution for ece1228 using Geometric Algebra. In order to justify this, I needed to add an appendix to that problem set that outlined enough of the ideas that such a solution might make sense to the grader.

I ended up changing my mind and reworked the problem entirely, removing any use of GA. Here’s the tutorial I initially considered submitting with that problem.

## Geometric Algebra in a nutshell.

Geometric Algebra defines a non-commutative, associative vector product

\label{eqn:gaTutorial:20}
\begin{aligned}
\Ba \Bb \Bc
&=
(\Ba \Bb) \Bc \\
&=
\Ba (\Bb \Bc),
\end{aligned}

where the square of a vector equals the squared vector magnitude

\label{eqn:gaTutorial:40}
\Ba^2 = \Abs{\Ba}^2,

In Euclidean spaces such a squared vector is always positive, but that is not necessarily the case in the mixed signature spaces used in special relativity.

There are a number of consequences of these two simple vector multiplication rules.

• Squared unit vectors have a unit magnitude (up to a sign). In a Euclidean space such a product is always positive

\label{eqn:gaTutorial:60}
(\Be_1)^2 = 1.

• Products of perpendicular vectors anticommute.

\label{eqn:gaTutorial:80}
\begin{aligned}
2
&=
(\Be_1 + \Be_2)^2 \\
&= (\Be_1 + \Be_2)(\Be_1 + \Be_2) \\
&= \Be_1^2 + \Be_2 \Be_1 + \Be_1 \Be_2 + \Be_2^2 \\
&= 2 + \Be_2 \Be_1 + \Be_1 \Be_2.
\end{aligned}

A product of two perpendicular vectors is called a bivector, and can be used to represent an oriented plane. The last line above shows an example of a scalar and bivector sum, called a multivector. In general Geometric Algebra allows sums of scalars, vectors, bivectors, and higher degree analogues (grades) be summed.

Comparison of the RHS and LHS of \ref{eqn:gaTutorial:80} shows that we must have

\label{eqn:gaTutorial:100}
\Be_2 \Be_1 = -\Be_1 \Be_2.

It is true in general that the product of two perpendicular vectors anticommutes. When, as above, such a product is a product of
two orthonormal vectors, it behaves like a non-commutative imaginary quantity, as it has an imaginary square in Euclidean spaces

\label{eqn:gaTutorial:120}
\begin{aligned}
(\Be_1 \Be_2)^2
&=
(\Be_1 \Be_2)
(\Be_1 \Be_2) \\
&=
\Be_1 (\Be_2
\Be_1) \Be_2 \\
&=
-\Be_1 (\Be_1
\Be_2) \Be_2 \\
&=
-(\Be_1 \Be_1)
(\Be_2 \Be_2) \\
&=-1.
\end{aligned}

Such “imaginary” (unit bivectors) have important applications describing rotations in Euclidean spaces, and boosts in Minkowski spaces.

• The product of three perpendicular vectors, such as

\label{eqn:gaTutorial:140}
I = \Be_1 \Be_2 \Be_3,

is called a trivector. In \R{3}, the product of three orthonormal vectors is called a pseudoscalar for the space, and can represent an oriented volume element. The quantity $$I$$ above is the typical orientation picked for the \R{3} unit pseudoscalar. This quantity also has characteristics of an imaginary number

\label{eqn:gaTutorial:160}
\begin{aligned}
I^2
&=
(\Be_1 \Be_2 \Be_3)
(\Be_1 \Be_2 \Be_3) \\
&=
\Be_1 \Be_2 (\Be_3
\Be_1) \Be_2 \Be_3 \\
&=
-\Be_1 \Be_2 \Be_1
\Be_3 \Be_2 \Be_3 \\
&=
-\Be_1 (\Be_2 \Be_1)
(\Be_3 \Be_2) \Be_3 \\
&=
-\Be_1 (\Be_1 \Be_2)
(\Be_2 \Be_3) \Be_3 \\
&=

\Be_1^2
\Be_2^2
\Be_3^2 \\
&=
-1.
\end{aligned}

• The product of two vectors in \R{3} can be expressed as the sum of a symmetric scalar product and antisymmetric bivector product

\label{eqn:gaTutorial:480}
\begin{aligned}
\Ba \Bb
&=
\sum_{i,j = 1}^n \Be_i \Be_j a_i b_j \\
&=
\sum_{i = 1}^n \Be_i^2 a_i b_i
+
\sum_{0 < i \ne j \le n} \Be_i \Be_j a_i b_j \\ &= \sum_{i = 1}^n a_i b_i + \sum_{0 < i < j \le n} \Be_i \Be_j (a_i b_j - a_j b_i). \end{aligned} The first (symmetric) term is clearly the dot product. The antisymmetric term is designated the wedge product. In general these are written $$\label{eqn:gaTutorial:500} \Ba \Bb = \Ba \cdot \Bb + \Ba \wedge \Bb,$$ where \label{eqn:gaTutorial:520} \begin{aligned} \Ba \cdot \Bb &\equiv \inv{2} \lr{ \Ba \Bb + \Bb \Ba } \\ \Ba \wedge \Bb &\equiv \inv{2} \lr{ \Ba \Bb - \Bb \Ba }, \end{aligned} The coordinate expansion of both can be seen above, but in \R{3} the wedge can also be written $$\label{eqn:gaTutorial:540} \Ba \wedge \Bb = \Be_1 \Be_2 \Be_3 (\Ba \cross \Bb) = I (\Ba \cross \Bb).$$ This allows for an handy dot plus cross product expansion of the vector product $$\label{eqn:gaTutorial:180} \Ba \Bb = \Ba \cdot \Bb + I (\Ba \cross \Bb).$$ This result should be familiar to the student of quantum spin states where one writes $$\label{eqn:gaTutorial:200} (\Bsigma \cdot \Ba) (\Bsigma \cdot \Bb) = (\Ba \cdot \Bb) + i (\Ba \cross \Bb) \cdot \Bsigma.$$ This correspondence is because the Pauli spin basis is a specific matrix representation of a Geometric Algebra, satisfying the same commutator and anticommutator relationships. A number of other algebra structures, such as complex numbers, and quaterions can also be modelled as Geometric Algebra elements.

• It is often useful to utilize the grade selection operator
$$\gpgrade{M}{n}$$ and scalar grade selection operator $$\gpgradezero{M} = \gpgrade{M}{0}$$
to select the scalar, vector, bivector, trivector, or higher grade algebraic elements. For example, operating on vectors $$\Ba, \Bb, \Bc$$, we have

\label{eqn:gaTutorial:580}
\begin{aligned}
&= \Ba \cdot \Bb \\
&=
\Ba (\Bb \cdot \Bc)
+
\Ba \cdot (\Bb \wedge \Bc) \\
&=
\Ba (\Bb \cdot \Bc)
+
(\Ba \cdot \Bb) \Bc

(\Ba \cdot \Bc) \Bb \\
\Ba \wedge \Bb \\
\Ba \wedge \Bb \wedge \Bc.
\end{aligned}

Note that the wedge product of any number of vectors such as $$\Ba \wedge \Bb \wedge \Bc$$ is associative and can be expressed in terms of the complete antisymmetrization of the product of those vectors. A consequence of that is the fact a wedge product that includes any colinear vectors in the product is zero.

## Example: Helmholz equations.

As an example of the power of \ref{eqn:gaTutorial:180}, consider the following Helmholtz equation derivation (wave equations for the electric and magnetic fields in the frequency domain.)

Application of \ref{eqn:gaTutorial:180} to
Maxwell equations in the frequency domain for source free simple media gives

\label{eqn:emtProblemSet1Problem6:340}
\label{eqn:emtProblemSet1Problem6:360}
\spacegrad \BE = -j \omega I \BB

\label{eqn:emtProblemSet1Problem6:380}
\spacegrad I \BB = -j \omega \mu \epsilon \BE.

These equations use the engineering (not physics) sign convention for the phasors where the time domain fields are of the form $$\boldsymbol{\mathcal{E}}(\Br, t) = \textrm{Re}( \BE e^{j\omega t}$$.

Operation with the gradient from the left produces the Helmholtz equation for each of the fields using nothing more than multiplication and simple substitution

\label{eqn:emtProblemSet1Problem6:400}
\label{eqn:emtProblemSet1Problem6:420}
\spacegrad^2 \BE = – \mu \epsilon \omega^2 \BE

\label{eqn:emtProblemSet1Problem6:440}
\spacegrad^2 I \BB = – \mu \epsilon \omega^2 I \BB.

There was no reason to go through the headache of looking up or deriving the expansion of $$\spacegrad \cross (\spacegrad \cross \BA )$$ as is required with the traditional vector algebra demonstration of these identities.

Observe that the usual Helmholtz equation for $$\BB$$ doesn’t have a pseudoscalar factor. That result can be obtained by just cancelling the factors $$I$$ since the \R{3} Euclidean pseudoscalar commutes with all grades (this isn’t the case in \R{2} nor in Minkowski spaces.)

## Example: Factoring the Laplacian.

There are various ways to demonstrate the identity

\label{eqn:gaTutorial:660}

such as the use of (somewhat obscure) tensor contraction techniques. We can also do this with Geometric Algebra (using a different set of obscure techniques) by factoring the Laplacian action on a vector

\label{eqn:gaTutorial:700}
\begin{aligned}
&=
&=
&=
+
%+
&=
+
\end{aligned}

Should we wish to express the last term using cross products, a grade one selection operation can be used
\label{eqn:gaTutorial:680}
\begin{aligned}
&=
&=
&=
&=
&=
\end{aligned}

Here coordinate expansion was not required in any step.

# References

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

[2] L. Dorst, D. Fontijne, and S. Mann. Geometric Algebra for Computer Science. Morgan Kaufmann, San Francisco, 2007.

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

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

## Update to old phy356 (Quantum Mechanics I) notes.

It’s been a long time since I took QM I. My notes from that class were pretty rough, but I’ve cleaned them up a bit.

The main value to these notes is that I worked a number of introductory Quantum Mechanics problems.

These were my personal lecture notes for the Fall 2010, University of Toronto Quantum mechanics I course (PHY356H1F), taught by Prof. Vatche Deyirmenjian.

The official description of this course was:

The general structure of wave mechanics; eigenfunctions and eigenvalues; operators; orbital angular momentum; spherical harmonics; central potential; separation of variables, hydrogen atom; Dirac notation; operator methods; harmonic oscillator and spin.

This document contains a few things

• My lecture notes.
Typos, if any, are probably mine(Peeter), and no claim nor attempt of spelling or grammar correctness will be made. The first four lectures had chosen not to take notes for since they followed the text very closely.
• Notes from reading of the text. This includes observations, notes on what seem like errors, and some solved problems. None of these problems have been graded. Note that my informal errata sheet for the text has been separated out from this document.
• Some assigned problems. I have corrected some the errors after receiving grading feedback, and where I have not done so I at least recorded some of the grading comments as a reference.
• Some worked problems associated with exam preparation.