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

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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}
\spacegrad F = J.

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}
F = \spacegrad A,

is a solution to the Maxwell statics equation? If such a solution exists, then Maxwell’s equation is simply
\label{eqn:staticPotentials:80}
\spacegrad^2 A = J,

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}
\spacegrad^2 A’ &= J \\
\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
&= \spacegrad^2 (A’ + \tilde{A})
\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
&= \spacegrad A \\
&=
-\int_V dV’ \lr{ \spacegrad \inv{\Norm{\Bx – \Bx’} } } J(\Bx’)
– \spacegrad \tilde{A}(\Bx) \\
&=
\int_V dV’ \lr{ \spacegrad’ \inv{\Norm{\Bx – \Bx’} } } J(\Bx’)
– \spacegrad \tilde{A}(\Bx) \\
&=
\int_V dV’ \spacegrad’ \frac{J(\Bx’)}{\Norm{\Bx – \Bx’} } – \int_V dV’ \frac{\spacegrad’ J(\Bx’)}{\Norm{\Bx – \Bx’} }
– \spacegrad \tilde{A}(\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

## Generalizing Ampere’s law using geometric algebra.

The question I’d like to explore in this post is how Ampere’s law, the relationship between the line integral of the magnetic field to current (i.e. the enclosed current)
\label{eqn:flux:20}
\oint_{\partial A} d\Bx \cdot \BH = -\int_A \ncap \cdot \BJ,

generalizes to geometric algebra where Maxwell’s equations for a statics configuration (all time derivatives zero) is
\label{eqn:flux:40}
\spacegrad F = J,

where the multivector fields and currents are
\label{eqn:flux:60}
\begin{aligned}
F &= \BE + I \eta \BH \\
J &= \eta \lr{ c \rho – \BJ } + I \lr{ c \rho_\txtm – \BM }.
\end{aligned}

Here (fictitious) the magnetic charge and current densities that can be useful in antenna theory have been included in the multivector current for generality.

My presumption is that it should be possible to utilize the fundamental theorem of geometric calculus for expressing the integral over an oriented surface to its boundary, but applied directly to Maxwell’s equation. That integral theorem has the form
\label{eqn:flux:80}
\int_A d^2 \Bx \boldpartial F = \oint_{\partial A} d\Bx F,

where $$d^2 \Bx = d\Ba \wedge d\Bb$$ is a two parameter bivector valued surface, and $$\boldpartial$$ is vector derivative, the projection of the gradient onto the tangent space. I won’t try to explain all of geometric calculus here, and refer the interested reader to [1], which is an excellent reference on geometric calculus and integration theory.

The gotcha is that we actually want a surface integral with $$\spacegrad F$$. We can split the gradient into the vector derivative a normal component
\label{eqn:flux:160}
\spacegrad = \boldpartial + \ncap (\ncap \cdot \spacegrad),

so
\label{eqn:flux:100}
\int_A d^2 \Bx \spacegrad F
=
\int_A d^2 \Bx \boldpartial F
+
\int_A d^2 \Bx \ncap \lr{ \ncap \cdot \spacegrad } F,

so
\label{eqn:flux:120}
\begin{aligned}
\oint_{\partial A} d\Bx F
&=
\int_A d^2 \Bx \lr{ J – \ncap \lr{ \ncap \cdot \spacegrad } F } \\
&=
\int_A dA \lr{ I \ncap J – \lr{ \ncap \cdot \spacegrad } I F }
\end{aligned}

This is not nearly as nice as the magnetic flux relationship which was nicely split with the current and fields nicely separated. The $$d\Bx F$$ product has all possible grades, as does the $$d^2 \Bx J$$ product (in general). Observe however, that the normal term on the right has only grades 1,2, so we can split our line integral relations into pairs with and without grade 1,2 components
\label{eqn:flux:140}
\begin{aligned}
\oint_{\partial A} \gpgrade{d\Bx F}{0,3}
&=
\int_A dA \gpgrade{ I \ncap J }{0,3} \\
\oint_{\partial A} \gpgrade{d\Bx F}{1,2}
&=
\int_A dA \lr{ \gpgrade{ I \ncap J }{1,2} – \lr{ \ncap \cdot \spacegrad } I F }.
\end{aligned}

Let’s expand these explicitly in terms of the component fields and densities to check against the conventional relationships, and see if things look right. The line integrand expands to
\label{eqn:flux:180}
\begin{aligned}
d\Bx F
&=
d\Bx \lr{ \BE + I \eta \BH }
=
d\Bx \cdot \BE + I \eta d\Bx \cdot \BH
+
d\Bx \wedge \BE + I \eta d\Bx \wedge \BH \\
&=
d\Bx \cdot \BE
– \eta (d\Bx \cross \BH)
+ I (d\Bx \cross \BE )
+ I \eta (d\Bx \cdot \BH),
\end{aligned}

the current integrand expands to
\label{eqn:flux:200}
\begin{aligned}
I \ncap J
&=
I \ncap
\lr{
\frac{\rho}{\epsilon} – \eta \BJ + I \lr{ c \rho_\txtm – \BM }
} \\
&=
\ncap I \frac{\rho}{\epsilon} – \eta \ncap I \BJ – \ncap c \rho_\txtm + \ncap \BM \\
&=
\ncap \cdot \BM
+ \eta (\ncap \cross \BJ)
– \ncap c \rho_\txtm
+ I (\ncap \cross \BM)
+ \ncap I \frac{\rho}{\epsilon}
– \eta I (\ncap \cdot \BJ).
\end{aligned}

We are left with
\label{eqn:flux:220}
\begin{aligned}
\oint_{\partial A}
\lr{
d\Bx \cdot \BE + I \eta (d\Bx \cdot \BH)
}
&=
\int_A dA
\lr{
\ncap \cdot \BM – \eta I (\ncap \cdot \BJ)
} \\
\oint_{\partial A}
\lr{
– \eta (d\Bx \cross \BH)
+ I (d\Bx \cross \BE )
}
&=
\int_A dA
\lr{
\eta (\ncap \cross \BJ)
– \ncap c \rho_\txtm
+ I (\ncap \cross \BM)
+ \ncap I \frac{\rho}{\epsilon}
-\PD{n}{} \lr{ I \BE – \eta \BH }
}.
\end{aligned}

This is a crazy mess of dots, crosses, fields and sources. We can split it into one equation for each grade, which will probably look a little more regular. That is
\label{eqn:flux:240}
\begin{aligned}
\oint_{\partial A} d\Bx \cdot \BE &= \int_A dA \ncap \cdot \BM \\
\oint_{\partial A} d\Bx \cross \BH
&=
\int_A dA
\lr{
– \ncap \cross \BJ
+ \frac{ \ncap \rho_\txtm }{\mu}
– \PD{n}{\BH}
} \\
\oint_{\partial A} d\Bx \cross \BE &=
\int_A dA
\lr{
\ncap \cross \BM
+ \frac{\ncap \rho}{\epsilon}
– \PD{n}{\BE}
} \\
\oint_{\partial A} d\Bx \cdot \BH &= -\int_A dA \ncap \cdot \BJ \\
\end{aligned}

The first and last equations could have been obtained much more easily from Maxwell’s equations in their conventional form more easily. The two cross product equations with the normal derivatives are not familiar to me, even without the fictitious magnetic sources. It is somewhat remarkable that so much can be packed into one multivector equation:
\label{eqn:flux:260}
\oint_{\partial A} d\Bx F
=
I \int_A dA \lr{ \ncap J – \PD{n}{F} }.

# References

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

## Solving Maxwell’s equation in freespace: Multivector plane wave representation

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The geometric algebra form of Maxwell’s equations in free space (or source free isotopic media with group velocity $$c$$) is the multivector equation
\label{eqn:planewavesMultivector:20}
\lr{ \spacegrad + \inv{c}\PD{t}{} } F(\Bx, t) = 0.

Here $$F = \BE + I c \BB$$ is a multivector with grades 1 and 2 (vector and bivector components). The velocity $$c$$ is called the group velocity since $$F$$, or its components $$\BE, \BH$$ satisfy the wave equation, which can be seen by pre-multiplying with $$\spacegrad – (1/c)\PDi{t}{}$$ to find
\label{eqn:planewavesMultivector:n}
\lr{ \spacegrad^2 – \inv{c^2}\PDSq{t}{} } F(\Bx, t) = 0.

Let’s look at the frequency domain solution of this equation with a presumed phasor representation
\label{eqn:planewavesMultivector:40}
F(\Bx, t) = \textrm{Re} \lr{ F(\Bk) e^{-j \Bk \cdot \Bx + j \omega t} },

where $$j$$ is a scalar imaginary, not necessarily with any geometric interpretation.

Maxwell’s equation reduces to just
\label{eqn:planewavesMultivector:60}
0
=
-j \lr{ \Bk – \frac{\omega}{c} } F(\Bk).

If $$F(\Bk)$$ has a left multivector factor
\label{eqn:planewavesMultivector:80}
F(\Bk) =
\lr{ \Bk + \frac{\omega}{c} } \tilde{F},

where $$\tilde{F}$$ is a multivector to be determined, then
\label{eqn:planewavesMultivector:100}
\begin{aligned}
\lr{ \Bk – \frac{\omega}{c} }
F(\Bk)
&=
\lr{ \Bk – \frac{\omega}{c} }
\lr{ \Bk + \frac{\omega}{c} } \tilde{F} \\
&=
\lr{ \Bk^2 – \lr{\frac{\omega}{c}}^2 } \tilde{F},
\end{aligned}

which is zero if $$\Norm{\Bk} = \ifrac{\omega}{c}$$.

Let $$\kcap = \ifrac{\Bk}{\Norm{\Bk}}$$, and $$\Norm{\Bk} \tilde{F} = F_0 + F_1 + F_2 + F_3$$, where $$F_0, F_1, F_2,$$ and $$F_3$$ are respectively have grades 0,1,2,3. Then
\label{eqn:planewavesMultivector:120}
\begin{aligned}
F(\Bk)
&= \lr{ 1 + \kcap } \lr{ F_0 + F_1 + F_2 + F_3 } \\
&=
F_0 + F_1 + F_2 + F_3
+
\kcap F_0 + \kcap F_1 + \kcap F_2 + \kcap F_3 \\
&=
F_0 + F_1 + F_2 + F_3
+
\kcap F_0 + \kcap \cdot F_1 + \kcap \cdot F_2 + \kcap \cdot F_3
+
\kcap \wedge F_1 + \kcap \wedge F_2 \\
&=
\lr{
F_0 + \kcap \cdot F_1
}
+
\lr{
F_1 + \kcap F_0 + \kcap \cdot F_2
}
+
\lr{
F_2 + \kcap \cdot F_3 + \kcap \wedge F_1
}
+
\lr{
F_3 + \kcap \wedge F_2
}.
\end{aligned}

Since the field $$F$$ has only vector and bivector grades, the grades zero and three components of the expansion above must be zero, or
\label{eqn:planewavesMultivector:140}
\begin{aligned}
F_0 &= – \kcap \cdot F_1 \\
F_3 &= – \kcap \wedge F_2,
\end{aligned}

so
\label{eqn:planewavesMultivector:160}
\begin{aligned}
F(\Bk)
&=
\lr{ 1 + \kcap } \lr{
F_1 – \kcap \cdot F_1 +
F_2 – \kcap \wedge F_2
} \\
&=
\lr{ 1 + \kcap } \lr{
F_1 – \kcap F_1 + \kcap \wedge F_1 +
F_2 – \kcap F_2 + \kcap \cdot F_2
}.
\end{aligned}

The multivector $$1 + \kcap$$ has the projective property of gobbling any leading factors of $$\kcap$$
\label{eqn:planewavesMultivector:180}
\begin{aligned}
(1 + \kcap)\kcap
&= \kcap + 1 \\
&= 1 + \kcap,
\end{aligned}

so for $$F_i \in F_1, F_2$$
\label{eqn:planewavesMultivector:200}
(1 + \kcap) ( F_i – \kcap F_i )
=
(1 + \kcap) ( F_i – F_i )
= 0,

leaving
\label{eqn:planewavesMultivector:220}
F(\Bk)
=
\lr{ 1 + \kcap } \lr{
\kcap \cdot F_2 +
\kcap \wedge F_1
}.

For $$\kcap \cdot F_2$$ to be non-zero $$F_2$$ must be a bivector that lies in a plane containing $$\kcap$$, and $$\kcap \cdot F_2$$ is a vector in that plane that is perpendicular to $$\kcap$$. On the other hand $$\kcap \wedge F_1$$ is non-zero only if $$F_1$$ has a non-zero component that does not lie in along the $$\kcap$$ direction, but $$\kcap \wedge F_1$$, like $$F_2$$ describes a plane that containing $$\kcap$$. This means that having both bivector and vector free variables $$F_2$$ and $$F_1$$ provide more degrees of freedom than required. For example, if $$\BE$$ is any vector, and $$F_2 = \kcap \wedge \BE$$, then
\label{eqn:planewavesMultivector:240}
\begin{aligned}
\lr{ 1 + \kcap }
\kcap \cdot F_2
&=
\lr{ 1 + \kcap }
\kcap \cdot \lr{ \kcap \wedge \BE } \\
&=
\lr{ 1 + \kcap }
\lr{
\BE

\kcap \lr{ \kcap \cdot \BE }
} \\
&=
\lr{ 1 + \kcap }
\kcap \lr{ \kcap \wedge \BE } \\
&=
\lr{ 1 + \kcap }
\kcap \wedge \BE,
\end{aligned}

which has the form $$\lr{ 1 + \kcap } \lr{ \kcap \wedge F_1 }$$, so the solution of the free space Maxwell’s equation can be written
\label{eqn:planewavesMultivector:260}
\boxed{
F(\Bx, t)
=
\textrm{Re} \lr{
\lr{ 1 + \kcap }
\BE\,
e^{-j \Bk \cdot \Bx + j \omega t}
}
,
}

where $$\BE$$ is any vector for which $$\BE \cdot \Bk = 0$$.

## Transverse gauge

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Jackson [1] has an interesting presentation of the transverse gauge. I’d like to walk through the details of this, but first want to translate the preliminaries to SI units (if I had the 3rd edition I’d not have to do this translation step).

### Gauge freedom

The starting point is noting that $$\spacegrad \cdot \BB = 0$$ the magnetic field can be expressed as a curl

\label{eqn:transverseGauge:20}
\BB = \spacegrad \cross \BA.

Faraday’s law now takes the form
\label{eqn:transverseGauge:40}
\begin{aligned}
0
&= \spacegrad \cross \BE + \PD{t}{\BB} \\
&= \spacegrad \cross \BE + \PD{t}{} \lr{ \spacegrad \cross \BA } \\
&= \spacegrad \cross \lr{ \BE + \PD{t}{\BA} }.
\end{aligned}

Because this curl is zero, the interior sum can be expressed as a gradient

\label{eqn:transverseGauge:60}
\BE + \PD{t}{\BA} \equiv -\spacegrad \Phi.

This can now be substituted into the remaining two Maxwell’s equations.

\label{eqn:transverseGauge:80}
\begin{aligned}
\spacegrad \cdot \BD &= \rho_v \\
\spacegrad \cross \BH &= \BJ + \PD{t}{\BD} \\
\end{aligned}

For Gauss’s law, in simple media, we have

\label{eqn:transverseGauge:140}
\begin{aligned}
\rho_v
&=
\epsilon \spacegrad \cdot \BE \\
&=
\epsilon \spacegrad \cdot \lr{ -\spacegrad \Phi – \PD{t}{\BA} }
\end{aligned}

For simple media again, the Ampere-Maxwell equation is

\label{eqn:transverseGauge:100}
\inv{\mu} \spacegrad \cross \lr{ \spacegrad \cross \BA } = \BJ + \epsilon \PD{t}{} \lr{ -\spacegrad \Phi – \PD{t}{\BA} }.

Expanding $$\spacegrad \cross \lr{ \spacegrad \cross \BA } = -\spacegrad^2 \BA + \spacegrad \lr{ \spacegrad \cdot \BA }$$ gives
\label{eqn:transverseGauge:120}
-\spacegrad^2 \BA + \spacegrad \lr{ \spacegrad \cdot \BA } + \epsilon \mu \PDSq{t}{\BA} = \mu \BJ – \epsilon \mu \spacegrad \PD{t}{\Phi}.

Maxwell’s equations are now reduced to
\label{eqn:transverseGauge:180}
\boxed{
\begin{aligned}
\spacegrad^2 \BA – \spacegrad \lr{ \spacegrad \cdot \BA + \epsilon \mu \PD{t}{\Phi}} – \epsilon \mu \PDSq{t}{\BA} &= -\mu \BJ \\
\spacegrad^2 \Phi + \PD{t}{\spacegrad \cdot \BA} &= -\frac{\rho_v }{\epsilon}.
\end{aligned}
}

There are two obvious constraints that we can impose
\label{eqn:transverseGauge:200}
\spacegrad \cdot \BA – \epsilon \mu \PD{t}{\Phi} = 0,

or
\label{eqn:transverseGauge:220}
\spacegrad \cdot \BA = 0.

The first constraint is the Lorentz gauge, which I’ve played with previously. It happens to be really nice in a relativistic context since, in vacuum with a four-vector potential $$A = (\Phi/c, \BA)$$, that is a requirement that the four-divergence of the four-potential vanishes ($$\partial_\mu A^\mu = 0$$).

### Transverse gauge

Jackson identifies the latter constraint as the transverse gauge, which I’m less familiar with. With this gauge selection, we have

\label{eqn:transverseGauge:260}
\spacegrad^2 \BA – \epsilon \mu \PDSq{t}{\BA} = -\mu \BJ + \epsilon\mu \spacegrad \PD{t}{\Phi}

\label{eqn:transverseGauge:280}
\spacegrad^2 \Phi = -\frac{\rho_v }{\epsilon}.

What’s not obvious is the fact that the irrotational (zero curl) contribution due to $$\Phi$$ in \ref{eqn:transverseGauge:260} cancels the corresponding irrotational term from the current. Jackson uses a transverse and longitudinal decomposition of the current, related to the Helmholtz theorem to allude to this.

That decomposition follows from expanding $$\spacegrad^2 J/R$$ in two ways using the delta function $$-4 \pi \delta(\Bx – \Bx’) = \spacegrad^2 1/R$$ representation, as well as directly

\label{eqn:transverseGauge:300}
\begin{aligned}
– 4 \pi \BJ(\Bx)
&=
\int \spacegrad^2 \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’ \\
&=
\int \spacegrad \cdot \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’
+
\int \spacegrad \wedge \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’ \\
&=
\int \BJ(\Bx’) \cdot \spacegrad’ \inv{\Abs{\Bx – \Bx’}} d^3 x’
+
\int \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’
} \\
&=
\int \spacegrad’ \cdot \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’
\int \frac{\spacegrad’ \cdot \BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’

\int \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’
}
\end{aligned}

The first term can be converted to a surface integral

\label{eqn:transverseGauge:320}
\int \spacegrad’ \cdot \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’
=
\int d\BA’ \cdot \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}},

so provided the currents are either localized or $$\Abs{\BJ}/R \rightarrow 0$$ on an infinite sphere, we can make the identification

\label{eqn:transverseGauge:340}
\BJ(\Bx)
=
-\spacegrad \inv{4 \pi} \int \frac{\spacegrad’ \cdot \BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’
+
\spacegrad \cross \spacegrad \cross \inv{4 \pi} \int \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’
\equiv
\BJ_l +
\BJ_t,

where $$\spacegrad \cross \BJ_l = 0$$ (irrotational, or longitudinal), whereas $$\spacegrad \cdot \BJ_t = 0$$ (solenoidal or transverse). The irrotational property is clear from inspection, and the transverse property can be verified readily

\label{eqn:transverseGauge:360}
\begin{aligned}
\spacegrad \cdot \lr{ \spacegrad \cross \lr{ \spacegrad \cross \BX } }
&=
-\spacegrad \cdot \lr{ \spacegrad \cdot \lr{ \spacegrad \wedge \BX } } \\
&=
-\spacegrad \cdot \lr{ \spacegrad^2 \BX – \spacegrad \lr{ \spacegrad \cdot \BX } } \\
&=
&= 0.
\end{aligned}

Since

\label{eqn:transverseGauge:380}
\Phi(\Bx, t)
=
\inv{4 \pi \epsilon} \int \frac{\rho_v(\Bx’, t)}{\Abs{\Bx – \Bx’}} d^3 x’,

we have

\label{eqn:transverseGauge:400}
\begin{aligned}
&=
\inv{4 \pi \epsilon} \spacegrad \int \frac{\partial_t \rho_v(\Bx’, t)}{\Abs{\Bx – \Bx’}} d^3 x’ \\
&=
\inv{4 \pi \epsilon} \spacegrad \int \frac{-\spacegrad’ \cdot \BJ}{\Abs{\Bx – \Bx’}} d^3 x’ \\
&=
\frac{\BJ_l}{\epsilon}.
\end{aligned}

This means that the Ampere-Maxwell equation takes the form

\label{eqn:transverseGauge:420}
\spacegrad^2 \BA – \epsilon \mu \PDSq{t}{\BA}
= -\mu \BJ + \mu \BJ_l
= -\mu \BJ_t.

This justifies the transverse in the label transverse gauge.

# References

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

## Line charge field and potential.

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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}
\frac{\rhocap}{\rho} = \spacegrad \ln \rho,

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.

## Electric and magnetic fields at an interface

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As pointed out in [1] the fields at an interface that is not a perfect conductor on either side are related by

\label{eqn:fieldsAtInterface:20}
\begin{aligned}
\ncap \cdot \lr{ \BD_2 – \BD_1 } &= \rho_{es} \\
\ncap \cross \lr{ \BE_2 – \BE_1 } &= -\BM_s \\
\ncap \cdot \lr{ \BB_2 – \BB_1 } &= \rho_{ms} \\
\ncap \cross \lr{ \BH_2 – \BH_1 } &= \BJ_s.
\end{aligned}

Given the fields in medium 1, assuming that boths sets of media are linear, we can use these relationships to determine the fields in the other medium.

\label{eqn:fieldsAtInterface:40}
\begin{aligned}
\ncap \cdot \BE_2 &= \inv{\epsilon_2} \lr{ \epsilon_1 \ncap \cdot \BE_1 + \rho_{es} } \\
\ncap \wedge \BE_2 &= \ncap \wedge \BE_1 -I \BM_s \\
\ncap \cdot \BB_2 &= \ncap \cdot \BB_1 + \rho_{ms} \\
\ncap \wedge \BB_2 &= \mu_2 \lr{ \inv{\mu_1} \ncap \wedge \BB_1 + I \BJ_s}.
\end{aligned}

Now the fields in interface 2 can be obtained by adding the normal and tangential projections. For the electric field

\label{eqn:fieldsAtInterface:60}
\begin{aligned}
\BE_2
&=
\ncap (\ncap \cdot \BE_2 )
+ \ncap \cdot (\ncap \wedge \BE_2) \\
&=
\inv{\epsilon_2} \ncap \lr{ \epsilon_1 \ncap \cdot \BE_1 + \rho_{es} }
+
\ncap \cdot (\ncap \wedge \BE_1 -I \BM_s).
\end{aligned}

Note that this manipulation can also be done without Geometric Algebra by writing $$\BE_2 = \ncap (\ncap \cdot \BE_2 ) – \ncap \cross (\ncap \cross \BE_2)$$).
Expanding $$\ncap \cdot (\ncap \wedge \BE_1) = \BE_1 – \ncap (\ncap \cdot \BE_1)$$, and $$\ncap \cdot (I \BM_s) = -\ncap \cross \BM_s$$, that is

\label{eqn:fieldsAtInterface:80}
\boxed{
\BE_2
=
\BE_1
+ \ncap (\ncap \cdot \BE_1) \lr{ \frac{\epsilon_1}{\epsilon_2} – 1 }
+ \frac{\rho_{es}}{\epsilon_2}
+ \ncap \cross \BM_s.
}

For the magnetic field

\label{eqn:fieldsAtInterface:100}
\begin{aligned}
\BB_2
&=
\ncap (\ncap \cdot \BB_2 )
+
\ncap \cdot (\ncap \wedge \BB_2) \\
&=
\ncap \lr{ \ncap \cdot \BB_1 + \rho_{ms} }
+
\mu_2 \ncap \cdot \lr{ \lr{ \inv{\mu_1} \ncap \wedge \BB_1 + I \BJ_s} },
\end{aligned}

which is

\label{eqn:fieldsAtInterface:120}
\boxed{
\BB_2
=
\frac{\mu_2}{\mu_1} \BB_1
+
\ncap (\ncap \cdot \BB_1) \lr{ 1 – \frac{\mu_2}{\mu_1} }
+ \ncap \rho_{ms}
– \ncap \cross \BJ_s.
}

These are kind of pretty results, having none of the explicit angle dependence that we see in the Fresnel relationships. In this analysis, it is assumed there is only a transmitted component of the ray in question, and no reflected component. Can we do a purely vectoral treatment of the Fresnel equations along these same lines?

# References

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

## Plane wave solution directly from Maxwell’s equations

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Here’s a problem that I thought was fun, an exercise for the reader to show that the plane wave solution to Maxwell’s equations can be found with ease directly from Maxwell’s equations. This is in contrast to the what seems like the usual method of first showing that Maxwell’s equations imply wave equations for the fields, and then solving those wave equations.

## Problem. $$\xcap$$ oriented plane wave electric field ([1] ex. 4.1)

A uniform plane wave having only an $$x$$ component of the electric field is traveling in the $$+ z$$ direction in an unbounded lossless, source-0free region. Using Maxwell’s equations write expressions for the electric and corresponding magnetic field intensities.

The phasor form of Maxwell’s equations for a source free region are

\label{eqn:ExPlaneWave:40}
\spacegrad \cross \BE = -j \omega \BB

\label{eqn:ExPlaneWave:60}
\spacegrad \cross \BH = j \omega \BD

\label{eqn:ExPlaneWave:80}
\spacegrad \cdot \BD = 0

\label{eqn:ExPlaneWave:100}
\spacegrad \cdot \BB = 0.

Since $$\BE = \xcap E(z)$$, the magnetic field follows from \ref{eqn:ExPlaneWave:40}

\label{eqn:ExPlaneWave:120}
-j \omega \BB
= \spacegrad \cross \BE
=
\begin{vmatrix}
\xcap & \ycap & \zcap \\
\partial_x & \partial_y & \partial_z \\
E & 0 & 0
\end{vmatrix}
=
\ycap \partial_z E(z)
– \zcap \partial_y E(z),

or

\label{eqn:ExPlaneWave:140}
\BB =
-\inv{j \omega} \partial_z E.

This is constrained by \ref{eqn:ExPlaneWave:60}

\label{eqn:ExPlaneWave:160}
j \omega \epsilon \xcap E
=
\inv{\mu} \spacegrad \cross \BB
=
-\inv{\mu j \omega}
\begin{vmatrix}
\xcap & \ycap & \zcap \\
\partial_x & \partial_y & \partial_z \\
0 & \partial_z E & 0
\end{vmatrix}
=
-\inv{\mu j \omega}
\lr{
-\xcap \partial_{z z} E
+ \zcap \partial_x \partial_z E
}

Since $$\partial_x \partial_z E = \partial_z \lr{ \partial_x E } = \partial_z \inv{\epsilon} \spacegrad \cdot \BD = \partial_z 0$$, this means

\label{eqn:ExPlaneWave:180}
\partial_{zz} E = -\omega^2 \epsilon\mu E = -k^2 E.

This is the usual starting place that we use to show that the plane wave has an exponential form

\label{eqn:ExPlaneWave:200}
\BE(z) =
\xcap
\lr{
E_{+} e^{-j k z}
+
E_{-} e^{j k z}
}.

The magnetic field from \ref{eqn:ExPlaneWave:140} is

\label{eqn:ExPlaneWave:220}
\BB
= \frac{j}{\omega} \lr{ -j k E_{+} e^{-j k z} + j k E_{-} e^{j k z} }
= \inv{c} \lr{ E_{+} e^{-j k z} – E_{-} e^{j k z} },

or

\label{eqn:ExPlaneWave:240}
\BH
= \inv{\mu c} \lr{ E_{+} e^{-j k z} – E_{-} e^{j k z} }
= \inv{\eta} \lr{ E_{+} e^{-j k z} – E_{-} e^{j k z} }.

A solution requires zero divergence for the magnetic field, but that can be seen to be the case by inspection.

# References

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

## Coupled wave equation in cylindrical coordinates

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In [1], for a sourceless configuration, it is noted that the electric field equations $$\spacegrad^2 \BE = -\beta^2 \BE$$ have the form

\label{eqn:cylindricalFieldSolution:20}
\spacegrad^2 E_\rho – \frac{E_\rho}{\rho^2} – \frac{2}{\rho^2} \PD{\phi}{E_\phi} = -\beta^2 E_\rho

\label{eqn:cylindricalFieldSolution:60}
\spacegrad^2 E_\phi – \frac{E_\phi}{\rho^2} + \frac{2}{\rho^2} \PD{\phi}{E_\rho} = -\beta^2 E_\phi

\label{eqn:cylindricalFieldSolution:80}
\spacegrad^2 E_z = -\beta^2 E_z,

where

\label{eqn:cylindricalFieldSolution:100}
\inv{\rho} \PD{\rho}{} \lr{ \rho \PD{\rho}{\psi}} + \inv{\rho^2}\PDSq{\phi}{\psi} + \PDSq{z}{\psi}.

He applies separation of variables to the last equation, ending up with the usual Bessel function solution, but the first two coupled equations are dismissed as coupled and difficult. It looks like separation of variables works for this too, but we have to prep the system slightly by writing $$\psi = E_\rho + j E_\phi$$, which gives

\label{eqn:cylindricalFieldSolution:120}
\spacegrad^2 \psi – \frac{\psi}{\rho^2} + \frac{2 j}{\rho^2} \PD{\phi}{\psi} = -\beta^2 \psi,

or

\label{eqn:cylindricalFieldSolution:140}
\inv{\rho} \PD{\rho}{} \lr{ \rho \PD{\rho}{\psi}} + \inv{\rho^2}\PDSq{\phi}{\psi} + \PDSq{z}{\psi}
– \frac{\psi}{\rho^2} + \frac{2 j}{\rho^2} \PD{\phi}{\psi} = -\beta^2 \psi.

With a separation of variables substitution $$\psi = f(\rho) g(\phi) h(z)$$ this gives

\label{eqn:cylindricalFieldSolution:160}
\inv{\rho f} \PD{\rho}{} \lr{ \rho \PD{\rho}{f}}
+ \inv{\rho^2 g}\PDSq{\phi}{g}
+ \inv{z} \PDSq{z}{h}
– \frac{1}{\rho^2} + \frac{2 j}{\rho^2 g} \PD{\phi}{g} = -\beta^2.

Assuming a solution for the function $$h$$ of

\label{eqn:cylindricalFieldSolution:180}
\inv{z} \PDSq{z}{h} = -\alpha^2,

the PDE is reduced to an equation in two functions

\label{eqn:cylindricalFieldSolution:200}
\inv{\rho f} \PD{\rho}{} \lr{ \rho \PD{\rho}{f}}
+ \inv{\rho^2 g}\PD{\phi}{} \lr{ g + 2 j g}
+ \beta^2 -\alpha^2
– \frac{1}{\rho^2}
= 0,

or

\label{eqn:cylindricalFieldSolution:220}
\frac{\rho}{f} \PD{\rho}{} \lr{ \rho \PD{\rho}{f}}
+ \inv{g}\PD{\phi}{} \lr{ g + 2 j g}
+ \lr{ \beta^2 -\alpha^2 }\rho^2
= 1.

With the term in $$g$$ having only $$\phi$$ dependence, we can assume

\label{eqn:cylindricalFieldSolution:240}
\inv{g}\PD{\phi}{} \lr{ g + 2 j g} = 1 – \gamma^2,

for

\label{eqn:cylindricalFieldSolution:260}
\frac{\rho}{f} \PD{\rho}{} \lr{ \rho \PD{\rho}{f}}
– \gamma^2
+ \lr{ \beta^2 -\alpha^2 }\rho^2
= 0.

I’m not sure off hand if these can be solved in known special functions, especially since the constants in the mix are complex.

# References

[1] Constantine A Balanis. Advanced engineering electromagnetics, volume 20. Wiley New York, 1989.

## Updated notes for ece1229 antenna theory

I’ve now posted a first update of my notes for the antenna theory course that I am taking this term at UofT.

Unlike most of the other classes I have taken, I am not attempting to take comprehensive notes for this class. The class is taught on slides which go by faster than I can easily take notes for (and some of which match the textbook closely). In class I have annotated my copy of textbook with little details instead. This set of notes contains musings of details that were unclear, or in some cases, details that were provided in class, but are not in the text (and too long to pencil into my book), as well as some notes Geometric Algebra formalism for Maxwell’s equations with magnetic sources (something I’ve encountered for the first time in any real detail in this class).

The notes compilation linked above includes all of the following separate notes, some of which have been posted separately on this blog:

## Parallel projection of electromagnetic fields with Geometric Algebra

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When computing the components of a polarized reflecting ray that were parallel or not-parallel to the reflecting surface, it was found that the electric and magnetic fields could be written as

\label{eqn:gaFieldProjection:280}
\BE = \lr{ \BE \cdot \pcap } \pcap + \lr{ \BE \cdot \qcap } \qcap = E_\parallel \pcap + E_\perp \qcap

\label{eqn:gaFieldProjection:300}
\BH = \lr{ \BH \cdot \pcap } \pcap + \lr{ \BH \cdot \qcap } \qcap = H_\parallel \pcap + H_\perp \qcap.

where a unit vector $$\pcap$$ that lies both in the reflecting plane and in the electromagnetic plane (tangential to the wave vector direction) was

\label{eqn:gaFieldProjection:340}
\pcap = \frac{\kcap \cross \ncap}{\Abs{\kcap \cross \ncap}}

\label{eqn:gaFieldProjection:360}
\qcap = \kcap \cross \pcap.

Here $$\qcap$$ is perpendicular to $$\pcap$$ but lies in the electromagnetic plane. This logically subdivides the fields into two pairs, one with the electric field parallel to the reflection plane

\label{eqn:gaFieldProjection:240}
\begin{aligned}
\BE_1 &= \lr{ \BE \cdot \pcap } \pcap = E_\parallel \pcap \\
\BH_1 &= \lr{ \BH \cdot \qcap } \qcap = H_\perp \qcap,
\end{aligned}

and one with the magnetic field parallel to the reflection plane

\label{eqn:gaFieldProjection:380}
\begin{aligned}
\BH_2 &= \lr{ \BH \cdot \pcap } \pcap = H_\parallel \pcap \\
\BE_2 &= \lr{ \BE \cdot \qcap } \qcap = E_\perp \qcap.
\end{aligned}

Expressed in Geometric Algebra form, each of these pairs of fields should be thought of as components of a single multivector field. That is

\label{eqn:gaFieldProjection:400}
F_1 = \BE_1 + c \mu_0 \BH_1 I

\label{eqn:gaFieldProjection:460}
F_2 = \BE_2 + c \mu_0 \BH_2 I

where the original total field is

\label{eqn:gaFieldProjection:420}
F = \BE + c \mu_0 \BH I.

In \ref{eqn:gaFieldProjection:400} we have a composite projection operation, finding the portion of the electric field that lies in the reflection plane, and simultaneously finding the component of the magnetic field that lies perpendicular to that (while still lying in the tangential plane of the electromagnetic field). In \ref{eqn:gaFieldProjection:460} the magnetic field is projected onto the reflection plane and a component of the electric field that lies in the tangential (to the wave vector direction) plane is computed.

If we operate only on the complete multivector field, can we find these composite projection field components in a single operation, instead of working with the individual electric and magnetic fields?

Working towards this goal, it is worthwhile to point out consequences of the assumption that the fields are plane wave (or equivalently far field spherical waves). For such a wave we have

\label{eqn:gaFieldProjection:480}
\begin{aligned}
\BH
&= \inv{\mu_0} \kcap \cross \BE \\
&= \inv{\mu_0} (-I)\lr{ \kcap \wedge \BE } \\
&= \inv{\mu_0} (-I)\lr{ \kcap \BE – \kcap \cdot \BE} \\
&= -\frac{I}{\mu_0} \kcap \BE,
\end{aligned}

or

\label{eqn:gaFieldProjection:520}
\mu_0 \BH I = \kcap \BE.

This made use of the identity $$\Ba \wedge \Bb = I \lr{\Ba \cross \Bb}$$, and the fact that the electric field is perpendicular to the wave vector direction. The total multivector field is

\label{eqn:gaFieldProjection:500}
\begin{aligned}
F
&= \BE + c \mu_0 \BH I \\
&= \lr{ 1 + c \kcap } \BE.
\end{aligned}

Expansion of magnetic field component that is perpendicular to the reflection plane gives

\label{eqn:gaFieldProjection:540}
\begin{aligned}
\mu_0 H_\perp
&= \mu_0 \BH \cdot \qcap \\
&= \gpgradezero{ \lr{-\kcap \BE I} \qcap } \\
&= -\gpgradezero{ \kcap \BE I \lr{ \kcap \cross \pcap} } \\
&= \gpgradezero{ \kcap \BE I I \lr{ \kcap \wedge \pcap} } \\
&= -\gpgradezero{ \kcap \BE \kcap \pcap } \\
&= \gpgradezero{ \kcap \kcap \BE \pcap } \\
&= \BE \cdot \pcap,
\end{aligned}

so

\label{eqn:gaFieldProjection:560}
F_1
= (\pcap + c I \qcap ) \BE \cdot \pcap.

Since $$\qcap \kcap \pcap = I$$, the component of the complete multivector field in the $$\pcap$$ direction is

\label{eqn:gaFieldProjection:580}
\begin{aligned}
F_1
&= (\pcap – c \pcap \kcap ) \BE \cdot \pcap \\
&= \pcap (1 – c \kcap ) \BE \cdot \pcap \\
&= (1 + c \kcap ) \pcap \BE \cdot \pcap.
\end{aligned}

It is reasonable to expect that $$F_2$$ has a similar form, but with $$\pcap \rightarrow \qcap$$. This is verified by expansion

\label{eqn:gaFieldProjection:600}
\begin{aligned}
F_2
&= E_\perp \qcap + c \lr{ \mu_0 H_\parallel } \pcap I \\
&= \lr{\BE \cdot \qcap} \qcap + c \gpgradezero{ – \kcap \BE I \kcap \qcap I } \lr{\kcap \qcap I} I \\
&= \lr{\BE \cdot \qcap} \qcap + c \gpgradezero{ \kcap \BE \kcap \qcap } \kcap \qcap (-1) \\
&= \lr{\BE \cdot \qcap} \qcap + c \gpgradezero{ \kcap \BE (-\qcap \kcap) } \kcap \qcap (-1) \\
&= \lr{\BE \cdot \qcap} \qcap + c \gpgradezero{ \kcap \kcap \BE \qcap } \kcap \qcap \\
&= \lr{ 1 + c \kcap } \qcap \lr{ \BE \cdot \qcap }
\end{aligned}

This and \ref{eqn:gaFieldProjection:580} before that makes a lot of sense. The original field can be written

\label{eqn:gaFieldProjection:620}
F = \lr{ \Ecap + c \lr{ \kcap \cross \Ecap } I } \BE \cdot \Ecap,

where the leading multivector term contains all the directional dependence of the electric and magnetic field components, and the trailing scalar has the magnitude of the field with respect to the reference direction $$\Ecap$$.

We have the same structure after projecting $$\BE$$ onto either the $$\pcap$$, or $$\qcap$$ directions respectively

\label{eqn:gaFieldProjection:660}
F_1 = \lr{ \pcap + c \lr{ \kcap \cross \pcap } I} \BE \cdot \pcap

\label{eqn:gaFieldProjection:680}
F_2 = \lr{ \qcap + c \lr{ \kcap \cross \qcap } I} \BE \cdot \qcap.

The next question is how to achieve this projection operation directly in terms of $$F$$ and $$\pcap, \qcap$$, without resorting to expression of $$F$$ in terms of $$\BE$$, and $$\BB$$. I’ve not yet been able to determine the structure of that operation.