magnetic field

Transverse gauge

November 16, 2016 math and physics play No comments , , , , , , , , , , , , , , , , ,

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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

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

Faraday’s law now takes the form
\begin{equation}\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}
\end{equation}

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

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

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

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

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

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

For simple media again, the Ampere-Maxwell equation is

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

Expanding \( \spacegrad \cross \lr{ \spacegrad \cross \BA } = -\spacegrad^2 \BA + \spacegrad \lr{ \spacegrad \cdot \BA } \) gives
\begin{equation}\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}.
\end{equation}

Maxwell’s equations are now reduced to
\begin{equation}\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}
}
\end{equation}

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

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

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

\begin{equation}\label{eqn:transverseGauge:260}
\spacegrad^2 \BA – \epsilon \mu \PDSq{t}{\BA} = -\mu \BJ + \epsilon\mu \spacegrad \PD{t}{\Phi}
\end{equation}
\begin{equation}\label{eqn:transverseGauge:280}
\spacegrad^2 \Phi = -\frac{\rho_v }{\epsilon}.
\end{equation}

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

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

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

The first term can be converted to a surface integral

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

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

\begin{equation}\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,
\end{equation}

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

\begin{equation}\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 } } \\
&=
-\spacegrad \cdot \lr{\spacegrad^2 \BX} + \spacegrad^2 \lr{ \spacegrad \cdot \BX } \\
&= 0.
\end{aligned}
\end{equation}

Since

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

we have

\begin{equation}\label{eqn:transverseGauge:400}
\begin{aligned}
\spacegrad \PD{t}{\Phi}
&=
\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}
\end{equation}

This means that the Ampere-Maxwell equation takes the form

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

This justifies the transverse in the label transverse gauge.

References

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

Calculating the magnetostatic field from the moment

November 14, 2016 math and physics play No comments , , , , , ,

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The vector potential, to first order, for a magnetostatic localized current distribution was found to be

\begin{equation}\label{eqn:magneticFieldFromMoment:20}
\BA(\Bx) = \frac{\mu_0}{4 \pi} \frac{\Bm \cross \Bx}{\Abs{\Bx}^3}.
\end{equation}

Initially, I tried to calculate the magnetic field from this, but ran into trouble. Here’s a new try.

\begin{equation}\label{eqn:magneticFieldFromMoment:40}
\begin{aligned}
\BB
&=
\frac{\mu_0}{4 \pi}
\spacegrad \cross \lr{ \Bm \cross \frac{\Bx}{r^3} } \\
&=
-\frac{\mu_0}{4 \pi}
\spacegrad \cdot \lr{ \Bm \wedge \frac{\Bx}{r^3} } \\
&=
-\frac{\mu_0}{4 \pi}
\lr{
(\Bm \cdot \spacegrad) \frac{\Bx}{r^3}
-\Bm \spacegrad \cdot \frac{\Bx}{r^3}
} \\
&=
\frac{\mu_0}{4 \pi}
\lr{
-\frac{(\Bm \cdot \spacegrad) \Bx}{r^3}
– \lr{ \Bm \cdot \lr{\spacegrad \inv{r^3} }} \Bx
+\Bm (\spacegrad \cdot \Bx) \inv{r^3}
+\Bm \lr{\spacegrad \inv{r^3} } \cdot \Bx
}.
\end{aligned}
\end{equation}

Here I’ve used \( \Ba \cross \lr{ \Bb \cross \Bc } = -\Ba \cdot \lr{ \Bb \wedge \Bc } \), and then expanded that with \( \Ba \cdot \lr{ \Bb \wedge \Bc } = (\Ba \cdot \Bb) \Bc – (\Ba \cdot \Bc) \Bb \). Since one of these vectors is the gradient, care must be taken to have it operate on the appropriate terms in such an expansion.

Since we have \( \spacegrad \cdot \Bx = 3 \), \( (\Bm \cdot \spacegrad) \Bx = \Bm \), and \( \spacegrad 1/r^n = -n \Bx/r^{n+2} \), this reduces to

\begin{equation}\label{eqn:magneticFieldFromMoment:60}
\begin{aligned}
\BB
&=
\frac{\mu_0}{4 \pi}
\lr{
– \frac{\Bm}{r^3}
+ 3 \frac{(\Bm \cdot \Bx) \Bx}{r^5} %
+ 3 \Bm \inv{r^3}
-3 \Bm \frac{\Bx}{r^5} \cdot \Bx
} \\
&=
\frac{\mu_0}{4 \pi}
\frac{3 (\Bm \cdot \ncap) \ncap -\Bm}{r^3},
\end{aligned}
\end{equation}

which is the desired result.

Magnetostatic force and torque

October 18, 2016 math and physics play No comments , , , , , , ,

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In Jackson [1], the following equations for the vector potential, magnetostatic force and torque are derived

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:20}
\Bm = \inv{2} \int \Bx’ \cross \BJ(\Bx’) d^3 x’
\end{equation}
\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:40}
\BF = \spacegrad( \Bm \cdot \BB ),
\end{equation}
\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:60}
\BN = \Bm \cross \BB,
\end{equation}

where \( \BB \) is an applied external magnetic field and \( \Bm \) is the magnetic dipole for the current in question. These results (and a similar one derived earlier for the vector potential \( \BA \)) all follow from
an analysis of localized current densities \( \BJ \), evaluated far enough away from the current sources.

For the force and torque, the starting point for the force is one that had me puzzled a bit. Namely

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:80}
\BF = \int \BJ(\Bx) \cross \BB(\Bx) d^3 x
\end{equation}

This is clearly the continuum generalization of the point particle Lorentz force equation, which for \( \BE = 0 \) is:

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:100}
\BF = q \Bv \cross \BB
\end{equation}

For the point particle, this is the force on the particle when it is in the external field \( BB \). i.e. this is the force at the position of the particle. My question is what does it mean to sum all the forces on the charge distribution over all space.
How can a force be applied over all, as opposed to a force applied at a single point, or against a surface?

In the special case of a localized current density, this makes some sense. Considering the other half of the force equation \( \BF = \ddt{}\int \rho_m \Bv dV \), where \( \rho_m \) here is mass density of the charged particles making up the continuous current distribution. The other half of this \( \BF = m\Ba \) equation is also an average phenomena, so we have an average of sorts on both the field contribution to the force equation and the mass contribution to the force equation. There is probably a centre-of-mass and centre-of-current density interpretation that would make a bit more sense of this continuum force description.

It’s kind of funny how you can work through all the detailed mathematical steps in a book like Jackson, but then go right back to the beginning and say “Hey, what does that even mean”?

Force

Moving on from the pondering of the meaning of the equation being manipulated, let’s do the easy part, the derivation of the results that Jackson comes up with.

Writing out \ref{eqn:magnetostaticsJacksonNotesForceAndTorque:80} in coordinates

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:320}
\BF = \epsilon_{ijk} \Be_i \int J_j B_k d^3 x.
\end{equation}

To first order, a slowly varying (external) magnetic field can be expanded around a point of interest

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:120}
\BB(\Bx) = \BB(\Bx_0) + \lr{ \Bx – \Bx_0 } \cdot \spacegrad \BB,
\end{equation}

where the directional derivative is evaluated at the point \( \Bx_0 \) after the gradient operation. Setting the origin at this point \( \Bx_0 \) gives

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:340}
\begin{aligned}
\BF
&= \epsilon_{ijk} \Be_i
\lr{
\int J_j(\Bx’) B_k(0) d^3 x’
+
\int J_j(\Bx’) (\Bx’ \cdot \spacegrad) B_k(0) d^3 x’
} \\
&=
\epsilon_{ijk} \Be_i
\Bk_0 \int J_j(\Bx’) d^3 x’
+
\epsilon_{ijk} \Be_i
\int J_j(\Bx’) (\Bx’ \cdot \spacegrad) B_k(0) d^3 x’.
\end{aligned}
\end{equation}

We found

earlier
that the first integral can be written as a divergence

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:140}
\int J_j(\Bx’) d^3 x’
=
\int \spacegrad’ \cdot \lr{ \BJ(\Bx’) x_j’ } dV’,
\end{equation}

which is zero when the integration surface is outside of the current localization region. We also found

that

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:160}
\int (\Bx \cdot \Bx’) \BJ
= -\inv{2} \Bx \cross \int \Bx’ \cross \BJ = \Bm \cross \Bx.
\end{equation}

so
\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:180}
\begin{aligned}
\int (\spacegrad B_k(0) \cdot \Bx’) J_j
&= -\inv{2} \lr{ \spacegrad B_k(0) \cross \int \Bx’ \cross \BJ}_j \\
&= \lr{ \Bm \cross (\spacegrad B_k(0)) }_j.
\end{aligned}
\end{equation}

This gives

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:200}
\begin{aligned}
\BF
&= \epsilon_{ijk} \Be_i \lr{ \Bm \cross (\spacegrad B_k(0)) }_j \\
&= \epsilon_{ijk} \Be_i \lr{ \Bm \cross \spacegrad }_j B_k(0) \\
&= (\Bm \cross \spacegrad) \cross \BB(0) \\
&= -\BB(0) \cross (\Bm \cross \lspacegrad) \\
&= (\BB(0) \cdot \Bm) \lspacegrad – (\BB \cdot \lspacegrad) \Bm \\
&= \spacegrad (\BB(0) \cdot \Bm) – \Bm (\spacegrad \cdot \BB(0)).
\end{aligned}
\end{equation}

The second term is killed by the magnetic Gauss’s law, leaving to first order

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:220}
\BF = \spacegrad \lr{\Bm \cdot \BB}.
\end{equation}

Torque

For the torque we have a similar quandary at the starting point. About what point is a continuum torque integral of the following form

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:240}
\BN = \int \Bx’ \cross (\BJ(\Bx’) \cross \BB(\Bx’)) d^3 x’?
\end{equation}

Ignoring that detail again, assuming the answer has something to do with the centre of mass and parallel axis theorem, we can proceed with a constant approximation of the magnetic field

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:260}
\begin{aligned}
\BN
&= \int \Bx’ \cross (\BJ(\Bx’) \cross \BB(0)) d^3 x’ \\
&=
-\int (\Bx’ \cdot \BJ(\Bx’)) \BB(0) d^3 x’
+\int (\Bx’ \cdot \BB(0)) \BJ(\Bx’) d^3 x’ \\
&=
-\BB(0) \int (\Bx’ \cdot \BJ(\Bx’)) d^3 x’
+\int (\Bx’ \cdot \BB(0)) \BJ(\Bx’) d^3 x’.
\end{aligned}
\end{equation}

Jackson’s trick for killing the first integral is to transform it into a divergence by evaluating

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:280}
\begin{aligned}
\spacegrad \cdot \lr{ \BJ \Abs{\Bx}^2 }
&=
(\spacegrad \cdot \BJ) \Abs{\Bx}^2
+
\BJ \cdot \spacegrad \Abs{\Bx}^2 \\
&=
\BJ \cdot \Be_i \partial_i x_m x_m \\
&=
2 \BJ \cdot \Be_i \delta_{im} x_m \\
&=
2 \BJ \cdot \Bx,
\end{aligned}
\end{equation}

so

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:300}
\begin{aligned}
\BN
&=
-\inv{2} \BB(0) \int \spacegrad’ \cdot \lr{ \BJ(\Bx’) \Abs{\Bx’}^2 } d^3 x’
+\int (\Bx’ \cdot \BB(0)) \BJ(\Bx’) d^3 x’ \\
&=
-\inv{2} \BB(0) \oint \Bn \cdot \lr{ \BJ(\Bx’) \Abs{\Bx’}^2 } d^3 x’
+\int (\Bx’ \cdot \BB(0)) \BJ(\Bx’) d^3 x’.
\end{aligned}
\end{equation}

Again, the localized current density assumption kills the surface integral. The second integral can be evaluated with \ref{eqn:magnetostaticsJacksonNotesForceAndTorque:160}, so to first order we have

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:360}
\BN
=
\Bm \cross \BB.
\end{equation}

References

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

Magnetic moment for a localized magnetostatic current

October 13, 2016 math and physics play No comments , , , , , , , , ,

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Motivation.

I was once again reading my Jackson [2]. This time I found that his presentation of magnetic moment didn’t really make sense to me. Here’s my own pass through it, filling in a number of details. As I did last time, I’ll also translate into SI units as I go.

Vector potential.

The Biot-Savart expression for the magnetic field can be factored into a curl expression using the usual tricks

\begin{equation}\label{eqn:magneticMomentJackson:20}
\begin{aligned}
\BB
&= \frac{\mu_0}{4\pi} \int \frac{\BJ(\Bx’) \cross (\Bx – \Bx’)}{\Abs{\Bx – \Bx’}^3} d^3 x’ \\
&= -\frac{\mu_0}{4\pi} \int \BJ(\Bx’) \cross \spacegrad \inv{\Abs{\Bx – \Bx’}} d^3 x’ \\
&= \frac{\mu_0}{4\pi} \spacegrad \cross \int \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’,
\end{aligned}
\end{equation}

so the vector potential, through its curl, defines the magnetic field \( \BB = \spacegrad \cross \BA \) is given by

\begin{equation}\label{eqn:magneticMomentJackson:40}
\BA(\Bx) = \frac{\mu_0}{4 \pi} \int \frac{J(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’.
\end{equation}

If the current source is localized (zero outside of some finite region), then there will always be a region for which \( \Abs{\Bx} \gg \Abs{\Bx’} \), so the denominator yields to Taylor expansion

\begin{equation}\label{eqn:magneticMomentJackson:60}
\begin{aligned}
\inv{\Abs{\Bx – \Bx’}}
&=
\inv{\Abs{\Bx}} \lr{1 + \frac{\Abs{\Bx’}^2}{\Abs{\Bx}^2} – 2 \frac{\Bx \cdot \Bx’}{\Abs{\Bx}^2} }^{-1/2} \\
&\approx
\inv{\Abs{\Bx}} \lr{ 1 + \frac{\Bx \cdot \Bx’}{\Abs{\Bx}^2} } \\
&=
\inv{\Abs{\Bx}} + \frac{\Bx \cdot \Bx’}{\Abs{\Bx}^3}.
\end{aligned}
\end{equation}

so the vector potential, far enough away from the current source is
\begin{equation}\label{eqn:magneticMomentJackson:80}
\BA(\Bx)
=
\frac{\mu_0}{4 \pi} \int \frac{J(\Bx’)}{\Abs{\Bx}} d^3 x’
+\frac{\mu_0}{4 \pi} \int \frac{(\Bx \cdot \Bx’)J(\Bx’)}{\Abs{\Bx}^3} d^3 x’.
\end{equation}

Jackson uses a sneaky trick to show that the first integral is killed for a localized source. That trick appears to be based on evaluating the following divergence

\begin{equation}\label{eqn:magneticMomentJackson:100}
\begin{aligned}
\spacegrad \cdot (\BJ(\Bx) x_i)
&=
(\spacegrad \cdot \BJ) x_i
+
(\spacegrad x_i) \cdot \BJ \\
&=
(\Be_k \partial_k x_i) \cdot\BJ \\
&=
\delta_{ki} J_k \\
&=
J_i.
\end{aligned}
\end{equation}

Note that this made use of the fact that \( \spacegrad \cdot \BJ = 0 \) for magnetostatics. This provides a way to rewrite the current density as a divergence

\begin{equation}\label{eqn:magneticMomentJackson:120}
\begin{aligned}
\int \frac{J(\Bx’)}{\Abs{\Bx}} d^3 x’
&=
\Be_i \int \frac{\spacegrad’ \cdot (x_i’ \BJ(\Bx’))}{\Abs{\Bx}} d^3 x’ \\
&=
\frac{\Be_i}{\Abs{\Bx}} \int \spacegrad’ \cdot (x_i’ \BJ(\Bx’)) d^3 x’ \\
&=
\frac{1}{\Abs{\Bx}} \oint \Bx’ (d\Ba \cdot \BJ(\Bx’)).
\end{aligned}
\end{equation}

When \( \BJ \) is localized, this is zero provided we pick the integration surface for the volume outside of that localization region.

It is now desired to rewrite \( \int \Bx \cdot \Bx’ \BJ \) as a triple cross product since the dot product of such a triple cross product has exactly this term in it

\begin{equation}\label{eqn:magneticMomentJackson:140}
\begin{aligned}
– \Bx \cross \int \Bx’ \cross \BJ
&=
\int (\Bx \cdot \Bx’) \BJ

\int (\Bx \cdot \BJ) \Bx’ \\
&=
\int (\Bx \cdot \Bx’) \BJ

\Be_k x_i \int J_i x_k’,
\end{aligned}
\end{equation}

so
\begin{equation}\label{eqn:magneticMomentJackson:160}
\int (\Bx \cdot \Bx’) \BJ
=
– \Bx \cross \int \Bx’ \cross \BJ
+
\Be_k x_i \int J_i x_k’.
\end{equation}

To get of this second term, the next sneaky trick is to consider the following divergence

\begin{equation}\label{eqn:magneticMomentJackson:180}
\begin{aligned}
\oint d\Ba’ \cdot (\BJ(\Bx’) x_i’ x_j’)
&=
\int dV’ \spacegrad’ \cdot (\BJ(\Bx’) x_i’ x_j’) \\
&=
\int dV’ (\spacegrad’ \cdot \BJ)
+
\int dV’ \BJ \cdot \spacegrad’ (x_i’ x_j’) \\
&=
\int dV’ J_k \cdot \lr{ x_i’ \partial_k x_j’ + x_j’ \partial_k x_i’ } \\
&=
\int dV’ \lr{J_k x_i’ \delta_{kj} + J_k x_j’ \delta_{ki}} \\
&=
\int dV’ \lr{J_j x_i’ + J_i x_j’}.
\end{aligned}
\end{equation}

The surface integral is once again zero, which means that we have an antisymmetric relationship in integrals of the form

\begin{equation}\label{eqn:magneticMomentJackson:200}
\int J_j x_i’ = -\int J_i x_j’.
\end{equation}

Now we can use the tensor algebra trick of writing \( y = (y + y)/2 \),

\begin{equation}\label{eqn:magneticMomentJackson:220}
\begin{aligned}
\int (\Bx \cdot \Bx’) \BJ
&=
– \Bx \cross \int \Bx’ \cross \BJ
+
\Be_k x_i \int J_i x_k’ \\
&=
– \Bx \cross \int \Bx’ \cross \BJ
+
\inv{2} \Be_k x_i \int \lr{ J_i x_k’ + J_i x_k’ } \\
&=
– \Bx \cross \int \Bx’ \cross \BJ
+
\inv{2} \Be_k x_i \int \lr{ J_i x_k’ – J_k x_i’ } \\
&=
– \Bx \cross \int \Bx’ \cross \BJ
+
\inv{2} \Be_k x_i \int (\BJ \cross \Bx’)_j \epsilon_{ikj} \\
&=
– \Bx \cross \int \Bx’ \cross \BJ

\inv{2} \epsilon_{kij} \Be_k x_i \int (\BJ \cross \Bx’)_j \\
&=
– \Bx \cross \int \Bx’ \cross \BJ

\inv{2} \Bx \cross \int \BJ \cross \Bx’ \\
&=
– \Bx \cross \int \Bx’ \cross \BJ
+
\inv{2} \Bx \cross \int \Bx’ \cross \BJ \\
&=
-\inv{2} \Bx \cross \int \Bx’ \cross \BJ,
\end{aligned}
\end{equation}

so

\begin{equation}\label{eqn:magneticMomentJackson:240}
\BA(\Bx) \approx \frac{\mu_0}{4 \pi \Abs{\Bx}^3} \lr{ -\frac{\Bx}{2} } \int \Bx’ \cross \BJ(\Bx’) d^3 x’.
\end{equation}

Letting

\begin{equation}\label{eqn:magneticMomentJackson:260}
\boxed{
\Bm = \inv{2} \int \Bx’ \cross \BJ(\Bx’) d^3 x’,
}
\end{equation}

the far field approximation of the vector potential is
\begin{equation}\label{eqn:magneticMomentJackson:280}
\boxed{
\BA(\Bx) = \frac{\mu_0}{4 \pi} \frac{\Bm \cross \Bx}{\Abs{\Bx}^3}.
}
\end{equation}

Note that when the current is restricted to an infintisimally thin loop, the magnetic moment reduces to

\begin{equation}\label{eqn:magneticMomentJackson:300}
\Bm(\Bx) = \frac{I}{2} \int \Bx \cross d\Bl’.
\end{equation}

Refering to [1] (pr. 1.60), this can be seen to be \( I \) times the “vector-area” integral.

References

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

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

Electric and magnetic fields at an interface

October 9, 2016 math and physics play No comments , , ,

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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

\begin{equation}\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}
\end{equation}

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.

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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.
}
\end{equation}

For the magnetic field

\begin{equation}\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}
\end{equation}

which is

\begin{equation}\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.
}
\end{equation}

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.

Green’s function inversion of the magnetostatic equation

September 27, 2016 math and physics play No comments , , , , , , , , ,

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A previous example of inverting a gradient equation was the electrostatics equation. We can do the same for the magnetostatics equation, which has the following Geometric Algebra form in linear media

\begin{equation}\label{eqn:biotSavartGreens:20}
\spacegrad I \BB = – \mu \BJ.
\end{equation}

The Green’s inversion of this is
\begin{equation}\label{eqn:biotSavartGreens:40}
\begin{aligned}
I \BB(\Bx)
&= \int_V dV’ G(\Bx, \Bx’) \spacegrad’ I \BB(\Bx’) \\
&= \int_V dV’ G(\Bx, \Bx’) (-\mu \BJ(\Bx’)) \\
&= \inv{4\pi} \int_V dV’ \frac{\Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 } (-\mu \BJ(\Bx’)).
\end{aligned}
\end{equation}

We expect the LHS to be a bivector, so the scalar component of this should be zero. That can be demonstrated with some of the usual trickery
\begin{equation}\label{eqn:biotSavartGreens:60}
\begin{aligned}
-\frac{\mu}{4\pi} \int_V dV’ \frac{\Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 } \cdot \BJ(\Bx’)
&= \frac{\mu}{4\pi} \int_V dV’ \lr{ \spacegrad \inv{ \Abs{\Bx – \Bx’} }} \cdot \BJ(\Bx’) \\
&= -\frac{\mu}{4\pi} \int_V dV’ \lr{ \spacegrad’ \inv{ \Abs{\Bx – \Bx’} }} \cdot \BJ(\Bx’) \\
&= -\frac{\mu}{4\pi} \int_V dV’ \lr{
\spacegrad’ \cdot \frac{\BJ(\Bx’)}{ \Abs{\Bx – \Bx’} }

\frac{\spacegrad’ \cdot \BJ(\Bx’)}{ \Abs{\Bx – \Bx’} }
}.
\end{aligned}
\end{equation}

The current \( \BJ \) is not unconstrained. This can be seen by premultiplying \ref{eqn:biotSavartGreens:20} by the gradient

\begin{equation}\label{eqn:biotSavartGreens:80}
\spacegrad^2 I \BB = -\mu \spacegrad \BJ.
\end{equation}

On the LHS we have a bivector so must have \( \spacegrad \BJ = \spacegrad \wedge \BJ \), or \( \spacegrad \cdot \BJ = 0 \). This kills the \( \spacegrad’ \cdot \BJ(\Bx’) \) integrand numerator in \ref{eqn:biotSavartGreens:60}, leaving

\begin{equation}\label{eqn:biotSavartGreens:100}
\begin{aligned}
-\frac{\mu}{4\pi} \int_V dV’ \frac{\Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 } \cdot \BJ(\Bx’)
&= -\frac{\mu}{4\pi} \int_V dV’ \spacegrad’ \cdot \frac{\BJ(\Bx’)}{ \Abs{\Bx – \Bx’} } \\
&= -\frac{\mu}{4\pi} \int_{\partial V} dA’ \ncap \cdot \frac{\BJ(\Bx’)}{ \Abs{\Bx – \Bx’} }.
\end{aligned}
\end{equation}

This shows that the scalar part of the equation is zero, provided the normal component of \( \BJ/\Abs{\Bx – \Bx’} \) vanishes on the boundary of the infinite sphere. This leaves the Biot-Savart law as a bivector equation

\begin{equation}\label{eqn:biotSavartGreens:120}
I \BB(\Bx)
= \frac{\mu}{4\pi} \int_V dV’ \BJ(\Bx’) \wedge \frac{\Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 }.
\end{equation}

Observe that the traditional vector form of the Biot-Savart law can be obtained by premultiplying both sides with \( -I \), leaving

\begin{equation}\label{eqn:biotSavartGreens:140}
\BB(\Bx)
= \frac{\mu}{4\pi} \int_V dV’ \BJ(\Bx’) \cross \frac{\Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 }.
\end{equation}

This checks against a trusted source such as [1] (eq. 5.39).

References

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

Constant magnetic solenoid field

September 24, 2015 phy1520 No comments , , , , ,

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In [2] the following vector potential

\begin{equation}\label{eqn:solenoidConstantField:20}
\BA = \frac{B \rho_a^2}{2 \rho} \phicap,
\end{equation}

is introduced in a discussion on the Aharonov-Bohm effect, for configurations where the interior field of a solenoid is either a constant \( \BB \) or zero.

I wasn’t able to make sense of this since the field I was calculating was zero for all \( \rho \ne 0 \)

\begin{equation}\label{eqn:solenoidConstantField:40}
\begin{aligned}
\BB
&= \spacegrad \cross \BA \\
&= \lr{ \rhocap \partial_\rho + \zcap \partial_z + \frac{\phicap}{\rho}
\partial_\phi } \cross \frac{B \rho_a^2}{2 \rho} \phicap \\
&= \lr{ \rhocap \partial_\rho + \frac{\phicap}{\rho} \partial_\phi } \cross
\frac{B \rho_a^2}{2 \rho} \phicap \\
&=
\frac{B \rho_a^2}{2}
\rhocap \cross \phicap \partial_\rho \lr{ \inv{\rho} }
+
\frac{B \rho_a^2}{2 \rho}
\frac{\phicap}{\rho} \cross \partial_\phi \phicap \\
&=
\frac{B \rho_a^2}{2 \rho^2} \lr{ -\zcap + \phicap \cross \partial_\phi \phicap}.
\end{aligned}
\end{equation}

Note that the \( \rho \) partial requires that \( \rho \ne 0 \). To expand the cross product in the second term let \( j = \Be_1 \Be_2 \), and expand using a Geometric Algebra representation of the unit vector

\begin{equation}\label{eqn:solenoidConstantField:60}
\begin{aligned}
\phicap \cross \partial_\phi \phicap
&=
\Be_2 e^{j \phi} \cross \lr{ \Be_2 \Be_1 \Be_2 e^{j \phi} } \\
&=
– \Be_1 \Be_2 \Be_3
\gpgradetwo{
\Be_2 e^{j \phi} (-\Be_1) e^{j \phi}
} \\
&=
\Be_1 \Be_2 \Be_3 \Be_2 \Be_1 \\
&= \Be_3 \\
&= \zcap.
\end{aligned}
\end{equation}

So, provided \( \rho \ne 0 \), \( \BB = 0 \).

The errata [1] provides the clarification, showing that a \( \rho > \rho_a \) constraint is required for this potential to produce the desired results. Continuity at \( \rho = \rho_a \) means that in the interior (or at least on the boundary) we must have one of

\begin{equation}\label{eqn:solenoidConstantField:80}
\BA = \frac{B \rho_a}{2} \phicap,
\end{equation}

or

\begin{equation}\label{eqn:solenoidConstantField:100}
\BA = \frac{B \rho}{2} \phicap.
\end{equation}

The first doesn’t work, but the second does

\begin{equation}\label{eqn:solenoidConstantField:120}
\begin{aligned}
\BB
&= \spacegrad \cross \BA \\
&= \lr{ \rhocap \partial_\rho + \zcap \partial_z + \frac{\phicap}{\rho}
\partial_\phi } \cross \frac{B \rho}{2 } \phicap \\
&=
\frac{B }{2 } \rhocap \cross \phicap
+
\frac{B \rho}{2 }
\frac{\phicap}{\rho} \cross \partial_\phi \phicap \\
&= B \zcap.
\end{aligned}
\end{equation}

So the vector potential that we want for a constant \( B \zcap \) field in the interior \( \rho < \rho_a \) of a cylindrical space, we need

\begin{equation}\label{eqn:solenoidConstantField:140}
\BA =
\left\{
\begin{array}{l l}
\frac{B \rho_a^2}{2 \rho} \phicap & \quad \mbox{if \( \rho \ge \rho_a \) } \\
\frac{B \rho}{2} \phicap & \quad \mbox{if \( \rho \le \rho_a \).}
\end{array}
\right.
\end{equation}

An example of the magnitude of potential is graphed in fig. 1.

solenoidPotentialFig1

fig. 1. Vector potential for constant field in cylindrical region.

 

References

[1] Jun John Sakurai and Jim J Napolitano. \emph{Errata: Typographical Errors, Mistakes, and Comments, Modern Quantum Mechanics, 2nd Edition}, 2013. URL http://www.rpi.edu/dept/phys/Courses/PHYS6520/Spring2015/ErrataMQM.pdf.

[2] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

Plane wave solution directly from Maxwell’s equations

May 6, 2015 math and physics play No comments , , , ,

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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.

Answer

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

\begin{equation}\label{eqn:ExPlaneWave:40}
\spacegrad \cross \BE = -j \omega \BB
\end{equation}
\begin{equation}\label{eqn:ExPlaneWave:60}
\spacegrad \cross \BH = j \omega \BD
\end{equation}
\begin{equation}\label{eqn:ExPlaneWave:80}
\spacegrad \cdot \BD = 0
\end{equation}
\begin{equation}\label{eqn:ExPlaneWave:100}
\spacegrad \cdot \BB = 0.
\end{equation}

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

\begin{equation}\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),
\end{equation}

or

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

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

\begin{equation}\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
}
\end{equation}

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

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

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

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

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

\begin{equation}\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} },
\end{equation}

or

\begin{equation}\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} }.
\end{equation}

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.

Updated notes for ece1229 antenna theory

March 16, 2015 ece1229 No comments , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

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

March 8, 2015 ece1229 No comments , , , ,

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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

\begin{equation}\label{eqn:gaFieldProjection:280}
\BE = \lr{ \BE \cdot \pcap } \pcap + \lr{ \BE \cdot \qcap } \qcap = E_\parallel \pcap + E_\perp \qcap
\end{equation}
\begin{equation}\label{eqn:gaFieldProjection:300}
\BH = \lr{ \BH \cdot \pcap } \pcap + \lr{ \BH \cdot \qcap } \qcap = H_\parallel \pcap + H_\perp \qcap.
\end{equation}

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

\begin{equation}\label{eqn:gaFieldProjection:340}
\pcap = \frac{\kcap \cross \ncap}{\Abs{\kcap \cross \ncap}}
\end{equation}
\begin{equation}\label{eqn:gaFieldProjection:360}
\qcap = \kcap \cross \pcap.
\end{equation}

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

\begin{equation}\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}
\end{equation}

and one with the magnetic field parallel to the reflection plane

\begin{equation}\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}
\end{equation}

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

\begin{equation}\label{eqn:gaFieldProjection:400}
F_1 = \BE_1 + c \mu_0 \BH_1 I
\end{equation}
\begin{equation}\label{eqn:gaFieldProjection:460}
F_2 = \BE_2 + c \mu_0 \BH_2 I
\end{equation}

where the original total field is

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

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

\begin{equation}\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}
\end{equation}

or

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

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

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

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

\begin{equation}\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}
\end{equation}

so

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

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

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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}
\end{equation}

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

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

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

\begin{equation}\label{eqn:gaFieldProjection:660}
F_1 = \lr{ \pcap + c \lr{ \kcap \cross \pcap } I} \BE \cdot \pcap
\end{equation}
\begin{equation}\label{eqn:gaFieldProjection:680}
F_2 = \lr{ \qcap + c \lr{ \kcap \cross \qcap } I} \BE \cdot \qcap.
\end{equation}

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.