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# Posts Tagged ‘planewave’

## Comparing phasor and geometric transverse solutions to the Maxwell equation (continued)

Posted by peeterjoot on August 9, 2009

Continuing with previous post.

# Explicit split of geometric phasor into advanced and receding parts

For a more general split of the geometric phasor into advanced and receding wave terms, will there be interdependence between the electric and magnetic field components? Going back to (16), and rearranging, we have

\begin{aligned}2 e^{ -I \hat{\mathbf{k}} \omega t } e^{ I \mathbf{k} \cdot \mathbf{x} }&=(C_{-} -I S_{-})+\hat{\mathbf{k}} (C_{-} -I S_{-} )+(C_{+} +I S_{+})-\hat{\mathbf{k}} (C_{+} +I S_{+}) \\ \end{aligned}

So we have

\begin{aligned}e^{ -I \hat{\mathbf{k}} \omega t } e^{ I \mathbf{k} \cdot \mathbf{x} }&=\frac{1}{{2}}(1 + \hat{\mathbf{k}})e^{-I(\omega t - \mathbf{k} \cdot \mathbf{x})}+\frac{1}{{2}}(1 - \hat{\mathbf{k}})e^{I(\omega t + \mathbf{k} \cdot \mathbf{x})} \end{aligned} \quad\quad\quad(19)

As observed if we have $\hat{\mathbf{k}} \mathcal{F} = \mathcal{F}$, the result is only the advanced wave term

\begin{aligned}e^{ -I \hat{\mathbf{k}} \omega t } e^{ I \mathbf{k} \cdot \mathbf{x} } \mathcal{F} = e^{-I(\omega t - \mathbf{k} \cdot \mathbf{x})} \mathcal{F} \end{aligned}

Similarly, with absorption of $\hat{\mathbf{k}}$ with the opposing sign $\hat{\mathbf{k}} \mathcal{F} = -\mathcal{F}$, we have only the receding wave

\begin{aligned}e^{ -I \hat{\mathbf{k}} \omega t } e^{ I \mathbf{k} \cdot \mathbf{x} } \mathcal{F} = e^{I(\omega t + \mathbf{k} \cdot \mathbf{x})} \mathcal{F} \end{aligned}

Either of the receding or advancing wave solutions should independently satisfy the Maxwell equation operator. Let’s verify both of these, and verify that for either the $\pm$ cases the following is a solution and examine the constraints for that to be the case.

\begin{aligned}F = \frac{1}{{2}}(1 \pm \hat{\mathbf{k}}) e^{\pm I(\omega t \pm \mathbf{k} \cdot \mathbf{x})} \mathcal{F} \end{aligned} \quad\quad\quad(20)

Now we wish to apply the Maxwell equation operator $\boldsymbol{\nabla} + \sqrt{\mu\epsilon}\partial_0$ to this assumed solution. That is

\begin{aligned}0 &= (\boldsymbol{\nabla} + \sqrt{\mu\epsilon}\partial_0) F \\ &= \sigma_m \frac{1}{{2}}(1 \pm \hat{\mathbf{k}}) (\pm I \pm k^m) e^{\pm I(\omega t \pm \mathbf{k} \cdot \mathbf{x})} \mathcal{F}+ \frac{1}{{2}}(1 \pm \hat{\mathbf{k}}) (\pm I \sqrt{\mu\epsilon}\omega/c) e^{\pm I(\omega t \pm \mathbf{k} \cdot \mathbf{x})} \mathcal{F} \\ &= \frac{\pm I}{2}\left(\pm \hat{\mathbf{k}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right)(1 \pm \hat{\mathbf{k}}) e^{\pm I(\omega t \pm \mathbf{k} \cdot \mathbf{x})} \mathcal{F} \end{aligned}

By left multiplication with the conjugate of the Maxwell operator $\nabla - \sqrt{\mu\epsilon}\partial_0$ we have the wave equation operator, and applying that, we have as before, a magnitude constraint on the wave number $\mathbf{k}$

\begin{aligned}0 &= (\boldsymbol{\nabla} - \sqrt{\mu\epsilon}\partial_0) (\boldsymbol{\nabla} + \sqrt{\mu\epsilon}\partial_0) F \\ &= (\boldsymbol{\nabla}^2 - {\mu\epsilon}\partial_{00}) F \\ &= \frac{-1}{2}(1 \pm \hat{\mathbf{k}}) \left( \mathbf{k}^2 - \mu\epsilon\frac{\omega^2}{c^2}\right) e^{\pm I(\omega t \pm \mathbf{k} \cdot \mathbf{x})} \mathcal{F} \end{aligned}

So we have as before ${\left\lvert{\mathbf{k}}\right\rvert} = \sqrt{\mu\epsilon}\omega/c$. Substitution into the the first order operator result we have

\begin{aligned}0 &= (\boldsymbol{\nabla} + \sqrt{\mu\epsilon}\partial_0) F \\ &= \frac{\pm I}{2}\sqrt{\mu\epsilon}\frac{\omega}{c}\left(\pm \hat{\mathbf{k}} + 1\right)(1 \pm \hat{\mathbf{k}}) e^{\pm I(\omega t \pm \mathbf{k} \cdot \mathbf{x})} \mathcal{F} \end{aligned}

Observe that the multivector $1 \pm \hat{\mathbf{k}}$, when squared is just a multiple of itself

\begin{aligned}(1 \pm \hat{\mathbf{k}})^2 = 1 + \hat{\mathbf{k}}^2 \pm 2 \hat{\mathbf{k}} = 2 (1 \pm \hat{\mathbf{k}}) \end{aligned}

So we have

\begin{aligned}0 &= (\boldsymbol{\nabla} + \sqrt{\mu\epsilon}\partial_0) F \\ &= {\pm I}\sqrt{\mu\epsilon}\frac{\omega}{c}(1 \pm \hat{\mathbf{k}}) e^{\pm I(\omega t \pm \mathbf{k} \cdot \mathbf{x})} \mathcal{F} \end{aligned}

So we see that the constraint again on the individual assumed solutions is again that of absorption. Separately the advanced or receding parts of the geometric phasor as expressed in (20) are solutions provided

\begin{aligned}\hat{\mathbf{k}} F = \mp F \end{aligned} \quad\quad\quad(21)

The geometric phasor is seen to be curious superposition of both advancing and receding states. Independently we have something pretty much like the standard transverse phasor wave states. Is this superposition state physically meaningful. It is a solution to the Maxwell equation (without any constraints on $\boldsymbol{\mathcal{E}}$ and $\boldsymbol{\mathcal{B}}$).

## Comparing phasor and geometric transverse solutions to the Maxwell equation

Posted by peeterjoot on August 8, 2009

# Motivation

In ([1]) a phasor like form of the transverse wave equation was found by considering Fourier solutions of the Maxwell equation. This will be called the “geometric phasor” since it is hard to refer and compare it without giving it a name. Curiously no perpendicularity condition for $\mathbf{E}$ and $\mathbf{B}$ seemed to be required for this geometric phasor. Why would that be the case? In Jackson’s treatment, which employed the traditional dot and cross product form of Maxwell’s equations, this followed by back substituting the assumed phasor solution back into the equations. This back substitution wasn’t done in ([1]). If we attempt this we should find the same sort of additional mutual perpendicularity constraints on the fields.

Here we start with the equations from Jackson ([2], ch7), expressed in GA form. Using the same assumed phasor form we should get the same results using GA. Anything else indicates a misunderstanding or mistake, so as an intermediate step we should at least recover the Jackson result.

After using a more traditional phasor form (where one would have to take real parts) we revisit the goemetric phasor found in ([1]). It will be found that the perpendicular constraints of the Jackson phasor solution lead to a representation where the geometric phasor is reduced to the Jackson form with a straight substitution of the imaginary $i$ with the pseudoscalar $I = \sigma_1\sigma_2\sigma_3$. This representation however, like the more general geometric phasor requires no selection of real or imaginary parts to construct a “physical” solution.

# With assumed phasor field

Maxwell’s equations in absence of charge and current ((7.1) of Jackson) can be summarized by

\begin{aligned}0 &= (\boldsymbol{\nabla} + \sqrt{\mu\epsilon}\partial_0) F \end{aligned} \quad\quad\quad(1)

The $F$ above is a composite electric and magnetic field merged into a single multivector. In the spatial basic the electric field component $\mathbf{E}$ is a vector, and the magnetic component $I\mathbf{B}$ is a bivector (in the Dirac basis both are bivectors).

\begin{aligned}F &= \mathbf{E} + I \mathbf{B}/\sqrt{\mu\epsilon} \end{aligned} \quad\quad\quad(2)

With an assumed phasor form

\begin{aligned}F = \mathcal{F} e^{ i(\mathbf{k} \cdot \mathbf{x} - \omega t) } = (\boldsymbol{\mathcal{E}} + I\boldsymbol{\mathcal{B}}/\sqrt{\mu\epsilon}) e^{ i(\mathbf{k} \cdot \mathbf{x} - \omega t) } \end{aligned} \quad\quad\quad(3)

Although there are many geometric multivectors that square to -1, we do not assume here that the imaginary $i$ has any specific geometric meaning, and in fact commutes with all multivectors. Because of this we have to take the real parts later when done.

Operating on $F$ with Maxwell’s equation we have

\begin{aligned}0 = (\boldsymbol{\nabla} + \sqrt{\mu\epsilon}\partial_0) F = i \left( \mathbf{k} - \sqrt{\mu\epsilon}\frac{\omega}{c} \right) F \end{aligned} \quad\quad\quad(4)

Similarly, left multiplication of Maxwell’s equation by the conjugate operator $\boldsymbol{\nabla} - \sqrt{\mu\epsilon}\partial_0$, we have the wave equation

\begin{aligned}0 &= \left(\boldsymbol{\nabla}^2 - \frac{\mu\epsilon}{c^2}\frac{\partial^2}{\partial t^2}\right) F \end{aligned} \quad\quad\quad(5)

and substitution of the assumed phasor solution gives us

\begin{aligned}0 = (\boldsymbol{\nabla}^2 - {\mu\epsilon}\partial_{00}) F = -\left( \mathbf{k}^2 - {\mu\epsilon}\frac{\omega^2}{c^2} \right) F \end{aligned} \quad\quad\quad(6)

This provides the relation between the magnitude of $\mathbf{k}$ and $\omega$, namely

\begin{aligned}{\left\lvert{\mathbf{k}}\right\rvert} = \pm \sqrt{\mu\epsilon}\frac{\omega}{c} \end{aligned} \quad\quad\quad(7)

Without any real loss of generality we can pick the positive root, so the result of the Maxwell equation operator on the phasor is

\begin{aligned}0 = (\boldsymbol{\nabla} + \sqrt{\mu\epsilon}\partial_0) F = i \sqrt{\mu\epsilon}\frac{\omega}{c} \left( \hat{\mathbf{k}} - 1\right) F \end{aligned} \quad\quad\quad(8)

Rearranging we have the curious property that the field $F$ can “swallow” a left multiplication by the propagation direction unit vector

\begin{aligned}\hat{\mathbf{k}} F = F \end{aligned} \quad\quad\quad(9)

Selection of the scalar and pseudoscalar grades of this equation shows that the electric and magnetic fields $\mathbf{E}$ and $\mathbf{B}$ are both completely transverse to the propagation direction $\hat{\mathbf{k}}$. For the scalar grades we have

\begin{aligned}0 &= \left\langle{{\hat{\mathbf{k}} F - F}}\right\rangle \\ &= \hat{\mathbf{k}} \cdot \mathbf{E} \end{aligned}

and for the pseudoscalar

\begin{aligned}0 &= {\left\langle{{\hat{\mathbf{k}} F - F}}\right\rangle}_{3} \\ &= I \hat{\mathbf{k}} \cdot \mathbf{B} \end{aligned}

From this we have $\hat{\mathbf{k}} \cdot \mathbf{B} = \hat{\mathbf{k}} \cdot \mathbf{B} = 0$. Because of this transverse property we see that the $\hat{\mathbf{k}}$ multiplication of $F$ in (9) serves to map electric field (vector) components into bivectors, and the magnetic bivector components into vectors. For the result to be the same means we must have an additional coupling between the field components. Writing out (9) in terms of the field components we have

\begin{aligned}\mathbf{E} + I\mathbf{B}/\sqrt{\mu\epsilon} &= \hat{\mathbf{k}} (\mathbf{E} + I\mathbf{B}/\sqrt{\mu\epsilon} ) \\ &= \hat{\mathbf{k}} \wedge \mathbf{E} + I (\hat{\mathbf{k}} \wedge \mathbf{B})/\sqrt{\mu\epsilon} \\ &= I \hat{\mathbf{k}} \times \mathbf{E} + I^2 (\hat{\mathbf{k}} \times \mathbf{B})/\sqrt{\mu\epsilon} \end{aligned}

Equating left and right hand grades we have

\begin{aligned}\mathbf{E} &= -(\hat{\mathbf{k}} \times \mathbf{B})/\sqrt{\mu\epsilon} \\ \mathbf{B} &= \sqrt{\mu\epsilon} (\hat{\mathbf{k}} \times \mathbf{E}) \end{aligned} \quad\quad\quad(10)

Since $\mathbf{E}$ and $\mathbf{B}$ both have the same phase relationships we also have

\begin{aligned}\boldsymbol{\mathcal{E}} &= -(\hat{\mathbf{k}} \times \boldsymbol{\mathcal{B}})/\sqrt{\mu\epsilon} \\ \boldsymbol{\mathcal{B}} &= \sqrt{\mu\epsilon} (\hat{\mathbf{k}} \times \boldsymbol{\mathcal{E}}) \end{aligned} \quad\quad\quad(12)

With phasors as used in electrical engineering it is usual to allow the fields to have complex values. Assuming this is allowed here too, taking real parts of $F$, and separating by grade, we have for the electric and magnetic fields

\begin{aligned}\begin{pmatrix}\mathbf{E} \\ \mathbf{B}\end{pmatrix}=\text{Real}\begin{pmatrix}\boldsymbol{\mathcal{E}} \\ \boldsymbol{\mathcal{B}}\end{pmatrix}\cos(\mathbf{k} \cdot \mathbf{x} - \omega t)+\text{Imag}\begin{pmatrix}\boldsymbol{\mathcal{E}} \\ \boldsymbol{\mathcal{B}}\end{pmatrix}\sin(\mathbf{k} \cdot \mathbf{x} - \omega t) \end{aligned} \quad\quad\quad(14)

We will find a slightly different separation into electric and magnetic fields with the geometric phasor.

# Geometrized phasor.

Translating from SI units to the CGS units of Jackson the geometric phasor representation of the field was found previously to be

\begin{aligned}F = e^{ -I \hat{\mathbf{k}} \omega t } e^{ I \mathbf{k} \cdot \mathbf{x} } (\boldsymbol{\mathcal{E}} + I\boldsymbol{\mathcal{B}}/\sqrt{\mu\epsilon}) \end{aligned} \quad\quad\quad(15)

As above the transverse requirement $\boldsymbol{\mathcal{E}} \cdot \mathbf{k} = \boldsymbol{\mathcal{B}} \cdot \mathbf{k} = 0$ was required. Application of Maxwell’s equation operator should show if we require any additional constraints. That is

\begin{aligned}0 &= (\boldsymbol{\nabla} + \sqrt{\mu\epsilon}\partial_0) F \\ &=(\boldsymbol{\nabla} + \sqrt{\mu\epsilon}\partial_0) e^{ -I \hat{\mathbf{k}} \omega t } e^{ I \mathbf{k} \cdot \mathbf{x} } (\boldsymbol{\mathcal{E}} + I\boldsymbol{\mathcal{B}}/\sqrt{\mu\epsilon}) \\ &=\sum \sigma_m e^{ -I \hat{\mathbf{k}} \omega t } (I k^m) e^{ I \mathbf{k} \cdot \mathbf{x} } (\boldsymbol{\mathcal{E}} + I\boldsymbol{\mathcal{B}}/\sqrt{\mu\epsilon}) -I \hat{\mathbf{k}} \sqrt{\mu\epsilon} \frac{\omega}{c} e^{ -I \hat{\mathbf{k}} \omega t } e^{ I \mathbf{k} \cdot \mathbf{x} } (\boldsymbol{\mathcal{E}} + I\boldsymbol{\mathcal{B}}/\sqrt{\mu\epsilon}) \\ &=I \left(\mathbf{k} - \hat{\mathbf{k}} \sqrt{\mu\epsilon} \frac{\omega}{c} \right) e^{ -I \hat{\mathbf{k}} \omega t } e^{ I \mathbf{k} \cdot \mathbf{x} } (\boldsymbol{\mathcal{E}} + I\boldsymbol{\mathcal{B}}/\sqrt{\mu\epsilon}) \end{aligned}

This is zero for any combinations of $\boldsymbol{\mathcal{E}}$ or $\boldsymbol{\mathcal{B}}$ since $\mathbf{k} = \hat{\mathbf{k}} \sqrt{\mu\epsilon} \omega/c$. It therefore appears that this geometric phasor has a fundamentally different nature than the non-geometric version. We have two exponentials that commute, but due to the difference in grades of the arguments, it doesn’t appear that there is any easy way to express this as an single argument exponential. Multiplying these out, and using the trig product to sum identities helps shed some light on the differences between the geometric phasor and the one using a generic imaginary. Starting off we have

\begin{aligned}e^{ -I \hat{\mathbf{k}} \omega t } e^{ I \mathbf{k} \cdot \mathbf{x} }&=(\cos(\omega t) -I\hat{\mathbf{k}} \sin(\omega t)) (\cos(\mathbf{k} \cdot \mathbf{x}) +I\sin(\mathbf{k} \cdot \mathbf{x})) \\ &=\cos(\omega t)\cos(\mathbf{k} \cdot \mathbf{x}) + \hat{\mathbf{k}} \sin(\omega t)\sin(\mathbf{k} \cdot \mathbf{x})-I\hat{\mathbf{k}} \sin(\omega t)\cos(\mathbf{k} \cdot \mathbf{x})+I \cos(\omega t) \sin(\mathbf{k} \cdot \mathbf{x}) \\ \end{aligned}

In this first expansion we see that this product of exponentials has scalar, vector, bivector, and pseudoscalar grades, despite the fact that we have only
vector and bivector terms in the end result. That will be seen to be due to the transverse nature of $\mathcal{F}$ that we multiply with. Before performing that final multiplication, writing $C_{-} = \cos(\omega t - \mathbf{k} \cdot \mathbf{x})$, $C_{+} = \cos(\omega t + \mathbf{k} \cdot \mathbf{x})$, $S_{-} = \sin(\omega t - \mathbf{k} \cdot \mathbf{x})$, and $S_{+} = \sin(\omega t + \mathbf{k} \cdot \mathbf{x})$, we have

\begin{aligned}e^{ -I \hat{\mathbf{k}} \omega t } e^{ I \mathbf{k} \cdot \mathbf{x} }&=\frac{1}{{2}}\left( (C_{-} + C_{+})+\hat{\mathbf{k}} (C_{-} - C_{+})-I \hat{\mathbf{k}} (S_{-} + S_{+})-I (S_{-} - S_{+})\right) \end{aligned} \quad\quad\quad(16)

As an operator the left multiplication of $\hat{\mathbf{k}}$ on a transverse vector has the action

\begin{aligned}\hat{\mathbf{k}} ( \cdot ) &= \hat{\mathbf{k}} \wedge (\cdot) \\ &= I (\hat{\mathbf{k}} \times (\cdot)) \\ \end{aligned}

This gives

\begin{aligned}e^{ -I \hat{\mathbf{k}} \omega t } e^{ I \mathbf{k} \cdot \mathbf{x} }&=\frac{1}{{2}}\left( (C_{-} + C_{+})+(C_{-} - C_{+}) I \hat{\mathbf{k}} \times+(S_{-} + S_{+}) \hat{\mathbf{k}} \times-I (S_{-} - S_{+})\right) \end{aligned} \quad\quad\quad(17)

Now, lets apply this to the field with $\mathcal{F} = \boldsymbol{\mathcal{E}} + I\boldsymbol{\mathcal{B}}/\sqrt{\mu\epsilon}$. To avoid dragging around the $\sqrt{\mu\epsilon}$ factors, let’s also temporarily
work with units where $\mu\epsilon = 1$. We then have

\begin{aligned}2 e^{ -I \hat{\mathbf{k}} \omega t } e^{ I \mathbf{k} \cdot \mathbf{x} } \mathcal{F}&= (C_{-} + C_{+}) (\boldsymbol{\mathcal{E}} + I\boldsymbol{\mathcal{B}}) \\ &+ (C_{-} - C_{+}) (I (\hat{\mathbf{k}} \times \boldsymbol{\mathcal{E}}) - \hat{\mathbf{k}} \times \boldsymbol{\mathcal{B}}) \\ &+ (S_{-} + S_{+}) (\hat{\mathbf{k}} \times \boldsymbol{\mathcal{E}} +I (\hat{\mathbf{k}} \times \boldsymbol{\mathcal{B}})) \\ &+ (S_{-} - S_{+}) (-I \boldsymbol{\mathcal{E}} + \boldsymbol{\mathcal{B}}) \end{aligned}

Rearranging explicitly in terms of the electric and magnetic field components this is

\begin{aligned}2 e^{ -I \hat{\mathbf{k}} \omega t } e^{ I \mathbf{k} \cdot \mathbf{x} } \mathcal{F}&= (C_{-} + C_{+}) \boldsymbol{\mathcal{E}} -(C_{-} - C_{+}) (\hat{\mathbf{k}} \times \boldsymbol{\mathcal{B}})+(S_{-} + S_{+}) (\hat{\mathbf{k}} \times \boldsymbol{\mathcal{E}})+(S_{-} - S_{+}) \boldsymbol{\mathcal{B}} \\ &+{I}\left( (C_{-} + C_{+}) \boldsymbol{\mathcal{B}}+(C_{-} - C_{+}) (\hat{\mathbf{k}} \times \boldsymbol{\mathcal{E}})+(S_{-} + S_{+}) (\hat{\mathbf{k}} \times \boldsymbol{\mathcal{B}})-(S_{-} - S_{+}) \boldsymbol{\mathcal{E}} \right) \\ \end{aligned}

Quite a mess! A first observation is that the application of the perpendicularity conditions (12) we have a remarkable reduction in complexity. That is

\begin{aligned}2 e^{ -I \hat{\mathbf{k}} \omega t } e^{ I \mathbf{k} \cdot \mathbf{x} } \mathcal{F}&= (C_{-} + C_{+}) \boldsymbol{\mathcal{E}} +(C_{-} - C_{+}) \boldsymbol{\mathcal{E}}+(S_{-} + S_{+}) \boldsymbol{\mathcal{B}}+(S_{-} - S_{+}) \boldsymbol{\mathcal{B}}\\ &+{I}\left( (C_{-} + C_{+}) \boldsymbol{\mathcal{B}}+(C_{-} - C_{+}) \boldsymbol{\mathcal{B}}-(S_{-} + S_{+}) \boldsymbol{\mathcal{E}}-(S_{-} - S_{+}) \boldsymbol{\mathcal{E}} \right) \\ \end{aligned}

This wipes out the receding wave terms leaving only the advanced wave terms, leaving

\begin{aligned}e^{ -I \hat{\mathbf{k}} \omega t } e^{ I \mathbf{k} \cdot \mathbf{x} } \mathcal{F}&= C_{-} \boldsymbol{\mathcal{E}} +S_{-} (\hat{\mathbf{k}} \times \boldsymbol{\mathcal{E}})+{I}\left( C_{-} \boldsymbol{\mathcal{B}} +S_{-} \hat{\mathbf{k}} \times \boldsymbol{\mathcal{B}} \right) \\ &= C_{-} (\boldsymbol{\mathcal{E}} + I\boldsymbol{\mathcal{B}})+S_{-} (\boldsymbol{\mathcal{B}} -I\boldsymbol{\mathcal{E}}) \\ &=( C_{-} -I S_{-} ) (\boldsymbol{\mathcal{E}} + I\boldsymbol{\mathcal{B}}) \\ \end{aligned}

We see therefore for this special case of mutually perpendicular (equ-magnitude) field components, our geometric phasor has only the advanced wave term

\begin{aligned}e^{ -I \hat{\mathbf{k}} \omega t } e^{ I \mathbf{k} \cdot \mathbf{x} } \mathcal{F} &= e^{-I(\omega t - \mathbf{k} \cdot \mathbf{x})} \mathcal{F} \end{aligned} \quad\quad\quad(18)

If we pick this as the starting point for the assumed solution, it is clear that the same perpendicularity constraints will follow as in Jackson’s treatment, or the GA version of it above. We have something that is slightly different though, for we have no requirement to take real parts of this simpified geometric phasor, since the result already contains just the vector and bivector terms of the electric and magnetic fields respectively.

A small aside, before continuing. Having made this observation that we can write the assumed phasor for this transverse field in the form of (18) an easier way to demonstrate that the product of exponentials reduces only to the advanced wave term is now clear. Instead of using (12) we could start back at (16) and employ the absorption property $\hat{\mathbf{k}} \mathcal{F} = \mathcal{F}$. That gives

\begin{aligned}e^{ -I \hat{\mathbf{k}} \omega t } e^{ I \mathbf{k} \cdot \mathbf{x} } \mathcal{F}&=\frac{1}{{2}}\left( (C_{-} + C_{+})+(C_{-} - C_{+})-I (S_{-} + S_{+})-I (S_{-} - S_{+})\right) \mathcal{F} \\ &=\left( C_{-} -I S_{-} \right) \mathcal{F} \end{aligned}

That’s the same result, obtained in a slicker manner. What is perhaps of more interest is examining the general split of our geometric phasor into advanced and receding wave terms, and examining the interdependence, if any, between the electric and magnetic field components. Since this didn’t lead exactly to where I expected, that’s now left as a project for a different day.

# References

[1] Peeter Joot. {Space time algebra solutions of the Maxwell equation for discrete frequencies} [online]. http://sites.google.com/site/peeterjoot/math2009/maxwellVacuum.pdf.

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

## (A POSSIBLY WRONG) Superposition of transverse electromagnetic field solutions.

Posted by peeterjoot on August 3, 2009

# Disclaimer

FIXME: I’M NOT CONFIDENT THAT i GOT THIS RIGHT. An attempt to verify the final result didn’t work. I’ve either messed up along the way (perhaps right at the beginning), or my verification itself was busted. Am posting these working notes for now, and will revisit later after thinking it through again (or trying the verification again).

# Motivation

In ([1]), a Geometric Algebra solution for the transverse components of the electromagnetic field was found. Here we construct a superposition of these transverse fields, keeping the propagation direction fixed, and allowing for continuous variation of the wave number and angular velocity. Evaluation of this superposition integral, first utilizing a contour integral, then utilizing an inverse Fourier transform allows for the determination of the functional form of a general wave packet moving along a fixed direction in space. This wave packet will be seen to have two time dependencies, an advanced time term, and a retarded time term. The phasors will be eliminated from both the propagation and the transverse fields and will provide operators with which the transverse field can be calculated from a general propagation field wave packet.

# Superposition of phasor solutions.

When the field is required to have explicit sinusoidal time and propagation direction, the field components in the transverse (to the propagation) direction were found to be

\begin{aligned}F_t &= \frac{1}{{i \left( \pm k \hat{\mathbf{z}} - \sqrt{\mu\epsilon}\frac{\omega}{c}\right) }} \boldsymbol{\nabla}_t F_z \\ &= \frac{i}{k^2 - \mu\epsilon\frac{\omega^2}{c^2}} \left( \pm k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right) \boldsymbol{\nabla}_t F_z \end{aligned} \quad\quad\quad(1)

Here $F_z = F_z(x,y) = \mathbf{E}_z(x,y) + I\mathbf{B}_z(x,y)/\sqrt{\mu\epsilon}$ is required by construction to commute with $\hat{\mathbf{z}}$, but is otherwise arbitrary, except perhaps for boundary value constraints not considered here.

Removing the restriction to fixed wave number and angular velocity we can integrate over both to express a more general transverse wave propagation

\begin{aligned}F_t &= \int d\omega e^{-i\omega t} \int dk e^{ikz} \frac{i}{k^2 - \mu\epsilon\frac{\omega^2}{c^2}} \left(k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right) \boldsymbol{\nabla}_t {F_z}^{k,\omega} \end{aligned} \quad\quad\quad(3)

Inventing temporary notation for convenience we write for the frequency dependent weighting function ${F_z}^{k,\omega} = F_z(x,y,k,\omega)$. Also note that explicit $\pm k$ has been dropped after allowing wave number to range over both positive and negative values.

Observe that each of these integrals has the form of a Fourier integral (ignoring constant factors that can be incorporated into the weighting function). Also observe that the integral kernel has two poles on the real $k$-axis (or real $\omega$-axis). These can be utilized to evaluate one of the integrals using an upper half plane semicircular contour. FIXME: picture here.

Assuming that ${F_z}^{k,\omega}$ is small enough at $\infty$ (on the large contour) to be neglected, we have for the integral after integrating around the poles at $k = \pm \sqrt{\mu\epsilon}{\left\lvert{\omega/c}\right\rvert}$

\begin{aligned} F_t &= \pi i \int d\omega e^{-i\omega t} {\left.e^{ikz} \frac{i}{k - \sqrt{\mu\epsilon}\frac{{\left\lvert{\omega}\right\rvert}}{c}} \left(k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right) \boldsymbol{\nabla}_t {F_z}^{k,\omega}\right\vert}_{k=-\sqrt{\mu\epsilon}\frac{{\left\lvert{\omega}\right\rvert}}{c}} \\ &+ \pi i \int d\omega e^{-i\omega t} {\left.e^{ikz} \frac{i}{k + \sqrt{\mu\epsilon}\frac{{\left\lvert{\omega}\right\rvert}}{c}} \left(k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right) \boldsymbol{\nabla}_t {F_z}^{k,\omega}\right\vert}_{k=\sqrt{\mu\epsilon}\frac{{\left\lvert{\omega}\right\rvert}}{c}} \end{aligned}

Writing ${F_z}^{\omega+} = F_z(x,y,\omega, k=\sqrt{\mu\epsilon}{\left\lvert{\omega}\right\rvert}/c)$, and ${F_z}^{\omega-} = F_z(x,y,\omega, k=-\sqrt{\mu\epsilon}{\left\lvert{\omega}\right\rvert}/c)$, for the values of our field at the poles, we have

\begin{aligned} F_t &= \frac{\pi}{2}\int d\omega e^{-i\omega t} e^{-i \sqrt{\mu\epsilon}\frac{{\left\lvert{\omega}\right\rvert}}{c} z}\frac{\omega -\hat{\mathbf{z}}{\left\lvert{\omega}\right\rvert}}{{\left\lvert{\omega}\right\rvert}}\boldsymbol{\nabla}_t {F_z}^{\omega-} \\ &- \frac{\pi}{2}\int d\omega e^{-i\omega t} e^{i \sqrt{\mu\epsilon}\frac{{\left\lvert{\omega}\right\rvert}}{c} z} \frac{\omega +\hat{\mathbf{z}}{\left\lvert{\omega}\right\rvert}}{{\left\lvert{\omega}\right\rvert}}\boldsymbol{\nabla}_t {F_z}^{\omega+} \end{aligned}

Can one ignore the singularity at $\omega = 0$ since it divides out? If so, then we have

\begin{aligned} F_t &= \frac{\pi}{2}\int d\omega e^{-i\omega t} e^{-i \sqrt{\mu\epsilon}\frac{\left\lvert{\omega}\right\rvert}{c} z}(\text{sgn}(\omega) - \hat{\mathbf{z}})\boldsymbol{\nabla}_t {F_z}^{\omega-} \\ &- \frac{\pi}{2}\int d\omega e^{-i\omega t} e^{i \sqrt{\mu\epsilon}\frac{\left\lvert{\omega}\right\rvert}{c} z} (\text{sgn}(\omega) + \hat{\mathbf{z}})\boldsymbol{\nabla}_t {F_z}^{\omega+} \end{aligned}

Writing this out separately for the regions greater and lesser than zero we have

\begin{aligned} F_t &= \frac{\pi}{2}\int_{0+}^\infty d\omega e^{-i\omega t} e^{-i \sqrt{\mu\epsilon}\frac{\omega}{c} z}(1 - \hat{\mathbf{z}})\boldsymbol{\nabla}_t {F_z}^{\omega-} - \frac{\pi}{2}\int_{0+}^\infty d\omega e^{-i\omega t} e^{i \sqrt{\mu\epsilon}\frac{\omega}{c} z} (1 + \hat{\mathbf{z}})\boldsymbol{\nabla}_t {F_z}^{\omega+} \\ &- \frac{\pi}{2}\int_{-\infty}^{0-} d\omega e^{-i\omega t} e^{i \sqrt{\mu\epsilon}\frac{\omega}{c} z}(1 + \hat{\mathbf{z}})\boldsymbol{\nabla}_t {F_z}^{\omega-} - \frac{\pi}{2}\int_{-\infty}^{0-} d\omega e^{-i\omega t} e^{-i \sqrt{\mu\epsilon}\frac{\omega}{c} z} (-1 + \hat{\mathbf{z}})\boldsymbol{\nabla}_t {F_z}^{\omega+} \\ \end{aligned}

Grouping by exponentials of like sign, and integrating over a region that omits some neighborhood around the origin, we have eliminated the ${\left\lvert{\omega}\right\rvert}$ factors

\begin{aligned}F_t &= (1 - \hat{\mathbf{z}}) \boldsymbol{\nabla}_t\int d\omega e^{-i\omega \left(t + \sqrt{\mu\epsilon}\frac{z}{c}\right)}\frac{\pi}{2}({F_z}^{\omega-}\theta(\omega) + {F_z}^{\omega+}\theta(-\omega)) \\ &-(1 + \hat{\mathbf{z}}) \boldsymbol{\nabla}_t\int d\omega e^{-i\omega \left(t - \sqrt{\mu\epsilon}\frac{z}{c}\right)}\frac{\pi}{2}({F_z}^{\omega-}\theta(\omega) + {F_z}^{\omega+}\theta(-\omega)) \end{aligned}

The Heaviside unit step function has been used to group things nicely over the doubled integration ranges, and what is left now can be observed to be Fourier transforms from the frequency to time domain. The transformed functions are to be evaluated at a shifted time in each case.

\begin{aligned}F_t &= (1 - \hat{\mathbf{z}}) \boldsymbol{\nabla}_t\frac{\pi}{2}{\left.\mathcal{F}\left( ({F_z}^{\omega-}\theta(\omega) + {F_z}^{\omega+}\theta(-\omega)) \right)\right\vert}_{t=t + \sqrt{\mu\epsilon}\frac{z}{c}} \\ &-(1 + \hat{\mathbf{z}}) \boldsymbol{\nabla}_t\frac{\pi}{2}{\left.\mathcal{F}\left( ({F_z}^{\omega-}\theta(\omega) + {F_z}^{\omega+}\theta(-\omega)) \right)\right\vert}_{t=t - \sqrt{\mu\epsilon}\frac{z}{c}} \end{aligned}

Attempting to actually evaluate this Fourier transform shouldn’t actually be necessary since we the original angular velocity and wave number domain function $F(x,y,\omega,k)$ was arbitrary (except for its commutation with $\hat{\mathbf{z}}$). Instead we just suppose (once again overloading the symbol $F_z$) that we have a function $F_z(x,y,u)$ that at time $u$ takes the value

\begin{aligned}F_z(x,y,u) = \frac{\pi}{2}\mathcal{F}\left( ({F_z}^{\omega-}\theta(\omega) + {F_z}^{\omega+}\theta(-\omega)) \right) \end{aligned}

The transform will not change the grades of the propagation field $F_z$ so this still commutes with $\hat{\mathbf{z}}$. We can now write

\begin{aligned}F_t &= (1 - \hat{\mathbf{z}}) \boldsymbol{\nabla}_t F_z(x,y, t + \sqrt{\mu\epsilon}{z}/{c})-(1 + \hat{\mathbf{z}}) \boldsymbol{\nabla}_t F_z(x,y, t - \sqrt{\mu\epsilon}{z}/{c}) \end{aligned} \quad\quad\quad(4)

So, after a bunch of manipulation, we find exactly how the transverse component of the field is related to the propagation direction field, and have eliminated the phasor description supplied our single frequency/wave-number field relations.

What we have left is a propagation field that has the form of an arbitrary unidirectional wave packet, traveling in either direction through the medium with speed $\pm c/\sqrt{\mu\epsilon}$. If all this math is right, the transverse field for an advancing wave is generated by application onto the propagation field of the operator

\begin{aligned}-(1 + \hat{\mathbf{z}}) \boldsymbol{\nabla}_t \end{aligned} \quad\quad\quad(5)

Similarly, the transverse field for a receding wave is given by application of the operator

\begin{aligned}(1 - \hat{\mathbf{z}}) \boldsymbol{\nabla}_t \end{aligned} \quad\quad\quad(6)

# Verification.

Given the doubt about the $\omega =0$ point in the integral, and the possibility for sign error and other algebraic mistakes it seems worthwhile now to go back to Maxwell’s equation and verify the correctness of these results. Then, presuming everything worked out okay, it would also be good to relate things back to the electric and magnetic field components of the field.

Maxwell’s equation in these units is

\begin{aligned}\left(\boldsymbol{\nabla}_t \pm i k \hat{\mathbf{z}} - \sqrt{\mu\epsilon}\frac{i\omega}{c}\right) F(x,y) = 0 \end{aligned} \quad\quad\quad(7)

and we want to compute the transverse direction projection from the propagation direction term

\begin{aligned}F_t &= \mathbf{E}_t + I \mathbf{B}_t = \frac{1}{{2}} (F - \hat{\mathbf{z}} F \hat{\mathbf{z}}) \\ F_z &= \mathbf{E}_z + I \mathbf{B}_z = \frac{1}{{2}} (F + \hat{\mathbf{z}} F \hat{\mathbf{z}}) \end{aligned} \quad\quad\quad(8)

Picking one of the terms of (4), we have

\begin{aligned}F_t = (1 + \pm (1 \mp \hat{\mathbf{z}}) \nabla_t) F_z(t \pm \sqrt{\mu\epsilon} z/c) \end{aligned}

Substiting into the left hand operator of Maxwell’s equation (7) and reducing I get

\begin{aligned}((\hat{\mathbf{z}} \pm 1) {\nabla_t}^2 + \nabla_t) F_z + \frac{\mu\epsilon}{c}(1 \pm \hat{\mathbf{z}}) F_z' \end{aligned}

I don’t see how this would neccessarily equal zero?

# References

[1] Peeter Joot. Transverse electric and magnetic fields [online]. http://sites.google.com/site/peeterjoot/math2009/transverseField.pdf.

## Sumarizing: Transverse electric and magnetic fields

Posted by peeterjoot on August 1, 2009

There’s potentially a lot of new ideas in the previous transverse field post (some for me even with previous exposure to the Geometric Algebra formalism). There was no real attempt to teach GA here, but for completeness the GA form of Maxwell’s equation was developed from the traditional divergence and curl formulation of Maxwell’s equations. That was mainly due to use of CGS units which differ since this makes Maxwell’s equation take a different form from the usual (see [1]).

This time a less exploratory summary of the previous results above is assembled.

In these CGS units our field $F$, and Maxwell’s equation (in absence of charge and current), take the form

\begin{aligned}F &= \mathbf{E} + \frac{I\mathbf{B}}{\sqrt{\mu\epsilon}} \\ 0 &= \left(\boldsymbol{\nabla} + \frac{\sqrt{\mu\epsilon}}{c}\partial_t\right) F \end{aligned} \quad\quad\quad(30)

The electric and magnetic fields can be picked off by selecting the grade one (vector) components

\begin{aligned}\mathbf{E} &= {\left\langle{{F}}\right\rangle}_{1} \\ \mathbf{B} &= {\left\langle{{-I F}}\right\rangle}_{1} \end{aligned} \quad\quad\quad(32)

With an explicit sinusoidal and $z$-axis time dependence for the field

\begin{aligned}F(x,y,z,t) &= F(x,y) e^{\pm i k z - i \omega t} \end{aligned} \quad\quad\quad(34)

and a split of the gradient into transverse and $z$-axis components $\boldsymbol{\nabla} = \boldsymbol{\nabla}_t + \hat{\mathbf{z}} \partial_z$, Maxwell’s equation takes the form

\begin{aligned}\left(\boldsymbol{\nabla}_t \pm i k \hat{\mathbf{z}} - \sqrt{\mu\epsilon}\frac{i\omega}{c}\right) F(x,y) = 0 \end{aligned} \quad\quad\quad(35)

Writing for short $F = F(x,y)$, we can split the field into transverse and $z$-axis components with the commutator and anticommutator products respectively. For the $z$-axis components we have

\begin{aligned}F_z \hat{\mathbf{z}} \equiv E_z + I B_z = \frac{1}{{2}} (F \hat{\mathbf{z}} + \hat{\mathbf{z}} F) \end{aligned} \quad\quad\quad(36)

The projections onto the $z$-axis and and transverse directions are respectively

\begin{aligned}F_z &= \mathbf{E}_z + I \mathbf{B}_z = \frac{1}{{2}} (F + \hat{\mathbf{z}} F \hat{\mathbf{z}}) \\ F_t &= \mathbf{E}_t + I \mathbf{B}_t = \frac{1}{{2}} (F - \hat{\mathbf{z}} F \hat{\mathbf{z}} ) \end{aligned} \quad\quad\quad(37)

With an application of the transverse gradient to the $z$-axis field we easily found the relation between the two
field components

\begin{aligned}\boldsymbol{\nabla}_t F_z &= i \left( \pm k \hat{\mathbf{z}} - \sqrt{\mu\epsilon}\frac{\omega}{c}\right) F_t \end{aligned} \quad\quad\quad(39)

A left division of the exponential factor gives the total transverse field

\begin{aligned}F_t &= \frac{1}{{i \left( \pm k \hat{\mathbf{z}} - \sqrt{\mu\epsilon}\frac{\omega}{c}\right) }} \boldsymbol{\nabla}_t F_z \end{aligned} \quad\quad\quad(40)

Multiplication of both the numerator and denominator by the conjugate normalizes this

\begin{aligned}F_t &= \frac{i}{k^2 - \mu\epsilon\frac{\omega^2}{c^2}} \left( \pm k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right) \boldsymbol{\nabla}_t F_z \end{aligned} \quad\quad\quad(41)

From this the transverse electric and magnetic fields may be picked off using the projective grade selection operations of (32), and are

\begin{aligned}\mathbf{E}_t &= \frac{i}{\mu\epsilon\frac{\omega^2}{c^2} -k^2} \left( \pm k \boldsymbol{\nabla}_t E_z - \frac{\omega}{c} \hat{\mathbf{z}} \times \boldsymbol{\nabla}_t B_z \right) \\ \mathbf{B}_t &= \frac{i}{\mu\epsilon\frac{\omega^2}{c^2} -k^2} \left( {\mu\epsilon}\frac{\omega}{c} \hat{\mathbf{z}} \times \boldsymbol{\nabla}_t E_z \pm k \boldsymbol{\nabla}_t B_z \right) \end{aligned} \quad\quad\quad(42)

# References

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

## Transverse electric and magnetic fields

Posted by peeterjoot on July 31, 2009

# Motivation

In Eli’s Transverse Electric and Magnetic Fields in a Conducting Waveguide blog entry he works through the algebra calculating the transverse components, the perpendicular to the propagation direction components.

This should be possible using Geometric Algebra too, and trying this made for a good exercise.

# Setup

The starting point can be the same, the source free Maxwell’s equations. Writing $\partial_0 = (1/c) \partial/{\partial t}$, we have

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{E} &= 0 \\ \boldsymbol{\nabla} \cdot \mathbf{B} &= 0 \\ \boldsymbol{\nabla} \times \mathbf{E} &= - \partial_0 \mathbf{B} \\ \boldsymbol{\nabla} \times \mathbf{B} &= \mu \epsilon \partial_0 \mathbf{E} \end{aligned} \quad\quad\quad(1)

Multiplication of the last two equations by the spatial pseudoscalar $I$, and using $I \mathbf{a} \times \mathbf{b} = \mathbf{a} \wedge \mathbf{b}$, the curl equations can be written in their dual bivector form

\begin{aligned}\boldsymbol{\nabla} \wedge \mathbf{E} &= - \partial_0 I \mathbf{B} \\ \boldsymbol{\nabla} \wedge \mathbf{B} &= \mu \epsilon \partial_0 I \mathbf{E} \end{aligned} \quad\quad\quad(5)

Now adding the dot and curl equations using $\mathbf{a} \mathbf{b} = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \wedge \mathbf{b}$ eliminates the cross products

\begin{aligned}\boldsymbol{\nabla} \mathbf{E} &= - \partial_0 I \mathbf{B} \\ \boldsymbol{\nabla} \mathbf{B} &= \mu \epsilon \partial_0 I \mathbf{E} \end{aligned} \quad\quad\quad(7)

These can be further merged without any loss, into the GA first order equation

\begin{aligned}\left(\boldsymbol{\nabla} + \frac{\sqrt{\mu\epsilon}}{c}\partial_t\right) \left(\mathbf{E} + \frac{I\mathbf{B}}{\sqrt{\mu\epsilon}} \right) = 0 \end{aligned} \quad\quad\quad(9)

We are really after solutions to the total multivector field $F = \mathbf{E} + I \mathbf{B}/\sqrt{\mu\epsilon}$. For this problem where separate electric and magnetic field components are desired, working from (7) is perhaps what we want?

Following Eli and Jackson, write $\boldsymbol{\nabla} = \boldsymbol{\nabla}_t + \hat{\mathbf{z}} \partial_z$, and

\begin{aligned}\mathbf{E}(x,y,z,t) &= \mathbf{E}(x,y) e^{\pm i k z - i \omega t} \\ \mathbf{B}(x,y,z,t) &= \mathbf{B}(x,y) e^{\pm i k z - i \omega t} \end{aligned} \quad\quad\quad(10)

Evaluating the $z$ and $t$ partials we have

\begin{aligned}(\boldsymbol{\nabla}_t \pm i k \hat{\mathbf{z}}) \mathbf{E}(x,y) &= \frac{i\omega}{c} I \mathbf{B}(x,y) \\ (\boldsymbol{\nabla}_t \pm i k \hat{\mathbf{z}}) \mathbf{B}(x,y) &= -\mu \epsilon \frac{i\omega}{c} I \mathbf{E}(x,y) \end{aligned} \quad\quad\quad(12)

For the remainder of these notes, the explicit $(x,y)$ dependence will be assumed for $\mathbf{E}$ and $\mathbf{B}$.

An obvious thing to try with these equations is just substitute one into the other. If that’s done we get the pair of second order harmonic equations

\begin{aligned}{\boldsymbol{\nabla}_t}^2\begin{pmatrix}\mathbf{E} \\ \mathbf{B} \end{pmatrix}= \left( k^2 - \mu \epsilon \frac{\omega^2}{c^2} \right)\begin{pmatrix}\mathbf{E} \\ \mathbf{B} \end{pmatrix} \end{aligned} \quad\quad\quad(14)

One could consider the problem solved here. Separately equating both sides of this equation to zero, we have the $k^2 = \mu\epsilon \omega^2/c^2$ constraint on the wave number and angular velocity, and the second order Laplacian on the left hand side is solved by the real or imaginary parts of any analytic function. Especially when one considers that we are after a multivector field that of intrinsic complex nature.

However, that is not really what we want as a solution. Doing the same on the unified Maxwell equation (9), we have

\begin{aligned}\left(\boldsymbol{\nabla}_t \pm i k \hat{\mathbf{z}} - \sqrt{\mu\epsilon}\frac{i\omega}{c}\right) \left(\mathbf{E} + \frac{I\mathbf{B}}{\sqrt{\mu\epsilon}} \right) = 0 \end{aligned} \quad\quad\quad(15)

Selecting scalar, vector, bivector and trivector grades of this equation produces the following respective relations between the various components

\begin{aligned}0 = \left\langle{{\cdots}}\right\rangle &= \boldsymbol{\nabla}_t \cdot \mathbf{E} \pm i k \hat{\mathbf{z}} \cdot \mathbf{E} \\ 0 = {\left\langle{{\cdots}}\right\rangle}_{1} &= I \boldsymbol{\nabla}_t \wedge \mathbf{B}/\sqrt{\mu\epsilon} \pm i I k \hat{\mathbf{z}} \wedge \mathbf{B}/\sqrt{\mu\epsilon} - i \sqrt{\mu\epsilon}\frac{\omega}{c} \mathbf{E} \\ 0 = {\left\langle{{\cdots}}\right\rangle}_{2} &= \boldsymbol{\nabla}_t \wedge \mathbf{E} \pm i k \hat{\mathbf{z}} \wedge \mathbf{E} - i \frac{\omega}{c} I \mathbf{B} \\ 0 = {\left\langle{{\cdots}}\right\rangle}_{3} &= I \boldsymbol{\nabla}_t \cdot \mathbf{B}/\sqrt{\mu\epsilon} \pm i I k \hat{\mathbf{z}} \cdot \mathbf{B}/\sqrt{\mu\epsilon} \end{aligned} \quad\quad\quad(16)

From the scalar and pseudoscalar grades we have the propagation components in terms of the transverse ones

\begin{aligned}E_z &= \frac{\pm i}{k} \boldsymbol{\nabla}_t \cdot \mathbf{E}_t \\ B_z &= \frac{\pm i}{k} \boldsymbol{\nabla}_t \cdot \mathbf{B}_t \end{aligned} \quad\quad\quad(20)

But this is the opposite of the relations that we are after. On the other hand from the vector and bivector grades we have

\begin{aligned}i \frac{\omega}{c} \mathbf{E} &= -\frac{1}{{\mu\epsilon}}\left(\boldsymbol{\nabla}_t \times \mathbf{B}_z \pm i k \hat{\mathbf{z}} \times \mathbf{B}_t\right) \\ i \frac{\omega}{c} \mathbf{B} &= \boldsymbol{\nabla}_t \times \mathbf{E}_z \pm i k \hat{\mathbf{z}} \times \mathbf{E}_t \end{aligned} \quad\quad\quad(22)

# A clue from the final result.

From (22) and a lot of messy algebra we should be able to get the transverse equations. Is there a slicker way? The end result that Eli obtained suggests a path. That result was

\begin{aligned}\mathbf{E}_t = \frac{i}{\mu\epsilon \frac{\omega^2}{c^2} - k^2} \left( \pm k \boldsymbol{\nabla}_t E_z - \frac{\omega}{c} \hat{\mathbf{z}} \times \boldsymbol{\nabla}_t B_z \right) \end{aligned} \quad\quad\quad(24)

The numerator looks like it can be factored, and after a bit of playing around a suitable factorization can be obtained:

\begin{aligned}{\left\langle{{ \left( \pm k + \frac{\omega}{c} \hat{\mathbf{z}} \right) \boldsymbol{\nabla}_t \hat{\mathbf{z}} \left( \mathbf{E}_z + I \mathbf{B}_z \right) }}\right\rangle}_{1}&={\left\langle{{ \left( \pm k + \frac{\omega}{c} \hat{\mathbf{z}} \right) \boldsymbol{\nabla}_t \left( E_z + I B_z \right) }}\right\rangle}_{1} \\ &=\pm k \boldsymbol{\nabla} E_z + \frac{\omega}{c} {\left\langle{{ I \hat{\mathbf{z}} \boldsymbol{\nabla}_t B_z }}\right\rangle}_{1} \\ &=\pm k \boldsymbol{\nabla} E_z + \frac{\omega}{c} I \hat{\mathbf{z}} \wedge \boldsymbol{\nabla}_t B_z \\ &=\pm k \boldsymbol{\nabla} E_z - \frac{\omega}{c} \hat{\mathbf{z}} \times \boldsymbol{\nabla}_t B_z \\ \end{aligned}

Observe that the propagation components of the field $\mathbf{E}_z + I\mathbf{E}_z$ can be written in terms of the symmetric product

\begin{aligned}\frac{1}{{2}} \left( \hat{\mathbf{z}} (\mathbf{E} + I\mathbf{B}) + (\mathbf{E} + I\mathbf{B}) \hat{\mathbf{z}} \right)&=\frac{1}{{2}} \left( \hat{\mathbf{z}} \mathbf{E} + \mathbf{E} \hat{\mathbf{z}} \right) + \frac{I}{2} \left( \hat{\mathbf{z}} \mathbf{B} + \mathbf{B} \hat{\mathbf{z}} + I \right) \\ &=\hat{\mathbf{z}} \cdot \mathbf{E} + I \hat{\mathbf{z}} \cdot \mathbf{B} \end{aligned}

Now the total field in CGS units was actually $F = \mathbf{E} + I \mathbf{B}/\sqrt{\mu\epsilon}$, not $F = \mathbf{E} + I \mathbf{B}$, so the factorization above isn’t exactly what we want. It does however, provide the required clue. We probably get the result we want by forming the symmetric product (a hybrid dot product selecting both the vector and bivector terms).

# Symmetric product of the field with the direction vector.

Rearranging Maxwell’s equation (15) in terms of the transverse gradient and the total field $F$ we have

\begin{aligned}\boldsymbol{\nabla}_t F = \left( \mp i k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{i\omega}{c}\right) F \end{aligned} \quad\quad\quad(25)

With this our symmetric product is

\begin{aligned}\boldsymbol{\nabla}_t ( F \hat{\mathbf{z}} + \hat{\mathbf{z}} F) &= (\boldsymbol{\nabla}_t F) \hat{\mathbf{z}} - \hat{\mathbf{z}} (\boldsymbol{\nabla}_t F) \\ &=\left( \mp i k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{i\omega}{c}\right) F \hat{\mathbf{z}}- \hat{\mathbf{z}} \left( \mp i k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{i\omega}{c}\right) F \\ &=i \left( \mp k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right) (F \hat{\mathbf{z}} - \hat{\mathbf{z}} F) \\ \end{aligned}

The antisymmetric product on the right hand side should contain the desired transverse field components. To verify multiply it out

\begin{aligned}\frac{1}{{2}}(F \hat{\mathbf{z}} - \hat{\mathbf{z}} F) &=\frac{1}{{2}}\left( \left(\mathbf{E} + I \mathbf{B}/\sqrt{\mu\epsilon}\right) \hat{\mathbf{z}} - \hat{\mathbf{z}} \left(\mathbf{E} + I \mathbf{B}/\sqrt{\mu\epsilon}\right) \right) \\ &=\mathbf{E} \wedge \hat{\mathbf{z}} + I \mathbf{B}/\sqrt{\mu\epsilon} \wedge \hat{\mathbf{z}} \\ &=(\mathbf{E}_t + I \mathbf{B}_t/\sqrt{\mu\epsilon}) \hat{\mathbf{z}} \\ \end{aligned}

Now, with multiplication by the conjugate quantity $-i(\pm k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\omega/c)$, we can extract these transverse components.

\begin{aligned}\left( \pm k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right) \left( \mp k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right) (F \hat{\mathbf{z}} - \hat{\mathbf{z}} F) &=\left( -k^2 + {\mu\epsilon}\frac{\omega^2}{c^2}\right) (F \hat{\mathbf{z}} - \hat{\mathbf{z}} F) \end{aligned}

Rearranging, we have the transverse components of the field

\begin{aligned}(\mathbf{E}_t + I \mathbf{B}_t/\sqrt{\mu\epsilon}) \hat{\mathbf{z}} &=\frac{i}{k^2 - \mu\epsilon\frac{\omega^2}{c^2}} \left( \pm k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right) \boldsymbol{\nabla}_t \frac{1}{{2}}( F \hat{\mathbf{z}} + \hat{\mathbf{z}} F) \end{aligned} \quad\quad\quad(26)

With left multiplication by $\hat{\mathbf{z}}$, and writing $F = F_t + F_z$ we have

\begin{aligned}F_t &= \frac{i}{k^2 - \mu\epsilon\frac{\omega^2}{c^2}} \left( \pm k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right) \boldsymbol{\nabla}_t F_z \end{aligned} \quad\quad\quad(27)

While this is a complete solution, we can additionally extract the electric and magnetic fields to compare results with Eli’s calculation. We take
vector grades to do so with $\mathbf{E}_t = {\left\langle{{F_t}}\right\rangle}_{1}$, and $\mathbf{B}_t/\sqrt{\mu\epsilon} = {\left\langle{{-I F_t}}\right\rangle}_{1}$. For the transverse electric field

\begin{aligned}{\left\langle{{ \left( \pm k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right) \boldsymbol{\nabla}_t (\mathbf{E}_z + I \mathbf{B}_z/\sqrt{/\mu\epsilon}) }}\right\rangle}_{1} &=\pm k \hat{\mathbf{z}} (-\hat{\mathbf{z}}) \boldsymbol{\nabla}_t E_z + \frac{\omega}{c} \underbrace{{\left\langle{{I \boldsymbol{\nabla}_t \hat{\mathbf{z}}}}\right\rangle}_{1}}_{-I^2 \hat{\mathbf{z}} \times \boldsymbol{\nabla}_t } B_z \\ &=\mp k \boldsymbol{\nabla}_t E_z + \frac{\omega}{c} \hat{\mathbf{z}} \times \boldsymbol{\nabla}_t B_z \\ \end{aligned}

and for the transverse magnetic field

\begin{aligned}{\left\langle{{ -I \left( \pm k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right) \boldsymbol{\nabla}_t (\mathbf{E}_z + I \mathbf{B}_z/\sqrt{\mu\epsilon}) }}\right\rangle}_{1} &=-I \sqrt{\mu\epsilon}\frac{\omega}{c} \boldsymbol{\nabla}_t \mathbf{E}_z+{\left\langle{{ \left( \pm k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right) \boldsymbol{\nabla}_t \mathbf{B}_z/\sqrt{\mu\epsilon} }}\right\rangle}_{1} \\ &=- \sqrt{\mu\epsilon}\frac{\omega}{c} \hat{\mathbf{z}} \times \boldsymbol{\nabla}_t E_z\mp k \boldsymbol{\nabla}_t B_z/\sqrt{\mu\epsilon} \\ \end{aligned}

Thus the split of transverse field into the electric and magnetic components yields

\begin{aligned}\mathbf{E}_t &= \frac{i}{k^2 - \mu\epsilon\frac{\omega^2}{c^2}} \left( \mp k \boldsymbol{\nabla}_t E_z + \frac{\omega}{c} \hat{\mathbf{z}} \times \boldsymbol{\nabla}_t B_z \right) \\ \mathbf{B}_t &= \frac{i}{k^2 - \mu\epsilon\frac{\omega^2}{c^2}} \left( - {\mu\epsilon}\frac{\omega}{c} \hat{\mathbf{z}} \times \boldsymbol{\nabla}_t E_z \mp k \boldsymbol{\nabla}_t B_z \right) \end{aligned} \quad\quad\quad(28)

Compared to Eli’s method using messy traditional vector algebra, this method also has a fair amount of messy tricky algebra, but of a different sort.