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## 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 $\boldsymbol{\mathcal{B}}$).