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# Archive for September, 2012

## Derivation of Fresnel equations for mixed polarization (using Geometric Algebra)

Posted by peeterjoot on September 25, 2012

[Click here for a PDF of this post with nicer formatting and figures]

# Motivation

In [2] we have a derivation of the Fresnel equations for the TE and TM polarization modes. Can we do this for an arbitrary polarization angles?

# Setup

The task at hand is to find evaluate the boundary value constraints. Following the interface plane conventions of [1], and his notation that is

\begin{aligned}\epsilon_1 ( \mathbf{E}_i + \mathbf{E}_r )_z = \epsilon_2 ( \mathbf{E}_t )_z\end{aligned} \hspace{\stretch{1}}(1.2.1a)

\begin{aligned}( \mathbf{B}_i + \mathbf{B}_r )_z = ( \mathbf{B}_t )_z\end{aligned} \hspace{\stretch{1}}(1.2.1b)

\begin{aligned}( \mathbf{E}_i + \mathbf{E}_r )_{x,y} = ( \mathbf{E}_t )_{x,y}\end{aligned} \hspace{\stretch{1}}(1.2.1c)

\begin{aligned}\frac{1}{{\mu_1}} ( \mathbf{B}_i + \mathbf{B}_r )_{x,y} = \frac{1}{{\mu_2}} ( \mathbf{B}_t )_{x,y}\end{aligned} \hspace{\stretch{1}}(1.2.1d)

I’ll work here with a phasor representation directly and not bother with taking real parts, or using tilde notation to mark the vectors as complex.

Our complex magnetic field phasors are related to the electric fields with

\begin{aligned}\mathbf{B} = \frac{1}{{v}} \hat{\mathbf{k}} \times \mathbf{E}.\end{aligned} \hspace{\stretch{1}}(1.2.2)

Referring to figure (see pdf) shows the geometrical task to tackle, since we’ve got to express all the various unit vectors algebraically. I’ll use Geometric Algebra here to do that for its compact expression of rotations. With

Figure: See pdf: Reflection and transmission of light at an interface

\begin{aligned}j = \mathbf{e}_3 \mathbf{e}_1,\end{aligned} \hspace{\stretch{1}}(1.2.3)

we can express each of the $k$ vector directions by inspection. Those are

\begin{aligned}\hat{\mathbf{k}}_i = \mathbf{e}_3 e^{j \theta_i} = \mathbf{e}_3 \cos\theta_i + \mathbf{e}_1 \sin\theta_i\end{aligned} \hspace{\stretch{1}}(1.2.4a)

\begin{aligned}\hat{\mathbf{k}}_r = -\mathbf{e}_3 e^{-j \theta_r} = -\mathbf{e}_3 \cos\theta_r +\mathbf{e}_1 \sin\theta_r\end{aligned} \hspace{\stretch{1}}(1.2.4b)

\begin{aligned}\hat{\mathbf{k}}_t = \mathbf{e}_3 e^{j \theta_t} = \mathbf{e}_3 \cos\theta_t + \mathbf{e}_1 \sin\theta_t.\end{aligned} \hspace{\stretch{1}}(1.2.4c)

Similarly, the perpendiculars $\hat{\mathbf{m}}_p = \mathbf{e}_2 \times \hat{\mathbf{k}}_p$ are

\begin{aligned}\hat{\mathbf{m}}_i = \mathbf{e}_{1} e^{j \theta_i}= \mathbf{e}_1 \cos\theta_i - \mathbf{e}_3 \sin\theta_i= \mathbf{e}_3 j e^{j \theta_i}\end{aligned} \hspace{\stretch{1}}(1.2.5a)

\begin{aligned}\hat{\mathbf{m}}_r = -\mathbf{e}_{1} e^{-j \theta_r}= -\mathbf{e}_1 \cos\theta_r - \mathbf{e}_3 \sin\theta_r= -\mathbf{e}_3 j e^{-j \theta_r}\end{aligned} \hspace{\stretch{1}}(1.2.5b)

\begin{aligned}\hat{\mathbf{m}}_t = \mathbf{e}_{1 } e^{j \theta_t} = \mathbf{e}_1 \cos\theta_t - \mathbf{e}_3 \sin\theta_t= \mathbf{e}_3 j e^{j \theta_t}\end{aligned} \hspace{\stretch{1}}(1.2.5c)

In [1] problem 9.14 we had to show that the polarization angles for normal incident ($\mathbf{E} \parallel \mathbf{e}_1$) must be the same due to the boundary constraints. Can we also tackle that problem for both this more general angle of incidence and a general polarization? Let’s try so, allowing temporarily for different polarizations of the reflected and transmitted components of the light, calling those polarization angles $\phi_i$, $\phi_r$, and $\phi_t$ respectively. Let’s set the $\phi_i = 0$ polarization aligned such that $\mathbf{E}_i$, $\mathbf{B}_i$ are aligned with the $\mathbf{e}_2$ and $-\hat{\mathbf{m}}_i$ directions respectively, so that the generally polarized phasors are

\begin{aligned}\begin{bmatrix}\mathbf{E}_p \\ \mathbf{B}_p \\ \end{bmatrix}=\begin{bmatrix}\mathbf{e}_2 \\ -\hat{\mathbf{m}}_p \\ \end{bmatrix}e^{ \hat{\mathbf{m}}_p \mathbf{e}_2 \phi_p }\end{aligned} \hspace{\stretch{1}}(1.2.6)

We are now set to at least express our boundary value constraints

\begin{aligned}\epsilon_1 \left( \mathbf{e}_2 E_i e^{ \hat{\mathbf{m}}_i \mathbf{e}_2 \phi_i } + \mathbf{e}_2 E_r e^{ \hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r } \right) \cdot \mathbf{e}_3 = \epsilon_2 \left( \mathbf{e}_2 E_t e^{ \hat{\mathbf{m}}_t \mathbf{e}_2 \phi_t } \right) \cdot \mathbf{e}_3\end{aligned} \hspace{\stretch{1}}(1.2.7a)

\begin{aligned}\frac{1}{{v_1}} \left( -\hat{\mathbf{m}}_i E_i e^{ \hat{\mathbf{m}}_i \mathbf{e}_2 \phi_i } - \hat{\mathbf{m}}_r E_r e^{ \hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r } \right) \cdot \mathbf{e}_3 = \frac{1}{{v_2}} \left( -\hat{\mathbf{m}}_t E_t e^{ \hat{\mathbf{m}}_t \mathbf{e}_2 \phi_t } \right) \cdot \mathbf{e}_3\end{aligned} \hspace{\stretch{1}}(1.2.7b)

\begin{aligned}\left( \mathbf{e}_2 E_i e^{ \hat{\mathbf{m}}_i \mathbf{e}_2 \phi_i } + \mathbf{e}_2 E_r e^{ \hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r } \right) \wedge \mathbf{e}_3 = \left( \mathbf{e}_2 E_t e^{ \hat{\mathbf{m}}_t \mathbf{e}_2 \phi_t } \right) \wedge \mathbf{e}_3\end{aligned} \hspace{\stretch{1}}(1.2.7c)

\begin{aligned}\frac{1}{{\mu_1 v_1}} \left( -\hat{\mathbf{m}}_i E_i e^{ \hat{\mathbf{m}}_i \mathbf{e}_2 \phi_i } - \hat{\mathbf{m}}_r E_r e^{ \hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r } \right) \wedge \mathbf{e}_3 = \frac{1}{{\mu_2 v_2}} \left( -\hat{\mathbf{m}}_t E_t e^{ \hat{\mathbf{m}}_t \mathbf{e}_2 \phi_t } \right) \wedge \mathbf{e}_3\end{aligned} \hspace{\stretch{1}}(1.2.7d)

Let’s try this in a couple of steps. First with polarization angles set so that one of the fields lies in the plane of the interface (with both variations), and then attempt the general case, first posing the problem in the tranditional way to see what equations fall out, and then using superposition.

Before doing so, let’s introduce a bit of notation to be used throughout. When we wish to refer to all the fields or angles, for example, $\mathbf{E}_i, \mathbf{E}_r, \mathbf{E}_t$ then we’ll write $\mathbf{E}_p$ where $p \in \{i, r, t\}$. Similarily, to refer to just the incident and transmitted components (or angles) we’ll use $\mathbf{E}_q$ where $q \in \{i, t\}$. Following [1] we’ll also write

\begin{aligned}\beta = \frac{\mu_1 v_1} {\mu_2 v_2} \end{aligned} \hspace{\stretch{1}}(1.2.8)

\begin{aligned}\alpha = \frac{\cos\theta_t}{\cos\theta_i},\end{aligned} \hspace{\stretch{1}}(1.2.8)

## Question: Sanity check. Verify for $\mathbf{E}$ parallel to the interface.

For the $\mathbf{E}_p \parallel \mathbf{e}_2$ polarization ($\phi_i = \phi_r = \phi_t$) our phasors are

\begin{aligned}\mathbf{E}_p = \mathbf{e}_2 E_p\end{aligned} \hspace{\stretch{1}}(1.2.9)

\begin{aligned}\mathbf{B}_p = -\frac{1}{{v_p}} \hat{\mathbf{m}}_p E_p\end{aligned} \hspace{\stretch{1}}(1.2.9)

Our boundary value constraints then become

\begin{aligned}\epsilon_1 \left( \mathbf{e}_2 E_i + \mathbf{e}_2 E_r \right) \cdot \mathbf{e}_3 = \epsilon_2 \left( \mathbf{e}_2 E_t \right) \cdot \mathbf{e}_3\end{aligned} \hspace{\stretch{1}}(1.2.10a)

\begin{aligned}\frac{1}{{v_1}} \left( \hat{\mathbf{m}}_i E_i + \hat{\mathbf{m}}_r E_r \right) \cdot \mathbf{e}_3 = \frac{1}{{v_2}} \left( \hat{\mathbf{m}}_t E_t \right) \cdot \mathbf{e}_3\end{aligned} \hspace{\stretch{1}}(1.2.10b)

\begin{aligned}\left( \mathbf{e}_2 E_i + \mathbf{e}_2 E_r \right) \wedge \mathbf{e}_3 = \left( \mathbf{e}_2 E_t \right) \wedge \mathbf{e}_3\end{aligned} \hspace{\stretch{1}}(1.2.10c)

\begin{aligned}\frac{1}{{\mu_1 v_1}} \left( \hat{\mathbf{m}}_i E_i + \hat{\mathbf{m}}_r E_r \right) \wedge \mathbf{e}_3 = \frac{1}{{\mu_2 v_2}} \left( \hat{\mathbf{m}}_t E_t \right) \wedge \mathbf{e}_3.\end{aligned} \hspace{\stretch{1}}(1.2.10d)

With $\hat{\mathbf{m}}_p$ substitution this is

\begin{aligned}\epsilon_1 \left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_i + \mathbf{e}_2 E_r \right) }}\right\rangle = \epsilon_2 \left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_t \right) }}\right\rangle\end{aligned} \hspace{\stretch{1}}(1.2.11a)

\begin{aligned}\frac{1}{{v_1}} \left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_1 e^{j \theta_i} E_i - \mathbf{e}_1 e^{-j \theta_r} E_r \right) }}\right\rangle = \frac{1}{{v_2}} \left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_1 e^{j \theta_t} E_t \right) }}\right\rangle\end{aligned} \hspace{\stretch{1}}(1.2.11b)

\begin{aligned}{\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_i + \mathbf{e}_2 E_r \right) }}\right\rangle}_{2} = {\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_t \right) }}\right\rangle}_{2}\end{aligned} \hspace{\stretch{1}}(1.2.11c)

\begin{aligned}\frac{1}{{\mu_1 v_1}} {\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_1 e^{j \theta_i} E_i -\mathbf{e}_1 e^{-j \theta_r} E_r \right) }}\right\rangle}_{2} = \frac{1}{{\mu_2 v_2}} {\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_1 e^{j \theta_t} E_t \right) }}\right\rangle}_{2}.\end{aligned} \hspace{\stretch{1}}(1.2.11d)

Evaluating the grade selections we have a separation into an analogue of real and imaginary parts for

\begin{aligned}0 = 0\end{aligned} \hspace{\stretch{1}}(1.2.12a)

\begin{aligned}\frac{1}{{v_1}} \left( -\sin\theta_i E_i - \sin\theta_r E_r \right) = \frac{1}{{v_2}} \left( -\sin\theta_t E_t \right)\end{aligned} \hspace{\stretch{1}}(1.2.12b)

\begin{aligned}E_i + E_r = E_t\end{aligned} \hspace{\stretch{1}}(1.2.12c)

\begin{aligned}\frac{1}{{\mu_1 v_1}} \left( \cos{\theta_i} E_i - \cos{\theta_r} E_r \right) = \frac{1}{{\mu_2 v_2}} \left( \cos{ \theta_t} E_t \right).\end{aligned} \hspace{\stretch{1}}(1.2.12d)

With $\theta_i = \theta_r$ and $\sin\theta_t/\sin\theta_i = n_1/n_2$ 1.2.12b becomes

\begin{aligned}E_i + E_r = \frac{n_1 v_1}{n_2 v_2} E_t = \frac{v_2 v_1}{v_1 v_2} E_t = E_t,\end{aligned} \hspace{\stretch{1}}(1.2.12d)

so that we find 1.2.12b and 1.2.12c are dependent. We are left with a pair of equations

\begin{aligned}E_i + E_r = E_t\end{aligned} \hspace{\stretch{1}}(1.2.14)

\begin{aligned}E_i - E_r = \frac{\mu_1 v_1}{\mu_2 v_2} \frac{\cos{ \theta_t}}{\cos\theta_i} E_t,\end{aligned} \hspace{\stretch{1}}(1.2.14)

\begin{aligned}2 E_i = \left( 1 + \frac{\mu_1 v_1}{\mu_2 v_2} \frac{\cos{ \theta_t}}{\cos\theta_i} \right) E_t\end{aligned} \hspace{\stretch{1}}(1.2.15)

\begin{aligned}2 E_r = \left( 1 - \frac{\mu_1 v_1}{\mu_2 v_2} \frac{\cos{ \theta_t}}{\cos\theta_i} \right) E_t,\end{aligned} \hspace{\stretch{1}}(1.2.15)

with a final rearrangement to yield

\begin{aligned}\frac{E_t}{E_i}=\frac{2 \mu_2 v_2 \cos\theta_i}{\mu_2 v_2 \cos\theta_i+\mu_1 v_1 \cos\theta_t}\end{aligned} \hspace{\stretch{1}}(1.2.16)

\begin{aligned}\frac{E_r}{E_i}=\frac{\mu_2 v_2 \cos\theta_i-\mu_1 v_1 \cos\theta_t}{\mu_2 v_2 \cos\theta_i+\mu_1 v_1 \cos\theta_t}\end{aligned} \hspace{\stretch{1}}(1.2.16)

Using the $\alpha$ and $\beta$ notation above we have

\begin{aligned}\frac{E_t}{E_i}=\frac{2 }{1 + \alpha \beta}\end{aligned} \hspace{\stretch{1}}(1.2.17)

\begin{aligned}\frac{E_r}{E_i}=\frac{1 - \alpha \beta}{1 + \alpha \beta}\end{aligned} \hspace{\stretch{1}}(1.2.17)

## Question: Sanity check. Verify for $\mathbf{B}$ parallel to the interface.

As a second sanity check let’s rotate our field polarizations by applying a rotation $e^{\mathbf{e}_2 \hat{\mathbf{m}}_p \pi/2} = \mathbf{e}_2 \hat{\mathbf{m}}_p$ ($\phi_i = \phi_r = \phi_t = -\pi/2$) so that

\begin{aligned}-\hat{\mathbf{m}}_p \rightarrow -\hat{\mathbf{m}}_p \mathbf{e}_2 \hat{\mathbf{m}}_p = \mathbf{e}_2\end{aligned} \hspace{\stretch{1}}(1.2.18)

\begin{aligned}\mathbf{e}_2 \rightarrow \mathbf{e}_2 \mathbf{e}_2 \hat{\mathbf{m}}_p = \hat{\mathbf{m}}_p\end{aligned} \hspace{\stretch{1}}(1.2.18)

This time we have $\mathbf{E}_p \parallel \hat{\mathbf{m}}_p$ and $\mathbf{B}_p \parallel \mathbf{e}_2$. Our boundary value equations become

\begin{aligned}\epsilon_1 \left\langle{{ \mathbf{e}_3 \left( \hat{\mathbf{m}}_i E_i + \hat{\mathbf{m}}_r E_r \right) }}\right\rangle = \epsilon_2 \left\langle{{ \mathbf{e}_3 \left( \hat{\mathbf{m}}_t E_t \right) }}\right\rangle\end{aligned} \hspace{\stretch{1}}(1.2.19a)

\begin{aligned}\frac{1}{{v_1}} \left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_i + \mathbf{e}_2 E_r \right) }}\right\rangle = \frac{1}{{v_2}} \left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_t \right) }}\right\rangle\end{aligned} \hspace{\stretch{1}}(1.2.19b)

\begin{aligned}{\left\langle{{ \mathbf{e}_3 \left( \hat{\mathbf{m}}_i E_i + \hat{\mathbf{m}}_r E_r \right) }}\right\rangle}_{2} = {\left\langle{{ \mathbf{e}_3 \left( \hat{\mathbf{m}}_t E_t \right) }}\right\rangle}_{2}\end{aligned} \hspace{\stretch{1}}(1.2.19c)

\begin{aligned}\frac{1}{{\mu_1 v_1}} {\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_i + \mathbf{e}_2 E_r \right) }}\right\rangle}_{2} = \frac{1}{{\mu_2 v_2}} {\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_t \right) }}\right\rangle}_{2}.\end{aligned} \hspace{\stretch{1}}(1.2.19d)

This second equation 1.2.19b is a $0 = 0$ identity, and the remaining after $\hat{\mathbf{m}}_p$ substitution are

\begin{aligned}\epsilon_1 \left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_3 j e^{j \theta_i} E_i + (-\mathbf{e}_3) j e^{-j \theta_r} E_r \right) }}\right\rangle = \epsilon_2 \left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_3 j e^{j \theta_t} E_t \right) }}\right\rangle\end{aligned} \hspace{\stretch{1}}(1.2.20a)

\begin{aligned}{\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_3 j e^{j \theta_i} E_i + (-\mathbf{e}_3) j e^{-j \theta_r} E_r \right) }}\right\rangle}_{2} = {\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_3 j e^{j \theta_t} E_t \right) }}\right\rangle}_{2}\end{aligned} \hspace{\stretch{1}}(1.2.20b)

\begin{aligned}\frac{1}{{\mu_1 v_1}} {\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_i + \mathbf{e}_2 E_r \right) }}\right\rangle}_{2} = \frac{1}{{\mu_2 v_2}} {\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_t \right) }}\right\rangle}_{2}.\end{aligned} \hspace{\stretch{1}}(1.2.20c)

Simplifying we have

\begin{aligned}\epsilon_1 \left( -\sin \theta_i E_i - \sin{\theta_r} E_r \right) = - \epsilon_2 \sin{\theta_t} E_t \end{aligned} \hspace{\stretch{1}}(1.2.21a)

\begin{aligned} \cos{ \theta_i} E_i - \cos{ \theta_r} E_r = \cos{ \theta_t} E_t\end{aligned} \hspace{\stretch{1}}(1.2.21b)

\begin{aligned}E_i + E_r = \frac{\mu_1 v_1} {\mu_2 v_2} E_t\end{aligned} \hspace{\stretch{1}}(1.2.21c)

We expect an equality

\begin{aligned}\frac{\epsilon_2 \sin\theta_t}{\epsilon_1 \sin\theta_i} = \frac{\mu_1 v_1} {\mu_2 v_2},\end{aligned} \hspace{\stretch{1}}(1.2.21c)

Noting that $\epsilon_p v_p = 1/(v_p \mu_p)$ we find that to be true

\begin{aligned}\frac{\epsilon_2 \sin\theta_t}{\epsilon_1 \sin\theta_i} = \frac{\epsilon_2 n_1}{\epsilon_1 n_2} = \frac{\epsilon_2 v_2}{\epsilon_1 v_1} = \frac{\mu_1 v_1}{\mu_2 v_2} \end{aligned} \hspace{\stretch{1}}(1.2.21c)

we see that 1.2.21a and 1.2.21c are dependent. We are left with the system

\begin{aligned}E_i - E_r = \alpha E_t\end{aligned} \hspace{\stretch{1}}(1.2.24a)

\begin{aligned}E_i + E_r = \beta E_t,\end{aligned} \hspace{\stretch{1}}(1.2.24b)

with solution

\begin{aligned}\frac{E_t}{E_i} = \frac{2 }{\beta + \alpha}\end{aligned} \hspace{\stretch{1}}(1.2.25)

\begin{aligned}\frac{E_r}{E_i} = \frac{\beta - \alpha}{\beta + \alpha}\end{aligned} \hspace{\stretch{1}}(1.2.25)

## Question: General case. Arbitrary polarization angle.

Determine the set of simulaneous equations that would have to be solved for if the incident polarization angle was allowed to be neither TE nor TM mode.

Substituting our $\hat{\mathbf{m}}_p$ vector expressions into the boundary value constraints we have

\begin{aligned}\epsilon_1 \left\langle{{ \mathbf{e}_3 \mathbf{e}_2 \left( E_i e^{ \hat{\mathbf{m}}_i \mathbf{e}_2 \phi_i } + E_r e^{ \hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r } \right) }}\right\rangle = \epsilon_2 \left\langle{{ \mathbf{e}_3 \mathbf{e}_2 E_t e^{ \hat{\mathbf{m}}_t \mathbf{e}_2 \phi_t } }}\right\rangle \end{aligned} \hspace{\stretch{1}}(1.2.26a)

\begin{aligned}\frac{1}{{v_1}} \left\langle{{ j e^{j \theta_i} E_i e^{ \hat{\mathbf{m}}_i \mathbf{e}_2 \phi_i } - j e^{-j \theta_r} E_r e^{ \hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r } }}\right\rangle = \frac{1}{{v_2}} \left\langle{{ j e^{j \theta_t} E_t e^{ \hat{\mathbf{m}}_t \mathbf{e}_2 \phi_t } }}\right\rangle\end{aligned} \hspace{\stretch{1}}(1.2.26b)

\begin{aligned}{\left\langle{{ \mathbf{e}_3 \mathbf{e}_2 \left( E_i e^{ \hat{\mathbf{m}}_i \mathbf{e}_2 \phi_i } + E_r e^{ \hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r } \right) }}\right\rangle}_{2} = {\left\langle{{ \mathbf{e}_3 \mathbf{e}_2 E_t e^{ \hat{\mathbf{m}}_t \mathbf{e}_2 \phi_t } }}\right\rangle}_{2}\end{aligned} \hspace{\stretch{1}}(1.2.26c)

\begin{aligned}\frac{1}{{\mu_1 v_1}} {\left\langle{{ j e^{j \theta_i} E_i e^{ \hat{\mathbf{m}}_i \mathbf{e}_2 \phi_i } - j e^{-j \theta_r} E_r e^{ \hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r } }}\right\rangle}_{2} = \frac{1}{{\mu_2 v_2}} {\left\langle{{ j e^{j \theta_t} E_t e^{ \hat{\mathbf{m}}_t \mathbf{e}_2 \phi_t } }}\right\rangle}_{2}\end{aligned} \hspace{\stretch{1}}(1.2.26d)

With $\alpha \in \{i,r\}$ we want to expand some intermediate multivector products

\begin{aligned}\mathbf{e}_{32} e^{\hat{\mathbf{m}}_q \mathbf{e}_2 \phi_q}=\mathbf{e}_{32} \cos \phi_q+\mathbf{e}_{32} \hat{\mathbf{m}}_q \mathbf{e}_2 \sin{\phi_q}=\mathbf{e}_{32} \cos \phi_q+\mathbf{e}_{32} \mathbf{e}_3 j e^{j \theta_q} \mathbf{e}_2 \sin{\phi_q}=\mathbf{e}_{32} \cos \phi_q-j e^{j \theta_q} \sin{\phi_q}=\mathbf{e}_{32} \cos \phi_q - \mathbf{e}_{31} \cos\theta_q \sin\phi_q+ \sin\theta_q \sin{\phi_q}\end{aligned} \hspace{\stretch{1}}(1.2.26d)

\begin{aligned}\mathbf{e}_{32} e^{\hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r}=\mathbf{e}_{32} \cos \phi_r+\mathbf{e}_{32} \hat{\mathbf{m}}_r \mathbf{e}_2 \sin{\phi_r}=\mathbf{e}_{32} \cos \phi_r+\mathbf{e}_{32} (-\mathbf{e}_3) j e^{-j \theta_r} \mathbf{e}_2 \sin{\phi_r}=\mathbf{e}_{32} \cos \phi_r+ j e^{-j \theta_r} \sin{\phi_r}=\mathbf{e}_{32} \cos \phi_r + \mathbf{e}_{31} \cos\theta_r \sin{\phi_r}+ \sin \theta_r \sin{\phi_r}\end{aligned} \hspace{\stretch{1}}(1.2.26d)

\begin{aligned}j e^{j \theta_q} e^{\hat{\mathbf{m}}_q \mathbf{e}_2 \phi_q}=j e^{j \theta_q} \left( \cos \phi_q+ \hat{\mathbf{m}}_q \mathbf{e}_2 \sin{\phi_q}\right)=j e^{j \theta_q} \left( \cos \phi_q+ \mathbf{e}_3 j e^{j \theta_q} \mathbf{e}_2 \sin{\phi_q}\right)=j e^{j \theta_q} \left( \cos \phi_q- j e^{-j \theta_q} \mathbf{e}_{32} \sin{\phi_q}\right)=j e^{j \theta_q} \cos \phi_q+ \mathbf{e}_{32} \sin{\phi_q}=\mathbf{e}_{31} \cos {j \theta_q} \cos \phi_q+ \mathbf{e}_{32} \sin{\phi_q}- \sin{ \theta_q} \cos \phi_q\end{aligned} \hspace{\stretch{1}}(1.2.26d)

\begin{aligned}-j e^{-j \theta_r} e^{\hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r}=-j e^{-j \theta_r} \left(\cos\phi_r + \hat{\mathbf{m}}_r \mathbf{e}_2 \sin\phi_r\right)=-j e^{-j \theta_r} \left(\cos\phi_r - \mathbf{e}_3 j e^{-j \theta_r} \mathbf{e}_2 \sin\phi_r\right)=-j e^{-j \theta_r} \left(\cos\phi_r + j e^{j \theta_r} \mathbf{e}_{32} \sin\phi_r\right)=-j e^{-j \theta_r} \cos\phi_r + \mathbf{e}_{32} \sin\phi_r=-\mathbf{e}_{31} \cos{\theta_r} \cos\phi_r + \mathbf{e}_{32} \sin\phi_r- \sin{ \theta_r} \cos\phi_r \end{aligned} \hspace{\stretch{1}}(1.2.26d)

Our boundary value conditions are then

\begin{aligned}\epsilon_1 \left( E_i \sin\theta_i \sin{\phi_i}+ E_r \sin \theta_r \sin{\phi_r}\right) = \epsilon_2 E_t \sin\theta_t \sin{\phi_t}\end{aligned} \hspace{\stretch{1}}(1.2.31)

\begin{aligned}\frac{1}{{v_1}}\left(E_i \sin{ \theta_i} \cos \phi_i+E_r \sin{ \theta_r} \cos\phi_r \right)=\frac{1}{{v_2}}E_t \sin{ \theta_t} \cos \phi_t\end{aligned} \hspace{\stretch{1}}(1.2.31)

\begin{aligned}E_i \cos \phi_i + E_r \cos \phi_r =E_t \cos \phi_t \end{aligned} \hspace{\stretch{1}}(1.2.31)

\begin{aligned}-E_i \cos\theta_i \sin\phi_i+ E_r \cos\theta_r \sin{\phi_r}=-E_t \cos\theta_t \sin\phi_t\end{aligned} \hspace{\stretch{1}}(1.2.31)

\begin{aligned}\frac{1}{{\mu_1 v_1}}\left(E_i\cos { \theta_i} \cos \phi_i -E_r\cos{\theta_r} \cos\phi_r \right)=\frac{1}{{\mu_2 v_2}}E_t \cos { \theta_t} \cos \phi_t \end{aligned} \hspace{\stretch{1}}(1.2.31)

\begin{aligned}\frac{1}{{\mu_1 v_1}}\left(E_i\sin{\phi_i}+E_r\sin\phi_r\right)=\frac{1}{{\mu_2 v_2}}E_t\sin{\phi_t}\end{aligned} \hspace{\stretch{1}}(1.2.31)

Note that the wedge product equations above have been separated into $\mathbf{e}_3 \mathbf{e}_1$ and $\mathbf{e}_3 \mathbf{e}_2$ components, yielding two equations each. Because of 1.2.21c, we see that 1.2.31 and 1.2.31 are dependent. Also, as demonstrated in 1.2.12d we see that 1.2.31 and 1.2.31 are also dependent. We can therefore consider only the last four equations (and still have additional linear dependencies to be discovered.)

Let’s write these as

\begin{aligned}E_i \cos \phi_i + E_r \cos \phi_r =E_t \cos \phi_t \end{aligned} \hspace{\stretch{1}}(1.2.32)

\begin{aligned}-E_i \sin\phi_i+ E_r \sin{\phi_r}=-E_t \alpha\sin\phi_t\end{aligned} \hspace{\stretch{1}}(1.2.32)

\begin{aligned}E_i\cos \phi_i -E_r\cos\phi_r =\alpha \beta E_t\cos \phi_t \end{aligned} \hspace{\stretch{1}}(1.2.32)

\begin{aligned}E_i\sin{\phi_i}+E_r\sin\phi_r=\beta E_t\sin{\phi_t}\end{aligned} \hspace{\stretch{1}}(1.2.32)

Observe that if $\phi_i = \phi_r = \phi_t = 0$ (killing all the sine terms) we recover 1.2.14, and with $\phi_i = \phi_r = \phi_t = \pi/2$ (killing all the cosines) we recover 1.2.24.

Now, if $\phi_i n \pi/2$ we’ve got a different story. Specifically it appears that should we wish to solve for the reflected and transmitted magnitudes, we also have to simulaneously solve for the polarization angles in the reflected and transmitted directions. This is now a problem of solving four simulaneous equations in two linear and two non-linear variables.

Does it make sense that we would have polarization rotation should our initial polarization angle be rotated? I think so. In dicusssing this problem with Prof Thywissen, he strongly suggested treating the problem as a superposition of two light waves. If we consider that, even without attempting to solve the problem, we see that we must have different reflected and transmitted magnitudes associated with the pair of incident waves since we have to calculate each of these with different Fresnel equations. This would have an effect of scaling and rotating the superimposed reflected and transmitted waves.

## Question: General case using using superposition

Using superposition determine the Fresnel equations for an arbitrary incident polarization angle. This should involve solving for both the magnitude and the polarization angle of the reflected and transmitted rays.

For a polarization of $\phi = 0$ and $\phi = \pi/2$ respectively, we have from problems \ref{fresnelAlternatePolarization:pr1-Answer} and \ref{fresnelAlternatePolarization:pr2-Answer}, or from 1.2.32 we have

\begin{aligned}\frac{E_{r \parallel}}{E_{i \parallel}} = \frac{1 - \alpha \beta}{1 + \alpha \beta}\end{aligned} \hspace{\stretch{1}}(1.2.33)

\begin{aligned}\frac{E_{t \parallel}}{E_{i \parallel}} = \frac{2 }{1 + \alpha \beta}\end{aligned} \hspace{\stretch{1}}(1.2.33)

\begin{aligned}\frac{E_{r \perp}}{E_{i \perp}} = \frac{ \beta - \alpha }{\beta + \alpha}\end{aligned} \hspace{\stretch{1}}(1.2.33)

\begin{aligned}\frac{E_{t \perp}}{E_{i \perp}} = \frac{ 2 }{\beta + \alpha}\end{aligned} \hspace{\stretch{1}}(1.2.33)

We can use these results to consider a polarization of $\phi < \pi/2$ as illustrated in figure (see pdf)

Figure: see pdf: Polarization of incident field to be considered

Our incident, reflected, and transmitted fields are

\begin{aligned}\mathbf{E}_i = E_{i} \mathbf{e}_2 e^{\mathbf{e}_2 \hat{\mathbf{m}}_i \phi}\end{aligned} \hspace{\stretch{1}}(1.2.34)

\begin{aligned}\mathbf{E}_r = E_{i \parallel}\frac{1 - \alpha\beta}{1 + \alpha\beta} \mathbf{e}_2 + E_{i \perp}\frac{\beta - \alpha}{\beta + \alpha} \hat{\mathbf{m}}_r\end{aligned} \hspace{\stretch{1}}(1.2.34)

\begin{aligned}\mathbf{E}_t = E_{i \parallel}\frac{2}{1 + \alpha\beta}\mathbf{e}_2+ E_{i \perp}\frac{2}{\beta + \alpha}\hat{\mathbf{m}}_i\end{aligned} \hspace{\stretch{1}}(1.2.34)

However, $E_{i \parallel} = E_i \cos \phi$ and $E_{i \perp} = E_i \sin\phi$ leaving us with

\begin{aligned}\mathbf{E}_i = E_{i} \left( \mathbf{e}_2 \cos\phi + \mathbf{e}_1 e^{j \theta_i} \sin\phi \right)\end{aligned} \hspace{\stretch{1}}(1.2.35)

\begin{aligned}\mathbf{E}_r = E_i \left(\cos\phi\frac{1 - \alpha\beta}{1 + \alpha\beta} \mathbf{e}_2 - \sin\phi\frac{\beta - \alpha}{\beta + \alpha} \mathbf{e}_1 e^{-j \theta_r}\right)\end{aligned} \hspace{\stretch{1}}(1.2.35)

\begin{aligned}\mathbf{E}_t = E_i\left(\cos\phi\frac{2}{1 + \alpha\beta}\mathbf{e}_2+ \sin\phi\frac{2}{\beta + \alpha}\mathbf{e}_1 e^{j \theta_t}\right)\end{aligned} \hspace{\stretch{1}}(1.2.35)

We find that the reflected and transmitted polarization angles are respectively

\begin{aligned}\tan \phi_r = \tan \phi\frac{\beta - \alpha}{\beta + \alpha} \frac{1 + \alpha \beta}{1 - \alpha \beta}\end{aligned} \hspace{\stretch{1}}(1.2.36)

\begin{aligned}\tan \phi_t = \tan \phi \frac{ 1 + \alpha \beta}{ \beta + \alpha}\end{aligned} \hspace{\stretch{1}}(1.2.36)

where the associated magnitudes are

\begin{aligned}\frac{E_r}{E_i}= \sqrt{\left(\cos\phi\frac{1 - \alpha\beta}{1 + \alpha\beta} \right)^2+ \left( \sin\phi\frac{\beta - \alpha}{\beta + \alpha} \right)^2}\end{aligned} \hspace{\stretch{1}}(1.2.37)

\begin{aligned}\frac{E_t}{E_i}=\sqrt{\left(\cos\phi\frac{2}{1 + \alpha\beta}\right)^2+\left(\sin\phi\frac{2}{\beta + \alpha}\right)^2}\end{aligned} \hspace{\stretch{1}}(1.2.37)

# References

[1] D.J. Griffith. Introduction to Electrodynamics. Prentice-Hall, 1981.

[2] E. Hecht. Optics. 1998.

Posted in Math and Physics Learning. | Tagged: , , , , | 5 Comments »

## Poking around to see how much stack to corrupt to alter a local variable.

Posted by peeterjoot on September 10, 2012

I’ve got a scenerio where it appears that the last stack variable declared appears to be have been corrupted (the highest order 32-bits of this 64-bit integer look like they’ve been zeroed). That got me wondering how far a calling function would have to corrupt to muck up this variable. Here’s what I wrote to test this:

#include <stdio.h>

int foo( int r )
{
Uint64         x ;
Uint64         y ;
Uint64         z ;
Uint64         w = 0 ;

w = 1 ;

printf( "&x: 0x%0lx\n", (long)&x ) ;
printf( "&w: 0x%0lx\n", (long)&w ) ;

if ( r )
{
foo( r - 1 ) ;
}

return w ;
}

int main()
{
foo( 2 ) ;

return 0 ;
}


and the results on this (linuxamd64) system:

&x: 0x7fffffffd490
&w: 0x7fffffffd488
&x: 0x7fffffffd450
&w: 0x7fffffffd448
&x: 0x7fffffffd410
&w: 0x7fffffffd408


So, it looks like I need about at least a (0x88-0x50 =) 56 byte corruption to do the job.

A quirk: Also see how the variables in my function actually got laid out in reverse address order on the stack. I’d not have expected that. However, since I don’t really have any reason to expect any specific stack layout so perhaps I shouldn’t be surprised.

## How to find exported symbols in Windows dlls

Posted by peeterjoot on September 5, 2012

One liner:

WSDB::E:\snap\> dumpbin /exports db2app64.dll | grep DiagWhat
510  212 00BD1152 pdDiagWhatIsRc = pdDiagWhatIsRc


This isn’t an nm equivalent, instead is more like the AIX command to dump just the exported symbols from a shared object (dump -TvHX32_64), but enough to tell me that I shouldn’t have a link error this iteration of my build.

I found it curious that ‘dumpbin /symbols’ didn’t produce any output for this dll, as is does for a .obj file, and don’t really know what the reason for that is.

## Plane wave solutions of Maxwell’s equation using Geometric Algebra

Posted by peeterjoot on September 3, 2012

# Motivation

Study of reflection and transmission of radiation in isotropic, charge and current free, linear matter utilizes the plane wave solutions to Maxwell’s equations. These have the structure of phasor equations, with some specific constraints on the components and the exponents.

These constraints are usually derived starting with the plain old vector form of Maxwell’s equations, and it is natural to wonder how this is done directly using Geometric Algebra. [1] provides one such derivation, using the covariant form of Maxwell’s equations. Here’s a slightly more pedestrian way of doing the same.

# Maxwell’s equations in media

We start with Maxwell’s equations for linear matter as found in [2]

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{E} = 0\end{aligned} \hspace{\stretch{1}}(1.2.1a)

\begin{aligned}\boldsymbol{\nabla} \times \mathbf{E} = -\frac{\partial {\mathbf{B}}}{\partial {t}}\end{aligned} \hspace{\stretch{1}}(1.2.1b)

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{B} = 0\end{aligned} \hspace{\stretch{1}}(1.2.1c)

\begin{aligned}\boldsymbol{\nabla} \times \mathbf{B} = \mu\epsilon \frac{\partial {\mathbf{E}}}{\partial {t}}.\end{aligned} \hspace{\stretch{1}}(1.2.1d)

We merge these using the geometric identity

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{a} + I \boldsymbol{\nabla} \times \mathbf{a} = \boldsymbol{\nabla} \mathbf{a},\end{aligned} \hspace{\stretch{1}}(1.2.2)

where $I$ is the 3D pseudoscalar $I = \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3$, to find

\begin{aligned}\boldsymbol{\nabla} \mathbf{E} = -I \frac{\partial {\mathbf{B}}}{\partial {t}}\end{aligned} \hspace{\stretch{1}}(1.2.3a)

\begin{aligned}\boldsymbol{\nabla} \mathbf{B} = I \mu\epsilon \frac{\partial {\mathbf{E}}}{\partial {t}}.\end{aligned} \hspace{\stretch{1}}(1.2.3b)

We want dimensions of $1/L$ for the derivative operator on the RHS of 1.2.3b, so we divide through by $\sqrt{\mu\epsilon} I$ for

\begin{aligned}-I \frac{1}{{\sqrt{\mu\epsilon}}} \boldsymbol{\nabla} \mathbf{B} = \sqrt{\mu\epsilon} \frac{\partial {\mathbf{E}}}{\partial {t}}.\end{aligned} \hspace{\stretch{1}}(1.2.4)

This can now be added to 1.2.3a for

\begin{aligned}\left(\boldsymbol{\nabla} + \sqrt{\mu\epsilon} \frac{\partial {}}{\partial {t}} \right) \left( \mathbf{E} + \frac{I}{\sqrt{\mu\epsilon}} \mathbf{B} \right)= 0.\end{aligned} \hspace{\stretch{1}}(1.2.5)

This is Maxwell’s equation in linear isotropic charge and current free matter in Geometric Algebra form.

# Phasor solutions

We write the electromagnetic field as

\begin{aligned}F = \left( \mathbf{E} + \frac{I}{\sqrt{\mu\epsilon}} \mathbf{B} \right),\end{aligned} \hspace{\stretch{1}}(1.3.6)

so that for vacuum where $1/\sqrt{\mu \epsilon} = c$ we have the usual $F = \mathbf{E} + I c \mathbf{B}$. Assuming a phasor solution of

\begin{aligned}\tilde{F} = F_0 e^{i (\mathbf{k} \cdot \mathbf{x} - \omega t)}\end{aligned} \hspace{\stretch{1}}(1.3.7)

where $F_0$ is allowed to be complex, and the actual field is obtained by taking the real part

\begin{aligned}F = \text{Real} \tilde{F} = \text{Real}(F_0) \cos(\mathbf{k} \cdot \mathbf{x} - \omega t)-\text{Imag}(F_0) \sin(\mathbf{k} \cdot \mathbf{x} - \omega t).\end{aligned} \hspace{\stretch{1}}(1.3.8)

Note carefully that we are using a scalar imaginary $i$, as well as the multivector (pseudoscalar) $I$, despite the fact that both have the square to scalar minus one property.

We now seek the constraints on $\mathbf{k}$, $\omega$, and $F_0$ that allow this to be a solution to 1.2.5

\begin{aligned}0 = \left(\boldsymbol{\nabla} + \sqrt{\mu\epsilon} \frac{\partial {}}{\partial {t}} \right) \tilde{F}.\end{aligned} \hspace{\stretch{1}}(1.3.9)

As usual in the non-geometric algebra treatment, we observe that any such solution $F$ to Maxwell’s equation is also a wave equation solution. In GA we can do so by right multiplying an operator that has a conjugate form,

\begin{aligned}\begin{aligned}0 &= \left(\boldsymbol{\nabla} + \sqrt{\mu\epsilon} \frac{\partial {}}{\partial {t}} \right) \tilde{F} \\ &= \left(\boldsymbol{\nabla} - \sqrt{\mu\epsilon} \frac{\partial {}}{\partial {t}} \right) \left(\boldsymbol{\nabla} + \sqrt{\mu\epsilon} \frac{\partial {}}{\partial {t}} \right) \tilde{F} \\ &=\left( \boldsymbol{\nabla}^2 - \mu\epsilon \frac{\partial^2}{\partial t^2} \right) \tilde{F} \\ &=\left( \boldsymbol{\nabla}^2 - \frac{1}{{v^2}} \frac{\partial^2}{\partial t^2} \right) \tilde{F},\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.3.10)

where $v = 1/\sqrt{\mu\epsilon}$ is the speed of the wave described by this solution.

Inserting the exponential form of our assumed solution 1.3.7 we find

\begin{aligned}0 = -(\mathbf{k}^2 - \omega^2/v^2) F_0 e^{i (\mathbf{k} \cdot \mathbf{x} - \omega t)},\end{aligned} \hspace{\stretch{1}}(1.3.11)

which implies that the wave number vector $\mathbf{k}$ and the angular frequency $\omega$ are related by

\begin{aligned}v^2 \mathbf{k}^2 = \omega^2.\end{aligned} \hspace{\stretch{1}}(1.3.12)

Our assumed solution must also satisfy the first order system 1.3.9

\begin{aligned}\begin{aligned}0 &= \left(\boldsymbol{\nabla} + \sqrt{\mu\epsilon} \frac{\partial {}}{\partial {t}} \right) F_0e^{i (\mathbf{k} \cdot \mathbf{x} - \omega t)} \\ &=i\left(\mathbf{e}_m k_m - \frac{\omega}{v}\right) F_0e^{i (\mathbf{k} \cdot \mathbf{x} - \omega t)} \\ &=i k ( \hat{\mathbf{k}} - 1 ) F_0 e^{i (\mathbf{k} \cdot \mathbf{x} - \omega t)}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.3.13)

The constraints on $F_0$ must then be given by

\begin{aligned}0 = ( \hat{\mathbf{k}} - 1 ) F_0.\end{aligned} \hspace{\stretch{1}}(1.3.14)

With

\begin{aligned}F_0 = \mathbf{E}_0 + I v \mathbf{B}_0,\end{aligned} \hspace{\stretch{1}}(1.3.15)

we must then have all grades of the multivector equation equal to zero

\begin{aligned}0 = ( \hat{\mathbf{k}} - 1 ) \left(\mathbf{E}_0 + I v \mathbf{B}_0\right).\end{aligned} \hspace{\stretch{1}}(1.3.16)

Writing out all the geometric products, noting that $I$ commutes with all of $\hat{\mathbf{k}}$, $\mathbf{E}_0$, and $\mathbf{B}_0$ and employing the identity $\mathbf{a} \mathbf{b} = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \wedge \mathbf{b}$ we have

\begin{aligned}\begin{array}{l l l l l}0 &= \hat{\mathbf{k}} \cdot \mathbf{E}_0 & - \mathbf{E}_0 & + \hat{\mathbf{k}} \wedge \mathbf{E}_0 & I v \hat{\mathbf{k}} \cdot \mathbf{B}_0 \\ & & + I v \hat{\mathbf{k}} \wedge \mathbf{B}_0 & + I v \mathbf{B}_0 &\end{array}\end{aligned} \hspace{\stretch{1}}(1.3.17)

This is

\begin{aligned}0 = \hat{\mathbf{k}} \cdot \mathbf{E}_0 \end{aligned} \hspace{\stretch{1}}(1.3.18a)

\begin{aligned}\mathbf{E}_0 =- \hat{\mathbf{k}} \times v \mathbf{B}_0 \end{aligned} \hspace{\stretch{1}}(1.3.18b)

\begin{aligned}v \mathbf{B}_0 = \hat{\mathbf{k}} \times \mathbf{E}_0 \end{aligned} \hspace{\stretch{1}}(1.3.18c)

\begin{aligned}0 = \hat{\mathbf{k}} \cdot \mathbf{B}_0.\end{aligned} \hspace{\stretch{1}}(1.3.18d)

This and 1.3.12 describe all the constraints on our phasor that are required for it to be a solution. Note that only one of the two cross product equations in are required because the two are not independent. This can be shown by crossing $\hat{\mathbf{k}}$ with 1.3.18b and using the identity

\begin{aligned}\mathbf{a} \times (\mathbf{a} \times \mathbf{b}) = - \mathbf{a}^2 \mathbf{b} + a (\mathbf{a} \cdot \mathbf{b}).\end{aligned} \hspace{\stretch{1}}(1.3.19)

One can find easily that 1.3.18b and 1.3.18c provide the same relationship between the $\mathbf{E}_0$ and $\mathbf{B}_0$ components of $F_0$. Writing out the complete expression for $F_0$ we have

\begin{aligned}\begin{aligned}F_0 &= \mathbf{E}_0 + I v \mathbf{B}_0 \\ &=\mathbf{E}_0 + I \hat{\mathbf{k}} \times \mathbf{E}_0 \\ &=\mathbf{E}_0 + \hat{\mathbf{k}} \wedge \mathbf{E}_0.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.3.20)

Since $\hat{\mathbf{k}} \cdot \mathbf{E}_0 = 0$, this is

\begin{aligned}F_0 = (1 + \hat{\mathbf{k}}) \mathbf{E}_0.\end{aligned} \hspace{\stretch{1}}(1.3.21)

Had we been clever enough this could have been deduced directly from the 1.3.14 directly, since we require a product that is killed by left multiplication with $\hat{\mathbf{k}} - 1$. Our complete plane wave solution to Maxwell’s equation is therefore given by

\begin{aligned}\begin{aligned}F &= \text{Real}(\tilde{F}) = \mathbf{E} + \frac{I}{\sqrt{\mu\epsilon}} \mathbf{B} \\ \tilde{F} &= (1 \pm \hat{\mathbf{k}}) \mathbf{E}_0 e^{i (\mathbf{k} \cdot \mathbf{x} \mp \omega t)} \\ 0 &= \hat{\mathbf{k}} \cdot \mathbf{E}_0 \\ \mathbf{k}^2 &= \omega^2 \mu \epsilon.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.3.22)

# References

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

[2] D.J. Griffith. Introduction to Electrodynamics. Prentice-Hall, 1981.