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\myskip
Now suppose $\bf{E}_0$ is complex and of the form
\[ \bf{E}_0 = \bf{a} - i\bf{b} \]
where $\bf{a}$ and $\bf{b}$ are linearly
independent real vectors in the plane
perpendicular to $\bf{k}$. Then the direction of
\[ \bf{E} = \bf{a} \cos(\bf{k} \cdot \bf{x} -
\omega t) + \bf{b} \sin(\bf{k} \cdot \bf{x} -
\omega t) \]
is not fixed but rotates in the plane
perpendicular to $\bf{k}$. In general it traces
out an ellipse and the wave is said to be
\emph{elliptically polarized}. The special case of
\emph{circular polarization} occurs when $|\bf{b}| =
|\bf{a}|$. Then $\bf{E}$ traces out a circle.

\myskip
The polarization if \emph{right-handed} if $\bf{b}
= \hat{\bf{k}} \times \bf{a}$ and
\emph{left-handed} if $\bf{b} = -\hat{\bf{k}}
\times \bf{a}$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/80ac98ceb35a11ed.png}
\end{center}
Exercise: work out how $\bf{B}$ evolves.

\subsubsection*{Reflection from a conductor}

Suppose a perfect conductor occupies the
half-space $x > 0$. An \emph{incident wave}
approaching the surface $x = 0$ in the $+x$
direction has the form
\[ \bf{E}_{\text{inc}} = \Re \left( E_0 e^{i(kx-
\omega t)} \bf{e}_y \right) \]
where we take $k > 0$, $\omega = ck > 0$ and $E_0
\in \RR$.

\myskip
Inside the conductor, we have $\bf{E} = \bf{0}$.
Since the tangential components of $\bf{E}$ are
continuous at the surface ($[\bf{n} \times \bf{E}]
= \bf{0}$) we require $E_y = 0$ at $x = 0$. This
can be satisfied by adding a \emph{reflected wave}
\[ \bf{E}_{\text{ref}} = \Re \left( -E_0
e^{i(-kx-\omega t)} \bf{e}_y \right) \]
propagating in the $-x$ direction.

\myskip
The combined solution is
\begin{align*}
    \bf{E}
    &= \bf{E}_{\text{inc}} + \bf{E}_{\text{ref}}
    \\
    &= \Re \left( 2iE_0 \sin(kx) e^{-i\omega t}
    \bf{e}_y \right) \\
    &= 2E_0 \sin(kx) \sin(\omega t) \bf{e}_y
\end{align*}
which is a \emph{standing wave}. The corresponding
field is
\begin{align*}
    \bf{B}
    &= \Re \left( \frac{E_0}{c} e^{i(kx - \omega
    t)} \bf{e}_z \right) + \Re \left(
    \frac{E_0}{c} e^{i(-kx -\omega t)} \bf{e}_z
    \right) \\
    &= \Re \left( \frac{2E_0}{c} \cos(kx)
    e^{-i\omega t} \bf{e}_z \right) \\
    &= \frac{2E_0}{c} \cos(kx) \cos(\omega t)
    \bf{e}_z
\end{align*}
Inside the conductor, $\bf{B} = \bf{0}$ (since
$\pfrac{\bf{B}}{t} = -\nabla \times \bf{E}$ and
$\bf{B}$ is harmonic in time). So there is a
tangential discontinuity in $\bf{B}$ and a surface
current
\[ \bf{K} = \frac{1}{\mu_0}[\bf{n} \times \bf{B}]
= \frac{2E_0}{\mu_0 c} \cos(\omega t) \bf{e}_y \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/852bc98ab35d11ed.png}
\end{center}

\subsection{Electromagnetic Energy}

In electrostatics, we saw that the energy per unit
volume in $\bf{E}$ is $\half \eps_0 |\bf{E}|^2$.

\myskip
In magnetostatics, we saw that the energy per unit
volume in $\bf{B}$ is $\frac{1}{2\mu_0}
|\bf{B}|^2$.

\myskip
ow derive a general conservation law from
Maxwell's equations. Consider the rate of change
of the electric and magnetic energy per unit
volume:
\begin{align*}
    \pfrac{}{t} \left( \half \eps_0 |\bf{E}|^2 +
    \frac{1}{2\mu_0} |\bf{B}|^2 \right)
    &= \bf{E} \cdot \left( \eps_0
    \pfrac{\bf{E}}{t} \right) + \bf{B} \cdot
    \left( \frac{1}{\mu_0} \pfrac{\bf{B}}{t}
    \right) \\
    &= \bf{E} \cdot \left( \frac{1}{\mu_0} \nabla
    \times \bf{B} - \bf{J} \right) + \bf{B} \cdot
    \left( -\frac{1}{\mu_0} \nabla \times \bf{E} \right)
\end{align*}
An identity from vector calculus:
\[ \nabla \cdot (\bf{E} \times \bf{B}) = \bf{B}
\cdot (\nabla \times \bf{E}) - \bf{E} \cdot
(\nabla \times \bf{B}) \]
Divide by $\mu_0$ and add to previous equation:
\[ \pfrac{}{t} \left( \half \eps_0 |\bf{E}|^2 +
\frac{1}{2\mu_0} |\bf{B}|^2 \right) + \nabla
\cdots \left( \frac{\bf{E} \times \bf{B}}{\mu_0}
\right) = -\bf{E} \cdot \bf{J} \]
This is the energy equation for the EM field.
Write it as
\[ \boxed{\pfrac{w}{t} + \nabla \cdot S = -\bf{E}
\cdot \bf{J}} \tag{1} \]
where
\[ \boxed{w = \half \eps_0 |\bf{E}|^2 + \frac{1}{2\mu_0}
|\bf{B}|^2} \]
is the \emph{energy density} (energy / unit
volume) and
\[ \boxed{\bf{S} = \frac{\bf{E} \times
\bf{B}}{\mu_0}} \]
is the \emph{energy flux density} (energy flux /
area). $\bf{S}$ is known as the \emph{Poynting
vector}. In the absence of currents, equation (1)
becomes
\[ \pfrac{w}{t} + \nabla \cdot \bf{S} = 0 \]
and expresses the conservation of EM energy.
Integrate over a time-independent volume $V$ with
$\partial V = S$ and use the divergence theorem:
\[ \dfrac{}{t} \int_V w \dd V + \int_S \bf{S}
\cdot \dd \bf{S} = 0 \]
(rate of change of energy in $V$ equals $-$ flux
of energy through $S$). On the RHS of equation
(1), $\bf{E} \cdot \bf{J}$ is the rate at which
the EM field loses energy (per unit volume) by
doin gwork on charged particles via the Lorentz
force. Recall
\begin{align*}
    \rho
    &= \sum_{i = 1}^N q_i \delta(\bf{x} -
    \bf{x}_i) \\
    \bf{J}
    &= \sum_{i = 1}^N q_i \dot{\bf{x}}_i
    \delta(\bf{x} - \bf{x}_i) \\
    \bf{F}_i
    &= q_i (\bf{E}(\bf{x}_i, t) + \dot{\bf{x}}_i
    \times \bf{B} (\bf{x}_i, t))
\end{align*}
so the rate of working is
\begin{align*}
    \sum_{i = 1}^N \bf{F}_i \cdot \dot{\bf{x}}_i
    &= \sum_{i = 1}^N \bf{E}(\bf{x}_i, t) \cdot
    (q_i \dot{\bf{x}}_i) \\
    &= \int \bf{E} \cdot \bf{J} \dd V
\end{align*}
In a conductor, Ohm's law $\bf{J} = \sigma \bf{E}$
implies $\bf{E} \cdot \bf{J} =
\frac{|\bf{J}|^2}{\sigma}$ which is positive
definite, and is the rate of Joule heating /
volume. (EM energy $\to$ heat)

\subsubsection*{Energy density and flux for an EM wave}

For
\begin{align*}
    \bf{E}
    &= \Re(\bf{E}_0 e^{i(\bf{k} \cdot \bf{x} -
    \omega t)}) \\
    \bf{B}
    &= \Re(\bf{B}_0 e^{i(\bf{k} \cdot \bf{x} -
    \omega t)})
\end{align*}
then using the identity
\[ \Re(A) \Re(B) = \Re \left( \frac{AB +
A\ol{B}}{2} \right), \]
we have
\begin{align*}
    w
    &= \half \eps_0 |\bf{E}|^2 + \frac{1}{2\mu_0}
    |\bf{B}|^2 \\
    &= \frac{1}{4}\eps_0 \Re ((\bf{E}_0 \cdot
    \bf{E}_0) e^{2i(\bf{k} \cdot \bf{x} - \omega
    t)} + |\bf{E}_0|^2)
    + \frac{1}{4\mu_0} \Re ((\bf{B}_0 \cdot
    \bf{B}_0) e^{2i(\bf{k} \cdot \bf{x} - \omega
    t)} + |\bf{B}_0|^2)
\end{align*}
The $e^{2i(\bf{k} \cdot \bf{x} - \omega t)}$ terms
average to zero, leaving the average energy
density
\[ \langle w \rangle = \frac{1}{4} \eps_0
|\bf{E}_0|^2 + \frac{1}{4\mu_0} |\bf{B}_0|^2 \]
Since $\bf{B}_0 = \frac{1}{c} \hat{\bf{k}} \times
Bf{E}_0$ and $\hat{\bf{k}} \cdot \bf{E}_0 = 0$,
the two contributions are equal and we have
\[ \langle w \rangle = \half \eps_0 |\bf{E}_0|^2 .\]
Similarly, the Poynting vector is
\begin{align*}
    \bf{S}
    &= \frac{\bf{E} \times \bf{B}}{\mu_0} \\
    &= \frac{1}{2\mu_0} \Re \left( (\bf{E}_0
    \times \bf{B}_0) e^{2i(\bf{k} \cdot \bf{x}
    - \omega t)} + (\bf{E}_0 \times \ol{\bf{B}}_0)
    \right)
\end{align*}
Average flux density
\[ \langle \bf{S} \rangle =
\frac{|\bf{E}_0|^2}{2\mu_0 c} \hat{\bf{k}} = \half
\eps_0 |\bf{E}_0|^2 c \hat{\bf{k}} = \langle w
\rangle c \hat{\bf{k}} \]
Implies that the EM wave transports energy at
speed $c$ in the direction $\hat{\bf{k}}$.