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\subsection{Electromagnetic waves}

\subsubsection*{The Wave Equation}

Consider freely evolving electric and magnetic
fields in a vaccuum, in the absence of charges and
currents:
\begin{align*}
    \nabla \cdot \bf{E}
    &= 0 \tag{M1} \\
    \nabla \cdot \bf{B}
    &= 0 \tag{M2} \\
    \nabla \times \bf{E}
    &= -\pfrac{\bf{B}}{t} \tag{M3} \\
    \nabla \times \bf{B}
    &= \mu_0 \eps_0 \pfrac{\bf{E}}{t} \tag{M4}
\end{align*}
Eliminate $\bf{B}$ by taking $\pfrac{}{t}$ (M4)
and substituting (M3):
\begin{align*}
    \mu_0 \eps_0 \pfrac{\bf{E}}{t}
    &= \pfrac{}{t} (\nabla \times \bf{B}) \\
    &= \nabla \times \pfrac{\bf{B}}{t} \\
    &= -\nabla \times (\nabla \times \bf{E}) \\
    &= \nabla^2 \bf{E}
\end{align*}
where we use the identity
\[ \nabla \times (\nabla \times \bf{E}) =
\nabla(\nabla \cdot \bf{E}) - \nabla^2 \bf{E} \]
and (M1).

\myskip
Alternatively, eliminate $\bf{E}$ by taking
$\pfrac{}{t}$ (M3) and substituting (M4):
\begin{align*}
    \pfrac[2]{\bf{B}}{t}
    &= -\pfrac{}{t} (\nabla \times \bf{E}) \\
    &= -\nabla \times \pfrac{\bf{E}}{t} \\
    &= -\frac{1}{\mu_0 \eps_0} \nabla \times
    (\nabla \times \bf{B}) \\
    &= \frac{1}{\mu_0 \eps_0} \nabla^2 \bf{B}
\end{align*}
using the same identity and (M2). So each
(Cartesian) component of $\bf{E}$ and $\bf{B}$
satisfies the \emph{wave equation}
\[ \boxed{\pfrac[2]{u}{t} = c^2 \nabla^2 u} \]
with wave speed
\[ \boxed{c = \frac{1}{\sqrt{\mu_0 \eps_0}}} \]
which is the \emph{speed of light} (in a vaccuum),
\[ c = 2.997923458 \times 10^8 \mathsf{ms^{-1}} .\]
This is of course because light is an
\emph{electromagnetic wave} involving oscillations
of $\bf{E}$ and $\bf{B}$. Depending on the
wavelength, EM waves can be radio waves,
microwaves, infrared, ultraviolet, X-rays, gamma
rays, etc.

\subsubsection*{Plane Electromagnetic Waves}

Consider a \emph{plane wave} in which $\bf{E}$ and
$\bf{B}$ depend only on $(x, t)$ and not on $(y,
z)$. A simple example is
\[ \bf{E} = E(x, t) \bf{e}_y \]
where $E(x, t)$ satisfies the 1D wave equation
\[ \pfrac[2]{E}{t} = c^2 \pfrac[2]{E}{t} \]
The general solution
\[ E(x, t) = f(x - ct) + g(x + ct) \]
is the sum of a wave travelling without change of
form in the $+x$ direction and another travelling
in the $-x$ direction. What is the corresponding
$\bf{B}$? We have
\begin{align*}
    \pfrac{\bf{B}}{t}
    &= -\nabla \times \bf{E} \\
    &= -\pfrac{E}{x} \bf{e}_z \\
    &= (-f'(x - ct) - g'(x + ct))\bf{e}_z
\end{align*}
and so
\[ \bf{B} = B(x, t) \bf{e}_z \]
with
\[ B(x, t) = \frac{1}{c} (f(x - ct) - g(x + ct)) \]
This also satisfies $\nabla \times \bf{B} = \mu_0
\eps_0 \pfrac{\bf{E}}{t}$ (exercise).

\myskip
Of particular importance is a \emph{monochromatic
wave} of a single \emph{angular frequency}
$\omega$, for example
\[ E = E_0 \cos(kx - \omega t), \qquad B =
\frac{E_0}{c} \ cos(kx - \omega t) \]
where $E_0$ is a constant amplitude and $h =
\frac{\omega}{c}$ is the \emph{wavenumber},
related to the wavelength $\lambda$ by $k =
\frac{2\pi}{\lambda}$. (The frequency is $\nu =
\frac{\omega}{2\pi}$ and the period is
$\frac{1}{\nu} = \frac{2\pi}{\omega}$.)

\myskip
Notes:
\begin{itemize}
    \item The (angular) frequency and wavenumber
        are related by the \emph{dispersion
        relation}
        \[ \omega^2 = c^2 k^2, \qquad \text{i.e. }
        \omega = \pm ck \]
    \item The oscillations and $\bf{E}$ and
        $\bf{B}$ are in phase but in orthogonal
        directions.
    \item The waves are \emph{transverse}: the
        oscillating fields are orthogonal to the
        direction in which the wave varies (and
        propagates).
\end{itemize}
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/7f66e1f2b1cc11ed.png}
\end{center}
Because Maxwell's equations are linear, EM waves
of different amplitudes, frequencies and
directions can be \emph{superposed}. (Light rays
can pass through each other!)

\subsubsection*{Polarization}

A more general approach to plane EM waves is to
seek solutions of the form
\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*}
where $\bf{E}_0$, $\bf{B}_0$ are constant
(complex) vector amplitudes, $\bf{k}$ is the
(real, constant) \emph{wavevector} and $\omega$ is
the (real, constant) angular frequency. The
wavenumer is $k = |\bf{k}|$.

\myskip
The wave equation is satisfied by $\bf{E}$ and
$\bf{B}$ if $\omega$ and $k$ satisfy the
dispersion relation
\[ \boxed{\omega^2 = c^2 k^2} \]
The individual Maxwell equations reduce to
algebraic conditions:
\begin{align*}
    \nabla \cdot \bf{E}
    &= 0
    &&\implies&
    \bf{k} \cdot \bf{E}_0
    &= 0 \\
    \nabla \cdot \bf{B}
    &= 0
    &&\implies&
    \bf{k} \cdot \bf{B}_0
    &= 0 \\
    \nabla \times \bf{E}
    &= -\pfrac{\bf{B}}{t}
    &&\implies&
    \bf{k} \times \bf{E}_0
    &= \omega \bf{B}_0 \\
    \nabla \times \bf{B}
    &= \mu_0 \eps_0 \pfrac{\bf{E}}{t}
    &&\implies&
    \bf{k} \times \bf{B}_0
    &= -\frac{\omega}{c^2} \bf{E}_0
\end{align*}
The fourth equation is redundant because the first
and third, together with the dispersion relation,
imply
\[ \bf{k} \times \bf{B}_0 = \frac{1}{\omega}
\bf{k} \times (\bf{k} \times \bf{E}_0) =
-\frac{k^2}{\omega^2} \bf{E}_0 =
-\frac{\omega}{c^2} \bf{E}_0 \]
Suppose $\bf{E}_0$ is real. Then $\bf{B}_0$ is
also real and the vectors $\bf{k}$, $\bf{E}_0$ and
$\bf{B}_0$ form an orthogonal triad. So $\bf{E}$
and $\bf{B}$ oscillated in fixed directions, which
are perpendicular to each other and to the
direction of propagation.

\myskip
This is similar to the 1D wave considered
previously and corresponds to a \emph{linearly
polarized wave}.