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%TCIDATA{<META NAME="Title" CONTENT="Reference/Physics/Electromagnetic Waves in a Vacuum">}
%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
%TCIDATA{LastRevised=Thu Feb 13 03:40:20 1997}
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\section{Electromagnetic Waves in a Vacuum\label%
{Electromagnetic waves in a vacuum}}

The wave equation $\Box \Psi (\vec{r},t)=-f(\vec{r},t)$ has the general
solution, with $c=\left( \varepsilon _{0}\mu _{0}\right) ^{-1/2}$: 
\[
\Psi (\vec{r},t)=\int \frac{f\left( \vec{r},t-\left| \vec{r}-\vec{r}%
~^{\prime }\right| /c\right) }{4\pi \left| \vec{r}-\vec{r}~^{\prime }\right| 
}d^{3}r^{\prime } 
\]
If $\vec{J}\left( \vec{r},t\right) =\vec{J}\left( \vec{r}\right) \exp \left(
-i\omega t\right) $ and $\vec{A}\left( \vec{r},t\right) =\vec{A}(\vec{r}%
)\exp \left( -i\omega t\right) $ with: 
\begin{eqnarray*}
\vec{A}\left( \overrightarrow{r}\right) &=&\frac{\mu }{4\pi }\int \vec{J}%
\left( \vec{r}~^{\prime }\right) \frac{\exp \left( ik\left| \vec{r}-\vec{r}%
~^{\prime }\right| \right) }{\left| \vec{r}-\vec{r}~^{\prime }\right| }d^{3}%
\vec{r}~^{\prime } \\
V\left( \overrightarrow{r}\right) &=&\frac{1}{4\pi \varepsilon }\int \rho
\left( \vec{r}~^{\prime }\right) \frac{\exp \left( ik\left| \vec{r}-\vec{r}%
~^{\prime }\right| \right) }{\left| \vec{r}-\vec{r}~^{\prime }\right| }d^{3}%
\vec{r}~^{\prime }
\end{eqnarray*}
a derivation via multipole expansion will show that for the radiated energy,
if $d,\lambda \gg r$: 
\[
\frac{dP}{d\Omega }=\frac{k^{2}}{32\pi ^{2}\varepsilon _{0}c}\left| \int
J_{\perp }\left( \vec{r}~^{\prime }\right) e^{i\vec{k}\cdot \vec{r}%
}d^{3}r^{\prime }\right| ^{2} 
\]
The energy density of the electromagnetic wave of a vibrating dipole at a
large distance is: 
\begin{eqnarray*}
w &=&\varepsilon _{0}E^{2}=\frac{p_{0}^{2}\omega ^{4}\sin ^{2}\theta }{16\pi
^{2}\varepsilon _{0}r^{2}c^{4}}\sin ^{2}\left( kr-\omega t\right) \\
\left\langle w\right\rangle _{t} &=&\frac{p_{0}^{2}\omega ^{4}\sin
^{2}\theta }{32\pi ^{2}\varepsilon _{0}r^{2}c^{4}} \\
P &=&\frac{ck^{4}\left| \vec{p}\right| ^{2}}{12\pi \varepsilon _{0}}
\end{eqnarray*}
\[
~ 
\]
The radiated energy can be derived from the \label{Poynting vector}\textit{%
Poynting vector} $\vec{S}$: $\vec{S}=\vec{E}\times \vec{H}=cW\vec{e}_{v}$.
The\label{Irradiance} \textit{irradiance} is the time-average of the
Poynting vector: $I=\left\langle |\vec{S}|\right\rangle _{t}$. The radiation
pressure $p_{\mathrm{s}}$ is given by 
\[
p_{\mathrm{s}}=\frac{(1+R)|\vec{S}|}{c} 
\]
where $R$ is the coefficient of reflection.

\subsection{\ Related Reference Topics}

{\small \hyperref{Calculus}{}{}{../Mathematics/CAL_contents.tex}}

{\small \hyperref{Differential Equations}{}{}{../Mathematics/ODE_contents.tex%
}}

{\small \hyperref{Matrix Algebra}{}{}{../Mathematics/MA_contents.tex}}

{\small \hyperref{Table of Integrals}{}{}{%
../Mathematics/TAB_integrals.tex#table of integrals}}

{\small \hyperref{Table of Derivatives}{}{}{%
../Mathematics/TAB_derivatives.tex}}

{\small \hyperref{Table of Physical Constants}{}{}{PhysicalConstants.tex}}

{\small \hyperref{Table of SI Units}{}{}{SIUnits.tex}}

\subsection{\ Related Scientific Notebook Topics}

{\small \hyperref{Numbers, Functions, and Units}{}{}{../../Help/DM2.tex}}

{\small \hyperref{Calculus}{}{}{../../Help/DM7.tex}}

{\small \hyperref{Matrix Algebra}{}{}{../../Help/DM8.tex}}

{\small \hyperref{Vector Calculus}{}{}{../../Help/DM9.tex}}

\noindent \hrulefill

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