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Maxwell's Equations in Vacuum

We assume a working knowledge of Maxwell's equations. Their form in vacuum will be sufficient for our purposes:


\begin{displaymath}
\vec{\nabla}\cdot\vec{E} = 4\pi k_1 \rho ,
\end{displaymath} (2.2)

where


\begin{displaymath}
k_1 = \left\{
\begin{array}{cl}
\frac{1}{4\pi\varepsilon_0} ...
...\pi} & \textrm{Heaviside-Lorentz system.}
\end{array} \right.
\end{displaymath} (2.3)


\begin{displaymath}
\vec{\nabla}\cdot\vec{B} = 0 .
\end{displaymath} (2.4)


\begin{displaymath}
\vec{\nabla}\times\vec{E} = -\frac{k_2}{c}
\frac{\partial\vec{B}}{\partial t} ,
\end{displaymath} (2.5)

where


\begin{displaymath}
k_2 = \left\{
\begin{array}{cl}
c & \textrm{MKSA system,} \\...
...\\
c & \textrm{Heaviside-Lorentz system.}
\end{array} \right.
\end{displaymath} (2.6)


\begin{displaymath}
k_2\vec{\nabla}\times\vec{B} = \frac{4\pi k_1}{c}\vec{j} +
\frac{1}{c}\frac{\partial\vec{E}}{\partial t} .
\end{displaymath} (2.7)

For Maxwell's equations, the Heaviside-Lorentz system of units is the simplest to work with, since all factors of $4\pi$ vanish.



Douglas M. Gingrich (gingrich@ ualberta.ca)
2004-03-18