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\begin{flushright}
\today
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\begin{center}
{\LARGE \bf Notes for Lecture \#1}
\end{center}

\section{Introduction}
\begin{enumerate}
\item Textbook and course structure
\item Motivation
\item Chapters I and 1 and Appendix of Jackson
\begin{enumerate}
\item Units - SI vs Gaussian
\item Laplace and Poisson Equations
\item Green's Theorm
\end{enumerate}
\end{enumerate}
\section{Units - SI vs Gaussian}

Coulomb's law has the form:
\begin{equation}
F = K_C \frac{q_1 q_2}{r_{12}^2}.
\end{equation}
Ampere's law has the form:
\begin{equation}
F = K_A \frac{i_1 i_2}{r_{12}^2} \; d{\bf{s}_1} \times d{\bf{s}_2} \times {\bf{\hat{r}}_{12}},
\end{equation}
where the current and charge are related by $i_1 = dq_1/dt$ for all unit systems. The two constants $K_C$ and $K_A$ are related so that their ratio $K_C/K_A$ has the units of $(m/s)^2$ and it is {\em{experimentally}} known that in both the SI and CGS (Gaussian) unit systems, it the value $K_C/K_A = c^2$, where $c$ is the speed of light.

The choices for these constants in the SI and Gaussian units are given below:


\begin{tabular}{||c|c|c||}
\hline \hline
 & CGS (Gaussian) &SI \\ \hline \hline
&&\\
$K_C$ & 1 & $\frac {1}{4 \pi \epsilon_0}$ \\ 
&&\\
$K_A$ & $\frac{1}{c^2}$ & $\frac {\mu_0}{4 \pi}$ \\
&&\\ \hline \hline 
\end{tabular}

Here, $\frac {\mu_0}{4 \pi} \equiv 10^{-7} N/A^2$ and $\frac {1}{4 \pi \epsilon_0} = c^2 \cdot 10^{-7} N/A^2 = 8.98755\times10^9 N \cdot m^2/C^2$.

Below is a table comparing SI and Gaussian unit systems. The fundamental units for each system are so labeled  and are used to define the derived units.

\begin{tabular}{||c||c|c||c|c||c||} \hline \hline
Variable & \multicolumn{2}{c||}{SI} & \multicolumn{2}{c||}{Gaussian} &  SI/Gaussian \\ \hline
& Unit & Relation & Unit & Relation &\\ \hline
&&&&&\\
length & $m$ & fundamental & $cm$ & fundamental & $100$ \\
&&&&&\\
mass & $kg$ & fundamental & $gm$  & fundamental & $1000$ \\
&&&&&\\
time & $s$  & fundamental & $s$  & fundamental & $1$ \\
&&&&&\\
force & $N$ & $kg \cdot m^2/s$ & $dyne$ & $gm \cdot cm^2/s$ & $10^5$\\
&&&&&\\
current & $A$ & fundamental & $statampere$ & $statcoulomb/s$  & $\frac{1}{10 c}$ \\
&&&&&\\
charge & $C$ & $ A \cdot s$ & $statcoulomb$ & $\sqrt{dyne \cdot cm^2}$ & $\frac{1}{10 c}$ \\ 
&&&&&\\ \hline \hline
\end{tabular}

One advantage of the Gaussian system is that all of the field vectors: ${\bf{E,D,B,H,P,M}}$ have the same dimensions, and in vacuum, ${\bf{B=H}}$ and ${\bf{E=D}}$ and the dielectric and permittivity constants $\epsilon$ and $\mu$ are unitless.

\begin{tabular}{||c|c||}
\hline \hline
CGS (Gaussian) & SI \\ \hline \hline
&\\
$\nabla \cdot {\bf{D}} = 4 \pi \rho$ &$\nabla \cdot {\bf{D}} = \rho$ \\ 
&\\
$\nabla \cdot {\bf{B}} = 0$ & $\nabla \cdot {\bf{B}} = 0$ \\ 
&\\
$\nabla \times {\bf{E}} = - \frac{1}{c} \frac{\partial{\bf{B}}}{\partial t}$ & $\nabla \times {\bf{E}} = -  \frac{\partial{\bf{B}}}{\partial t}$ \\ 
&\\
$\nabla \times {\bf{H}} = \frac{4 \pi}{c} {\bf{J}} + \frac{1}{c} \frac{\partial{\bf{D}}}{\partial t}$ & $\nabla \times {\bf{H}} =  {\bf{J}} +  \frac{\partial{\bf{D}}}{\partial t}$ \\ 
&\\
${\bf{F}}  = q ({\bf{E}} + \frac{{\bf{v}}}{c} \times {\bf{B}}$ & ${\bf{F}}  = q ({\bf{E}} + {\bf{v}} \times {\bf{B}}$\\ 
&\\
$u = \frac{1}{8 \pi} ({\bf{E}}\cdot{\bf{D}}+{\bf{B}}\cdot{\bf{H}})$& $u = \frac{1}{2} ({\bf{E}}\cdot{\bf{D}}+{\bf{B}}\cdot{\bf{H}})$\\ 
&\\
${\bf{S}} = \frac{c}{4 \pi} ({\bf{E}} \times {\bf{H}})$& ${\bf{S}} =  ({\bf{E}} \times {\bf{H}})$ \\ 
&\\ \hline \hline
\end{tabular}



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