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\begin{document}

\noindent
\href{http://www.math.toronto.edu/~drorbn}{Dror Bar-Natan}:
\href{http://www.math.toronto.edu/~drorbn/classes}{Classes}:
\href{http://www.math.toronto.edu/~drorbn/classes/\#2021}{2020-21}:
\href{http://drorbn.net/2021-257}{2021-257 Analysis II}:
\hfill \qrcode[height=3em,level=L,nolink]{drorbn.net/2021-257}

\noindent\parbox{2.4in}{
  {\large\bf A Bit on Maxwell's Equations}
  \vskip 5mm
  \par\noindent{\bf Prerequisites.}
  \begin{myitemize}
  \item Poincar\'e's Lemma, which says that on $\bbR^n$, every closed form is
    exact. That is, if $d\omega=0$, then there exists $\eta$ with
    $d\eta=\omega$.
  \item Integration by parts: $\int_{\bbR^n}\omega\wedge d\eta =
    -(-1)^{\deg\omega}\int_{\bbR^n}(d\omega)\wedge\eta$ for compactly supported forms.
  \item The Hodge star operator $\star$ which satisfies
    $\omega\wedge\star\eta=\langle\omega,\eta\rangle dx_1\cdots dx_n$ whenever $\omega$ and $\eta$
    are of the same degree.
  \item The simplesest least action principle: the extremes of $q \mapsto S(q) = \int_a^b\left(\frac12m\dot{q}^2(t)-V(q(t))\right)dt$ occur when $m\ddot{q}=-V'(q(t))$. That is, when $F=ma$.
  \end{myitemize}
}\hfill
\parbox{5.0in}{\centering
  $\includegraphics[width=\linewidth]{../../2011-11/ClassicalPhysicsAccordingToFeynman.png}$\newline
  {\small The Feynman Lectures on Physics vol.\ II, page 18-2}
}

\vskip 3mm \par\noindent{\bf The Action Principle. }
The {\em 4-Vector Potential} is a compactly supported 1-form $A$ on $\bbR^4$ which extremizes the {\em action}
\[ \fbox{$\ds
  S_J(A) := \int_{\bbR^4}\frac12\left\lVert dA\right\rVert^2dtdxdydz + J\wedge A
$} \]
where the 3-form $J$ is the {\em charge-current}.

\vskip 3mm \par\noindent{\bf The Euler-Lagrange Equations} in this case
are $d\star dA=J$, meaning that there's no hope for a solution unless
$dJ=0$, and that we might as well (think Poincar\'e's Lemma!) change
variables to $F:=dA$.  We thus get
\[ \fbox{$\ds
  dJ=0
  \qquad dF=0
  \qquad d\star F=J
$} \]

\vskip 3mm \par\noindent{\bf These are the Maxwell equations! } Indeed,
writing $F=(E_xdxdt+E_ydydt+E_zdzdt)+(B_xdydz+B_ydzdx+B_zdxdy)$ and
$J=\rho dxdydz-j_xdydzdt-j_ydzdxdt-j_zdxdydt$, we find:

\vskip 3mm
\begin{center} \renewcommand{\arraystretch}{2}
  \begin{tabular}{|rcl|}
  \hline
    $\ds dJ=0 \Longrightarrow$ &
    $\ds \opdiv j = -\frac{\partial\rho}{\partial t}$ &
    \qquad ``conservation of charge''
  \\
    $\ds dF=0 \Longrightarrow$ &
    $\ds \opdiv B=0$ &
    \qquad ``no magnetic monopoles''
  \\
    &
    $\ds \opcurl E = -\frac{\partial B}{\partial t}$ &
    \qquad that's how generators work!
  \\
    $\ds d*F=J \Longrightarrow$ &
    $\ds \opdiv E = -\rho$ &
    \qquad ``electrostatics''
  \\
    &
    $\ds \opcurl B = j - \frac{\partial E}{\partial t}$ &
    \qquad that's how electromagnets work!
  \\
  \hline
  \end{tabular}
\end{center}

\vskip 3mm \par\noindent{\bf Exercise.} Use the Lorentz metric to fix the sign errors.

\vskip 1mm \par\noindent{\bf Exercise.} Use pullbacks along Lorentz transformations to figure out
  how $E$ and $B$ (and $j$ and $\rho$) appear to moving observers.

\vskip 1mm \par\noindent{\bf Exercise.} With $ds^2=c^2dt^2-dx^2-dy^2-dz^2$ use $S=mc\int_{e_1}^{e_2}(ds+eA)$ to derive Feynman's ``law of motion'' and ``force law''.

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