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\begin{document} \beamertemplatenavigationsymbolsempty

\begin{frame}
\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}
\newline{\large\bf A Bit on Maxwell's Equations}

\parbox{1.9in}{There's also a handout at \url{http://drorbn.net/2021-257/ap/Maxwell.pdf}}\hfill\adjustbox{valign=m}{\includegraphics[width=2.4in]{Maxwell.pdf}}
\end{frame}

\begin{frame}
\includegraphics[width=4.2in]{../../2011-11/ClassicalPhysicsAccordingToFeynman.png}
\newline{\small The Feynman Lectures on Physics vol.\ II, page 18-2}
\end{frame}

\begin{frame}{\bf Prerequisites.}
\begin{itemize}
  \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{itemize}
\end{frame}

\begin{frame}{\bf Prerequisite 1.}
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 $d\eta=\omega$.
\end{frame}

\begin{frame}{\bf Prerequisite 2.}
Integration by parts: $\int_{\bbR^n}\omega\wedge d\eta = -(-1)^{\deg\omega}\int_{\bbR^n}(d\omega)\wedge\eta$ for compactly supported forms.
\end{frame}

\begin{frame}{\bf Prerequisite 3.}
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.
\end{frame}

\begin{frame}{\bf Prerequisite 4.}
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{frame}

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

\begin{frame}{\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
\[ \fbox{$\ds dJ=0 \qquad dF=0 \qquad d\star F=J$} \]
\end{frame}

\begin{frame}{\bf These are the Maxwell equations!}
\[ \fbox{$\ds dJ=0 \qquad dF=0 \qquad d\star F=J$} \]
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}
\end{frame}

\begin{frame}{\bf $dJ=0$}
\[ \fbox{$\ds dJ=0 \qquad dF=0 \qquad d\star F=J$} \]
with $J=\rho dxdydz-j_xdydzdt-j_ydzdxdt-j_zdxdydt$.

\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'' \\
  \hline
  \end{tabular}
\end{center}
\end{frame}

\begin{frame}{\bf $dF=0$}
\[ \fbox{$\ds dJ=0 \qquad dF=0 \qquad d\star F=J$} \]
with $F=(E_xdxdt+E_ydydt+E_zdzdt)+(B_xdydz+B_ydzdx+B_zdxdy)$ and
$J=\rho dxdydz-j_xdydzdt-j_ydzdxdt-j_zdxdydt$.

\vskip 3mm
\begin{center} \renewcommand{\arraystretch}{2}
  \begin{tabular}{|rcl|}
  \hline
    $\ds dF=0 \Longrightarrow$ &
    $\ds \opdiv B=0$ &
    \qquad ``no magnetic monopoles'' \\
    &
    $\ds \opcurl E = -\frac{\partial B}{\partial t}$ &
    \qquad that's how generators work! \\
  \hline
  \end{tabular}
\end{center}
\end{frame}

\begin{frame}{\bf $d*F=J $}
\[ \fbox{$\ds dJ=0 \qquad dF=0 \qquad d\star F=J$} \]
with $F=(E_xdxdt+E_ydydt+E_zdzdt)+(B_xdydz+B_ydzdx+B_zdxdy)$ and
$J=\rho dxdydz-j_xdydzdt-j_ydzdxdt-j_zdxdydt$.

With $\omega\wedge*\omega=|\omega|^2dtdxdydz$ we have
\[ *dxdt=-dydz, \qquad *dydt=-dzdx, \qquad *dzdt=-dxdy, \]
\[ *dydz=-dxdt, \qquad *dzdx=-dydt, \qquad *dxdy=-dxdt, \]
so $*F=(-B_xdxdt-B_ydydt-B_zdzdt)+(-E_xdydz-E_ydzdx-E_zdxdy)$.

\vskip 3mm
\begin{center} \renewcommand{\arraystretch}{2}
  \begin{tabular}{|rcl|}
  \hline
    $\ds d*F=J \Longrightarrow$ &
    $\ds \opdiv E = -\rho$ &
    \qquad ``electrostatics'' \\
    &
    $\ds \opcurl B = j - \frac{\partial E}{\partial t}$ &
    \qquad that's how electromagnets work! \\
  \hline
  \end{tabular}
\end{center}
\end{frame}

\begin{frame}
\includegraphics[width=4.2in]{../../2011-11/ClassicalPhysicsAccordingToFeynman.png}
\newline{\small Feynman again. But wait, in our last two equations the sign of $E$ is wrong!}
\end{frame}

\begin{frame}
  {\bf Exercise 1.} Use the Lorentz metric to fix the sign errors.
\end{frame}

\begin{frame}{\bf Exercise 2.}
  Use pullbacks along Lorentz transformations to figure out how $E$ and $B$ (and $j$ and $\rho$) appear to moving observers.
\end{frame}

\begin{frame}{\bf Exercise 3.}
  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''.
\end{frame}

\end{document}