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\subsubsection*{Gauss' Law}

Integrate these vector fields over a sphere $S$ of
radius $r$
\[ \int_S \bf{g} \cdot \bf{\dd s} = -4\pi GM \]
and
\[ \int_S \bf{E} \cdot \bf{\dd s} =
\frac{Q}{\eps_0} \]

\begin{note*}
    The result is independent of $r$! The flux of
    the field tells us the total mass / charge
    inside the sphere.
\end{note*}

\noindent
This is Gauss' law (in integrated form). \myskip
In fact, Gauss' law is equivalent to the force
laws. Consider a sphere  of radius $R$ and mass
$M$, with some spherically symmetric mass
distribution.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/cf9821c899c111ec.png}
\end{center}
\[ \text{symmetry } \implies \bf{g}(\bf{x}) = g(f)
\hat{\bf{r}} \]
Consider a surface $S$ which is a sphere of
radius $r > R$.
\begin{align*}
    \int_S \bf{g}(\bf{x}) \cdot \bf{\dd s}
    &= \int_S g(r) \dd s \\
    &= 4\pi r^2 g(r) \\
    &= -4\pi GM
    &&\text{by Gauss} \\
    \implies g(r)
    &= -\frac{GM}{r^2} \hat{\bf{r}}
\end{align*}
which is Newton's law.

\begin{note*}
    Don't need a point mass $M$. It holds for any
    spherically symmetric mass density.
\end{note*}

\noindent
This is known as the \emph{Gauss flux method}. \\
We can also use Gauss' law in other situations.
Consider an infinite wire with charge per unit
length $\sigma$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/645993b499c211ec.png}
\end{center}
By symmetry $\bf{E}(r) = E(r) \hat{\bf{r}}$ ($r$
cylindrical polar $r^2 = x^2 + y^2$)
\[ \int_S \bf{E} \cdot \bf{\dd s} = 2\pi r L E(r)
= \frac{Q}{\eps_0} = \frac{\sigma L}{\eps_0} \]

\begin{note*}
    $\bf{E} \cdot \bf{n} = 0$ on end caps so no
    contribution.
\end{note*}

\[ \implies \bf{E}(r) = \frac{\sigma}{2\pi \eps_0
r} \hat{\bf{r}} \]
is electric field due to wire.

\begin{note*}
    Now $\frac{1}{r}$ instead of $\frac{1}{r^2}$
    as field spreads out in $\RR^2$ instead of
    $\RR^3$.
\end{note*}

\noindent
In $\RR^n$, the electric / gravitational field
would be $\bf{F} \sim \frac{\hat{\bf{r}}}{r^{n -
1}}$. \myskip
There's a different way to write Gauss' law. If
the mass density is $\rho(\bf{x})$ then from
Gauss' theorem
\begin{align*}
    \int_V \nabla \cdot \bf{g} \dd V
    &= \int_S \bf{g} \cdot \bf{\dd s}
    &&\text{by divergence theorem} \\
    &= -4\pi GM
    &&\text{by Gauss' law} \\
    &= -4\pi G \int_V \rho(\bf{x}) \dd V \\
    \implies \int_V (\nabla \cdot \bf{g} + 4\pi G
    \rho(\bf{x})) \dd V
    &= 0
\end{align*}
But this is true for all volumes $V$
\[ \implies \nabla \cdot \bf{g} = -4\pi G
\rho(\bf{x}) .\]
This is Gauss' law in differential form. It is the
more sophisticated version of Newton's force law.
Similarly
\[ \nabla \cdot \bf{E} =
\frac{\rho_e(\bf{x})}{\eps_0} \]
($\rho_e$ is electric charge density). This is the
grown-up version of Coulomb's law.

\subsubsection*{Potentials}

It is a fact that the fields $\bf{g}$ and $\bf{E}$
are conservative, i.e.
\[ \bf{g} = -\nabla \Phi \qquad \text{and} \qquad
Ebf = -\nabla \phi \]
for potentials $\Phi$ and $\phi$. We've seen some
consequences of this. We have
\[ \nabla \times \bf{g} = \nabla \times \bf{E} = 0 \]
and
\[ \oint_C \bf{g} \cdot \bf{\dd x} = \oint_C
\bf{E} \cdot \bf{\dd x} = 0 \]
and most importantly, it means that there is a
conserved energy
\[ \text{Energy } = \half m \dot{\bf{x}}^2 +
m\Phi(\bf{x}) + q\phi(\bf{x}) \]
Gauss' law then becomes
\[ \nabla^2 \Phi = 4\pi G \rho(x) \]
and
\[ \nabla^2 \phi = -\frac{\rho_e(\bf{x})}{\eps_0} \]
This is the \emph{Poisson equation}. The goal is
to solve for $\Phi(\bf{x})$ for some fixed
``source'' $\rho(\bf{x})$. \\
If $\rho = 0$, then we solve
\[ \nabla^2 \Phi = 0 .\]
This is the \emph{Laplace equation}.

\subsection{The Laplace and Poisson Equations}

We write
\[ \nabla^2 \psi(\bf{x}) = -\rho(\bf{x}) \]
Solve for $\psi(\bf{x})$ given a \emph{source}
$\rho(\bf{x})$ \emph{and} suitable boundary
conditions.

\begin{note*}
    $\nabla^2 \psi = 0$ is linear, so if $\psi_1$
    and $\psi_2$ are solutions, then so is $\psi_1
    + \psi_2$.
\end{note*}

\noindent
Solutions to the Laplace equation act as
complementary solutions to Poisson.

\subsubsection*{Isotropic Solutions}

$\nabla^2 \psi = 0$ is a PDE. But with suitable
symmetry, it becomes an ODE. For example,
spherical symmetry $\implies \psi = \psi(r)$ ($r^2
= x^2 + y^2 + z^2$).
\begin{align*}
    \nabla^2 \psi
    &= \dfrac[2]{\psi}{r} + \frac{2}{r} \\
    &= \frac{1}{r^2} \dfrac{}{r} \left( r^2
    \dfrac{\psi}{r} \right) \\
    &= 0 \\
    \implies \psi
    &= \frac{A}{r} + B
\end{align*}
($A$, $B$ constants).
