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\begin{note*}
    Here are some useful tips when evaluating
    derivatives of radial fields. We use
    \[ r^2 = x^i x^i \implies 2r \pfrac{r}{x^i} =
    2x^i \]
    \[ \implies \pfrac{r}{x^i} = \frac{x^i}{r} \]
    Then we have
    \begin{align*}
        \nabla r^p
        &= \bf{e}_i \pfrac{r^p}{x^i} \\
        &= pr^{p - 1} \frac{x_i}{r} \bf{e}_i \\
        &= pr^{p - 1} \hat{\bf{r}}
    \end{align*}
    The vector $\bf{x} = x^i \bf{e}_i$ can also be
    written as $\bf{r} = r \hat{\bf{r}}$ to
    highlight that it points outwards. We have
    \[ \nabla \cdot \bf{r} = \pfrac{x^i}{x^i} =
    \delta_ii = n \]
    (in $\RR^n$). \\
    Also in $\RR^3$,
    \[ \nabla \times \bf{r} = \eps_{ijk}
    \pfrac{x^j}{x^i} \bf{e}_k = 0 \]
\end{note*}

\subsubsection*{Some Basic Properties}

For constant $\alpha$, scalar fields $\phi$ and
$\psi$, and vector fields $\bf{F}$ and $\bf{G}$,
we have
\begin{align*}
    \nabla(\alpha \phi + \psi)
    &= \alpha \nabla \phi + \nabla \psi \\
    \nabla \cdot (\alpha \bf{F} + Gbg)
    &= \alpha \nabla \cdot \bf{F} + \nabla \cdot
    \bf{G} \\
    \nabla \times (\alpha \bf{F} + \bf{G})
    &= \alpha \nabla \times \bf{F} + \nabla \times
    \bf{G}
\end{align*}
This is the statement that $\nabla$ is a
\emph{linear operator}. \myskip
$\nabla$ obeys a generalised product rule (known
as Leibniz property):
\begin{align*}
    \nabla(\phi\psi)
    &= \phi \nabla \psi + \psi \nabla \phi \\
    \nabla \cdot (\phi \bf{F})
    &= (\nabla \phi) \cdot \bf{F} +
    \phi (\nabla \cdot \bf{F}) \\
    \nabla \times (\phi \bf{F})
    &= (\nabla \phi) \times \bf{F} + \phi( \nabla
    \times \bf{F})
\end{align*}
The proofs of these follow from the definitions.
For example
\begin{align*}
    \nabla \cdot (\phi \bf{F})
    &= \pfrac{}{x^i} (\phi F_i) \\
    &= \pfrac{\phi}{x^i} F_i + \phi
    \pfrac{F_i}{x^i} \\
    &= \nabla \phi \cdot \bf{F} + \phi \nabla
    \cdot \bf{F}
\end{align*}
There are some further properties
\begin{align*}
    \nabla \cdot (\bf{F} \times \bf{G})
    &= (\nabla \times \bf{F}) \cdot \bf{G} - \bf{F}
    \cdot (\nabla \times \bf{G}) \\
    \nabla (\bf{F} \cdot \bf{G})
    &= \bf{F} \times (\nabla \times \bf{G}) +
    \bf{G} \times (\nabla \times \bf{F}) + (\bf{F}
    \cdot \nabla) \bf{G} + (\bf{G} \cdot \nabla)
    \bf{F} \\
    \nabla \times (\bf{F} \times \bf{G})
    &= (\nabla \cdot \bf{G})\bf{F} - (\nabla \cdot
    \bf{F})\bf{G} + (\bf{G} \cdot \nabla)\bf{F} -
    (\bf{F} \cdot \nabla)\bf{G}
\end{align*}
All of these are proven using index notation. IN
the last two identities, we have introduced the
notation
\[ \bf{F} \cdot \nabla = F_i \pfrac{}{x^i} \]

\subsubsection*{Definitions}

\begin{itemize}
    \item A vector field $\bf{F}$ is
        \emph{conservative} if it can be written
        \[ \bf{F} = \nabla \phi \]
        for some scalar $\phi$.
    \item A vector field is called
        \emph{irrotational} if
        \[ \nabla \times \bf{F} = 0 \]
    \item A vector field is \emph{divergence free}
        or \emph{solenoidal} if
        \[ \nabla \cdot \bf{F} = 0 \]
\end{itemize}

\begin{theorem*}[A baby version of the Poincar\'e lemma]
    For fields defined everywhere on $\RR^3$,
    conservative $\iff$ irrotational, i.e.
    \[ \nabla \times \bf{F} = 0 \iff \bf{F} =
    \nabla \phi \]
\end{theorem*}

\begin{proof}[(sketch)]
    If $F_i = \pfrac{\phi}{x^i}$ then
    \[ (\nabla \times \bf{F})_k = \eps_{ijk
    } \frac{\partial^2 \phi}{\partial x^i \partial
    x^j} = 0 \]
    by symmetry. We will show $\nabla \times
    \bf{F} = 0 \implies \bf{F} = \nabla \phi$ when
    we prove Stokes' theorem in section 4.
\end{proof}

\begin{theorem*}
    For $\bf{F}$ defined everywhere on $\RR^3$,
    \[ \nabla \cdot \bf{F} = 0 \iff \bf{F} =
    \nabla \times \bf{A} \]
    for some vector field $\bf{A}$.
\end{theorem*}

\begin{proof}[(sketch)]
    If $F_i = \eps_{ijk} \partial_j A_k$
    \[ \implies \nabla \cdot \bf{F} = \partial_i
    (\eps_{ijk} \partial_j A_k) = 0 \]
    by symmetry. The other way is an optional
    question on Example Sheet 2.
\end{proof}

\begin{definition*}
    The \emph{Laplacian} is a second order
    differential operator
    \[ \nabla^2 = \nabla \cdot \nabla =
    \pfrac{\partial^2}{\partial x^i \partial x^i} \]
    for example in $\RR^3$, we have
    \[ \nabla^2 = \pfrac[2]{}{x} + \pfrac[2]{}{y}
    + \pfrac[2]{}{z} \]
    Acting on a scalar field $\phi$, $\nabla^2$
    gives back another scalar field $\nabla^2
    \phi$. \\
    It acts component-wise on a vector field
    $\bf{F}$ to give another vector field
    $\nabla^2 \bf{F}$.
\end{definition*}

\begin{claim*}
    $\nabla^2 \bf{F} = \nabla(\nabla \cdot \bf{F})
    - \nabla \times (\nabla \times \bf{F})$
\end{claim*}

\begin{proof}
    Use triple product formula for $\nabla \times
    (\nabla \times \bf{F})$.
\end{proof}

\myskip
The Laplacian occurs in many places in maths and
physics. For example, the \emph{heat equation}
\[ \pfrac{T}{t} = D \nabla^2 T \]
and tells us how temperature $T(\bf{x}, t)$ evolve
in time. ($D$ is called the diffusion constant).
\myskip
The linear operator $\nabla$ also appears in many
laws of physics. For example, the electric field
$\bf{E}(\bf{x}, t)$ and magnetic field
$\bf{B}(\bf{x}, t)$ are governed by the
\emph{Maxwell equations}
\begin{align*}
    \nabla \cdot \bf{E}
    &= \frac{\rho}{\eps_0} \\
    \nabla \cdot \bf{B}
    &= 0 \\
    \nabla \times \bf{E}
    &= -\pfrac{\bf{B}}{t} \\
    \nabla \times \bf{B}
    &= \mu_0 \left( \bf{J} + \eps_0
    \pfrac{\bf{E}}{t} \right)
\end{align*}
where $\rho(\bf{x}, t)$ is the electric charge
density and$\bf{J}(\bf{x}, t)$ is the electric
current density, and $\mu_0$ and $\eps_0$ are
constants of nature.