% vim: tw=50
% 22/02/2023 11AM

\noindent
Recall from last time the quadratic equations:
\[ (*) \begin{cases}
    \dfrac{}{t} (E\dot{u} + F \dot{v}) = \half
    (E_u \dot{u}^2 + 2F_u \dot{u} \dot{v} + G_u
    \dot{v}^2) \\
    \dfrac{}{t} (F\dot{u} + G \dot{v}) = \half
    (E_v \dot{u}^2 + 2F_v \dot{u} \dot{v} + G_v
    \dot{v}^2)
\end{cases} \]
Probably the best way to think about these is via
the \emph{Euler-Lagrange} (E-L) equations of the
Lagrangian:
\[ L(u, v, \dot{u}, \dot{v}) = \half (E\dot{u}^2 +
2F \dot{u} \dot{v} + G \dot{v}^2) \]
(purely kinetic energy). Recall from Variational
Principles that the (E-L) equations are:
\[ \dfrac{}{t} \pfrac{L}{\dot{q}_i} =
\pfrac{L}{\dot{q}_i} \]
and here $q_1 = u$, $q_2 = v$. (\dag) = ($*$).

\begin{flashcard}[geodesic-in-R3]
\begin{proposition}
    Let $\Sigma \subset \RR^3$ a smooth surface. A
    smooth curve $\gamma : [a, b] \to \Sigma$ is a
    geodesic if and only if \cloze{$\ddot{\gamma}(t)$ is
    \emph{everywhere} normal to $\Sigma$.}
\end{proposition}
\end{flashcard}

\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/04d64940b2a411ed.png}
\end{center}

\begin{proof}
    The statement is purely local so we can work
    in a parametrisation $\sigma : V \to U \subset
    \Sigma$ and as usual, $\gamma(t) =
    \sigma(u(t), v(t))$, $\dot{\gamma}(t) =
    \sigma_u \dot{u} + \sigma_v \dot{v}$. So
    $\ddot{\gamma}(t)$ is normal to $\Sigma$
    exactly when it is orthogonal to
    $T_{\gamma(t)} \Sigma$ spanned by $\{\sigma_u,
    \sigma_v\}$. In other words:
    \begin{align*}
        \left\langle \dfrac{}{t} (\sigma_u
        \dot{u} + \sigma_v \dot{v}), \sigma_u
        \right\rangle
        &= 0 \tag{$*_1$} \\
        \left\langle \dfrac{}{t} (\sigma_u \dot{u}
        + \sigma_v \dot{v}), \sigma_v \right\rangle
        &= 0 \tag{$*_2$}
    \end{align*}
    ($*_1$) is equivalent to
    \[ \dfrac{}{t} \langle \sigma_u \dot{u} +
    \sigma_v \dot{v}, \sigma_u \rangle - \left\langle
    \sigma_u \dot{u} + \sigma_v \dot{v},
    \dfrac{}{t} \sigma_u \right\rangle = 0 \]
    Noting that $E = \langle \sigma_u, \sigma_u
    \rangle$ and $F = \langle \sigma_u, \sigma_v
    \rangle$, this is:
    \[ \dfrac{}{t} (E\dot{u} + F\dot{v}) - \langle
    \sigma_u \dot{u} + \sigma_v \dot{v},
    \sigma_{uu} \dot{u} + \sigma_{vv} \dot{v}
    \rangle = 0 \]
    This second term is:
    \[ \dot{u}^2 \langle \sigma_u, \sigma_{uu}
    \rangle + \dot{u} \dot{v} (\langle \sigma_u,
    \sigma_{uv} \rangle + \langle \sigma_v,
    \sigma_{uu} \rangle) + \dot{v}^2 \langle
    \sigma_u, \sigma_{uv} \rangle \]
    But
    \begin{align*}
        E
        &= \langle \sigma_u, \sigma_u \rangle
        &&\implies&
        E_u
        &= 2\langle \sigma_u, \sigma_{uu} \rangle
        \\
        F
        &= \langle \sigma_u, \sigma_v \rangle
        &&\implies&
        F_u
        &= \langle \sigma_u, \sigma_{uv} \rangle +
        \langle \sigma_v, \sigma_{uu} \rangle
        \\
        G
        &= \langle \sigma_v, \sigma_v \rangle
        &&\implies&
        G_u
        &= 2\langle \sigma_v, \sigma_{uv} \rangle
    \end{align*}
    Thus ($*_1$) becomes:
    \[ \dfrac{}{t} (E\dot{u} + F\dot{v}) = \half
    (E_u \dot{u}^2 + 2F_u \dot{u} \dot{v} + G_u
    \dot{v}^2) \]
    which is the first geodesic equation.
    Similarly for ($*_2$).
\end{proof}

\begin{flashcard}[constant-speed-geodesic]
\begin{corollary}
    If $\gamma : [a, b] \to \Sigma$ is a geodesic,
    then $|\dot{\gamma}(t)|$ is constant.
\end{corollary}

\begin{proof}
\cloze{
    \[ \dfrac{}{t} \langle \dot{\gamma},
    \dot{\gamma} \rangle = 2\langle \ddot{\gamma},
    \dot{\gamma} \rangle = 0 \]
    Since $\ddot{\gamma} \perp
    T_{\gamma(t)}\Sigma$ and $\dot{\gamma}(t) \in
    T_{\gamma(t)} \Sigma$.
}
\end{proof}
\end{flashcard}

\myskip
Thus geodesics are parametrised with
\emph{constant speed} (i.e. proporitonal to
arc-length).

\subsubsection*{Length vs Energy}

Energy \emph{is} sensitive to
\emph{parametrisations}. Given $\gamma : [a, b]
\to \RR^3$, we always have:
\[ (L(\gamma))^2 \le (b - a)E(\gamma) \]
with \emph{equality} if and only if
$|\dot{\gamma}|$ is constant. Indeed, using
Cauchy-Schwarz:
\[ \left( \int_a^b \ub{|\dot{\gamma}(t)|}_{=
|\dot{\gamma}(t) \cdot 1} \dd t
\right)^2 \le \ub{\left( \int_a^b |\dot{\gamma}(t)|^2
\dd t \right)}_{=E(\gamma)} \ub{\left( \int_a^b 1
\dd t \right)}_{= (b - a)} \]
with equality if and only if $|\dot{\gamma}(t)|$
is constant.

\begin{corollary}
    A smooth curve $\gamma : [a, b] \to \Sigma
    \subset \RR^3$ that minimises length and has
    constant speed is a \emph{geodesic}.
\end{corollary}

\begin{hiddenflashcard}[geodesic-suff-condition]
Sufficient condition for $\gamma$ to be a
geodesic? \\
\cloze{$\gamma$ minimises length and has constant
speed. (Proof is by using Cauchy-Schwarz).}
\end{hiddenflashcard}

\begin{proof}
    Need to prove $\gamma$ is a critical point of
    $E$. $\tau : [a, b] \to \Sigma$ any other
    curve connecting $\gamma(a)$ to $\gamma(b)$.
    \begin{align*}
        E(\gamma)
        &= \frac{(L(\gamma))^2}{b - a}
        &&\text{$|\dot{\gamma}|$ constant} \\
        &\le \frac{(L(\tau))^2}{b - a}
        &&\text{$\gamma$ minimises length} \\
        &\le E(\tau)
    \end{align*}
    Hence $\gamma$ is critical for $E$ and hence a
    geodesic.
\end{proof}

\noindent
But geodesics might not be global minimisers
(examples coming up) but they are always
\emph{local} minimisers. (No proof in this course,
see Wilson's book).

\subsubsection*{Examples}

\begin{enumerate}[(1)]
    \item The plane $\RR^2$ has $\sigma(u, v) =
        (u, v, 0)$ and FFF $\dd u^2 + \dd v^2$.
        Geodesic equations are
        \[ \dfrac{}{t}(\dot{u}) = 0, \qquad
        \dfrac{}{t} (\dot{v}) = 0 \]
        So $u(t) = \alpha t + \beta$, $v(t) = \gamma
        t + \delta$, which is a straight line
        parametrised by constant speed.
    \item Take the unit sphere with $\sigma$ given
        by spherical coordinates:
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/f143b4fab2a611ed.png}
        \end{center}
        \begin{align*}
            \sigma(\varphi, \theta)
            &= (\sin\theta \cos\varphi, \sin
            \theta \sin\varphi, \cos\theta) \\
            \sigma_\varphi
            &= (-\sin\theta
            \sin\varphi, \sin\theta \cos\varphi,
            0) \\
            \sigma_\theta
            &= (\cos\theta \cos\varphi,
            -\cos\theta \sin\varphi, -\sin\theta)
            \\
            \implies&E = \sin^2\theta, F = 0, G =
            1
        \end{align*}
        $L = \half(\sin^2 \theta \dot{\varphi}^2 +
        \dot{\theta}^2)$, $L(\theta, \varphi,
        \dot{\theta}, \dot{\varphi})$.
        Euler-Lagrange:
        \[ \pfrac{L}{\dot{\theta}} = \dot{\theta},
        \qquad \pfrac{L}{\dot{\varphi}} =
        \sin^2\theta \dot{\varphi} \]
        $\pfrac{L}{\varphi} = 0$.
        \[ (*) \begin{cases}
            \dfrac{}{t} (\sin^2 \theta
            \dot{\varphi}) = 0 \\
            \ddot{\theta} = \sin\theta \cos\theta
            \dot{\varphi}^2
        \end{cases} \]
        ($*$) gives right away that the equator $t
        \mapsto \left( t, \frac{\pi}{2} \right)$
        is a geodesic.

        \myskip
        In fact all great circles parametrised
        with constant speed are geodesics. We can
        prove this by integrating ($*$), but we
        can see this geometrically by notiving
        that such curves have $\ddot{\gamma} \perp
        T_{\gamma(t)} S^2$. Since geodesics solve
        a second order ODE (ordinary differential
        equation) prescribing $v \in T_p\Sigma$
        determines the geodesic completely (proof
        next time). Thus great circles are
        \emph{all possible} geodesics.

        \myskip
        Note that $\gamma$ between $p$ and $q$
        as in the picture does \emph{not} minimise
        length.
        \begin{center}
            \includegraphics[width=0.3\linewidth]
            {images/2d363306b2a811ed.png}
        \end{center}
\end{enumerate}
