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\myskip
Things change when Luke turns around. Suppose
instead that Luke meets his friend Han who's just
leaving $P$ towards Leia. On this journey, Han
thinks Leia ages by $\frac{T}{\gamma^2}$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/ac4c18a4c59111ec.png}
\end{center}
But he thinks Leia is at point $Z$ when he leaves
\[ \left( t = 2T - \frac{T}{\gamma^2} \right) \]
The asymmetry arises (between Luke and Leia) when
Luke turns around. At this point, he sees Leia age
rapidly from $X \to Z$.

\subsubsection*{Length Contraction}

A rod of length $L'$ is stationary in frame $S'$.
\myskip
Question: What is its length in frame $S$? \\
In $S'$
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/357f7ca6c59211ec.png}
\end{center}
The length of the rod is the distance between the
two end points \emph{at equal times}. In $S$
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/57ccc78cc59211ec.png}
\end{center}
(Lorentz transforms $x' = f(x, t)$, $t' = g(x,
t)$, Inverse transformations $x = f'(x', t')$, $t
= g'(x', t')$). $P_2$ is at $t = \frac{\gamma v
L'}{c^2}$ and $x = \gamma L'$. Need to measure $L$
at equal time in $S$. Follow $P_2$ back to $Q$,
$Q$ sits at
\[ x = \gamma L' - vt = \gamma L' - \frac{\gamma
v^2 L'}{c^2} = \frac{L'}{\gamma} \]
The length of the rod in $S$ is $L =
\frac{L'}{\gamma}$, i.e. moving rods are shorter.

\begin{example*}
    Consider a barn of length $L$, ladder of
    length $2L$. Question: does it fit?
    \begin{itemize}
        \item From perspective of barn, ladder
            length is $\frac{2L}{\gamma}$, so if
            $\gamma \ge 2$, ladder firs.
        \item From perspective of ladder holder,
            barn length $= \frac{L}{\gamma}$, so
            doesn't fit.
    \end{itemize}
    So whether ladder fits or not depends on the
    frame. (So this is a matter of simultaneity:
    events about the front and back being in the
    barn).
\end{example*}

\subsubsection*{Addition of Velocities}

A particle moves with velocity $u'$ in $S'$, which
moves with respect to $S$. What is the velocity
$\bf{u}$ in $S$? \myskip
The worldline of the particle in $S'$
\[ x' = u't' \tag{$*$} \]
In $S$,
\[ u = \frac{x}{t} = \frac{\gamma \left( x' +
vt' \right) }{\gamma \left( t' + \frac{vx'}{c^2}
\right)} \]
(inverse Lorentz transformations)
\[ (*) \implies \boxed{u = \frac{u' + v}{1 +
\frac{u'v}{c^2}}} \]
\begin{itemize}
    \item $u'v \ll c^2 \implies$ Galilean addition
        of velocities
    \item $u' = c \implies u = c$, consistent with
        speed of light being constant in any
        frame.
\end{itemize}

\subsection{Geometry of Spacetime}

There is a quantity that all observer agree on.
Consider two events $P_1$ with coordinates $(ct_1,
x_1)$, $P_2$ with coordinates $(ct_2, x_2)$.
\[ \Delta t := t_1 - t_2 \qquad \Delta x := x_1 -
x_2 \]
The \emph{invariant interval} is defined as
\[ \Delta s^2 = c^2 (\Delta t)^2 - (\Delta x)^2 \]
Claim: All observers agree on value of $\Delta
s^2$: have
\begin{align*}
    \Delta s^2
    &= \gamma^2 \left( c\Delta t' + \frac{v\Delta
    x'}{c} \right)^2 - \gamma^2 (\Delta x' + v
    \Delta t')^2
    &&\text{inverse Lorentz transforms} \\
    &= \gamma^2 (c^2 - v^2) \Delta t'^2 - \gamma^2
    \left( 1 - \frac{v^2}{c^2} \right) \Delta x'^2
    \\
    &= c^2 \Delta t'^2 - \Delta x'^2
\end{align*}
In three spatial dimensions, the invariant
intervals is
\[ \Delta s^2 = c^2 \Delta t^2 - \Delta x^2 -
\Delta y^2 - \Delta z^2 \]
Spacetime is parametrised by 4 numbers, i.e.
$\RR^4$. When endowed with the (non-positive
definite) measure of distance $\Delta s^2$, it is
called \emph{Minkowski space}. We say it has
dimension $1 + 3$, to stress the difference with
Euclidean space.
\begin{itemize}
    \item Events with $\Delta s^2 > 0$ are
        \emph{timlike separated}:
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/f17c9beac59311ec.png}
        \end{center}
    \item Events with $\Delta s^2 < 0$ are
        \emph{spacelike separated}:
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/fc8f8a88c59311ec.png}
        \end{center}
    \item Events with $\Delta s^2 = 0$ are
        \emph{lightlike-} or \emph{null-}
        separated:
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/0bc55ef6c59411ec.png}
        \end{center}
        \begin{note*}
            $\Delta s^2 = 0$ doesn't mean same point.
        \end{note*}
\end{itemize}

\subsubsection*{The Lorentz Group}

\begin{note*}
    3D rotations
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/63a36a0ac59411ec.png}
    \end{center}
    distance of $H$ to origin
    \[ S_{\text{Euclidean}}^2 = x^2 + y^2 + z^2 \]
    In a rotated frame,
    \[
    \begin{pmatrix}
        x' \\
        y' \\
        z'
    \end{pmatrix}
    = R
    \begin{pmatrix}
        x \\
        y \\
        z
    \end{pmatrix}
    \]
    $3 \times 3$ rotation matrix such that $R^\top
    R = I_{3 \times 3}$.
    \[ S_{\text{Euclidean}}^2 = x'^2 + y'^2 + z'^2 \]
\end{note*}

\noindent
It's best to think of $\Delta s^2$ as primary
object in Minkowsky space. We'll use this now to
redefine Lorentz transformation. \myskip
The coordinates of an event $P$ in frame $S$ can
be written as a \emph{four-vector} $X$:
\[ X^\mu =
\begin{pmatrix}
    ct \\
    x \\
    y \\
    z
\end{pmatrix}
\]
(we label the entries with $\mu = 0, 1, 2, 3$
respectively.) The invariant distance from the
origin and $P$ can be written as an inner product
\[ X \cdot X := X^\top \eta X \]
where
\[ \eta :=
\begin{pmatrix}
    1 & 0 & 0 & 0 \\
    0 & -1 & 0 & 0 \\
    0 & 0 & -1 & 0 \\
    0 & 0 & 0 & -1
\end{pmatrix}
\]
is the \emph{Minkowski metric}.

\begin{note*}
    $X \cdot X := X^\top \eta X := X^\mu \eta_{\mu
    \nu} X^\nu$ by Einstein summation convention.
\end{note*}