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\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/927e3e50c58e11ec.png}
\end{center}
Light rays travel at $\frac{\pi}{4}$ angle with
respect to the axes. We'll see later that
particles can't travel with $v > c$. (i.e.
particle worldlines are steeper than
$\frac{\pi}{4}$). We could also draw the axes of
$S'$, moving with velocity $\bf{v} = (v, 0, 0)$
with respect to $S$. The $t'$ axis sits at $x' =
0$ or $x = vt$. The $x'$ axis sits at $t' = 0$ so,
from the Lorentz transformation, $ct =
\frac{vx}{c}$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/b5a1c67cc58e11ec.png}
\end{center}
Axes are symmetrical about the light ray diagonal,
reflecting that the speed of light is also $c$ in
$S'$.

\subsection{Relativistic Physics}

Speeding train
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/250ab03cc58f11ec.png}
\end{center}
Observer on train sees light hit front and back
simultaneously. Station master on ground: sees
light rays going forward and back at same speed
$c$, but while the light is travelling, the train
moves, so the station master sees the light hit
the back of the train first.

\subsubsection*{Simultaneity}

An observer in $S$ decides that events $P_1$ and
$P_2$ occur simultaneously if $t_1 = t_2$
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/42ed729cc58f11ec.png}
\end{center}
But in $S'$, simultaneous events occur with equal
$t'$, or $t - \frac{vx}{c^2} = \text{constant}$
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/7aeb183ec58f11ec.png}
\end{center}
$\cdots = \text{lines of simultaneity in $S'$}$.
$P_1$ and $P_2$ occur at \emph{different} $t'$ (as
measured in $S'$).

\subsubsection*{Causality}

Question: If observers disagree on temporal
ordering of events, where does that leave the idea
of cause and effect? Thankfully, this still holds:
there are only some events which observers
disagree about. \myskip
Because Lorentz boosts are only possible for $v <
c$, the lines of simultaneity are at an angle $<
\frac{\pi}{4}$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/eb48da76c58f11ec.png}
\end{center}
All observers agree that $Q$ occurred after $P$.
But they can disagree on the temporal ordering of
$P$ and $R$. If nothing travels faster than $c$,
then $P$ can only be influenced by events in its
past light cone (the reflection of the future
light cone about $P$).

\subsubsection*{Time Dilation}

A clock sitting in $S'$ ticks at intervals $T'$,
for example tick 1 occurs at coordinates $(ct',
x') = (ct_1', 0)$, tick 2 occurs at coordinates
$(ct', x') = (c(t_1' + T'), 0)$ etc. \myskip
Question: What are the intervals between the ticks
in $S$? \\
Clock sits at $x' = 0$. Lorentz transformations
implies that $\gamma = \left( t' + \frac{vx'}{c^2}
\right) \implies T = \gamma T'$. Since $\gamma >
1$, time runs slower in $S$.

\subsubsection*{Twin Paradox}

Two twins: Luke \& Leia.
\begin{itemize}
    \item Leia stays at home
    \item Luke heads to a planet $P$
    \item When he arrives at $P$, he turns around
        and comes back at the same speed.
\end{itemize}
Leia's perspective:
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/232c7578c59111ec.png}
\end{center}
When Luke returns, Leia has aged $2T$. Luke has
aged $\frac{2T}{\gamma}$, i.e. he's younger than
Leia. \myskip
Question: can't we use the same argument from
Luke's perspective to say that Leia is younger
than him? Why aren't things symmetric? Resolution
is that Luke's not in an inertial frame at $P$ -
spoils symmetry. \\
When Luke arrives at $P$, he thinks Leia is at
simultaneous event $X$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/396e85f6c59111ec.png}
\end{center}
Point $P$ is $(ct, x) = (cT, vT)$. Time Luke
thinks it takes to make journey is
\[ T' = \gamma \left( T - \frac{v^2}{c^2}T \right)
= \frac{T}{\gamma} .\]
Point $X$ occurs at $x = 0$ and $t' = T' =
\frac{1}{\gamma}$, since it's simultaneous from
Luke's perspective, with his arrival at $P$. So
\[ t' = \gamma \left( t - \frac{v^2x}{c^2} \right)
\implies t = \frac{T'}{\gamma} =
\frac{T}{\gamma^2} .\]
Leia's age from Luke's perspective when he arrives
at $P$ is $t$, i.e. Luke thinks Leia is younger
than him by a factor $\frac{1}{\gamma}$ $\to$ so
far, it's symmetric.
