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\section{Special Relativity}

When particles travel very fast, Newtonian
mechanics breaks down and is replaced by
Einstein's theory of special relativity. The
effects of special relativity only become apparent
when the speed of the particles approaches the
speed of light (in a vacuum)
\[ c = 3.00 \times 10^8 \mathsf{ms^{-1}} \]
\[ \text{speed of sound in air} \approx 300
\mathsf{ms^{-1}} \]
\[ \text{escape velocity} \approx 10^4
\mathsf{ms^{-1}} \]
The theory of special relativity rests on 2
postulates:
\begin{enumerate}[(1)]
    \item The principle of relativity - the laws
        of physics are the same in all inertial
        frames (Galilean principle of relativity).
    \item The speed of light in vacuous (= in a
        vacuum) is the same in all inertial
        frames.
\end{enumerate}
Postulate 2 is weird - `common sense' says that a
car moving at speed $v$ should have its light
travelling at speed $c + v$ with respect to the
ground. (This is not compatible with Galilean
relativity). \myskip
Consider 2 inertial frames, $S$ and $S'$, moving
relative to each other with $\bf{v} = (v, 0, 0)$.
Galilean transformations would relate the
coordinates as:
\[ x' = x - vt \quad y' = y, \quad z' = z \quad t'
= t .\]
A ray of light travels in the $x$-direction in
frame $S$ with speed $c$ $\implies$ it traces out
trajectory $\frac{x}{t} = c$.
\[ \text{Galilean transformation} \implies
\frac{x'}{t'} = \frac{x - vt}{t} = c - v \]
Only way to incorporate postulate 2 (and 1) is to
change the definition of time.

\subsection{Lorentz Transformations}

$S$, $S'$ moving relative speed $v$ in
$x$-direction. Ignore $y$ and $z$ for now. Want to
know relationship between $x'$, $t'$, $x$ and $t$.
The most general form is
\[ x' = f(x, t) \qquad t' = g(x, t) \]
for some functions $f$ and $g$. \myskip
Left alone, a particle should travel in a straight
line viewed in $S$ and $S'$. (both inertial
frames). So, the map $(x, t) \mapsto (x', t')$
takes straight lines to straight lines (i.e. it is
linear, for example $x' = \alpha_1 x + \alpha_2
t$, $t' = \alpha_3 x + \alpha_4 t$; $\alpha_i$ are
functions of $v$). $S$, $S'$ move at relative
speed $v$. Assume that origins coincide $\implies$
lines $x = vt$ must map to
\[ x' = 0 \implies x' = \gamma(x - vt) \tag{8.1} \]
for some $\gamma = \gamma(v)$ (it shouldn't depend
on 1D direction of $\bf{v}$, i.e. $\gamma(v) =
\gamma(-v)$; from rotational invariance).
(position of observer sitting origin of $S'$).
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/c6cbe1669a3d11ec.png}
\end{center}
But from the perspective of $S'$, $S$ moves with
velocity $-v$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/d3b3cd309a3d11ec.png}
\end{center}
Same argument as (8.1)
\[ \implies x = \gamma(x' + vt') \tag{8.2} \]
so
\[ ct' = \gamma(c - v)t \qquad \text{and} \qquad
ct = \gamma(c + v) t' \]
so
\[ \boxed{\gamma = \sqrt{\frac{1}{1 -
\frac{v^2}{c^2}}}} \]
- important!

\begin{note*}
    \begin{itemize}
        \item $v \ll c$: $\gamma \approx 1$
            $\implies$ Special Relativity $\to$
            Galilean transformations
        \item $v \to c$: $\gamma \to \infty$
        \item $v > c$: $\gamma$ is imaginary -
            unphysical
    \end{itemize}
\end{note*}

\noindent
(8.1) reads $\boxed{x' = \gamma(x - vt)}$ while
\[ x = \gamma(x' + vt') = \gamma [\gamma(x - vt) +
vt'] \]
\[ \implies \boxed{t' = \gamma \left( t -
\frac{v}{c^2}x \right)} \]
(this is the \emph{Lorentz transformations} (=
\emph{Lorentz boosts})). They tend to Galilean
transformations in $\lim \frac{v}{c} \to 0$. ($t'
= t$). Inverting,
\[ x = \gamma(x' + vt') \]
\[ t = \gamma \left( t' + \frac{v}{c^2}x \right) \]
(note same, with $v \mapsto -v$). The directions
perpendicular to direction of motion have the
trivial transformation law $y' = y$, $z' = z$.

\subsubsection*{Checking the Constancy of the Speed of Light}

A light ray travelling in the $y$-direction in $S$
has trajectory $x = 0$, $y = ct$. In $S'$,
trajectory is
\[ x' = -vt' \qquad \text{and} \qquad y' =
\frac{ct'}{\gamma} \]
and the speed of light is
\[ v'^2 = \left( \frac{x'}{t'} \right)^2 = v^2 +
\frac{c^2}{\gamma^2} = c^2 \]
i.e. the same, so postulate 2 holds.

\subsubsection*{Space-time Diagrams}

In a fixed inertial frame $S$, draw a direction of
space - say $x$ coordinate along horizontal
coordinate and $ct$ along vertical coordinate.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/101a85ce9a3f11ec.png}
\end{center}
The union of space and time is called
\emph{Minkowski spacetime}. Each point $P$
represents an \emph{event}. We label the
coordinates of $P$ as $(ct, x)$ (note backwards
convention)!
