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\noindent
We see that
\[ X \cdot X = ct^2 - x^2 - y^2 - z^2 \]
is the invariant interval $\Delta s^2$ between the
origin and point $P$. 4-vectors with $X \cdot X >
0$ are called timelike, $X \cdot X < 0$ are called
spacelike and $X \cdot X = 0$ are called null.

\myskip
We redefine a Lorentz transformation as a $4
\times 4$ matrix $\Lambda$, rotating the coords in
frame $S$ to those in $S'$ such that
\[ \ub{X'}_{\text{4-vector in $S$'}} =
\ub{\Lambda}_{4 \times 4 \text{ matrix}}
\ub{X}_{\text{4-vector in $S$}} \]
Lorentz transformations are now defined to be any
matrix which leaves the inner product invariant,
\[ X' \cdot X' = X \cdot X  \qquad \forall\,\, X \]
i.e.
\[ \Lambda^\top \eta \Lambda = \eta \tag{\ddag} \]
There are 2 classes of solution to this equation:
\begin{itemize}
    \item If $R$ is a $3 \times 3$ matrix obeying
        $R^\top R = I_{3 \times 3}$ i.e. $R$ is a
        rotation matrix (+ reflection). (one can
        parametrise $R$ by three angles of
        rotation - one around each axis).
        \[ \Lambda =
        \begin{pmatrix}
            1 & 0 & 0 & 0 \\
            0 & & & \\
            0 & & R & \\
            0 & & &
        \end{pmatrix}
        \]
    \item A boost in the $x$-direction:
        \[ \Lambda =
        \begin{pmatrix}
            \gamma & -\frac{\gamma v}{c} & 0 & 0 \\
            -\frac{\gamma v}{c} & \gamma & 0 & 0 \\
            0 & 0 & 1 & 0 \\
            0 & 0 & 0 & 1
        \end{pmatrix}
        \]
        (Two further ones in $y$- and
        $z$-directions).
\end{itemize}
The set of all matrices solving (\ddag) form the
\emph{Lorentz group} (a group under matrix
multiplication) denoted $O(1, 3)$. \myskip
Taking $\det(\ddag)$, $+(\det \Lambda)^2 = +1$
hence $\det \Lambda = \pm 1$. \\
The subgroup with $\det \Lambda = +1$ is called
the \emph{proper Lorentz group}. Additionally
requiring $\lambda_0^0 > 0$ leads to another
subgroup, the \emph{proper orthochronous Lorentz
group}, denoted $SO(1, 3)^\uparrow$ or $SO(1,
3)^+$. (the $S$ means special, and means that
$\det \Lambda = +1$)

\subsubsection*{Rapidity}

Focus on upper $2 \times 2$ block of $\Lambda$,
boost in $x$-direction.
\[ \Lambda [v] =
\begin{pmatrix}
    \gamma & -\frac{\gamma v}{c} \\
    -\frac{\gamma v}{c} \gamma
\end{pmatrix}
\]
Combining two successive boosts in the
$x$-direction,
\[ \Lambda [v_1] \Lambda [v_2] =
\begin{pmatrix}
    \gamma_1 & -\frac{\gamma_1 v_1}{c} \\
    -\frac{\gamma_1 v_1}{c} & \gamma_1
\end{pmatrix}
\begin{pmatrix}
    \gamma_2 & -\frac{\gamma_2 v_2}{c} \\
    -\frac{\gamma_2 v_2}{c} & \gamma_2
\end{pmatrix}
= \cdots = \Lambda \left[ \frac{v_1 + v_2}{1 +
\frac{v_1 v_2}{c}} \right] \]
i.e. compatible with our velocity addition formula
from last lecture. \myskip
There's a much nicer way of doing this. \\
Recall that, for 2D rotations,
\[ R(\theta) =
\begin{pmatrix}
    \cos\theta & \sin\theta \\
    -\sin\theta & \cos\theta
\end{pmatrix}
\implies R(\theta_1)R(\theta_2) = R(\theta_1 +
\theta_2) \]
- do an analogous thing for Lorentz boosts.
\myskip
For Lorentz boosts, we define the \emph{rapidity},
$\phi$, defined by $\gamma = \cosh \phi$, so
\[ \sinh \phi = \sqrt{\cosh^2 \phi - 1} =
\sqrt{\gamma^2 - 1} = \frac{\gamma v}{c} \]
\[ \implies \Lambda [\phi] =
\begin{pmatrix}
    \cosh \phi & -\sinh \phi \\
    -\sinh \phi & \cosh \phi
\end{pmatrix}
\]
so that rapidities just add, like angles of
rotation:
\[ \Lambda [\phi_1] \Lambda [\phi_2] = \Lambda
[\phi_1 + \phi_2] \]
- shows the relationships between rotations and
boosts.

\begin{remark*}
    $c \approx 3.00 \times 10^8 \mathsf{ms^{-1}}$.
    In fact $c = 299792458 \mathsf{ms^{-1}}$
    exactly because this is how the metre is
    defined:
    \[ 1 \mathsf{m} = \text{distance travelled by
    light in $\frac{1}{299792458}$ seconds.} \]
    Could have defined $c$ to be 1 (light second)
    $\mathsf{s^{-1}}$. (later on in tripos, you'll
    redefine units such that $c = 1$)
\end{remark*}

\subsubsection*{4-velocity}

\[ \Delta s^2 = c^2 \Delta t'^2 \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/a04679eec59811ec.png}
\end{center}
We define \emph{propertime $\tau$} as
\[ \Delta \tau = \frac{\Delta s}{c} \]
the time experienced by the particle, but $\Delta
\tau = \frac{\Delta s}{c}$ holds in all inertial
frames, since $\Delta s$ is invariant under
Lorentz transformations.
