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\noindent
$\tau$ provides a way to parametrise the
world-line of a particle in a way that all
inertial observers agree on. \\
In frame $S$, the wordline of any particle can be
parametrised
\[ \bf{x}(\tau), \qquad t(\tau) \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/25a1cd00c59911ec.png}
\end{center}
Along a small segment of worldline
\begin{align*}
    \dd \tau
    &= \sqrt{\dd t^2 - \frac{(\dd \bf{x})^2}{c^2}}
    \\
    &= \dd t \sqrt{1 - \frac{\bf{u}^2}{c^2}}
\end{align*}
where $\bf{u} := \dfrac{\bf{x}}{t}$
\[ \implies \dfrac{t}{\tau} = \gamma_u \tag{$*$} \]
with
\[ \gamma_u := \frac{1}{\sqrt{1 -
\frac{\bf{u}^2}{c^2}}} .\]
The total time experienced by the particle along
its worldline
\[ T = \int \dd \tau \stackrel{(*)}{=} \int
\frac{\dd t}{\gamma_u} \]

\subsubsection*{4-velocity}

The general trajectory of a particle traces out a
4-vector
\[ X(\tau) =
\begin{pmatrix}
    ct(\tau) \\
    \bf{x}^\top (\tau)
\end{pmatrix}
.\]
The \emph{4-velocity} is defined as
\[ U = \dfrac{X}{\tau} =
\begin{pmatrix}
    c \dfrac{t(\tau)}{\tau} \\
    \left( \dfrac{\bf{x}(\tau)}{\tau} \right)^\top
\end{pmatrix}
= \dfrac{t}{\tau}
\begin{pmatrix}
    c \\
    \bf{u}^\top
\end{pmatrix}
\stackrel{(*)}{=} \gamma_u
\begin{pmatrix}
    c \\
    \bf{u}^\top
\end{pmatrix}
\]
Because $\dd \tau$ is invariant for inertial
observers, we easily write down the 4-velocity in
other frames: If $S'$ is related to $S$ by $X' =
\Lambda X$ then the 4-velocity of the particle in
$S'$ is
\[ U' = \Lambda U \]
(take $\dfrac{}{\tau}$).

\begin{note*}
    Wouldn't work for $\dfrac{X}{t}$ since both
    $X$ \emph{and} $t$ transform under Lorentz
    boosts.
\end{note*}

\noindent
We know that
\[ U \cdot U = U' \cdot U' \]
i.e. is the same for all observers. In the
rest-frame of the particle,
\[ U =
\begin{pmatrix}
    c \\
    \bf{0}^\top
\end{pmatrix}
\implies U \cdot U = c^2 .\]
i.e. 4-velocities have 3 independent parameters.

\begin{note*}
    In general, a \emph{4-vector} is a 4-component
    vector that transforms as $A \to \Lambda A$
    under Lorentz transformations.
\end{note*}

\subsubsection*{Addition of velocities at an angle}

In frame $S$, a particle travels with 4-velocity
\[ U = \gamma_u
\begin{pmatrix}
    c \\
    u \cos \alpha \\
    u \sin \alpha \\
    0
\end{pmatrix}
\]
Frame $S'$ moves with velocity $v$ in the
$x$-direction (with respect to $S$).
\begin{align*}
    \implies U'
    &= \Lambda [v] U \\
    &=
    \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}
    \begin{pmatrix}
        \gamma_u c \\
        U \gamma_u \cos \alpha \\
        U \gamma_u \sin \alpha \\
        0
    \end{pmatrix}
    \\
    &=
    \begin{pmatrix}
        c \gamma \gamma_u \left( 1 - \frac{uv}{c^2}
        \cos\alpha \right) \\
        \gamma \gamma_u (u \cos\alpha - v) \\
        u \gamma_u \sin\alpha \\
        0
    \end{pmatrix}
    \\
    &:=
    \begin{pmatrix}
        \gamma_{u'} c \\
        u' \gamma_{u'} \cos\alpha' \\
        u' \gamma_{u'} \sin\alpha' \\
        0
    \end{pmatrix}
\end{align*}
Divide first component by zeroth on each side:
\[ u' \cos \alpha' = \frac{u\cos\alpha - v}{1 -
\frac{uv}{c^2} \cos\alpha} \]
Divide second component by first:
\[ \tan\alpha' = \frac{u \sin\alpha}{\gamma(u
\cos\alpha - v)} \]

\subsubsection*{4-momentum}

The \emph{4-momentum} of a particle of mass $m$ is
\[ \bf{P} = mU =
\begin{pmatrix}
    mc \gamma \\
    m \gamma \bf{u}^\top
\end{pmatrix}
\]
\[ \gamma = \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} \]
where $m$ is the \emph{rest-mass}. In relativity,
4-momentum is conserved. Spatial components give
the 3-momenta $\bf{p} = m\gamma \bf{u}$.

\begin{note*}
    As $|\bf{u}| \to c$, $|\bf{p}| \to \infty$.
\end{note*}

\noindent
What's the interpretation of $\bf{P}^0 = \gamma
mc$? Taylor expand for $|\bf{u}| \ll c$:
\[ \bf{P}^0 = \frac{mc}{\sqrt{1 -
\frac{\bf{u}^2}{c^2}}} = \frac{1}{c} \left( mc^2 +
\half m \bf{u}^2 + \cdots \right) \]
suggesting we should interpret $\bf{P}$ as
\[ \bf{P} =
\begin{pmatrix}
    \frac{E}{c} \\
    \bf{p}^\top
\end{pmatrix}
\]
where $E = \text{energy}$. (Noether's theorem). \\
We learn that both mass and kinetic energy
contribute to the energy $\boxed{E = \gamma
mc^2}$.

\begin{note*}
    As $|\bf{u}| \to c$, $E \to \infty$, i.e.
    can't get particles to exceed the speed of
    light.
\end{note*}

\noindent
For a stationary particle, $\gamma = 1$ so $E =
mc^2$ $\to$ mass is a form of energy. \myskip
We can also write the energy in terms of momentum
since
\[ \bf{P} \cdot \bf{P} = \frac{E^2}{c^2} -
\bf{p}^2 ,\]
since same in all frames, calculate in rest frame,
$\bf{p} = 0$, $E = mc^2$.
\[ \implies \bf{P} \cdot \bf{P} =
\frac{m^2c^4}{c^2} \implies \boxed{E^2 = \bf{p}^2
c^2 + m^2 c^4} \]
In Newtonian physics, mass and energy are
conserved separately. In relativity, mass is just
another form of energy and we can convert kinetic
energy $\leftrightarrow$ mass.

\begin{note*}
    $E$ and $\bf{p}$ are frame-dependent! Lorentz
    transform:
    \[ U \to \Lambda U \]
    \[ U \cdot U = U^\top \eta U
    \stackrel{\text{LT}}{\to} U^\top \Lambda^\top
    \eta \Lambda U = u^\top \eta U = U \cdot U \]
\end{note*}