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\subsubsection*{Massless Particles}

Proper time $\tau$ is only defined for massive
particles that move slowed than $c$. But from last
lecture, have
\[ \bf{P} \cdot \bf{P} = \frac{E^2}{c^2} -
\bf{p}^2 = \bf{p}^2 - \bf{p}^2 = 0 \]
i.e. the 4-momentum is null - and lies on a light
ray i.e. $\exists$ a maximum speed for massless
particles (and it happens that photons - particles
of light - are massless). (Gravitons are the other
ones). \myskip
For $m = 0$, $E^2 = \bf{p}^2 c^2$ so
\[ \bf{P} = \frac{E}{c}
\begin{pmatrix}
    1 \\
    \hat{\bf{p}}
\end{pmatrix}
\]
($\omega$ is angular frequency, and $\lambda$ is
wavelength of light)
To get an expression for the energy, we take it
from quantum mechanics
\[ E = \hbar \omega = \frac{2\phi\hbar c}{\lambda} \]
$h = \text{Planck's constant} = 6.6 \times
10^{-340} \mathsf{m^2 kgs^{-1}}$. ($\hbar :=
\frac{h}{2\pi}$)

\subsubsection*{Newton's Laws of Motion}

In special relativity, N2 becomes
\[ \dfrac{\bf{P}^\mu}{\tau} = F^\mu \]
where $F^\mu$ is a 4-vector force, related to the
3-vector force $\bf{f}$ by
\[ F^\mu =
\begin{pmatrix}
    F^0 \\
    \gamma \bf{f}^\top
\end{pmatrix}
\]
so
\[ \dfrac{\bf{p}}{t} = \dfrac{\tau}{t}
\dfrac{\bf{p}}{\tau} = \frac{1}{\gamma}
\dfrac{\bf{p}}{\tau} = \bf{f} \]
(agrees with N2). The time component is related to
the power
\[ F^0 = \dfrac{\bf{P}^0}{\tau} = \frac{\gamma}{c}
\dfrac{E}{t} \]

\subsection{Particle Physics}

We'll discuss problems which just need
conservation of 4-momentum
\[ \bf{P} =
\begin{pmatrix}
    \frac{E}{c} \\
    \bf{p}
\end{pmatrix}
\]
Recall
\[ \bf{P} \cdot \bf{P} = m^2 c^2 \]
and $E^2 = \bf{p}^2 c^2 + m^2 c^4$. \myskip
Often make problems easy by choosing a handy
inertial frame - almost always this is the centre
of mass (should be ``centre of momentum'', $\sum_i
\bf{p}_i = 0$). Sometimes, $\exists$ a
$\bf{P}^\mu$ we don't know. Can use conservation
of 4-momentum, re-write equations as $\bf{P}^\mu =
\cdots$ and then form $\bf{P} \cdot \bf{P} =
\cdots$.

\subsubsection*{Particle Decay}

A particle of mass $m$, decays to particles of
mass $m_2$ and $m_3$.
\begin{align*}
    \bf{P}_1
    &= \bf{P}_2 + \bf{P}_3
    &&\text{(4-momentum conserved)} \\
    \implies E_1
    &= E_2 + E_3
    &&\text{zeroth component} \\
    \bf{p}_1
    &= \bf{p}_2 + \bf{p}_3
    &&\text{first, second, third components}
\end{align*}
In the rest-frame of decaying particle,
\[ E_1 = m_1 c^2 = \sqrt{\bf{p}^2 c^2 + m_2^2 c^4}
+ \sqrt{\bf{p}_3^2 c^2 + m_3^2 c^4} \ge m_2 c^2 +
m_3 c^2 \]
i.e. decay only happens if $m_1 \ge m_2 + m_3$.

\begin{example*}
    $h \to \gamma \gamma$ (higgs boson decaying
    into 2 photons). In $h$ rest-frame,
    \[ \bf{P}_h =
    \begin{pmatrix}
        m_h c \\
        \bf{0}^\top
    \end{pmatrix}
    ; \qquad \bf{P}_h = \bf{P}_\gamma +
    \bf{P}_{\gamma'} \]
    4-momentum conserved (first, second, third
    components):
    \[ \bf{p}_\gamma = -\bf{p}_{\gamma'} \]
    \[ \implies E_\gamma = E_{\gamma'} = \frac{m_h
    c^2}{2} \]
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/7759c43ec70d11ec.png}
    \end{center}
    \[ (\bf{P}_\gamma + \bf{P}_{\gamma'}) \cdot
    (\bf{P}_\gamma + \bf{P}_{\gamma'}) = m_h^2 \]
\end{example*}

\subsubsection*{Colliders}

Collide 2 particles of mass $m$:
\[ \ub{\bf{P}_1 + \bf{P}_2}_{\text{incoming}} =
\ub{\bf{P}_3 + \bf{P}_4}_{\text{outgoing}} \]
In the centre of mass frame, $\bf{p}_1 + \bf{p}_2
= 0$. Pick axes so that
\[ \bf{P}_1^\mu = (mc \gamma_v, mv \gamma_v, 0, 0) \]
\[ \bf{P}_2^\mu = (mc \gamma_v, -mv \gamma_v, 0,
0) \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/255fd88ec70e11ec.png}
\end{center}
After collision, particles must still have equal
and opposite 3-momenta. Choose axes such that
\[ \bf{P}_3^\mu = (mc \gamma_v, mv \gamma_v
\cos\theta, mv \gamma_v \sin\theta, 0) \]
\[ \bf{P}_4^\mu = (mc \gamma_v, -mv \gamma_v
\cos\theta, -mv \gamma_v \sin\theta, 0) \]
Let's say that in the lab frame, one particle is
at rest. Velocity addition formula implies that
other particles has speed
\[ u = \frac{2v}{1 + \frac{v^2}{c^2}} \]

\subsubsection*{Particle Creation}

Collide 2 particles of mass $m$ energetically
enough to create an extra particle of mass $M$.
4-momentum conserved:
\[ \bf{P}_1 + \bf{P}_2 = \bf{P}_3 + \bf{P}_4 +
\bf{P}_5 \tag{\dag} \]
\[ \bf{P}_1^2 = \bf{P}_2^2 = \bf{P}_3^2 =
\bf{P}_4^2 = m^2 c^2 \]
\[ \bf{P}_5^2 = M^2 c^2 \]
Question: what is $v$? \\
Centre of mass frame
\[ \bf{P}_1 = (\gamma mc, \gamma m \bf{v}) \]
\[ \bf{P}_2 = (\gamma mc, -\gamma m \bf{v}) \]
$(\dag) \cdot (\dag)$:
\[ (\bf{P}_1 + \bf{P}_2)^2 := (\bf{P}_1 +
\bf{P}_2) \cdot (\bf{P}_1 + \bf{P}_2) = (\bf{P}_3
+ \bf{P}_4 + \bf{P}_5)^2 \tag{$*$} \]

\begin{lemma*}
    If $\bf{P}^2 = m_1^2 c^2$ and $Q^2 = m_2^2
    c^2$ then $\bf{P} \cdot Q \ge m_1 m_2 c^2$.
\end{lemma*}

\begin{proof}
    In the rest frame of $m$,
    \begin{align*}
        \bf{P} \cdot Q
        &= (m_1, c \bf{0}) \cdot
        \begin{pmatrix}
            \frac{E_2}{c} \\
            |\bf{p}_2|
        \end{pmatrix}
        \\
        &= m_1 E_2 \\
        m_1 \sqrt{m_2^2 c^4 + \bf{p}^2 c^2} \\
        &\ge m_1 m_2 c^2
    \end{align*}
\end{proof}

\myskip
Expand ($*$) and use lemma:
\[ \bf{P}_1 + \bf{P}_2 = (\gamma mc, \bf{0}) \]
so LHS is
\begin{align*}
    4 \gamma^2 m^2 c^2
    &= \bf{P}_3^2 + \bf{P}_4^2 + \bf{P}_5^2 +
    2(\bf{P}_3 \cdot \bf{P}_4 + \bf{P}_3 \cdot
    \bf{P}_5 + \bf{P}_4 \cdot \bf{P}_5) \\
    &= 2m^2 c^2 + M^2 c^2 + 2(\bf{P}_3 \cdot
    \bf{P}_4 + \bf{P}_3 \cdot \bf{P}_5 + \bf{P}_4
    \cdot \bf{P}_5) \\
    &\ge 2m^2 c^2 + M^2 c^2 + 2(m^2 c^2 + mMc^2 +
    mMc^2) \\
    \implies \Aboxed{\gamma
    &\ge 1 + \frac{M}{2m}}
\end{align*}

\begin{note*}
    Kinetic energy of incoming particle
    \[ T = \gamma mc^2 - mc^2 \ge \half Mc^2 \]
\end{note*}