🧠 Remember: Massless particles have $E=c|\vec p|$ and $E=hc/\lambda=hf$ and de-Broglie wavelength is $\lambda_\text{dB}=h/|\vec p|$
We can express massless particles in terms of wavenumber $|\vec k|=2\pi/\lambda$ and the angular frequency $\omega=2\pi f$ as $\vec p=\hbar \vec k$ and $E=\hbar c|\vec k|=\hbar \omega$
Consider a massless particle with energy $E=c|\vec p|$ then the contravariant components of the 4-moment are $P^\mu=(|\vec p|,\vec p)$. For a particle moving in the $xy$ plane we can write in some frame $S$ that
$$ P^\mu=p(1,\cos\theta,\sin\theta)=\frac Ec (1,\cos\theta,\sin\theta) $$
where $\theta$ is the angle relative to the $x$-axis and $p=|\vec p|$.
We can apply a Lorentz transformation to a moving frame at velocity $v$ in the $x$-direction
$$ P'^\mu=\begin{pmatrix} E'/c \\ p'_x \\ p'_y \\ p'z \end{pmatrix}=\Lambda^\mu\upsilon P^\upsilon=\begin{bmatrix} \gamma & -\gamma\beta & 0 & 0 \\ -\gamma \beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\begin{pmatrix} E/c \\ p_x \\ p_y \\ p_z \end{pmatrix} $$
Using this we can write
$$ \begin{aligned} E'&=\gamma(E-vp\cos\theta) \\ p'\cos\theta'=p'_x&=\gamma \left ( p\cos\theta-\frac{vE}{c^2}\right ) \\ p'\sin\theta'=p'_y&=p\sin\theta \end{aligned} \\
\Rightarrow p'=\gamma p\left ( 1-\frac{v}{c}\cos\theta \right ) $$
🗒️ Note: We are studying two frames $S$ and $S'$ which are moving relative to each other at $v$ in $x$-direction. Consider a source at rest at some position $(x_s,y_s)$ in $S$ with the observer located at some point $(x_o,y_s)$ in $S'$. $x_0>x_s$ if $v$ is positive they move apart otherwise together. The source is emitting radiation isotropically with $\cos\theta=1$ corresponding to a photon emitted to the right and $\cos(-\theta)$ to the left.
If we assume that photons are emitted with a momentum $p=hf/c$ then
$$ \frac {f'}f=\gamma\left ( 1-\frac vc \cos\theta\right ) $$
if $v>0$ and $\cos\theta=1$ then the source is moving away and this is called longitudinal relativistic Doppler effect
$$ \frac{f'}{f}=\gamma\left ( 1-\frac vc \right ) =\left ( \frac {1-\frac vc}{1+\frac vc} \right )^\frac 12 \approx 1-\frac vc+\cdots $$
This leads to red shifting of the photon with $f$ decreasing and $\lambda$ increasing
If $v<0$ the source is moving towards the photon is blue shifted with f increased and $\lambda$ decreased
The apparent angle $\theta'$ measured in $S'$ can be deduces to satisfy
$$ \begin{aligned} \cos\theta'&=\frac{\cos\theta-\frac vc}{1-\frac vc \cos\theta} \\ \sin\theta'&=\frac{\sin\theta}{\gamma(1-\frac vc\cos\theta)}
\end{aligned} $$
When $v\to-c$, $S'$ and $S$ getting closer then $\cos\theta'=1$ and $\sin\theta=0$
This leads to a “relativistic headlight effect”
If both frames are coming together at velocity $v=-c+\Delta v$ we get
$$ \cos\theta'=1-\frac{\Delta v}{c}\tan^2\left ( \frac 12 \theta \right ) +\mathcal O \left [ \left ( \frac {\Delta v}c \right )^2\right ] $$
When $v\approx 0$ the source appears like an isotropic source - we are assume that somehow light is being reflected isotropically off the spaceship towards the observer. $As |v|$ increases towards $-c$ the number of rays pointing leftward in the diagram is reduced and the number pointing rightward are increased.
The flux caused by an isotropic source emitting $\dot N$ photons per unit time in some frame $S$ and therefore the number per unit time and solid angle is
$$ \frac{\text dN}{\text d\Omega\text dt}=\frac{\dot N}{4\pi} $$
Consider another frame $S'$ to move along the $x$-direction as before but with the $y$-direction rotated into the page by azimuthal angle which is unaffected by the Lorentz transformation that is $\phi'=\phi$.
$$ \frac{\text dN}{\text d\Omega'\text dt}=\frac{\dot N}{4\pi}\frac{\left [ 1-\left ( \frac vc \right )^2 \right ]}{\left ( 1+\frac vc\cos\theta' \right )^2} $$
For $\theta'\approx 0$ we get
$$ \frac{\text dN}{\text d\Omega'\text dt}=\frac{\dot N}{4\pi}\left ( \frac{1-\frac vc}{1+\frac vc}\right ) $$
which is $\sim \dot Nc/(2\pi\Delta v)$ when $v=-c+\Delta v$. So we see that the flux $\to\infin$ as $\Delta v\to 0$
Consider 2 particle 1 and 2 initially which become A and B. Consider kinematic constraints ( the constraints imposed by the conservation of 4-momentum )
$$ \overset{↝}{\bold P}_1+ \overset{↝}{\bold P}_2=\overset{↝}{\bold P}_A+ \overset{↝}{\bold P}_B $$
This implies that $E_1+E_2=E_A+E_B$ and $\vec p_1+\vec p_2=\vec p_A+\vec p_B$
The on-shell condition implies $\overset{↝}{\bold P}_i^2=(m_i c)^2$ for $i= 1,2,A,B$
🗒️ Note: If massless particle $\overset{↝}{\bold P}_i^2=0$
By taking the inner product on each side we find that
$$ \overset{↝}{\bold P}_A\cdot \overset{↝}{\bold P}_B=\overset{↝}{\bold P}_1\cdot \overset{↝}{\bold P}_2+\frac 12\Delta c^2 $$
where $\Delta=m_1^2+m_2^2-m^2_A-m^2_B$ is the change in the mass squared
🗒️ Note: if we know $E_1$, $E_2$, $\vec p_1$, $\vec p_2$ the masses $m_1$ and $m_2$ can be inferred from the on shell condition. To find $E_A,E_B,\vec p_A, \vec p _B,m_A,m_B$ we have $2n+4$ unknowns where $n$ are the spatial dimensions. Now there are $n+3$ constraints due to 4-momentum and on-shell condition. Thus we can solve with $n=1$ only if with have at least 2 more information, like $m_A ,m_B$
If we have a 2 collinear photon (same direction) we have
$$ P^\mu_1=\frac{E_1}{c}(1,1,0,0) \quad ; \quad P^\mu_2=\frac{E_2}{c}(1,1,0,0) $$
which implies total 4-momentum is
$$ P^\mu_\text{tot}=P^\mu_1+P^\mu_2=\left ( \frac{E_1+E_2}{c} \right )(1,1,0,0) $$
🗒️ Note: this cannot be distinguished from that of a single photon with energy $E_1+E_2$. This is general to $N$ collinear photons and $\overset{↝}{\bold P}\cdot \overset{↝}{\bold P}\equiv 0$.
Consider the following total 4-momentum in a situation
$$ P^\mu_\text{tot}=\left ( \frac Ec,p,0,0 \right ) $$
If we make a Lorentz transformation along the $x$-direction
$$ E'=\gamma(E-vp) \quad ;\quad p'=\gamma \left ( p-\frac{vE}{c^2} \right )=0 $$
which requires $v/c=cp/E<1$, this defines the Centre of Momentum frame, that is, one where $P'^\mu_\text{tot}=(E'/c,0,0,0)$. In this frame we can define
$$ \overset{↝}{\bold P}\text{tot}\cdot \overset{↝}{\bold P}\text{tot}=\eta_{\mu \upsilon} P^\mu_\text{tot} P^\upsilon_\text{tot}=(m_\text{inv} c)^2>0 $$
where $m_\text{inv}$ is the invariant mass of the system.
🗒️ Note: if this is non-zero (not a single photon) then $v/c<1$ and we can define this frame using Lorentz transformation
<aside> <img src="https://s3-us-west-2.amazonaws.com/secure.notion-static.com/5a3d6cb9-f49a-4a98-85bc-0957ad7ba24d/Invariant_mass.png" alt="https://s3-us-west-2.amazonaws.com/secure.notion-static.com/5a3d6cb9-f49a-4a98-85bc-0957ad7ba24d/Invariant_mass.png" width="40px" /> Invariant mass is conserved in elastic collision and is also the same in all frames.
$$ (m_\text{inv}c)^2=\left ( \frac 1c \sum_i E_i\right )^2-\left ( \sum_i p_i \right )^2 $$
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In the Centre of Momentum frame we have
$$ m_\text{inv}c=\frac 1c\sum_iE_i=\sum_i\sqrt{|\vec p_i|^2+(m_ic)^2}\ge c\sum_i m_i $$
where the inequality is saturated when $p_i=0\;\forall i$. This state is useful to check if there is enough energy and momentum to produce the required particle at rest.
