# Special Relativity and High-Energy Astrophysical Phenomena
## Special Relativity Principles
In classical mechanics, the **Newton laws of motion** states**:**
1\) Every body continues in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by the forces impressed upon it.
2\) The change of motion is proportional to the motive force impressed, and is made in the direction of the right line in which that force is impressed.
3\) To every action there is always opposed an equal and opposite reaction; or, the mutual actions of two bodies upon each other are always equal, and directed to contrary parts.
They express the **classical mechanical principle of relativity,** meaning the equivalence between the conditions of "rest" and "uniform motion in a right line”. Corollary 5 of Newton’s Principia states: The motions of bodies included in a given space are the same among themselves, whether that space is at rest or moves uniformly forwards in a straight line without circular motion.
From the first principle, the concept of **inertial reference frames** emerges**:** they are reference frames in which the free motion of a body not subject to any external force happens at constant velocity.
Einstein adopted a more comprehensive interpretation of inertia, basing the special theory of relativity on the proposition that: The laws by which the states of physical systems undergo changes are not affected, whether these changes of state be referred to the one or the other of two systems of \[inertial] coordinates in uniform translatory motion.
Special relativity is based on two principles:
1. *The Principle of Relativity:* Physical laws are the same in all inertial frames moving with constant speed one with respect to the other. If one inertial reference frame moves with uniform linear motion with respect to another frame, this is also inertial.
2. *The Principle of Invariance of the speed of light:* The velocity of light in vacuum $$c = 2.99793 \times 10^8 {\rm ; m/s}$$ has the same value in all inertial frames. **It is a universal constant.**
These 2 principles introduce relevant consequences:
1. a big change in our perception of space and time: time and length intervals are no longer absolute;
* if we measure a ruler at rest and then when it moves, this last is shorter.
* If we synchronise two clocks at rest and then let one move, we find that the moving clock is delaying.
* Galilean transformations had to be replaced by new ones, Lorentz transformations.
2. Another striking consequence of this revolution was that mass is just a particular form of energy;
3. kinematics at velocities near c is quite different from the usual one and (astro)particle physics is the laboratory to test it.
## Galileo's Transformations
In the simplified case of two inertial frames S and S', with observers O and O', and with S' traveling at spead V along the $$x \equiv x'$$ axis, the set of coordinates (x,y,z) and (x',y,z') transform according to:
x' = x-Vt; y' = y; z' = z and, obviously, the time of events remains unchanged between the 2 reference frames: t' = t.
Consequently, deriving with respect to t one obtains: $$v'\_x = v\_x - V; v'\_y = v\_y; v'\_z = v\_z$$ and $$a'\_x = a\_x ; a'\_y = a\_y; a'\_z = a\_z$$
### Spacetime
Each point in spacetime is labeled by a set of four coordinates : $$x^\mu = (x^0, \vec{x}),$$ with time-component $$x^0 = ct$$ and space components $$x^1 = x, x^2 = y, x^3 = z$$. These are the coordinates of a 4-vector that sometimes we will indicate also with a capital letter (X) or with $$x^\mu$$, where $$\mu = 1, 2, 3$$. The locus of events occupied by a particle is a curve in spacetime called *worldline*, described by $$x^\mu (\tau)$$, where $$\tau$$ is the proper time. In an inertial system, worldlines of unaccelerated particles are described by linear relationships among coordnates. The distance between two events P and Q on a worldline is $$\Delta x^\mu = x^\mu (Q) - x^\mu (P)$$ and $$\frac{\Delta x^{i}}{\Delta x^0}$$ are independent of P and Q and $$\Delta x^{i}$$ is independent of $$x^0$$ .
**Controvariant and covariant components**: we can write in matrix form the 4 vector $$x^\mu = g^{\mu \nu} x\_\nu$$ , where define the matrix$$g\_{\mu\nu}$$, where $$\mu,\nu = 0,1,2,3$$ :
$$
g\_{00} = + 1, g\_{11} = g\_{22} = g\_{33} = - 1 , g\_{\mu\nu} = 0, ;\rm{for}; \mu \neq\nu
$$
namely
$$g\_{\mu\nu} \equiv \begin{pmatrix} 1& 0& 0 & 0 \ 0 &-1 & 0 & 0 \ 0 & 0 & -1 & 0 \ 0 & 0 & 0 & -1 \end{pmatrix}$$
This matrix is called a metric and the space-time with this metric is called the **Minkowski space.**
We can define $$x\_\mu \equiv g\_{\mu \nu} x^\nu$$, which is called *covariant component* of the four vector **x** and $$x^\mu$$ is called controvariant component. In other words:
$$x\_0 = x^0 = ct; x\_1 = -x^1 = -x ; x\_2 = -x^2 = - y; x\_3 =- x^3 = -z$$ .
While it is not true for a 4 vector that the components witn index upstairs and downstairs are equal, this is not true for the metric tensor which enjoys the property: $$g\_{\mu\nu} = g^{\mu \nu}$$. From the definition of covariant components, we can write:
$$
X^2 = x^\mu x\_\mu = g\_{\mu\nu} x^\mu x^\nu = (x^0)^2 -(x^1)^2 -(x^2)^2 -(x^3)^2
$$
(of course the symbol of sum over the components is omitted). $$X^2 >0 (<0)$$ is time(space)-like; $$X^2 =0$$is a null-vector (eg. for a photon).
We define the *interval* between two events: $$\Delta s^2 \equiv g\_{\mu\nu} \Delta x^\mu \Delta x^\nu = \Delta x^\mu \Delta x\_\nu$$. We can define the proper time as the time between events measured in the rest frame of the object: $$c\Delta \tau = \sqrt{\Delta s^2}$$.
We show below that $$\Delta s^2$$ is an invariant due to the ivariance of c.
In general, the product of two 4-vectors $$X^2$$ or in general $$A^\mu B\_\mu$$are invariants. Notice that $$A^\mu B\_\mu = A\_\mu B^\mu$$, meaning $$A^0B\_0+A^1B\_1+A^2 B\_2 + A^3 B\_3 = A^0B^0-A^1B1-A^2 B2 +-A^3 B^3.$$
**Invariance of the interval:** As a matter of fact, the second principle imposes that if we follow the same light-ray traveling over a distance across two inertial frames S and S', with observers O and O' and set of coordinates $$x^\mu$$ and $$x'^\mu$$ the intervals of time $$\Delta t$$ and $$\Delta t'$$ it takes to cross the distance $$|\Delta r|$$ and $$|\Delta r'|$$in the 2 reference frames satisfies the invariance of the velocity of light :
$$c = \frac{|\Delta r|}{\Delta t} = \frac{|\Delta r'|}{\Delta t'}$$. We can express this also with the line element being an invariant between two inertial frames:
$$\Delta s^2 = c^2 \Delta t^2 - \Delta x^2 - \Delta y^2 -\Delta z^2 = c^2 \Delta t^2 \left( 1 - \frac{v^2}{c^2}\right),$$ where we used the definition of speed $$v^2 = \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2$$ and for the photon v = c. In general $$\Delta s^2$$ an **invariant scalar**.
It is easy to see, that for a light-ray (v = c) and so $$\Delta s^2 = 0$$ and the separation in space-time is said *light-like*. Unlike the tridimensional Euclidean case, where the distance between two events is positively-defined, for space time we can have:
$$\Delta s^2 >0$$ *time-like* separation;
$$\Delta s^2 =0$$ *light-like* separation;
$$\Delta s^2 >0$$ *space-like* separation.
Two events can be casually connected only if $$\Delta s^2\ge 0$$, meaning that $$c\Delta t \ge |\Delta r|$$. Hence they fall inside the **light cone** (see image below with copytight [here](https://en.wikipedia.org/wiki/Light_cone#/media/File:World_line.svg)) defined by $$ds^2 = c^2dt^2 -d x^2 -d y^2 -dz^2 = 0$$.
### Lorentz Transformations
A Lorentz transformation or *boost* is the coordinate transformation from one inertial reference frame to another. At the end of 1800, Lorentz determined them as the transformations that leave invariant Maxwell equations, condition that was not satisfied by Galileo's transformations. In general it is a $$(4\times4)$$ matrix which encapsulates the system of equations describing the transformation (in natural units).
$$
\begin{aligned}
ct' &= \gamma(ct - \beta x) \\
x' &= \gamma(x - \beta t) \\
y' &= y \\
z' &= z
\end{aligned}
$$
The matrix form of the transformation from a rest frame to a moving frame with speed v in the direction of the $$x^1$$-axis is a Lorentz boost along the $$x^1$$-axis and the 4-dimensional rotation is given by the following matrix:
$$
x'^\mu = {\Lambda^\mu}*\nu x^\nu, ; {\Lambda^\mu}*\nu =
\begin{pmatrix}
\gamma & -\beta\gamma & 0 & 0 \\
-\beta\gamma & \gamma & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
$$
The inverse transformation is obtained repacing the - signs with +.
This transformation leaves $$ds^2 = (dx^0)^2-(dx^1)^2- (dx^2)^2-(dx^3)^2= (cdt)^2 -dx^2-dy^2-dz^2$$ invariant due to invariance of c. Because $$ds^2$$ is invariant under Lorentz transformation it is a *four-scalar* or *Lorentz scalar*. From this property, we find:
$$\Delta s'^2 = \Delta x'^\mu \Delta x'*\mu = \Lambda^\mu*{, , \nu} \Delta x^\nu g\_{\mu \rho} \Delta x'^\rho = g\_{\mu \rho}\Lambda^\mu\_{, , \nu}\Lambda^\rho\_{, , \sigma}\Delta x^\nu\Delta x^\sigma$$
and the property $$g\_{\mu \rho}\Lambda^\mu\_{, ,\nu} \Lambda^\rho\_{, , \sigma} = g\_{\nu \sigma}$$ holds since it is also true that $$\Delta s'^2 = \Delta s^2 = \Delta x^\nu \Delta x\_\nu = g\_{\nu \sigma} \Delta x^\sigma \Delta x^\nu$$ . The derived property can be written in matrix form as $$\bf \Lambda^T g \Lambda$$ , which *means that Lorentz boosts are symmetric matrixes*: $$\bf \Lambda^T = \Lambda$$. Symmetric matrices describe boosts of frames in relative motion with aligned axes.
From Lorentz transformations one can immediately see that *a signal propagating at speed c in an inertial system will propagate in any other inertial system with this speed*. Also note that for $$v \rightarrow 0$$ we obtain the galileian transformations.
Lorentz transformation have the following (some of them really bizarre) consequences:
* **Simultaneity.** For galileian transformations whatever is the position of 2 events, if they are simultaneous in a reference frame they are in the other: t'= t. But in relativity, if two events happen at positions $$x\_1, x\_2$$ and at instant of time $$t\_0$$ , they will be simultaneous with respect to the spaceship:$$\Delta t = 0$$ for the observer on the spaceship . But this doesn't imply that$$\Delta t' = 0$$ for the observer O, steady on the Earth because:
$$
t'\_1 = \gamma (t\_0 - \beta x\_1/c)\\
t'\_2 = \gamma (t\_0 - \beta x\_2/c)
$$
Hence, they are not simultaneous but separated by the time interval which depends on their distance:
$$
\Delta t' = t'\_2 - t'\_1 = \gamma (x\_2 - x\_1)\beta/c = \gamma \Delta x/c
$$
In order to appear simultaneous in all Lorentz frames, a pair of events must coincide in both space and time — in which case they are really just one event!
Hence in general in relativity:
$$
\Delta t' = \gamma \left(\Delta t - \frac{\beta}{c} \Delta x \right)
$$
* **Time dilation.** Let's immagine a clock at rest in reference frame S ($$\Delta x = 0$$) which sends a periodic signal with period $$\Delta t$$. For the observer in the moving frame S' (moving with respect to S with speed $$\vec{\beta}$$ along the x axis - and x and x' are parallel axis for the two frames) the period is longer:
$$
\Delta t' = \gamma \Delta t
$$
* **Length contraction.** Let's consider a rod of length $$\Delta x$$ in the rest frame S (proper frame of the object). The length of the rod with respect to the observer in the moving frame S' (given that observations are simultaneous in S' : $$\Delta t' = 0$$) is: $$\Delta t' = 0 = \gamma(\Delta t - \beta \Delta x/c) \Rightarrow \Delta t = \beta x/c$$. Hence, we obtain $$\Delta x' = \gamma \Delta x - \gamma v \beta \Delta x/c = \gamma \Delta x (1-\beta^2) = \frac{\gamma}{\gamma^2}\Delta x$$ , and so in the moving frame length is contracted:
$$
\Delta x' = \frac{\Delta x}{\gamma}
$$
**The twin paradox**: the twin Elena travels on a fast spacecraft while Giacomo stays on Earth. Her clock (including her biological one) runs slower than the clock of earth-bound Giacomo. When she returns, she finds that Giacomo has aged more than she has, since her clock runs slower. But from Elena's point of view it is Giacomo who moved away! And so she expects to age much more than him! This non-sense can be solved realising that the situation is not symmetric since Giacomo has been all the time in an *inertial frame* while Elena must change inertial frames at the half point and accelerates and decelerates during a fraction of her trip.
Let’s specifically suppose that the distance between Earth and the distant space station STAR, where Elena sets off, is 25 ly (measured in the system in which Earth and the station are at rest). The clock on STAR and the one on Earth are synchronised. She takes spacecraft A to reach STAR and to come back she almost instantaneously jumps on spacecraft B to come back. Almost immediately reaches 0.9998c, so that Elena's Lorentz factor is 50. The distance Earth-space STAR is then contracted by 50, so it takes to her about 25/50 = 0.5 yr to go and to come back, and her biological clock will have advanced by 1 yr. From the point of stay-on-earth Giacomo, Elena's clock is 50 times slower than his own and, since she has to cover 50 lyr, she will have aged 1 yr compared to his 50 yrs. So Elena and Giacomo agree in what they observe.
The difference resides in the fact that the clocks at rest on the Earth and on STAR can be synchronised but they cannot be synchronised to the clock on the moving spacecrafts A and B. When she will reach the station, despite having synchronised the clock on Earth, her clock will be 25 yrs ahead and a similar effect is found on the return travel. Indeed the traveling person ages slower.
There is experimental evidence for time dilation in special relativity and also in general relativity. In 1941 Rossi and Hall detected cosmic-ray muons at the summit and base of Mount Washington in New Hampshire. The muon and anti-muons have mean lifetime of $$\tau =$$2.2 μs, when the decay happens at rest, and decay in the channels: $$\mu^- \rightarrow e^- \bar{\nu}*e \nu*\mu$$ and $$\mu^+ \rightarrow e^+ \nu\_e \bar{\nu}*\mu$$. The time of flight at the speed of light between the top and bottom of the mountain, across the distance of $$\Delta x\sim$$ 2 km for vertical muons, is $$\Delta x/c$$ \~7 μs. In the absence of relativistic effects, the ratio = flux at the top of the mountain/flux at the bottom should have been 22, while they found 1.4. This meant that many more muons than expected where surving and that the “clock” of this nuclear decay process was dramatically slowed down by the motion of the muons. As a matter of fact, supposing that the energy of muons is $$E*\mu \sim 2$$GeV, the muon Lorentz factor is $$\gamma = \frac{E\_\mu}{m\_\mu c^2} = \frac{2 \times 10^3}{105} \sim 19.$$ Hence, they can reach the Earth without decaying over a distance $$\Delta x' = \gamma c \tau = 12.5$$km.
**The Hafele-Keating Experiment**: In 1971, J.C. Hafele and R.E. Keating of the U.S. Naval Observatory brought atomic clocks on board commercial airliners and went around the world, once from East to West and once from West to East. They observed that there was a discrepancy between the times measured by the traveling clocks and the times measured by similar clocks that stayed at the lab in Washington. The result was that the East-going clock lost an amount of time $$Δt\_E = −59 \pm 10 \rm ns$$ , while the West-going one gained $$Δt\_W = 273 \pm 7 \rm ns$$ . The East going airplane travels in the same direction of Earth and so its velocity relative to Earth is greater and hence was more delayed. This establishes that time is not universal and absolute. Causality is preserved (no room for paradox like killing one own’s grandmother) since the travel of three days is much longer than these effects. On the other hand, the asymmetry of the results, with $$∣Δt\_E∣ \neq ∣Δt\_W∣,$$ implies that time's rate of flow also changes with height in a gravitational field. Technology has now progressed to the point where such effects have everyday consequences. The satellites of the Global Positioning System (GPS) orbit at a speed of $$1.9 \times 10^3$$ m/s, an order of magnitude faster than a commercial jet, and at altitude of 20'000 km i much greater than that of an aircraft. The atomic clocks aboard the satellites are tuned to a frequency of 10.22999999543 MHz, which is perceived on the ground as 10.23 MHz due to Doppler shift.
### Lorentz transformations of velocity
The velocity transformations are obtained differentiating the above coordinates:
$$dt' = \frac{dt - \frac{v}{c^2}dx}{\sqrt{1-\frac{v^2}{c^2}}}\ dx' = \frac{dx - vdt}{\sqrt{1-\frac{v^2}{c^2}}} \ dy' = dy\ dz' = dz\\$$ and then from the velocity definition:
$$V'\_x = \frac{dx'}{dt'} = \frac{dx - vdt}{dt - \frac{v}{c^2}dx} = \frac{V\_x-v}{1-\frac{vV\_x}{c^2}}\ V'\_y = \frac{dy'}{dt'} = \frac{V\_y \sqrt{1-\frac{v^2}{c'^2}}}{1-\frac{vV\_x}{c^2}}\ V'\_z =\frac{dz'}{dt'} = \frac{V\_z\sqrt{1-\frac{v^2}{c^2}}}{1-\frac{vV\_x}{c^2}}\\$$
The above Lorentz transformation of velocity was calculated for K and K' with parallel x-axis and velocity $$\vec{v}$$of the K' frame with respect to K along it. We can generalize this to any direction of velocity $$\vec{v}$$ . We call$$V\_{||}$$ and $$V\_{\perp}$$the parallel and perpendicular components to $$\vec{v}$$ , respectively, of an object with velocity$$\vec{V}$$and ask ourselves how$$\vec{V}$$ will transform in the moving frame K':
The equations that provide this transformation are:
$$V'*{||} = \frac{V*{||}-v}{1- \frac{vV\_{||}}{c^2}} = \frac{V \cos\theta-v}{1- \frac{vV\cos\theta}{c^2}}$$ and $$V'*{\perp} = \frac{V*{\perp}}{\gamma \left(1- \frac{vV\_{||}}{c^2} \right)} = \frac{V\sin\theta}{\gamma \left(1- \frac{vV\cos\theta}{c^2} \right)}$$ .
V can be obtained reversing the sign of above equations. Hence, the directions of the velocities are given by:
$$\cos\theta = \frac{\cos\theta' + \beta}{1+\beta \cos\theta'}$$ and $$\sin\theta = \frac{\sin\theta'}{\gamma (1+\beta \cos\theta')}$$
and so $$\tan\theta =\frac{V\_{\perp}}{V\_{||}} =\frac{V'\sin\theta'}{\gamma (V'\cos\theta' + v)}$$. For light this formula reads: $$\tan\theta =\frac{c\sin\theta'}{\gamma (c \cos\theta' + v)} = \frac{\sin\theta'}{\gamma(\cos\theta' + \beta)}$$ .
**Aberration of light and beaming**
Aberration is the apparent change in the direction of a moving object when the observer is also moving. For instance, immagine that we are at rest in the spacecraft and we see light coming from every direction from stars. If the spacecraft where we are starts speeding, at relativistic speeds the whole field of view shrinks and even photons coming from behind, look as coming from the forward direction. Also in our common experience if we rush while raining, we need to incline our ombrella since rain will come not anymore from the vertical but against us.\
For a photon emitted in K', the comovig reference frame with the source, at $$\theta' = 90^\circ$$ we obtain in K that: $$\tan\theta = \frac{1}{\gamma \beta}$$ and for highly relativistic speeds $$\beta \sim 1$$ and $$\gamma >> 1$$ we have $$\theta \sim \frac{1}{\gamma}$$, so that in K photons appear as if the come from a small cone with half-opening $$\theta \sim \frac{1}{\gamma}$$. Light aberration causes most of the photons to be emitted along the object's direction of motion. In K, an isotropic source in K’, would appear emitting photons in a cone of semi-aperture 1/γ. Notice that relativistic aberration can modify the apparent luminosity of jets from Active Galactic Nuclei and Gamma-Ray Bursts.
{% content-ref url="../acceleration-of-cosmic-rays/source-of-extragalactic-cosmic-rays" %}
[source-of-extragalactic-cosmic-rays](https://astroparticle.gitbook.io/docs/acceleration-of-cosmic-rays/source-of-extragalactic-cosmic-rays)
{% endcontent-ref %}
### Classical and relativistic Doppler shift
The Doppler effect is the change in frequency or wavelength of the emitted waves by a moving source with respect to an observer. When the speed of the source and of the receiver are non relativistic $$v\_s, v\_r <<$$ c, the relationship between emitted and observed frequency is: $$f\_{obs} = \frac{c+v\_r}{c+v\_s} f\_{source}$$ , where in general c is the speed of waves in the medium, $$v\_r$$ is the velocity of the receiver relative to the medium (>0 if receiver goes towards the source, <0 if the receiver receeds) and $$v\_s$$ is the velocity of the source relative to the medium (>0 if the source receeds from the receiver, <0 if the source goes towards the receiver). Notice that $$f\_{obs} < f\_{source}$$ if either is moving away ( $$v\_r < 0$$ if the receiver moves away or $$v\_s > 0$$ if the source moves away). If $$v\_s, v\_r <<$$ c, one can consider the series developments: $$f\_{obs} = (1+\frac{v\_r}{c})(1 + \frac{v\_s}{c})^{-1} f\_{source} \sim (1+\frac{v\_r}{c})(1 - \frac{v\_s}{c}) = \left( 1 + \frac{v\_r - v\_s}{c} \right)f\_{source} = \left( 1 + \frac{\Delta v}{c} \right)f\_{source}$$where if the source and receiver move away (towards) the other $$\Delta v = (v\_r - v\_s) < 0$$ (>0). Nice graphical representations are in [Wikipedia](https://en.wikipedia.org/wiki/Doppler_effect).
### Relativistic Doppler shift
A source at rest in $$K$$ emits periodic pulses with angular frequency $$\omega\_s$$ and period $$\Delta t\_s = \frac{2\pi}{\omega\_s} =1/\nu\_s$$, where $$\nu\_s$$ is the frequecy in Herz (sometimes also indicated with f ) in Hz in the MKS or CGS systems of units. The source moves at speed V with respect to the an observer in the referece frame $$K$$(the telescope). What is the frequency perceived by the observer ? The period is streched by **time dilation** in the frame where the source moves: $$\Delta t = \gamma \Delta t\_s = \frac{2\pi \gamma}{\omega\_s}$$ . Looking at the image, in the time $$\Delta t$$ the source moves from point 1 to 2 across a distance $$V\Delta t$$ . The light emitted at 1 has to travel the additional distance $$d = V \Delta t \cos\theta$$ with respect to the light ray emitted in 2. For the observer, the first signal appears to be emitted at $$t\_{1,A}$$ and has to travel longer by d/c and the second signal (happening fter a period) is emitted at $$t\_{2,A} = t\_{1,A} + \Delta t$$. Hence, the difference in apparent times is: $$\Delta t\_A = t\_{1,A} + \Delta t - t\_{1,A} - \frac{V\Delta t\cos\theta}{c} = \Delta t (1 - \frac{V}{c}\cos\theta) = \gamma \Delta\_s (1 - \frac{V}{c}\cos\theta).$$
Hence, the apparente frequency to the observer is:
$$\omega\_o = \frac{2\pi}{\Delta t\_A} = \frac{2\pi}{\gamma \Delta t' (1-\beta \cos\theta)} = \frac{\omega'}{\gamma (1-\beta \cos\theta)}$$
The relativistic Doppler formula can be written as
$$
\omega\_{o}= \frac{\omega\_{s}}{\gamma(1-\beta \cos\theta)}
$$
where *s* stands for '*source*' (equivalent to primed variables) and *o* stands for '*observer*'. We can define the so called 'Doppler factor': $$\delta = \frac{1}{\gamma (1-\beta \cos\theta)}$$ so that $$\omega\_o = \delta \omega\_s$$ and its inverse is
$$
\omega\_{s}= \omega\_{o}{\gamma(1-\beta \cos\theta)} = \frac{\omega\_o}{\delta}
$$
Given that $$\omega = 2\pi \nu \Rightarrow \nu\_o = \delta \nu\_{s}$$. Notice that in general if we call with $$\vec{n}$$ the unit vector going out of the telescope along the line of sight , $$\theta$$ is the angle between the line of sight and the speed vector $$\vec{\beta}$$ and so $$1+ \vec{n} \cdot \beta = 1 -\beta \cos\theta.$$ For $$\gamma >> 1, \theta$$ is a small angle and $$\cos\theta \sim 1 - \frac{\theta^2}{2}$$ and $$\beta = \sqrt{1-\frac{1}{\gamma^2}} \sim 1 -\frac{1}{2\gamma^2} \Rightarrow 1-\beta \cos\theta \sim 1- (1-\frac{\theta^2}{2})(1-\frac{1}{2\gamma^2}) \sim \theta^2 + \frac{1}{2\gamma^2} \Rightarrow \delta \sim 2\gamma \Rightarrow \nu\_s \sim 2 \gamma \nu\_o .$$
All this reasoning simplifies a lot for unidimensional motion of observer and receiver. Below we show a simple graph showing the effect: let's imagine a galaxy moving away from an observer galaxy at speed $$v$$, its emitted light wavelength is redshifted.
If the motion is alog the line of sight and source moving away from observer we are in the case above of $$\theta = 180^\circ$$and the problem can be treated very simply as we remind below: After a time $$t\_s$$ the source has receeded by $$vt\_s$$. The crest of the wave emission is at $$\lambda\_s+v t\_s=ct\_s$$. The period in the reference system of the source is given by:
$$
t\_s = \frac{\lambda\_s}{c-v} = \frac{c}{(c-v)f\_s} = \frac{1}{(1-\beta)f\_s},
$$
For an observer however time passes differently. Remember that when a reference system, $$O\_{s}$$, is moving at speed $$\beta$$ from a another reference system, $$O\_{o}$$, the time relation is given by the relativistic expression,:
$$
\Delta t\_s = \gamma(\Delta t\_{o} - \beta \Delta x\_{o}),
$$
Since $$\Delta x\_{o} =0$$ (we are not moving in the observable reference system, we are just measuring when the crests of the waves arrive), then the time observed $$t\_{o}$$ in the reference system of the observer is given:
$$
t\_o = \frac{t\_s}{\gamma},
$$
and the corresponding observed frequency:
$$
f\_o = \frac{1}{t\_o} = \gamma (1-\beta) f\_s = \sqrt{\frac{1-\beta}{1+\beta}},f\_s,
$$
An important aspect related to the redshift is addressed in another section and concerns
###
For $$\theta = 0$$ the source moves towards the observer: $$\omega\_{s}= \omega\_{o}\sqrt{\frac{1-\beta}{1+\beta}}$$ , the source moves towards the observer and $$\omega\_{o} > \omega\_{s}$$ (**blue shift** since the frequency appears higher than the emitted one and the wavelength smaller);
For $$\theta = 180^\circ$$ the source moves away from the observer $$\omega\_{s}= \omega\_{o}\sqrt{\frac{1+\beta}{1-\beta}}$$ and $$\omega\_{o} < \omega\_{s}$$(**red shift** since the frequency appears lower than the emitted one and the wavelength larger). Similar equations can be written in terms of the frequency and wavelength.
### Redshift, Blueshift
> You ain't never been blue; no, no, no,\
> You ain't never been blue.
>
> *Duke Ellington*
We call redshift and blueshift to the shift of light towards longer or shorter wavelengths. In astronomy, it is often used the quantity, called redshift, defined as:
$$
z=\frac{\lambda\_{o} - \lambda\_{s}}{\lambda\_{s}} ,
$$
where $$\lambda\_{s}$$ is the wavelength emitted by a source of light, and $$\lambda\_{o}$$ is wavelength observed by any particular observer. Light shifts towards red when $$z>0$$, for negative values of the redshift, in really light is shifting towards blue ($$z<0$$). The same quantity, rather being expressed in wavelength, can be expressed using frequency:
$$
z = \frac{\lambda\_{o} - \lambda\_{s}}{\lambda\_{s}} = \frac{f\_{s} - f\_{o}}{f\_{o}} = \sqrt{\frac{1+\beta}{1-\beta}} - 1.
$$
Hence, we can use the definition to rewrite for the Doppler effect for the source and observer moving away one from the other and write the *Doppler factor*:
$$
\frac{\lambda\_{o}}{\lambda\_{s}} = \frac{f\_{s}}{f\_{o}} = \sqrt{\frac{1+\beta}{1-\beta}}
$$
For the non relativistic case ($$\beta << 1$$ or $$v \ll c$$):
$$
\lambda\_{o} = \sqrt{\frac{1+\beta}{1-\beta}}\lambda\_{s} \simeq (1+\beta)\lambda\_{s}
$$
{% hint style="success" %}
Note that in the non-relativistic limit ($$v \ll c$$) the redshift can be approximated by $$z \simeq \beta = v/c$$ which is the classical Doppler effect.
{% endhint %}
### Some relativistic invariants
These are quantities that do not change under Lorentz transforations between inertial frames.
1. **Total emitted power**: Because the 4-momentum $$(\frac{E}{c}, \vec{p})$$, where $$E = \gamma m c^2$$ (see below and $$\vec{p} = \gamma m \vec{v}$$) has time-component E/c and the time-component of the four vector displacement $$(ct, \vec{r})$$ is ct, boh components transform in the same way. Hence, taking corresponding infinitesimal quantities, dE/dt is an invariant.
2. **Phase space volume**: we consider $$d^3 \vec{x} = dx , dy , dz$$ and $$d^3 \vec{p} = dp\_x , dp\_y , dp\_z$$and $$dV\_{ps} = d^3\vec{x} d^3 \vec{p}$$ . Let's consider particles with small spread in position and momentum but not in energy in K': dE' = 0. We know that for Lorentz contraction of distances along x (K' moves with respect to K with velocity parallel to x): $$d^3 \vec{x} = \gamma^{-1} d^3\vec{x'}$$ . And that $$dp\_x = \gamma (dp'*x + v dE'/c^2) = \gamma dp'*x \Rightarrow dV*{ps} = d^3\vec{x} d^3\vec{p} = \frac{d^3\vec{x'}}{\gamma} \gamma d^3\vec{p'} = dV'*{ps}$$
3. **Phase space distribution of particles**: the number of particles within a phase element, and so in $$dV\_{ps}$$, is a conserved quantity: dN' = dN and we know that $$dV\_{ps} = dV'*{ps}$$ , hence $$\rho = \frac{dN}{dV*{ps}} = \frac{dN'}{dV'\_{ps}} = \rho'$$ .
4. **Intensity:** Finally, we define the intensity as $$I\_\nu = \frac{2h\nu^3}{c^2}\rho$$.
5. Because $$\rho$$ is an invariant under Lorentz transfrmations between two reference frames in relative motions one with respect to the other with constant velocity, also $$\frac{I\_\nu}{\nu^3}$$ is an invariant and so $$\frac{I\_\nu(\nu)}{\nu^3}=\frac{I'\_\nu(\nu')}{\nu'^3}$$ .
### Apparent luminosity of active galactic nuclei (AGN)
The luminosity of an active galactic nuclei observed at Earth will be higher than the intrinsic luminosity of the jet measured in its rest frame due to **relativistic aberration**, the **Doppler effect** and the **time dilation**. Let's consider a relativistic blob emitted in a jet of an AGN (see [Source of Extragalactic Cosmic Rays](https://astroparticle.gitbook.io/docs/acceleration-of-cosmic-rays/source-of-extragalactic-cosmic-rays) and see the notes [Ghisellini, Radiative processes in high energy astrophysics](http://arxiv.org/pdf/1202.5949.pdf). We compare to a light bulb. In the comoving frame of the emitting light bulb, it is at rest and its emission is isotropic. In the frame of the observer at rest its emission will look to happen is the solid angle indicated in the figure below namely around an angle$$\sin\theta = 1/\gamma$$and and the emitted light is blue-shifted (since the jet goes towards the observer) $$\nu = \delta \nu'.$$ The solid angle can be obtained from the aberrated angle formulas: $$\cos\theta = \frac{\cos\theta' + \beta}{1+\beta\cos\theta}\Rightarrow d\cos\theta = \frac{d\cos\theta' (1+\beta \cos\theta') - \beta d\cos\theta' (\cos\theta'+\beta)}{(1+\beta \cos\theta')^2} = d\cos\theta' \frac{1-\beta^2}{(1 +\beta\cos\theta')^2}$$ . Hence, $$d\Omega = d\Omega' \frac{1-\beta^2}{\left\[1+ \beta \frac{\cos\theta - \beta}{(1-\beta\cos\theta)}\right]^2} = d\Omega' \frac{(1-\beta^2)(1-\beta\cos\theta)^2}{1-\beta\cos\theta+\beta\cos\theta-\beta^2} = d\Omega' \gamma^2 (1-\beta\cos\theta)^2 = \frac{d\Omega'}{\delta^2}$$
Given the definition of monochromatic intensity from the source: $$I\_\nu(\nu) = \frac{dN h\nu}{d\nu dt dA\_\perp d\Omega},$$ which is the photon energy per unit surface, time, frequency and solid angle in $$\rm erg , cm^{-2} s^{-1} Hz^{-1} sr^{-1}$$ , we can estimate how it transforms fom K to K': dN is invariant, $$dA\_{\perp}$$is invariant as well because is the perpendicular area to the direction of motion, $$d\Omega = \frac{d\Omega'}{\delta^2}$$ and $$d\nu = \delta d\nu'$$. So we obtain: $$I\_\nu(\nu) = \delta^3 I'*\nu(\nu').$$ This result can also be obtained from the fact that that $$\frac{I*\nu}{\nu^3}$$ is an invariant (see previous section).
Notice that the integrated intensity over the full speftrum of frequencie transforms as $$I(\nu) = \delta^4 I'(\nu').$$
The luminosity of a source \[erg/s] is isotropically emitted and the observed flux at distance d is (see [Introduction to astronomical quantities](https://astroparticle.gitbook.io/docs/basic-concepts/astro-intro)): $$F = \frac{L}{4\pi d^2}$$ in \[ $$\rm erg cm^{-2} s^{-1}$$ ]. The flux and the luminosity will transform as the Intensity.
### Relativistic Kinematics
Once we know all the formulas describing special relativity we can study particular cases in which particles hit other particles, or particles decay, and things like that. For that we are going to introduce a new notation in dealing with momentum, energy and velocity, the four-vectors.
#### Four-vectors and four-momentum
As already said, the *controvariant four-vector* is $$A^\mu = (A^0, \vec{A})$$, with four components $$A^0, A^1, A^2, A^3$$, which are four quantities that transform in the same way as $$x^\mu$$ under Lorentz transformations. The *covariant four-vector* is $$A\_\mu = (A^0, -\vec{A})$$. The A quantity, like $$ds^2$$ or $$A^2 = A^\mu A\_\mu$$$$= (A^0)^2 -(A^1)^2 -(A^2)^2 -(A^3)^2$$ , invariant under Lorentz transformation, is a *four-scalar* or *Lorentz scalar*.
Remember also that the product of to 4-vectors is:
$$A^\mu B\_\mu = A^0 B\_0 + A^1 B\_1 + A^2 B\_2 + A^3 B\_3$$
Likewise the *momentum 4-vector* has $$p^0 = E/c$$ time componet, and space components $$p^1 = p\_x$$, $$p^2 = p\_y$$, $$p^3 = p\_z$$ or we can write
$$
{\bf P} = (\frac{E}{c}, \vec{p})
$$
where the total energy of a particle of mass m is $$E =E\_{tot} = \gamma m c^2$$ and the momentum is
$${\vec p}= \gamma m \vec{v} = \gamma m \vec{\beta} c = \frac{E \vec{\beta}}{c}$$ .
The 4-momentum definition comes from the *4-velocity* definition. In a 4D space a moving particle is described by 4 coordinates and its trajectory (the *world line*) is as a function of a parameter that changes along it. A possible choice of this parameter is the *proper time* that gives the spacetime distance τ measured positively or negatively from some arbitrary starting point: $$x^\mu = x^\mu(\tau)$$. The proper time is related to the time by the relation of the dilatin of time, with the proper time in the reference frame at rest: $$\tau = \frac{t}{\gamma}$$. The proper time is measured by clocks carried along the world line.
The 4-velocity **u** is defined through its components which are the derivative of the position along the world line with respect to the proper time: $$u^\mu \equiv \frac{dx^\mu}{d\tau}$$. The four-vector u(τ) at any point along the particle’s world line is the unit timelike tangent four-vector at that point. Notice that $$u^1 = \frac{dt}{d\tau} = \gamma$$ and for the space part $$u^2 = \frac{dx}{d\tau} = \gamma v\_x,$$ and so on for the other two components. Hence, $$u^\mu =(\gamma, \gamma \vec{v})$$. From this we obtaint he 4-momentum defined above.
The 4-momentum squared is a Lorentz invariant is:
$$P^2 = p\_\mu p^{\mu} = m^2 c^2 = \left( \frac{E}{c} \right)^2 -{\vec p\cdot \vec p} \rightarrow E\_{tot} = \sqrt{p^2c^2 +m^2 c^4}$$
The kinetic energy is
$$E\_{kin} \equiv E\_{tot} - m c^2 = (\gamma -1) m c^2$$
and the rest energy of is $$E\_0 = mc^2$$ , which indicates the **equivalence between mass and energy.**
Notice that : $$P^2 = \left( \frac{E}{c} \right)^2 -p^2 = m^2 \gamma^2 c^2 -m^2 \gamma^2c'2 \beta^2 = m^2 \gamma^2 c^2 (1-\beta^2) = m^2 c^2$$ which is an invariant scalar.
#### Transformation between the laboratory and the Center-of-Momentum (COM) frames
The center-of-momentum frame (COM) is the *unique* inertial coordinate frame where the total three-momentum $$P' \equiv (\frac{E'}{c}, {\bf 0})$$, where quantities with the apex are in the COM frame. To have an idea of what are the COM frame and the laboratory one we can consider the scattering process bewteen two particles in the image:
The Lorentz transformation provides the energy and momentum in the COM frame moving with velocity $$\vec{\beta}$$ with respect to the laboratory frame:
$$\begin{pmatrix} E'/c\ p'*{||} \end{pmatrix} = \begin{pmatrix} \gamma & -\beta\gamma \ -\beta\gamma & \gamma \end{pmatrix} \begin{pmatrix} E/c\ p*{||} \end{pmatrix}$$ and $$p'*T = p\_T$$ , where $$p*{||}$$ and $$p\_{T}$$ are the parallel and perpendicular components of the momentum with respect to the speed $$\vec{\beta}$$ also called the *boost*.
In the COM frame the invariant square of the 4-momentum is equal to the total COM energy squared : $$P'^2 = p'^\mu p'\_{\mu} = \left( \frac{E^2}{c^2}' \right)$$. Because this is a scalar invariant, it will be equal to the square of the 4-momentum in the laboratory frame:
$$P'^2 = P^2 \Rightarrow \frac{E'^2}{c^2} = \frac{E'^2}{c^2} - \vec{p}^2$$
Notice that the Lorentz transformation from the laboratory frame to the COM is: $$p^\mu = \Lambda^\mu\_\nu p'^\nu = (\gamma \frac{E'}{c}, \gamma \frac{\vec{\beta}E'}{c})$$. Notice that from these transformations we also get that $$\gamma = \frac{E}{E'}$$ and the *boost* is $$\vec{\beta} = \frac{\vec{p}c}{E}$$.
#### Comparison of a fixed target and collider collision
Let's consider the collision between two particles where particle 2 is at rest. In the lab frame: $$P\_1 = (\frac{E\_1}{c}, \vec{p}*1)$$ and $$P\_2 = (m\_2 c, 0)$$. Hence, the total 4-momentum in the lab is: $$P*{tot} = (\frac{E\_1}{c}+m\_2 c, \vec{p}\_1)$$.
Becasue of the 4-momentum-squared invariance and because of the definition of COM frame $$P'\_{tot} = (\frac{E'}{c},0)$$, we obtain:
$$(\frac{E'}{c})^2 = (\frac{E\_1}{c}+m\_2c)^2 - p\_1^2$$(\*) $$\Rightarrow E'^2 = E\_1^2 + m\_2^2 c^4 + 2m\_2 c^2 E\_1 - p\_1^2 = (m\_1^2 + m\_2^2)c^4 + 2m\_2 c^2 E\_1$$
We can see that E' grows with the energy of the projectile particle $$E\_1$$. The boost is given by:
$$\vec{\beta} = \frac{\vec{p}\_1 c}{E} = \frac{\vec{p}\_1 }{\frac{E\_1}{c} + m\_2 c} = \frac{m\_1 \gamma\_1 \vec{v\_1} }{m\_1 \gamma\_1 c + m\_2 c}$$.
We can also see that if $$E\_1 >> m\_1 c^2, m\_2 c^2$$ , then $$E' \sim \sqrt{2 m\_2 c^2 E\_1}$$, this is the case for a **fixed target experiment** (like for a cosmic ray hitting a nucleus in the atmosphere with energy >> of the mass of the nucleus!). Notice that we can also immediately find what is the minimum energy of particle 1 (threshold energy) to produce a new particle of mass $$m\_3$$ at rest: from equation (\*) :
$$m\_3^2 c^2 = (\frac{E'}{c})^2 = (\frac{E\_1}{c}+m\_2c)^2 - p\_1^2 \Rightarrow$$ $$m\_3^2 c^4 = (m\_1^2 + m\_2^2) c^4 + 2m\_2 E\_1c^2$$
Hence, $$E\_1 = \frac{(m\_3^2-m\_1^2 - m\_2^2)c^2}{2m\_2 }$$ . The threshold energy is given by the difference of the square-mass of the final state particles minus the sum of the projectile and target masses squared, divided by two times the mass of the target particle.
**Application:** *What is the equivalent energy to the LHC 14 TeV run for cosmic rays?* We consider a proton hitting an atmospheric nucleus, ofr which we take $$\ = 14.5$$, the average number of nucleons considering the atmospheric composition, dominated by Nytrogen atoms. In LHC the energy in the COM for the beam of protons and antiprotons is E' = 14 TeV. Because 7 TeV >> 1 GeV (the proton mass) and of the nucleus mass of about $$\ m\_p\sim 14.5$$ GeV, we estimate that the equivalent energy to LHC COM energy is $$E' \sim \sqrt{2 \ m\_p E}$$ $$\Rightarrow E \sim \frac{E'^2}{2 \ m\_p}\sim \frac {(14 \times 10^3)^2}{2 \times 14.5 GeV} \sim 7 \times 10^6 GeV$$. This can be seen in the following plot by T.K. Gaisser \[]
For a **collider experiment**, instead, we consider two beams with particles 1 and 2 with energies $$E\_1$$ and $$E\_2$$ , and 3-momenta $$\vec{p}\_1$$ and $$\vec{p}\_2$$. Their 4-momenta in the lab frame are: $$P\_1 = (\frac{E\_1}{c}, \vec{p}\_1)$$ and $$P\_2 = (\frac{E\_2}{c}, \vec{p}*2)$$ and the two beams are equal and opposite so that: $$\vec{p}*1 = -\vec{p\_2}$$ . But this means that the total momentum is zero both in the lab frame and in the COM one by definition. Hence, the lab and COM frame coincide and $$P'*{tot} = P*{tot} = (\frac{E\_1+E\_2}{c}, {\bf 0})$$. Hence, if the two colliding beams have the same energy $$E\_1 = E\_2 = E \Rightarrow$$ E' = 2 E. So this is a more efficient way to accelerate particles since $$E' \propto 2 E$$ and not $$E' \propto (2E')^{½}$$. Notice that the threshold energy to produce the particle of mass $$m\_3$$ is immediately given by the condition : $$E\_1+E\_2 = m\_3 c^2$$ .
We can solve the same problems in the general case :
**Threshold energy:** Which is the energy needed for particle 1 to **create a new particle at rest of mass** $$m\_3$$**?**
In general for two particless we have for c = 1:$$\begin{aligned} s = (E\_1 + E\_2)^2 - (\vec{p}\_1 + \vec{p}\_2)^2 = m\_1^2 + m\_2^2 + 2E\_1E\_2 - 2(\vec{p}\_1 \cdot \vec{p}*1)= \ = m\_1^2 + m\_2^2 + 2E\_1E\_2( 1 - \beta\_A\beta\_B\cos \theta)= E*{COM}^2 \end{aligned}$$
where we used the fact that: $$\vec{p}\_1 \cdot \vec{p}\_2 = |\vec{p}\_1||\vec{p}\_2|\cos\theta = E\_1 E\_2 \beta\_1 \beta\_2 \cos\theta$$
since $$|\vec{p}| = E\beta$$.
The energy available for new particle creation is $$\epsilon = E\_{COM} - m\_1c^2 - m\_2 c^2.$$ If $$E\_1 \gg m\_1 c^2$$ and $$E\_2 \gg m\_2 c^2$$ then $$\epsilon^2 \approx 2 (E\_1 E\_2 - \vec{p}\_1 \cdot \vec{p}\_2 )$$
#### Fixed-target experiment
If the target particle 2 is at rest in the laboratory system then $$E\_2 = m\_2 c^2$$ and $$\vec{p}\_2 = 0$$ . In this case,
$$E\_{COM}^2 = m\_1^2 c^4 + m\_2^2 c^4 + 2E\_1m\_2 c^2$$
which in the ultra-relativistic limit where energies are much higher than the masses $$E\_1 \gg m\_1, m\_2$$ simplifies to
$$\epsilon = E\_{COM} -m\_1 c^2 -m\_2 c^2 \simeq \sqrt{2m\_2 c^2 E\_1}$$.
Equivalently, the threshold energy of the beam particle 1 needed to produce a particle of mass $$m\_3$$ at rest in the boosted frame is:
$$E\_{1,\mathrm{thresh}} = \frac{m\_3^2 c^4}{2m\_2 c^2}$$
#### Collider experiment
In the case of a collider experiment where beam particles 1 and 2 are counter-circulating in an accelerator and collide head-on, then
$${\vec p}*{1}\cdot{\vec p}*{2} = -|{\vec p}*{1}||{\vec p}*{2}|$$
and $$\begin{align} s = E\_{COM}^2 = m\_1^2 + m\_2^2 + 2(E\_1E\_2 + |{\vec p}*{1}||{\vec p}*{2}|) \rightarrow \epsilon^2 \simeq 4E\_1E\_2 \end{align}$$
in the relativistic limit where mass can be ignored. This in turn has the consequence that in a collider experiment the energy available in the COM to produce new particles rises linearly with beam energy when $$E\_1 = E\_2$$.
**Example 1:** **Photo-pion production** : $$\gamma + p \rightarrow p + \pi^0$$ or : $$\gamma + p \rightarrow n + \pi^+$$ . The intermediate state of this reaction is a $$\Delta$$ resonance that does not live long and desintegrates again into a pion and nucleon. This is the reaction that causes the *GZK cut-off* of the cosmic rays at the extreme high energy range above $$10^{19}$$ eV when they interact with the cosmic microwave background. We estimate the threshold energy considering that the pion and the nucleon are produced at rest. The COM frame is the one where proton and photon are one against the other with total momentum : $$P'*{tot} = (E', {\bf 0})$$ . The lab is the rest frame of the proton: $$P*{tot} = (m\_p + E\_\gamma, \vec{p}\_\gamma)$$ .
As derived above, the energy in the COM frame is:
$$\begin{align} E' = \sqrt{m\_p^2 + 2E\_\gamma m\_p} = m\_p + m\_{\pi^0}\Rightarrow m\_p^2 + 2 E\_\gamma m\_p = m\_p^2 + m^2\_{\pi^0} + 2 m\_p^2 m^2\_{\pi^0} \end{align}$$
$$
E\_{\gamma,thresh} =m\_{\pi^0} c^2 + \frac{m^2\_{\pi^0}c^4}{2 m\_p c^2} \approx 145 {\rm ; MeV}
$$
This is the energy threshold energy of photons in the rest frame of the proton (the lab). What should be the minimum energy of a cosmic ray proton to photoproduce? We need to boost this energy to the frame of the traveling cosmic ray and we can derive the Lorentz factor from the energy of the radiation field of the cosmic microwave radiation which has an effective temperature of $$T^{eff}*{CMB} \sim 2.73 ^\circ K.$$ The effective energy (the average energy of the CMB radiation that follows a black body spectrum with $$T^{eff}*{CMB}$$) is $$E'*\gamma \sim 3 k T^{eff}*{CMB} \sim 3 \times 8.62 \times 10^{-5} \times 2.73 \sim 7 \times 10^{-4}$$ eV, where we used the value of k = Boltzmann constant. Hence, from the Lorentz transformation of the energy of the photons we obtain: $$\gamma = \frac{E\_{thresh,\gamma}}{E'*\gamma} = \frac{145 \times 10^6}{7 \times 10^{-4}}\sim 2 \times 10^{11}.$$ Hence, the traveling proton (COM frame), which interacts with the CMB radiation field with energy $$E'*\gamma$$ , will need to have an energy larger than
$$E'\_p = \gamma m\_p \sim 2 \times 10^{11} \times 10^9 \sim 2 \times 10^{20}$$ eV.
This value is a bit higher than what we would find if we wuold use not the average value 3 k T of the black body spectrum but the exact distribution of energy. See also below an alternative formulation.
#### Example 2: creation of particle 3 from particle 1 and 2 (no photons)
E.g. consider the reaction happening when cosmic rays interacto with the interstellar matter (mostly hydrogen) in the galactic plane $$p + p \rightarrow p + p + \pi^0$$ when one initial p is at rest ( $$\vec{p}\_2 = 0, \gamma\_2 = 1$$ ).
The threshold energy in the COM must be $$E\_{COM} = m\_1c^2 + m\_2 c^2 + m\_3 c^2$$
For invariance:
$$
(m\_1c^2 + m\_2 c^2 + m\_3 c^2)^2/c^2 =m\_1^2 c^2 + m\_2^2 c^2 + 2 (E\_1 E\_2/c^2 - \vec{p}\_1 \cdot \vec{p}\_2)
$$
$$
m\_3^2c^2 + 2m\_1 m\_2 c^2 +2 m\_3 ( m\_1 + m\_2) c^2 = 2 (E\_1 E\_2/c^2 - \vec{p}\_1 \cdot \vec{p}\_2)
$$
We obtain the formula (\*):
$$
E\_1E\_2/c^2 - p\_1p\_2 \cos\theta = m\_1m\_2 c^2 + m\_3 c^2 (m\_1+m\_2 + m\_3/2)
$$
Notice that:
$$
E\_i = m\_i \gamma\_i c^2; p\_i = m\_i \beta\_i \gamma\_i c = m\_i c \sqrt{\gamma\_i^2-1}; \beta\_i = \frac{\sqrt{\gamma\_i^2-1}}{\gamma\_i}
$$
Hence:
$$
\gamma\_1 \gamma\_2 - \sqrt{\gamma\_1^2-1}\sqrt{\gamma\_2^2-1} \cos\theta =1 + m\_3 \left(\frac{1}{m\_1}+\frac{1}{m\_2}+ \frac{m\_3}{2m\_1m\_2}\right)
$$
and for our case ( $$\gamma\_2 = 1$$ ), $$\gamma\_1 = m\_{\pi^0} \left(\frac{2}{m\_p} + \frac{m\_{\pi^0}}{2m\_p^2}\right) \sim 1.3$$ since $$m\_p \sim 938 \rm MeV/c^2, m\_{\pi^0} \sim 135 \rm MeV/c^2$$ and $$E\_{p, kin th} =(\gamma\_1 - 1) m\_p c^2 \sim 280 \rm MeV.$$
#### Example 2: creation of particle 3 from particle 1 and 2 (a photon)
If particle 2 is a photon ( $$E\_2 /c= p\_2, m\_2 = 0)$$ , formula (\*) becomes :
$$
E\_2/c (E\_1/c - p\_1\cos\theta) = m\_3 c^2 (m\_1+ m\_3/2)
$$
$$
E\_2/c \cdot m\_1c (\gamma\_1 - \sqrt{\gamma\_1^2-1} \cos\theta) = m\_3 c^2 (m\_1+m\_3/2)
$$
We reconsider the case of the GZK interaction : $$p+\gamma \rightarrow p+\pi^0$$ (through the delta production), on CMB photons with $$E\_2 \sim 3 k T^{eff}\_{CMB} \sim 7 \times 10^{-4} \rm eV$$ and we obtain for most favourable collisions (head-on $$\cos\theta = -1$$ and $$\gamma\_1 >> 1$$ ):
$$
2\gamma\_1 \sim \frac{m\_{\pi^0}c^2}{3KT\_{CMB}^{eff}}\left(1 + \frac{m\_{\pi^0}}{2m\_p}\right) \Rightarrow \gamma\_1 \sim 2 \cdot 10^{11}
$$
This reaction is responsible for the attenuation of UHECRs above $$E\_{CR} = (\gamma\_1 -1) m\_pc^2 \sim 10^{20}\rm eV$$ as depicted in [Bietenholz, 2013](https://arxiv.org/pdf/1305.1346.pdf).
We can define the cooling time, to derive the **horizon** accessible to UHECRs:
$$
t\_c(E) \equiv \frac{E}{{\rm d}E/{\rm d}t} \sim \frac{E}{{\Delta}E/{\Delta}t} = \frac{E}{\frac{\Delta E}{\Lambda/c}} = \frac{E \Lambda}{k\_{p\gamma}E c}=\frac{1}{k\_{p\gamma} n\_\gamma \sigma\_{p\gamma} c}
$$
where $$\Lambda = c\Delta \tau = \frac{1}{n\sigma}$$ is the mean free path and $$k\_{p\gamma} = \Delta E/E \sim 0.2$$ is the inelasticity or fraction for pions.
#### Example 2: Pair production
The process can happen with 2 photons producing a pair of electrons: $$\gamma + \gamma \rightarrow e^+ + e^-$$
The threshold for this reaction (from eq. (\*) above):
$$
E\_1E\_2/c^2 - p\_1p\_2 \cos\theta = m\_1m\_2 c^2 + m\_3 c^2 (m\_1+m\_2 + m\_3/2)
$$
with $$m\_1 = m\_2 = 0; m\_3 = 2m\_e; E\_{1,2}/c = p\_{1,2}$$
$$
\Rightarrow E\_1E\_2 (1-\cos\theta) = 2m\_e^2 c^4
$$
For head-on collisions and producing the pair at rest : $$E\_1 E\_2 = m\_e^2 c^4$$. For a TeV photon $$E\_1 = 10^{12} \rm eV$$, the interaction is possible with $$E\_2 \ge (0.511)^2 \rm eV \sim 0.26 \rm eV$$(infrared photons).
We consider the reaction: $$\gamma + R \rightarrow e^+ + e^- + R$$ with R = nucleus or proton initially at rest.
From formula (\*):
$$
E\_1E\_2/c^2 - p\_1p\_2 \cos\theta = m\_1m\_2 c^2 + m\_3 c^2 (m\_1+m\_2 + m\_3/2)
$$
with $$E\_1 = E\_\gamma = p\_\gamma c; m\_2 = m\_R, E\_2 = m\_R c^2\gamma\_R; m\_3 = 2m\_e$$ :
$$
E\_{\gamma} m\_R \gamma\_R - E\_\gamma m\_R \sqrt{\gamma\_R^2 -1}\cos\theta = 2m\_e c^2 (m\_R + m\_e)
$$
For R almost at rest $$\gamma\_R \sim 1 \Rightarrow E\_\gamma \ge 2m\_e c^2 (1+ m\_e/m\_R)$$
Therefore, when the interaction involves heavy atoms $$m\_R >> m\_e$$ , and thus in a good approximation photon energies $$E\_\gamma \ge 2 m\_e c^2 \sim 1.022 \rm keV$$ allow the creation of electron-positron pairs. However, for a proton as recoil particle the calculated threshold energy is increased by $$2 \times 511^2/938 = 557 \rm eV$$.
#### Example 3: Decays
In this case, we consider the invariant mass: $$E'^2 = m^2 = p^\mu p\_\mu$$ .
For instance for the pion decay: $$\pi^- \rightarrow \mu^- + \bar{\nu}\_\mu$$ , we have:
$$m\_\pi^2 = (E\_{\mu}+E\_\nu)^2 - (\vec{p}*\mu +\vec{p}*\nu)^2$$
Being in the rest frame of the pion, the second term is null since $$\vec{p}*\mu = -\vec{p}*\nu$$. $$\Rightarrow m\_\pi = E\_\mu + p\_\mu$$ , since $$E\_\nu \sim p\_\nu = p\_\mu$$ given that we consider negligible the mass of neutrinos $$m\_\nu \sim 0$$. Hence, $$m\_\pi = E\_\mu + \sqrt{E\_\mu^2 -m\_\mu^2}$$ $$\Rightarrow m\_\pi^2 + E\_\mu^2 - 2 m\_\pi E\_\mu = E\_\mu^2 - m\_\mu^2$$ $$\Rightarrow E\_\mu = \frac{m\_\pi^ 2 + m\_\mu^2}{2m\_\pi}$$ $$\sim \frac{3}{4} m\_\pi$$ and
$$p\_\mu^2 = p\_\nu^2 \Rightarrow E\_\nu^2 = E\_\mu^2-m\_\mu^2 =$$ Hence, $$E\_\nu^2 = \frac{m\_\pi^4 + m\_\mu^4 +2 m\_\pi^2 m\_\mu^2 - 4m\_\pi^2 m\_\mu^2}{4m\_\pi^2} = \frac{(m\_\pi^2-m\_\mu^2)^2}{4m\_\pi^2}$$
$$\Rightarrow E\_\nu = \frac{m\_\pi^2 - m\_\nu^2}{2m\_\pi} \sim \frac{1}{4} m\_\pi.$$
We find that neutrinos take about 1/4 of the pion energy. You can see that for a typical 2-body decay the energy of the resulting 2 particels is monochromatic. Instead for a 3-body decay the spectrum is a continuuom of energy values. For the muon decay which follows the muon production in the pion deacy ( $$\pi^+ \rightarrow \mu^+ \rightarrow e^+ \nu\_e + \bar{\nu}*\mu$$ or $$\pi^- \rightarrow\mu^- \rightarrow e^- \bar{\nu}*e + \nu*\mu$$ ) the neutrinos in the final state take on average a similar fraction of energy ( $$\sim \frac{1}{4} E*\pi$$) as shown in the plot below.
![Fraction of energy carried by the resulting neutrinos from the charged pion chain of decays. Plot from Kelner, Aharonian and Bugayov, 2006 \[https://arxiv.org/pdf/astro-ph/0606058.pdf\] ](https://978429123-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LL-TjWvtGhAa4RFZygl%2F-L_rV6uRNwEviJbTke3u%2F-L_rXmcAmHxRapxQrpC1%2Fmuondecay.png?alt=media\&token=c5b89958-ef0d-4aee-8071-a23575397690)
#### Some useful reference
* J.B. Hartle Gravity Ch 1-3
* L. Bergstrom and A. Goobar, Cosmology and Particle Astrophysics (2nd edition), Springer 2004 Ch 1-2
* High-Energy Astrophysics S. Rosswog and M. Bruggen Ch 1
* R. Hagedorn, Relativistic kinematics, W\.A. Benjamin INC,1963 Ch 1
* G.B. Rybicki and A.P. Lightman, Radiative processes in astrophysics, Wiley & Sons, 1979
* [Special relativity in Wikipedia](https://en.wikipedia.org/wiki/Special_relativity)
* Math Pages: [Reflection on relativity](http://www.mathpages.com/rr/rrtoc.htm) (Kevin Brown)