Cosmic Rays Interactions and Radiative processes

Charge particles undergo several energy loss processes when accelerated in sources or when they propagate in the atmosphere. Additionally, photons are not only radiated but also interact.

Interactions of particles

A very useful reference to understand interaction of particles in the atmosphere or in a detector is the Review in the Data Particle Book (DPP) on the Passage of Particles through matter.

Charged particles undergo multiple interactions of the follwoing kinds: ionization, bremsstrahlung and Cherenkov energy losses. Neutrons and other hadrons undergo nuclear interactions. Photons undergo the following interactions for increasing energy: photoelectric effect, Compton scattering and pair production, when the energy becomes larger than $\sim 2 m_e \sim 1 \rm MeV$. Neutrinos interact only weakly.

Ionization and Excitation

Ionisation is the phenomenon of anelastic collisions of a particle with electrons of much smaller mass. During ionization a charge particle of mass $m >> m_e$, velocity $v = \beta c$, Lorentz factor $\gamma = (1-\beta^2)^{-½}$ and charge $ze$ (in our case a cosmic ray or one of the charge particles generated in a shower) collides with the atomic electrons of the material, where it is passing through. The particle kicks out of their orbits elecrons of mass $m_e$. The ionization energy loss per unit distance $x$ through a medium of density $\rho$, mass number $A$ and atomic number $Z$ for the high energy charged particle in collisions with atomic electrons is described by the Bethe-Bloch formula:

$$-\left\langle\frac{{\textrm d} E}{{\textrm d} x}\right\rangle_{ion}= K z^2 \rho \left(\frac{Z}{A}\right) \frac{1}{\beta^2}\left\{\frac{1}{2} \ln\left[\frac{2m_e c^2 \beta^2 \gamma^2 T_{max}}{I}\right]-\beta^2 - \frac{\delta}{2}\right\}$$

where $K = 4\pi N_A r_e^2 m_e c^2$, with $r_e = \frac{e^2}{4 \pi \epsilon_0 m_e c^2} \sim 2.82 \times 10^{-15} \rm m$ is the classical radius of the electron, $N_A$ is the Avogadro's number, $I$ is the ionization potential of the medium, which was found to be approximatively $I \sim 10 \textrm{\; eV} \cdot Z$, although today we have precise tabulated values for this (e.g. we find I/Z = 13.6 eV for H, 24.6 eV for Helium and 7.9 eV for Fe from Wikipedia). $T_{max} = \frac{2m_e c^2 \beta^2 \gamma^2}{1 + 2\gamma m_e/m + (m_e/m)^2}$ is the maximum energy transfer in a single collision. The low energy approximation is the energy transferred in head on collisions: $T_{max} = 2 m_e c^2 \beta^2 \gamma^2,$ valid for $2\gamma m_e << m$ If we use $X = \rho x$ in $\rm g/cm^2$ we get rid of the density dependence.

By looking at the formula there are some conclusions that we can already take:

• Approximately the relevant terms in the formula are $\propto \frac{1}{\beta^2} \ln(const \cdot \beta^2 \gamma^2)$

• It is independent of the incident particle mass $m$(aside from the weak dependency in the log term).

• Since $Z/A \sim \frac{1}{2}$ in most materials, the energy losses via ionization do not depend much on the medium. Most of the dependence is through I in the logarithmic term.

• At low speed, $- \left\langle\frac{{\textrm d} E}{{\textrm d} x}\right\rangle_{ion} \sim 1/v^2$ until it reaches a minimum, that depends on the incident particle mass. The minimum is at $\beta \gamma = p/mc \sim 3-4$ and corresponds to about $1-2 \rm \,MeV/(g cm^{-2})$. A particle with this energy losses is called a MIP (minimum ionizing particle). In practical cases, most of the cosmic rays can be considered MIPs.

• Then the energy losses increases logarithmically until they reach a plateau value or saturation due to the$\delta$ density correction term. At high energy the transverse component of the electric field increases ( $\gamma E_y$ ) but a polarization effect (density dependent) shields the electrical field far from the particle path, effectively cutting the long range contribution and leading to saturation.

• At low energy the graph below show that ionization losses become very important at low speeds. This has an important impact in the development of an air shower as once the shower particle loose much energy in particle production, ionisation begins to dominate. This typically marks the end of the development of the shower.

Energy loss of an anti-muon in copper from the DPP

Particle identification (PID)

The PID can be realized by the simultaneous measurement of $\frac{{\rm d}E}{{\rm d} x}$ and momentum. See e.g. the figure 2 in this ALICE detector paper.

Left: PID with dE/dx in the TPC gas detector and measurement from the TOF scintillation detectors

Range of particles

The average range is easily obtained integrating the inverse of energy losses along the particles path on energy from the initial energy $E_0$ until the energy is exhausted:

$$R = \int_{E_{0}}^{0} -\left\langle\frac{{\textrm d} x }{{\textrm d} E}\right\rangle_{ion} dE$$

Fluorescence:

So far we only talked about ionization, but soft collision or excitation (an atomic electron jumping into a higher energy state) is also possible. Excitation has a less important role in the development of the air shower, however in these soft collisions cosmic rays can excite fluorescence from nitrogen molecules in the atmosphere. This excitation of nitrogen (called fluorescence) typically produces 5 photons per m track in the blue wavelength region which will make it possible to detect with ground observatories above energies of about $10^{17} \rm eV$ .

Radiation processes

Bremmsstrahlung

In addition to ionization losses, charged particles travelling through a material can also undergo bremsstrahlung, which is a german word meaning braking radiation. Bremsstrahlung, unlike ionization, is a radiative process because energy is lost as photons are emitted. The energy losses due to bremsstrahlung are described by the following equation:

$$\left\langle\frac{{\textrm d} E}{{\textrm d} x}\right\rangle_{rad} = -\frac{E}{X_0}$$

where $X_0$ is the radiation length, where for an incident electron this is given by:

$$\frac{1}{X_0} = 4 \alpha\left(\frac{Z}{A}\right)(Z + 1)^2r_e^2N_0\log\left(\frac{183}{Z^{1/3}}\right)$$

where $r_e$ is the classical electron radius and $\alpha = \frac{e^2}{4\pi\epsilon_0 \bar{h}c} =\frac{1}{137}$ is the fine structure constant. Again there are some considerations we can make:

• Bremsstrahlung is proportional to $\frac{1}{X0} \propto r^2_{e} \propto e^2/(m_e c)^2$. The radiation length of a proton will be $(m_p/m_e)^2 \sim 4 \cdot 10^6$ times that for an electron.

• Bremsstrahlung is proportional to energy. The main message is that while ionization is almost energy independent (except for very low energies), bremsstrahlung is proportional to the energy. Therefore there is an energy in which both energy losses should be the same. This is called the critical energy, $E_c$, and is defined as the energy when the following equality is satisfied:

$$\left\langle\frac{{\rm d}E}{{\rm d}x}\right\rangle_{ion} = \left\langle\frac{{\rm d}E}{{\rm d}x}\right\rangle_{bremss}$$

Above this energy the radiation process dominates, below is the ionization that dominates. For electrons this is roughly $E_c \sim 600/Z \textrm{\;MeV}$, and for the atmosphere this is $E_c \sim 85 \textrm{\;MeV}$. From the DPP, we show the meaning of critical energy and also the critical energy trand for various materials of different $Z.$

The critical energy for copper when Bremsstrahlung dominates over ionisation and its its for solids and liquids (solid line) and gases (dashed line). The rms deviation is 2.2% for the solids and 4.0% for the gases.

Synchrotron radiation

Synchrotron radiation is extremely important for astrophysics as it was realized by Shklovskii in 1957 when studying the non-thermal emission of the Crab remnant. In order to understand better the synchrotron radiation we will have to dig in a bit in electromagnetism. We already saw that a charged particle of charge $q$, for example an electron, moving with velocity $v$ in a magnetic field $\vec{B}$ feels an an external force:

$$\vec{F} = \frac{q}{c}(\vec{v}\times \vec{B})$$

Because of the force on the particle is perpendicular to the motion, the magnetic field cannot do work on the particle, and so its speed does not change, i.e. $|v| = {\rm ct}$, but there is an acceleration since the direction will change. On the other hand an accelerated electrical charge radiates electromagnetic waves which will slow down the particle... so the speed should change... so, what is going on?

The reason of this apparent inconsistency comes from the fact that we treat the electric field lines as the purely Coulombic action-at-a-distance. One has take into account that as a particle moves, the electric field lines need to re-arrange and this re-arrangement cannot happen at a speed faster than the speed of light. Here are going to derive the radiation emission for a particle with an acceleration and we are going to do it by messing around with the Coulomb fields.

For a more formal argument derived from Maxwell equations you can see the book of Longair.

Let's assume that the particle is at rest at the moment $t=0$. The electric field lines clearly point away from the origin. At that moment the particle accelerates which brings the velocity of the particle to $\Delta v$ in a time $\Delta t$, after that the particle continues with uniform velocity.

After a certain time, the particle will be in the position $t\Delta v$. In a sphere located far way (with a radius larger than $c t$), the electric field lines are still those of the stationary particle, since they field lines cannot know yet that the particle has moved, so they point radially towards the origin at $t=0$. Inside a sphere of radius $c(t - \Delta t)$, the electric field lines are already those from the electron that moves at a constant velocity.

The perturbation of the electric field lines needs to propagate radially, this kink is nothing more than a radiation! From simple geometry relations we can get that:

$$\frac{E_\theta}{E_{r}} = \frac{\Delta v t \sin\theta}{c\Delta t} = \frac{r \sin \theta}{c^2}\frac{\Delta v}{\Delta t}$$

where $r = c t$, and $E_r$ is simply the Coulomb field $E_r = q/4\pi\epsilon_0 r^2$ or in gaussian units since the kink is really small and the electric field along the kink remains the same $E_r = q/r^2$ and therefore the transverse component becomes:

$$E_\theta = \frac{q a \sin\theta}{c^2 r}$$

where we used the fact that $\Delta v/\Delta t$ is just the acceleration $a$. An interesting fact appears here, $E_\theta$ depends on $r^{-1}$ and not $r^{-2}$ so for larger $r$, $E_\theta$is going to dominate over $E_r$. Accompanying this transverse electric field there will be a magnetic field, which is a property of an electromagnetic wave. In other words, an electromagnetic pulse is generated by the accelerated charged particle. Since this is an electromagnetic radiation there is an energy flow per unit area, per second and the direction is given by the Poynting vector (with $|E|=|B|$ as in electromagnetic wave):

$$\vec{S} = \frac{c}{4\pi}(\vec{E}\times\vec{B})$$

Which points in the radial direction. In this case can be reduced to:

$$|S| \equiv \frac{ {\rm d}E}{{\rm d}t {\rm d}A} = \frac{c}{4\pi} E^2_\theta$$

Which is the energy flow per unit area per second. The unit area ${\rm d}A$ can be rewritten in terms of the solid angle as ${\rm d} A= r^2{\rm d}\Omega$, and so the rate of energy loss through the area subtended by the solid angle ${\rm d} \Omega$ at distance $r$ is given by:

$$\left(\frac{{\rm d}E}{{\rm d}t {\rm d} \Omega}\right)_{rad} = \frac{q^2a^2 \sin^2 \theta}{4\pi c^3}$$

This energy loss rate follows a dipole pattern ${\rm d}P/{\rm d}\Omega \propto \sin^2 \theta$ and that $\theta$ is defined along the acceleration line. If we now integrate over all solid angles we obtain that the emitted power is given by:

$$P \equiv -\left(\frac{{\rm d}E}{{\rm d}t}\right)_{rad} = \frac{2q^2 a^2 }{3c^3}$$

Which is call the Larmor's formula.

Note that the Larmor's formula is valid for any form of acceleration $a$, perpendicular or parallel.

In the relativistic case the Larmor's formula can be rewritten as:

$$P = \frac{2q^2}{3c^3} \gamma^4 [ \gamma^2 a_{\parallel}^2 + a^2_{\perp}]$$

Case study: Single electron in uniform magnetic field

In a uniform magnetic field, a high energy charged particle, for example an electron, moves in spiral path at a constant pitch angle, $\alpha$.

Its velocity along the field lines is constant $v_{\parallel} = v \cos\alpha = {\rm const.}$, but its circular component $v_\perp = v\sin\alpha$. Let's first attack the non-relativistic case.

Non relativistic case: Cyclotron radiation

From Newton's law and the Lorentz force we have that:

$$m_e a_\perp = m_e \frac{v_\perp^2}{r_{gyr}} = \frac{e v_\perp B}{c}$$

where $r_{gyr}$ is the gyroradius which can be written as:

$$r_{gyr} = \frac{v_\perp}{\omega_{gyr}} = \frac{v\sin\alpha}{\omega_{gyr}}$$

where we can define the gyrofrecuency as:

$$\omega_{gyr} \equiv \frac{eB}{m_e c}$$

Therefore according to the Larmor's equation of power emitted we can write:

$$P = \frac{2 e^2}{3c^3} \omega_{gyr}^2 v^2\sin^2\alpha$$

Note the Larmor's formula does not tell us frequency spectrum, but if a particle is moving in a circular motion, then from an observer far way, the "apparent" motion will be sinusoidal as illustrated in the figure below.

These processes are relevant to understand the spectral emission distribution of a cosmic source (see Radiation Mechanism).

Energy losses of photons

From DPB: Photon total cross sections as a function of energy in carbon (top) and lead (bottom)

The plot above shows the cross section for the contributing processes to the photon energy losses. At low energy the dominant process is photoelectric absorption and electron ejection ( $\sigma_{p.e.}$ ), although Compton scattering ( $\sigma_{Compton}$), Rayleigh scattering ($\sigma_{Rayleigh}$ ), and pair production in the nuclear ( $\kappa_{nucl}$ ) and electron fields ( $\kappa_e$ ) dominate above 1 MeV also contribute. The photoelectric cross section is characterized by discontinuities (absorption edges) as thresholds for photoionization of various atomic levels are reached. The high energy limit of the pair production cross section is $\sigma = \frac{7}{9}(A/(X_0N_A))$. Pair prodution can also happen in a radition field. E.g. the cross section for pair production with photons is shown below as a function of the interacting photon energies. It is in units of Thompson cross section (the cross-section for classical elastic scattering of electromagnetic radiation on free electrons given by $\sigma_T = \frac{8\pi}{3} r_e^2 = 6.65 \rm \, fm^2$ ). The maximum of the cross section is at $\epsilon_1 \epsilon_2 = 4 m_e^2 c^4$, where $\sigma_{\gamma\gamma} \sim 0.25 \sigma_T$.

Pair production cross section in units of Thompson cross section

Particle Physics elements

As discussed in the Particle Physics elements section, the absorption length (shown below) or mean free path $\lambda = 1/(\mu/\rho)$, is obtained as the inverse of the above cross section and depends also on the densitiy of the material $\rho$ where photons propagate. The mass attenuation coefficient is μ. The intensity of the photon beam I remaining after traversal of a thickness t (in mass/unit area) is given by $I = I_0 exp(-t/\lambda)$. Above 1 MeV the threshold of pair production is surpassed and this process dominates (see its derivation in the link below).

Special Relativity and High-Energy Astrophysical Phenomena

From the DPB.