Particle Physics elements

While an astroparticle physics course should come after a particle physics course, in case you did not have this opportunity we remind some relevant concepts.

Short reminders on the Standard Model of Particles

There is an enormous amount of online material for education on particle physics (for instance see the Particle Physics and Information sites). Moreover more than 200 particles are listed with their properties (such as mass and how it is measured, lifetime before decay, decay channels and their corresponding relative probability to happen - called branching ratio) in the data particle book. An example is an extract from the listing of the muon ($\mu$), where we can see its spin, mass with its error of about $m_\mu \sim$ 105 MeV/c$^2$ and its decay lifetime of about 2.2 μs. Also it can be understood that it almost always decays into the electron (which has considerably lighter mass of about $m_\mu\sim$ 0.5 MeV/c$^2$.

Notice that muons are negative particles and in their decay they produce electrons, conserving the charge. Also, muon is a lepton of muonic flavor and the electron into which it decays has electron flavor so you need to conserve also the single flavor leptonic numbers.

Another quantum number is conserved the lepton number. To understand this we need to remind what are leptons and quarks, the building blocks of matter.

There are four interactions, summarized in the following table:

Interaction	Acts on	Relative Strength	Range	Gauge boson	Spin of boson
Strong	Hadrons	1	$10^{-15}$	Gluon	1
Electromagnetic	Electric charges	$10^{-2}$	$\infty$ $(1/r^2)$	Photon	1
Weak	Hadrons and leptons	$10^{-5}$	$10^{-18}$m	Z, W$^{\pm}$	1
Gravity	Mass	$10^{-39}$	$\infty$ $(1/r^2)$	Graviton	2

The carriers of forces are bosons. Gluons come in 3 flavors (red, blue and gree) and for anti-quarks there are anti-colours. The carrier of forces (bosons), fermions and the recently discovered Higgs compose the Standard Model of particles.

We first consider Leptons, which do not interact through the strong force are fermions with spin 1/2 and are organised in 3 families according to the scheme below:

Neutrinos have small masses

Quarks are the constituents of hadrons. They are organised in 3 families as leptons and have fractionay charge and their properties are summarized below with all relevant quantum numbers:

Name	Symbol	Mass (MeV/c $^2$)	Charge Q	Spin J	Baryon number B
Up	u	1.8-2.7	+2/3	1/2	1/3
Down	d	4.4-5.2	-1/3	1/2	1/3
Charm	c	1275	+2/3	1/2	1/3
Strange	s	92-104	-1/3	1/2	1/3
Top	t	173	+2/3	1/2	1/3
Beauty	b	4.190	-1/3	1/2	1/3

The fact that quarks are always confined inside hadrons is a property of the strong force (confinement), which increases in strength the more 2 coloured particles are pulled away from each other. At a certain distance enough potential energy is built up to create another pair of quark-antiquark which recombine in composite particles. In the opposite extreme, the more two quarks get closer together the more the force becomes weaker (asymptotic freedom). In this state a quark-gluon plasma is formed.

Mesons are composed of $q\bar{q}$ and baryons are composed of $qqq$. Mesons can be organised in an octet considering charge vs the spin. In it corresponding to zero spin and charge there is the neutral pion which is: $\pi^0 = \frac{u\bar{u} + d\bar{d}}{\sqrt{2}}$ , $\pi^- =\bar{u}d$ and $\pi^+ =u\bar{d}$ . Baryons are organised in an octet for particles with spin 1/2 (including the neutron n = udd and the proton p = uud) the and a decuplet for particles with spin 3/2. The particles in the meson octet and baryon decuplet are arranged according to their strangeness (s), a property associated with the strange quark, and electric charge (q). The particles in the baryon octet are organized according to electric charge and isospin, a property related to the strong interaction described.

Source: © Wikimedia Commons, GNU license version 1.2. Authors: Laurascudder, 2007 (Meson octet and Baryon decuplet) and Dr_Eric_Simon, 2006 (Baryon octet).

Neutrinos (a summary on their properties is here) do not interact through the strong or electromagnetic force (being neutral). Weak interactions create neutrinos in three leptonic flavors: the electron neutrino $(\nu_e)$, the muon neutrino $(\nu_\mu)$ and the tau neutrino $(\nu_\tau)$. It was discovered through the neutrino oscillation process, which is a consequence of lepton number violation and flavor neutrino mixing, that they are not all massless. Neutrinos are massless in the Standard Model, so the neutrino sector is the first to indicate the need of a theory that surpasses it.

Neutrino oscillations were first discovered using solar neutrinos (Davies - Nobel prize 2002- with the Homestake experiment first measured a deficit of electron neutrinos from the sun thermonuclear reactions and McDonald - Nobel prize 2015 - with the SNO experiment determined the channels of these oscillations). In 1998 it was comminicated at the Neutrino 1998 conference by SuperKamiokande and by the MACRO experiment, that neutrino oscillations were observed in the atmospheric neutrinos as well (Nobel prize 2015 as well to T. Kajita). Neutrino oscillation experiments measure the difference between the square masses of the neutrino mass eigenstates and the mixing angles (that determine the elements of the mixing matrix). Flavor eigenstates can be expressed as a linear combination of the mass eigenstates: $\nu_\ell (x)= \sum_j U_{\ell j } \nu_j (x)$, where $\ell = e, \mu, \tau$. The states with a mass are $\nu_j$.

We know that there are two discrete mass differences between the three neutrinos, the smallest is measured in the solar neutrino sector and the other one in the atmospheric neutrino sector. This is illustrated in the plot below where the three masses of mass eigenstates are shown for normal or inverted mass ordering. One of the challenges of this coming decade will be to determine this order. The coloros indicate the flavor states composing the mass eigenstates. It can be seen

The figure is taken from S.F. King, J.Phys. G: Nucl. Part. Phys. 42 (2015) 12300

To fully know neutrino masses all the following parameters need to be measured and current status in given below:

From F. Capozzi, E. Di Valentino, E. Lisi, A. Marrone, A. Melchiorri, and A. Palazzo, Phys. Rev. D 95, 096014 and the Data Particle Book [http://pdg.lbl.gov/2018/reviews/rpp2018-rev-neutrino-mixing.pdf] and

Notice that a neutrino is created with a specific flavor in weak interactions. For example an electron neutrino is created in a beta decay when a neutron transform into a proton. As a matter of fact this is the reaction from which Pauli deduced their own existence in 1930. His famous letter on Dec 4, 1930 read like this:

Dear Radioactive Ladies and Gentlemen,

As the bearer of these lines, to whom I graciously ask you to listen, will explain to you in more detail, how because of the "wrong" statistics of the N and Li $^6$ nuclei and the continuous beta spectrum, I have hit upon a desperate remedy to save the "exchange theorem" of statistics and the law of conservation of energy. Namely, the possibility that there could exist in the nuclei electrically neutral particles, that I wish to call neutrons, which have spin 1/2 and obey the exclusion principle and which further differ from light quanta in that they do not travel with the velocity of light. The mass of the neutrons should be of the same order of magnitude as the electron mass and in any event not larger than 0.01 proton masses. The continuous beta spectrum would then become understandable by the assumption that in beta decay a neutron is emitted in addition to the electron such that the sum of the energies of the neutron and the electron is constant...I agree that my remedy could seem incredible because one should have seen these neutrons much earlier if they really exist. But only the one who dare can win and the difficult situation, due to the continuous structure of the beta spectrum, is lighted by a remark of my honoured predecessor, Mr Debye, who told me recently in Bruxelles: "Oh, It's well better not to think about this at all, like new taxes". From now on, every solution to the issue must be discussed. Thus, dear radioactive people, look and judge. Unfortunately, I cannot appear in Tubingen personally since I am indispensable here in Zurich because of a ball on the night of 6/7 December. With my best regards to you, and also to Mr Back. Your humble servant, W. Pauli

The $\beta$ decay reaction is possible since the neutron, with mass a bit smaller than 1 GeV and about 1000 times heavier than an electron, is about 1 MeV heavier than the proton. The proton instead is the lightest baryon and should be in principle stable. Nonetheless searches for proton decy are happening and constrain its lifeto be longer than $2 \times 10^{39}$yrs . and In 1932 James Chadwick discovered the more massive neutron and the word "neutrino" was first used by Enrico Fermi during a conference in Paris in July 1932 and at the Solvay Conference in October 1933 Pauli also employed it. The name is the Italian equivalent of "little neutral one" and was jokingly coined by Edoardo Amaldi during a conversation with Fermi at the Institute of Physics of via Panisperna in Rome, in order to distinguish this light neutral particle from Chadwick's heavy neutron.

The beta decay reaction is: $n \rightarrow p + e^- + \bar{\nu}_e$. This reaction is possible because the neutron has mass of about 1 GeV as the proton, but it is about a MeV heavier ( $m_p = 931.5$ $MeV/c^2$ and $m_n = 939.6 MeV/c^2$). Neutrinos are elusive and were not observed, but the spectrum in energy of the electrons did not look like a line such as in a 2-body decay but it looked like a continuous spectrum like an at least 3-body decay. Pauli got soon scared of having found a particles that is almost no measurable! later it was named neutrino The flavor state is a quantum linear superposition of three mass states. As a result, neutrinos oscillate from one flavor state to another when propagating across a distance (or baseline). For example, an electron neutrino produced in a beta decay reaction may interact in a distant detector as a muon or a tau neutrino.

Cross sections, number density, lifetime, and interaction lengths

The cross-section of an interaction is a very important parameter expressing its "strength" or probability to happen. It can be considered as the effective area for a collision between a target (populated by scattering cenres) and a projectile beam. The cross-section of an interaction depends on the potential associated to a force, the energy of the projectile particles, ... Cross-sections are typically measured using surface units, cm$^2$ or barns:

$$1\;{\rm barn} = 10^{-24} {\rm cm}^2$$

The unit barn comes from the expression big as a barn as in the past physicists studing interactions with atomic nuclei, saw with surprise that interactions were more frequent than expected, and they thought the nucleus was in fact bigger than they thought... big as a barn.

We consider a flux of projectile particles, namely N = number of incident particles/(s cm$^2$), are crossing a volume of target particles of thickness dx (in cm) and $N_c$ scattering centres in the target per unit volume = $\frac{\rho}{A m_p}$ (cm$^{-3}$), where $\rho$ is the density of the target material and $A m_p$is its atomic mass. The number of interacting or scattered particles per unit time and surface is:

$$-dN = \sigma N N_c dx \, [\rm cm^2 cm^{-2} s^{-1} cm^{-3} cm = cm^{-2} s^{-1}]$$

where cross section $\sigma$ is the cross section. Solving this equation we obtain the exponential decrease of the beam intensity with distance: $N = N_0 e^{-N_c \sigma x}$ , where $N_0$is the initial flux (*).

The differential cross section is the rate of scattered particles $dN_s = \frac{dN}{N_c dx}$in the solid angle $d\Omega = 2 \pi sin\theta d\theta$(see image below) divided by the incident flux:

$\frac{d\sigma(E,\Omega)}{d\Omega} = \frac{1}{N}\frac{dN_s}{d\Omega}$ and the total cross section is obtained integrating on the full angle $\sigma(E) = \int \frac{d\sigma}{d\Omega} d\Omega$.

Figure from Wikipedia.

For a reaction a + b $\rightarrow$ c+d, the incident flux is $N = n_a v_i$, where $n_a =$density of incident particles with speed $v_i$. The interaction rate on the single particle target b or transition rate is usually defined as $W = N\sigma = n_a v_i \sigma$and it can be calculated knowing the potential of the interaction U: $W = \left( \frac{2\pi}{\hbar} \right) |T_{if}|^2 \rho_f$, where $T_{if}$ is the transition amplitude between the initial and final state which contains the coupling strength of the interaction according to the Fermi's Second Golden Rule $\int \psi_f U \psi_i dV$, and $\rho_f$is the energy density of final states.

Interaction length:

$$\lambda = \frac{1}{N_c\sigma}$$

We soon see from the formula above (*), that $N = N_0 e^{-x/\lambda}$.

We can prove that $\lambda$is the average distance that a particle travels in a medium between two collisions (clearly inversely proportional to the cross section). As a matter of fact, we consider the interaction probability in the target distance dx: $w dx$= $N_c \sigma dx$. Hence we define:

$P(x) =$ the probability that a particle does not interact in a medium for a distance x and

$P(x+dx) = P(x) + \frac{dP}{dx} dx = P(x) (1-w dx)$ = probability that the particle has no interactions between x and x+dx.

Hence $\frac{dP}{P} = -w dx$$\Rightarrow P(x) = Ce^{-wx}$and we find the constant from the fact that for x=0 surely the particle has not interacted P(0) = 1 = C. Given this we find the mean path between two collisions as: $\lambda = \frac{\int x P(x) dx}{\int P(x) dx} = \frac{1}{w} = \frac{1}{N_c\sigma}$= $\frac{Am_p}{\rho\sigma}$. We can express this interaction length in cm but we can also get rid of the dependency on the density and define the interaction length in g/cm$^2$ by: $\lambda_I = \lambda \rho = \frac{A m_p}{\sigma} = \frac{\rho}{N_c \sigma}$.

Likewise if projectile particles are travelling at speed $v$, the length travelled can be expressed as ${\rm d}x = v {\rm d} t$ giving a similar expression with a time constant:

$$\tau = \frac{1}{N_c v\sigma}$$

Known as the lifetime. If several processes are taking place in several media, we need to replace $N_c \sigma_N$ as $\sum N_{c,i}\sigma_i$, which gives:

$$\frac{1}{\tau_{total}} = \frac{1}{\tau_1} + \frac{1}{\tau_2} + ... + \frac{1}{\tau_n}$$

Examples

Atmospheric extinction:

In the 'A little bit of Astronomy' section we defined the atmospheric exctintion. The atmosphere is opaque to X-rays and UV radiation ( ~3300 Å due to absorption by ozone $O_3$, and also by other isotopes of oxygen $O$, $O_2$, nytrogen and its isotopes $N$, $N_2$, and H2O. In the near IR (1-20μm), absorption by $H_2 O$ and $CO_2$ dominate. At longer wavelengths, $H_2O$ is opaque to the sub-mm band. Photons are both absorbed and scattered from the path. The absorption coefficient for constituent $i$ of the atmosphere $k_i$ is given by:

$k_i = \frac{\sigma_i n_i}{r_i \rho_0}$

where $\sigma_i$ is the cross section for process $i$(as a function of energy or wavelength of photons), $n_i$ = is the number density of the elements in the atmosphere (that in the section above was called $N_{c,i}$), $r_i$ is the fractional abundance of element $i$ and $\rho_0$ is the density of air. The optical depth (or interaction length) $\tau_i$ through the atmosphere is given by the integral, from our elevation (defined in the section on the Atmosphere) $l_0$ to infinity, of the product of $r_i(l)$, $k_i(l)$, and $\rho_0(l)$ which are all functions of elevation in the atmosphere.

The optical depth (or interaction length) τ through the atmosphere is given by the integral, from your elevation $l_0$ to infinity, of the product of the absorption coefficient and it is a sort of `opacity' of the atmosphere: $\tau =$ $\sum_i k_i = \int_{l_0}^{\infty} k(l) dl$. Hence, the attenuation of light at an elevation $l_0$and for a zenith angle $\theta$ is given by:

$I(l_0)/I(\infty) = e^{-sec(\theta) \tau}$, where $I(\infty)$is the intensity of light when entering at the top of the atmosphere and $sec(\theta)$ is the airmass.

Interaction length of cosmic rays in the atmosphere:

cosmic rays are mostly protons. When they enter the atmosphere they collide with other neutrons. Protons will interact with the atmospheric atoms, mostly nitrogen N. An average value of the atomic number of the atmosphere is about $<A_{atm}> \sim 14.5$. According to the superposition principle (explained below), we treat the target particles as A independent protons or nucleons of mass of about 1 $GeV/c^2$.

The proton-proton cross section is measured in accelerator and cosmic ray experiments at different energies. In the plot below it can be seen that the inelastic cross section at 1 TeV is $\sigma_{pp,inel}\sim 50$mb.

We can extrapolate this value considering that for a nucleus the relevant area for the interaction is given by $\sigma_{pN} = \pi r_N^2$$\sim \pi (r_0 A^{⅓})^2$. Essentially the nuclear cross section scales with the square of the nuclear radius which depends on $A^{2/3}$. This translates into $\sigma_{p,Air} \sim 50 \times 14.5^{2/3}\sim$300 mb.

From the cross section we then estimate the proton interaction length in the atmosphere:

$\lambda_p = \frac{Am_p}{\sigma_{p,Air}} \sim \frac{14.5\times 1.67 \times 10^{-24}}{300 \times 10^{-27}} \sim$80 g/cm$^2$

Notice that since a proton is compsed by 3 quarks (uud) and a pion by 2 (quark and anti-quark) the secondary pion cross section is $\sigma_{\pi} \sim 2/3 \sigma_p$ and hence the mean free path in the atmosphere is $\lambda_{\pi} \sim 120$ g/cm$^2$.

Finally, for a heavy cosmic ray nucleus, such as Fe ($A \sim 56$), using the fact that $\sigma_{Fe}\sim \pi r_0^2 (A_{Fe}^{2/3}+A_{Atm}^{2/3})$, we obtain:

$\frac{\sigma_{Fe}}{\sigma_{p}} \sim \frac{(A_{Fe}^{2/3}+A_{Atm}^{2/3})}{A_{Atm}^{2/3}}$= 3.5 the interaction length is about 23 g/cm$^2$.

Proton-proton total and inelastic cross section from D.A. Fagundes et al, PRD 2015 [https://journals.aps.org/prd/abstract/10.1103/PhysRevD.91.114011]