Absorption curve

γ-radiation includes electromagnetic waves, the wavelength of which is much less than the interatomic distances, i.e. λ< а, где а ~ 10 -8 см. Таким образом, нижний предел энергии γ-квантов получается Е = hν = hc/λ. = 12 кэВ.
Like charged particles, the photon flux is absorbed by matter mainly due to electromagnetic interaction. However, the mechanism of this absorption is essentially different. There are two reasons for this:
1) photons do not have an electric charge and, therefore, are not affected by long-range Coulomb forces. Therefore, when passing through matter, photons relatively rarely collide with electrons and nuclei, but on the other hand, upon collision, as a rule, they sharply deviate from their path, i.e. practically drop out of the beam;
2) photons have zero rest mass and, therefore, cannot have a speed different from the speed of light. And this means that in the environment they cannot slow down. They are either absorbed or scattered, mostly at large angles. When a photon beam passes through a substance, the intensity of this beam is gradually weakened as a result of interactions with the medium. Let us find the law according to which this weakening occurs, i.e. curve of absorption of photons in matter.

Let a photon flux J 0 cm -2 s -1 fall on the surface of a flat target perpendicular to it (Fig. 3.1), and the target thickness x (cm) is so small that only a single interaction occurs. The change in the intensity of this flux dJ when photons pass through a layer of matter dx is proportional to the value of the flux J at the depth of this layer, the layer thickness dx (cm), the density of atoms n (cm -3) and the effective photon interaction cross section σ (cm 2):

Solving this equation gives the absorption curve

J x \u003d J 0 e -σnx.

Usually, two concepts are associated with the absorption of photons in matter.

1. Linear absorption coefficient τ = nσ; [τ] = cm -1 and J x = J0e -τx . Thus, τ is the thickness of the substance in centimeters at which the photon flux is attenuated by a factor of e.
2. The mass absorption coefficient μ = τ/ρ = σn/ρ, where ρ (g/cm) is the density of the substance. The dimension of μ is obtained as follows: [μ] = cm 2 /g. In this case, the change in the photon flux takes the form:

J x \u003d J 0 e -μxρ,

where xρ (g / cm 2) is the thickness of the substance, measured in mass units. The meaning is the same - this is such a thickness of the substance in g / cm 2, on which the flow is weakened by e times.

The absorption coefficient completely characterizes the passage of photons through matter. It depends on the properties of the medium and the photon energy. If absorption occurs due to several different processes, each of which has its own absorption coefficient, μ i , τ i ,..., then the total absorption coefficient μ = ∑μ i and τ = ∑τ i
The absorption of photons by matter mainly occurs due to three processes: the photoelectric effect, the Compton effect, and the production of electron-positron pairs in the Coulomb field of the nucleus.

3.2 Photoelectric effect

The photoelectric effect is the release of electrons that are in a substance in a bound state, under the influence of photons. Distinguish between internal and external photoelectric effect.
The internal photoelectric effect is the transition of electrons under the influence of electromagnetic radiation inside a semiconductor or dielectric from bound states to free ones without escaping to the outside.
The external photoelectric effect is observed in solids, gases, on individual atoms and molecules - this is the emission of electrons outward when photons are absorbed. In these lectures, only the external photoelectric effect will be discussed. The photoelectric effect is the process in which an atom absorbs a photon and emits an electron. In this case, the incident photon interacts with the electron bound in the atom and transfers its energy to it. The electron receives kinetic energy Te and leaves the atom, while the atom remains in an excited state. Therefore, the photoelectric effect is always accompanied by the characteristic X-ray emission of an atom or the emission of Auger electrons. With the Auger effect, there is a direct transfer of the excitation energy of an atom to one of its electrons, which, as a result, leaves the atom. The laws of conservation of energy and momentum in the photoelectric effect can be represented as:

hν = T e + I i + T i, and

where , is the kinetic energy of the recoil nucleus; I i - ionization energy
i-th shell of the atom; . Since usually hν >> I i + T i, then the energy of photoelectrons is T e ≈ hν, and, consequently, the energy spectrum of photoelectrons is close to monochromatic.
It follows from the laws of conservation of energy and momentum that the photoelectric effect cannot occur on a free electron. Let us prove this "by contradiction": suppose that such a process is possible. Then the conservation laws will look like this

From here we obtain the equation 1 − β = √1 − β 2 , which has two roots β = 0 and β = 1. The first of them corresponds to T e = hν = 0, and the second has no physical meaning for particles with a mass other than zero.
This proof looks even clearer for the nonrelativistic case: hν = m e v 2 /2 and hν/c = m e v. The solution of the system leads to the expression v = 2c, which cannot be.
Thus, a free electron cannot absorb a photon. For the photoelectric effect, the connection of an electron with an atom is essential, to which a part of the photon momentum is transferred. The photoelectric effect is possible only on a bound electron. The lower the binding energy of an electron with an atom compared to the energy of a photon, the less likely the photoelectric effect. This circumstance determines all the main properties of the photoelectric effect:

A) the course of the cross section with the photon energy − σ f (hν) , b) the ratio of the probabilities of the photoelectric effect on different electron shells, c) the dependence of the cross section on the Z of the medium.

Fig.3.2. Dependence of the effective cross section of the photoelectric effect on the photon energy

a) Figure 3.2 shows the dependence of the effective cross section of the photoelectric effect on the photon energy. If the photon energy is large compared to the binding energy of the electrons in the atom, then the cross section of the photoelectric effect φ decreases rapidly with increasing photon energy. For I i<< hν < m e c 2 σ ф ~ (hν) -3.5 .
When hν > m e c 2 σ f ~ (hν) -1 .
As hν decreases, i.e. As the electron connectivity I k /hν increases, the cross section of the process grows rapidly until the photon energy becomes equal to the energy I k . For hν< I k фотоэффект на K-оболочке атома станет невозможным, сечение фотоэффекта будет определяться только взаимодействием фотонов с электронами L, М и др. оболочек. Но эти электроны связаны с ядром слабее, чем
K-electrons. Therefore, at equal photon energies, the probability of a photoelectric effect on L-electrons is much less than on K-electrons. Depending on σ f (hν) there will be a sharp jump. Then at
hν< I k снова σ ф начинает расти с убыванием hν, так как возрастает относительная связность электрона L/hν, и т.д.
b) The formulas for the cross section of the photoelectric effect on K-electrons, obtained by the methods of quantum electrodynamics and confirmed by experiment, are:

The ratios of the cross sections of the photoelectric effect on different shells are obtained as follows:

Therefore, when calculating the total cross section of the photoelectric effect, the relation is usually used:

c) From the same formula one can see a strong dependence of σ f on Z of the medium: σ f ~ Z . This is understandable, since in light elements the electrons are weaker bound by the Coulomb forces of the nucleus than in heavy ones. In heavy substances, the photoelectric effect is the main reason for the absorption of soft photons.
The angular distribution of photoelectrons is obtained by calculation from the formula for the differential cross section. It follows from it that the photoelectrons are distributed symmetrically according to the law ~ cos 2 φ with respect to the direction of the electric vector of the incident electromagnetic wave. In addition, the angular distribution essentially depends on the photoelectron energy. In the nonrelativistic case T e<< m е c 2 (β << 1) интенсивность фотоэлектронов максимальна в плоскости поляризации векторов и фотона, т.е. в плоскости, перпендикулярной направлению движения фотона. При больших энергиях Т е >m e c 2 the angle at which the intensity of photoelectrons is maximum decreases, and the greater the energy of electrons, the smaller the angle of their departure compared to the direction of photon motion, the angular distribution is elongated forward.

3.3. Compton effect

The interaction of photons with matter can lead to their scattering without absorption. Scattering can be of two types: 1) without changing the wavelength (coherent scattering, Thomson, classical) and 2) with changing the wavelength (incoherent, Compton scattering).

1. Thomson scattering happens if hν< I i (λ ~10 -8 см). В этом случае атом воспринимается фотоном "как единое целое", и фотон обменивается энергией и импульсом со всем атомом. Так как масса атома очень велика по сравнению с эквивалентной массой фотона hν/c , то отдача в этом случае практически отсутствует. Поэтому рассеяние фотонов происходит без изменения их энергии, т.е. когерентно.
It can be considered that the source of scattered radiation is the bound electrons of the atom, which come into resonant vibrations under the action of the incident radiation and, as a result, emit photons of the same frequency. The Thomson scattering cross section depends on the photon scattering angle 0:

σ(θ) = 0.5re 2 (l + cos 2 θ),

where r e 2 = e 2 /m e c 2 = 2.8 10 -13 cm is the classical radius of the electron. Integrating over all θ, one can obtain the cross section for total Thomson scattering. The effective cross section of Thomson scattering, calculated per 1 electron, is equal to:

σ T = (8/3)πr e 2 = 0.66 barn,

where σ T is a universal constant and does not depend on the frequency of the incident radiation.

2. Compton scattering occurs when hν >> I i . In this case, all the electrons of the atom can be considered free.

Compton scattering occurs as a result of an elastic collision of a photon with an electron, and the photon transfers part of its energy and momentum to the electron. Therefore, the energy and angular characteristics of the phenomenon are completely determined by the laws of conservation of energy and momentum for an elastic impact (Fig. 3.3):

hν = hν " + T e,

where and are the kinetic energy and momentum of the recoil electron.

The joint solution of these equations makes it possible to obtain the energies of the scattered photon hν " and recoil electron Te depending on the photon scattering angle θ:

A number of important consequences follow from these relations.

1. From the first relation, it is easy to find how much the electromagnetic wave length has changed during Compton scattering (Compton formula):

where λ 0 \u003d h / m e c \u003d 2.426 10 -10 cm is the Compton wavelength of the electron. From the Compton formula it follows that:

A) wave shift Δλ does not depend on the magnitude of the wavelength; b) shift Δλ, determined only by the photon scattering angle θ: at θ = 0 Δλ = 0 (i.e. no scattering), at θ = π/2 Δλ = λ 0 and at θ = π, Δλ = 2λ 0 (maximum a possible shift occurs during backscattering).

2. The energy spectrum of photons obtained as a result of Compton scattering of a beam of monoenergetic γ-quanta turns out to be continuous in the energy range from

at θ = π to hν max = hν at θ = 0.

3. As a result of Compton scattering of monoenergetic γ-quanta, a continuous energy spectrum of recoil electrons is obtained in the range from

T e min = 0 at θ = 0 up to for θ = π.

4. The relationship between the escape angles of the scattered photon θ and the recoil electron φ (Fig. 3.3) can be found from the momentum conservation law written for the longitudinal and transverse components (relative to the direction of motion of the primary photon):

Let's transform the second equation:

From here we find:

It can be seen from the relation obtained that a change in the photon scattering angle in the interval 0 ≤ θ ≤ π corresponds to a change in the angle of emission of a recoil electron in the interval π/2 ≥ φ ≥ 0. primary photon.
The differential effective cross section for Compton scattering was first calculated by O. Klein and I. Nishina in 1929, and in 1930 I.E. Tamm obtained the same formula in a different way. The Klein-Nishina-Tamm formula has the form:

where dσ K /dΩ is the differential effective cross section of photon scattering at an angle θ into the solid angle dΩ, and r e is the classical electron radius. After substituting the values ​​hν " the dependence of the differential cross section of Compton scattering only on hν and on θ is obtained, and the form of the dependence of the cross section on θ changes with the photon energy. For small values ​​of hν:
dσ K /dΩ ~ 1 + cos 2 θ. As hν increases, an increasing number of photons are scattered in the "forward" direction, and with an increase in the primary energy hν, the probability of scattering at small angles increases (Fig. 3.4).
The total cross section is found after integration over all θ:

where σ T = (8π/3)r e 2 is the Thomson scattering cross section, and ƒ(hν/m e c 2)< 1 и возрастает с увеличением hν.
For small values ​​of hν (I K<< hν/m e c 2 <<1), σ K ~ σ T (1 − 2hν/m e c 2) → σ T with decreasing hν.

Since there are Zn electrons in 1 cm of the medium, then the total probability of Compton scattering on a 1 cm path in the substance (Z, A, ρ) will be:

Thus, the probability of Compton scattering per 1 cm of the path is inversely proportional to the photon energy and proportional to the Z of the substance (the cross section per 1 electron does not depend on the Z of the substance, and each atom contains Z electrons). Figure 3.5 shows a plot of σ K /σ T versus photon energy. This figure shows in the same units the cross section of the photoelectric effect in various substances. A comparison of the dependences shows that with an increase in the photon energy, the probability of the Compton effect becomes much larger than the cross sections of the photoelectric effect.

Fig.3.5. Dependence of the total cross sections of Compton scattering (solid curve) and the photoelectric effect in terms of 1 electron (dotted line for C, Al, Cu and Pb) on the photon energy

Compton scattering can occur not only on electrons, but also on other particles that have an electric charge. However, the likelihood of such an effect is very small. For example, Compton scattering on the nuclei of atoms is negligible due to the fact that the nuclei have a very small value of their classical electromagnetic radius Ze 2 /m i s 2 .
There is another phenomenon called the inverse Compton effect. It occurs during elastic scattering of photons by relativistic electrons. In this case, the energy and momentum of photons will increase due to the energy and momentum of target electrons.

3.4. Birth of electron-positron pairs

At a sufficiently high photon energy (hν > 2m e c 2), the process of pair formation becomes possible, in which a photon is absorbed in the field of the nucleus, and an electron and a positron are born. Calculation by QED and experience indicate that this process does not occur inside the nucleus, but near it, in a region having the size of the Compton wavelength λ 0 = 2.4 10 -10 cm. Since this interaction of a photon with the field of the nucleus produces an electron and positron, then this process has an energy threshold, i.e. it occurs if hν > 2m e c 2 . The laws of conservation of energy and momentum can be written as:

hν = 2m e c 2 + Т − + Т + + Т i,

where β − and β + are the relative velocities of the electron and positron, T − and T + are their kinetic energies, and T i and p i are the energy and momentum of the recoil nucleus.
Based on the laws of conservation of energy and momentum, it can be shown that the formation of an electron-positron pair by a photon in vacuum is impossible: energy and momentum must necessarily be distributed between three particles: an electron, a positron, and, for example, a nucleus. If we assume that the birth of a pair can occur in a vacuum (T i = p i = 0), then the conservation laws take the form:

hν = 2m e c 2 + Т − + Т + and

The first of these equations can be written in the form:

and its incompatibility with the second equation immediately becomes apparent.
In the particular case when T − = T + = 0, a system of contradictory equations is obtained: hν = 2m e c 2 and
hν/c = 0. Thus, for the conservation laws to be satisfied, a third particle is needed, in the field of which the process of pair production takes place and which takes on the excess momentum. Such a particle can be not only a nucleus, but also, for example, an electron. But if the nucleus T i \u003d p i 2 / 2m i is a small value, then the electron will have a very large recoil, and the recoil electron can receive energy of the same order as the components of the pair. In this case, the process threshold will significantly exceed 2m e c 2 . The threshold photon energy for pair formation in the electron field is 4m e c 2 =2.044 MeV.
Theoretical calculations of the dependence of the pair production cross section on the γ-ray energy lead to a rather complicated form. However, for the energy range 5m e c 2< hν < 50m e c 2 эта зависимость может быть представлена в виде:

At photon energy hν< 5m e c 2 и hν >50m e c 2 the cross section grows more slowly. For hν > 50m e c 2, the growth of the cross section is limited by screening of the Coulomb field of the nucleus by atomic electrons. In the limiting relativistic case, for hν > 10 3 m e c 2, the cross section does not depend on the energy:

σ P ~ 0.08 Z 2 r e 2 = 0.63 10 -26 Z 2 cm 2.

The general character of the dependence of the cross section on the photon energy is shown in Fig. . 3.6.

Fig.3.6. Dependence of the cross section of pair production on photon energy

The process of pair production is similar to the process of bremsstrahlung. Therefore, the expressions describing these two processes are very similar in their structure: in the case of complete screening, the probability that a photon with energy E " = hν on a path of 1 cm forms an electron with energy E in the interval (E, E + dE) and a positron with energy (E " − E) will:

The probability of pair formation does not depend on the energy of the electron E and positron E " − E, and this is understandable, since in the process of their formation the photon disappears and the distribution of energy between the components of the pair is equally probable. Knowing w n , we can find the total probability of pairing on a path of 1 cm:

Thus, in the case of complete screening, the total pair production cross section does not depend on the photon energy.

3.5. Other processes of interaction of photons with matter

1. Nuclear photoelectric effect - absorption of a γ-quantum by a nucleus and emission of a nucleon, i.e. (γ,n)-reaction. The threshold of the nuclear photoelectric effect is -6-10 MeV i.e. order of the binding energy of nucleons in nuclei. The cross section of the nuclear photoelectric effect σ yf ~ Z and is much smaller in magnitude than the cross sections of the three effects considered.

2. If the energy of photons is much greater than the binding energy of nucleons in nuclei, then photodisintegration of nuclei can occur with the emission of several particles. For example, (γ,2р), (γ,n,2р) are reactions. The cross section of such a process is σ i ~ 10 -26 cm.

3. If hν > 2m μ s 2 , i.e. hν > 200 MeV, then in the field of the nucleus γ-quanta can form μ − μ + -pairs, similarly to e − e + -pairs.

4. If hν > m π с 2 , i.e. hν >140 MeV, photogeneration of pions with a cross section of ~10 -28 A cm 2 may occur.

Thus, the absorption of γ-quanta due to all the listed processes is negligibly small compared to σ P.

3.6. The total cross section for the interaction of photons with matter

The weakening of the photon flux when passing through matter is determined mainly by three processes: the photoelectric effect, the Compton effect, and the formation of pairs in the Coulomb field of atomic nuclei. As a result, in the formula J = J0 e -σnx, the section o is the sum of the sections of these processes:
σ = σ f + σ K + σ P, and the linear and mass absorption coefficients, respectively, are:
τ = σn = τ f + τ K + τ P and μ = σn/ρ = μ f + μ K + μ P. Each of the terms depends differently on the photon energy and properties of the substance, so the relative role of individual terms can vary greatly . So, in aluminum (Fig. 3.7) in a wide range of photon energies of 50 keV< hν <15 МэВ преобладает комптон-эффект, а при hν >15 MeV - pair production. In lead, however, the photoelectric effect (Fig. 3.7) is dominant up to an energy of 0.5 MeV, and for hν >5 MeV, the process of pair production plays the main role.

Fig.3.7. The dependence of the mass absorption coefficient of photons on their energy in aluminum, copper and lead

In conclusion, an important circumstance should be noted: all three types of interaction of photons with matter lead to the appearance of fast electrons.

3.7. Positron annihilation in matter

The word "annihilation" means "disappearance", "turning into nothing". This is a process in which a particle and its antiparticle are converted into electromagnetic radiation (photons) or other elementary particles (leptons, quarks). This is a process opposite to the production of pairs by γ-quanta. Both processes are simply mutual transformations.
These mutual transformations are controlled by fundamental conservation laws: the law of conservation of energy, momentum, angular momentum, electric charge, etc.
The processes of particle creation and annihilation were theoretically predicted in 1931 by P.A. Dirac. They followed from the theory of the electron he created. According to Dirac, it is possible to combine quantum mechanics (by that time already confirmed by experiment) with the theory of relativity only if, along with the state of an electron with positive energy, we introduce the state of an electron with negative energy (or a positive "electron" with positive energy).
In 1932 K.D. Anderson, investigating the composition of cosmic rays with a cloud chamber placed in a magnetic field, obtained experimental evidence for the existence of the positron (Nobel Prize, 1936). By the sign of the curvature of the particle track, it was found that the particle is positive, and by the change in curvature (after it passed 6 mm of lead) and by the density of grains in the track, the mass and momentum of the particle were determined. In 1933, Frederic and Irene Joliot-Curie first obtained a photograph of a cloud chamber with traces of an electron and a positron produced by a gamma quantum, and in the same year, F. Joliot-Curie first observed the annihilation of electrons and positrons into two photons.
How does positron annihilation occur? Once in matter, fast positrons behave in the same way as electrons, i.e. at T e > ε they experience radiative drag, and at T e< ε − ионизационные потери и, как правило, почти полностью теряют свою скорость. В дальнейшем начинается их диффузия в веществе до встречи со свободными или связанными в атомах электронами и последующая аннигиляция позитронов. Перед аннигиляцией обе частицы (электрон и позитрон) чаще всего находятся в состоянии, когда их моменты количества движения равны нулю (S-состояние). Дальнейшая судьба их зависит от взаимной ориентации внутренних моментов количества движения (спинов) и от того, свободен ли электрон или находится в связанном состоянии.
When an electron and a positron meet, their total energy, including the rest energy, is almost entirely converted into the energy of electromagnetic radiation (a process opposite to the production of pairs) and is partially transferred to some third body, for example, the nucleus. If positron annihilation occurs on an electron that is part of an atom, then annihilation with the formation of one photon is possible, since the momentum of the resulting photon will be compensated by the recoil of the atom or nucleus, and the law of conservation of momentum will be fulfilled. The laws of conservation of energy and momentum for this situation can be written as follows: + = ∑ t /c.

A positron slowed down to a thermal velocity can annihilate with a free electron, for example, with one of the conduction electrons in a metal or with one of the outer electrons of an atom. If we assume that the electron and positron were at rest before annihilation, then the conservation laws take the form:

2m e c 2 = ∑ t and 0 = ∑ t /c,

i.e. annihilation on a free electron is possible only if at least two photons are emitted simultaneously in opposite directions. Since both annihilating particles are most likely in the S-state, the result of annihilation will depend on the mutual orientation of the internal momentum of the particles, i.e. their spins.
If the spins of an electron and a positron are directed in opposite directions (+1/2ћ and -1/2ћ), and, consequently, their total spin is zero, then as a result of annihilation (according to the law of conservation of charge parity), only an even number of photons with spins , also directed in opposite directions, since the spin of each photon is equal to l ћ. Since the probability of annihilation is w ~ α n , where n is the number of photons, two photons are most likely to be born (w ~ α 2) - the so-called two-photon annihilation , less likely - four photons (w ~ α 4), etc.
Since the momenta of the electron and positron are close to zero, the total momentum of the system is also zero, and, consequently, the photons formed during annihilation fly in opposite directions, each of them taking half the energy of the system, i.e. by 0.511 MeV.
If the spins of an electron and a positron are parallel, then their total spin is 1 ћ. In this case, the formation of an odd number of photons, most likely three, is possible, since one photon cannot arise due to the violation of the momentum conservation law. The probability of three-photon annihilation ~ a 3 , i.e. much smaller (by a factor of 1/137) than a two-photon one. On average, three-photon annihilation occurs in (0.2 - 0.3)% of cases.
If the annihilation occurs "on the fly", i.e. in the case when the positron has not yet lost speed, then the photons scatter at an angle, and the angle of photon expansion depends on their speed. At high energies of annihilating positrons, the resulting photons are emitted predominantly "forward" and "backward" relative to the direction of positron motion. A photon flying forward carries away most of the energy of the positron. The fraction of a photon flying backwards has the minimum energy, ie 0.511 MeV. Therefore, when fast positrons pass through matter, a beam of gamma rays flying in one direction is formed, which is used to obtain monochromatic beams of high-energy photons.
The positron is a stable particle, in vacuum it exists indefinitely, but in matter the positron annihilates very quickly. The average lifetime of a positron with respect to the annihilation process in solids is τ ~ 10 -10 s, and in air under normal conditions τ ~ 10 -5 s.
Sometimes annihilation goes through an intermediate stage, through the formation of a bound state of an electron and a positron, which is called positronium . Positronium, in which the spins of the positron and electron are antiparallel (parapositronium), annihilates into two gamma quanta with a lifetime
τ ~ 1.25 10 -10 s. Positronium with parallel particle spins (orthopositronium) produces three gamma quanta with a lifetime of τ ~ 1.4·10 -7 s.
The phenomenon of positron annihilation is now widely used to study the properties of elementary particles. On colliding beams of positrons and electrons in the vacuum of the accelerator chamber, an annihilation process occurs, in which a precisely defined energy is released. Point interaction and knowledge of its energy is used to prove the existence of quarks and determine their mass.

Questions and tasks for chapter 3

1. A monochromatic photon beam passing through an aluminum plate 2.9 cm thick is attenuated by a factor of 2.6. Determine τ, μ and σ.

Feynman diagram for photon-photon scattering. Photons themselves cannot interact with each other, since they are neutral particles. Therefore, one of the photons turns into a particle-antiparticle pair, with which the other photon interacts.

Physicists from the ATLAS collaboration for the first time registered the effect of scattering of light quanta, photons, on photons. This effect is one of the oldest predictions of quantum electrodynamics, it was described theoretically more than 70 years ago, but has not yet been observed experimentally. Interestingly, it violates the classical Maxwell equations, being a purely quantum phenomenon. The study was published this week in the journal Nature Physics, however, the preprint of the article came out as early as February 2017. Details about it were reported by the Elements.ru portal.

One of the main properties of classical Maxwellian electrodynamics is the principle of superposition for electromagnetic fields in vacuum. It allows you to directly add the fields from different charges. Since photons are field excitations, they cannot interact with each other within the framework of classical electrodynamics. Instead, they should freely pass through each other.

ATLAS detector magnets

Quantum electrodynamics extends the action of the classical theory to the motion of charged particles with near-light velocities, in addition, it takes into account the quantization of field energy. Due to this, in quantum electrodynamics it is possible to explain unusual phenomena associated with high-energy processes - for example, the creation of pairs of electrons and positrons from vacuum in high-intensity fields.

In quantum electrodynamics, two photons can collide with each other and scatter. But this process does not go directly - light quanta are uncharged and cannot interact with each other. Instead, there is an intermediate formation of a virtual particle-antiparticle pair (electron-positron) from one photon, with which the second photon interacts. Such a process is very unlikely for visible light quanta. This can be estimated from the fact that light from quasars 10 billion light years away reaches the Earth. But with increasing photon energy, the probability of a process with the birth of virtual electrons increases.

Until now, the intensity and energy of even the most powerful lasers were not enough to see the scattering of photons directly. However, researchers have already found a way to see this process indirectly, for example, in the decay of a single photon into a pair of lower-energy photons near the heavy nucleus of an atom.

It was possible to see directly the scattering of a photon by a photon only in the Large Hadron Collider. The process became visible in experiments after increasing the particle energy in the accelerator in 2015 - with the launch of Run 2. Physicists of the ATLAS collaboration investigated the processes of "ultraperipheral" collisions between heavy lead nuclei accelerated by the collider to energies of 5 teraelectronvolts per nucleus nucleon. In such collisions, the nuclei themselves do not directly collide with each other. Instead, their electromagnetic fields interact, in which photons of enormous energies arise (this is due to the proximity of the speed of the nuclei to the speed of light).

Photon-photon scattering event (yellow beams)

Ultra-peripheral collisions are distinguished by great purity. In them, in the case of successful scattering, only a pair of photons with transverse momenta directed in different directions arises. In contrast, ordinary collisions of nuclei form thousands of new fragment particles. Among the four billion events collected by ATLAS in 2015, scientists were able to select 13 corresponding to scattering using the statistics of collisions of lead nuclei. This is about 4.5 times more than the background signal that physicists expected to see.

Scheme of the scattering process in the collider. Two nuclei fly close - their electromagnetic fields interact

The ATLAS Collaboration

The collaboration will continue to explore the process at the end of 2018, when the collider will again host a session of collisions of heavy nuclei. Interestingly, it was the ATLAS detector that turned out to be suitable for searching for rare events of photon-photon scattering, although another experiment, ALICE, was specially developed to analyze collisions of heavy nuclei.

Now at the Large Hadron Collider a set of statistics of proton-proton collisions. Recently, scientists about the discovery of the first doubly charmed baryon at the accelerator, and back in the spring of the physics of the ATLAS collaboration about an unusual excess of production events of two weak interaction bosons in the high energy region (about three teraelectronvolts). It may indicate a new superheavy particle, but the statistical significance of the signal does not yet exceed three sigma.

Vladimir Korolev

The distribution function in this case, according to (5.7), will take the form:

where: m is the mathematical expectation, s is the standard deviation.

The normal distribution is also called Gaussian after the German mathematician Gauss. The fact that a random variable has a normal distribution with parameters: m,, is denoted as follows: N (m, s), where: m =a =M ;

Quite often, in formulas, the mathematical expectation is denoted by a . If a random variable is distributed according to the law N(0,1), then it is called a normalized or standardized normal value. The distribution function for it has the form:

.

The graph of the density of the normal distribution, which is called the normal curve or Gaussian curve, is shown in Fig. 5.4.

Rice. 5.4. Normal distribution density

Determining the numerical characteristics of a random variable by its density is considered on an example.

Example 6.

A continuous random variable is given by the distribution density: .

Determine the type of distribution, find the mathematical expectation M(X) and the variance D(X).

Comparing the given distribution density with (5.16), we can conclude that the normal distribution law with m =4 is given. Therefore, mathematical expectation M(X)=4, variance D(X)=9.

Standard deviation s=3.

The Laplace function, which has the form:

,

is related to the normal distribution function (5.17), by the relation:

F 0 (x) \u003d F (x) + 0.5.

The Laplace function is odd.

Ф(-x)=-Ф(x).

The values ​​of the Laplace function Ф(х) are tabulated and taken from the table according to the value of x (see Appendix 1).

The normal distribution of a continuous random variable plays an important role in the theory of probability and in the description of reality; it is very widespread in random natural phenomena. In practice, very often there are random variables that are formed precisely as a result of the summation of many random terms. In particular, the analysis of measurement errors shows that they are the sum of various kinds of errors. Practice shows that the probability distribution of measurement errors is close to the normal law.

Using the Laplace function, one can solve problems of calculating the probability of falling into a given interval and a given deviation of a normal random variable.

Consider a uniform continuous distribution. Let's calculate the mathematical expectation and variance. Let's generate random values ​​using the MS EXCEL functionRAND() and the Analysis Package add-in, we will evaluate the mean and standard deviation.

evenly distributed on the interval, the random variable has:

Let's generate an array of 50 numbers from the range )