1 Multi–messenger Astrophysics

Definition for the field of research discussed here is “multi–messenger astrophysics”, thatis the study of the sources and environments in the cosmos with the (simultaneous) ob-servations of different particles that act as “messengers”, or carriers of information.

Four “cosmic messengers” exist: photons, neutrinos, cosmic rays and gravitationalwaves.

PhotonsThe Photon of course has been the fundamental “Nuncius Sidereus” (“Messenger fromthe stars”) since the prehistory of humankind, and nearly everything we know about theuniverse has been obtained from observations of light of different wavelengths.

This is reflected in the fact that in common language the words “observation” and“seeing” describe decoding of the information contained in the light we detect with outeyes.

Crucial steps have been the development of the optical telescope at the beginningof the 17th century, and then the opening of different “windows” in photon wavelength(or equivalently energy), observed with always more sensitive instruments. The recentdevelopment of γ–ray telescopes for observation of very high energy photons represents arecent important chapter in this history.

Cosmic RaysCosmic rays are relativistic charged particles, that arrive at the Earth with an approxi-mately isotropic flux, because their trajectories are bent by (galactic and extragalactic)magnetic fields. They have a very broad energy range that extends more than ten ordersof magnitude, from particles that are only moderately relativistic, to particles with anenergy of order 1020 eV.

The discovery of cosmic rays approximately hundred years ago played a crucial rolein the early development of Particle Physics when these particles were the only source ofvery high energy particles before the development of particle accelerators. This discoveryrevealed the existence of a hidden “High Energy Universe”, an ensemble of astrophysicalobjects, environments and mechanisms capable of producing particles of extraordinarilyhigh energy.

NeutrinosThe idea of “neutrino astronomy”, that is the use of neutrinos as cosmic messengersfrom astrophysical sources, is as natural as profoundly fascinating, and was immediatelyformulated as soon as the neutrino was discovered. Neutrinos, as photons, travel alongstraight lines, and their detection allows a direct imaging of their sources. Because of theirvery small interaction cross section, neutrinos can emerge from deep inside astrophysicalobjects, and carry information that is very different from what can be inferred from thephotons emitted by the same source.

The “dream” of neutrino astronomy is now a reality. Three sources have been identi-fied: The Sun, Supernova 1987A and the Earth (with the detection of geo (anti)–neutrinosgenerated by the radioactive decays of unstable nuclei inside the Earth). More recently (in2013), at much higher energy (Eν & 30 TeV) the IceCube detector at the South Pole hasdetected, a diffuse flux of extraterrestrial neutrinos generated by astrophysical sources,

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that have not yet been identified.

Gravitational WavesThe existence of gravitational waves, ripples in curvature of space time that propagateat the speed of light, is a key prediction of General Relativity. Large mass astrophysicalbodies undergoing large accelerations should emit waves that are in principle directlyobservable by detector at the Earth. The existing detectors have now the sensitivity toobserve the first and their results are eagerly awaited.

The study of Gravitational Waves will not be discussed here, What is important tostress is that the sources of gravitational waves are very likely also copious emitters of highenergy particles in the form of photons, neutrinos and cosmic rays, and therefore thesefields are intimately related. One example of several possible multi–messenger studies isthe search for the simultaneous emission of gravitational waves from Gamma Ray Burstsevents (very energetic explosive events observed in other galaxies that are also predictedto be observable as high energy neutrino emitters).

2 “Low energy” neutrino detection

The detection of neutrinos with energy between 0.1–100 MeV (that here will be consideredas “low”) have given results that are extraordinarily interesting.

The most remarkable results are:

• The successful observations of solar neutrinos (Eν ' 0.1–8 MeV)

• The observations of supernova neutrinos (Eν ∼ 10–30 MeV).

• The observations of geophysical neutrinos (Eν . 1–3.272 MeV, the endpoint of thespectrum of decay of radioactive nuclei).

The importance of the solar and supernova observations has been recognized by the Nobelprizes of Raymond Davis and Masatoshi Koshiba in 2002, and the Nobel prize of ArthurMcDonald and Takaaki Kajita in 2015.

The observations of solar neutrinos, has also resulted in the discovery of neutrino flavoroscillations. The field of solar neutrino detection is now mature, but far from exhausted,and several studies remain interesting. Perhaps the most interesting challenge is themeasurement of the neutrinos created in the CNO-cycle [a subdominant (in the Sun) setof reactions for the fusion of hydrogen into helium].

The handful of neutrinos events from SuperNova SN 1987A observed by Kamiokandeand IMB (and perhaps also Baksan) [One should also remember the puzzling and contro-versial results of the LSD detector in the Mont Blanc Tunnel] has given us an extraordinaryamount of information on the physics of gravitational collapse. The existing (and evenmore, the future) detectors for SN neutrinos (will) have much higher sensitivity and there-fore will yield a much more detailed and valuable information. The obvious problem isthat one has to wait for such an explosion to occur, but the results will be so interestingthat there are no doubts that one has to be well prepared.

Also of the highest interest has been the detection (performed by KamLAND in Japan,and BOREXINO in the Gran Sasso laboratory) of geo–neutrinos, the νe generated by

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radioactive decays of unstable nuclei inside the Earth. This study has the fascinatingpotential to give us important information about the internal structure of the Earth.

Solar, SuperNova and geo–neutrinos have energies that are approximately in the sameenergy range, that also overlaps with the energy range of the anti–neutrinos generatedin nuclear reactors Reactor neutrinos have played a crucial role in the study of neutrinoflavor oscillations, and have a very important potential for future studies. This has allowedseveral detectors (Kamiokande, Super–Kamiokande, BOREXINO) to obtain importantresults for different problems, including measurements of higher energy atmospheric andaccelerator neutrinos. The possibility of developing multi–purpose detectors is clearly afundamental consideration for the design of future instruments.

3 The “High Energy Universe”

Three “particle messengers”: cosmic rays, gamma–rays, and neutrino can give us infor-mation can give us information about the “High Energy Universe”. In most cases theastrophysical sources directly accelerate to very high energy charged particles (protons,nuclei and electrons). These particles can then generate photons and neutrinos when theyinteract with matter or radiation fields inside or near the accelerators. The observationsof the three particle messengers give complementary information about the nature andstructure of the astrophysical sources and the mechanisms that operate in them.

[.....]

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4 Hadronic mechanism of production

Hadronic mechanism for the production of high energy gamma rays and neutrinos:

(i) protons (or ionized nuclei) are accelerated to relativistic energies by some electro-magnetic mechanism.

(ii) These relativistic particles interact with some target (gas of radiation fields) creatinghadrons (such as charged and neutral pions, and kaons kaons).

(iiia) The decays of neutral pions generate gamma rays vias π◦ → γγ. Additional photonsare created in the decay of other mesons (such as η and η′).

(iiib) The Weak decays of the hadrons generate neutrinos directly (as in π+ → νµ µ+) or

indirectly in the subsequent decay of muons produced in the same hadronic decays(µ+ → νµ νe e

+ and charge conjugate channel). Any environment where relativis-tic hadrons are accelerated or stored by magnetic fields is therefore a high energyneutrino source.

5 Two body decay of a particle

Let is consider the two body decay π◦ → γγDecay of π◦ in center of mass system (isotropic decay) Energy in the c.m.

E∗γ =mπ0

2(1)

Isotropic distribution in c.m. frame:

dNγ

d cos θ∗= 1 (2)

Normalization: 2 particles in the final state∫ +1

−1d cos θ∗

dNγ

d cos θ∗= 2 (3)

Calculation of the spectrum of the photons in the laboratory frame (where the pion hasenergy Eπ). The expression of the energy of the photon in th lab system can be calculatedwith a Lorentz boost:

Eγ = γ [E∗γ + βπ E∗γ cos θ∗] (4)

(The angle θ∗ is with respect to the boost direction). The Lorentz parameter γ = Eπ/mπ,and E∗ = mπ/2):

Eγ =Eπmπ

[1 + βπ cos θ∗]mπ

2(5)

Eγ =Eπ2

[1 + βπ cos θ∗] (6)

The limits of the photon energy can be set putting cos θ∗ = ∓1:

Emin,maxγ =

Eπ2

[1∓ βπ] (7)

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In the limit of an ultrarelativistic pion

βπ =

√1− m2

π

E2π

→ 1 (8)

one finds that the limit are

Eminγ = 0, Emax

γ = Eπ . (9)

The spectrum can be calculated simply with a change of variable:

dNγ

dEγ=

dNγ

d cos θ∗

[dEγ

d cos θ∗

]−1(10)

with the result:dNγ

dEγ=

2

Eπ βπ(11)

That is the spectrum is flat, between the kinematical limits of Eq. (??). It is simple tocheck that the Normalization: ∫

dEγdNγ

dEγ= 2 (12)∫

dEγ EγdNγ

dEγ= Eπ (13)

Ultrarelativistic limit:dNγ

dEγ=

2

Eπ(14)

between limits :Eminγ = 0 (15)

Eminγ = Eπ (16)

Useful to introduce a scaling variable

y =EγEπ

(17)

dNγ

dy=

2

βπ(18)

ymin =1− βπ

2(19)

ymax =1 + βπ

2(20)

Ultrarelativistic limit:dNγ

dy= 2 (21)

ymin = 0 (22)

ymax = 1 (23)

Note: scaling function:

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6 Charged pion decay. Branching ratios

The charged pion has two modes of decay possible:

π+ → µ+ νµ

π+ → e+ νe

The decay happens nearly entireli into muons, with the electron mode havinbg a branchingratio of order 1.23× 10−4.

The expression for the rate of decay into the mode π+ → `+ ν` is:

Γπ→µν =G2F |Vud|2

8πf2πmπm

2`

(1−

m2`

m2π

)2

(24)

Note the factor m2` that is the consequence of the (V-A) stricture of the Weak interactions.

The strong suppression of the electron mode is an outstanding manifestation of theV − A nature of the charged current weak interactions, and a clear illustration of thedifference between the chirality and the helicity.

Let us consider (see fig. ??) the decay of pions at rest: π+ → `+ + ν`.

/++e+ie

pp eie

S Sie e+

Figure 1: Charged pion decay in the rest frame of the pion.

The V-A structure of weak interactions requires the emitted ν` to be of left handedchirality. For mν ' 0 this also means that it has the left handed (or negative) helicity (spinantiparallel to its momentum). Conservation of total angular momentum then requires`+ to have negative helicity. However, the `+ is an antiparticle, and again for the V-Astructure of weak interactions it must be produced in a state of right–handed chirality.Therefore the amplitude of the process must be proportional to the admixture of lefthanded (negative) helicity for a right–handed chirality charged–lepton, that is proportionalto its mass: A(π+ → `+ + ν`) ∝ m`.

Including phase space effects ∝ (m2π −m2

` ) one has the expectation:

Rπ ≡Γ(π+ → e+ + νe)

Γ(π+ → µ+ + νµ)=

(me

mµ

)2(m2π −me

m2π −m2

µ

)2

= 1.28× 10−4 (25)

In agreement (after including a 4% radiative correction) with the experimental value:Rπ = (1.230± 0.004)× 10−4.

7 Charged pion decay. Kinematics

General case of a two body decay:x→ a+ b

6

Energy in the c.m. frame:

E∗a =(m2

x +m2a −m2

b)

2mx(26)

E∗b =(m2

x +m2b +m2

a)

2mx(27)

It is easy to check that energy and momentum are conserved (the two particles are pro-duced in the c.m. frame with equal and opposite momenta).

E∗a + E∗b = mx (28)√E∗a −m2

a =√E∗b −m2

b (29)

Appplying this general rules to the pion decay (and using the approximation mν ' 0)one finds:

E∗µ =mπ

2

(1 +

m2µ

m2π

)=mπ

2(1 + rπ) (30)

and

E∗ν = p∗ =mπ

2

(1−

m2µ

m2π

)=mπ

2(1− rπ) (31)

The factor rπ is:

rπ =m2µ

m2π

= 0.573 (32)

The important point is that the energy of the pion mass is not shared equally betweenthe two particles in the final state. The muon, that has a mass close to the pion, takes awaya fraction (1 + rπ)/2 ' 0.79, while the neutrino carries away a fraction (1− rπ)/2 ' 0.21.

To compute the spectrum in the laboratory frame (where the pion has energy Eπ) onecan perform a Lorentz transformation: whith γ = Eπ/mπ and β = pπ/Eπ:

Elab = γ (E∗ + β p∗ cos θ∗) (33)

The decay is isotropic in the rest frame (because the pion is a scalar particle of spinzero), and accordingly the distribution of the angle cos θ∗ is flat:

dN

d cos θ∗=

1

2(34)

The distribution in Elab can be calculated as:

dN

dElab=

dN

d cos θ∗d cos θ∗

dElab. (35)

One obtains the result:dN

dElab=

1

2

1

γ βπ p∗(36)

That is the distribution is flat.

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We still have to compute the kinematical limits that is the minimum and maximumenergy of the particles in the lab. frame. The limits are obviously obtained for cos θ∗ = ∓1:

Emin,maxµ = γ

(E∗µ ∓ βπ p∗

)=

Eπmπ

mπ

2[(1 + rπ)∓ βπ (1− rπ)]

=Eπ2

[(1 + rπ)∓ βπ (1− rπ)]

Similarly:

Emin,maxν = γ (p∗ ∓ βπ p∗)

=Eπmπ

mπ

2[(1− rπ)∓ βπ (1− rπ)]

=Eπ2

[(1− rπ)∓ βπ (1− rπ)]

It is again possible to study the spectra in term of the fractional energy yν,µ = Eν,µ/Eπ.In the limit of an ultrarelativistic pion, one has that the kinamtically allowed range of theyν,µ are:

yν ∈ [0, (1− rπ)] (37)

yµ ∈ [rπ, 1] (38)

The spectra of neutrinos and muons created in pion decay are shown in Fig. ??. Onecan see how the spectra take an asymptotic form for Eπ � mπ.

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EΠ = 0.2, 0.5, ¥ GeV

0.0 0.2 0.4 0.6 0.8 1.00

1

2

3

4

y = EΝ,Μ�EΠ

dN�d

y

Figure 2: Spectra of muon and neutrinos in the laboratory in terms of the scaling variable y

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8 Muon decay

Decay of unpolarized muons Matrix element:

µ− → νµ + e− + νe (39)

To compute the distribution of energy (and direction) of the particles in the final state,one has to take into account the dynamics of the decay, that is controlled by the matrixelement.

Note that the muon has spin 1/2. One can consider the decay of a polarized muon(with spin described by the 4–vector sµ). Summing over the different polarizations of thefinal state particles, the matrix element can be expressed in terms of the 4–momenta ofthe particles and the 4–vector of the muon spin.

|M |2 ∝ (pνµ · pe−) (pµ− · pνe) (40)

For the decay of a polarized muon with spin 4-vector sµ:

|M |2 ∝ (pνµ · pe−) [(pµ− · pνe)−mµ (sµ · pνe)] (41)

The differential rate for muon decay can be calculated in the standard way:

dΓ =1

2mµ|M |2 dΦ (42)

In this equation M is the matrix element for the decay, and dΦ is the differential phasespace element for three bodies in the final state, that written explicitely is:

dΦ =

[d3p1

(2π)3 2E1

] [d3p2

(2π)3 2E2

] [d3p3

(2π)3 2E3

](2π)4 δ(4)[pµ − p1 − p2 − p3] (43)

Integrating over the entire phase space one obtains the total decay rate:

Γµ =1

τµ=G2F m

5µ

192π3(44)

Numerically:τµ ' 2.197× 10−6 s (45)

`µ = c τµ βµEµmµ' 6.23

(Eµ

GeV

)km (46)

The inclusive spectra of the neutrinos can be calculated obtaining dΓ/dEj after inte-gration in all other kinematical variables.

Note: the phase space for a final state of three particles is described by 5 variables(one has three 3–momenta, that is 9 variables, and 4 constraints that come from theconservation of energy and momentum).

For an inclusive spectrum, one choose the variable in the rest frame of the parentparticle (the muon) the variables: {E, cos θ, ϕ}, that is the energy of the particle, he angleθ with respect to the spin direction, and an azimuth angle, that by symmetry is clearly

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dynamically irrelevant. To obtain dΓ/(dE d cos θ) one needs to integrate over the rest ofthe phase space.

An outline of the integration follows here: One can perform the integration over d3p3using the delta function (obtaining ~p3 = ~P − ~p1 − ~p2)

dΦ3 body =1

(2π)51

8E1E2E3d3p1 d

3p2 δ[E − E1 − E2 − E3]

=1

(2π)51

8E1E2E3d3p1 d

3p2 δ

[E −

√p21 +m2

1 −√p22 +m2

2 −√p23 +m2

3

](47)

writing

d3p1 = dp1 p21 d cos θ1 dϕ1

d3p2 = dp2 p22 d cos θ12 dϕ12

One can perform the integral over d cos θ12 using the delta over energy with the result:∫d cos θ12[. . .] δ[E −

√p21 +m2

1 −√p22 +m2

2 −√p21 + p22 + 2p1p2 cos θ12 +m2

3] =E3

p1 p2(48)

Let us consider the decay of a muon (a µ−) at rest, and polarized with the spin alongthe +z direction. The spectra of the final state particles will depend on the direction ofemission (with respect to the axis of the muon spin).

One can define the variable x:

x =2E∗νmµ

(49)

where E∗ is energy in the rest frame of the muon. The variable x can vary in the interval:(0 ≤ x ≤ 1). Neglecting the neutrino and the electron mass one has the distributions:

dN

dx d cos θ

∣∣∣∣µ−→ νµ

(x, cos θ) = x2 (3− 2x) + cos θ x2(1− 2x) (50)

dN

dx d cos θ

∣∣∣∣µ−→νe

(x, cos θ) = 6x2 (1− x) + cos θ 6x2(1− x) (51)

Note that in the approximation of neglecting the electron mass, the spectrum of theelectron is identical to the spectrum of the νµ. This can be understood simply inspectingthe structure of the matrix element (that is symmetric for the exchange pe ↔ pνµ).

Integrating in angle, the cos θ term gives zero, and one finds the same result as in thedecay of an upolarized muon:

dN

dx

∣∣∣∣µ−→ νµ

(x) = 2x2 (3− 2x) (52)

dN

dx

∣∣∣∣µ−→νe

(x) = 12x2 (1− x) (53)

These two spectra are shown in Fig. ??.

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The averages of these distributions are:

〈xνµ〉 =7

10(54)

〈xνe〉 =6

10(55)

Note that more energy goes into the νµ (and less energy goes into the νe).Summing over all particles (and using the fact that 〈xe〉 = 〈xνµ〉) one obtains:

〈xνµ〉+ 〈xνe〉+ 〈xe〉 = 2 (56)

(so all energy is accounted for).Integrating in energy (that is in x) one finds the results:

dN

d cos θ

∣∣∣∣µ−→νµ

(x, cos θ) =1

2− cos θ

6(57)

dN

d cos θ

∣∣∣∣µ−→νe

(x, cos θ) =1

2+

cos θ

2(58)

These angular distributions have average:

〈cos θνµ〉 = −1

9(59)

〈cos θνe〉 = +1

3(60)

This indicated that the νµ is emitted preferentially in the direction opposite to thespin of the muon, while the νe is emitted in the direction of the spin of the muon.

[Note that in the decay of of a µ+, one has the opposite, the νµ is emitted in thedirection of the spin of the µ+ and the νe is emitted preferentially in the opposite direction.This can be immediately understood on the basis of CP symmetry]The qualitative results are the following

• The νµ carries more energy than the νe

• The νµ emitted preferentially in the direction of the µ− spin.The νe is emitted preferentially in the direction opposite to the µ− spin.

These results are easy to understand qualitatively in the basis of simple considerationsabout the the spins of the different particles in the final state In the decay of µ− →νµ + νe + e−, the νµ and the e− have helicity −1, while the νe has helicity +1.

Figure ?? wants to illustrate qualitatively why in µ− decay the νµ has a harder spec-trum that the νe (and similarly in µ+ decay the νµ has a harder spectrum that the νe).

One can inspect a configuration where, in the c.m. frame, one particle (the νµ inthe left panel and the νe in the right panel) is emitted with the maximum kinematicallyallowed energy (E∗ = mµ/2). The other two particles must then be antiparallel, and the

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0.0 0.2 0.4 0.6 0.8 1.00.0

0.5

1.0

1.5

2.0

x = 2 E*�mΜ

dN�d

x

Figure 3: Spectra of neutrinos in muon decay. The spectra are shown in the muon rest frame. The solidline is for the spectrum of νµ in µ− decay, The dashed line is for the spectrum of νe in µ− decay.

sum of their energy must also be equal to mµ/2. Because of the (V − A) structure ofthe Weak interaction, one has that the helicity of the e− and the νe are negative, andthe helicity of the νe is positive. One can see that when the particle with the maximumenergy mµ/2 is the νµ (left panel), the spins of the three particles combine to give a totalangular momentum of 1/2 (in units of ~). This is of course allowed. However when theparticle with the maximum energy is the νe (right panel), the spins of the three particlescombine to give a total angular momentum of 3/2, and this is forbidden (because theangular momentum of the initial state is 1/2.

It follows that the emission of νe at high energy in the c.m. frame is more suppressedthan the production of νµ, and the spectrum is softer.

Note that this argument remains valid also in the case of the decay of µ+, because thehelicity of all particles in the final state is reversed (and the spectrum of the νe is softerthan the spectrum of the νµ).

Figure ?? illustrates qualitatively why in the decay µ− → νµ e− νe, in the rest frame

of the µ−, the νe is emitted prevalently in the direction of the muon spin, and the νµ ismostly emitted in the direction that is opposite to the muon spin.

Note that the effect is inverted in the decay of µ+, where the νµ is emitted mostlyparallel to the spin, and the νe is emitted mostly anti–parallel to the spin.

The qualitative argument is simply that, because of the V −A structure of Weak decays,the neutrinos are emitted with negative helicity, (spin anti-paralle to 3-momentum vector)and the anti–neutrinos have positive helicity (spin parallel to 3-momentum vector).

Conservation of the angular momentum favors emission with the spin of the final state

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Μ- decay ΝΜ

e-

Νe

Allowed

Μ- decay Νe

e-

ΝΜ

Forbidden

Figure 4: Qualitative discussion of the neutrino spectra in muon decay.

particle parallel to the spin of the parent particle. Since the spin direction and the 3–momentum direction are perfectly correlated, the angular distribution of the emission isdistorted.

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Μ- decay Νe Μ- decay

ΝΜ

Figure 5: Qualitative discussion of the neutrino spectra in muon decay.

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8.1 Laboratory frame distributions

To compute the spectrum in the laboratory frame, one can start from the expression ofthe energy of the a particle in that frame:

Eν = γ (1 + βµ cos θ∗) E∗ν (61)

In the ultrarelativistic limit: βµ → 1, and using γ = Eµ/mµ, E∗ = mµ x/2, and with thescaling variable y = Eν/Eµ, one has:

y =x

2(1 + cos θ∗) (62)

To obtain the energy distribution of the neutrinos in the laboratory frame one has tointegrate over x and cos θ∗ (with the appropriate constraint).

dN

dy

∣∣∣∣µ→ν

=

∫ 1

0dx

∫ +1

−1d cos θ∗

{dN

dx d cos θ∗(x, cos θ∗)

}δ[y − x

2(1 + cos θ∗)

](63)

One can perform one integration using the delta function.Integrating in x:

dN

dy

∣∣∣∣µ→ν

=

∫ 1

y

dx

x

{dN

dx d cos θ∗(x, cos θ∗ = 2y/x− 1)

}(64)

Note that in this integration θ∗ is the angle with respect to the boost direction, howeverwe have calculated the energy–angle distribution of the neutrino with the angle θ withrespect to the spin of the muon.

It is however straightforward to compute the two special cases where the muon is inan eigenstate of helicity h = ±1, because in these cases

cos θspin = ± cos θboost (65)

and therefore one can use equations (??) and (??) to insert them in the integration ofequation (??).

The result for the general case of a muon that has helicity 〈h〉 (with range of possiblevalues 〈h〉 ∈ [−1, 1]) One obtains the distributions:

dN

dy

∣∣∣∣µ−→ νµ

(y) =5

3− 3 y2 +

4

3y3 + 〈helicity〉

[1

3− 3 y2 +

8

3y3]

(66)

dN

dy

∣∣∣∣µ−→ νe

(y) = 2 +−6 y2 + 4 y3 − 〈helicity〉[2 (1− 4y) (1− y)2

](67)

The results can be summarized as follows: In the decay of a µ± the muon (anti)–neutrino and the electron neutrinos carry away an fractional energy:

〈yνµ〉 =7

20− 〈helicity〉 1

20(68)

〈yνe〉 =6

20+ 〈helicity〉 1

10(69)

In the case of unpolarized parent muons, the muon neutrino (or anti–neutrino) carriesa larger energy than the electron anti–neutrino (or neutrino).

The sharing of energy between the two neutrino types depends however on the helicitystate of the parent muon.

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0.0 0.2 0.4 0.6 0.8 1.00.0

0.5

1.0

1.5

2.0

y = EΝ�EΜ

dN�d

y

Figure 6: Spectra of neutrinos in the decay of unpolarized µ−. The spectra are shown in a frame wherethe muon is ultrarelativistic. The solid line is for νν , The dashed line is for νe..

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0.0 0.2 0.4 0.6 0.8 1.00.0

0.5

1.0

1.5

2.0

y = EΝ�EΜ

dN�d

y

Figure 7: Spectra of νµ in µ− decay. The spectra are shown in a frame where the muon is ultrarealtivistic.The solid line is for an unpolarized muon. The dashed line is for helicity +1. The dot–dashed line is forhelicity −1.

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0.0 0.2 0.4 0.6 0.8 1.00

1

2

3

4

y = EΝ�EΜ

dN�d

y

Figure 8: Spectra of νe in µ− decay. The spectra are shown in a frame where the muon is ultrarealtivistic.The solid line is for an unpolarized muon. The dashed line is for helicity +1. The dot–dashed line is forhelicity −1.

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8.2 Chain decay of pions

As discussed in the previous subsection, the spectrum of the neutrinos created in the decayof a muon depends on its polarization state.

Muons that are created in the decay of charged pions are polarized, and therefore incalculating the spectrum of neutrinos generated in the chain decay πµν one has to takeinto account this fact.

The reason why the muon created in the decay of a charged pion is polarized hasbeen already discussed. The muon polarization is a simple consequence of fact of theconservation of angular momentum and that the neutrino is left–handed.

The problem of calculating the muon polarization is very simple in the rest frame ofthe pion. In this frame the muon and the neutrino are emitted back to back. In the decayπ− → µ− νµ, the ν is right–handed, and to conserve angular momentum, also the µ− mustbe right–handed (so that the total angular momentum of the final state is zero).

In a general frame, the calculation of the polarization of the muon is more compli-cated, and the helicity of the muon depends on its energy. The problem is illustrated infig. ??. Inspecting the figure one can understand that in a frame where the parent pion isrelativistic, the µ− with highest energy have helicity +1, and the muons with the lowestenergy have helicity −1.

[Note that in the case of the charged–conjugate decay π+ → µ+νµ, all spins must bereversed. The µ+ is left–handed in the rest frame. In a frame where the π+ is relativistic,high energy µ+ have helicity −1 and low energy muons have helicity +1].

20

-1.5 -1.0 -0.5 0.0 0.5 1.0 1.5-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

pz�p*

p x�p

*

-1 0 1 2 3-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

pz�p*

p x�p

*

-1 0 1 2 3-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

pz�p*

p x�p

*

5 6 7 8 9 10 11 12-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

pz�p*

p x�p

*

Figure 9: Muon polarization in charged pion decay (For the π− → µ− νµ. In all panels the ellipse showsthe kinematically allowed values of p⊥ and pz for the muon in the final state. The red arrows shows thedirection of the muon spin. In the first panel the decay is seen in the pion rest frame, and the µ− hasalways helicity +1. In the other panels the decay is seen in frames where the muon has different energies(0.143, 0.1450 and 0.350 GeV). In general the muon helicity depends on the angle of emission of the muon(or equivalently on the muon momentum). Note that when the pion has energy Eπ = 0.1450, the particleshas β ' 0.27138, that is equal to the velocity of the muon is the pion rest frame, and a muon produced“backward” in the pion rest frame is at rest in the lab. frame.

21

The spin 4–momentum for the muon s, is the 4–vector that in the rest frame of themuon is purely spatial

s = {0, n̂} (70)

with n̂ is the versor of the spin direction. Note that one has:

s · s = −1 (71)

The spin 4–vector is orthogonal to the 4–momentum p:

s · p = 0 (72)

If a particle has 4–momentum

p = mγ {1, β n̂}

then the spin 4–vector in the case of helicity +1 is:

s = γ {β, n̂}

Let us now consider the decay of a pion (π− → µ−νµ) in the pion rest frame, where themuon is emitted in the direction cos θ∗ (and azimuth angle 0, so that the 3–momentumis in the (x, z) plane). In the rest frame the µ− is polarized 100% with helicity +1. Themomentum and spin 4-vectors are:

pµ =mπ

2{(1 + r), (1− r) sin θ∗, 0, (1− r) cos θ∗} (73)

The spin 4-vector is then:

sµ =mπ

2mµ{(1− r), (1 + r) sin θ∗, 0, (1 + r) cos θ∗} (74)

Transforming in the laboratory frame, where the pion has energy Eπ (and momentumin the +z direction), one finds that the muon energy, and the 0–component of the spin4–vector are:

Elabµ =

Eπ2

[(1 + r) + βπ (1− r) cos θ∗] (75)

slab0 =Eπ

2mµ[(1− r) + βπ (1 + r) cos θ∗] (76)

To compute the polarization, we can now rotate the coordinates, so that the 4–momentum of the muon has form:

plab,rotµ = {E, 0, 0, E βµ} (77)

The spin 4–vectors has components:

slab,rotµ = {s0, sx, 0, sz} = {s0, sx, s0/βµ} (78)

The last equality is obtained requiring that the 4–vectors s and p are orthogonal.

22

We can now perform a boost along the new z axis that brings the muon at rest. Thez–component of the spin 4–vectors becomes:

sr.f.z = γµ (sz − βµ s0)

= γµ

(s0βµ− βµ s0

)=

γµβµ

s0(1− β2µ

)=

s0γµ βµ

(79)

The quantity sr.f.z is in fact the helicity of the muon. This is because is simply equal tocos θ with θ the angle with respect to the muon momentum. The helicity h of a particleis equal to

h = P+ − P− (80)

where P+ and P− are the probability of having spin +1/2 and spin −1/2 in the directionof the momentum. If the spin is pointing in the direction θ with respect to the momentum,then P+ = cos2 θ/2 P− = sin2 θ/2 and

h = cos2θ

2− sin2 θ

2= cos θ (81)

Helicity =1

βµ

(1− r) + (1 + r) cos θ∗ βπ(1 + r) + (1− r) cos θ∗ βπ

(82)

In the ultarelativistic limit one has βπ → 1, βµ → 1, and the helicity of the muonbecomes

Helicity =(1− r) + (1 + r) cos θ∗

(1 + r) + (1− r) cos θ∗(83)

this can be also expressed a function of the quantity x = Eµ/Eπ using the fact that:

x =1

2[(1 + r) + (1− r) cos θ∗] (84)

Helicity =(1− r) + (1 + r) cos θ∗

(1 + r) + (1− r) cos θ∗(85)

The result can therefore be written as:

Helicity =x− r (2− x)

x(1− r)(86)

This function is shown in Fig. ??.

23

0.6 0.7 0.8 0.9 1.0-1.0

-0.5

0.0

0.5

1.0

EΜ�EΠ

Hel

icity

HΜ+

L=-

Hel

icity

HΜ-

L

Figure 10: Helicity of the muon generated in the decay of a pion. The helicity is calculated in a framewhere the particles are ultrarelativistic, and is plotted as a function of the ratio Eµ/Eπ.

24

-1.0 -0.5 0.0 0.5 1.0-1.0

-0.5

0.0

0.5

1.0

EΜ�HEΜ,maxL-EΜ,minL

Hel

icity

HΜ+

L=-

Hel

icity

HΜ-

L

Figure 11: Helicity of the muon generated in the decay of a pion. The helicity is calculated in a framewhere the pion has energy Eπ and is plotted as a function of the ratio Eµ/(Eµ,max − Eµ,min). The pionenergy are: Eπ = 0.142 GeV, Eπ = E∗π = (m2

π +m2µ)/(2mπ), Eπ = 0.2 GeV, Eπ = 0.3 GeV and Eπ →∞.

25

The final step of the calculation (to obtain the spectrum of neutrinos in the chaindecay πµ→ ν requires the convolution of the spectrum of muons in pion decay, and thenthe decay of the ploarized muon.

Defining: x = Eµ/Eπ, y = Eν/Eµ and z = Eν/Eπ One obviously has z = x y, and

dN

dz

∣∣∣∣π→µ→ν

(z) =

∫ 1

rdx

1

1− r

∫dy

dNµ → mu

dy(y;h[x]) δ[z − x y] (87)

With the appropriate helicityThe spectra are shown in the following figures.The average energies are: ⟨

yνµ⟩

=3 + 4 rπ

20' 0.2648 (88)

〈yνe〉 =2 + rπ

10' 0.2573 (89)

Note that

2⟨yνµ⟩

+ 〈yνe〉 =1 + rπ

2(90)

that is the fraction of the pion energy carried by the muon (conservation of energy issatisfied).

0.0 0.2 0.4 0.6 0.8 1.00.0

0.5

1.0

1.5

2.0

2.5

y = EΝ�EΠ

dN�d

y

Figure 12: Spectra of neutrinos in the chain decay of charged pions. The spectra are shown in a framewhere the pion is ultrarelativistic. The solid (dashed) line is the spectrum of νµ (νe) in the chain decayπ− → µ− → νµ.

It is easy to understand qualitatively these results. In the decay of π− → µ−+ νµ, theµ− is created in a state of helicity +1. This is because the anti–neutrino is right–handed,and one has to conserve the angular momentum.

26

0.0 0.2 0.4 0.6 0.8 1.00.0

0.5

1.0

1.5

2.0

y = Eî�EΠ

dN�dy

Figure 13: Spectra of neutrinos in the chain decay of charged pions. The spectra are shown in a framewhere the pion is ultrarealtivistic. The solid line is the spectrum of νµ in the chain decay π− → µ− → νµ.The dashed line is the same spectrum calculated neglecting the muon polarization effects.

0.0 0.2 0.4 0.6 0.8 1.00.0

0.5

1.0

1.5

2.0

2.5

3.0

y = EΝ�EΠ

dN�d

y

Figure 14: Spectra of neutrinos in the chain decay of charged pions. The spectra are shown in a framewhere the pion is ultrarealtivistic. The solid line is the spectrum of νe in the chain decay π− → µ− → νe.The dashed line is the same spectrum calculated neglecting the muon polarization effects.

27

8.3 Decay of Kaons

In most circumstances (the only exception is interaction in very dense media) the dominantparticles for neutrino production are the charged pions π± that, together with their neutralisospin partner, are the particles most abundantly produced in hadronic interactions,accounting for over one half of the total energy in the final state. Charged pions decaywith approximately 100% branching ratio in the mode

π+ → µ+ + νµ (91)

(and charge conjugate mode). In most astrophysical environments the medium where themuons propagates has a density sufficiently low, so that the µ± decay before loosing anappreciably amount of energy in the well known mode

µ+ → e+ + νe + νµ (92)

Therefore the chain decay of a π+ results in the emission of three neutrinos with flavors(νµ, νµ, νe), and the decay of a π− in the charge conjugate triplet (νµ, νµ, νe). This,together with a knowledge of the π+/π− production ratio essentially determines the flavorcomposition before propagation for most neutrino sources. The neutral isospin partner ofthe charged pions decays electromagnetically in the mode: π◦ → γ+γ generating photons.The dynamics of production of π+, π− and π◦ are connected by isospin symmetry, andthis is the origin of a strong correlation between the neutrino and photon fluxes.

The second source of neutrinos in order of importance is the decay of kaons. Theimportant decay modes for charged kaons are:

K+ → µ+ + νµ (BR = 0.634) (93)

K+ → π◦ + e+ + νe (BR = 0.0482) (94)

K+ → π◦ + µ+ + νµ (BR = 0.0318) (95)

(and charge conjugate channels). For neutral kaons, only the KL has significant semi–leptonic branching ratios in the modes:

KL → π∓ + e± + νe(νe) (BR = 0.194) (96)

KL → π∓ + µ± + νµ(νµ) (BR = 0.135) (97)

The amount of energy that goes into kaon production is only approximately 10% of theenergy that goes into producing pions, however in the case of atmospheric neutrinos theshorter lifetime of kaons results in a larger decay probability and an enhanced contributionof kaons. Additional sources of neutrinos are the decay of hadrons containing charm (andheavier quarks).

28

9 Decay spectra in the ultrarelativistic limit

The inclusive energy spectra of a secondary particle f generated in the decay of a primaryparticle i, calculated in a frame where the parent particle is ultrarelativistic has a scalingform:

In general the spectrum is a function of two variables: (the energies of the initial andfinal particles)

dN

dEf

∣∣∣∣i→f

= fi→f (Ef ;Ei) (98)

but in the limit of large Ei, then the spectrum takes the scaling form:

dN

dEf

∣∣∣∣i→f

=1

EiFi→f

(EiEf

)(99)

[Note that the inclusive spectrum is normalized so that after integration over all energiesone obtains the average number of particles of type f created in a decay] Equation (??)can be also rewritten in the form:

dNi→fdy

= Fi→f (y) (100)

where we have introduced the fractional energy

y =EiEf

. (101)

The inclusive spectrum of particle f in the rest frame can be described in the directionof emission cos θ∗ in terms of the variable x:

x =2E∗

mi(102)

The variable xi, in general can only vary in the interval

0 ≤ x ≤ 1 (103)

(depending on the decay mode, the limits on the range of x will be in general morerestricted with xmin > 0 and xmax < 1.)

The quantities Elab and E∗ (the energy of particle f in the laboratory frame and inthe c.m. frame) are related by a Lorentz transformation:

Ef,lab = γi [E∗ + βi p∗ cos θ∗]

= γi [1 + βi βf cos θ∗] E∗

=Eimi

[1 + βi βf cos θ∗]mi x

2

=Ei x

2[1 + βi βf cos θ∗]

=Ei x

2[1 + βi βf cos θ∗]

=⇒ Ei x

2[1 + cos θ∗] (104)

29

The bottom line is that asymptotically one has the simple relation

y =EfEi' x

2[1 + cos θ∗] (105)

Note that the energy Ei of the parent particle is not present in the asymptotic expression.The spectrum dN/dy can be obtained summing (integrating) on all combinations of x

and cos θ∗ that yield y:

dN

dy

∣∣∣∣i→f

=

∫ 1

0dx

∫ +1

−1d cos θ∗

dN

dx d cos θ∗(x, cos θ∗) δ

[y − x

2(1 + cos θ∗)

](106)

One can perform one integration using the delta function. For example one can inte-grate in x:

dN

dy

∣∣∣∣i→f

=

∫ 1

y

dx

x

dN

dx d cos θ∗(x, cos θ∗ = 2y/x− 1) (107)

This is explicitely independent from the energy of the parent particles

30

10 Decay of a power law spectrum of parent particles

Let us consider the decay of an ensemble of primary particles that have a power lawspectrum. For example an ensemble of charged pions with spectrum:

Nπ(Eπ) = Kπ E−απ . (108)

After the complete decay of all of these particles, one finds a spectrum of neutrinos thatcan be calculted performing the integral:

Nν(Eν) =

∫ ∞Eν

dEπ Nπ(Eπ)dN

Eν

∣∣∣∣π→ν

(Eν ;Eπ) (109)

Using the scaling relation

dN

Eν

∣∣∣∣π→ν

(Eν ;Eπ) =1

EπFπ→ν

(EνEπ

)(110)

and the power law form of the pion spectrum one can express the integration (??) as:

Nν(Eν) =

∫ ∞Eν

dEπ Nπ(Eπ)dN

Eν

∣∣∣∣π→ν

(Eν ;Eπ)

=

∫ ∞Eν

dEπ Kπ E−απ

1

EπFπ→ν

(EνEπ

)(111)

One can now change variable, passing from the integration in Eπ to the integration in thevariable y = Eν/Eπ. Using:

Eπ =Eνy

(112)

dy = −EνE2π

= −ydEπEπ

(113)

or:dy

y= −dEπ

Eπ(114)

One can rewrite Eq. (??) as:

Nν(Eν) =

∫ ∞Eν

dEπ Kπ E−απ

1

EπFπ→ν

(EνEπ

)

=

∫ 1

0

dy

yKπ

(Eνy

)−αFπ→ν(y)

= Kπ E−αν

∫ 1

0dy yα−1 Fπ→ν(y)

= Kπ Zπ→ν(α) E−αν (115)

where we have defined the so called Z factor:

Zπ→ν(α) =

∫ 1

0dy yα−1 Fπ→ν(y) (116)

The (α− 1) momentum of the (scaling) decay distribution.

31

11 Z factors for Weak decays

The Z–factors for Weak decays can be calculated from their definition:

Za→b(p) =

∫ 1

0dx xp−1 Fa→b(x) (117)

For the two body decay of charged pions one has for the neutrino:

Zπ+→νµ(p) =(1− rπ)p−1

p(118)

and for the muon:

Zπ+→µ+(p) =(1− rpπ)

p (1− rπ)(119)

For the decay of muons one has:

Zµ−→νµ(p) =2(p+ 5)

p (p2 + 5p+ 6)+ h

2− 2p

p3 + 5p2 + 6p(120)

and:

Zµ−→νe(p) =12

p3 + 5p2 + 6p+ h

12(p− 1)

p (p3 + 6p2 + 11p+ 6)(121)

(where 〈h〉 is the helicity of the muon in the laboratory frame.The Z factors for the decay of µ+ are given by the same expressions but inverting the

sign of the helicity (that is with the replacement h→ −h).For the chain decay π+ → µ+ → ν one has:

Zπ+→µ+→νµ(p) =4 ((p(r − 1) + 2r − 3)rp − 2r + 3)

p2(p+ 2)(p+ 3)(r − 1)2(122)

and

Zπ+→µ+→νe(p) =24 (r (rp − 1) + p(−r) + p)

p2(p+ 1)(p+ 2)(p+ 3)(r − 1)2(123)

32

Π+ ® Μ+ + ΝΜ ® e+ Νe ΝΜ + ΝΜ

ΝΜ

Νe

ΝΜ

1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.00.0

0.1

0.2

0.3

0.4

0.5

Exponent Α

ZHΑ

L

Figure 15: The figure shows the Z factors for the decays π+ → νµ, π+ → µ+ → νµ and π+ → µ+ → νeas a function of the exponent α.

K+ ® Μ+ + ΝΜ ® e+ Νe ΝΜ + ΝΜ

ΝΜ

Νe

ΝΜ

1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.00.0

0.1

0.2

0.3

0.4

Exponent Α

ZHΑ

L

Figure 16: The figure shows the Z factors for the decays K+ → νµ, K+ → µ+ → νµ and K+ → µ+ → νeas a function of the exponent α (considering only the two–body mode K+ → µ+νµ).

33

Π+ ® Μ+ + ΝΜ ® e+ Νe ΝΜ + ΝΜ

ΝΜ

Νe

ΝΜ

1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.00.8

1.0

1.2

1.4

1.6

1.8

Exponent Α

Rel

ativ

eN

eutr

ino

prod

uctio

n

Figure 17: Relative size of the Z–factors for the decays π+ → νµ, π+ → µ+ → νµ and π+ → µ+ → νeas a function of the exponent α. One can see that the νµ (created in the first decay) has the smallestZ–factor. The difference between the three neutrino types grows with the exponent α.

34

K+ ® Μ+ + ΝΜ ® e+ Νe ΝΜ + ΝΜ

ΝΜ

Νe

ΝΜ

1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.00.6

0.7

0.8

0.9

1.0

1.1

1.2

Exponent Α

Rel

ativ

ene

utri

nopr

oduc

tion

Figure 18: Relative size of the Z–factors for the decays K+ → νµ, K+ → µ+ → νµ and K+ → µ+ → νeas a function of the exponent α (only the two–body decay K+ → µ+νµ is taken into account).

35

Π+ ® Μ+ ® e+ Νe ΝΜ

Νe

ΝΜ

1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0

-0.15

-0.10

-0.05

0.00

0.05

0.10

0.15

Exponent Α

DZ

pola

riza

tion

HΑL�

ZHΑ

L

Figure 19: This figure shows the effect of taking into account the polarization of the muon in thecalculation of the neutrino spectra for the chain decay of charged pions (π+ → µ+ + νµ → (e+νeνµ) + νµ).The figure shows the quantity ∆Z/Zpol = (Zpol − Z0)/Zpol, where Zpol and Z0 are the momenta of thenetrino calculated including and neglecting the effects of the muon polarization. The polarization effectsuppresses (enhances) the flux for the muon (electron) anti–neutrino produced in the muon decay.

36

K+ ® Μ+ ® e+ Νe ΝΜ

Νe

ΝΜ

1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0-0.4

-0.2

0.0

0.2

0.4

Exponent Α

DZ

pola

riza

tion

HΑL�

ZHΑ

L

Figure 20: Same in fig. ??, but for the chain decay of charged kaons. Note that the effect of polarizationis larger.

37

Neutrinos from pion decay

Π+ = Π- = Π0

1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.00.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

Exponent Α

Γ�

Ν

Figure 21: This figure shows the quantity Rπνγ(α) = 2(Zπ+→νµ + Zπ+→µ+→νe + Zπ+→µ+→νµ)/Zπ0→γ .This quantity is an approximation of the ratio φν/φγ if the source of neutrinos and the photons is a powerlaw of pions with equal numbers for the three charge states (π± and π◦).

38