Basdevant J.-L., Rich J., Spiro M. Fundamentals in Nuclear Physics: From Nuclear Structure to Cosmology

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4.3 Weak interactions 211

RRRR

detecto

tritium

source

solenoid

coils

solenoid

coils

B field lines

electrodes

18.55

retarding energy (keV)

counts per second

0.04

0.03

0.02

0.01

18.56 18.57 18.58

fit for m=0

Fig. 4.18. The Mainz tritium β-decay spectrometer [45]. Electrons emerging from

a thin tritium source enter a region with an electric and magnetic ﬁeld. Electrons

spiral around the magnetic ﬁeld lines and the electric ﬁeld reﬂects all but the most

energetic ones. Those passing this barrier then pass through a symmetric electric

ﬁeld to recover their initial kinetic energy before being detected. The counting rate

a a function of the retarding potential reﬂects the neutrino energy spectrum near

the end-point. The curve shows the prediction for vanishing neutrino mass (after

correction for experimental resolution).

212 4. Nuclear decays and fundamental interactions

appropriately corrected for experimental resolution, results in an upper limit

on the mass of the

< 3eV . (4.103)

Note that the limit is comparable to atomic binding energies so to correctly

analyze the experiment, it is necessary to consider the spectrum of the pro-

duced helium atom.

A second important characteristic of neutrinos (antineutrinos) is that they

are produced in β-decay with negative (positive) helicity:

s · p

|p|

=+¯h/2(

ν)=−¯h/2(ν) , (4.104)

where s is the neutrino spin. This is a manifestation of parity violation in

the weak interactions.

There is only one experiment where one has been able to deduce the

neutrino helicity, essentially by creating a situation where the neutrino helic-

ity is correlated with a photon helicity and then measuring the latter. The

experiment [46] uses the electron-capture decay of

152m

Eu shown in Fig. 4.19

− 152m

Eu(0

−

) →

152

∗

−

) ν

= 840 keV

152

∗

−

) →

152

Sm(0

) γ (961 keV) . (4.105)

As shown in the ﬁgure, if the neutrino and photon are emitted in opposite

directions, conservation of angular momentum requires that the photon spin

have the opposite sign of that of the neutrino spin. Since they go in oppo-

site directions, the photon helicity must have the same sign as the neutrino

helicity.

Because the directions of the photon and neutrino momenta are nearly

uncorrelated, the problem is to select only those photons that happen to be

emitted opposite to the neutrino. This is done by scattering the photons in

a sample of

152

Sm before they enter the detector. Only photons emitted op-

posite to the direction of the neutrino have suﬃcient energy to be resonantly

scattered

152

Sm(0

) →

152

∗

−

) → γ

152

Sm(0

) . (4.106)

The fact that they were emitted opposite to the neutrino nearly compensates

for the energy taken by the

152

Sm in the radiative decay (4.105) (Exercise

4.9). Photons of energy 961 keV detected in the NaI scintillator must then

have been emitted opposite to the neutrino. (All photons can Compton scat-

ter but this lowers their energy below 961 keV.)

Once one has selected the photons emitted opposite to the neutrinos,

it is only necessary to measure their helicity. This can be done by placing

magnetized iron in their path. Because the Compton cross-section depends

on the relative orientations of the photon and electron spins, photons are

scattered out of their initial direction at rates that depend on the relative

4.3 Weak interactions 213

NaI

961 keV

ν 840 ke

152

−

961 keV837 keV

−

13 y

9.3 h

j=0 j=1

j=0

−

shield

Fig. 4.19. Experiment used to measure the helicity of the neutrino [46]. The ex-

periment uses the electron capture decay of the 0

−

isomer of

152

Eu decaying with

a half-life of 9.3 h to the 1

−

excited state of

152

Sm. In the experimental apparatus

shown on the upper right, the 961 keV de-excitation photons can be scattered in

152

Sm and be detected in a NaI scintillator. The de-excitation photons are emitted

in directions that are uncorrelated with the neutrino direction, but only photons

emitted opposite to the neutrinos have suﬃcient energy to excite the 1

−

state of

152

Sm and therefore be resonantly scattered into the NaI. This is because in this

case, the recoil energy of the 1

−

nucleus compensates for the recoil of the 0

nucleus

so the photon has the entire 961 keV transition energy (Exercise 4.9). As shown on

the bottom of the ﬁgure, in order to conserve angular momentum, photons emitted

opposite to the neutrino must have the same helicity as the neutrino. The neutrino

helicity can therefore be deduced by measuring the helicity of the photon.

214 4. Nuclear decays and fundamental interactions

orientation of their spins and the magnetization vector. Measurement of the

photon counting rate for two magnetizations then allows one to deduce the

photon polarization.

4.3.5 Neutrino detection

Because of their tiny cross-sections, neutrino detection presents the challenge

of using detectors that are large enough to have a suﬃcient counting rate yet

intelligent enough to distinguish neutrino events from much more common

backgrounds coming from cosmic-rays and natural radioactivity. These diﬃ-

culties explain the long delay between Pauli’s neutrino hypothesis in 1934 and

their experimental detection by Reines and Cowan in 1956. Since their pio-

neering experiment, observations of solar neutrinos (Sect. 8.4.1), supernova

neutrinos (Sect. 8.4.2) and of “artiﬁcial” neutrinos from reactors and particle

accelerators have lead not only to important advances in astrophysics but

also the surprising observation of “neutrino oscillations” caused by neutrino

masses (Sect. 4.4).

The reaction used by Reines and Cowan was

p → e

n . (4.107)

This reaction was chosen for several reasons. First, nuclear-ﬁssion reactors are

intense sources of

because most ﬁssion products are short-lived β

−

-emitters

(Chap. 6). Second, massive targets can be made since ordinary organic com-

pounds contain large numbers of protons. Finally, the reaction gives a distinct

“signature” of a positron produced in the neutrino reactions followed, some

time later, by the capture of the thermalized neutron. (The neutron is ther-

malized by elastic scatters on nuclei, especially hydrogen nuclei.)

The

production rate of a nuclear reactor is easily estimated since each

ﬁssion yields about 200 MeV of thermal energy and, on average, 5

coming

from the cascade of β-decays of the two ﬁssion products. The neutrino pro-

duction rate rate is therefore simply related to the thermal power P of the

reactor

200 MeV

∼ 1.5 × 10

−1



1GW



. (4.108)

The mean

energy is ∼ 4 MeV so the mean

cross-section, from Table

3.1 is σ ∼ 10

−47

. At a distance R from the reactor, this gives a rate per

proton of

λ(

p → e

n) =

4πR

=10

−30

sec

−1



10m



. (4.109)

To get a rate of 1 event per hour at a distance of 10 m, we need about 2×10

protons corresponding to about 2 kg of organic material (CH).

A typical realization of such a detector is shown schematically in Fig.

4.20. The organic target is in the form of a liquid scintillator that emits light

4.3 Weak interactions 215

neutron−proton

elastic scatters

neutron capture

compton scatte

ν p

n e

photo−electric absorption

photomultipliers

liquid scintillator

photoelectric

absorption

annihilation

pair production

Fig. 4.20. A schematic of the standard method of detecting

through the re-

action

p → e

n. The detector consists of liquid scintillator instrumented with

photomultipliers. The

scatters on a proton contained in the scintillator, an or-

ganic compound. The positron stops through ionization loss (Sect. 5.3) and then

annihilates, e

−

→ γγ. The neutron thermalizes though elastic scatters on pro-

tons and is eventually captured on a nucleus, n(A, Z) → γγ(A + 1, Z). The photons

produced in the capture and in the annihilation either convert to e

−

pairs or

lose energy through Compton scattering, eventually being absorbed photoelectri-

cally. Scintillation light is produced by the electrons and positrons slowing down in

the scintillator. The scintillation light is detected by the photomultipliers. The light

comes in two ﬂashes, the ﬁrst from the positron produced in the original interaction,

and the second from the Compton and photo-electrons after the thermalization and

capture of the neutron.

216 4. Nuclear decays and fundamental interactions

E (MeV)

reactor on

reactor off

Events

100

200

300

Fig. 4.21. Spectrum of positrons created by the reaction

p → ne

as observed

by the Chooz neutrino experiment [47].

in response to the passage of charged particles. A ﬂash of light is then pro-

duced by the positron created in the

interaction. A certain fraction of this

light is detected by photomultipliers deployed in the liquid. The neutron is

thermalized by elastic collisions with the protons in the scintillator (Exer-

cise 4.10). It is eventually captured, either by a proton or a nucleus with a

high capture cross-section that is added to the scintillator. The photons pro-

duced in the capture then interact in the scintillator via Compton scattering,

electron–positron pair production, and photoelectric absorption (Sect. 5.3.4,

Exercise 4.11). Each of these processes create electrons and their total eﬀect

is to create a second pulse of scintillation light. The time between the ﬁrst

and second pulse is determined by the neutron-capture rate and is gener-

ally ∼ 1 ms. The double pulse then distinguishes neutrino events from more

common events due to natural radioactivity.

Figure 4.21 shows the positron energy spectrum deduced from the quan-

tity of scintillation light produced in the initial pulse. The positron en-

4.3 Weak interactions 217

ergy is just equal to the neutrino energy minus the reaction threshold,

+ m

− m

=1.8 MeV. The distribution of neutrino energies agrees

with that expected from the β-decays of the ﬁssion products.

The detection of ν

from β

-decay is more diﬃcult than detection of

because the analogous reaction ν

n → e

−

p requires a target consisting of free

neutrons. A good approximation to a free neutron target uses deuterons:

H → e

−

pp. (4.110)

Suﬃciently massive neutrino detectors can be constructed from heavy water

and have been used to detect solar neutrinos (Sect. 8.4.1). In these detectors,

the ﬁnal-state electron is seen through its emission of “Cherenkov” light that

is detected by photomultipliers. Because there are no neutrons produced in

the event, there is no distinctive double pulse to signal a neutrino interac-

tion so the reaction can only be used for E

> 5 MeV where there is little

background coming from radioactivity.

The SNO experiment using heavy water also observes the deuteron

breakup reaction

H → ν pn. (4.111)

This reaction has a rather nondistinctive signature, the liberation of a neutron

and its subsequent capture. It has the advantage of being a neutral current

reaction so it is equally sensitive to all neutrino species, ν

, ν

and ν

Heavy water (and light water) detectors are also sensitive to neutrino-

electron elastic scattering

ν e

−

→ ν e

−

. (4.112)

This is a mixed neutral-current/charged reaction so it is sensitive to all species

but with a larger cross-section for ν

Low-energy neutrinos can be also detected through capture reactions

(A, Z) → e

−

(A, Z+1). (4.113)

Important examples that have been used for solar neutrino detection are

Cl → e

− 37

Ar Q = −0.81569 MeV , (4.114)

and

Ga → e

− 71

Ge Q = −0.22194 MeV . (4.115)

The magnitudes of the (negative) Q values give the threshold neutrino energy.

In both cases radiochemical techniques are used where the ﬁnal state nucleus

is extracted from a multi-ton target and then observed to decay in a small

counter.

Finally, we mention that high-energy neutrinos are copiously produced

at high-energy accelerators and by cosmic rays in the Earth’s atmosphere.

The most important production mechanism is the decay of pions produced

in nucleon-nucleon interactions

218 4. Nuclear decays and fundamental interactions

→ µ

(

) . (4.116)

Mostly ν

are produced in pion decay but some ν

and ν

are produced by

the decays of heavier particles. While still requiring massive detectors, their

observation is simpliﬁed by the fact that their energy is much higher than

that of natural radioactivity.

4.3.6 Muon decay

We have already presented the µ lepton, or muon, in Sect. 1.8, when we

studied muonic atoms.

The muon is elementary in the same sense as the electron. It has the same

charge, the same spin, but it is 200 times heavier, m

= 206.8m

, and it is

unstable. It decays into an electron and two neutrinos : µ

−

→ e

−

with a

lifetime τ =2× 10

−6

The existence of the muon was an enigma for nearly 40 years. When it

was discovered, Rabi said “Who ordered that?” Why a heavy electron? All

the matter we know around us can be built with protons, neutrons, electrons

and neutrinos, or, in terms of fundamental constituents, with the family of

quarks and leptons {u, d, e, ν}. Why should there be a heavy electron, with

which one can achieve the dreams of Gulliver? Because the size of atomic

systems are inversely proportional to the system’s reduced mass, one could

imagine atoms, molecules, a chemistry, a biology 200 times smaller but 200

more energetic than the beings we know! We have found many applications

of the muon as probes of nuclei, of crystal structure, and of pyramids,

but

why does it exist? What is its use in nature?

Fig. 4.22. Decay process of the muon.

A ﬁrst clue comes from the calculation of the muon lifetime µ → e

νν (we

do not specify the charges or lepton numbers since one of them determines

all the others). This is also a weak decay governed by the Fermi constant!

For the ﬁrst time, we are facing the universality of weak interactions.

Muons are by far the most penetrating charged particles since they have no

strong interactions and, because of their large mass, radiate (bremsstrahlung)

much less eﬃciently than electrons.

4.3 Weak interactions 219

This reaction is represented on Fig. 4.22. It occurs via the virtual pro-

duction of a W boson, i.e. to a very good approximation it is a point-like

interaction.

We denote as p

, p

the four-momenta of the 4 particles. The

square of the matrix element for this reaction is

T |

=4G

· p

)(p

· p

)

. (4.117)

We can make a few comments on this result:

• By inserting this into (4.16) we notice that the decay rate is written in an

explicitly Lorentz invariant form: besides scalar products of four-vectors

there is a factor d

p/E for each ﬁnal-state particle. It is readily shown that

this factor is Lorentz invariant. This leaves only the factor 1/E for the

initial state particle which simply reﬂects the fact that the mean lifetime

is aﬀected by relativistic time dilation.

• One can check that with the same notations, if we change the muon into a

neutron and the ν

into a proton, and if we neglect the recoil of the ﬁnal

proton, this expression boils down to the same as in (4.67) for neutron

decay except that the factor 2 becomes 2.4. This factor is due to the fact

that the neutron is not an elementary particle (as opposed to a muon).

To simplify the calculation, we shall neglect the electron mass (it is easy

to take it into account, which is necessary in an accurate calculation) so that

 p

c where p

is the electron momentum.

In order to simplify the notations, we set:

≡ p

and we place ourselves in the rest frame of the muon. By energy-momentum

conservation, we have p

= p

+ p

, i.e. p

− p

= p

+ p

, and, by

squaring

· p

− p

)

c (m

c − 2p

)

always in the approximation m

 0.

Altogether, the muon decay rate is

¯h

(2π¯h)

I (4.118)

with

I =



c − 2p

)

δ(p

+ p

)

× δ(m

− p

c − p

c)d

(4.119)

We use the momentum conservation δ function to integrate over p

−(p

+ p

)).

220 4. Nuclear decays and fundamental interactions

Transforming to spherical coordinates for p

and p

, we obtain (d

p =

dpdΩ)

I =



c − 2p

)

δ(m

− p

c − p

c)p

dΩ

Let θ be the angle between p

and p

, the integration on the other angles

gives a factor of 4π×2π =8π

. We now use the energy conservation δ function

δ(m

−p

c −p

c)=(1/c)δ(m

c −p

−p

) in order to integrate

over cos θ, thanks to the relation δ(f(x)) =



δ(x − x

) ,f(x

)=0 .

We have p

(cos θ)=(p

+ p

+2p

cos θ)

1/2

therefore



dcosθ



i.e.

I =

8π



c − 2p

)dp

. (4.120)

The integration bounds are obtained by noticing that in a three-body

system of zero mass particles of zero total momentum and of total energy

• the maximum and minimum values of the energy of any of these particles

are obtained when the three momenta are aligned along the same axis;

• the maximum energy of one of the particles is Mc

/2 ;

• the minimum invariant mass m

of a two-body subsystem (m

=(p

)

)isM/2.

For a ﬁxed value of p

, which therefore varies between 0 and m

c/2, the

integration bounds in p

are m

c/2 − p

and m

c/2. A simple calculation

leads to

I =

8π

(

)

, (4.121)

and to a decay rate



(¯hc)



)

192π

¯h

. (4.122)

Numerically, this gives a lifetime τ

=2.187 10

−6

s, in excellent agreement

with the experimental value τ

=2.197 10

−6

s. Note that we recover in this

formula the ﬁfth power of the maximum energy of the ﬁnal electron.

Actually, muon decay is the cleanest way to determine the value of the

Fermi constant. The muon and electron are both point-like particles, so that

there are no wavefunction corrections to care about, and there are no ﬁnal

state interactions. The further corrections to the matrix element and to the

above result are well understood.