Chapter 10 — Summary and Problems · Elementary Particle Physics

Almost nothing here asks you to derive. It asks how much matter, how wide a beam, how long a wait — the problem set of a field whose difficulty is entirely in the counting.

🎯 Why this matters

Once the three factors are in hand you can size an experiment that does not exist yet. DUNE, Hyper-Kamiokande and JUNO are each one line of this arithmetic away, which is why these problems will outlast the results they were written about.

Thirteen problems, and unusually few of them are formalism. Nine are variations on a single calculation — flux × cross-section × targets = rate — applied to sources ranging from a reactor at 100 m to the relic neutrino background filling the Universe. That repetition is the point: the same three-factor arithmetic spans thirty orders of magnitude in rate, and what changes is only which of the three is the hard one.

Four are detector sizing — 10.1, 10.5, 10.9 and 10.10 — where you are handed a flux and asked how much matter you need. The trap in all four is which targets count: electrons for elastic scattering, free protons for inverse beta decay, oxygen nuclei for a nuclear cross-section.
Three are beam design — 10.2, 10.6 and 10.7 — and 10.6(c) is the one worth doing even if you skip the rest, because it shows that a long-baseline beam is kilometres wide when it arrives.
Two are kinematics with no oscillation in them at all — 10.3 and 10.8 — and both underpin real analyses: quasi-elastic energy reconstruction and the $E_\nu = E_{\rm vis} + 0.78$ MeV of every reactor experiment.
10.11 and 10.13 are order-of-magnitude astrophysics, and both end with a number that closes off a possibility rather than opening one.
10.12 is three lines and is the whole of §10.7’s experimental problem.

The book gives no solutions. All thirteen are worked below.

What the chapter established

Bettini p. 480, expanded. What Chapter 10 settled, what it left open, and where each result came from.

Question↕	Answer↕	How↕	Section↕
Do neutrinos have mass?		Flavour change requires unequal m², since a common mass gives no relative phase	§10.1
How do they mix?		Global fit to solar, atmospheric, reactor and accelerator data	§10.1
How do they change flavour?	Two distinct mechanisms		§10.2, §10.3
What are the splittings?		KamLAND for the first, atmospheric and accelerator disappearance for the second	§10.2b, §10.4
Where did the solar neutrinos go?	Into ν_μ and ν_τ; the total was never wrong		§10.4
Is CP violated in leptons?		ν_e appearance in ν_μ and ν̄_μ beams, T2K and NOvA	§10.5
How heavy are they?		Cosmology from above, the splittings from below	§10.5–10.6
Is the neutrino its own antiparticle?	Unknown, after 87 years		§10.7

One picture of the whole chapter

Every experiment in Chapter 10 is a point on a single axis. The vacuum probability depends on $L$ and $E$ only through their ratio, so $L/E$ is the one coordinate that matters — and the field covers eleven orders of magnitude of it.

Every source of Chapter 10 on the one axis that matters, from the values computed in §10.1. Daya Bay at 1.5 km and T2K at 295 km land on the same point — 500 km/GeV — because their energies differ by the same factor their baselines do. OPERA sits deliberately short of it, forced there by the tau threshold. KamLAND is a hundred atmospheric maxima out, which is why it sees the solar oscillation with reactors. And the Sun is 3×10⁸ maxima away, so everything fast has averaged to a constant before the neutrinos arrive.

The other picture is the one the chapter cannot finish. Three observables reach the absolute mass scale, they weight the three masses differently, and no measurement has yet landed inside the window the splittings and cosmology leave open.

Σ mᵢ — cosmology
m_νₑ — β decay
M_ee — 0ν2β, largest allowed

Normal ordering, computed from the splittings of (10.8) and the |U_ei|² of (10.73) — the same formulas the PyBlocks on §10.5 and §10.7 verify. Three observables, three different weightings, and they converge only at the right-hand edge where all three masses are nearly equal. The M_ee curve is the upper edge of its band: with the Majorana phases free it can fall to zero near a lightest mass of 4 meV. Everything to the left of about 30 meV is currently unreachable by all three techniques, and that is where the answer probably is.

The numbers, all at once

every numeric answer in the chapter's problems

import numpy as np
NA, YR, ln2 = 6.02214076e23, 3.1557e7, np.log(2)
mpi, mmu, mn, mp, me, mtau, MZ = .13957, .105658, .939565, .938272, .000511, 1.77686, 91.1876

Ne = 1e8/(9*12.011+12*1.008)*NA*66; r0 = 4.6e13*.6e-48*Ne*86400
P = 1-.5*np.sin(np.deg2rad(68))**2
print(f"10.1  no oscillation {r0:.0f}/day;  P_ee = {P:.3f};  with conversion {r0*(P+(1-P)/6):.0f}/day")

ppi = np.sqrt(25-mpi**2); k = mpi**2-mmu**2
E2 = k/(2*(5-ppi*np.cos(np.deg2rad(2.5)))); ct = (E2/((5/mpi)*k/(2*mpi))-1)/(ppi/5)
ph = 1.27*2.5e-3*295/E2
print(f"10.2  E_nu(2.5deg) = {E2:.3f} GeV, theta* = {np.rad2deg(np.arccos(ct)):.0f} deg;"
      f" E_nu(0) = {k/(2*(5-ppi)):.2f} GeV")
print(f"      {2e11*3e-43*22.5e9*NA:.0f} CC/yr;  P_dis = {np.sin(ph)**2:.3f};"
      f"  P(nu_e) = {.5*np.sin(np.deg2rad(10))**2*np.sin(ph)**2:.4f}")

pmu = np.sqrt(.25-mmu**2)
print(f"10.3  E_nu = {(mp**2-mn**2-mmu**2+2*mn*.5)/(2*(mn-.5+pmu*np.cos(np.deg2rad(30)))):.3f} GeV")

NO = 22.5e9/18*NA; R, h = 6371., 15.; Lh = np.sqrt((R+h)**2-R**2)
print(f"10.4  {130*1e-42*NO*YR:.1f} interactions/yr;  upward: phase {1.27*2.5e-3*(2*R+h):.0f} rad"
      f" -> P = 0.50;  horizontal L = {Lh:.0f} km -> P = {1-np.sin(1.27*2.5e-3*Lh)**2:.2f}")

r5 = 1e10*1e-47*NO*10*.5*YR
print(f"10.5  expected {r5:.2g}/yr;  observed 0.84x0.304+0.16 = {.84*.304+.16:.2f} of it"
      f" = {r5*(.84*.304+.16):.2g}/yr")

for Epi, L, nm in ((80., 730., 'CNGS'), (7., 295., 'T2K ')):
    p, g = np.sqrt(Epi**2-mpi**2), Epi/mpi; th = np.arctan(1/(g*p/Epi))
    print(f"10.6  {nm}: decay length {g*7.8045/1e3:.2f} km;  E_nu {k/(2*(Epi-p)):.2f} GeV to"
          f" {k/(2*(Epi+p))*1e3:.3f} MeV;  theta*=90 -> {th*1e3:.2f} mrad, r = {L*np.tan(th):.2f} km")

Rmu = 4.3e8*1e-41*2e9*NA; P7 = np.sin(1.27*2.5e-3*730/18)**2
print(f"10.7  E_thr = {mtau*(mtau+2*mn)/(2*mn):.2f} GeV;  {Rmu:.1f} nu_mu CC/yr;"
      f"  P(tau) = {P7:.4f} -> {Rmu*P7:.3f}/yr;  nu_e {Rmu*.5*np.sin(np.deg2rad(14))**2*P7:.1e}/yr")

Ee = 3-(mn-mp)*1e3; pe = np.sqrt(Ee**2-(me*1e3)**2)
print(f"10.8  T_n(max) = {(3+pe)**2/(2*mn*1e3)*1e3:.0f} keV;  E_nu = E_vis + "
      f"{((mn-mp)-me)*1e3:.2f} MeV;  threshold {((mn+me)**2-mp**2)/(2*mp)*1e3:.3f} MeV")

for L in (100., 2000.):
    Np = (100/86400)/(6e20/(4*np.pi*L**2)*1e-47)
    print(f"10.9  L = {L:6.0f} m: N_p = {Np:.2e}, hydrogen mass {Np/NA/1e3:.3g} kg")

Np10 = (1000/YR)/(3.5e10*.6*.05*1e-47)
fH = .2*(18*1.008)/210.32 + .8*(26*1.008)/170.34
print(f"10.10 N_p = {Np10:.2e} -> {Np10/NA/1e6:.0f} t of hydrogen; blend is {fH:.3f} H by mass"
      f" -> {Np10/NA/fH/1e6:.2g} t of scintillator")

print(f"10.11 lambda = 1/(rho N_A sigma) = {1/(5e6*NA*3e-36)/1e3:.0f} km  (Earth 12742 km -> opaque)")
print("10.12 " + ";  ".join(f"{i} {1e6/A*NA*ln2/1e27:.1f}/yr" for i, A in
                            (('76Ge', 76), ('130Te', 130), ('136Xe', 136))))
v = np.sqrt(2*.25e-3/.1); lam = 1/(5.6e7*2.1e-37)
print(f"10.13 v_rms = {v:.3f} c;  E_res = M_Z^2/2m = {MZ**2*1e18/.2:.1e} eV;"
      f"  lambda = {lam:.1e} m = {lam/9.461e15:.0e} ly")

prints

10.1  no oscillation 79/day;  P_ee = 0.570;  with conversion 51/day
10.2  E_nu(2.5deg) = 0.620 GeV, theta* = 115 deg; E_nu(0) = 2.13 GeV
    813 CC/yr;  P_dis = 0.996;  P(nu_e) = 0.0150
10.3  E_nu = 0.537 GeV
10.4  3.1 interactions/yr;  upward: phase 41 rad -> P = 0.50;  horizontal L = 437 km -> P = 0.03
10.5  expected 1.2e+04/yr;  observed 0.84x0.304+0.16 = 0.42 of it = 4.9e+03/yr
10.6  CNGS: decay length 4.47 km;  E_nu 34.15 GeV to 0.026 MeV;  theta*=90 -> 1.74 mrad, r = 1.27 km
10.6  T2K : decay length 0.39 km;  E_nu 2.99 GeV to 0.297 MeV;  theta*=90 -> 19.94 mrad, r = 5.88 km
10.7  E_thr = 3.46 GeV;  5.2 nu_mu CC/yr;  P(tau) = 0.0165 -> 0.085/yr;  nu_e 2.5e-03/yr
10.8  T_n(max) = 11 keV;  E_nu = E_vis + 0.78 MeV;  threshold 1.806 MeV
10.9  L =    100 m: N_p = 2.42e+28, hydrogen mass 40.3 kg
10.9  L =   2000 m: N_p = 9.70e+30, hydrogen mass 1.61e+04 kg
10.10 N_p = 3.02e+33 -> 5011 t of hydrogen; blend is 0.140 H by mass -> 3.6e+04 t of scintillator
10.11 lambda = 1/(rho N_A sigma) = 111 km  (Earth 12742 km -> opaque)
10.12 76Ge 5.5/yr;  130Te 3.2/yr;  136Xe 3.1/yr
10.13 v_rms = 0.071 c;  E_res = M_Z^2/2m = 4.2e+22 eV;  lambda = 8.5e+28 m = 9e+12 ly

Four of those repay a second look.

10.1 and 10.5 are the same trap, twice. Both use elastic scattering, and in both the naive answer is to multiply by the survival probability. That is wrong: the converted $\nu_\mu$ and $\nu_\tau$ still scatter, at 1/6 the cross-section, so the observed rate is $0.84P + 0.16$ of the expectation and never falls below 16 % however complete the conversion. In 10.1 that is the difference between 45 and 51 events per day; in 10.5 it is the difference between 0.30 and 0.42, and historically it is why Kamiokande’s deficit looked shallower than Homestake’s for a decade.

10.4(2) has no exact answer, and knowing that is the point. The phase for a vertically upward 1 GeV neutrino is 41 radians. Computing $\sin^2(41)$ gives 0.11; shift the energy by 2 % and it gives something else entirely. When the phase is tens of radians the only physical answer is the average, $P = 0.50$ — which is exactly the number Super-Kamiokande measured and read $\theta_{23}$ off.

10.6(c) explains the economics of the whole field. Half the neutrinos land inside a circle 1.3 km across at Gran Sasso and 5.9 km across at Kamioka, against detectors tens of metres wide. A long-baseline experiment intercepts about one part in $10^6$ of its own beam — which is why these things need megawatt proton drivers and kilotonne targets to collect a few hundred events.

10.13 kills an idea rather than testing one. The Z-resonance on the relic background is real physics, it is calculable, and the answer is that it needs neutrinos of $4\times10^{22}$ eV that do not exist, with an interaction length two hundred times the size of the observable Universe. Both facts are single-line estimates, and together they close the question.

📝 Chapter 10 — thirteen problems, all worked

0/13 solved

10.1solar neutrinostheory
Borexino detects the monoenergetic $^7$ Be solar neutrinos at $E = 0.862$ MeV through $\nu_e e^- \to \nu_e e^-$ in 100 t of pseudocumene, C $_9$ H $_{12}$ . (1) If electron neutrinos did not change flavour, how many events per day would you expect? (2) Which mechanism converts Be neutrinos — vacuum oscillation or the MSW effect? (3) How many events per day are actually expected?
- $\Phi = 4.6\times10^{13}$ m⁻² s⁻¹
- $\sigma(\nu_e e) = 0.6\times10^{-48}$ m², $\sigma(\nu_{\mu,\tau} e) = \sigma(\nu_e e)/6$
- $\theta_{12} = 34°$ , $\delta m^2 = 80$ meV²
10.2accelerator beamstheory
In T2K the J-PARC proton beam makes pions that are focused and allowed to decay; everything but the neutrinos is then absorbed. The far detector is Super-Kamiokande at $L = 295$ km. (1) Which flavour dominates the beam? (2) For $E_\pi = 5$ GeV, what is $E_\nu$ at a lab angle of 2.5°, and what CM angle $\theta^*$ does that correspond to? (3) What is $E_\nu$ at 0°? (4) Treating the beam as monoenergetic at the energy of (2) and ignoring oscillation, how many CC interactions per year occur in the fiducial volume? (5) What is the disappearance probability at that energy? (6) What is the $\nu_e$ appearance probability if $\theta_{13} = 5°$ ?
- $\Phi = 2\times10^{11}$ ν m⁻² yr⁻¹, fiducial mass 22.5 kt
- $\sigma = 3\times10^{-43}$ m² per nucleon
- $\theta_{23} = 45°$ , $\theta_{13} = 0$ for (5), $\Delta m^2 = 2500$ meV²
10.3kinematicstheory
A $\nu_\mu$ from J-PARC interacts quasi-elastically in Super-Kamiokande, $\nu_\mu n \to \mu^- p$ . The direction of the incoming neutrino is known but its energy is not; the muon energy and its angle to the beam are measured. Find $E_\nu$ for $E_\mu = 0.5$ GeV and $\theta = 30°$ .
- $m_n = 939.565$ MeV, $m_p = 938.272$ MeV, $m_\mu = 105.658$ MeV
- the neutron is at rest
10.4atmospheric neutrinostheory
Atmospheric muon neutrinos have a spectrum falling roughly as $E^{-3}$ . Around the zenith at $E = 1$ GeV the flux is about 130 m⁻² s⁻¹ sr⁻¹ GeV⁻¹. Super-Kamiokande detects them through $\nu_\mu\,{}^{16}\mathrm{O} \to \mu^-X$ . (1) How many interactions per year come from within 1 sr around the zenith in a 1 GeV interval? (2) What fraction of muon neutrinos arriving vertically upwards survives? (3) And at 90° to the zenith?
- fiducial mass 22.5 kt of water, $\sigma(\nu_\mu\,{}^{16}\mathrm{O}) = 10^{-42}$ m²
- $\theta_{23} = 45°$ , $\theta_{13} = 0$ , $\Delta m^2 = 2500$ meV²
- Earth radius 6371 km, production height ≈ 15 km
10.5solar neutrinostheory
$^8$ B neutrinos dominate the solar spectrum near 10 MeV. In the interval $9 < E < 11$ MeV their flux at Earth would be $10^{10}$ m⁻² s⁻¹ in the absence of oscillations. Super-Kamiokande detects them through $\nu_e e^- \to \nu_e e^-$ . How many events per year are expected? How many are observed, and why the difference?
- fiducial mass 22.5 kt of water, $\sigma(\nu_e e) = 10^{-47}$ m²
- detection efficiency $\varepsilon = 50\%$
10.6beam designtheory
In a long-baseline beam, pions are focused forward, decay in a vacuum pipe, and everything but the neutrinos is absorbed. Take $E_\pi = 80$ GeV for the CNGS beam ( $L = 730$ km) and $E_\pi = 7$ GeV for T2K ( $L = 295$ km). For each: (a) the pion decay length; (b) the maximum and minimum neutrino energy; (c) neutrinos are emitted isotropically in $\theta^*$ , so half go forward — find the lab angle corresponding to $\theta^* = 90°$ and the resulting beam radius at the far detector.
- $c\tau_\pi = 7.80$ m, $m_\pi = 139.57$ MeV, $m_\mu = 105.66$ MeV
10.7ν_τ appearancetheory
OPERA looked for $\nu_\tau$ appearance in a $\nu_\mu$ beam from CERN, $L = 730$ km, through $\nu_\tau n \to \tau^- p$ . Assume all neutrinos have $E = 18$ GeV. (a) What is the energy threshold for the reaction? (b) With a yearly integrated flux $N = 4.3\times10^8$ m⁻² and a 2000 t lead target, how many $\nu_\mu$ CC interactions occur per year? (c) How many $\nu_\tau$ CC interactions are expected with oscillations? (d) And how many $\nu_e$ if $\theta_{13} = 7°$ ?
- $\sigma_{CC} = 10^{-41}$ m², $m_\tau = 1777$ MeV, $m_N = 939$ MeV
- $\theta_{23} = 45°$ , $\theta_{13} = 0$ for (c), $\Delta m^2 = 2500$ meV²
10.8inverse beta decaytheory
Electron antineutrinos of a few MeV are detected through $\bar\nu_e p \to e^+ n$ , followed by $e^+e^- \to 2\gamma$ and the deposit of the gamma energy. The total deposited energy is the 'visible energy' $E_{\rm vis}$ . (a) What are the dominant processes by which the gammas deposit their energy? (b) For $E_\nu = 3$ MeV, what is the maximum kinetic energy of the recoiling neutron? (c) Given (b), relate $E_\nu$ to $E_{\rm vis}$ . (d) What is the minimum detectable neutrino energy?
- $m_n - m_p = 1.293$ MeV, $m_e = 0.511$ MeV
10.9reactor experimentstheory
You are planning a reactor experiment using a 3 GW complex that emits $6\times10^{20}$ electron antineutrinos per second, observing $\bar\nu_e p \to e^+n$ in a detector containing free protons. You want to measure the flux at $L_1 = 100$ m and $L_2 = 2$ km. Calculate the proton mass needed at each distance for 100 counts per day.
- $\sigma \approx 10^{-47}$ m² (a rough average over the spectrum)
10.10geoneutrinostheory
The Earth radiates about 40 TW, of which the Bulk Silicate Earth model attributes half to radioactivity. Only $^{238}$ U and $^{232}$ Th produce antineutrinos above the 1.8 MeV inverse-beta threshold. Their total surface flux is expected to be $3.5\times10^{10}$ m⁻² s⁻¹; a fraction $P_{ee} \approx 0.6$ survives oscillation and a fraction $f \approx 0.05$ is above threshold. Using a blend of 20 % PXE (C $_{16}$ H $_{18}$ ) and 80 % dodecane (C $_{12}$ H $_{26}$ ) by mass, what scintillator mass is needed for 1000 events?
- $\sigma \approx 10^{-47}$ m², 1000 events in one year
10.11ultra-high-energy neutrinostheory
The highest-energy cosmic-ray protons make pions on the microwave background through $p\gamma \to n\pi^+$ — the Greisen–Zatsepin–Kuzmin effect. The neutrinos from those pion decays reach about 10 EeV. Their cross-section on nucleons is large because of the number of open channels. Calculate the average distance to interact inside the Earth.
- $\sigma_{NC} = 3\times10^{-36}$ m², $\rho = 5\times10^{3}$ kg m⁻³
10.12double beta decaytheory
$^{76}$ Ge, $^{130}$ Te and $^{136}$ Xe are very stable nuclei, but if the neutrino is a Majorana particle they may decay by the neutrinoless double-beta mechanism. Assuming a half-life $T_{1/2} = 10^{27}$ yr for each, what is the average number of decays expected in one year in 1 t of each isotope?
- $N_A = 6.022\times10^{23}$ mol⁻¹
10.13relic neutrinostheory
The cosmic neutrino background fills the Universe. At extremely high energies an incoming neutrino can annihilate on a relic antineutrino through the resonant process $\nu_x\bar\nu_x \to Z \to$ anything. (a) Estimate the mean square speed of the relic neutrinos. (b) Estimate the incoming neutrino energy needed for the resonance. (c) Estimate the mean free path.
- $\langle E_k\rangle = 0.25$ meV, $n = 5.6\times10^{7}$ m⁻³ per flavour, $m = 100$ meV
- $\sigma \approx 2.1$ nb at the Z peak, $M_Z = 91.19$ GeV

🔑 If you remember only three things

Elastic scattering never drops below 16 % of its expectation. The converted neutrinos still scatter, and forgetting it made Kamiokande’s deficit look shallower than Homestake’s for a decade.
A phase of tens of radians has no value, only an average. Once 2 % in energy moves the answer anywhere at all, the average is the physical result — and it is what Super-Kamiokande read θ₂₃ from.
Geometry sets the cost of the field. A beam kilometres wide arriving at a detector tens of metres across is why these experiments need megawatt drivers and kilotonne targets.

Where this goes next

Chapter 10 is where the Standard Model stops being right. Everything before it — the quark model, QCD, the electroweak unification, the Higgs — is a theory that fits. This chapter is a set of measurements the theory cannot accommodate without being extended, and the extension is not known.

Three things are genuinely open, and they are not independent:

the mass ordering — JUNO will read the sign of $\Delta m^2$ off the interference between the two oscillations, and DUNE will get it from matter effects over 1300 km;
the CP-violating phase $\delta$ — a $3\sigma$ hint, and the only known candidate for the matter–antimatter asymmetry that the quark sector cannot supply;
Dirac or Majorana — which is really the question of where the mass comes from, since a Majorana mass term needs no right-chirality neutrino and the BEH mechanism cannot supply one.

Chapter 11 turns to gravity — the one interaction the Standard Model omits entirely — and to the gravitational waves that have made it an experimental subject. The Epilogue then takes the open questions in turn, neutrino mass among them.

✅ Check yourself — the chapter, end to end

0/6 answered · 0 correct

1.Problems 10.1 and 10.5 both use ν e⁻ → ν e⁻. Why is the answer not simply the flux times the survival probability?
2.Problem 10.4 asks for the survival probability of vertically upward-going 1 GeV neutrinos. The phase comes out at 41 radians. What is the right answer?
3.In Problem 10.6(c), half the neutrinos from a 80 GeV pion beam land inside a circle 1.3 km across at Gran Sasso. What does that imply?
4.Which of these are correct readings of the chapter's open questions? (Select all that apply.)
5.Problem 10.13 finds a Z resonance on the relic neutrino background at 4×10²² eV with a mean free path of 9×10¹² light years. What kind of result is that?
6.Nine of the thirteen problems reduce to flux × cross-section × targets. What varies between them?