The Longer Period Oscillation II — Three Flavours, and the Experiments

Three flavours turn out to change only the depth of the curve, never its shape. What is left to measure is a set of depths, and four unrelated technologies measured them.

🎯 Why this matters

An 8 % effect with one unknown in it beat a 5 % effect entangled with three. Choosing what a measurement does not depend on is worth more than choosing what it does, and here that choice decided who got there first.

§10.2a derived $P = \sin^2 2\theta \sin^2(1.27\,\Delta m^2 L/E)$ for two flavours. There are three, and none of the numbers in §10.1 came from a formula — they came from detectors. This section is the evidence.

The order is the historical one, and it is a good one: a deficit nobody was looking for (1998), the accelerator beams built to confirm it, the reactor experiments that cornered the last mixing angle (2012), and finally somebody catching the particle the missing neutrinos had turned into.

Three flavours collapse back to two

With three mass states there are three frequencies and nine transition probabilities. Two facts flatten that almost completely.

One frequency is redundant. $m_3^2 - m_2^2$ and $m_3^2 - m_1^2$ differ by $\delta m^2$ , which is 3 % of either. To any practical accuracy both are $\Delta m^2$ , so there are two frequencies, not three.

The two that remain are 33 times apart. An experiment sitting at the fast one sees the slow oscillation as not yet started; an experiment at the slow one sees the fast one as already averaged away. Neither ever has to solve the full problem.

💡 What this really says — and where the approximation is deliberately broken

Separated time scales are the reason neutrino physics is tractable. It is the same move as analysing a fast inner loop and a slow outer loop separately, and it works for the same reason.

But the crossover carries information the separated regimes throw away. A detector placed where both oscillations are developing sees their interference, and the interference pattern is not symmetric under $\Delta m^2 \to -\Delta m^2$ — so it carries the sign, which nothing else in this section can. That is JUNO: 20 kt of liquid scintillator at 50 km from a reactor complex, under construction in China, built to answer the mass-ordering question by deliberately refusing the approximation everything else relies on.

P(\nu_x\to\nu_y) = \htmlClass{t-A}{A(\nu_x\to\nu_y)}\; \sin^2\!\left[\htmlClass{t-p}{1.27\,\Delta m^2\left(L/E\right)}\right]

(10.21)

Bettini p. 446. In the two-frequency approximation, every flavour pair oscillates with the SAME phase — only the amplitude in front differs.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

💡 What this really says — three flavours cost you almost nothing but a change of height

Three flavours cost you almost nothing. The shape of every curve is the same $\sin^2$ you already have; only the height changes. So an experiment measures a depth and a position, exactly as in the two-flavour case, and the three-flavour structure shows up only in how the depth is built from the angles.

The nine amplitudes are not independent — unitarity ties them together, and the next three equations are a worked example of that.

Three amplitudes, and one of them does not add up

A(\nu_\mu\to\nu_x) = \htmlClass{t-w}{4\sin^2\theta_{23}\cos^2\theta_{13}} \left(1-\sin^2\theta_{23}\cos^2\theta_{13}\right) \approx 1

(10.22)

Bettini p. 446 — the total ν_μ disappearance amplitude, WITH THE LEADING FACTOR CORRECTED. The book prints sin²2θ₂₃cos²θ₁₃ and evaluates it to ½; see the erratum below.

Every symbol, one at a time

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💡 What this really says — a 50 % effect needs an honest experiment, not a precise one

The muon neutrino very nearly vanishes. $A \approx 1$ means that at the oscillation maximum essentially none survive, and past the washout the flux settles at $1 - A/2 \approx \tfrac12$ .

That factor of two is what Super-Kamiokande measured, and it is why the discovery was possible at all with 1998 statistics: a 50 % effect does not need a precision experiment, it needs an honest one.

A(\nu_\mu\to\nu_e) = \htmlClass{t-a}{\sin^2\theta_{23}}\,\htmlClass{t-b}{\sin^2 2\theta_{13}} \approx 2\theta_{13}^2

(10.23)

Bettini p. 446. The ν_e appearance amplitude in a ν_μ beam — the channel §10.5 will use to chase δ.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

💡 What this really says — the most valuable 5 % in the field

$A(\nu_\mu\to\nu_e) = 0.046$ — under 5 %. Small, and the smallness is the reason Super-Kamiokande’s electron-like data show no deficit at all, which is how the 1998 experiment could tell it was seeing $\nu_\mu \to \nu_\tau$ and not $\nu_\mu \to \nu_e$ .

It is also the most valuable 5 % in the field. This is the only channel where $\delta$ appears, so every experiment in §10.5 is built to measure it.

A(\nu_\mu\to\nu_\tau) = \htmlClass{t-a}{\sin^2 2\theta_{23}}\,\htmlClass{t-b}{\cos^4\theta_{13}} \approx 1

(10.24)

Bettini p. 446. Where the disappeared muon neutrinos actually go.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

💡 What this really says — unitarity being used rather than assumed

Add (10.23) and (10.24): $0.046 + 0.948 = 0.994$ , which is exactly (10.22). The missing muon neutrinos are accounted for, to four decimal places, with nothing left over. That is unitarity being used rather than assumed, and it is the prediction OPERA was built to test directly.

🔢 Worked example — the three amplitudes must close, and one of them does not

Every amplitude here is $4|U_{\alpha3}|^2|U_{\beta3}|^2$ for the appearance channels and $4|U_{\mu3}|^2(1-|U_{\mu3}|^2)$ for the total. Compute all of them from (10.3) alone:

The atmospheric amplitudes, and the sum rule

import numpy as np

s23s, s13s = 0.551, 0.0214                  # (10.3)
c13s = 1 - s13s
U3 = np.array([s13s, s23s*c13s, (1-s23s)*c13s])   # |U_e3|^2, |U_mu3|^2, |U_tau3|^2
print(f"|U_e3|^2 = {U3[0]:.4f}   |U_mu3|^2 = {U3[1]:.4f}   |U_tau3|^2 = {U3[2]:.4f}"
      f"   sum = {U3.sum():.4f}")

A_e   = 4*U3[1]*U3[0]                       # (10.23)
A_tau = 4*U3[1]*U3[2]                       # (10.24)
A_x   = 4*U3[1]*(1 - U3[1])                 # total disappearance
print(f"\nA(mu->e)   = {A_e:.4f}    (10.23) sin^2(th23) sin^2(2th13) = "
      f"{s23s*4*s13s*c13s:.4f}")
print(f"A(mu->tau) = {A_tau:.4f}    (10.24) sin^2(2th23) cos^4(th13) = "
      f"{4*s23s*(1-s23s)*c13s**2:.4f}")
print(f"A(mu->x)   = {A_x:.4f}    and A(mu->e) + A(mu->tau) = {A_e + A_tau:.4f}")

printed = 4*s23s*(1-s23s) * c13s * (1 - s23s*c13s)          # (10.22) as printed
print(f"\n(10.22) as printed evaluates to {printed:.4f}, and the page says it is about 0.5")
print(f"the flux reduction that implies is 1 - {printed:.4f}/2 = {1-printed/2:.3f}"
      f" of the expectation, not the 1/2 the book reads off Fig. 10.8")
print(f"with A = {A_x:.4f} the reduction is 1 - A/2 = {1-A_x/2:.3f}  <-- the observed one")

prints

|U_e3|^2 = 0.0214   |U_mu3|^2 = 0.5392   |U_tau3|^2 = 0.4394   sum = 1.0000

A(mu->e)   = 0.0462    (10.23) sin^2(th23) sin^2(2th13) = 0.0462
A(mu->tau) = 0.9477    (10.24) sin^2(2th23) cos^4(th13) = 0.9477
A(mu->x)   = 0.9939    and A(mu->e) + A(mu->tau) = 0.9939

(10.22) as printed evaluates to 0.4462, and the page says it is about 0.5
the flux reduction that implies is 1 - 0.4462/2 = 0.777 of the expectation, not the 1/2 the book reads off Fig. 10.8
with A = 0.9939 the reduction is 1 - A/2 = 0.503  <-- the observed one

(10.23) and (10.24) reproduce the book’s expressions exactly. Their sum is 0.9939 — and (10.22) as printed gives 0.4462.

Erratum — Eq. (10.22) contradicts the two equations under it

The book prints the total $\nu_\mu$ disappearance amplitude as

A(\nu_\mu\to\nu_x) = \sin^2(2\theta_{23})\cos^2(\theta_{13})\left(1-\sin^2\theta_{23}\cos^2\theta_{13}\right) \approx \tfrac12

The leading factor should be $4\sin^2\theta_{23}\cos^2\theta_{13}$ , not $\sin^2 2\theta_{23}\cos^2\theta_{13}$ — the printed version carries a spurious $\cos^2\theta_{23}$ , which is a factor 0.449. Three independent checks, all inside the same two pages:

A subset cannot exceed the whole. (10.23) and (10.24) give the two channels the disappeared $\nu_\mu$ can go into, and they sum to 0.994. The total cannot be ½.
The book’s own reading of Fig. 10.8. p. 448 says the upward-going flux is “one half of the expected flux” and concludes $\theta_{23}\approx\pi/4$ . Past the washout the flux is $1 - A/2$ , so a half requires $A = 1$ . With $A = \tfrac12$ the flux would be 78 % of expectation and there would be no discovery to report.
The approximations do not give ½ either. Substituting $\cos^2\theta_{13}\to1$ and $\sin^2\theta_{23}\to\tfrac12$ , as the book says it does, into the corrected form gives $4\times\tfrac12\times\tfrac12 = 1$ .

The correct form is the standard one, $A = 4|U_{\mu3}|^2\left(1-|U_{\mu3}|^2\right)$ , which is what the equation above shows. Everything else on the page is right, including both appearance amplitudes.

The discovery: Super-Kamiokande, 1998

🔬 Experiment card — Super-Kamiokande and the atmospheric ν_μ deficit

Apparatus

A 22 500 t fiducial-mass water Cherenkov detector under the Japanese Alps, built to look for proton decay and instead used as a neutrino telescope. Atmospheric neutrinos are detected by their charged-current scattering, which produces a charged lepton, which produces a Cherenkov ring on the phototube wall.

What is measured

For each single-ring event: the ring’s sharpness (muon or electron), its direction (and therefore the zenith angle, and therefore the flight length), and roughly the energy. Four categories: e-like and μ-like, each split at 1.33 GeV of visible energy. The charge is not measured, so neutrinos and antineutrinos are not separated.

The result

In a 328 kt·yr exposure the electron-like distributions match the no-oscillation prediction. The high-energy muon-like events match it for downward-going neutrinos and fall to half of it for upward-going ones. The low-energy muon-like events are depleted at every angle.

What it proved

Muon neutrinos disappear, as a function of $L/E$ and not of anything else. The position of the transition gives $\Delta m^2$ ; the depth, one half, gives $\theta_{23}\approx45^\circ$ ; and the electron-like data being flat says $\theta_{13}$ is small, so the muon neutrinos are not turning into electron neutrinos. First evidence of physics beyond the Standard Model at an accelerator laboratory scale, and the 2015 Nobel Prize.

\htmlClass{t-m}{\nu_\mu + N \to \mu + N'} \qquad\qquad \htmlClass{t-e}{\nu_e + N \to e + N'}

(10.25)

Bettini p. 447. The two charged-current reactions Super-Kamiokande sees. The outgoing lepton carries the flavour of the neutrino — that is the entire flavour-tagging method.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

💡 What this really says — a flavour measurement is really a lepton-identification measurement

A neutrino’s flavour is not observable; the charged lepton it makes is. So a flavour measurement is a lepton identification measurement, and in water it comes down to whether the edge of a ring of light is sharp or fuzzy.

That is worth pausing on. The 1998 result rested on classifying rings by crispness, and it held up against twenty-five years of far more sophisticated detectors.

Bettini Figs. 10.7 and 1.20, side by side because they are the two halves of one measurement. The zenith angle fixes the flight length over three decades, from 15 km straight down to 12 757 km through the Earth — and neutrinos cross the planet without noticing it. The ring on the right decides which flavour arrived.

🔢 Worked example — the zenith angle IS the baseline

Production happens about 15 km up. The flight length is then a chord, and it depends on the zenith angle alone:

Flight length and survival versus zenith angle

import numpy as np

R, h = 6371.0, 15.0                     # km: Earth radius, production height
Dm2, A = 2455e-6, 0.9939                # eV^2, and the disappearance amplitude

def baseline(cos_t):                    # chord from the production shell to the detector
    return np.sqrt((R + h)**2 - (R*np.sin(np.arccos(cos_t)))**2) - R*cos_t

def survival(cos_t, E):                 # averaged over a +-30 % energy acceptance
    w = np.linspace(0.7, 1.3, 601)
    return np.mean([1 - A*np.sin(1.27*Dm2*baseline(cos_t)/(E*f))**2 for f in w])

print(" cos(zenith)      L (km)    <P> at 1 GeV   <P> at 10 GeV")
for ct in (1.0, 0.5, 0.0, -0.5, -1.0):
    print(f"   {ct:+4.1f}      {baseline(ct):9.1f}      {survival(ct, 1.0):.3f}"
          f"          {survival(ct, 10.0):.3f}")

print(f"\nthe washed-out asymptote 1 - A/2 = {1 - A/2:.3f}")
print(f"first maximum of the atmospheric oscillation: L/E = "
      f"{np.pi/2/(1.27*Dm2):.0f} km/GeV, i.e. {np.pi/2/(1.27*Dm2)*10:.0f} km at 10 GeV")

prints

 cos(zenith)      L (km)    <P> at 1 GeV   <P> at 10 GeV
 +1.0           15.0      0.998          1.000
 +0.5           29.9      0.991          1.000
 +0.0          437.4      0.092          0.980
 -0.5         6400.9      0.532          0.284
 -1.0        12757.0      0.523          0.526

the washed-out asymptote 1 - A/2 = 0.503
first maximum of the atmospheric oscillation: L/E = 504 km/GeV, i.e. 5038 km at 10 GeV

Every feature of Fig. 10.8 is in that table.

High energy, downward ( $\cos\theta = +1$ , 10 GeV): survival 1.000. 15 km is nothing; no deficit, and the prediction is checked against reality with the oscillation switched off.

High energy, upward ( $\cos\theta = -1$ ): 0.526 — the factor of two, arriving by itself from the geometry. It is the washed-out asymptote $1 - A/2 = 0.503$ , and the fact that it lands there is the measurement of $\theta_{23}$ .

Low energy, everywhere: even the horizontal direction is past the first maximum at 1 GeV, so the low-energy muon sample is depleted at all angles. The book notes this in one sentence; here it is arithmetic.

The transition sits where $L/E$ crosses 504 km/GeV — around 5000 km at 10 GeV, which is $\cos\theta \approx -0.4$ . Move $\Delta m^2$ and the transition slides. That is how one detector measures a mass splitting from geometry.

Bettini Fig. 10.8, redrawn as a ratio and computed rather than traced: the red curves come from the survival probability in the worked example above, evaluated at each zenith angle. The electron panels are flat at 1 because A(ν_μ→ν_e) = 0.046 and θ₁₃ is small — that flatness is itself a measurement. The high-energy muon panel is the discovery: full flux from overhead, half from below, and a transition whose position gives Δm².

Confirmation, four times

Bettini pp. 448–449. The muon-neutrino deficit, checked by four independent experiments in three countries with three different technologies.

Source↕	L↕	E↕	L/E (km/GeV)↕	Years↕
atmospheric	15–12 757 km	0.1–10 GeV	spans it	1998–
atmospheric	spans	GeV	spans it	1990s
atmospheric	spans	GeV	spans it	1990s
KEK accelerator	250 km	1.5 GeV		1999–2004
FNAL NuMI beam	735 km	2–3 GeV	245–368	2005–2012

L/E is the design value; the atmospheric first maximum is at 504 km/GeV. Atmospheric experiments do not have a single L/E — they span it.

📏 Were K2K and MINOS really on the oscillation maximum?

The book says K2K’s 1.5 GeV peak was “chosen to have the first maximum of the oscillation amplitude at the far detector”, and that MINOS’s 2–3 GeV was “chosen with the same criterion”. By the book’s own (10.19) and $\Delta m^2$ that does not work out.

The first maximum is at $L/E = 504$ km/GeV. K2K at 250 km therefore wants 0.50 GeV, not 1.5 — at 1.5 GeV it sits at $L/E = 167$ , a third of the way, with a disappearance probability of 0.24 rather than ~1. MINOS at 735 km and 2.5 GeV reaches $L/E = 294$ , about 58 % of the way.

Neither was a mistake: a neutrino cross-section grows with energy and a beam flux falls with it, so an early experiment took the rate it could get. Being genuinely on the maximum is what the off-axis trick buys, and it was invented afterwards — T2K at 295 km / 0.6 GeV reaches $L/E = 492$ , 98 % of the maximum, and NOvA reaches 80 %. That is the actual design story, and it is more interesting than the one the sentence tells.

The last angle: θ₁₃, and reactors

A(\bar\nu_e\to\nu_x) = \htmlClass{t-s}{\sin^2 2\theta_{13}} \approx 4\theta_{13}^2

(10.26)

Bettini p. 449. The reactor disappearance amplitude — the cleanest possible handle on θ₁₃, because it depends on nothing else.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

💡 What this really says — the reactors won because their channel has only one unknown in it

Compare with (10.23), the accelerator route to the same angle: $\sin^2\theta_{23}\sin^2 2\theta_{13}$ , which is entangled with $\theta_{23}$ , and in reality with $\delta$ and matter effects as well. The reactor channel is $\sin^2 2\theta_{13}$ alone.

That is the trade the field made in 2012: accelerators had the bigger apparatus and a signal contaminated by three other unknowns; reactors had an 8 % effect and a clean one. The reactors won, and the clean number then made the accelerator measurement interpretable.

There is a second advantage the book flags in one sentence. The relevant variable is $L/E$ , and reactor antineutrinos are a thousand times less energetic than beam neutrinos — so the same $L/E$ is reached at 1.5 km instead of 300 km. You can put the whole experiment inside one hillside.

🔬 Experiment card — Daya Bay and the discovery of θ₁₃ ≠ 0

Apparatus

Six commercial reactor cores totalling 17.4 GW, and eight antineutrino detectors of 20 t of gadolinium-loaded liquid scintillator each, in three underground halls at flux-averaged baselines of 500 m, 500 m and 1650 m. A few hundred metres of rock overhead handles the cosmic rays.

What is measured

The $\bar\nu_e$ energy spectrum at every hall, through inverse beta decay: a prompt positron signal, then a neutron capture on gadolinium about 100 μs later releasing 8 MeV of gammas. The delayed coincidence is what makes the measurement possible — it is why the background is almost invisible on the plot.

The far spectrum is compared against the near ones, so the reactor flux and the detector response largely cancel.

The result

A clear deficit at the far hall, developing with $L_{\text{eff}}/\langle E\rangle$ exactly as $\sin^2$ predicts. From 3158 days and 5.55 million candidates, $\theta_{13} = 8.5^\circ \pm 0.1^\circ$ — about 1 % accuracy.

What it proved

$\theta_{13} \neq 0$ , announced in 2012 on 55 days of data and confirmed within weeks by RENO in Korea, with an indication from Double Chooz in France. Because the CP-violating phase enters only alongside $\sin\theta_{13}$ , this result is what made leptonic CP violation an experimental question rather than a hypothetical one.

🛠️ Daya Bay — six cores, eight detectors, three halls

Click a numbered marker for what that piece does.

Bettini Fig. 10.9(a). Two near halls at 500 m sample the flux before the oscillation develops; the far hall at 1650 m sits at the first maximum for a 3 MeV antineutrino. Click a callout.

\htmlClass{t-r}{\bar\nu_e + p \to e^+ + n}

(10.27)

Bettini p. 450. Inverse beta decay — the same reaction Reines and Cowan used in 1956, still the only practical way to see a reactor antineutrino.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

💡 What this really says — gadolinium is the whole trick

Gadolinium is the whole trick. Natural Gd has a neutron capture cross-section of 49 kb — five orders of magnitude above a typical nuclear cross-section — and the capture releases a distinctive 8 MeV cascade. Doping the scintillator with 0.1 % of it converts a single flash, indistinguishable from ambient radioactivity, into a timed pair that essentially nothing else produces.

⚙️ Engineer’s bridge — Daya Bay is two textbook noise-rejection tricks

Nothing about this experiment is exotic instrumentation. It is two techniques you have used, applied at an absurd scale.

1. Coincidence gating instead of a quieter detector. The prompt positron alone is one flash among a sea of radioactivity. Requiring a second flash, of the right energy, in the right place, about 100 μs later, is a coincidence gate — and the accidental rate is the product of two small numbers, so demanding the pair buys orders of magnitude that no amount of extra shielding would. This is why Fig. 10.9(b)‘s background is the grey sliver in the corner rather than the whole plot.

2. A differential measurement, not an absolute one. The signal is an 8 % change in flux. The reactor power, the antineutrino yield per fission, the detector efficiency and the target mass are each known to a few per cent at best, so an absolute measurement would drown. Putting near halls on the same reactors turns it into a ratio, and everything common to numerator and denominator cancels — the classic common-mode rejection you get from a differential pair or a ratiometric ADC reading. Eight nominally identical detectors, moved between halls during the run, push the cancellation further still.

Where it breaks: a reference arm is supposed to carry no signal, and this one does. The near halls already lose 2.1 % of their antineutrinos, which is a quarter of the effect being measured. So the “reference” has to be modelled rather than subtracted — the fit uses all three halls at once, and treating the near halls as unoscillated would bias θ₁₃ badly.

🔢 Worked example — reproduce Fig. 10.9(c) from θ₁₃ alone

Daya Bay survival probabilities

import numpy as np

Dm2, s2_2th13 = 2455e-6, 0.0841         # eV^2, and sin^2(2 theta_13) at theta_13 = 8.4 deg

def P(L_m, E_MeV):                      # (10.20) with the amplitude (10.26);
    return 1 - s2_2th13*np.sin(1.27*Dm2*(L_m/E_MeV))**2   # L/E in m/MeV = km/GeV

for hall, L in (("EH1", 500), ("EH2", 500), ("EH3", 1650)):
    print(f"{hall} at {L:5d} m:  P(3 MeV) = {P(L,3):.4f}   P(6 MeV) = {P(L,6):.4f}")

far, near = P(1650, 3), P(500, 3)
print(f"\nfar/near ratio at 3 MeV = {far/near:.4f}  ->  a {100*(1-far/near):.1f} % deficit")
print(f"the near halls are NOT unoscillated: they already lose {100*(1-near):.1f} %")

x = np.linspace(1, 900, 9001)           # L_eff / <E> in m/MeV — the axis of Fig. 10.9(c)
p = 1 - s2_2th13*np.sin(1.27*Dm2*x)**2
print(f"\ndeepest point of Fig. 10.9(c): P = {p.min():.4f} at {x[p.argmin()]:.0f} m/MeV")
print(f"and 1 - sin^2(2 theta_13) = {1 - s2_2th13:.4f} — the same number")

prints

EH1 at   500 m:  P(3 MeV) = 0.9793   P(6 MeV) = 0.9944
EH2 at   500 m:  P(3 MeV) = 0.9793   P(6 MeV) = 0.9944
EH3 at  1650 m:  P(3 MeV) = 0.9176   P(6 MeV) = 0.9519

far/near ratio at 3 MeV = 0.9371  ->  a 6.3 % deficit
the near halls are NOT unoscillated: they already lose 2.1 %

deepest point of Fig. 10.9(c): P = 0.9159 at 504 m/MeV
and 1 - sin^2(2 theta_13) = 0.9159 — the same number

The minimum of the published curve, 0.916 at 504 m/MeV, is $1 - \sin^2 2\theta_{13}$ — the depth is the amplitude, read off directly.

Note the units coincidence the reactor community lives on: because $L/E$ in m/MeV and in km/GeV are the same number, the 504 that governs a 300 km accelerator also governs a 500 m tunnel.

best fit, θ₁₃ = 8.4°
the three halls at ⟨E⟩ ≈ 3.3 MeV

Bettini Fig. 10.9(c), computed from θ₁₃ = 8.4° and Δm² = 2455 meV². The two near halls sit at the left, barely oscillated; the far hall sits almost exactly on the first maximum. The whole measurement is the 8 % between the dashed line and 1.

Erratum — Daya Bay has eight detectors, and Fig. 10.9(b) carries the fit

Two slips on p. 450, both catchable without leaving the chapter.

“The six detectors provide an overall target mass of 120 t.” p. 449 says “eight antineutrino detectors”, Fig. 10.9(a) draws eight grey rectangles (two in EH1, two in EH2, four in EH3), and the same paragraph says each holds 20 t. Eight × 20 t = 160 t. Daya Bay did start with six in 2012 and complete the array to eight later that year, but the 3158-day, 5.55-million-event sample being described used all eight.

“Fig. 10.9(a) shows also the best-fit curve.” Panel (a) is the map of tunnels and halls. The spectrum with the best-fit and no-oscillation curves is panel (b), which the same sentence pair cites correctly immediately before and immediately after.

(A third, harmless one: the text abbreviates inverse beta decay as “IVB”, twice. The field’s abbreviation is IBD, and the initials of the words are too.)

Where did they go? OPERA, and a photograph of a tau

Everything so far is a disappearance. The chain of reasoning says the missing $\nu_\mu$ became $\nu_\tau$ — but that is an inference from unitarity, and inferences get tested.

🔬 Experiment card — OPERA and ν_τ appearance

Apparatus

A $\nu_\mu$ beam from the CERN SPS aimed through the Earth’s crust at the Gran Sasso laboratory, 737 km away. The detector is 1250 t of “bricks” — sandwiches of photographic emulsion and 1 mm lead sheets, about 150 000 of them, 110 000 m² of film in all — read out by electronic trackers that say which brick to develop.

What is measured

$\nu_\tau + N \to \tau + N'$ , identified geometrically. The τ lives about 0.3 ps, so it travels a fraction of a millimetre before decaying: the signature is a kink or a multi-prong vertex within a millimetre of the interaction point. Nothing but emulsion resolves that, and nothing but kilotons of target gets the rate.

The result

From 2008 to 2012, $1.8\times10^{20}$ protons on target, 10 $\nu_\tau$ candidates on an estimated background of $2.0\pm0.2$ — mostly charm decays, which have the same topology.

What it proved

The disappeared muon neutrinos really do become tau neutrinos. Super-Kamiokande has since seen the same appearance in atmospheric neutrinos. Ten events sounds thin; against a background of two it is a five-sigma statement, and it closes the unitarity argument experimentally rather than by assumption.

\htmlClass{t-r}{\nu_\tau + N \to \tau + N'}

(10.28)

Bettini p. 451. The reaction OPERA was built to see — and the one whose threshold forced every other design choice.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

💡 What this really says — every design choice follows from the tau’s rest mass

Producing a tau costs 1.777 GeV of rest mass, and the neutrino has to bring it. Everything else about OPERA follows from that one constraint: a high-energy beam, therefore a small oscillation probability, therefore a huge target, therefore a target that is cheap per kilogram — lead — and therefore a readout that can find a 1 mm decay inside it, which means emulsion.

Emulsion in 2008 is not nostalgia. Nothing else resolves a micrometre over a kiloton, and that is what the measurement needs.

🔢 Worked example — the trade OPERA could not avoid

Threshold against oscillation maximum, at 737 km

import numpy as np

Dm2, A_tau = 2455e-6, 0.9477          # eV^2 and (10.24)
L, m_tau, m_N = 737.0, 1.777, 0.939   # km, GeV, GeV

E_thr = m_tau*(m_tau + 2*m_N)/(2*m_N)                 # nu_tau N -> tau N'
E_max = L/(np.pi/2/(1.27*Dm2))                        # E that puts L on the first maximum
print(f"tau production threshold      E > {E_thr:.2f} GeV")
print(f"first oscillation maximum at  E = {E_max:.2f} GeV  <-- below the threshold\n")

def P(E): return A_tau*np.sin(1.27*Dm2*L/E)**2
for E in (E_max, E_thr, 10.0, 17.0):
    print(f"E = {E:5.2f} GeV:  P(nu_mu -> nu_tau) = {P(E):.5f}"
          f"   = {P(E)/P(E_max)*100:5.2f} % of the maximum")
print(f"\nOPERA ran at about 17 GeV, so it paid a factor "
      f"{P(E_max)/P(17.0):.0f} in probability to be able to make a tau at all")

prints

tau production threshold      E > 3.46 GeV
first oscillation maximum at  E = 1.46 GeV  <-- below the threshold

E =  1.46 GeV:  P(nu_mu -> nu_tau) = 0.94770   = 100.00 % of the maximum
E =  3.46 GeV:  P(nu_mu -> nu_tau) = 0.36032   = 38.02 % of the maximum
E = 10.00 GeV:  P(nu_mu -> nu_tau) = 0.04917   =  5.19 % of the maximum
E = 17.00 GeV:  P(nu_mu -> nu_tau) = 0.01721   =  1.82 % of the maximum

OPERA ran at about 17 GeV, so it paid a factor 55 in probability to be able to make a tau at all

The two requirements are incompatible at this baseline. Sitting on the oscillation maximum at 737 km means running at 1.46 GeV, which is below the tau threshold — the probability would be 95 % and the cross-section zero. Running above threshold means running far off the maximum.

The book says this in a sentence: “the 737 km flight length is small compared with the distance of the oscillation maximum, corresponding to a small expected number of tau neutrinos.” The factor is 55, and it is the reason a 1250 t detector running for five years yielded ten events.

Bettini Fig. 10.10 redrawn — the second OPERA ν_τ candidate, in which the τ decays to three charged particles. The book prints the emulsion photograph; this site never embeds a scan, so here is what it shows. The whole event is two millimetres across, and the entire measurement is that the two vertices are distinguishable.

Erratum — the ν_τ conclusion is implied by (10.24), not (10.26)

p. 451 says: “the vast majority of disappeared $\nu_\mu$ should appear as $\nu_\tau$ . This fact is implied in Eq. (10.26).”

(10.26) is $A(\bar\nu_e\to\nu_x) = \sin^2 2\theta_{13}$ — the reactor amplitude, for electron antineutrinos, which says nothing about muon neutrinos. The statement is implied by (10.24), $A(\nu_\mu\to\nu_\tau) \approx 1$ , five pages earlier.

(A second, minor one on the same page: the CNGS baseline is given as 737 km here and as 730 km in Problems 10.6 and 10.7. The accepted CERN → Gran Sasso distance is 730 km, and MINOS’s 735 km is quoted correctly two pages earlier, so 737 is the outlier. It changes nothing — 1 % of a baseline is 1 % of a phase that is 0.14 radians.)

Two-flavour oscillation — Eqs. (10.19) and (10.20)

survival — still ν_μ
appearance — now ν_τ

mixing angle θ = 48.0° · the measured valuebeam energy spread σ(E)/E = 0 % · Fig. 10.6 — no real source is monochromatic

Δm²

2455 meV²

first maximum

504 km/GeV

sin²2θ — the depth

0.989

washed-out average

0.505

Super-K / MINOS ν_μ ↔ ν_τ. Push the energy spread to 30 % and read the plateau: 1 − ½sin²2θ₂₃ ≈ 0.5. That is the number Super-Kamiokande measured from below the horizon, and it is where θ₂₃ ≈ 45° comes from.

Survival and appearance always sum to 1 — unlike the meson case there is nothing to decay into, so mixing only moves probability between flavours. Push the spread slider and watch the wiggles die into the dashed mean: past a few maxima a real beam measures a constant deficit, and the frequency information is gone.

The two measurements of this section on one axis. They share Δm², so the curves have the same shape and the same first maximum; only the depth differs, and the depth is the angle. Compare with §10.2a, where the same widget derived the formula.

🔑 If you remember only three things

The deficit was found by a detector built for something else. Super-Kamiokande was hunting proton decay, and atmospheric neutrinos were the background it had to understand first.
Every parameter here was measured twice, by unrelated technologies. Water Cherenkov, streamer tubes, iron calorimeters, liquid scintillator and photographic emulsion agree, and none of them shares another’s systematics.
A disappearance is an inference until something appears. The chain of reasoning said the missing muon neutrinos became tau neutrinos; it took 110 000 m² of film to watch one arrive.

Where this goes next

The atmospheric oscillation is settled: $\Delta m^2$ to 1.5 %, $\theta_{23}$ near-maximal with the octant open, $\theta_{13}$ to 1 %, and the disappeared $\nu_\mu$ caught in the act of arriving as $\nu_\tau$ .

That leaves the other oscillation — the slow one, driven by $\delta m^2 = 73.4$ meV² and $\theta_{12}$ . It was seen first, in 1968, and it took thirty-four years to understand, because in the Sun it is not an oscillation at all.

§10.3 builds the machinery: matter gives $\nu_e$ an extra potential, the effective mass eigenvalues cross, and (10.17) — the innocuous ratio from §10.2a — turns a small vacuum angle into maximal mixing inside a star. §10.4 is the experimental story that machinery explains.

✅ Check yourself — three flavours and the experiments

0/6 answered · 0 correct

1.Eq. (10.22) gives the total ν_μ disappearance amplitude as about ½. Why can that not be right, using only what is on the same two pages?
2.In Fig. 10.8 the electron-like distributions match the no-oscillation prediction while the muon-like ones do not. What does the flat electron data establish?
3.Which features of Super-Kamiokande's zenith-angle result come out of the geometry and the formula alone, with no free parameters beyond Δm² and θ₂₃? (Select all that apply.)
4.Reactor experiments beat accelerators to θ₁₃ despite measuring an 8 % effect against a much larger accelerator signal. Why?
5.Daya Bay puts near halls at 500 m and a far hall at 1650 m. What is each distance for?
6.OPERA ran at about 17 GeV over 737 km, where the ν_τ appearance probability is 1.7 %. Sitting on the oscillation maximum would have given 95 %. Why not do that?