Atmospheric neutrinos

Atmospheric neutrinos are the natural neutrino beam that never switches off. Every second, cosmic-ray protons slam into nuclei at the top of the atmosphere, some fifteen kilometres overhead, and start hadronic cascades whose decay products are neutrinos of electron and muon flavour together with their antiparticles. Counting the decay chain is almost trivial — a charged pion decays to a muon and a muon-neutrino, and the muon in turn decays to an electron and two more neutrinos — so the source delivers about two muon-type neutrinos for every electron-type one. That clean prediction made atmospheric neutrinos, alongside solar neutrinos, the setting for the first compelling evidence of neutrino oscillations. The genius of the experiment is geometry the Earth provides for free. Neutrinos made directly overhead travel only tens of kilometres to a detector; those made on the far side of the planet travel the full diameter, nearly thirteen thousand kilometres, before arriving from below. The production physics is identical in every direction, so any up-down asymmetry in the flavour content can only come from something happening in flight. Super-Kamiokande found exactly such an asymmetry in 1998: upward-going muon-neutrinos were depleted by half, while electron-neutrinos were not — the muon-neutrinos were oscillating into tau-neutrinos over the long baseline. This article assembles that story and shows how kinematic thresholds decide which oscillation signals a detector can actually see.

Key takeaways

Cosmic rays striking the atmosphere start hadronic cascades whose pion and muon decays make neutrinos in a fixed flavour ratio (νμ + ν̄μ)/(νe + ν̄e) ≈ 2.
The Earth supplies baselines from ~15 km (overhead) to ~12,742 km (upward through the planet), so any up-down asymmetry is a clock ticking during flight.
Two-flavour oscillation gives survival P = 1 − sin²(2θ₂₃) sin²(1.267 Δm²₃₂ L/E), with near-maximal mixing sin²2θ₂₃ ≈ 1 and Δm²₃₂ ≈ 2.5×10⁻³ eV².
The first minimum sits at L/E ≈ 496 km/GeV; upward neutrinos average over many cycles, so exactly half the muon-neutrinos survive.
Oscillation shows up as disappearance of right-flavour leptons or appearance of wrong-flavour ones, but appearance is gated by charged-current thresholds: νe ≈ 0, νμ ≈ 110 MeV, ντ ≈ 3.5 GeV on a nucleon.
Key idea: the same neutrino beam seen at every baseline turns the Earth into an oscillation experiment, but you can only detect a wrong-flavour charged lepton if the neutrino carries enough energy to make it — which is why ντ appearance demands multi-GeV neutrinos while νe appearance has essentially no threshold.

The core insight

The Earth turns one well-understood neutrino source into an oscillation experiment spanning four decades of baseline, and the muon-neutrino deficit seen only on the long upward paths is the signature of νμ → ντ oscillation — visible as a disappearance because the ντ usually lacks the energy to make a detectable tau.

Where this lives today

Atmospheric neutrinos remain a frontier tool. Super-Kamiokande keeps sharpening the parameters and has reported tau-neutrino appearance above 4σ from the rare above-threshold events. IceCube's DeepCore array and the KM3NeT/ORCA detector in the Mediterranean instrument cubic kilometres of ice and water to catch the same flux at higher statistics, targeting the neutrino mass ordering through the matter effect atmospheric neutrinos pick up crossing the Earth's dense core — a directional modulation that separates normal from inverted ordering. The value of $\theta_{23}$ , and whether it sits just above or below 45° (the "octant" question), is still open and is a headline goal for Hyper-Kamiokande and DUNE. The same L/E physics the Earth demonstrated for free is now engineered in long-baseline beams like T2K and NOvA, tuned to sit right at the first oscillation dip.

The physical picture

A cosmic-ray proton arriving at the top of the atmosphere collides with an oxygen or nitrogen nucleus, spraying out pions and kaons in a hadronic shower. The charged pions are the workhorses: a positive pion decays to an antimuon and a muon-neutrino, and the antimuon then decays to a positron, an electron-neutrino, and an antimuon-neutrino. Tallying one chain gives one νe against two muon-type states (νμ and ν̄μ) — the celebrated factor of two. The first figure walks through that bookkeeping, because the whole anomaly is a departure from this number.

Figure

Cosmic-ray protons strike air nuclei ≈ 15 km up and start a hadronic cascade; each charged-pion chain π⁺ → μ⁺ + νμ followed by μ⁺ → e⁺ + νe + ν̄μ yields two muon-type neutrinos for every one electron-type, fixing the production ratio (νμ + ν̄μ)/(νe + ν̄e) ≈ 2

The moment you look at where the neutrinos come from, the experiment writes itself. Because neutrinos barely interact, the atmosphere and rock are transparent to them: one made in the sky over Argentina sails straight through the planet and is caught coming upward out of the ground in Japan. A single detector thus samples the same source over baselines from about fifteen kilometres, for neutrinos born overhead, to the Earth's diameter of 12,742 kilometres for those born on the antipode. The production is up-down symmetric, so if the arriving flavour content depends on direction, the neutrinos must be changing en route — oscillation, whose size depends on L/E. The second figure is the punchline: the survival probability versus L/E, the curve Super-Kamiokande carved out of the sky.

Figure

Two-flavour muon-neutrino survival P(νμ→νμ) = 1 − sin²(2θ₂₃) sin²(1.267 Δm²₃₂ L/E) for maximal mixing sin²2θ₂₃ = 1 and Δm²₃₂ = 2.5×10⁻³ eV²; the first minimum is at L/E ≈ 496 km/GeV, and for long upward baselines the rapid oscillation averages to a survival of exactly 1/2

Down-going neutrinos have a tiny baseline, so L/E is small, the phase is negligible, and essentially all muon-neutrinos survive — the left edge of the curve. Up-going neutrinos of a few GeV sit well past the first dip, so the rapid oscillations wash out under finite resolution and the survival averages to one-half: half the upward muon-neutrinos vanish, reappearing as tau-neutrinos that mostly cannot be detected. That up-down asymmetry — full flux down, half flux up — is the atmospheric neutrino anomaly.

The core equations

Everything begins with the flavour ratio the source hands us. Because a charged-pion chain makes two muon-type neutrinos per electron-type one, the production ratio is fixed by counting, and it is the denominator against which any deficit is measured. At higher energies the muons increasingly reach the ground before decaying, suppressing their electron-neutrinos and pushing the ratio above two, but near a GeV the clean value holds.

R \equiv \frac{\Phi(\nu_\mu)+\Phi(\bar\nu_\mu)}{\Phi(\nu_e)+\Phi(\bar\nu_e)} \approx 2 \qquad (1)

where: $\Phi$ — the arriving flux of each flavour; the ratio is $\approx 2$ near 1 GeV and rises with energy as decay-in-flight muons are lost. Reading: the source is muon-rich by a factor two, and that is the number oscillations must spoil. The disappearance is governed by the two-flavour oscillation formula. In the atmospheric sector the relevant mixing is between the second and third neutrinos, controlled by the angle $\theta_{23}$ and the mass-squared splitting $\Delta m^2_{32}$ . The survival probability of a muon-neutrino after travelling a distance $L$ with energy $E$ is a single oscillating term subtracted from one.

P(\nu_\mu\!\to\!\nu_\mu)=1-\sin^2(2\theta_{23})\,\sin^2\!\left(\frac{\Delta m^2_{32}\,L}{4E}\right) \qquad (2)

where: $\theta_{23}$ — atmospheric mixing angle, near maximal ( $\theta_{23}\approx 45°$ , so $\sin^2 2\theta_{23}\approx 1$ ); $\Delta m^2_{32}=m_3^2-m_2^2\approx 2.5\times10^{-3}\ \mathrm{eV^2}$ ; $L$ — baseline; $E$ — energy. The amplitude $\sin^2 2\theta_{23}$ sets how deep the dip goes; the phase $\Delta m^2_{32} L/4E$ sets where it sits. For numbers, the phase is written in the experimenter's units, where $\Delta m^2$ is in electron-volts squared, $L$ in kilometres, and $E$ in GeV. Collecting the factors of $\hbar c$ into a single constant gives the form every neutrino talk quotes.

P(\nu_\mu\!\to\!\nu_\mu)=1-\sin^2(2\theta_{23})\,\sin^2\!\left(1.267\,\frac{\Delta m^2_{32}[\mathrm{eV^2}]\,L[\mathrm{km}]}{E[\mathrm{GeV}]}\right) \qquad (3)

where: the constant $1.267=\Delta m^2 L/(4E)$ evaluated with $\hbar c=197.3\ \mathrm{MeV\,fm}$ and the stated units. Reading: the first minimum ( $\text{phase}=\pi/2$ ) is at $L/E=496\ \mathrm{km/GeV}$ ; the argument shows at a glance where a baseline and energy sit on the curve. Whether an oscillated flavour can be seen is a separate, kinematic question. A neutrino is detected mostly through the charged-current reaction $\nu_\ell + N \to \ell^- + N'$ , which cannot proceed without enough energy to make the charged lepton. On a nucleon at rest this sets a threshold energy.

E_\nu^{\rm th}=\frac{(m_\ell+m_{N'})^2-m_N^2}{2\,m_N} \qquad (4)

where: $m_\ell$ — mass of the produced charged lepton; $m_N,\,m_{N'}$ — target and recoil nucleon masses. Reading: because $m_\tau$ is large, the tau threshold is several GeV, while the electron threshold is essentially zero — this is the gate on appearance signals. Finally, why the nucleon, not the atomic electron, is the target that matters. The charged-current cross section grows with the centre-of-mass energy squared, and $s\approx 2 m E_\nu$ scales with the target mass. The nucleon is nearly two thousand times heavier than the electron, so nucleon interactions dominate wherever they are open, and the cross section rises linearly with energy — the signature of scattering off point-like quarks.

\sigma_{\rm CC}\propto s\approx 2\,m_N E_\nu,\qquad \frac{\sigma_{\rm CC}^{\nu N}}{E_\nu}\approx 0.68\times10^{-38}\ \mathrm{cm^2\,GeV^{-1}} \qquad (5)

where: $m_N$ — nucleon mass; the antineutrino slope is about half the neutrino value. Reading: heavier target and higher energy both raise the rate, which is why detection takes multi-kilotonne detectors and GeV neutrinos.

Derivation AThe two-to-one flavour ratio from the pion decay chain

The goal is to derive the source prediction $R=(\nu_\mu+\bar\nu_\mu)/(\nu_e+\bar\nu_e)\approx 2$ by tracking every neutrino emitted in the decay of a single charged pion and its daughter muon, so the atmospheric anomaly can be stated precisely as a deficit relative to this number. We count final-state neutrinos flavour by flavour, note the antiparticle content, and see what modifies the clean ratio at higher energy — fixing the denominator of the whole measurement.

Step 1. Start from the dominant hadron in the cascade, the charged pion, and write its leptonic decay: $\pi^+\to\mu^+ + \nu_\mu$ . This emits exactly one muon-type neutrino. Why: the pion is spin-0 and the weak interaction is left-handed, so the two-body decay to $\mu^+\nu_\mu$ dominates; it is the first neutrino of the chain.

Step 2. Follow the antimuon, which is itself unstable and decays as $\mu^+\to e^+ + \nu_e + \bar\nu_\mu$ . This adds one electron-type neutrino and one more muon-type (anti)neutrino. Why: muon decay is a three-body weak process that must conserve both electron and muon lepton number, forcing exactly one $\nu_e$ and one $\bar\nu_\mu$ into the final state.

Step 3. Tally the chain: muon-type states are $\nu_\mu$ and $\bar\nu_\mu$ (two), electron-type is $\nu_e$ (one). The negative-pion chain gives the charge conjugates, preserving the count, so $R=(\nu_\mu+\bar\nu_\mu)/(\nu_e+\bar\nu_e)=2/1=2$ . Why: detectors cannot separate neutrino from antineutrino event by event, so what matters is the summed muon-type versus electron-type flux — rigidly two to one.

Step 4. State the caveat that keeps the number honest: above a few GeV the muons travel far enough to reach the ground and stop before decaying, so their $\nu_e$ is never emitted and $R$ climbs above two. Detailed cascade simulations predict $R$ to a few percent across the relevant energies. Why: the clean value 2 is a low-energy limit; the experiments quote a double ratio $R_{\rm data}/R_{\rm MC}$ precisely so that these known corrections cancel.

Go deeper · Derivation B · 1 minThe muon-neutrino survival probability and the 1.267 L over E rule

Derivation BThe muon-neutrino survival probability and the 1.267 L over E rule

The aim is to obtain the two-flavour survival probability of equation 2 and to pin down the practical constant 1.267 that converts natural units into kilometres, GeV, and electron-volts squared. We rotate the muon-neutrino into the mass basis, propagate each mass eigenstate with its own quantum phase, project back onto the muon flavour, and restore the factors of $\hbar$ and $c$ — the engine behind the survival curve of the second figure.

Step 1. Write the flavour state as a rotation of two mass eigenstates, $|\nu_\mu\rangle=-\sin\theta\,|\nu_1\rangle+\cos\theta\,|\nu_2\rangle$ , with $\theta=\theta_{23}$ , each carrying energy $E_i\approx E+m_i^2/2E$ at fixed momentum. Why: neutrinos are made and detected as flavour states but propagate as mass states, so mixing is the entire origin of oscillation; the energy expansion holds because $m_i\ll E$ .

Step 2. Propagate a distance $L\approx t$ and form the relative phase $\Delta\phi=(E_2-E_1)L=\dfrac{(m_2^2-m_1^2)L}{2E}=\dfrac{\Delta m^2 L}{2E}$ . Why: only the phase difference between the two eigenstates is physical, and it is this accumulating difference that periodically rotates the state away from and back toward the original flavour.

Step 3. Project the propagated state back onto $|\nu_\mu\rangle$ and square to get the survival probability $P=1-\sin^2(2\theta)\,\sin^2(\Delta\phi/2)=1-\sin^2(2\theta)\,\sin^2\!\big(\Delta m^2 L/4E\big)$ . Why: the half-angle appears because the amplitude carries $\Delta\phi/2$ while the probability squares it; the mixing enters only through $\sin^2 2\theta$ , which is why maximal mixing gives the deepest possible dip.

Step 4. Restore units. With $\hbar c=197.3\ \mathrm{MeV\,fm}$ , evaluating $\Delta m^2[\mathrm{eV^2}]\,L[\mathrm{km}]/4E[\mathrm{GeV}]$ gives the conversion factor 1.267. Setting the argument to $\pi/2$ puts the first minimum at $L/E=(\pi/2)/(1.267\times2.5\times10^{-3})\approx 496\ \mathrm{km/GeV}$ . Why: this single constant lets an experimenter read the survival curve directly from measured baseline and energy, without returning to natural units.

Go deeper · Derivation C · 1 minThe charged-current threshold and why tau appearance needs multi-GeV neutrinos

Derivation CThe charged-current threshold and why tau appearance needs multi-GeV neutrinos

The goal is to derive the threshold energy of equation 4 for producing a charged lepton on a nucleon target, and to show numerically that a tau-neutrino below about 3.5 GeV cannot make a tau at all, so its appearance is kinematically forbidden however strongly it oscillated. We impose energy-momentum conservation at threshold, where the final particles are at rest in the centre of mass, then evaluate for the electron, muon, and tau — the distinction that separates a disappearance experiment from an appearance one.

Step 1. Write the reaction $\nu_\ell + N \to \ell^- + N'$ and demand that at threshold the final state has just enough energy to exist, i.e. the invariant mass squared equals $s_{\min}=(m_\ell+m_{N'})^2$ . Why: threshold is defined by the final particles being at rest relative to one another, the minimum possible invariant mass; below it the reaction cannot conserve four-momentum.

Step 2. For a neutrino of energy $E_\nu$ hitting a nucleon of mass $m_N$ at rest, the invariant is $s=m_N^2+2m_N E_\nu$ . Set $s=s_{\min}$ and solve for the neutrino energy: $E_\nu^{\rm th}=\big[(m_\ell+m_{N'})^2-m_N^2\big]/2m_N$ . Why: this isolates the one unknown, the beam energy, in terms of masses alone, giving a clean floor for each flavour.

Step 3. Evaluate for the muon on the reaction $\nu_\mu n\to\mu^- p$ : with $m_\mu=105.7$ , $m_p=938.3$ , $m_n=939.6$ MeV, $E^{\rm th}_{\nu_\mu}=\big[(105.7+938.3)^2-939.6^2\big]/(2\times939.6)\approx 110\ \mathrm{MeV}$ . For the electron the recoil proton is lighter than the target neutron, so the threshold is essentially zero. Why: the muon threshold is modest, so muon appearance and disappearance are both accessible above a few hundred MeV, while electron-neutrinos interact at any energy.

Step 4. Now the tau: with $m_\tau=1776.9$ MeV, $E^{\rm th}_{\nu_\tau}=\big[(1776.9+938.3)^2-939.6^2\big]/(2\times939.6)\approx 3.45\ \mathrm{GeV}\approx 3.5\ \mathrm{GeV}$ . Below this the oscillated $\nu_\tau$ flux is real but invisible in the charged-current channel. Why: the heavy tau mass pushes the threshold into the multi-GeV range, which is why atmospheric $\nu_\tau$ appearance is a hard, statistics-hungry measurement while $\nu_\mu$ disappearance was seen first.

Limiting cases

Case	Result	Meaning
Short baseline (down-going)	$L/E\to0$ , $P\to1$	phase negligible; no oscillation, full muon-neutrino flux survives
First survival minimum	$L/E\approx496\ \mathrm{km/GeV}$ , $P\to1-\sin^2 2\theta$	deepest dip; for maximal mixing the muon-neutrinos are fully depleted
Long baseline (up-going, averaged)	$\langle P\rangle=1-\tfrac12\sin^2 2\theta=\tfrac12$	many cycles wash out; exactly half survive for maximal mixing
Maximal mixing	$\sin^2 2\theta_{23}\approx1$ ( $\theta_{23}\approx45°$ )	dips reach zero; atmospheric mixing is near-maximal, unlike the small quark angles
Below tau threshold	$E_\nu<3.5\ \mathrm{GeV}$ : $\nu_\tau$ CC forbidden	oscillated $\nu_\tau$ present but undetectable as a tau; seen only as $\nu_\mu$ disappearance
High energy	$E\gg$ few GeV, phase $\to0$	oscillations shut off; flux arrives unoscillated, a control sample at the detector

Common trapAny oscillated neutrino flavour is automatically detectable.

It is not. Oscillation changes the flavour of the neutrino, but detecting that flavour through the charged-current channel requires making the corresponding charged lepton, and that costs energy. A tau-neutrino below about 3.5 GeV cannot produce a tau on a nucleon at all — equation 4 forbids it — so its charged-current appearance is kinematically impossible no matter how completely the muon-neutrino oscillated into it. This is why the atmospheric anomaly was first and most cleanly seen as a disappearance of multi-GeV muon-neutrinos rather than an appearance of tau-neutrinos, which were mostly below threshold and invisible until later dug out above 4σ. Flavour change and detectability are two separate questions, and thresholds decide the second.

Worked examples

1 · The tau appearance threshold (fully worked)

A muon-neutrino has oscillated into a tau-neutrino. What minimum energy must it have to produce a tau lepton on a nucleon, and what does that imply for atmospheric tau appearance?

Step 1. Identify the reaction and masses. The quasi-elastic charged-current process is $\nu_\tau n\to\tau^- p$ , with $m_\tau=1776.9$ MeV, target neutron $m_n=939.6$ MeV, recoil proton $m_p=938.3$ MeV. The threshold formula is $E_\nu^{\rm th}=[(m_\tau+m_p)^2-m_n^2]/2m_n$ .

Step 2. Compute the numerator: $(m_\tau+m_p)^2=(1776.9+938.3)^2=(2715.2)^2=7.372\times10^{6}\ \mathrm{MeV^2}$ , and $m_n^2=(939.6)^2=8.828\times10^{5}\ \mathrm{MeV^2}$ , so the numerator is $7.372\times10^6-0.883\times10^6=6.489\times10^{6}\ \mathrm{MeV^2}$ .

Step 3. Divide by $2m_n=1879.1$ MeV: $E_\nu^{\rm th}=6.489\times10^6/1879.1=3453\ \mathrm{MeV}$ . So the tau threshold is about 3.5 GeV.

Step 4. Interpret. Most atmospheric neutrinos sit in the sub-GeV to few-GeV range, so only the high-energy tail of the oscillated $\nu_\tau$ flux is above threshold. The tau, once made, decays almost instantly into a busy final state that is hard to tag against neutral-current background. This is why tau appearance took two decades longer to confirm than muon disappearance, even though both come from the same oscillation.

2 · Oscillation length and the up-down asymmetry (fully worked)

Take $\Delta m^2_{32}=2.5\times10^{-3}\ \mathrm{eV^2}$ and maximal mixing. For a 1 GeV muon-neutrino, find the baseline of the first survival minimum, and compare the survival for a down-going path (L ≈ 15 km) with an up-going path through the Earth (L ≈ 12,742 km).

Step 1. First minimum. Set the phase to $\pi/2$ : $1.267\,\Delta m^2 L/E=\pi/2$ , so $L/E=(\pi/2)/(1.267\times2.5\times10^{-3})=496\ \mathrm{km/GeV}$ . At $E=1$ GeV the first minimum is at L ≈ 496 km.

Step 2. Down-going. With $L=15$ km, $E=1$ GeV, the phase is $1.267\times2.5\times10^{-3}\times15=0.0475\ \mathrm{rad}$ . Then $\sin^2(0.0475)=2.3\times10^{-3}$ , so $P=1-2.3\times10^{-3}\approx\mathbf{0.998}$ — essentially every down-going muon-neutrino survives.

Step 3. Up-going. With $L=12{,}742$ km, the phase is $1.267\times2.5\times10^{-3}\times12742=40.4\ \mathrm{rad}$ , about $12.9\pi$ — many full oscillations. Detector resolution smears $\sin^2$ over these cycles to its average of $\tfrac12$ , so $P=1-\tfrac12=\mathbf{0.50}$ .

Step 4. The asymmetry. Down-going flux is full; up-going flux is halved. A single detector sees the same source deficient by a factor two from below but intact from above — the up-down asymmetry that established νμ oscillation, an up/down ratio of muon-like events near $0.5$ instead of $1.0$ .

3 · Reading the double ratio

Super-Kamiokande reports a double ratio $R_{\rm data}/R_{\rm MC}\approx0.65$ of muon-like to electron-like events. Show it is consistent with averaged νμ disappearance and no νe change.

Step 1. Start from the source ratio $R=(\nu_\mu+\bar\nu_\mu)/(\nu_e+\bar\nu_e)=2$ of Derivation A.

Step 2. Apply averaged oscillation: upward muon-neutrinos are halved while electron-neutrinos are essentially unaffected by $\theta_{23}$ mixing, so across a zenith-weighted sample the muon-type flux falls between its full value (down) and half (up), landing near $0.65$ of the unoscillated level.

Step 3. With the electron-type unchanged, this gives $R_{\rm data}/R_{\rm MC}\approx0.65$ . A double ratio well below 1 with the electron channel intact is the fingerprint of νμ → ντ, not νμ → νe — had the muon-neutrinos become electron-neutrinos, the electron-like sample would have risen instead.

Physics of the portrait

Atmospheric neutrinos are where a bookkeeping puzzle became a Nobel Prize. For years the muon deficit looked like a possible flux-model error; the person who turned it into a discovery did so by showing the deficit depended on the one thing no systematic error could fake — the direction the neutrino came from.

Origins · 1 minTakaaki Kajita and how this was discovered

Takaaki Kajita

1959– · atmospheric neutrino oscillations at Super-Kamiokande

Takaaki Kajita, born in 1959 near Tokyo, did his doctoral work under Masatoshi Koshiba and spent his career on the Kamiokande and Super-Kamiokande water-Cherenkov detectors under Mount Ikeno. Through the late 1980s Kamiokande and the American IMB experiment both reported fewer muon-like atmospheric events than expected — the "atmospheric neutrino anomaly" — but the effect could be blamed on uncertain cosmic-ray flux models, and the community stayed cautious. Kajita's decisive move was to bin the events by zenith angle, turning the Earth's diameter into a variable baseline. In 1998, at the Neutrino conference in Takayama, he presented Super-Kamiokande data showing down-going muon-neutrinos arriving at full strength while up-going ones, having crossed the planet, were depleted by half, with the electron-neutrinos untouched. No flux uncertainty can produce an up-down asymmetry in an isotropic source; only oscillation in flight can, and the measured L/E dependence matched $\Delta m^2\approx2.5\times10^{-3}\ \mathrm{eV^2}$ at near-maximal mixing. It was the first firm evidence that neutrinos have mass. Kajita shared the 2015 Nobel Prize with Arthur McDonald of SNO, whose solar results completed the case.

History

1957

— Pontecorvo proposes that neutrinos could oscillate

1962

— Maki, Nakagawa and Sakata formulate neutrino flavour mixing

1965

— first atmospheric neutrinos detected in deep mines in India and South Africa

1998

— Super-Kamiokande reports zenith-angle evidence for muon-neutrino disappearance

2015

— Kajita and McDonald share the Nobel Prize for neutrino oscillations

Going further

Texts. Thomson, Modern Particle Physics, §13.6 (neutrino oscillation experiments) and §13.6.1 (interaction thresholds), with the PMNS matrix at eq. 13.33 · Griffiths, Introduction to Elementary Particles, 2nd ed., Ch. 11 (neutrino oscillations) · Halzen & Martin, Quarks and Leptons, Ch. 12 (weak interactions and neutrino scattering) · Perkins, Particle Astrophysics, Ch. 9 (atmospheric and solar neutrinos) · Schwartz, Quantum Field Theory and the Standard Model, Ch. 29 (neutrino masses and mixing). The key numerical anchors used throughout this article are: the flavour ratio 2 from the $\pi\to\mu\to e$ chain; the constant 1.267 and the first survival minimum at $L/E=496$ km/GeV for $\Delta m^2=2.5\times10^{-3}$ eV²; the charged-current thresholds $E^{\rm th}_{\nu_\mu}\approx110$ MeV and $E^{\rm th}_{\nu_\tau}\approx3.45$ GeV from equation 4; and the averaged survival of $\tfrac12$ for maximal mixing over long baselines.

Prerequisites. Neutrino Flavours · Charged-Current Weak Interaction · Mandelstam Variables. Leads to. Two-Flavour Oscillations · Three-Flavour PMNS Mixing · Neutrino Oscillation Experiments.

Ask the tutor: ask me to work through the zenith-angle survival curve for a chosen energy, or to show how the Earth's matter effect on atmospheric neutrinos could reveal the neutrino mass ordering.

Why is the flavour ratio 2 and not 3, given that muon decay makes three neutrinos?

The chain $\pi^+\to\mu^+\nu_\mu$ then $\mu^+\to e^+\nu_e\bar\nu_\mu$ does make three neutrinos in total, but they are not three different flavours. Two of them are muon-type ( $\nu_\mu$ from the pion and $\bar\nu_\mu$ from the muon) and only one is electron-type ( $\nu_e$ ). Since a detector counts muon-type and electron-type events and cannot separate neutrino from antineutrino, the relevant number is the sum in each flavour, two to one. Above a few GeV the muons often reach the ground before decaying, so their $\nu_e$ is lost and the ratio climbs above two — one reason experiments quote a double ratio instead.

How does the Earth act as part of the experiment?

Neutrinos interact so weakly that ordinary matter is transparent to them, so one produced in the atmosphere on the opposite side of the planet passes straight through the rock and arrives at the detector travelling upward. Arrival direction thus becomes a proxy for baseline: downward events have travelled about fifteen kilometres, horizontal events a few hundred, and straight-up events the full Earth diameter of 12,742 kilometres. Because the source is up-down symmetric, any dependence of the flavour content on zenith angle is unambiguous evidence for oscillation in flight — the Earth converts one detector into a scan over four decades of L.

Why was the anomaly seen as muon disappearance rather than tau appearance?

Because of thresholds. The muon-neutrinos that oscillate are typically multi-GeV and interact readily, producing sharp muon tracks that are easy to count, so their deficit is visible directly. The tau-neutrinos they turn into need at least about 3.5 GeV to make a tau lepton, and most of the atmospheric flux is below that; even above threshold the tau decays instantly into a messy final state that mimics neutral-current background. So the same oscillation is loud in the disappearance channel and nearly silent in the appearance channel. Tau appearance was eventually confirmed, but only with dedicated selections and years of exposure, long after the disappearance had settled the physics.

What is the difference between appearance and disappearance experiments?

Disappearance looks for a deficit of the original flavour: you start with a known muon-neutrino flux and find fewer muons than expected. Appearance looks for a surplus of a new flavour — electrons or taus the source did not make. Disappearance is easier because you count only one flavour, but it tells you a neutrino left the beam without saying where it went; appearance is the definitive test of what the neutrino became, yet it is gated by the charged-current threshold for the new lepton. Atmospheric neutrinos gave a clean disappearance signal first, with appearance following once the threshold-limited tau events could be dug out.

Why does the cross section grow with energy, and why the nucleon over the electron?

Charged-current scattering off a point-like quark has a cross section that rises with the centre-of-mass energy squared, $\sigma\propto s$ , and $s\approx2mE_\nu$ grows with both the target mass $m$ and the beam energy. A nucleon is about 1836 times heavier than an electron, so at the same neutrino energy the available $s$ — and hence the cross section — is far larger on the nucleon, which is why nucleon interactions dominate whenever they are open. The linear rise with energy is itself a signature that the neutrino scatters off point charges inside the nucleon, the quarks.

Is the two-flavour formula enough, or do you need all three neutrinos?

For the leading atmospheric signal the two-flavour formula is an excellent approximation, because the atmospheric splitting $\Delta m^2_{32}\approx2.5\times10^{-3}$ eV² dwarfs the solar one and the $\nu_\mu\to\nu_\tau$ mixing is dominated by $\theta_{23}$ — over the L/E where the muon deficit lives, the physics is effectively one angle and one splitting. The full three-flavour treatment matters for subtler effects: $\theta_{13}$ opens a $\nu_\mu\to\nu_e$ channel, the CP phase tilts neutrino versus antineutrino rates, and the matter effect across the Earth's core imprints the mass ordering. Those refinements are the subject of the PMNS article.