University Physics IV · Particle Physics · 14.9
Neutrino Mass & Open Questions
Oscillation is the only laboratory evidence that neutrinos have mass at all, and it arrives as a deficit rather than as a mass. Learn to turn a survival probability into a splitting and a mixing angle — then to be honest about the three things that measurement still cannot tell you.
Build the model
Connect the measurement to the mechanism.
A neutrino is made at a weak vertex, so it is born in a flavour state — the partner of whichever charged lepton came with it. What propagates, though, is the mass basis, and the two bases are related by a rotation, the PMNS matrix. Each mass eigenstate carries its own phase, growing as mᵢ²c⁴L/(2ħcE), so after a flight of length L the superposition that was pure νμ has picked up a relative phase and is no longer pure anything.
The interference gives a survival probability that dips and returns as sin²(1.27 Δm² L/E): the observable is not a mass but a squared-mass difference, and it is only ever seen through the ratio L/E. That is the whole model, and it has a price. Interference is blind to a common offset — add m₀² to every squared mass and every probability is unchanged — so oscillation fixes Δm²₂₁ and |Δm²₃₂| and never the absolute scale.
In vacuum it is blind to the sign of Δm²₃₂ as well, so the ordering stays open. And the mass it proves must exist has no home in the Standard Model as written, which contains no right-handed neutrino to pair a Dirac mass with; supplying one adds seven measured numbers to the model's nineteen and still leaves dark matter, the baryon asymmetry and gravity outside it.
- Simple definition
- Neutrino oscillation is the periodic change of a travelling neutrino's flavour, caused by flavour states being superpositions of mass states that accumulate different quantum phases — so it can happen only if the masses are non-zero and unequal.
- Example
- A 4.0 MeV reactor antineutrino one kilometre from the core has a 4.3% chance of no longer being an electron antineutrino, taking Δm² = 2.5 × 10⁻³ eV² and sin²2θ₁₃ = 0.085 — the deficit Daya Bay measured.
Two independent knobs: θ fixes how deep each dip goes, Δm² fixes where along L it falls.
Δm² = m₂² − m₁² in eV², θ the vacuum mixing angle, ħc = 1.973 × 10⁻⁷ eV m
Converts a measured deficit straight into Δm² once the baseline and beam energy are known.
Δm² in eV², L in km, E in GeV. The 1.27 is 10³/(4 × 197.3) and holds in no other pairing.
Lets you pick a site before you build: L/E must land on the splitting you want to see.
km, with E in GeV and Δm² in eV². For 2.5 × 10⁻³ eV², L/E at the first minimum is 496 km/GeV.
Mixing must be non-zero as well as the splitting: with U = 1 every oscillation vanishes.
θ₁₂ ≈ 33.4°, θ₂₃ ≈ 45°, θ₁₃ ≈ 8.6° — all far larger than their CKM counterparts
Oscillation supplies a floor and no ceiling; cosmology's Σmν < 0.12 eV supplies the ceiling.
From Δm²₂₁ = 7.53 × 10⁻⁵ eV² and |Δm²₃₂| = 2.45 × 10⁻³ eV² with the lightest state set to zero
Every one is measured rather than derived, so the count is the honest size of the model's ignorance.
28 if neutrinos are Majorana, which adds two further CP-violating phases
Born in one basis, travelling in another
The weak charged current creates a neutrino alongside a definite charged lepton, so what leaves the vertex is a flavour state |νμ⟩. Propagation is governed by the Hamiltonian, whose eigenstates are the mass states |ν₁⟩, |ν₂⟩, |ν₃⟩. In the two-flavour shorthand, |νₑ⟩ = cos θ|ν₁⟩ + sin θ|ν₂⟩ and |νμ⟩ = −sin θ|ν₁⟩ + cos θ|ν₂⟩. Each mass state advances with its own phase, so after a distance L the two coefficients no longer stand in the ratio that made a pure νμ, and the projection onto |νₑ⟩ is non-zero. Two conditions are needed and each is physical. If θ = 0 the bases coincide and there is nothing to rotate. If m₁ = m₂ the two phases stay locked together forever and the superposition never changes shape. So a single confirmed oscillation is a proof of mass — a negative statement about the vacuum turned into a positive one about the particle.
Where the phase, and the constant 1.27, come from
For a neutrino of momentum p, Eᵢ = √(p²c² + mᵢ²c⁴) ≈ pc + mᵢ²c⁴/(2pc), because mic² is at most an electronvolt against MeV or GeV of energy. The common pc term is identical for every mass state and cancels out of any relative phase; only the m²/2E piece survives. Over a flight time t ≈ L/c the relative phase is Δm²c⁴L/(2ħcE), and the probability, being an amplitude squared, carries half of it: sin²(Δm²c⁴L/4ħcE). Now convert. With ħc = 1.973 × 10⁻⁷ eV m, Δm² in eV², L in kilometres and E in GeV, the bracket becomes 10³/(4 × 1.973 × 10⁻⁷ × 10⁹) = 1.267. That number is a unit conversion, not physics. It happens to hold equally for L in metres with E in MeV, since both scale by 10³; mix kilometres with MeV and the phase is out by a factor of a thousand.
The two deficits that forced the model
Super-Kamiokande's 1998 atmospheric result compared muon-like events arriving from overhead, L ≈ 15 km, with those that had crossed the Earth, L ≈ 13 000 km, at the same energies. Downward rates matched expectation; upward rates were down by about half. Both samples come from the same cosmic-ray flux, so no flux error can delete one hemisphere — only an L/E-dependent survival can. The zenith-angle fit gives |Δm²₃₂| ≈ 2.5 × 10⁻³ eV² and sin²2θ₂₃ ≈ 1, near-maximal mixing. The solar deficit took longer, because the answer could always have been a wrong Sun. SNO closed that door by measuring two channels at once: a charged-current rate sensitive to νₑ alone, and a neutral-current rate sensitive to all three flavours equally. The neutral-current total matched the standard solar model while the charged-current part was a third of it. The neutrinos had not vanished; they had changed flavour, with matter-enhanced (MSW) conversion driving the ⁸B survival probability towards sin²θ₁₂ ≈ 0.31.
Designing a baseline: L/E is the only dial you own
Δm² is given by nature; L and E are yours. The first minimum sits at L/E = 1.24/Δm², in km per GeV, so each splitting names its own site. For |Δm²₃₂| = 2.5 × 10⁻³ eV² that is 496 km/GeV: T2K fires a 0.6 GeV beam 295 km from Tokai to Kamioka, giving L/E = 492 km/GeV, essentially on the peak. For the same splitting with 4.0 MeV reactor antineutrinos the first minimum falls at 2.0 km, which is why the Daya Bay and RENO far halls sit between about 1.4 and 1.9 km from their cores. The solar splitting, thirty-three times smaller, pushes that to 66 km at the same 4 MeV, and KamLAND's 180 km flux-weighted baseline sits past it, where the spectrum shows a whole oscillatory shape rather than a single dip. Miss in either direction and you lose: at L ≪ Losc the probability grows only as L², while at L ≫ Losc the fringes wash out to the flat average ½sin²2θ and the splitting is no longer readable.
What oscillation is blind to
Three things survive every oscillation experiment ever run. The absolute scale: only differences of squares appear, so the family m₁ = m₀, m₂ = √(m₀² + Δm²₂₁) fits identically for any m₀ from zero up to the cosmological ceiling. The ordering: vacuum probabilities depend on sin², which is even, so the sign of Δm²₃₂ is invisible, and the sign of Δm²₂₁ is known only because matter effects in the Sun break that symmetry. And the nature of the particle: nothing in a flavour transition says whether a neutrino is its own antiparticle. Each gap needs a different instrument. KATRIN weighs the tritium beta endpoint and reports mβ < 0.45 eV. Cosmology reads the suppression of small-scale structure and gives Σmν < 0.12 eV. Neutrinoless double beta decay, unobserved past half-lives of 10²⁶ years, is the only handle on the Majorana question.
The bill the Standard Model cannot pay
Written without neutrino mass, the model carries 19 measured parameters. Admitting Dirac neutrino masses adds three masses, three PMNS angles and one CP phase, taking the bill to 26, or 28 if the neutrino is Majorana. Nothing predicts any of them. Worse, the mass term has no natural home: the model has no right-handed neutrino, and inserting one forces a Yukawa coupling near 3 × 10⁻¹³ for a 0.05 eV mass, twelve orders of magnitude below the top quark's and defensible by nobody. The alternative, a dimension-five Weinberg operator, points at a heavy Majorana scale near 6 × 10¹⁴ GeV — physics the model does not contain. Three larger holes stay open too. Neutrinos cannot be the dark matter: even at Σmν = 0.12 eV they supply Ων ≈ 0.003 against ΩDM ≈ 0.26, and they are relativistic when structure forms. CKM CP violation is far too small to make the observed baryon asymmetry, nB/nγ ≈ 6 × 10⁻¹⁰. Gravity is not in the model at all.
Change one variable at a time
Make the relationship visible.
Set θ to 45° and slide Δm²: the depth of every dip is fixed by the angle alone, while the whole pattern stretches or shrinks with E/Δm². Watch the 295 km dot climb out of its trough the moment Δm² leaves 2.5 — that is how one baseline measures a splitting.
DEPTH sin²2θ1.000
OSCILLATION LENGTH595 km
FIRST MINIMUM298 km
P AT L = 295 km0.000
Live interpretationDEPTH sin²2θ: 1.000. OSCILLATION LENGTH: 595 km. FIRST MINIMUM: 298 km. P AT L = 295 km: 0.000
Catch the common trap
Explain before calculating.
A two-flavour fit to a long-baseline disappearance experiment returns sin²2θ = 0.97 and Δm² = 2.4 × 10⁻³ eV². What has that fit established about the neutrino masses themselves?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA reactor emits antineutrinos of energy 4.0 MeV and a detector sits 1.0 km from the core. Using |Δm²₃₁| = 2.5 × 10⁻³ eV² and sin²2θ₁₃ = 0.085, find the fraction that has disappeared, then check that ignoring the solar splitting Δm²₂₁ = 7.5 × 10⁻⁵ eV² (sin²2θ₁₂ = 0.85) is safe at this baseline.
- Put the numbers in the units the constant 1.27 demands: E = 4.0 MeV = 4.0 × 10⁻³ GeV, L = 1.0 km, Δm² in eV².
- Phase: 1.27 × (2.5 × 10⁻³) × 1.0 ÷ (4.0 × 10⁻³) = 0.794 rad, so sin²(0.794) = 0.508.
- Disappearance: P = 0.085 × 0.508 = 0.0432, that is 4.3% of the antineutrinos gone.
- Now the solar term: 1.27 × (7.5 × 10⁻⁵) × 1.0 ÷ (4.0 × 10⁻³) = 0.0238 rad, and sin²(0.0238) = 5.67 × 10⁻⁴.
- Its contribution is 0.85 × 5.67 × 10⁻⁴ = 4.8 × 10⁻⁴, or 0.048% — ninety times smaller than the θ₁₃ effect, so a two-flavour treatment is fine here.
AnswerAbout 4.3% of the antineutrinos have disappeared at 1.0 km; the solar splitting adds only 0.048%, so the two-flavour θ₁₃ formula is adequate.
MediumSuper-Kamiokande compares muon neutrinos arriving from directly overhead (L ≈ 15 km) with those that have crossed the Earth (L ≈ 12 800 km), both at E = 2.0 GeV. Take |Δm²₃₂| = 2.5 × 10⁻³ eV² and sin²2θ₂₃ = 1.00. Find both survival probabilities, then say what the detector actually records.
- Oscillation length first: Losc = 2.48 × 2.0 ÷ (2.5 × 10⁻³) = 1.98 × 10³ km. The downward path of 15 km is under 1% of it.
- Downward: phase = 1.27 × (2.5 × 10⁻³) × 15 ÷ 2.0 = 0.0238 rad, sin² = 5.67 × 10⁻⁴, so Pₛᵤᵣᵥᵢᵥₐₗ = 0.9994 — no measurable loss.
- Upward: phase = 1.27 × (2.5 × 10⁻³) × 12 800 ÷ 2.0 = 20.3 rad, which is 12 800 ÷ 1984 = 6.45 oscillation lengths of travel.
- Formally sin²(20.3) = 0.990, giving Pₛᵤᵣᵥᵢᵥₐₗ = 0.010 — but that digit is fiction. A 1% shift in E or L moves the phase by 0.2 rad and swings P across most of its range, so no detector can resolve it.
- Average instead. Over the spread of energies and path lengths ⟨sin²⟩ → ½, so Pₛᵤᵣᵥᵢᵥₐₗ → 1 − ½ sin²2θ₂₃ = 1 − 0.50 = 0.50.
- The record is therefore a zenith-angle asymmetry: downward muon events at full rate, upward ones at about half. Same flux, same detector, different L — which is what makes it oscillation and not a flux error.
AnswerOverhead Pₛᵤᵣᵥᵢᵥₐₗ ≈ 0.999; through the Earth the monochromatic value 0.010 is unresolvable, and the measured, averaged value is 1 − ½sin²2θ₂₃ ≈ 0.50, so upward muon neutrinos arrive at about half strength.
HardGlobal fits give Δm²₂₁ = 7.53 × 10⁻⁵ eV² and |Δm²₃₂| = 2.45 × 10⁻³ eV². Find the smallest Σmν allowed by the normal ordering (m₁ = 0) and by the inverted ordering (m₃ = 0). Cosmology bounds Σmν < 0.12 eV: does that settle the ordering, and what is the largest m₁ the normal ordering still permits?
- Normal ordering with m₁ = 0: m₂ = √(7.53 × 10⁻⁵) = 8.68 × 10⁻³ eV, and m₃² = 7.53 × 10⁻⁵ + 2.45 × 10⁻³ = 2.525 × 10⁻³ eV², so m₃ = 5.03 × 10⁻² eV.
- Sum: Σmν = 0 + 0.00868 + 0.05025 = 0.0589 eV. That is the floor the normal ordering cannot go below.
- Inverted ordering with m₃ = 0: m₂ = √(2.45 × 10⁻³) = 4.950 × 10⁻² eV, and m₁² = 2.45 × 10⁻³ − 7.53 × 10⁻⁵ = 2.375 × 10⁻³ eV², so m₁ = 4.873 × 10⁻² eV and Σmν = 0.0982 eV.
- Both floors sit under 0.12 eV, so cosmology excludes neither ordering — but it leaves the normal ordering 0.061 eV of headroom and the inverted ordering only 0.022 eV.
- Largest m₁ in the normal ordering: solve m₁ + √(m₁² + 7.53 × 10⁻⁵) + √(m₁² + 2.525 × 10⁻³) = 0.12. At m₁ = 0.0300 the sum is 0.1198 and at m₁ = 0.0302 it is 0.1203, so m₁ ≈ 0.0301 eV.
- Every mass is now boxed into roughly 0.009–0.06 eV, yet the ordering, the absolute scale and the Dirac-or-Majorana question all remain open. KATRIN's direct limit, mβ < 0.45 eV, sits more than ten times above the heaviest value this box allows.
AnswerΣmν ≥ 0.059 eV (normal) and ≥ 0.098 eV (inverted). Σmν < 0.12 eV excludes neither, but caps m₁ at about 0.030 eV in the normal ordering; the ordering and absolute scale stay unmeasured.