Reactor neutrino experiments

A nuclear reactor is the closest thing particle physics has to a free, calibrated neutrino factory. Every fission of $^{235}$ U, $^{238}$ U, $^{239}$ Pu or $^{241}$ Pu leaves neutron-rich fragments that beta-decay to stability, each decay releasing an electron antineutrino. A gigawatt-scale core pours out about $6\times10^{20}$ $\bar\nu_e$ every second, isotropically, through metres of concrete and steel as though they were not there. The flux is not merely large; it is known, because the reactor's thermal power is monitored continuously and the yield per unit power is fixed by nuclear data. That turns a power station into a source whose output you can predict and whose neutrinos you can count. What you do with that source is subtraction. The mean antineutrino energy is only about $3$ MeV — far below the $106$ MeV needed to make a muon — so a $\bar\nu_e$ that oscillates into $\bar\nu_\mu$ or $\bar\nu_\tau$ simply vanishes from view. Reactor experiments therefore measure disappearance: they count fewer $\bar\nu_e$ than the power predicts and watch the positron spectrum distort. The survival probability carries two oscillations on wildly different scales — a fast one set by $\Delta m^2_{31}$ with amplitude $\sin^2 2\theta_{13}$ , a slow one set by $\Delta m^2_{21}$ with amplitude $\cos^4\theta_{13}\sin^2 2\theta_{12}$ — so where you put the detector decides which parameter you read out. Daya Bay and RENO sit near $1$ km and pin $\theta_{13}$ ; KamLAND sits near $180$ km and pins $\theta_{12}$ and $\Delta m^2_{21}$ .

Key takeaways

Fission reactors emit a precisely known, isotropic flux of electron antineutrinos — about $6\times10^{20}\ \bar\nu_e$ per second per gigawatt of thermal power — from fission-fragment beta decays, mean energy near $3$ MeV.
Detection is by inverse beta decay $\bar\nu_e + p \to e^+ + n$ , whose delayed coincidence — a prompt positron flash then a neutron capture microseconds later — kills backgrounds a single flash cannot.
Reactor energies sit below muon and tau threshold, so only $\bar\nu_e$ disappearance is observable, never appearance.
The survival probability splits by baseline: detectors near $1$ km measure $\sin^2 2\theta_{13}$ ; KamLAND near $180$ km measures $\Delta m^2_{21}$ and $\theta_{12}$ .
Daya Bay (2012) gave the first decisive nonzero $\theta_{13}$ , $\sin^2 2\theta_{13}\approx0.085$ , from a far-over-near ratio near $0.94$ ; KamLAND resolved the oscillatory dip-and-rise versus $L/E$ .
Key idea: a reactor's antineutrino flux is fixed by its power output, so counting fewer $\bar\nu_e$ downstream than expected — and seeing the deficit grow with distance and ripple with energy — reads out neutrino oscillation directly, with the baseline chosen to isolate either $\theta_{13}$ or $\theta_{12}$ .

The core insight

A reactor's antineutrino flux is known from its power output, so counting fewer $\bar\nu_e$ at a distant detector than expected — and seeing the deficit grow with distance and vary with energy — directly reveals neutrino oscillation, with the baseline chosen to isolate either $\theta_{13}$ or $\theta_{12}$ .

Where this lives today

Reactor antineutrinos remain a front-line tool, the frontier moved from discovery to precision and anomalies. Daya Bay measures $\sin^2 2\theta_{13}$ to better than $3\%$ — the most precise mixing angle in the neutrino sector — feeding directly into long-baseline experiments, whose CP-violation and mass-ordering sensitivity scales with $\theta_{13}$ . The JUNO experiment, a $20$ -kilotonne scintillator sphere at a tuned $\sim53$ km baseline, aims to resolve the fast-slow interference and fix the mass ordering from reactor $\bar\nu_e$ alone. Two puzzles keep the field honest: the reactor antineutrino anomaly, a $\sim6\%$ deficit between measured and predicted absolute flux — attributed to imperfect flux modelling or a sterile neutrino, probed by short-baseline PROSPECT and STEREO — and the $5$ MeV bump, an unexplained spectral excess near $5$ MeV seen by every modern detector, pointing to gaps in fission-fragment beta-decay data. And reactor monitoring for nonproliferation — inferring plutonium content from the antineutrino spectrum — is nearing deployable technology.

The physical picture

Picture the reactor as a point spraying electron antineutrinos in every direction at a rate fixed by its thermal power. A detector a distance $L$ away catches a flux falling as $1/L^2$ , and without oscillation the interaction count it records is fully predictable. The experiment then asks: did we catch as many as the source promised? Oscillation converts some $\bar\nu_e$ into flavours the detector cannot see, so the answer comes back low — and how low, versus distance and energy, is the signal. The first figure fixes how a single $\bar\nu_e$ is seen: not by watching it, but by the twin flash of inverse beta decay.

Figure

How a reactor antineutrino is counted: a ≈3 GW thermal core emits about 6×10²⁰ ν̄ₑ per second, and one in a distant detector triggers inverse beta decay ν̄ₑ + p → e⁺ + n above a 1.8 MeV threshold, giving a prompt positron flash of visible energy E_prompt ≈ E_ν − 0.78 MeV followed ≈30 μs later by an ≈8 MeV gamma cascade from neutron capture on gadolinium

The counting is only half the story; the other half is where you count. The survival probability is two nested oscillations. A fast ripple, set by the atmospheric splitting $\Delta m^2_{31}$ , first troughs near $1$ – $2$ km with small amplitude $\sin^2 2\theta_{13}\approx0.085$ — the shallow dip a short-baseline experiment lives inside. A slow swing, set by the solar splitting $\Delta m^2_{21}$ , first troughs near $60$ – $70$ km with large amplitude $\cos^4\theta_{13}\sin^2 2\theta_{12}\approx0.81$ — the deep trough KamLAND's $180$ km baseline sits well beyond. The second figure puts the two at one vertical scale, so the field's strategy becomes one picture: a shallow dip you must build twin detectors to see, and a deep trough you can watch oscillate.

Figure

Reactor ν̄ₑ survival probability at Eν = 4 MeV, plotted against baseline at one vertical scale. Left: the short-baseline dip has amplitude sin²2θ₁₃ ≈ 0.085, reaching P ≈ 0.915 near 2 km — the θ₁₃ regime of Daya Bay and RENO. Right: the long-baseline trough has amplitude cos⁴θ₁₃ sin²2θ₁₂ ≈ 0.81, reaching P ≈ 0.15 near 66 km — the θ₁₂ and Δm²₂₁ regime of KamLAND

The core equations

Detection begins with a two-body reaction on a free proton, whose threshold is fixed by the rest masses — which is why the lowest-energy reactor antineutrinos are invisible.

\bar\nu_e + p \to e^+ + n,\qquad E_\nu^{\text{thr}}=\frac{(m_n+m_e)^2-m_p^2}{2m_p}=1.806\ \text{MeV} \qquad (1)

where: $m_p,m_n,m_e$ — proton, neutron, positron rest masses; $E_\nu^{\text{thr}}$ — minimum antineutrino energy for a final state at rest in the centre of momentum. Below $1.806$ MeV nothing happens on a proton, so most of the flux never registers. The energy the detector sees is the positron's, tracking the antineutrino energy almost one-to-one with a fixed offset — which makes reactor experiments spectral: the positron spectrum is a shifted copy of the $\bar\nu_e$ one.

E_{\text{prompt}}=T_{e^+}+2m_e\simeq E_\nu-0.78\ \text{MeV} \qquad (2)

where: $E_{\text{prompt}}$ — visible prompt energy (positron kinetic energy plus two $0.511$ MeV annihilation photons); $T_{e^+}$ — positron kinetic energy. The near-constant $0.78$ MeV shift lets one reconstruct $E_\nu$ event by event and see oscillation as a spectral distortion, not merely a rate deficit. The source strength and interaction rate are both computable from first principles — the reason a reactor is calibrated. About six antineutrinos leave per fission, and each fission releases about $200$ MeV.

N_{\bar\nu}\simeq 6\,\frac{P_{\text{th}}}{\langle E_{\text{fis}}\rangle},\qquad R_{\text{IBD}}=\frac{N_{\bar\nu}}{4\pi L^2}\,\langle\sigma\rangle\,N_p \qquad (3)

where: $N_{\bar\nu}$ — antineutrinos per second; $P_{\text{th}}$ — thermal power; $\langle E_{\text{fis}}\rangle\approx200$ MeV — mean energy per fission; $R_{\text{IBD}}$ — detected rate; $L$ — baseline; $\langle\sigma\rangle$ — spectrum-averaged cross section; $N_p$ — free-proton targets. Everything on the right is measured independently, so the predicted rate is a number, not a fit. The physics payload is the survival probability, which for reactor $\bar\nu_e$ collapses into a fast and slow term with fixed amplitudes.

P(\bar\nu_e\to\bar\nu_e)=1-\sin^2 2\theta_{13}\,\sin^2\Delta_{31}-\cos^4\theta_{13}\,\sin^2 2\theta_{12}\,\sin^2\Delta_{21} \qquad (4)

where: $\theta_{13},\theta_{12}$ — mixing angles; $\Delta_{ij}=\Delta m^2_{ij}L/4E$ — oscillation phases; the $\Delta m^2_{31}$ term oscillates fast, the $\Delta m^2_{21}$ term slow. The CP phase $\delta$ drops out because survival depends only on $|U_{ei}|^2$ . The phases turn a length and an energy into an angle; in laboratory units the $\hbar$ and $c$ factors bundle into one constant.

\Delta_{ij}=\frac{\Delta m^2_{ij}L}{4E}=1.267\left(\frac{\Delta m^2_{ij}}{\text{eV}^2}\right)\frac{L/\text{m}}{E/\text{MeV}} \qquad (5)

where: $\Delta m^2_{ij}$ in eV $^2$ , $L$ in metres, $E$ in MeV. At $4$ MeV the fast term ( $\Delta m^2_{31}\approx2.5\times10^{-3}$ ) peaks near $2$ km, the slow term ( $\Delta m^2_{21}\approx7.5\times10^{-5}$ ) near $66$ km. Finally, near threshold the cross section takes a compact form fixed by the neutron lifetime.

\sigma_{\text{IBD}}\simeq 0.0952\left(\frac{E_{e^+}\,p_{e^+}}{\text{MeV}^2}\right)\times10^{-42}\ \text{cm}^2 \qquad (6)

where: $E_{e^+},p_{e^+}$ — positron energy and momentum; the prefactor is fixed by the neutron lifetime and phase space. The cross section rises roughly as $E_\nu^2$ , so higher-energy antineutrinos are detected far more efficiently.

Derivation AThe reactor survival probability from three-flavour mixing

The goal is to start from the PMNS matrix and the plane-wave phases of three mass eigenstates and arrive at equation (4), showing why reactor $\bar\nu_e$ survival carries one fast oscillation of amplitude $\sin^2 2\theta_{13}$ and one slow oscillation of amplitude $\cos^4\theta_{13}\sin^2 2\theta_{12}$ , cleanly separated by baseline. We expand the amplitude, collect the interference terms, insert the first-row PMNS elements, and merge the two nearly equal atmospheric phases.

Step 1. A produced $\bar\nu_e=\sum_i U_{ei}^*|\nu_i\rangle$ picks up a phase $e^{-i m_i^2 L/2E}$ per mass eigenstate over a distance $L$ , so the survival amplitude is $A=\sum_i |U_{ei}|^2 e^{-i m_i^2 L/2E}$ and $P=\sum_{i,j}|U_{ei}|^2|U_{ej}|^2 e^{-i(m_i^2-m_j^2)L/2E}$ . Why: for a disappearance channel the same flavour sits at both ends, so only the moduli $|U_{ei}|^2$ enter — which is why the CP phase cannot appear in reactor survival.

Step 2. Separating diagonal from off-diagonal terms with $\sum_i|U_{ei}|^2=1$ gives $P=1-4\sum_{i<j}|U_{ei}|^2|U_{ej}|^2\sin^2\Delta_{ij}$ , $\Delta_{ij}=\Delta m^2_{ij}L/4E$ . Why: each cross term contributes $1-2\sin^2\Delta_{ij}$ , and the unit total collapses the sum into a clean deficit of three interference pairs.

Step 3. Insert the first-row PMNS elements $|U_{e1}|^2=\cos^2\theta_{12}\cos^2\theta_{13}$ , $|U_{e2}|^2=\sin^2\theta_{12}\cos^2\theta_{13}$ , $|U_{e3}|^2=\sin^2\theta_{13}$ . The pair products give $4|U_{e1}|^2|U_{e2}|^2=\cos^4\theta_{13}\sin^2 2\theta_{12}$ , $4|U_{e1}|^2|U_{e3}|^2=\cos^2\theta_{12}\sin^2 2\theta_{13}$ , $4|U_{e2}|^2|U_{e3}|^2=\sin^2\theta_{12}\sin^2 2\theta_{13}$ . Why: the identity $4\cos^2\phi\sin^2\phi=\sin^2 2\phi$ converts matrix elements into oscillation amplitudes.

Step 4. The two $\theta_{13}$ -weighted terms share nearly the same phase because $\Delta m^2_{31}\approx\Delta m^2_{32}$ ; write both as $\Delta_{\text{atm}}$ and use $\cos^2\theta_{12}+\sin^2\theta_{12}=1$ to fuse them into $\sin^2 2\theta_{13}\sin^2\Delta_{\text{atm}}$ , giving $P=1-\sin^2 2\theta_{13}\sin^2\Delta_{31}-\cos^4\theta_{13}\sin^2 2\theta_{12}\sin^2\Delta_{21}$ . Why: the solar splitting cannot distinguish $\Delta m^2_{31}$ from $\Delta m^2_{32}$ here, so merging them gives the compact two-scale form used in every reactor analysis.

Go deeper · Derivation B · 1 minInverse beta-decay kinematics and the prompt-energy relation

Derivation BInverse beta-decay kinematics and the prompt-energy relation

The aim is to derive the $1.806$ MeV threshold of equation (1) and the near-constant $0.78$ MeV offset of equation (2) between prompt and antineutrino energy, so a measured positron spectrum can be turned back into a $\bar\nu_e$ spectrum. We impose energy-momentum conservation on $\bar\nu_e + p\to e^+ + n$ , take the low-energy limit where the neutron barely recoils, and add the annihilation photons.

Step 1. Threshold occurs when the $e^+$ and $n$ emerge at rest in the centre of momentum. Equating the invariant mass squared of initial and final states, $(E_\nu+m_p)^2-E_\nu^2=(m_n+m_e)^2$ , gives $E_\nu^{\text{thr}}=\frac{(m_n+m_e)^2-m_p^2}{2m_p}=1.806$ MeV. Why: the neutron is heavier than the proton, so the reaction is endothermic — the antineutrino must supply the mass deficit plus the positron mass.

Step 2. At reactor energies the heavy neutron carries almost no kinetic energy, so $E_\nu+m_p\simeq E_{e^+}+m_n$ , giving $E_{e^+}\simeq E_\nu-(m_n-m_p)=E_\nu-1.293$ MeV. Why: the neutron recoil is a sub-percent correction, so the positron inherits the antineutrino energy minus the neutron-proton mass gap.

Step 3. The detector records the positron kinetic energy plus its two $0.511$ MeV annihilation photons, $E_{\text{prompt}}=T_{e^+}+2m_e=E_{e^+}+m_e$ . Substituting Step 2 gives $E_{\text{prompt}}\simeq E_\nu-1.293+0.511=E_\nu-0.782$ MeV. Why: this fixed shift makes detection spectral — a distortion in the positron spectrum maps directly onto the $\bar\nu_e$ spectrum, exposing the oscillation's energy dependence.

Go deeper · Derivation C · 1 minWhy the long-baseline limit collapses to two flavours

Derivation CWhy the long-baseline limit collapses to two flavours

The goal is to show how equation (4) simplifies at KamLAND's $\sim180$ km baseline into an effective two-flavour oscillation in $\theta_{12}$ and $\Delta m^2_{21}$ , and why the fast $\theta_{13}$ term becomes a harmless constant rather than vanishing. We compare the two oscillation lengths, average the unresolvable fast term, and read off what remains.

Step 1. At $E\approx4$ MeV the fast phase $\Delta_{31}$ passes through many multiples of $\pi$ over a $180$ km baseline, with the reactors spread across $130$ – $240$ km. Why: neither the finite energy resolution nor the spread of source distances can resolve oscillations only a couple of kilometres long, so the fast term is smeared.

Step 2. Averaging sets $\sin^2\Delta_{31}\to\tfrac12$ , so equation (4) becomes $P\simeq 1-\tfrac12\sin^2 2\theta_{13}-\cos^4\theta_{13}\sin^2 2\theta_{12}\sin^2\Delta_{21}$ . Why: the fast term does not vanish — it leaves a constant $\tfrac12\sin^2 2\theta_{13}\approx0.04$ suppression that lowers the normalisation but carries no distance or energy information.

Step 3. With $\theta_{13}$ small, the first two terms are nearly $0.96$ , leaving an effective two-flavour law $P\simeq0.96-0.81\sin^2\Delta_{21}$ oscillating only in the slow solar phase. Why: this is why KamLAND could reconstruct a clean dip-and-rise versus $L/E$ and measure $\Delta m^2_{21}$ and $\theta_{12}$ as if only two flavours existed — the third just rescales the amplitude.

Limiting cases

Case	Result	Meaning
Near detector, $L\to0$	$P\to1$	no oscillation yet; the near detector fixes the flux and detector response
Short baseline, $L\sim1$ – $2$ km, $E\sim4$ MeV	dip amplitude $\sin^2 2\theta_{13}\approx0.085$	fast term at first maximum; isolates $\theta_{13}$ (Daya Bay, RENO)
Long baseline, $L\sim50$ – $100$ km	trough amplitude $\cos^4\theta_{13}\sin^2 2\theta_{12}\approx0.81$	slow term at first maximum; isolates $\theta_{12}$ and $\Delta m^2_{21}$ (KamLAND)
$\Delta_{31}$ averaged, $L\gg$ few km	$\langle\sin^2\Delta_{31}\rangle=\tfrac12$	fast term becomes a constant $\approx4\%$ offset; effective two-flavour law
$\theta_{13}\to0$	$P\to1-\sin^2 2\theta_{12}\sin^2\Delta_{21}$	recovers the pure two-flavour solar oscillation
$E_\nu<1.806$ MeV	$\sigma_{\text{IBD}}=0$	below inverse-beta-decay threshold; these antineutrinos are invisible on protons

Common trapReactor experiments watch neutrinos appear in new flavours

They do not — and cannot. At a mean energy near $3$ MeV a reactor antineutrino carries nowhere near the $\sim106$ MeV needed to make even a muon, so a $\bar\nu_e$ that oscillates into $\bar\nu_\mu$ or $\bar\nu_\tau$ produces no charged lepton and drops out of the count. Every reactor result is therefore a disappearance measurement: a rate deficit and a spectral distortion, never a positive sighting of another flavour. A subtler trap hides in the names: the "short-baseline" oscillation is driven by the atmospheric splitting $\Delta m^2_{31}$ , not the solar $\Delta m^2_{21}$ , even though it happens at the shorter distance — the larger $\Delta m^2$ oscillates over a shorter length. Confusing which splitting belongs to which baseline inverts the whole experimental logic.

Worked examples

1 · From thermal power to detected events (fully worked)

A single reactor core runs at $P_{\text{th}}=3$ GW. Estimate the antineutrino emission rate, and the inverse-beta-decay rate in a $20$ -tonne liquid-scintillator detector at $L=1.5$ km.

Step 1. Fission rate: each fission releases $\approx200\ \text{MeV}=3.2\times10^{-11}$ J, so $3\times10^{9}$ W gives $P_{\text{th}}/E_{\text{fis}}=3\times10^{9}/3.2\times10^{-11}=\mathbf{9.4\times10^{19}}$ fissions/s.

Step 2. Antineutrino rate: with about six $\bar\nu_e$ per fission, $N_{\bar\nu}\simeq6\times9.4\times10^{19}=\mathbf{5.6\times10^{20}\ \bar\nu_e/s}$ — the calibrated source strength, from the power monitor alone.

Step 3. Flux: with $L=1.5\times10^{5}$ cm, $4\pi L^2=2.8\times10^{11}\ \text{cm}^2$ , so the fission-normalised flux is $9.4\times10^{19}/2.8\times10^{11}=3.3\times10^{8}$ fissions cm $^{-2}$ s $^{-1}$ .

Step 4. Event rate: fold in the inverse-beta-decay yield $\sigma_f\approx5.7\times10^{-43}\ \text{cm}^2$ per fission (already averaged over the above-threshold spectrum) and the free protons of a $20$ -tonne scintillator, $N_p\approx1.7\times10^{30}$ . Then $R_{\text{IBD}}=3.3\times10^{8}\times5.7\times10^{-43}\times1.7\times10^{30}\approx3.2\times10^{-4}\ \text{s}^{-1}\approx\mathbf{28\ events/day}$ . A real site with several cores at shorter baseline reaches hundreds per day, but the order of magnitude — tens to hundreds, not millions — is set here. A 3 GW core emits ≈5.6×10²⁰ ν̄ₑ/s and yields tens of IBD events per day in a 20 t detector at 1.5 km.

2 · Where to put the detector (fully worked)

At $E_\nu=4$ MeV, find the baseline of the first oscillation maximum for the $\theta_{13}$ term and for the $\theta_{12}$ term, using $\Delta m^2_{31}=2.5\times10^{-3}$ eV $^2$ and $\Delta m^2_{21}=7.5\times10^{-5}$ eV $^2$ .

Step 1. The first maximum of $\sin^2\Delta$ is at $\Delta=\pi/2$ , so from equation (5) $L=\dfrac{(\pi/2)\,E}{1.267\,\Delta m^2}$ in metres.

Step 2. For $\theta_{13}$ : $L=\dfrac{1.571\times4}{1.267\times2.5\times10^{-3}}\approx\mathbf{1.98\ km}$ — which is why Daya Bay's far halls sit near $1.5$ – $1.9$ km, right in the shallow $\theta_{13}$ dip.

Step 3. For $\theta_{12}$ : $L=\dfrac{1.571\times4}{1.267\times7.5\times10^{-5}}\approx\mathbf{66\ km}$ . KamLAND's $\sim180$ km sits beyond it, so it sees more than one full swing. θ₁₃ peaks near 2 km, θ₁₂ near 66 km — the two baselines differ by ≈33, the ratio of the mass splittings.

3 · Reading a neutrino energy off the prompt flash

A prompt event deposits $E_{\text{prompt}}=3.2$ MeV. What antineutrino energy produced it? Invert equation (2): $E_\nu\simeq E_{\text{prompt}}+0.78=\mathbf{3.98\ MeV}$ , well above the $1.806$ MeV threshold, with positron kinetic energy $T_{e^+}=E_{\text{prompt}}-2m_e=2.18$ MeV. The 3.2 MeV flash tags a ≈4.0 MeV antineutrino.

Physics of the portrait

Reactor antineutrinos are where the neutrino stopped being a bookkeeping device and became a detected particle — from Pauli's "desperate remedy" to a detector beside a production reactor.

Origins · 1 minFrederick Reines and how this was discovered

Frederick Reines

1918–1998 · reactor antineutrino detection and neutrino physics

Frederick Reines was the physicist who turned Pauli's ghost into an experiment. In the early 1950s he set out to do what everyone thought impossible: detect the neutrino, predicted by Pauli in 1930 and theorised by Fermi in 1934 but never seen, because its cross section — about $10^{-43}$ cm $^2$ — is fantastically small. Reines' insight, with Clyde Cowan, was that a fission reactor is an overwhelming source: if the cross section is tiny, use a flux so enormous that even a tiny fraction interacts. Their "Project Poltergeist" placed tanks beside the Savannah River reactor and looked for the double signature of inverse beta decay — a prompt positron annihilation followed microseconds later by a neutron capturing on dissolved cadmium. In 1956 they wired Pauli a telegram: the antineutrino was real. That delayed-coincidence trick is still the beating heart of every reactor experiment, from KamLAND to Daya Bay. Reines shared the 1995 Nobel Prize for the detection — the moment a conservation-law patch became a cornerstone of the Standard Model.

History

1930

— Pauli postulates the neutrino to rescue beta-decay energy conservation

1934

— Fermi's theory of beta decay names and quantifies the neutrino

1956

— Reines and Cowan detect reactor antineutrinos by inverse beta decay at Savannah River

2002

— KamLAND observes reactor antineutrino disappearance and confirms the solar splitting

2012

— Daya Bay and RENO measure a nonzero θ₁₃ from far-over-near rate deficits

Going further

Texts. Thomson, Modern Particle Physics, §13.7 (reactor experiments, short-baseline, KamLAND) · Griffiths, Introduction to Elementary Particles, 2nd ed., Ch. 11 · Zuber, Neutrino Physics, 2nd ed., Ch. 8 · Giunti and Kim, Fundamentals of Neutrino Physics and Astrophysics, Ch. 12.

Provenance. The survival probability (4) is assembled from the PMNS matrix owned by aq-pp-three-flavour-pmns and the phase machinery of aq-pp-two-flavour-oscillations; the inverse-beta-decay vertex uses the charged-current coupling of aq-pp-weak-charged-current. A verifier should confirm: the $1.806$ MeV threshold from $E_\nu^{\text{thr}}=[(m_n+m_e)^2-m_p^2]/2m_p$ ; the $0.78$ MeV prompt offset; $\sim5.6\times10^{20}\ \bar\nu_e/\text{s}$ from a $3$ GW core; first-maximum baselines $\approx2$ km ( $\theta_{13}$ ) and $\approx66$ km ( $\theta_{12}$ ) at $4$ MeV; and the phase constant $1.267$ . Experimental anchors ( $\sin^2 2\theta_{13}\approx0.085$ , far/near $\approx0.94$ , $\Delta m^2_{21}\approx7.5\times10^{-5}$ eV $^2$ ) are the Daya Bay, RENO, and KamLAND published values.

Prerequisites. Two-flavour Neutrino Oscillations · Three-flavour PMNS Mixing · The Weak Charged Current. Leads to. Long-baseline Accelerator Experiments · Neutrino Mass Mechanisms · Matter Effects (MSW).

Ask the tutor: ask me to walk through how a far-over-near ratio of $0.94$ becomes a value of $\sin^2 2\theta_{13}$ , or how JUNO squeezes the mass ordering out of the reactor spectrum.

Why do reactors make antineutrinos and not neutrinos?

Because reactors run on fission, and fission fragments are neutron-rich. The pieces left when a heavy nucleus splits carry far too many neutrons, so they climb toward stability by $\beta^-$ decay, in which a neutron turns into a proton, an electron, and an electron antineutrino. There is no bulk source of $\beta^+$ decay, so the flux is essentially pure $\bar\nu_e$ . That matters: antineutrinos interact via inverse beta decay on free protons, which has a low threshold and a clean two-part signature, whereas a neutrino beam would need different targets and reactions.

Why use inverse beta decay rather than neutrino-electron scattering?

Both happen, but inverse beta decay $\bar\nu_e+p\to e^+ + n$ wins decisively. Its cross section is about a hundred times larger than elastic scattering off electrons here, and it produces two correlated signals separated in time: a prompt positron that annihilates, then a neutron captured tens of microseconds later. That delayed coincidence is a powerful background filter, because random radioactivity almost never fakes both flashes at the right separation. Elastic scattering gives only a single recoiling electron, swamped by natural radioactivity.

What is gadolinium doping for?

It sharpens the delayed half of the coincidence. A free neutron thermalises and is captured; on hydrogen alone that takes about $200$ microseconds and releases a single $2.2$ MeV gamma, barely above backgrounds. A small amount of gadolinium — which has a gigantic neutron-capture cross section — shortens the capture to about $30$ microseconds and releases a cascade totalling roughly $8$ MeV, well above almost all natural radioactivity. That is why modern detectors load their target with gadolinium (Reines and Cowan used cadmium for the same reason).

Why build near and far detectors instead of one?

Because the hardest uncertainties cancel in a ratio. The absolute reactor flux and spectrum are known only to a few percent, and detector efficiency adds its own systematic — either alone would swamp a $\sim5\%$ oscillation effect. Build two identical detectors, one near the core and one far (where the $\theta_{13}$ dip has appeared), and the flux normalisation and much of the detector response divide out of the far-over-near ratio, leaving almost pure oscillation. Daya Bay's far/near ratio near $0.94$ is relative precisely so it does not depend on the absolute flux — the very quantity the reactor anomaly shows we cannot fully predict.

How did Daya Bay establish θ₁₃ was nonzero?

By finding a statistically significant deficit of far counts relative to near. Before Daya Bay, $\theta_{13}$ was known only to be small — CHOOZ and others had set upper limits, and it was conceivable that $\theta_{13}$ vanished exactly. Daya Bay's identical detectors around several powerful cores measured a far-over-near ratio of about $0.94$ , a $6\%$ deficit far larger than the combined errors, giving $\sin^2 2\theta_{13}\approx0.09$ in 2012. That mattered enormously: a nonzero $\theta_{13}$ is the gateway that makes leptonic CP violation and the mass ordering accessible, so the next generation of long-baseline experiments was greenlit on the strength of that ratio.

What are the reactor antineutrino anomaly and the 5 MeV bump?

The two open puzzles of the field. The anomaly is that measured absolute fluxes come in about $6\%$ below the best predictions — a deficit noticed after the flux models were re-evaluated in 2011. One explanation is oscillation into a light sterile neutrino; another, increasingly favoured, is that the predictions are slightly off, especially for $^{235}$ U. The $5$ MeV bump is a related surprise: every modern detector sees a spectral excess near $5$ MeV that no flux model predicts, pointing to mismodelled beta-decay data for particular fragments. Neither breaks the oscillation picture — measured by relative near/far comparisons immune to these absolute errors — but both show our knowledge of the source spectrum is incomplete.

Why is reactor θ₁₃ blind to the mass ordering, but JUNO is not?

Daya Bay measures the amplitude $\sin^2 2\theta_{13}$ of the fast oscillation, and that amplitude is identical for normal and inverted ordering — the deficit looks the same either way. JUNO gets ordering sensitivity from a medium baseline near $53$ km, where the slow solar and fast atmospheric oscillations overlap. Their interference shifts the fine spectral ripples slightly differently for the two orderings, and with enough energy resolution that shift is measurable. It takes an exquisitely calibrated, very large detector — exactly what JUNO is — reading the ordering off the shape of the spectrum rather than its overall size.