The solar neutrino problem

For thirty years the Sun told two different stories about itself. The theory of stellar fusion, worked out in exquisite detail, said that the proton-proton chain burning in the solar core must flood the Earth with electron neutrinos — some $6\times10^{10}$ of them crossing every square centimetre each second, a number pinned not to any fragile model detail but to the Sun's measured brightness. Then the detectors reported back. Ray Davis's tank of cleaning fluid a mile underground counted barely a third of the predicted electron neutrinos. The gallium experiments, reaching all the way down to the dominant low-energy $pp$ neutrinos, saw about half. Something was eating the neutrinos, or the Sun was not shining the way we thought — and no one could say which. This is the solar neutrino problem, and its resolution is one of the cleanest detective stories in physics. The Sun can only make electron neutrinos: the fusion reactions that power it emit $\nu_e$ and nothing else. So the deficit had two possible cures — either the flux was genuinely smaller (the solar model was wrong), or the electron neutrinos were changing into something the detectors could not see. The Sudbury Neutrino Observatory settled it in 2002 by doing what no radiochemical experiment could: it measured the total neutrino flux of all three flavours as well as the electron-neutrino flux alone. The total matched the solar model perfectly. The electron-neutrino part was down by two-thirds. The missing $\nu_e$ had not vanished — they had turned into $\nu_\mu$ and $\nu_\tau$ in flight. Neutrinos oscillate, and therefore they have mass. This article rebuilds that argument from the fusion reactions to the flavour count.

Key takeaways

Solar fusion ( $4p\to{}^4\text{He}+2e^++2\nu_e$ ) produces a huge flux of electron neutrinos only, $\sim6.5\times10^{10}\,\text{cm}^{-2}\text{s}^{-1}$ , fixed by the Sun's luminosity.
Radiochemical experiments (Homestake $^{37}$ Cl, then SAGE and GALLEX $^{71}$ Ga) measured one-third to one-half the predicted $\nu_e$ rate — the deficit, and it was energy-dependent.
Super-Kamiokande confirmed the $^8$ B deficit via $\nu_e e^-\to\nu_e e^-$ and showed the recoils point back at the Sun ( $\cos\theta_{\rm sun}\to1$ ), proving the neutrinos were solar.
SNO's heavy-water target ran three reactions on the deuteron: charged current (CC, $\nu_e$ only), neutral current (NC, all flavours), and elastic scattering (ES).
The NC total flux $\phi_{\rm NC}=5.09\times10^{6}$ matched the solar model, while CC gave $\phi(\nu_e)=1.76\times10^{6}$ — a survival probability $P_{ee}=0.35$ and a large $\nu_\mu+\nu_\tau$ component the Sun cannot make.
Key idea: The Sun produces only electron neutrinos, yet detectors found a $\nu_e$ deficit; SNO showed the total flux is undiminished but partly converted to $\nu_\mu/\nu_\tau$ — so neutrinos change flavour, implying they are massive and oscillate.

Where this lives today

The solar neutrino problem is solved, but solar neutrinos have graduated from a puzzle into a set of precision instruments. The Borexino experiment in Italy pushed detection down to the individual branches, measuring the $^7$ Be, $pep$ , and finally, in 2020, the faint CNO-cycle neutrinos in real time — the first direct proof that a second fusion cycle burns in the Sun, and a new handle on the solar core's metallicity. The KamLAND reactor experiment nailed the oscillation parameters from the opposite direction, watching antineutrinos from Japanese power plants disappear and pinning $\Delta m^2_{21}$ and $\theta_{12}$ to the LMA-MSW values that the solar data demanded. Super-Kamiokande and SNO+ continue to hunt for the predicted low-energy "upturn" in the survival probability and for the day-night asymmetry from Earth-matter regeneration. Looking forward, the giant JUNO detector in China aims to measure $\theta_{12}$ and the solar $\Delta m^2$ to sub-percent precision and to determine the neutrino mass ordering, while DARWIN and other dark-matter detectors will soon see solar neutrinos as a background — the "neutrino floor" — turning the Sun's core into a calibration source. Every one of these rests on the flavour bookkeeping that SNO first balanced.

The physical picture

An electron neutrino born in the solar core has a quantum-mechanical identity crisis waiting for it. The flavour state $\nu_e$ that the weak interaction creates is not a state of definite mass; it is a superposition of the mass eigenstates that actually propagate. As the neutrino streams the $1.5\times10^{8}\,\text{km}$ to Earth, those mass components drift out of phase, and the probability of still finding a $\nu_e$ when it arrives falls below one. Because the Sun and the detectors are far apart compared with the oscillation length, the phase averages out and the survival probability settles to a fixed fraction — different for low- and high-energy neutrinos because matter in the Sun (the MSW effect) tilts the mixing. The first figure follows one $^8$ B neutrino from birth as a pure $\nu_e$ to arrival as a flavour mixture, and shows the two kinds of detector that see the two different things.

Why did SNO succeed where thirty years of experiments had only deepened the puzzle? Because heavy water gives you three different questions to ask the same neutrinos. The charged-current reaction $\nu_e+d\to p+p+e^-$ needs a $W$ boson to turn the neutrino into a charged lepton, and at solar energies only $\nu_e$ can do this (making a $\mu$ or $\tau$ would cost far more than the available $\sim10\,\text{MeV}$ ) — so CC counts electron neutrinos alone. The neutral-current reaction $\nu_x+d\to p+n+\nu_x$ just breaks the deuteron apart via a $Z$ boson, which couples identically to all three flavours — so NC counts every neutrino regardless of type. The gap between the two is the smoking gun. The second figure is the plot that ended the problem: the measured fluxes side by side against the solar-model prediction.

Figure

The SNO resolution: solar-model prediction φ(⁸B) = 5.05×10⁶ cm⁻²s⁻¹ (amber band) versus SNO measurements. The neutral-current flux (all flavours) is 5.09×10⁶ — it matches the model. The charged-current flux (ν_e only) is 1.76×10⁶, and elastic scattering gives 2.39×10⁶. The gap NC − CC = 3.33×10⁶ is the ν_μ + ν_τ the Sun cannot produce

The core equations

The starting point is not an oscillation formula but an accounting identity: the Sun's brightness fixes its neutrino output. Fusing four protons into a helium-4 nucleus releases $26.73\,\text{MeV}$ and, crucially, exactly two electron neutrinos. Divide the solar luminosity by the energy per reaction and you have the production rate, independent of the intricate nuclear-physics detail of which side chains carry the reactions.

4p \to {}^4_2\text{He} + 2e^+ + 2\nu_e + 26.73\ \text{MeV},\qquad \Phi_{\nu_e}^{\rm tot}\simeq\frac{2L_\odot}{Q\,4\pi d^2}\approx6.5\times10^{10}\ \text{cm}^{-2}\text{s}^{-1} \qquad (1)

where: $L_\odot=3.85\times10^{26}\,\text{W}$ — the solar luminosity; $Q=26.73\,\text{MeV}$ — energy released per helium-4 built (of which $\sim2\%$ leaves as neutrinos); $d=1.496\times10^{11}\,\text{m}$ — the Earth-Sun distance; the factor $2$ is the neutrinos per fusion. The total flux is robust; only its splitting among the $pp$ , $^7$ Be, and $^8$ B branches depends on the solar model, and it is the higher branches that most experiments actually see. The deficit is explained by neutrino oscillation. In the two-flavour picture the electron neutrino is a rotation of two mass eigenstates by a mixing angle $\theta$ , and the survival probability oscillates with the ratio of baseline to energy set by the mass-squared splitting.

P(\nu_e\to\nu_e)=1-\sin^2 2\theta\,\sin^2\!\left(\frac{\Delta m^2\,L}{4E}\right) \qquad (2)

where: $\theta$ — the mixing angle ( $\theta_{12}\approx33.4^\circ$ , $\sin^2\theta_{12}\approx0.31$ ); $\Delta m^2=m_2^2-m_1^2\approx7.5\times10^{-5}\,\text{eV}^2$ — the mass-squared difference; $L$ — the baseline; $E$ — the neutrino energy. For the enormous solar baseline the sine-squared oscillates many times and averages to $\tfrac12$ , leaving a smooth, energy-dependent survival that matter effects in the Sun push further down at high energy. SNO's power is that its three reactions weight the flavours differently, giving three linear equations in two unknowns — the electron flux and the combined muon-plus-tau flux.

\phi_{\rm CC}=\phi(\nu_e),\qquad \phi_{\rm ES}=\phi(\nu_e)+0.155\,\phi(\nu_{\mu\tau}),\qquad \phi_{\rm NC}=\phi(\nu_e)+\phi(\nu_{\mu\tau}) \qquad (3)

where: $\phi_{\rm CC}$ — charged-current flux $\nu_e d\to ppe^-$ , electron-only; $\phi_{\rm ES}$ — elastic-scattering flux $\nu_x e^-\to\nu_x e^-$ , with $\nu_{\mu\tau}$ entering at $0.155$ of the $\nu_e$ strength (they lack the charged-current contribution); $\phi_{\rm NC}$ — neutral-current flux $\nu_x d\to pn\nu_x$ , flavour-blind. The three channels overdetermine the two fluxes, so their consistency is itself a check. Solving the CC and NC lines gives the two numbers that closed the case: the fraction of electron neutrinos that survived, and the flux of the wrong-flavour neutrinos the Sun can never emit.

P_{ee}=\frac{\phi_{\rm CC}}{\phi_{\rm NC}}=\frac{1.76}{5.09}=0.35,\qquad \phi(\nu_{\mu\tau})=\phi_{\rm NC}-\phi_{\rm CC}=3.33\times10^{6}\ \text{cm}^{-2}\text{s}^{-1} \qquad (4)

where: $\phi_{\rm CC}=1.76\times10^{6}$ and $\phi_{\rm NC}=5.09\times10^{6}\,\text{cm}^{-2}\text{s}^{-1}$ are the SNO measurements. The survival probability $P_{ee}=0.35$ is the size of the deficit; the nonzero $\phi(\nu_{\mu\tau})$ , more than five standard deviations from zero, is direct evidence that the missing $\nu_e$ reappeared as other flavours rather than disappearing.

Derivation AThe solar neutrino flux fixed by the Sun's luminosity

The goal is to show that the total electron-neutrino flux at Earth is not a soft prediction of a detailed solar model but an almost model-independent consequence of the Sun's measured brightness together with the stoichiometry of hydrogen burning, and to reproduce the canonical figure $\sim6\times10^{10}\,\text{cm}^{-2}\text{s}^{-1}$ . We tie every unit of solar power to a fixed number of neutrinos, then spread that output over a sphere at the Earth's orbit.

Step 1. Fix the reaction stoichiometry. The net result of the $pp$ chain is $4p\to{}^4\text{He}+2e^++2\nu_e$ , releasing $Q=26.73\,\text{MeV}$ and, invariably, two electron neutrinos — one per $\beta^+$ -type step ( $p+p\to D+e^++\nu_e$ , occasionally $p+e^-+p\to D+\nu_e$ ). Why: whatever the branching among side chains, closing the cycle to $^4$ He always emits two $\nu_e$ and the same total energy, so energy output and neutrino output are rigidly linked.

Step 2. Convert luminosity to a reaction rate. Almost all of $Q$ eventually emerges as sunlight (the neutrinos carry off only $\sim2\%$ ), so the number of completed cycles per second is $R\simeq L_\odot/Q=(3.85\times10^{26}\,\text{W})/(26.73\times1.6\times10^{-13}\,\text{J})\approx9.0\times10^{37}\,\text{s}^{-1}$ . Why: the Sun is in steady state — power out equals power generated — so its brightness is a direct tally of fusion reactions.

Step 3. Count neutrinos and spread over the orbit. Each cycle makes two neutrinos, so the luminosity of neutrinos is $2R\approx1.8\times10^{38}\,\text{s}^{-1}$ . Dividing by the area of a sphere at $d=1.496\times10^{11}\,\text{m}$ , $4\pi d^2=2.81\times10^{23}\,\text{m}^2$ , gives $\Phi=1.8\times10^{38}/2.81\times10^{23}=6.4\times10^{14}\,\text{m}^{-2}\text{s}^{-1}=6.4\times10^{10}\,\text{cm}^{-2}\text{s}^{-1}$ . Why: this is the whole prediction — a number tied to the Sun's brightness and simple geometry, which is why the deficit could not be waved away as a modelling error.

Step 4. Note what is and is not model-dependent. The total is robust, but the split into $pp$ ( $E<0.42\,\text{MeV}$ , $\sim91\%$ ), $^7$ Be ( $0.86\,\text{MeV}$ , $\sim7\%$ ), and $^8$ B ( $E\lesssim15\,\text{MeV}$ , $\sim0.02\%$ ) neutrinos depends on the core temperature roughly as $T^{25}$ for $^8$ B. Why: detectors with high thresholds (chlorine, water) see mostly the rare, temperature-sensitive $^8$ B flux, which is exactly why an over-optimistic solar model was the leading suspect for so long.

Go deeper · Derivation B · 1 minDisentangling the flavours with three reactions on the deuteron

Derivation BDisentangling the flavours with three reactions on the deuteron

The aim is to prove that heavy water lets a single detector separately measure the electron-neutrino flux and the total flux of all flavours, so that a flavour-changing hypothesis can be tested without any reliance on the solar model's absolute normalization. We write each reaction's flavour sensitivity, assemble the linear system, and solve for the two physical fluxes.

Step 1. Write the charged-current sensitivity. $\nu_e+d\to p+p+e^-$ proceeds through a $W$ boson that must convert the neutrino into its charged partner; producing a $\mu$ or $\tau$ needs $\gtrsim106\,\text{MeV}$ , far above the solar $\sim10\,\text{MeV}$ , so only $\nu_e$ contributes: $\phi_{\rm CC}=\phi(\nu_e)$ . Why: the kinematic ceiling makes CC a clean electron-flavour counter, the same electron flux the radiochemical experiments measured.

Step 2. Write the neutral-current sensitivity. $\nu_x+d\to p+n+\nu_x$ proceeds through a $Z$ boson that couples with equal strength to all three flavours and merely dissociates the deuteron (threshold $2.22\,\text{MeV}$ , its binding energy), leaving the neutrino intact: $\phi_{\rm NC}=\phi(\nu_e)+\phi(\nu_\mu)+\phi(\nu_\tau)$ . Why: because the $Z$ is flavour-democratic, NC counts the total flux whatever the mixture — the quantity the solar model actually predicts.

Step 3. Add elastic scattering as a cross-check. $\nu_x+e^-\to\nu_x+e^-$ runs for all flavours but $\nu_e$ scatters more strongly (it has both $W$ and $Z$ contributions), giving $\phi_{\rm ES}=\phi(\nu_e)+0.155\,\phi(\nu_{\mu\tau})$ — the reaction Super-Kamiokande used, sitting between the pure CC and the flavour-blind NC. Why: ES provides a third, independent constraint whose intermediate flavour weighting must be consistent with the CC and NC extraction.

Step 4. Solve and interpret. With $\phi_{\rm CC}=1.76$ and $\phi_{\rm NC}=5.09$ (in $10^6\,\text{cm}^{-2}\text{s}^{-1}$ ), subtraction gives $\phi(\nu_{\mu\tau})=3.33$ and the ratio $P_{ee}=\phi_{\rm CC}/\phi_{\rm NC}=0.35$ ; $\phi_{\rm NC}$ agrees with the solar-model $\phi(^8\text{B})=5.05$ . Why: the total is undiminished and model-consistent while the electron part is suppressed — a result that requires flavour transformation and forbids the "the Sun is dimmer in neutrinos" escape.

Go deeper · Derivation C · 2 minThe energy-dependent deficit from vacuum averaging and the MSW effect

Derivation CThe energy-dependent deficit from vacuum averaging and the MSW effect

The purpose is to explain the single most puzzling feature of the pre-SNO data — that the deficit was not a constant factor but deeper for the higher-energy neutrinos (gallium $\sim0.5$ , chlorine and water $\sim0.3$ ) — by combining vacuum-averaged oscillation for low energies with matter-enhanced (MSW) conversion for high energies. We average the vacuum formula over the long baseline, then show how solar matter changes the answer above a crossover energy.

Step 1. Average the vacuum probability. Over the Earth-Sun baseline the phase $\Delta m^2 L/4E$ runs through many cycles and, smeared over the production region and energy resolution, $\sin^2(\cdot)\to\tfrac12$ , so $P_{ee}\to1-\tfrac12\sin^2 2\theta_{12}$ . With $\sin^2\theta_{12}=0.31$ , $\sin^2 2\theta_{12}=0.86$ and $P_{ee}\to0.57$ . Why: this is the low-energy plateau — the $pp$ and $^7$ Be neutrinos that gallium sees survive at $\sim55\%$ , matching the shallower gallium deficit.

Step 2. Introduce matter. Inside the dense solar core electron neutrinos feel an extra potential $V=\sqrt{2}\,G_F n_e$ from coherent forward scattering off electrons (a term $\nu_\mu,\nu_\tau$ do not share), which shifts the effective mixing in matter. Why: the medium breaks the symmetry between flavours, so the mixing that governs propagation in the Sun differs from the vacuum mixing — the essence of the MSW mechanism.

Step 3. Take the adiabatic high-energy limit. For $^8$ B neutrinos the matter term dominates at production; the neutrino is created essentially as the heavier mass eigenstate $\nu_2$ and, if it exits the falling-density Sun adiabatically, stays $\nu_2$ , arriving with $P_{ee}\to\sin^2\theta_{12}\approx0.31$ . Why: this is the deep high-energy plateau — the $^8$ B neutrinos that chlorine, Super-Kamiokande, and SNO see survive at only $\sim31\%$ , matching the larger deficit and the SNO ratio $0.35$ .

Step 4. Read off the energy dependence. Between the two plateaus (a few MeV) lies a smooth rise from $\sim0.31$ up to $\sim0.57$ as one goes from high to low energy — the "upturn" the LMA (large-mixing-angle) solution predicts. Why: a single set of parameters ( $\Delta m^2_{21}$ , $\theta_{12}$ ) reproduces every experiment's deficit at once, which is what elevated LMA-MSW from one option among several to the confirmed solution.

Limiting cases

Case	Result	Meaning
Low-energy $pp$ / $^7$ Be neutrinos	$P_{ee}\to1-\tfrac12\sin^2 2\theta_{12}\approx0.57$	vacuum-averaged oscillation; gallium sees $\sim0.5$ — a shallow deficit
High-energy $^8$ B neutrinos	$P_{ee}\to\sin^2\theta_{12}\approx0.31$	adiabatic MSW conversion; chlorine, Super-K, SNO see $\sim0.3$
No oscillation ( $\theta=0$ or $m_\nu=0$ )	$P_{ee}=1$ , $\phi_{\rm CC}=\phi_{\rm NC}$	no deficit — the solar-model-only expectation, decisively ruled out
Neutral-current channel	flux $=\phi(\nu_e)+\phi(\nu_\mu)+\phi(\nu_\tau)$	flavour-blind; undiminished and equal to the solar model
Short baseline (reactor near)	$P_{ee}\to1$	phase $\Delta m^2 L/4E\to0$ ; oscillation has not yet developed
Day versus night	slight $\nu_e$ regeneration at night	neutrinos crossing the Earth partly re-convert — a small LMA-MSW signature

Common trapThe deficit means solar neutrinos are absorbed or destroyed on the way to Earth

It does not. Nothing in the $1.5\times10^8\,\text{km}$ of near-vacuum between the Sun and Earth removes neutrinos — their interaction cross-section is so small that the entire flux passes through the Earth itself almost untouched. SNO's neutral-current measurement is the direct proof: the total flux of all three flavours, $5.09\times10^{6}\,\text{cm}^{-2}\text{s}^{-1}$ , matches the solar-model prediction $5.05\times10^{6}$ to within errors. Not a single neutrino went missing. What changed is only the label: an electron neutrino born in the Sun is a superposition of mass eigenstates, and by the time it arrives the superposition has rotated so that a measurement is likely to call it $\nu_\mu$ or $\nu_\tau$ instead. The $\nu_e$ -only experiments (chlorine, gallium, and the charged-current part of SNO) were counting one term in a sum that still adds to the full total. "Deficit" is therefore a statement about flavour composition, not about number. (A related trap: the deficit is not evidence that the solar model is wrong — SNO's NC flux vindicated the model to remarkable precision, and John Bahcall's decades of insistence on it.)

Worked examples

1 · Reading flavour change off the SNO fluxes (fully worked)

SNO measured a charged-current flux $\phi_{\rm CC}=1.76\times10^{6}\,\text{cm}^{-2}\text{s}^{-1}$ ( $\nu_e$ only) and a neutral-current flux $\phi_{\rm NC}=5.09\times10^{6}\,\text{cm}^{-2}\text{s}^{-1}$ (all flavours). Find the electron-neutrino survival probability, the muon-plus-tau flux, and state what a no-oscillation hypothesis would have required.

Step 1. Identify the two fluxes with the physical unknowns. By equation (3), $\phi_{\rm CC}=\phi(\nu_e)$ and $\phi_{\rm NC}=\phi(\nu_e)+\phi(\nu_{\mu\tau})$ . So $\phi(\nu_e)=1.76\times10^{6}$ directly, and the total is $5.09\times10^{6}$ . The electron-neutrino flux is $1.76\times10^{6}$ ; the total is $5.09\times10^{6}\,\text{cm}^{-2}\text{s}^{-1}$ .

Step 2. Compute the survival probability. $P_{ee}=\phi_{\rm CC}/\phi_{\rm NC}=1.76/5.09=0.346$ . Only about $35\%$ of the electron neutrinos arrived still labelled $\nu_e$ .

Step 3. Extract the wrong-flavour flux. $\phi(\nu_{\mu\tau})=\phi_{\rm NC}-\phi_{\rm CC}=5.09-1.76=3.33\times10^{6}\,\text{cm}^{-2}\text{s}^{-1}$ (the full SNO fit, which also folds in ES, gives $3.41\times10^{6}$ , nonzero at $5.3\sigma$ ). A large $\nu_\mu+\nu_\tau$ flux is present — flavours the Sun cannot emit.

Step 4. Compare with no oscillation. If neutrinos did not change flavour, $\phi(\nu_{\mu\tau})=0$ and $\phi_{\rm CC}=\phi_{\rm NC}$ , i.e. $P_{ee}=1$ . The data give $P_{ee}=0.35$ and a $\nu_{\mu\tau}$ flux five standard deviations from zero. The no-oscillation hypothesis is excluded; the NC total simultaneously confirms the solar model $\phi(^8\text{B})=5.05\times10^{6}$ .

2 · The solar neutrino flux from the luminosity (numerical)

Estimate the total solar electron-neutrino flux at Earth from the luminosity $L_\odot=3.85\times10^{26}\,\text{W}$ , given $26.73\,\text{MeV}$ and two neutrinos per helium-4 built.

Step 1. Energy per reaction in joules: $Q=26.73\times10^{6}\times1.602\times10^{-19}=4.28\times10^{-12}\,\text{J}$ . Reaction rate $R=L_\odot/Q=3.85\times10^{26}/4.28\times10^{-12}=9.0\times10^{37}\,\text{s}^{-1}$ . The Sun completes $\sim9\times10^{37}$ fusion cycles per second.

Step 2. Neutrino output: two per cycle, so $\dot N_\nu=2R=1.8\times10^{38}\,\text{s}^{-1}$ . The Sun emits $\sim1.8\times10^{38}$ neutrinos per second.

Step 3. Spread over the orbit: $4\pi d^2=4\pi(1.496\times10^{11})^2=2.81\times10^{23}\,\text{m}^2$ , so $\Phi=1.8\times10^{38}/2.81\times10^{23}=6.4\times10^{14}\,\text{m}^{-2}\text{s}^{-1}$ . Converting, $\Phi\approx6.4\times10^{10}\,\text{cm}^{-2}\text{s}^{-1}$ — matching the solar-model total and pinned to the Sun's brightness.

3 · Why the oscillations wash out (numerical)

For a $10\,\text{MeV}$ $^8$ B neutrino with $\Delta m^2_{21}=7.5\times10^{-5}\,\text{eV}^2$ , find the vacuum oscillation length and compare it with the Earth-Sun distance.

Step 1. Use $L_{\rm osc}=4\pi\hbar c\,E/\Delta m^2 c^4=\dfrac{4\pi(1.973\times10^{-7}\,\text{eV}\,\cdot\,\text{m})(10^{7}\,\text{eV})}{7.5\times10^{-5}\,\text{eV}^2}$ . This gives $L_{\rm osc}\approx3.3\times10^{5}\,\text{m}=330\,\text{km}$ .

Step 2. Compare with the baseline $L=1.5\times10^{8}\,\text{km}$ : the ratio $L/L_{\rm osc}\approx4.5\times10^{5}$ . The neutrino completes hundreds of thousands of oscillations en route.

Step 3. Fold in the finite production region ( $\sim0.1R_\odot$ ) and detector energy resolution: the phase is completely scrambled, so $\sin^2(\Delta m^2 L/4E)\to\tfrac12$ . This is why the survival probability is a smooth energy-dependent fraction, not a fringing pattern — the oscillation phase averages away and only the mixing angle (and matter effects) survives.

Physics of the portrait

The solar neutrino problem is inseparable from the man who spent half his career underground creating it and defending it — and from the theorist who told him what to expect. Ray Davis built the experiment that first saw fewer neutrinos than the Sun should send; John Bahcall built the model that said how many there should be.

Origins · 2 minRaymond Davis Jr. and how this was discovered

Raymond Davis Jr.

1914–2006 · the first detection of solar neutrinos and the discovery of the deficit

Raymond Davis Jr. was a chemist who became a physicist by refusing to accept that solar neutrinos were undetectable. Trained in physical chemistry and hired by Brookhaven National Laboratory in 1948, he spent the 1950s learning to coax individual radioactive atoms out of enormous volumes of liquid, a skill nobody else had industrialized. When the theorist John Bahcall computed, in the early 1960s, that the $^8$ B neutrinos from the Sun's core would occasionally convert chlorine-37 into argon-37 at a countable rate, Davis proposed to look. From 1967 he ran a tank of $615$ tons of perchloroethylene — dry-cleaning fluid — nearly a mile down in the Homestake gold mine in South Dakota, shielded from cosmic rays, and every few weeks he flushed out the few dozen argon atoms that a quarter-million-billion-billion chlorine atoms had produced, counting them one radioactive decay at a time. The result was steady and troubling: about a third of what Bahcall predicted. For twenty years most physicists assumed Davis had made an error, or Bahcall had, and the two men — who became close friends and lifelong collaborators — insisted together that neither had. They were right. The deficit was real, the solar model was right, and the neutrinos were changing flavour, as the Sudbury Neutrino Observatory proved in 2001–02. Davis, by then in his late eighties, shared the 2002 Nobel Prize in Physics for opening the field of neutrino astronomy; Bahcall, whose standard solar model made the whole enterprise quantitative, was never awarded it, an omission many regard as one of the prize's clearest injustices. "I just wanted to see the center of the Sun," Davis said. He was the first person ever to do so.

History

1939

— Bethe works out the proton-proton chain and CNO cycle that power the Sun

1964

— Bahcall and Davis propose the chlorine experiment and compute the expected capture rate

1968

— the Homestake experiment reports a deficit: far fewer ν_e than predicted

1986

— Mikheyev and Smirnov, extending Wolfenstein, describe matter-enhanced oscillation (MSW)

2002

— SNO's neutral-current result shows the total flux is undiminished; Davis shares the Nobel Prize

Going further

Texts. Thomson, Modern Particle Physics, §13.2–13.2.2 (solar neutrinos, Super-Kamiokande, SNO) · Griffiths, Introduction to Elementary Particles, 2nd ed., §11.7 (neutrino oscillations) · Halzen & Martin, Quarks and Leptons, §12.6 (neutrino masses and mixing) · Perkins, Particle Astrophysics, 2nd ed., Ch. 3 (solar neutrinos and the MSW effect) · Peskin & Schroeder and Schwartz, Quantum Field Theory and the Standard Model, on the weak charged and neutral currents that define the CC/NC channels. The key numerical results in this article are: $\Phi\approx6.4\times10^{10}\,\text{cm}^{-2}\text{s}^{-1}$ from $L_\odot$ , $Q=26.73\,\text{MeV}$ , and two $\nu_e$ per cycle; $P_{ee}=\phi_{\rm CC}/\phi_{\rm NC}=1.76/5.09=0.35$ and $\phi(\nu_{\mu\tau})=3.33\times10^{6}\,\text{cm}^{-2}\text{s}^{-1}$ ; $\phi_{\rm NC}=5.09\times10^{6}$ against the solar-model $5.05\times10^{6}\,\text{cm}^{-2}\text{s}^{-1}$ ; vacuum-averaged $P_{ee}=1-\tfrac12\sin^2 2\theta_{12}=0.57$ and adiabatic $P_{ee}=\sin^2\theta_{12}=0.31$ for $\sin^2\theta_{12}=0.31$ ; and $L_{\rm osc}\approx330\,\text{km}$ at $E=10\,\text{MeV}$ , $\Delta m^2=7.5\times10^{-5}\,\text{eV}^2$ . The measured fluxes are from Ahmad et al. (SNO, 2001–2002); Homestake ( $0.48\pm0.04$ Ar/day vs $1.7$ predicted) from Cleveland et al. (1998); Super-Kamiokande from Fukuda et al. (2001); the standard solar model from Bahcall and Pinsonneault (2004).

Prerequisites. Neutrino Flavours · The Weak Charged Current · The Neutral Current · Čerenkov Radiation. Leads to. Two-Flavour Oscillations · MSW Matter Effects · Neutrino Mass Mechanisms · Atmospheric Neutrinos.

Ask the tutor: Ask me to walk through how SNO's three deuteron reactions pull apart the electron flux from the total, or why the ⁸B deficit runs deeper than the pp deficit.

Why does the Sun produce only electron neutrinos and not muon or tau neutrinos?

Because the neutrinos come from nuclear beta-type reactions, and those emit a neutrino paired with a charged lepton at the same vertex. In the Sun's core the only charged lepton light enough to be produced is the positron (or, in electron capture, the electron), so the accompanying neutrino is always an electron neutrino. Making a $\nu_\mu$ would require producing a muon, which weighs $106\,\text{MeV}$ — hundreds of times more than the few MeV available in solar fusion — and a $\nu_\tau$ is even more out of reach. So the Sun is a pure $\nu_e$ source, which is exactly what makes the appearance of $\nu_\mu$ and $\nu_\tau$ at Earth such unambiguous evidence for flavour change: there is no way to make them at the source.

Couldn't the deficit just mean the solar model was wrong?

For thirty years that was the leading suspicion, and it was a reasonable one — the $^8$ B flux that most experiments measured depends on the core temperature to about the twenty-fifth power, so a few-percent temperature error could in principle slash it. Two things killed the idea. First, the deficit was energy-dependent: gallium (low energy) saw about half, chlorine and water (high energy) about a third, and no single tweak to the temperature can suppress the two differently — cooling the core reduces both. Second, and decisively, SNO's neutral-current channel measured the total flux of all flavours directly and found it equal to the solar model's prediction. The model was right all along; John Bahcall's long insistence on it was vindicated. The neutrinos, not the Sun, were doing something unexpected.

How can SNO count muon and tau neutrinos if they can't make muons or taus at these energies?

Through the neutral current. The charged-current reaction does require making the neutrino's charged partner, which is why it sees only $\nu_e$ . But the neutral-current reaction, $\nu_x+d\to p+n+\nu_x$ , works differently: a $Z$ boson is exchanged, the neutrino stays a neutrino, and all that happens is the deuteron is split into a free proton and neutron. The $Z$ couples to every neutrino flavour with exactly the same strength, so a $\nu_\mu$ or $\nu_\tau$ dissociates the deuteron just as readily as a $\nu_e$ — no charged lepton ever needs to appear. SNO detected the freed neutron (by the gamma rays from its capture), giving a flavour-blind count of the total flux. That is the whole trick that made the neutral current the decisive measurement.

Why is the deficit different for different experiments?

Because different experiments have different energy thresholds and therefore sample different parts of the solar spectrum, and the survival probability depends on energy. Gallium ( $^{71}$ Ga, threshold $0.23\,\text{MeV}$ ) reaches the dominant low-energy $pp$ neutrinos, which oscillate in the vacuum-averaged regime and survive at about $57\%$ — a shallow deficit. Chlorine ( $^{37}$ Cl, threshold $0.81\,\text{MeV}$ ) and the water detectors ( $\gtrsim5\,\text{MeV}$ ) see mostly the high-energy $^8$ B neutrinos, which undergo matter-enhanced (MSW) conversion in the Sun and survive at only about $31\%$ — a deep deficit. Far from being an inconsistency, this energy dependence is a fingerprint: a single pair of oscillation parameters (the LMA solution) reproduces every experiment's number at once.

Does a neutrino change flavour gradually, or does it flip suddenly?

Neither picture is quite right; the honest answer is quantum-mechanical. The neutrino travels as a coherent superposition of mass eigenstates, and "flavour" is just a particular combination of them. As it propagates, the mass components accumulate different phases, so the combination that started as pure $\nu_e$ smoothly rotates into a mixture with $\nu_\mu$ and $\nu_\tau$ amplitudes — the probability of measuring each flavour oscillates continuously with distance. Nothing flips; the state evolves smoothly. The apparent suddenness only appears at detection, when a measurement forces the neutrino to reveal one definite flavour, with the probabilities set by how far the superposition had rotated. For solar neutrinos the baseline is so long that this rotation has been scrambled many times over, leaving a fixed average survival probability.

How is the solar neutrino problem related to the atmospheric neutrino problem?

They are two faces of the same discovery — that neutrinos have mass and mix — but they probe different mixing angles and mass splittings. The solar problem involves $\nu_e$ disappearing over the $150$ -million-km baseline from the Sun, governed by the "12" parameters ( $\theta_{12}\approx33^\circ$ , $\Delta m^2_{21}\approx7.5\times10^{-5}\,\text{eV}^2$ ) with matter effects playing a starring role. The atmospheric problem, found by Super-Kamiokande in 1998, involves $\nu_\mu$ from cosmic-ray showers disappearing over baselines up to the Earth's diameter, governed by the "23" parameters ( $\theta_{23}\approx45^\circ$ , $\Delta m^2_{32}\approx2.5\times10^{-3}\,\text{eV}^2$ ). Together they established the three-flavour mixing picture, and their discoverers — McDonald for solar (SNO) and Kajita for atmospheric (Super-K) — shared the 2015 Nobel Prize.