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The neutron and Positron together have more rest mass than the initial proton, so a finite incident energy is necessary even before recoil is considered. In the proton's rest frame, use four-momentum conservation and a negligible neutrino mass. The initial squared center-of-mass energy is $m_p^2{c}^4+2m_p{c}^2{E}_{\nu }$; the least final squared center-of-mass energy is $(m_n+m_e{)}^2{c}^4$. Equating them gives the threshold above. At threshold the final particles share one center-of-mass velocity; they are not both at rest in the laboratory.

Writing $\Delta =m_n-m_p$, the result separates intoThe positive second term is the laboratory recoil correction. With rounded rest energies $m_p{c}^2=938.272\mathrm{MeV}$, $m_n{c}^2=939.565\mathrm{MeV}$ and $m_e{c}^2=0.511\mathrm{MeV}$, the calculated threshold is about $1.806\mathrm{MeV}$. These rounded inputs follow the recommended particle masses.^[ref]

As a further ideal energy balance, suppose the Positron stops and annihilates with an Electron, and all prompt annihilation gamma rays are contained. Its prompt visible energy is $T_{e^{+}}+2m_e{c}^2$. Neglecting neutron recoil gives $E_{\mathrm{vis}}\simeq E_{\nu }-(m_n-m_p)c^2+m_e{c}^2\simeq E_{\nu }-0.782\mathrm{MeV}$. Escaping gamma rays, detector response and recoil change that measured sum. The delayed capture signal is a separate energy deposit, not an extra contribution to the incident electron antineutrino's prompt energy reconstruction.

How can a detector establish that an unseen particle changes identity? The Super-Kamiokande atmospheric-neutrino experiment compares different travel paths; the Sudbury Neutrino Observatory compares different interaction channels. Both infer neutrino flavor from reaction products and calibrated responses. Their complementarity connects particle detection to neutrino oscillation, neutrino mass and neutrino astronomy.

Takaaki Kajita and Arthur B. McDonald shared the 2015 physics Nobel Prize. These were collaboration achievements: Yoichiro Suzuki's Super-Kamiokande work included the solar neutrino analysis, and George Ewan helped establish and lead the Sudbury Neutrino Observatory.^[ref][ref]

The older Kamiokande account explains water Cherenkov neutrino reconstruction; Raymond Davis Jr.'s Homestake experiment explains why a deficit in an electron neutrino channel alone leaves a source-versus-propagation ambiguity. Complementary channels can break that ambiguity.

A Majorana neutrino is its own antiparticle. A corresponding mass term can violate conventional lepton number by two. Observing a wrong-helicity charged-current process is nevertheless suppressed by the very small neutrino mass relative to its energy, rather than having the ordinary charge-conjugate branching fraction.

Neutrino oscillation is coherent conversion between neutrino flavors: different mass eigenstates acquire different propagation phases. It measures squared-mass differences and mixing, rather than an absolute neutrino mass. The Super-Kamiokande atmospheric-neutrino experiment tests path-dependent disappearance; the Sudbury Neutrino Observatory compares reactions with complementary neutrino flavor sensitivity. Solar neutrino propagation also requires matter effects; the two-flavour vacuum survival model is only an illustration.

In the pion rest frame, write its four-momentum as $P=(m_{\pi },0)$ and the charged lepton's as $p_{\ell }=(E_{\ell },\mathbf{p})$. By four-momentum conservation, the massless antineutrino has $k=P-p_{\ell }$, with $k^2=0$. ThereforeIts energy is $E_{\nu }=m_{\pi }-E_{\ell }$, and for a massless antineutrino, $E_{\nu }=|\mathbf{k}|$. The final momenta are equal and opposite, soThe expression is positive precisely when the two-body decay is energetically allowed, $m_{\pi }>m_{\ell }$, neglecting the neutrino mass.

All three processes are weak charged current transformations between neighboring nuclear isobars. Neglecting neutrino masses, and writing $M_a(A,Z)$ for a neutral atomic mass, their Q values are:

These formulae use the paper's natural units. With energies in MeV, beta-plus decay requires an atomic mass-energy difference exceeding $2m_e{c}^2=1.022\mathrm{MeV}$. The extra electron-mass subtraction arises from using neutral atomic rather than bare nuclear masses. Electron capture can occur when the atomic mass-energy difference is positive but below this threshold. For a particular capture shell or excited daughter, the available energy must also cover the electron-hole and nuclear excitation energies, and the small nuclear recoil. Small atomic binding corrections are suppressed in the displayed threshold formulae.

The question's fixed-$A$ family consists of nuclear isobars, rather than isotopes: isotopes have fixed $Z$. The semi-empirical mass formula gives a convex nuclear mass parabola as a function of $Z$ at fixed $A$. For odd $A$, nuclear pairing does not split the family into two branches, so there is typically one beta-stable isobar, at the discrete minimum. Neighboring members can lower their mass by changing $Z$ toward it.

For even $A$, nuclear pairing lowers even-even nuclei and raises odd-odd nuclei, producing two offset nuclear mass parabolas. There are often two beta-stable even-even isobars, separated by an odd-odd member to which neither can energetically decay. The number is not an exact universal rule: some even-$A$ chains have one or more than two such minima against single beta decay, and a few light odd-odd nuclei are stable. Stability against single beta decay also does not guarantee stability against double beta decay.

For a deliberately simplified neutrino oscillation calculation, take two neutrino flavors related to two mass eigenstates by a real rotation matrix with angle $\theta$. A quantum superposition acquires different phases during propagation; projecting back onto the initial neutrino flavor gives the probabilityHere $L$ is path length and $E$ is neutrino energy; the numerical coefficient incorporates unit conversion.

The paper states the corresponding conversion formula.^[ref]

The path dependence has a geometrical origin. In an original spherical-Earth model, let $R$ be the radius and $h$ a fixed production height, neglecting detector depth. Let $\Theta$ be the zenith distance of the direction in the sky from which the neutrino arrives, so $c=\cos \Theta =+1$ means downward propagation. The backward ray from detector to production obeys $R^2+L^2+2RLc=(R+h{)}^2$. Its positive solution isThus $L(+1)=h$ and $L(-1)=2R+h$. The calculation explains why opposite directions test very different baselines, rather than attributing every observed direction to the same travel distance. Actual production heights and the angle between the outgoing charged lepton and incoming neutrino smear that relationship.

For an original chosen example, set $\theta =\pi /4$ and $\Delta m^2=2.5\times 10^{-3}{\mathrm{eV}}^2$. At $E=1\mathrm{GeV}$, $L=20\mathrm{km}$ gives $P_{\mu \mu }\simeq 0.9960$, while $L=500\mathrm{km}$ gives $P_{\mu \mu }\simeq 0.000279$. The first minimum is at $L/E\simeq 494.7km/GeV$. These are stipulated inputs, not a fit to the 1998 data. A broad distribution of true $L/E$, together with energy and directional uncertainty, smooths the sharp minima.

A direct amplitude check makes the mechanism explicit. DefineThen $S_{11}={\cos }^2\theta +{\sin }^2\theta e^{-2i\phi }$. Taking its squared absolute value yields the displayed survival formula because $1-\cos (2\phi )=2{\sin }^2\phi$. Unitarity gives $|S_{11}{|}^2+|S_{21}{|}^2=1$: disappearance from one channel becomes conversion to the other within this closed model.

This illustration does not describe all solar neutrino propagation. Coherent forward scattering in matter changes the effective propagation operator, and three active neutrino flavors require a larger mixing description. Even in the simplified model, neutrino oscillation measures a squared-mass difference; it does not determine the absolute neutrino mass.