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In the reaction displayed here, an electron antineutrino interacts with a proton to produce a Positron and a neutron. It turns an otherwise hard-to-observe incident particle into two products with different detection signatures. A prompt Positron signal followed by neutron capture can identify a candidate much more selectively than a single energy deposit. Inverse beta decay is a weak interaction; the incident electron antineutrino is detected through its reaction products.

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.

The antineutrino associated with the muon flavour of the weak charged current has zero electric charge and lepton number $-1$ in the usual production-and-detection description. A negative muon decays to an Electron, an electron antineutrino and a muon neutrino; the charge-conjugate process instead produces the muon antineutrino.

Use the inverse beta decay threshold, with the charged-lepton mass generalized to $m_{\ell }$. For a stationary proton and a massless incident antineutrino, the Mandelstam invariant is $s=m_p^2+2m_p{E}_{\nu }$. At threshold the final particles have no relative momentum in their centre-of-momentum frame, so $s_{\mathrm{thr}}=(m_n+m_{\ell }{)}^2$. ThereforeUsing the supplied, rounded masses givesThe second term is the recoil correction; setting the threshold equal just to the final rest-mass excess would miss it.

Reactor electron antineutrinos have energies of a few million electronvolts. They can produce a Positron through inverse beta decay above the first threshold, but cannot produce a muon. Thus reactor neutrino oscillations are measured primarily as electron-antineutrino disappearance, through the event rate and energy spectrum relative to the unoscillated prediction. An oscillated muon antineutrino cannot be identified by the proposed muon-production channel at reactor energies; disappearance alone does not distinguish all possible final flavours.

In both semileptonic decays, an active quark emits a virtual W boson, which produces an Electron and an electron antineutrino:The unchanged $u,d$ spectator quarks combine with the produced $u$ into the final proton. The relevant Cabibbo-Kobayashi-Maskawa matrix elements are $V_{ud}\simeq \cos {\theta }_C$ and $V_{us}\simeq \sin {\theta }_C$.

Use the quoted masses in energy units, equivalently natural units $c=1$, and define the energy releases before subtracting the electron rest energy asThe prescribed density of states ansatz uses precisely these differences to the fifth power. A Cabibbo-angle extraction from semileptonic decay rates also needs a common reduced hadronic strength. In that approximation the partial decay rates, measured in ${\mathrm{s}}^{-1}$, areThe supplied mean lifetimes describe all decays, so multiply their inverses by the relevant branching fractions:Taking their ratio cancels $K$ and givesChoose the conventional first-quadrant Cabibbo angle:The Lambda baryon partial rate is much larger despite its smaller weak flavour coupling because the fifth-power phase space factor is enormous.

There is an assumption beyond the stated density of states ansatz: the neutron and Lambda reduced hadronic strengths have been set equal. If their ratio is $H=K_{\Lambda }/K_n$, the same data instead imply ${\tan }^2{\theta }_C=0.0588300/H$. Without specifying $H$, the angle is not uniquely determined. Unequal hadronic form factors, finite electron mass and recoil make the displayed common-strength result an exercise-level estimate, rather than a precision measurement.

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.

The original Kamiokande paper reported a low-energy event cluster on 23 February 1987 above surrounding background. Inverse beta decay, ${\bar{\nu }}_e+p⟶e^{+}+n$, provides an electron antineutrino channel through Positron light. Neutrino-Electron elastic scattering is more directional. The burst supported core-collapse supernova neutrino emission, with limited statistics and channel ambiguity.^[ref]

An original counting illustration shows why timing matters. Suppose an independently measured background is a Poisson process of rate $b$. In a prespecified window $\Delta t$, the probability of at least two background events isFor the synthetic choice $b\Delta t=0.1$, this is about $0.004679$, obtained by subtracting the zero- and one-event probabilities of the Poisson distribution. The Homestake experiment's delayed extraction cannot supply such capture-time clustering. This calculation is not a significance estimate for SN 1987A: searching many possible windows, varying acceptance and correlated backgrounds change the inference.

In the modern weak charged current interpretation, one allowed assignment isThe electron antineutrino, tau neutrino, muon neutrino and antineutrino of the tau neutrino carry away energy and momentum. There are four undetected particles in this decay assignment. Interchanging which parent decays electronically gives the opposite electric charge assignment. Each parent conserves electric charge and, in this approximation, total lepton number; four-momentum conservation applies separately to production and to each particle decay.

If $B_e$ and $B_{\mu }$ denote the corresponding branching fractions, independent parent decays give an inclusive mixed-flavour fraction $2B_e{B}_{\mu }$. The factor two counts the two assignments to opposite parent charges. For chosen $B_e=B_{\mu }=1/5$, the mixed fraction is $2/25$, before acceptance and detection efficiency. Correlations between the parents' spins can affect angular selection even when the fully inclusive branching fractions factorize.

The neutrinos in that equation are a physical interpretation to test. Neither a deficit in visible energy nor two unusual tracks alone establishes their identities or uniquely counts them. Even a compatible invariant mass and an angular distribution require particle identification, background checks and agreement across further particle decay channels. The direct detection milestone for the tau neutrino was in 2000, separate from the charged tau lepton discovery.