Inverse beta decay threshold ($E_{\nu ,\mathrm{thr}}=\frac{(m_n+m_e{)}^2-m_p^2}{2m_p}{c}^2$)

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.