Reissner-Nordstrom Hawking temperature ($T=(r_{+}-r_{-})/(4\pi r_{+}^2)$)

For $f=(r-r_{+})(r-r_{-})/r^2$, near-horizon Euclidean geometry has polar form $d{\rho }^2+[f^{\prime }(r_{+})/2{]}^2{\rho }^{2}d{\tau }^2$. Smoothness gives inverse temperature $4\pi /f^{\prime }(r_{+})$ and the displayed formula. This assumes a nondegenerate Killing horizon and time normalized at infinity. Its nonextremal limit is zero at $M=|Q|$.

In the ideal thermal-cutoff model where charged particles of mass $m$ are emitted only for $T>m$, a sufficiently large initial mass keeps charge fixed throughout evaporation exactly when $m$ is at least the fixed-charge Reissner-Nordstrom temperature maximum. Neutral radiation then approaches an extremal charged remnant of area $4\pi Q^2$. This conclusion does not include nonthermal electric-field discharge, nor establish finite-time attainment.

Set $y=r_{+}/|Q|\ge 1$. Then $M/|Q|=(y+y^{-1})/2$ and $|Q|T=(y^{-1}-y^{-3})/(4\pi )$. Differentiation gives a unique maximum at $y=\sqrt{3}$, proving the displayed mass and temperature. The curve is zero at extremality and asymptotic to $1/(8\pi M)$ at large mass.