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Let $f(r)=1-2M/r+Q^2/r^2$. A positive-radius horizon requires a root of $r^2-2Mr+Q^2=0$, givingFor the usual asymptotically flat Reissner-Nordstrom spacetime, the black-hole conditions areThe strict inequality gives a nonextremal outer event horizon at $r_{+}$; equality gives an extremal black hole. If $M<|Q|$ with $M>0$, there is no horizon and the central curvature singularity is naked. Nonpositive $M$ cannot supply a positive-radius black-hole horizon; $M=Q=0$ is Minkowski spacetime. For $Q=0$, the zero root $r_{-}=0$ is the singularity, not an additional regular horizon.

In the static coordinates the vector $k={\partial }_t$ has constant components and every metric component is independent of $t$. Its Lie derivative therefore satisfiesFor the torsion-free Levi-Civita connection, ${\mathcal{L}}_{k}g=2{\nabla }_{(a}{k}_{b)}$, proving the Killing equation directly. This also normalizes the Killing vector field to unit time translation at infinity.

At $r=r_{+}$, $k^2=-f=0$. Use the regular advanced chart $v=t+r_{\ast }$ with $d{r}_{\ast }/dr=f^{-1}$, in whichOn the horizon, $k_a=(dr{)}_a$. Thus the Killing vector field is both null and normal to the horizon, and its orbits generate that Killing horizon. The normal lines of a null hypersurface are null geodesics, possibly with nonaffine parameter. To see this without assuming a separate geodesic property, the Killing equation givesRestriction to the future horizon therefore givesThe acceleration is proportional to the tangent, which is exactly the geodesic equation with a nonaffine parameter. An affinely parametrized tangent can be obtained by rescaling $k$ along each generator; that rescaled vector need not itself extend to a spacetime Killing vector field. At a nonextremal bifurcation surface, $k$ vanishes, so a nonzero generator tangent must be described in a regular affine parametrization there.

On the past branch, use the retarded chart with radial-time cross term $-2dudr$. Now $k_a=-(dr{)}_a$ at the horizon, so the same calculation yields the opposite signed nonaffinity. Since $r_{+}$ is constant and the spacetime is spherically symmetric, $f^{\prime }(r_{+})/2$ is constant over the future horizon. This gives the two signs in the horizon relation without assuming their value beforehand.

Factor $f=(r-r_{+})(r-r_{-})/r^2$. Its derivative at the outer horizon gives the Reissner-Nordstrom horizon surface gravityIt is positive for $M>|Q|$, and at $Q=0$ reduces to the Schwarzschild surface gravity $1/(4M)$. For $M=|Q|>0$, the root is double and $\kappa =0$. Thus the horizon remains present when its surface gravity vanishes: it is a degenerate Killing horizon, and $k$ itself has affine parametrization along its nonzero horizon generators. The strictly positive-constant description applies only to the nonextremal case; the zero value is the legitimate extremal limit.

Let $\Delta M=M_1-M_0$ and retain only terms first order in $\mu /M_0$ and $\rho /M_0$. The outgoing event horizon equation expands toHere ${\kappa }_0$ is the initial Schwarzschild surface gravity. The integrating factor $e^{-{\kappa }_{0}v}$ gives $d(e^{-{\kappa }_{0}v}\rho )/dv=-e^{-{\kappa }_{0}v}\mu /(2M_0)$. Future stationarity imposes $\rho =2\Delta M$ for $v>v_0$ and removes the growing homogeneous solution. Integrating from $v$ to infinity yields the teleological response of a Vaidya event horizon:The integral is finite because $\mu$ is bounded. Integrating by parts also givesFor $v>v_0$, this reduces to $2\Delta M$. For $v<0$, the first form becomesConsequently $r_H\to 2M_0$ as $v\to -\infty$. The positive constant $C$ shows that this event horizon is already expanding in the initially vacuum region. For monotone accretion, $\rho -2\mu \ge 0$, so it lies outside the instantaneous apparent horizon to first order. All displayed radius corrections are understood up to $O((\Delta M{)}^2/M_0)$.

The stress-energy tensor is already first order, so its energy flux can be evaluated on the unperturbed Schwarzschild event horizon $r=2M_0$, with the infinity-normalized Killing vector field $\xi ={\partial }_v$. On this background $g^{rv}=1$, and henceThe conserved stress-energy current from a Killing vector is $J^a=-T^a{}_b{\xi }^b$. Its radial component is negative, so the positive inward energy flux is $-J^r$. Using the directed null-surface measure $r_H^{2}dvd\Omega$ givesThe initial and final event horizon areas are $A_0=16\pi M_0^2$ and $A_1=16\pi (M_0+\Delta M{)}^2$. Thus $\Delta A=32\pi M_0\Delta M+O((\Delta M{)}^2)$. With Schwarzschild surface gravity ${\kappa }_0=1/(4M_0)$, the Physical-process first law of black-hole mechanics follows:The flux is Killing energy, including gravitational redshift, rather than the energy measured by a sequence of static observers arbitrarily close to the event horizon.

Work initially in units $G=c=\hslash =k_B=1$ and consider gravitational collapse settling to a nonextremal Schwarzschild black hole. Classically, outgoing rays sufficiently close to the forming event horizon suffer an arbitrarily large redshift. Quantum mechanically, the meaning of positive frequency at past null infinity differs from that at future null infinity; a Bogoliubov transformation between these two mode descriptions can therefore mix creation operators and annihilation operators. Hawking's argument computes that mixing and obtains a thermal occupation, even when the incoming state has no particles.

To see why the frequency mixing is universal, use final Schwarzschild time $t$, the Schwarzschild tortoise coordinate $r_{\ast }$, and retarded time $u=t-r_{\ast }$. Near the final event horizon, $r_{\ast }\sim (2\kappa {)}^{-1}\log (r-2M)$, with Schwarzschild surface gravity $\kappa =1/(4M)$. On a fixed advanced-time section, $u=v-2r_{\ast }$, so $r-2M$ is proportional to $e^{-\kappa u}$. Tracing a ray back through the regular collapsing region maps this small separation smoothly to its separation from the last escaping incoming ray. If $U$ is an affine incoming null coordinate and $U_H$ labels that limiting ray, the Hawking exponential ray map isThe logarithm at a simple horizon root fixes the exponent. Smooth propagation through the earlier collapse changes $A$ and phases but not the leading exponential redshift.

For a massless bosonic test scalar field, an outgoing mode of frequency $\omega >0$ at future null infinity behaves as $e^{-i\omega u}$. Trace it backward in the high-frequency geometric optics approximation. Its relevant incoming dependence is, up to normalization and a phase,Here $\Theta$ is the Heaviside step function. Incoming rays with $U>U_H$ do not escape. This truncated logarithmic phase is not purely positive frequency with respect to $U$. Decomposing it into incoming positive-frequency solutions $e^{-i{\omega }^{\prime }U}$ and negative modes $e^{+i{\omega }^{\prime }U}$ gives the Bogoliubov transformation. With $x=U_H-U$, the two relevant Fourier transform integrals areThe former is the positive-frequency coefficient and the latter is the negative-frequency coefficient; the Klein-Gordon inner product introduces the same frequency normalization into their modulus ratio. By the defining integral of the gamma function and analytic continuation in its Laplace parameter,For a complex number $z$ in the right half-plane, $|z^{-1-ia}|=|z{|}^{-1}{e}^{a\arg z}$. As $ϵ\downarrow 0$, the two arguments tend to $+\pi /2$ and $-\pi /2$. This gives the thermal ratio of Hawking Bogoliubov coefficients:The regulator fixes the branches and is essential to the sign of the thermal exponent.

Expand the outgoing annihilator as $b=\int d{\omega }^{\prime }({\alpha }_{\omega {\omega }^{\prime }}^{\ast }{a}_{{\omega }^{\prime }}-{\beta }_{\omega {\omega }^{\prime }}^{\ast }{a}_{{\omega }^{\prime }}^{†})$, including a complete set of incoming channels. The canonical identities for a bosonic Bogoliubov transformation express $[b,b^{†}]=1$. For normalized outgoing wave packets, they give $\int (|\alpha {|}^2-|\beta {|}^2)d{\omega }^{\prime }=1$. In the incoming Fock vacuum, the outgoing mean of the particle number operator is $N_{\omega }=\int |\beta {|}^{2}d{\omega }^{\prime }$. Combining the two identities with the thermal ratio, in the late-time narrow-frequency packet limit, yieldsWave packets avoid interpreting the divergent normalization of infinitely long continuum modes as a finite particle count. The temperature is the Hawking temperature, and the occupation is the zero-chemical-potential Bose-Einstein distribution. Restoring constants, the Schwarzschild black hole temperature is $T_H=\hslash c^3/(8\pi GM{k}_B)$ for physical mass $M$.

The preceding thermal occupation describes the horizon-originating channel before exterior scattering. The exterior angular and curvature potential partly reflects it. For a scalar field, the number flux at future null infinity is consequentlywhere the greybody factor ${\Gamma }_{\ell }$ is the transmission probability of a partial wave. Multiplication by $\omega$ gives its energy flux. Thus the asymptotic spectrum has the thermal denominator but is not an exact blackbody spectrum. For fermionic fields the canonical anticommutation relations replace the minus sign in the occupation denominator by a plus sign, at the same Hawking temperature.

The relevant state is the collapse vacuum for Hawking radiation: no incoming thermal bath at past null infinity, and regular short-distance behavior in freely falling coordinates at the forming event horizon. It predicts an outgoing flux, unlike equilibrium with a thermal bath on an eternal black hole. The derivation uses quantum field theory in curved spacetime on a prescribed classical geometry; it does not require outgoing particles to follow forbidden classical paths out of the interior. Positive energy carried to infinity is accompanied, in the semiclassical description, by negative Killing-energy flux into the black hole. Including that flux in slow backreaction decreases its mass, giving black-hole evaporation.

The approximation applies to the late radiation of a large nonextremal black hole while its surface gravity changes slowly on a time scale ${\kappa }^{-1}$. Backward propagation involves very high locally measured frequencies, so the calculation assumes the usual regular short-distance quantum-field state. It determines the leading radiation and temperature, not the Planck-scale endpoint of black-hole evaporation. The exponential horizon redshift forces positive/negative frequency mixing with a thermal ratio, giving outgoing Hawking radiation at $T_H=\kappa /(2\pi )$.

For radial free fall, let $E=f{\dot{t}}_A>0$ be Alice's conserved Killing energy. The inward branch of the timelike geodesic has $\dot{r}=-\sqrt{E^2-f}$, henceWith $\delta =r-2M$, the near-horizon geodesic equations giveThe constants in the logarithms are understood to make their arguments dimensionless. Therefore $u_A=t_A-r_{\ast }=2t_A+O(1)$. For an outgoing signal reaching the fixed-radius Bob, $t_B=u_A+r_{\ast }(r_B)$, so $t_B=2t_A+O(1)$. The redshift consequently behaves asThis is the answer when $t$ is Bob's Schwarzschild time, as appropriate to the stated observation. It is the Schwarzschild surface gravity $a=\kappa =1/(4M)$. If $t$ instead means the emission Schwarzschild time $t_A$, the same redshift is proportional to $e^{t_A/(2M)}$ and gives $M=1/(2a)$. The two answers use different clocks, not different dynamics. If the measured exponential uses Bob's proper time ${\tau }_B$, its rate is $1/(4M\sqrt{f_B})$. In SI units the reception-time result is $M=c^3/(4Ga)$.

The reception-time exponent also holds for smooth radial infall crossing the future Schwarzschild event horizon with finite nonzero $dr/d\tau$: regular ingoing Eddington-Finkelstein coordinates give finite $v$ there, $u=v-2r_{\ast }\sim -4M\log ({\tau }_H-\tau )$. Thus no special value of $E$ is needed.