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DefineThe Bekenstein-Hawking entropy and Hawking temperature becomeAt fixed $J$,The fixed-$J$ first law gives $(\partial S_{BH}/\partial M{)}_J=1/T_H$, soTherefore the Kerr black-hole heat capacity at fixed angular momentum isAt $a=0$, this correctly reduces to $C_J=-8\pi M^2$.

The horizon generator must be null at $r=r_{+}$. Substitution of $\xi ={\partial }_v+{\Omega }_H{\partial }_{\chi }$ in the Kerr black hole metric givesThe surface gravity and Hawking temperature are

The four laws of black-hole mechanics closely parallel thermodynamics. The zeroth law says that the surface gravity $\kappa$ is constant over a stationary Killing horizon. The first law isThe second law is Hawking's area theorem, $\delta A\ge 0$ under the null energy condition, and the third law of black-hole mechanics says that a regular extremal horizon with $\kappa =0$ cannot be reached by a finite physical process. These match constancy of temperature in equilibrium, $dE=TdS+$ work terms, entropy increase, and unattainability of zero temperature.

Quantum field theory makes the analogy literal. A stationary horizon emits at the Hawking temperatureComparing $T_{H}dS$ with the area term in the first law gives the Bekenstein-Hawking entropyFor a Schwarzschild black hole, $\kappa =c^4/(4GM)$ and $T_H=\hslash c^3/(8\pi GM{k}_B)$.

To see why particles are produced, quantize a real scalar field using the conserved Klein-Gordon inner product. In the asymptotically Minkowski past choose positive-frequency modes $u_i^{\mathrm{in}}$ and writeThe asymptotically Minkowski future supplies another positive-frequency basis $u_j^{\mathrm{out}}$ and operators $a_j^{\mathrm{out}}$. In a nonstationary middle region there is no preferred timelike Killing vector and therefore no invariant positive-frequency split: the notion of particle is observer- and basis-dependent. The two mode bases are related by a Bogoliubov transformation,so the corresponding operators mix annihilation and creation operators. The in-vacuum then containsout-particles whenever $\beta \ne 0$.

For a black hole formed by collapse, late outgoing modes traced backwards toward the event horizon undergo an exponentially large blueshift. This produces a universal Bogoliubov mixing withthe Planck distribution at $T_H$. Greybody scattering outside the horizon modifies the flux reaching infinity but not its characteristic temperature.

Hawking radiation gives a black hole negative heat capacity: it heats up as it loses mass and can evaporate in finite time. The generalized second law assigns entropy $S_{\mathrm{BH}}$ to the hole and states that this plus exterior entropy does not decrease. The enormous area entropy suggests microscopic horizon degrees of freedom. If semiclassical evaporation ends with only thermal radiation, an initially pure state appears to become mixed, producing the black hole information paradox. Resolving the endpoint, information recovery, and the microscopic origin of the area law are central constraints on quantum gravity.

For a metric $d{s}^2=-f(r)d{v}^2+2dvdr+r^{2}d{\Omega }_2^2$ with a simple Killing horizon, the surface gravity of $k={\partial }_v$ is $\kappa =f^{\prime }(r_{+})/2$. HeresoThe Hawking temperature of the Schwarzschild--anti-de Sitter black hole is therefore

Let $k^a=(d/d\lambda {)}^a$ be an affine horizon generator, normalized so that the horizon Killing field is ${\xi }^a=\kappa \lambda k^a$. To first order in the small total injected energy, expansion and shear-squared terms are second order, and the Null Raychaudhuri equation on the horizon becomesThe flux of Killing energy through all the shells isSubstitute the linearized focusing equation and integrate by parts. Stationarity before and after the process makes the endpoint term $\lambda \theta$ vanish, while $dA/d\lambda =\theta A$ to first order. ThereforeUsing the Hawking temperature $T_H=\kappa /(2\pi )$ and Bekenstein-Hawking entropy $S_{BH}=A/4$ gives the Physical-process first law of black-hole mechanicsThe derivation is linear in the stress tensor, so separated shells simply contribute additively.

The four-dimensional Schwarzschild--anti-de Sitter black hole is a static, spherically symmetric vacuum spacetime with negative cosmological constant $\Lambda =-3/L^2$. In units $G=c=\hslash =k_B=1$, its metric tensor isHere $L$ is the Anti-de Sitter curvature length and $M>0$ is the mass parameter. Setting $f(r_{+})=0$ gives $M=r_{+}(1+r_{+}^2/L^2)/2$. The surface gravity $\kappa =f^{\prime }(r_{+})/2$ then gives the Hawking temperatureUnlike its asymptotically flat limit $L\to \infty$, its temperature eventually increases with horizon radius. Differentiation gives a minimum at $r_{+}=L/\sqrt{3}$, separating negative- and positive-heat capacity branches.^[ref]

For the future-directed central-time Killing vector field in a static patch of de Sitter spacetime, the outgoing null chart gives the future branch coefficient $+1/R$ in ${\nabla }_{k}k=\kappa k$. The ingoing chart gives $-1/R$ on the past branch. Both follow from $d(k^2)=-2\kappa k^{♭}$; the cross-term sign changes the horizon covector. The physical Hawking temperature uses $|\kappa |$, not a signed negative temperature.

Approximate an isolated Schwarzschild black hole as emitting photon blackbody radiation from an area $16\pi G^2{M}^2/c^4$ at Hawking temperature $\hslash c^3/(8\pi G{k}_{B}M)$. The Stefan–Boltzmann law gives $P=\hslash c^6/(15360\pi G^2{M}^2)$. Integrating $c^{2}dM/dt=-P$ gives the displayed lifetime, about $2\times 10^{67}$ years for a solar mass. Greybody factors and additional emitted species modify the coefficient; the robust large-mass scaling is $M_0^3$, and the semiclassical calculation does not determine the final quantum-gravity endpoint.

// Target: general-relativity.bigb

The time translation is normalized at infinity. For the roots $r_{\pm }^2=M\pm \sqrt{M^2-Q^2}$, evaluate $V^{\prime }(r)/2$ to get the signed horizon values. The outer Hawking temperature in natural units is $\sqrt{M^2-Q^2}/(\pi r_{+}^3)$. The extremal limit has zero surface gravity; the neutral case has only the outer positive-radius horizon and temperature $1/(2\pi \sqrt{2M})$.