 Solution

The two identities. Set $F_{\beta \gamma }={\nabla }_{\beta }{K}_{\gamma }$. The Killing equation says $F_{\beta \gamma }=-F_{\gamma \beta }$. The Levi-Civita connection has symmetric lower connection indices, so they cancel in the antisymmetric difference:Use normalized antisymmetrization, soPut $a^{\sigma }=K^{\tau }{\nabla }_{\tau }{K}^{\sigma }$. Contracting with $F^{\beta \gamma }$, the second term is $a^{\gamma }{F}_{\gamma \alpha }$ and the third is the same after using antisymmetry. ThusRearranging proves the Killing derivative contraction identity

Restriction to a Killing horizon. On a Killing horizon, the generator $K$ is normal to the horizon as well as tangent to its null generators. Consequently $K_{[\alpha }{\nabla }_{\beta }{K}_{\gamma ]}=0$ there. One can see this without assuming hypersurface orthogonality away from the horizon: if the horizon is locally $\Phi =0$, write its dual one-form as $K=fd\Phi +\Phi q$ near it; then $K\wedge dK=0$ on $\Phi =0$. By the definition of surface gravity, $a^{\sigma }=\kappa K^{\sigma }$ on the horizon, and lowering that relation gives $K^{\sigma }{\nabla }_{\sigma }{K}_{\alpha }=\kappa K_{\alpha }$. The contraction identity becomesAway from points where $K$ vanishes, cancel a nonzero component of $K_{\alpha }$; at a regular bifurcation surface extend by continuity. ThereforeThis is surface gravity from the Killing derivative; it concerns the horizon, not every point in the exterior.

Static spherical metric. Choose the Killing vector $K={\partial }_t$. Its dual one-form is $-Adt$, and the first identity immediately giveswith all other components zero. HenceUsing the horizon identity and the simultaneous simple zeros,Here the positive value is chosen for a regular outer Killing horizon with $A,B>0$ on the exterior side. This is surface gravity of a static spherical horizon. A different normalization $K\mapsto cK$ rescales surface gravity by $c$; when an asymptotically flat normalization is wanted, the time coordinate is chosen so $A\to 1$ at infinity.

Euclidean regularity and temperature. Let $\delta =r-r_{+}$, $a=A^{\prime }(r_{+})$ and $b=B^{\prime }(r_{+})$. On the exterior side, define the proper radial coordinate $R=2\sqrt{\delta /b}$. After Wick rotation, the near-horizon metric isThe radial-time plane is a polar plane with angular coordinate $\vartheta =\kappa \tau$. Its circumference-to-radius ratio is $\kappa \Delta \tau$; the conical singularity disappears precisely whenThis is the Euclidean black-hole regularity condition. If a signed surface gravity convention is used, the period uses $|\kappa |$.

In a thermal quantum field theory, imaginary-time periodicity is the inverse-temperature condition expressed by the KMS condition. Therefore smoothness identifies the Hawking temperaturein units $\hslash =k_B=1$, or $T_H=\hslash \kappa /(2\pi k_B)$ when $\kappa$ is measured as an inverse time. It provides the thermal interpretation of surface gravity in black-hole thermodynamics. The regular Euclidean construction describes an equilibrium thermal state; the collapse calculation of Hawking radiation in the last solution supplies the outgoing spectrum. The simple-zero hypothesis excludes an extremal black hole, for which this polar-plane argument changes.