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For a Killing vector field, set $F_{ab}={\nabla }_a{K}_b$ and $a^s=K^t{\nabla }_t{K}^s$. The Killing equation makes $F$ antisymmetric, and expansion of normalized three-index antisymmetrization gives the displayed identity. On a Killing horizon, hypersurface orthogonality removes its first right-hand term and relates $F^{ab}{F}_{ab}$ to surface gravity.

For a non-null Killing vector field $V$ of a Levi-Civita connection, hypersurface orthogonality implies that $\eta =V^{♭}/F$, with $F=g(V,V)$, is a closed differential form. The Killing equation and $V^{♭}\wedge d{V}^{♭}=0$ give $2F{\nabla }_{\mu }{V}_{\nu }=V_{\nu }{\partial }_{\mu }F-V_{\mu }{\partial }_{\nu }F$, whence $d\eta =0$. The Poincaré lemma gives $\eta =d\varphi$ locally. For a timelike $V$, $\varphi$ supplies local static time coordinates. This does not assert global exactness.

A nonzero vector field has hypersurface orthogonality when it is locally normal to a family of hypersurfaces. In terms of its associated differential one-form $K^{♭}=g_{\mu \nu }{K}^{\nu }d{x}^{\mu }$, the Frobenius theorem gives the equivalent conditionslocally with nonzero $f$ and $dh$; square brackets denote tensor index antisymmetrization.

All coefficients of the Reissner-Nordstrom metric are independent of $t$, and $K={\partial }_t$ has constant coordinate components. Thus the Lie derivative obeys ${\mathcal{L}}_{K}g={\partial }_{t}g=0$: $K$ is a Killing vector field. Moreover,Thus $K$ has hypersurface orthogonality, with the $t=\text{constant}$ hypersurfaces providing the orthogonal family wherever this chart is regular. It is timelike where $F>0$, since $g(K,K)=-F$. Its hypersurface orthogonality also holds in a regular extension across a horizon; singular static coordinates do not invalidate the Frobenius theorem test.

For relativistic dust, $p=0$, the relativistic Euler equation gives $\rho a_b=0$. The nowhere-vanishing density therefore makes the four-acceleration zero: $u$ is a timelike geodesic vector field. Let $\alpha =u^{♭}$ be the differential one-form obtained by the musical isomorphism. Its exterior derivative satisfiesHere ${\iota }_u$ is the interior product. By hypersurface orthogonality and the Frobenius theorem, $\alpha \wedge d\alpha =0$. Taking the interior product and using $\alpha (u)=-1$ givesThus $\boxed{d{u}^{♭}=0}$. By the Poincaré lemma, locally $\alpha =-dt$, and then $u(t)=-\alpha (u)=1$. Choose spatial coordinate functions $x^i$ on one level set of $t$ and extend them constantly along the integral curves of a vector field $u$, using its smooth flow. This constructs $u={\partial }_t$. Since $dt({\partial }_i)=0$,The metric on the orthogonal complement of a timelike vector is positive-definite, so the synchronous coordinates for irrotational dust haveThis construction is local; crossing geodesics can prevent these synchronous coordinates from covering a larger region.

Work on a regular level surface, so $d\beta \ne 0$. A nonzero normal covector has the local form $n_a=f{\nabla }_a\beta$. Extend it as a normal to neighbouring level surfaces. Since the Levi-Civita connection is torsion-free,Antisymmetrizing the product with $n_c=f{\beta }_{;c}$ over three indices kills the symmetric second-derivative term and the term containing two copies of $d\beta$. HenceEquivalently $n\wedge dn=0$, the Frobenius theorem condition for hypersurface orthogonality.

For the second covariant derivative of a Killing vector, the curvature sign needs explicit care. Denote the curvature defined on the cover by $C^a{}_{bcd}$, and obtain its other index positions by ordinary raising and lowering with the metric tensor. With $D_{abc}=k_{a;bc}$, the Killing equation gives $D_{abc}=-D_{bac}$, while the supplied Ricci identity givesSuccessive use of these two facts givesSubstitute $D_{acb}=D_{abc}-C_a{}^d{}_{bc}{k}_d$. The curvature symmetries and first Bianchi identity then implyThus the identity consistent with the cover convention isThe plus-sign version is valid for the opposite curvature tensor $\tilde{R}=-C$. With ordinary metric index raising, the printed plus sign and the cover's curvature convention cannot both be used. This is the curvature-sign convention in Killing derivative identities, not a failure of the Killing equation.

An explicit sign check uses $d{s}^2=e^{2x}d{t}^2-d{x}^2-d{y}^2-d{z}^2$ and the timelike Killing vector $k={\partial }_t$. Here $k_t=e^{2x}$, ${\Gamma }^t{}_{tx}=1$ and ${\Gamma }^x{}_{tt}=e^{2x}$. Direct differentiation gives $k_{t;xx}=e^{2x}$, whereas the cover definition gives $C^t{}_{xxt}=-1$ and $C_{txx}{}^d{k}_d=-e^{2x}$. This also verifies the corrected sign without relying on a remembered curvature convention.

To prove the static Killing field is a Ricci eigenvector, use the permitted alternative of adapted coordinates. Hypersurface orthogonality supplies hypersurfaces orthogonal to $k$. Choose spatial coordinates on one such hypersurface and carry them along the flow of the Killing vector, using its flow parameter as $t$. The flow preserves both the metric tensor and $k$, so it preserves orthogonality to $k$. Consequently the mixed metric components vanish, and the Killing equation makes all coefficients independent of $t$:The only nonzero Christoffel symbols involving time can bein particular ${\Gamma }^0{}_{ij}={\Gamma }^i{}_{0j}={\Gamma }^0{}_{00}=0$. In $C_{0i}$, the derivative of the connection trace vanishes because ${\Gamma }^a{}_{0a}=0$; the other derivative is either a time derivative of a static coefficient or a derivative of ${\Gamma }^j{}_{0i}=0$. Every connection-product term contains a vanishing coefficient of these same types. Thus $C_{0i}=0$. The one-form $C_{ab}{k}^b=C_{a0}$ has only a time component, and is proportional to $k_a$. ThereforeChanging the overall curvature sign leaves this zero identity unchanged.

Finally a perfect fluid in signature $(+---)$ has $T_{ab}=(\rho +p)u_a{u}_b-p{g}_{ab}$ and $u^a{u}_a=1$. Contract the Einstein field equations with $k^b$, and antisymmetrize with $k_a$. Terms proportional to $g_{ab}$ disappear, as does the Ricci term just proved. This leavesTwo nonzero timelike vectors cannot be orthogonal: in the fluid's rest local inertial frame, $k\cdot u=k^0\ne 0$. The hypothesis $\rho +p>0$ therefore forces $u_{[a}{k}_{b]}=0$. Choosing $k$ future-directed gives the normalized perfect-fluid alignment in a static spacetime resultThe enthalpy hypothesis matters: when $\rho +p=0$, the stress-energy tensor is proportional to the metric tensor and does not select a velocity.

The Frobenius theorem says that hypersurface orthogonality of the nonzero covector $U_a$ is equivalent to $U_{[a}{\nabla }_b{U}_{c]}=0$, or locally $U_a=h{\nabla }_{a}S$ for some nonzero $h$. This immediately makes the screen projection of ${\nabla }_{[b}{U}_{a]}$ zero, since its antisymmetric part is a wedge product with $\nabla S$, which is proportional to $U$. Thus hypersurface orthogonality gives zero null twist.

For the converse, put $B_{ab}={\nabla }_b{U}_a$. Nullness gives $U^a{B}_{ab}=\frac{1}{2}{\nabla }_b(U^2)=0$, and the affine geodesic equation gives $B_{ab}{U}^b=0$. Resolve both tensor slots in the basis consisting of $U,N$ and the screen. These two identities exclude covector factors of $N$. Consequentlyfor some covector $q$. If $\hat{\omega }=0$, wedging this identity with $U$ makes $U_{[a}{\nabla }_b{U}_{c]}=0$. Applying Frobenius theorem completes the proof:This is the null twist vanishes exactly for hypersurface-orthogonal geodesics criterion; the geodesic assumption ensures that the screen criterion implies the full Frobenius condition. The resulting hypersurfaces are null, and the congruence follows their null generators.

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.

Write $F=V_{\rho }{V}^{\rho }<0$ and $A_{\mu \nu }={\nabla }_{\mu }{V}_{\nu }$. The Killing equation makes $A_{\mu \nu }$ antisymmetric. Expanding hypersurface orthogonality and contracting with $V^{\alpha }$ givesHere metric compatibility gives ${\partial }_{\mu }F=2V^{\rho }{A}_{\mu \rho }$. Thereforeand the normalized Killing one-form satisfiesThus the required exponent is $\boxed{n=-2}$. The notation $|V{|}^2$ in this paper denotes the negative scalar $F$, rather than a positive squared norm; the answer uses only its real reciprocal. More generally, for a multiplier $h(F)$ the antisymmetric derivative is $(h/F+h^{\prime })(V_{\nu }{\partial }_{\mu }F-V_{\mu }{\partial }_{\nu }F)$, so $h=C/F$ is the universally valid choice. If $F$ is constant, other exponents can work in that particular metric.

Use an affine parameter $\lambda$ along a generator of a null hypersurface. Its null twist vanishes by hypersurface orthogonality. The screen metric is positive definite, so ${\hat{\sigma }}^{ab}{\hat{\sigma }}_{ab}\ge 0$. The Einstein field equations and null energy condition imply $R_{ab}{U}^a{U}^b\ge 0$, giving from the Null Raychaudhuri equationStarting with ${\theta }_0<0$, the null expansion stays negative as long as the congruence remains regular. Set $z=-1/\theta >0$. Then $z^{\prime }={\theta }^{\prime }/{\theta }^2\le -1/2$, soA finite negative null expansion cannot persist to $\lambda -{\lambda }_0=2/|{\theta }_0|$. Thus the null focusing theorem givesThis is a focal breakdown of the transverse congruence, provided the generator and regular geometry extend that far; it is not by itself a curvature singularity or a statement that the individual null geodesic terminates there.

Differentiate the normalization of the timelike unit normal with the metric-compatible covariant derivative:Thus the derivative of the normal is already orthogonal to the normal in its second index. Expanding the second spatial projection tensor in the definition of the extrinsic curvature of a spatial hypersurface givesTo prove symmetry using Frobenius theorem, set $F_{\alpha \beta }={\nabla }_{\alpha }{n}_{\beta }-{\nabla }_{\beta }{n}_{\alpha }$. Hypersurface orthogonality is equivalent toContract this with $n^{\alpha }{P}^{\beta }{}_{\mu }{P}^{\lambda }{}_{\nu }$. The last two terms vanish because a projected normal vanishes, while $n^{\alpha }{n}_{\alpha }=-1$ in the first term. HenceThe antisymmetric part of the doubly projected covariant derivative of $n$ is therefore zero. This proves that hypersurface orthogonality implies symmetric extrinsic curvature:No geodesic assumption for the normal congruence is needed; its normal acceleration may be nonzero.

A timelike Killing vector with hypersurface orthogonality admits local coordinates with $k={\partial }_t$ and $d{s}^2=N(\mathbf{x}{)}^{2}d{t}^2-h_{ij}(\mathbf{x})d{x}^{i}d{x}^j$. The only connection coefficients with time indices that can be nonzero are ${\Gamma }_{0i}^0={\partial }_i\log N$ and ${\Gamma }_{00}^i=N{h}^{ij}{\partial }_{j}N$. Substitution into the Ricci tensor gives $R_{0i}=0$. Thus $R_{ab}{k}^b$ has only its time component and is proportional to $k_a$. Equivalently $k_d{R}^d{}_{[a}{k}_{b]}=0$. This local statement does not require the Vacuum Einstein equations and is independent of the overall curvature-sign convention.

For relativistic dust with nowhere-zero density, stress-energy conservation implies zero four-acceleration. If its normalized four-velocity $u$ has hypersurface orthogonality, then $u^{♭}\wedge d{u}^{♭}=0$ and ${\iota }_{u}d{u}^{♭}=0$. The interior product of the first identity gives $d{u}^{♭}=0$. The Poincaré lemma supplies $u^{♭}=-dt$ locally; spatial coordinate functions constant along the smooth flow of $u$ give $u={\partial }_t$, $g_{tt}=-1$ and $g_{ti}=0$. The resulting synchronous coordinates describe the dust until the local coordinate construction fails.