Normal Killing Fields and Null Generators

A Killing horizon is where a continuous spacetime symmetry becomes a null normal. That single statement has a strong consequence: the symmetry orbits on the horizon are precisely its null geodesic generators, though the Killing parameter need not be affine. The result is local and geometric. It neither requires an event horizon nor says that every event horizon is stationary.

Required background. Killing Fields and Isometry Groups supplies the Killing equation and the relation between vector fields and one-parameter isometry groups.

Helpful background. Exact Solutions of Einstein’s Equations supplies the Schwarzschild and Kerr examples; Causality and Global Structure supplies global horizon language; The Einstein Equation clarifies which later laws use dynamics; Black-Hole Geometry and Mechanics locates Killing horizons among the other horizon notions.

From a null normal to a null generator

Let a smooth hypersurface $\mathcal H$ be given locally by $u=0$, with $du\neq0$ there. Every normal covector has the form

$$n_a=F\nabla_a u,$$

where $F$ is nowhere zero. If $\mathcal H$ is null, then $n^a n_a=0$ on $\mathcal H$. Its normal vector is consequently tangent as well, because

$$n^a\nabla_a u=F^{-1}n^a n_a=0.$$

This is the distinctive geometry of a null hypersurface. The induced metric has a one-dimensional null direction, and $n^a$ spans it. Integral curves of $n^a$ therefore remain in $\mathcal H$.

They are also null geodesics as unparametrized curves. Because $n=F\,du$, hypersurface orthogonality gives $dn=d(\ln F)\wedge n$. Contracting this identity with $n^a$ gives

$$n^b\nabla_b n_a-\frac12\nabla_a(n^2) =\bigl(n\cdot d\ln F\bigr)n_a \quad\text{on }\mathcal H.$$

The scalar $n^2$ vanishes everywhere on $\mathcal H$, so its gradient there is normal to $\mathcal H$ and hence proportional to $n_a$. Raising the remaining index therefore yields

$$n^b\nabla_b n^a=\kappa_{(n)}n^a.$$

Thus the acceleration has no component transverse to the tangent. The scalar $\kappa_{(n)}$ measures non-affinity, not a force deflecting the ray. A rescaling $n'^a=f n^a$ changes it according to

$$\kappa_{(n')}=f\left(\kappa_{(n)}+n^a\nabla_a\ln f\right).$$

Choosing $f$ along each generator can set the right-hand side to zero and produce an affine tangent. This construction and transformation law are derived directly in Gourgoulhon and Jaramillo 2006, arXiv v2, §2.5, pp. 21–24.

When the normal is a Killing field

A Killing horizon is a smooth null hypersurface $\mathcal H$ for which a nonzero Killing field $\chi^a$ is normal. On the horizon,

$$\chi^2=0, \qquad \chi^b\nabla_b\chi^a=\kappa\chi^a.$$

The first equation says that the symmetry becomes null. Normality says more: it identifies the symmetry orbits with the unique null direction in the tangent space of $\mathcal H$. The Killing equation then converts the non-affine geodesic equation into the useful gradient identity

$$\nabla_a(\chi^2)=-2\kappa\chi_a \quad\text{on }\mathcal H.$$

Indeed, $\chi^b\nabla_b\chi_a=-\chi^b\nabla_a\chi_b =-\tfrac12\nabla_a(\chi^2)$. No Einstein equation has entered. This is why Killing-horizon kinematics applies in any metric theory admitting the stated smooth symmetry.

The normalization is restricted more strongly than for a generic null normal. Multiplying a Killing field by a nonconstant function generally destroys the Killing equation. A constant rescaling $\chi'^a=c\chi^a$ remains Killing and gives $\kappa'=c\kappa$. Surface gravity is therefore not an absolute number until the symmetry parameter has been normalized.

Null is necessary but normality is decisive

Finding the zero set of a Killing norm does not by itself identify a Killing horizon. The candidate set must be a smooth null hypersurface, and the Killing field must span its null normal direction. This distinction is visible in Kerr: the stationary field $t^a$ becomes null on the stationary-limit surface, but that surface is generally not a Killing horizon of $t^a$. The event horizon is instead generated by

$$\chi^a=t^a+\Omega_H\phi^a,$$

where $\phi^a$ generates rotations and $\Omega_H$ is constant on the horizon. The explicit Kerr calculation is given in Gourgoulhon and Jaramillo 2006, arXiv v2, Appendix D.3, pp. 205–208. This is also the stationary generator used in the classical formulation of black-hole mechanics Bardeen, Carter, and Hawking 1973, pp. 161–170.

Two counterexamples keep the definition honest. A dynamical event horizon need not admit any Killing field, so it need not be a Killing horizon. Conversely, the Rindler horizons of a Lorentz boost in Minkowski spacetime are Killing horizons but are not black-hole event horizons relative to future null infinity. “Killing” describes symmetry; “event” describes causal escape.

Checks in Schwarzschild and Kerr

For Schwarzschild in ingoing Eddington–Finkelstein coordinates,

$$ds^2=-f(r)\,dv^2+2\,dv\,dr+r^2d\Omega_2^2, \qquad f(r)=1-\frac{2GM}{r}.$$

The field $\chi^a=(\partial_v)^a$ is Killing and has norm $-f$. At $r_H=2GM$, the normal covector to the level set is $dr$, while raising its index gives $(dr)^a=(\partial_v)^a$ on the horizon. Thus nullness, normality, and tangency agree. The gradient identity gives $\kappa=f'(r_H)/2=1/(4GM)$ for the normalization agreeing with unit time translation at infinity. The same regular-coordinate check appears in Gourgoulhon and Jaramillo 2006, arXiv v2, Example 2.5, pp. 24–28.

For four-dimensional Kerr, write

$$r_+=GM+\sqrt{(GM)^2-a^2}, \qquad \Omega_H=\frac{a}{r_+^2+a^2},$$

with $a=J/M$ and $|a|\le GM$. Direct evaluation on $r=r_+$ shows that $t^a+\Omega_H\phi^a$ is null and normal there. The $a\to0$ limit gives $\Omega_H\to0$ and recovers the Schwarzschild generator. This limit tests both the linear combination and its normalization.

Diagnose the candidate surface

Suppose a Killing field $k^a$ satisfies $k^2=0$ on a smooth three-surface $S$. What additional calculation decides whether $S$ is a Killing horizon of $k^a$?

Solution

First verify that $S$ itself is null. Then choose any nonzero normal covector $s_a$ to $S$ and test whether $k_a$ is proportional to $s_a$ everywhere on $S$. Equivalently, check that $k^a$ lies in the one-dimensional radical of the induced metric. The norm equation alone is insufficient: a Killing field may be null while tangent to a timelike surface, as the stationary field is on a generic Kerr stationary-limit surface. Once proportionality holds, the Killing equation implies $k^b\nabla_bk^a=\kappa k^a$, so its orbits are the null generators.

Surface Gravity and Normalization next compares the equivalent definitions of $\kappa$ and separates geometric rescaling from the physical choice of asymptotic time.

References

• Bardeen, James M., Brandon Carter, and Stephen W. Hawking. “The Four Laws of Black Hole Mechanics.” Communications in Mathematical Physics 31 (1973): 161–170. DOI.
• Gourgoulhon, Éric, and José Luis Jaramillo. “A 3+1 Perspective on Null Hypersurfaces and Isolated Horizons.” Physics Reports 423 (2006): 159–294. DOI. Open PDF.