Killing Horizons

A Killing horizon is a null hypersurface generated by a spacetime symmetry: some Killing field becomes both tangent and normal to it. This local geometric condition supplies a preferred generator and a natural surface gravity, but it does not by itself say that signals fail to reach infinity. In familiar stationary black holes the Killing and event horizons coincide only after the asymptotic region, generator normalization, and global causal structure have also been fixed.

Helpful background. Exact Solutions of Einstein’s Equations supplies Schwarzschild and Kerr; Causality and Global Structure distinguishes observer and event horizons; The Einstein Equation governs the examples; Timelike Congruences and Spatial Projection supplies expansion; Hypersurface Normals and Tangential Projectors supplies normal geometry; and Black-Hole Geometry and Mechanics gives the wider mechanics route.

Null hypersurfaces generated by a symmetry

Let $\chi^a$ be a nonzero Killing field,

$$\nabla_{(a}\chi_{b)}=0.$$

A null hypersurface $\mathcal H$ is a Killing horizon of $\chi^a$ if $\chi^a$ is normal to $\mathcal H$. A null normal is also tangent, so the Killing orbits on $\mathcal H$ are its null generators. Equivalently, $\chi^2=g_{ab}\chi^a \chi^b$ vanishes on $\mathcal H$ and, away from points where $\chi^a=0$, its gradient is normal to the same surface. The field need not be hypersurface orthogonal away from the horizon; rotating Killing horizons are the important counterexample.

Because the norm is constant on the horizon, there is a function $\kappa$ on $\mathcal H$ such that

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

Contracting the Killing equation gives the equivalent nonaffine geodesic equation

$$\chi^b\nabla_b\chi^a=\kappa\chi^a.$$

Thus $\kappa$ measures the failure of the symmetry parameter along a generator to be affine. Constancy of $\kappa$ over a connected component is the content of a zeroth-law theorem under additional hypotheses; it is not part of this definition. The null-hypersurface derivation and its relation to isolated horizons are developed in Gourgoulhon and Jaramillo 2006, §§2–5 and §7.

Normalization is physical data

If $\chi'^a=c\chi^a$ for a positive constant $c$, then the same hypersurface and generator curves are obtained, but

$$\kappa'=c\kappa.$$

Surface gravity is therefore not a number attached to a bare null surface. In an asymptotically flat static spacetime, the standard normalization chooses $\chi^2\to-1$ at the selected infinity. In a stationary axisymmetric black hole, the future horizon generator is instead

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

where $t^a$ is the asymptotically normalized time translation and $\phi^a$ has $2\pi$-periodic orbits. These choices fix the scale of $\chi^a$ and hence of $\kappa$. Spacetimes without a preferred asymptotic time require some other normalization prescription; comparing two quoted surface gravities without checking that prescription is meaningless.

A position-dependent rescaling is more delicate. It preserves the null direction but generally destroys the Killing equation, so it does not define another normal Killing field. Isolated-horizon formalisms deliberately use specified equivalence classes of null normals instead; that is a different geometric structure.

Schwarzschild calculation in a regular chart

Use ingoing Eddington–Finkelstein coordinates for the four-dimensional vacuum Schwarzschild geometry,

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

with physical mass $M\gt0$. The stationary Killing field is $\chi^a=(\partial_v)^a$, and its squared norm is $\chi^2=-f$. At $r_H=2GM$, $\chi^2=0$. Moreover,

$$\chi_a\,dx^a=-f\,dv+dr \quad\Longrightarrow\quad \chi_a\,dx^a=dr\quad\text{on }r=r_H,$$

so $\chi^a$ is normal to the null surface $r=r_H$. Applying the norm-gradient definition gives

$$-f'(r_H)\,dr=-2\kappa\,dr, \qquad \kappa=\frac12 f'(r_H)=\frac{1}{4GM}.$$

This computation also checks the sign: the future generator normalized at the right-hand infinity has positive $\kappa$. The usual Schwarzschild $t,r$ chart obscures the normal calculation because it degenerates at $r=2GM$; that coordinate failure is not a geometric singularity.

Rotation selects a different generator

For Kerr with $M\gt0$, write $a=J/M$ and

$$r_\pm=GM\pm\sqrt{G^2M^2-a^2}.$$

The outer horizon $r=r_+$ is generated by

$$\chi^a=t^a+\Omega_H\phi^a, \qquad \Omega_H=\frac{a}{r_+^2+a^2},$$

and its surface gravity is

$$\kappa=\frac{r_+-r_-}{2(r_+^2+a^2)}.$$

For nonzero spin, the asymptotic time translation $t^a$ alone is not null on the whole horizon. It becomes null at the static limit, whose equatorial location differs from $r_+$. The correct linear combination is selected by requiring the symmetry field to be normal to the horizon. At $a=0$, by contrast, $\chi^a=t^a$ and the static limit coincides with the horizon at every angle. The limits provide independent checks: $a\to0$ gives $r_+=2GM$ and $\kappa=1/(4GM)$, while $|a|\to GM$ gives $r_+=r_-$ and $\kappa\to0$. Booth compares Killing and other horizon notions in Booth 2005, §2.3 and §5, pp. 1081–1083 and 1092–1095.

Expansion along the Killing generators

The symmetry also explains why a regular Killing horizon is nonexpanding. On a cross-section with induced metric $q_{ab}$,

$$\theta_{(\chi)} =\frac12q^{ab}\mathcal L_\chi q_{ab}.$$

Because $\chi^a$ generates an isometry, its flow preserves the spacetime metric and hence the projected cross-sectional metric. Therefore $\theta_{(\chi)}=0$; the trace-free part of the same deformation vanishes, so the generator shear is zero as well. This argument applies on regular portions where $\chi^a$ is a nonzero null generator. At a bifurcation surface $\chi^a=0$, one reaches the result by continuity from either horizon branch. The calculation shows why stationary Killing horizons naturally provide nonexpanding examples, but it still does not supply the global causal identification required for an event horizon.

Why a Killing horizon need not be an event horizon

In Minkowski spacetime, the boost field

$$\chi=x\,\partial_t+t\,\partial_x$$

is Killing and becomes null on the planes $x=\pm t$. Those planes are Killing horizons bounding Rindler wedges for uniformly accelerated observers. Yet Minkowski spacetime has no black-hole region relative to its complete future null infinity. This is the clean counterexample to “Killing horizon implies black-hole event horizon.”

The reverse implication also fails: a black hole produced by nonspherical collapse can have a perfectly well-defined event horizon while no Killing field exists during its dynamical era. Coincidence requires a global event horizon together with an appropriate stationary symmetry and regularity conditions. Rigidity theorems can provide that symmetry in restricted classes of stationary black holes; the definition of a Killing horizon cannot.

Check your understanding

Suppose a source uses $\widetilde\chi^a=2\chi^a$ for the Schwarzschild horizon. What happens to the horizon, its generator curves, and the quoted surface gravity? Would the rescaled field still have unit norm at infinity?

Solution

The set $r=2GM$ and the unparametrized generator curves do not change. The norm gradient equation gives $\widetilde\kappa=2\kappa=1/(2GM)$. However, $\widetilde\chi^2\to-4$ rather than $-1$ at infinity, so the rescaled value is not the standard asymptotically normalized Schwarzschild surface gravity. A surface-gravity comparison must therefore carry the generator normalization.

Next, Apparent Horizons and Marginally Trapped Surfaces replaces symmetry by the null expansions of a two-surface on a chosen slice.

References

• Booth, Ivan. “Black-Hole Boundaries.” Canadian Journal of Physics 83 (2005): 1073–1099. arXiv:gr-qc/0508107.
• Gourgoulhon, Éric, and José Luis Jaramillo. “A 3+1 Perspective on Null Hypersurfaces and Isolated Horizons.” Physics Reports 423 (2006): 159–294. DOI.