Nonexpanding Horizons

A nonexpanding horizon is a null inner boundary whose intrinsic area geometry is carried unchanged along its generators. This quasi-local equilibrium condition does not require a stationary exterior or knowledge of the complete future. With the field equations and a causal energy-flux condition, Raychaudhuri’s equation upgrades vanishing expansion to vanishing shear and null energy flux, which in turn makes the induced two-metric time independent. Those consequences—and no stronger exterior symmetry—define the starting point for isolated-horizon mechanics.

Required background. Hypersurface Normals and Tangential Projectors supplies null-boundary pullbacks and screen metrics. Helpful background. Exact Solutions of Einstein’s Equations supplies stationary checks; Causality and Global Structure distinguishes null inner boundaries from event horizons; Einstein Equations turns focusing into flux restrictions; Black-Hole Geometry and Mechanics supplies the horizon hierarchy; Variational and Symplectic Geometry prepares the internal-boundary phase space used later.

Definition on a null boundary

In four-dimensional Einstein gravity, let $\Delta$ be a smooth null three-manifold with topology $S^2\times\mathbb R$. Its null generators define a future-directed normal field $\ell^a$, unique up to positive rescaling. A standard nonexpanding horizon requires:

1. every spacelike cross-section is compact and has finite area;
2. the expansion $\theta_{(\ell)}$ vanishes everywhere on $\Delta$;
3. the field equations hold at $\Delta$;
4. $-T^a{}_b\ell^b$ is future causal on $\Delta$.

The final condition implies $T_{ab}\ell^a\ell^b\ge0$. It is slightly more informative than the null energy condition along one generator and is useful for controlling flux components. Ashtekar and Krishnan state the boundary conditions and topology in Ashtekar and Krishnan 2004, §2.1.1, Definition 1.

If $\ell'^a=f\ell^a$ for a positive function $f$ on $\Delta$, then $\theta_{(\ell')}=f\theta_{(\ell)}$. Thus zero expansion is invariant under all positive rescalings, even though surface gravity and the normal-bundle connection will not be. The definition applies to the entire null hypersurface, not merely to one marginal cross-section.

Raychaudhuri eliminates shear and flux

Use a possibly nonaffine generator, $\ell^b\nabla_b\ell^a=\kappa_{(\ell)}\ell^a$. The twist of its congruence vanishes because $\ell^a$ is normal to a null hypersurface. In four dimensions Raychaudhuri’s equation is

$$\mathcal L_\ell\theta_{(\ell)} =\kappa_{(\ell)}\theta_{(\ell)} -\frac12\theta_{(\ell)}^2 -\sigma^{(\ell)}_{ab}\sigma_{(\ell)}^{ab} -8\pi G T_{ab}\ell^a\ell^b.$$

Both $\theta_{(\ell)}$ and its derivative along $\Delta$ vanish. The shear norm is nonnegative because it lives in the positive-definite screen space, and the energy term is nonnegative before the minus sign. Hence

$$\sigma^{(\ell)}_{ab}=0, \qquad T_{ab}\ell^a\ell^b=0.$$

A cosmological term does not alter this step: its null contraction vanishes, $\Lambda g_{ab}\ell^a\ell^b=0$. Thus only the non-cosmological stress tensor enters the displayed flux condition when $\Lambda g_{ab}$ is kept on the geometric side of Einstein’s equation.

The argument is pointwise: a sum of two nonpositive terms vanishes only when each vanishes. It would fail if zero expansion held only on one leaf, because $\mathcal L_\ell\theta$ could then be nonzero, or if negative null energy could balance shear. Gourgoulhon and Jaramillo give the null-hypersurface calculation and nonexpanding consequences in Gourgoulhon and Jaramillo 2006, §6.2 and §7.1.3, pp. 210–211 and 219.

The intrinsic metric is time independent

Let $q_{ab}$ be the positive-definite metric on a spacelike cross-section, pulled back to the quotient of $T\Delta$ by the null direction. The null deformation tensor decomposes as

$$\frac12\mathcal L_\ell q_{ab} =\sigma^{(\ell)}_{ab} +\frac12\theta_{(\ell)}q_{ab}.$$

Both terms on the right vanish, so

$$\mathcal L_\ell q_{ab}=0, \qquad \mathcal L_\ell\epsilon_{ab}=0,$$

where $\epsilon_{ab}$ is the area two-form. Every cross-section obtained by sliding along the generators has the same intrinsic metric and area. The horizon is in equilibrium in this precise intrinsic sense.

The result also rules out gravitational radiation crossing $\Delta$ in the usual shear-flux channel and matter energy crossing along the generators. It does not say that all components of $T_{ab}$ vanish, nor does it exclude radiation in the exterior. Boundary equilibrium is weaker than a Killing neighborhood.

Induced connection and normal-bundle data

For vector fields $X^a,Y^a$ tangent to $\Delta$, the nonexpanding conditions ensure that the spacetime derivative of $Y^a$ along $X^a$ has a well-defined tangent pullback. Denote the induced derivative by $D_a$. Because the null direction spans the degenerate kernel of the intrinsic metric,

$$D_a\ell^b=\omega_a\ell^b,$$

which defines the rotation one-form $\omega_a$. Its contraction is the surface gravity associated with this representative,

$$\kappa_{(\ell)}=\ell^a\omega_a.$$

Under $\ell'^a=f\ell^a$,

$$\omega'_a=\omega_a+D_a\ln f, \qquad \kappa_{(\ell')}=f\kappa_{(\ell)}+\mathcal L_\ell f.$$

Thus a nonexpanding horizon supplies invariant area geometry but no preferred normalization of surface gravity. Nor does it require $\omega_a$ or the full connection $D_a$ to be time independent. Weak isolation adds exactly the condition needed to control that remaining evolution.

What is proved and what is not

The implication is

$$\text{NEH conditions} \Longrightarrow \theta_{(\ell)}=\sigma^{(\ell)}_{ab} =T_{ab}\ell^a\ell^b=0 \Longrightarrow \mathcal L_\ell q_{ab}=0.$$

It does not imply that $\ell^a$ extends to a Killing field, that $\Delta$ is the event horizon, or that the exterior is stationary. Conversely, one marginally outer trapped surface is not a nonexpanding horizon: it supplies one zero expansion on one leaf, not a null tube with time-independent intrinsic metric. A dynamical horizon is spacelike and carries positive flux, so it intentionally fails the null nonexpansion conditions.

Schwarzschild supplies the equilibrium check. Its future Killing horizon at $r=2GM$ is null, has $S^2$ sections of fixed area $16\pi G^2M^2$, and has no shear or stress-energy flux. It is an NEH, but the definition itself never mentions Schwarzschild coordinates or global stationarity.

Derive the area conclusion

Assume $\theta_{(\ell)}=0$ on all of $\Delta$ and the stated energy condition. Show why nonzero shear cannot persist. Then explain why the same argument is invalid if expansion vanishes on only one cross-section.

Solution

On all of $\Delta$, both $\theta$ and $\mathcal L_\ell\theta$ vanish. Raychaudhuri then gives

$$0=-\sigma_{ab}\sigma^{ab}-8\pi G T_{ab}\ell^a\ell^b.$$

Both quantities after the minus signs are nonnegative, so each is zero. The deformation equation then gives $\mathcal L_\ell q_{ab}=0$ and hence constant area along the generators. On a single zero-expansion leaf, $\mathcal L_\ell\theta$ need not vanish; it can be negative as focusing begins. Raychaudhuri would then relate the derivative to shear and flux rather than forcing them to zero. One marginal leaf therefore does not establish an NEH.

Weakly Isolated and Isolated Horizons now adds controlled time-independence of the intrinsic connection.

References

• Ashtekar, Abhay, and Badri Krishnan. “Isolated and Dynamical Horizons and Their Applications.” Living Reviews in Relativity 7 (2004): article 10. DOI.
• Gourgoulhon, Éric, and José Luis Jaramillo. “A 3+1 Perspective on Null Hypersurfaces and Isolated Horizons.” Physics Reports 423 (2006): 159–294. DOI.