Apparent-Horizon Finding

An apparent-horizon finder turns a geometric condition on a spatial slice into a nonlinear two-dimensional boundary-value problem. Its output is useful only when the surface is the intended marginally outer trapped surface, not merely a root of a discretized equation. This page develops that distinction through the expansion residual, branch searches, analytic black-hole benchmarks, and resolution tests.

Helpful background. Formulations of General Relativity supplies the hypersurface variables and constraints; Differential Equations for Relativity supplies nonlinear boundary-value methods; Black-Hole Geometry and Mechanics distinguishes trapping horizons from event horizons; Gravitational Radiation and Waveforms explains why horizon data complement radiation; and Computational Relativity provides the volume-wide verification framework.

The marginal-surface equation

Let $n^a$ be the future-directed unit normal to a spatial slice $\Sigma$, and let $s^a$ be the outward unit normal within $\Sigma$ to a closed two-surface $S$. We use

$$K_{ij}=-h_i{}^a h_j{}^b\nabla_a n_b,$$

where $h_{ab}=g_{ab}+n_a n_b$. With the outgoing null direction proportional to $\ell^a=n^a+s^a$, the expansion vanishes when

$$\Theta_{(\ell)}=D_i s^i+K_{ij}s^i s^j-K=0.$$

Rescaling $\ell^a$ by a smooth positive function rescales the expansion but does not change its zero set. The equation is slice dependent: a solution is a marginally outer trapped surface (MOTS) on the supplied slice, not an event horizon located by local computation. In the usual numerical terminology the apparent horizon is the outermost MOTS bounding the trapped region. A local expansion or stability test alone does not establish global outermostness. The finder must also define “outward,” surface topology, and the region searched. For the usual asymptotically flat binary problem one seeks smooth surfaces of spherical topology, but neither topology nor existence should be silently assumed.

A common representation is a star-shaped surface $r=h(\theta,\phi)$ about a chosen center. Substitution makes $\Theta_{(\ell)}[h]=0$ a nonlinear elliptic equation on the sphere. Spherical-harmonic coefficients, finite differences on angular patches, or a triangulated surface can represent $h$. Star-shaped coordinates are efficient for mildly distorted horizons but can fail for a badly chosen center or a surface that intersects a coordinate ray more than once. Thornburg explains both the geometric equation and major algorithm families in Thornburg 2007, §§3.1.1–3.1.6 and §§3.2.3–3.2.7.

Solving without losing a branch

Newton iteration linearizes the discrete residual $F_A(h)=\Theta_{(\ell)}(x_A)$,

$$J_{AB}\,\delta h_B=-F_A, \qquad h_B\leftarrow h_B+\lambda\,\delta h_B,$$

with damping $0<\lambda\leq1$. A small final norm $\lVert F\rVert_2$ certifies only that the iteration found a discrete root. The initial guess and continuation strategy determine which basin of attraction is reached. In a binary, two individual horizons, an outer common horizon, and an inner common branch may coexist; a single search from one initial sphere can therefore miss physically relevant surfaces.

A robust search combines several controls. Continue each previously found surface to the next time slice. Seed additional searches around each compact object and around the system barycenter. Vary initial radius and center rather than accepting the first converged root. After finding a candidate, evaluate the expansion on a finer surface grid using interpolation and derivative operators independent of those in the nonlinear solve. Check that the outward expansion changes in the expected direction under small normal deformations when outermostness matters. These practices make explicit why the flow and elliptic methods tested in Schnetter 2003, §§2–4 and Tables 1–5 solve related but not identical numerical problems.

Benchmarks and convergence checks

Keep $c=1$ and distinguish physical charges from their geometrized lengths by $M_{\mathrm{geom}}=G M_{\mathrm{phys}}$ and $J_{\mathrm{geom}}=G J_{\mathrm{phys}}$. A time-symmetric Schwarzschild slice in isotropic radius has $K_{ij}=0$, horizon coordinate radius $r=M_{\mathrm{geom}}/2$, and physical area

$$A_{\mathrm{H}}=16\pi M_{\mathrm{geom}}^2.$$

This test separates three errors: the coordinate location error, the surface-equation residual, and the area-integration error. They need not converge at the same order. For Kerr with dimensionless spin $a_*=J_{\mathrm{geom}}/M_{\mathrm{geom}}^2=J_{\mathrm{phys}}/(G M_{\mathrm{phys}}^2)$, the exact area is

$$A_{\mathrm{H}}=8\pi M_{\mathrm{geom}}^2\left(1+\sqrt{1-a_*^2}\right).$$

Vary spin and horizon orientation so that a coordinate-aligned spherical answer cannot pass accidentally. If an angular discretization has formal order $p$, compare at least three resolutions $N$, $2N$, and $4N$. Report the observed order from a norm of the surface difference or invariant error; do not infer convergence from three visually overlapping shapes. The residual should be recomputed after interpolation from the three-dimensional evolution grid, because coarse volume data can cap the surface convergence even when the angular solver is spectral.

For a dynamical or binary test without an analytic surface, convergence alone is not independent evidence. Compare two surface representations or two implementations, and check geometric relations: positivity and smoothness of the induced metric, closure of the surface, consistency of area computed by two quadratures, and stable ordering of nested branches. A finder that produces the correct Schwarzschild area but misses a common binary horizon has passed a unit test, not the target scientific claim.

Deriving the Schwarzschild test

Write $m=GM_{\mathrm{phys}}$ and $h_{ij}dx^i dx^j=\psi^4(dr^2+r^2d\Omega^2)$ with $\psi=1+m/(2r)$ for $r\gt0$. A coordinate sphere has $s^r=\psi^{-2}$ and $\sqrt{h}=\psi^6r^2\sin\theta$. Since $K_{ij}=0$,

$$\begin{aligned} \Theta_{(\ell)} &=\frac{1}{\psi^6r^2}\frac{d}{dr}(\psi^4r^2)\\ &=\frac{2}{r\psi^2} \frac{1-m/(2r)}{1+m/(2r)}. \end{aligned}$$

The root is $r=m/2$. There $\psi=2$, so the areal radius is $R=\psi^2r=2m$ and $A=4\pi R^2=16\pi m^2$. Thus the coordinate radius and the physical radius differ by a factor of four at the same horizon. The expansion is positive just outside and negative just inside relative to this chosen asymptotic end. These three independently calculable outputs—root, sign change, and area—make a stronger elementary test than area alone.

A reproducible horizon record

The environment record should pin the spacetime-data format and checksum, interpolation library, angular representation, nonlinear and linear solvers, floating-point precision, tolerances, and software revision. Inputs include the slice time, coordinate chart, candidate centers and radii, maximum harmonic degree or angular grid, topology assumption, and the sign of $K_{ij}$. Machine-readable outputs should contain surface coefficients or vertices, $\lVert\Theta_{(\ell)}\rVert$ in named norms, iteration history, area, and branch labels.

A clean future rerun sequence would ingest a versioned Schwarzschild or Kerr slice, execute a declared sweep such as $N=24,48,96$, evaluate the surface with an independent residual operator, and emit a table of radius, area, residual, and observed order. This page specifies that interface; it does not claim that a repository program has performed the calculation.

Failure criteria must be fixed before looking at a production result. Reject a surface if the nonlinear residual stagnates, if its independent residual fails to decrease under refinement, if the area or coordinate location does not approach the analytic value, if the result changes discontinuously under modest initial-guess variations without a resolved branch explanation, or if interpolation error dominates. Near merger, do not discard an inner common branch merely because it is not outermost; label branches geometrically and track them separately.

Check your understanding

A binary evolution produces a candidate common surface with $\lVert\Theta_{(\ell)}\rVert_2=10^{-12}$ on every angular grid. Doubling the three-dimensional grid resolution moves its area by $4\%$, while changing the initial trial radius produces a second, larger root. Is the first number sufficient evidence for a resolved outer apparent horizon?

Solution

No. The tiny norm is the residual of the discrete angular solve, so it says neither that the volume interpolation is resolved nor that the selected root is outermost. The area change shows a material three-dimensional resolution error, and the second root demonstrates branch dependence. Continue both roots, evaluate each expansion with an independent operator, refine the volume and surface grids together, and use outward-deformation or nesting information to identify the outer branch. Only then can the outer-horizon claim be attached to a quantitative uncertainty.

Continue with Quasi-Local Mass, Spin, and Multipoles.

References

• Schnetter, Erik. 2003. “A Fast Apparent-Horizon Algorithm.” Classical and Quantum Gravity 20, 4719–4737. DOI.
• Thornburg, Jonathan. 2007. “Event and Apparent Horizon Finders for 3+1 Numerical Relativity.” Living Reviews in Relativity 10, 3. DOI.