Inner Boundaries and Trapped-Surface Data

An excised inner boundary can represent a marginally outer trapped surface only when its null expansion, orientation, and constraint boundary conditions are specified. With the volume’s extrinsic-curvature sign, the outgoing condition is a nonlinear Robin condition coupling the conformal factor to the normal and tangential parts of $K_{ij}$. In time symmetry it reduces exactly to the minimal-surface condition.

Required background. Analysis for Mathematical Relativity supplies elliptic boundary theory, The Einstein Equation supplies the constraints, Hypersurface Normals and Tangential Projectors supplies boundary normals and $K_{ij}$, and Differential Equations for Relativity supplies Robin conditions.

Helpful background. Causality and Global Structure distinguishes trapped surfaces from event horizons, while PDEs and Green Functions for Gravitation reviews elliptic boundary problems.

Null expansions on an inner boundary

Let $S\subset\Sigma$ be a closed two-surface. Choose the unit normal $s^i$ in $\Sigma$ to point toward the designated exterior end, and let

$$q_{ij}=h_{ij}-s_is_j$$

be the induced two-metric. With future unit normal $n^\mu$ to $\Sigma$, choose null normals

$$\ell^\mu=n^\mu+s^\mu, \qquad k^\mu=n^\mu-s^\mu, \qquad \ell\cdot k=-2.$$

For

$$K_{ij}=-h_i{}^\mu h_j{}^\nu\nabla_\mu n_\nu, \qquad H=D_is^i,$$

the outgoing expansion is

$$\theta_+ =q^{\mu\nu}\nabla_\mu\ell_\nu =H-q^{ij}K_{ij} =H-K+K_{ij}s^is^j.$$

The other future expansion is $\theta_-=-H-K+K_{ij}s^is^j$. A marginally outer trapped surface, or MOTS, satisfies $\theta_+=0$; this condition alone imposes no sign on $\theta_-$. A future trapped surface has both future null expansions negative. Neither definition refers to an event horizon, which is a global causal boundary. An apparent horizon is normally the outer boundary of a trapped region in a chosen slice and is slicing-dependent.

Orientation cannot be suppressed. Reversing $s^i$ interchanges which null direction is called outward and sends $H\mapsto-H$; it does not leave the same $\theta_+$ equation unchanged.

Boundary conditions in the conformal method

Write $h_{ij}=\phi^4\widetilde h_{ij}$. If $\widetilde s$ is the unit normal for $\widetilde h$, the physical mean curvature is

$$H =\phi^{-2} \left(\widetilde H+4\phi^{-1} \widetilde s^i\widetilde D_i\phi\right).$$

The trace-free curvature scales as $A^{ij}=\phi^{-10}\widetilde A^{ij}$, while $K=\tau$ is prescribed. Since $s^i=\phi^{-2}\widetilde s^i$, its normal contraction is $A_{ij}s^is^j=\phi^{-6}\widetilde A_{ij}\widetilde s^i\widetilde s^j$. Multiplying $\theta_+=0$ by $\phi^3$ gives the explicit nonlinear Robin condition

$$4\widetilde s^i\widetilde D_i\phi +\widetilde H\phi +\widetilde A_{ij}\widetilde s^i\widetilde s^j\phi^{-3} -\frac23\tau\phi^3=0.$$

At an excision boundary, $s$ points into the retained exterior, opposite to the outward normal of that domain as used in integration by parts. This orientation difference must be carried into an elliptic boundary estimate. The vector-potential equation requires complementary boundary data that fix the relevant normal momentum and tangential components. Together they must satisfy the complementing condition for the coupled elliptic system.

The normalization of the null vectors does not affect the zero set of an expansion: replacing $\ell$ by $f\ell$ with positive $f$ multiplies $\theta_+$ by $f$. It does affect nonzero prescribed-expansion data, stability operators, and comparisons between surfaces, so the choice $\ell=n+s$ must still be recorded.

A theorem therefore states at least:

• which side of $S$ is retained and which normal points toward its exterior end;
• whether $S$ is MOTS, minimal, weakly outer trapped, or supplied with a prescribed expansion;
• the conformal free data and mean curvature;
• boundary conditions for both $\phi$ and the vector unknown;
• positivity, Yamabe, kernel, barrier, and regularity hypotheses;
• whether the constructed MOTS is merely the chosen boundary or is proved outermost.

The distinction between deriving an apparent-horizon boundary condition and proving solvability for admissible free data is emphasized in Bartnik and Isenberg 2004, §4.1, discussion of interior boundary conditions. Carlotto reviews the later constructions and the barriers used to obtain positive conformal factors in Carlotto 2021, §§2–3.

Time symmetry and multiple ends

For time-symmetric data $K_{ij}=0$, the momentum constraint vanishes and

$$\theta_+=H.$$

Thus a MOTS boundary is a minimal surface. The Hamiltonian constraint becomes the scalar-curvature equation $R(h)=2\Lambda+16\pi G\rho$. In vacuum with $\Lambda=0$, one seeks a scalar-flat metric with minimal boundary. Doubling smoothly across a totally geodesic boundary can sometimes replace the boundary problem by a symmetric problem on a manifold with another end, but minimality alone does not guarantee the regularity required for that doubling.

Multiple asymptotically flat ends occur naturally in puncture and bridge data. Each end has its own orientation and charges, and a minimal throat need not be the outermost MOTS relative to every end. Topology also matters: existence and outermost-surface theorems impose energy and regularity hypotheses, while the constraint equations alone do not declare an excision sphere to be a black hole.

Worked example: the Schwarzschild throat

Take $h=\psi^4(dr^2+r^2d\omega^2)$, $\psi=1+m/(2r)$, $m=GM\gt0$, and $K_{ij}=0$. For the sphere $r=b$ with normal toward increasing $r$, $\widetilde H=2/b$. The boundary equation becomes

$$4\psi'(b)+\frac{2}{b}\psi(b) =\frac{2}{b}-\frac{m}{b^2}=0,$$

so $b=m/2$. Its areal radius is $\psi(b)^2b=2m$ and its area is $16\pi m^2=16\pi G^2M^2$. Both expansions vanish on this time-symmetric minimal sphere: it is the Schwarzschild bifurcation sphere, not a strictly future-trapped sphere. This example checks the normal orientation, coefficient $4$, conformal powers, and mass dimensions in one calculation.

Check your understanding

Show that time symmetry reduces the MOTS condition to minimality, and determine what happens if the normal $s^i$ is reversed.

Solution

When $K_{ij}=0$, $\theta_+=H$, so $\theta_+=0$ is exactly $H=0$. Under $s\mapsto-s$, the mean curvature changes sign, but zero remains zero. Thus an unoriented minimal surface is still minimal. Away from time symmetry, reversing $s$ changes $K_{ij}s^is^j$ only quadratically but changes $H$ and exchanges the outward null direction, so one must recompute the appropriate expansion rather than reuse the same label.

References

• Bartnik, Robert, and Jim Isenberg. “The Constraint Equations.” In The Einstein Equations and the Large Scale Behavior of Gravitational Fields, 1–38. Birkhäuser, 2004. arXiv:gr-qc/0405092.
• Carlotto, Alessandro. “The General Relativistic Constraint Equations.” Living Reviews in Relativity 24 (2021): 2. DOI.