Hyperboloidal Layers and Cauchy Matching

A hyperboloidal layer leaves a central Cauchy coordinate system unchanged and transforms only an exterior annulus into compactified hyperboloidal coordinates. The interface is not a physical boundary: it is a change of coordinates and, when conformal compactification is used, a change of metric representative. Smooth matching therefore requires the coordinate map, metric, evolved fields, and gauge data to describe one solution on both sides.

Required background. The Riemann Curvature Tensor supplies the tensorial transformation rules needed at the layer interface.

Helpful background. We use The Einstein Equation, Hypersurface Normals and Tangential Projectors, Formulations of General Relativity, and Causality and Global Structure.

A smooth exterior layer

Begin with physical Minkowski spacetime on a radial half-line,

$$d\widetilde s^2=-dt^2+dx^2+x^2d\omega_2^2.$$

Choose an interface $R$ and a finite outer coordinate $S\gt R$. In the interior $0\le\rho\le R$, set $x=\rho$, $\tau=t$, and $\Omega=1$. In the layer let $q=\rho-R$, $W=S-R$, and choose

$$\Omega(\rho)=1-\frac{q^2}{W^2}, \qquad x=R+\frac q\Omega, \qquad R\lt \rho\lt S.$$

The boundary $\rho=S$ represents infinity because $\Omega(S)=0$ and $\Omega'(S)=-2/W\ne0$. Define

$$L=\Omega-q\Omega' =1+\frac{q^2}{W^2}\gt 0, \qquad \frac{dx}{d\rho}=\frac{L}{\Omega^2}\gt 0.$$

Positivity of $L$ proves monotonicity and excludes a coordinate fold. At the interface,

$$x(R)=R, \qquad x'(R)=1, \qquad x''(R)=0.$$

These are the value, tangent, and curvature matching conditions used in the original layer construction Zenginoğlu 2011, § 2.5, equations (28)–(29).

Extended by $\Omega=1$ in the interior, this simple quadratic switch is $C^1$ at $R$; the coordinate $x(\rho)$ is $C^2$, so the metric and its first derivatives can be checked continuously. If a $C^\infty$ interface is required, replace it by any smooth nonincreasing $\Omega$ that equals $1$ near $R$, remains positive before $S$, and has a simple zero at $S$. Then $L=\Omega-q\Omega'\gt 0$, and every formula below is unchanged. The switch may be flat at $R$, but not at $S$, because a regular conformal boundary requires $d\Omega\ne0$.

Now introduce a height function,

$$\tau=t-h(x), \qquad H=\frac{dh}{dx},$$

with $H=0$ in the interior. Choose it in the layer by

$$1-H=\frac{\Omega^2}{L}.$$

Because $0\lt \Omega^2/L\le1$, one has $0\le H\lt 1$: every $\tau$ slice remains physically spacelike and becomes asymptotically null. Moreover, $H(R)=H'(R)=0$, so the time transformation also matches through first derivative.

Metric, characteristic, and phase checks

Since $dt=d\tau+H x'd\rho$, the physical metric becomes

$$d\widetilde s^2 =-d\tau^2-2Hx'\,d\tau\,d\rho +(1-H^2)(x')^2d\rho^2+x^2d\omega_2^2.$$

Define the unphysical metric by the same convention used throughout the chapter,

$$g_{\mu\nu}=\Omega^2\widetilde g_{\mu\nu}.$$

The layer metric is then

$$ds^2 =-\Omega^2d\tau^2-2HL\,d\tau\,d\rho +(1+H)L\,d\rho^2 +\left(\Omega R+q\right)^2d\omega_2^2.$$

At $R$, its coefficients and first derivatives equal those of the interior metric because $\Omega=1$, $H=0$, $L=1$, and all three first derivatives vanish there except the correctly matched derivative of the area radius. At $S$, $\Omega=0$, $H=1$, and $L_S=-W\Omega'(S)\gt0$; the quadratic example has $L_S=2$. The $(\tau,\rho)$ block has determinant $-L^2$, which remains nonzero. Thus the flat-space example is regular at its conformal outer end and has a $C^1$-matched metric at the quadratic interface. The smooth switch described above gives the corresponding $C^\infty$ interface; that smoother choice is required when a theorem assumes smooth coefficients throughout the domain.

Radial characteristic speeds are

$$c_+=\frac{\Omega^2}{L(1-H)}=1, \qquad c_-=-\frac{\Omega^2}{L(1+H)}.$$

They match $+1$ and $-1$ at the interface. At $\rho=S$, the outgoing speed stays finite while $c_-\to0$; no characteristic enters from beyond future null infinity. This choice and its characteristic interpretation are derived in Zenginoğlu 2011, § 2.5, equations (30)–(31).

The outgoing physical phase gives an even sharper test. Let $\widetilde u=t-x$. In the layer,

$$\frac{\partial\widetilde u}{\partial\rho} =(H-1)x'=-1.$$

Matching at $R$ therefore yields $\widetilde u=\tau-\rho$ on both sides. An outgoing waveform has continuous phase and spatial frequency across the interface; spatial compactification without the height transformation would instead create an infinite blueshift.

What must be matched in Einstein evolution

For a general spacetime, the transformation should be sufficiently differentiable that the metric and its first derivatives are continuous. Formulation variables must be transformed from the same metric and derivative data rather than copied component by component. Lapse, shift, gauge-source functions, constraint fields, and any rescaled matter or curvature variables must satisfy their defining relations through the transition. A jump in a first derivative acts like an artificial source even when the metric itself is continuous.

The layer construction is not, by itself, a proof of conformal regularity. A height function and compact radial coordinate can be used for a test-field problem without asserting that $g=\Omega^2\widetilde g$ solves regular conformal Einstein equations. For nonlinear gravity one additionally needs compatible hyperboloidal constraints, a smooth physical asymptotic structure, and a formulation whose variables have finite limits. Zenginoğlu’s Einstein discussion explicitly assumes such a smooth completion while allowing the conformal factor to equal unity in an interior region Zenginoğlu 2008, § 4, pp. 7–8.

Common pitfalls

Checking only coordinate continuity. Equality of $x$ and $\tau$ at $R$ does not match the metric derivatives. At minimum, inspect the Jacobian and the derivatives that enter the evolved formulation.

Calling every layer a conformal completion. The coordinate map places $x=\infty$ at finite $\rho$. A Penrose-type completion additionally requires a smooth nondegenerate $g$, $d\Omega\ne0$, and regular conformal fields.

Exercises

For the layer above, compute $L$, the interface derivatives of $x$, and the outgoing phase. Show explicitly why the outgoing characteristic speed is unity.

Solution

With $q=\rho-R$,

$$\Omega'= -\frac{2q}{W^2}, \qquad L=\Omega-q\Omega'=1+\frac{q^2}{W^2}\gt 0.$$

Thus $x'=L/\Omega^2$, giving $x'(R)=1$; differentiating once more gives $x''(R)=0$. Since $1-H=\Omega^2/L$,

$$c_+=\frac{\Omega^2}{L(1-H)}=1,$$

and

$$\partial_\rho(t-x)=(H-1)\frac{L}{\Omega^2}=-1.$$

The value at the interface is $t-x=\tau-R$, so the unique integrated phase is $t-x=\tau-\rho$ throughout the layer.

References

• Zenginoğlu, Anıl. “Hyperboloidal Evolution with the Einstein Equations.” Classical and Quantum Gravity 25 (2008): 195025. DOI. Open PDF.
• Zenginoğlu, Anıl. “Hyperboloidal Layers for Hyperbolic Equations on Unbounded Domains.” Journal of Computational Physics 230 (2011): 2286–2302. DOI. Open preprint.