Metric Reconstruction and Radiative Observables

A complete master field tells us more than its oscillation frequency: in a vacuum Schwarzschild region, the Zerilli–Moncrief and Cunningham–Price–Moncrief fields determine the corresponding metric perturbation modulo gauge at each multipole $\ell\geq2$. Their boundary limits then determine radiation. The reconstruction is local; a physical boundary condition, an asymptotic frame, and the separate low multipoles are additional information. This page develops the inverse and follows its normalization through strain, curvature, outward radiation and horizon absorption.

Required background. PDEs and Green Functions for Gravitation supplies boundary data and distributional solution concepts.

Helpful background. Black-Hole Geometry and Mechanics fixes regular horizon charts and asymptotic regions; Formulations of General Relativity supplies the constraint reconstruction; Presymplectic Potentials and Currents prepares energy flux; Black-Hole Perturbation Theory supplies invariant observables.

The metric information carried by a master field

Work to first order in $g_{\alpha\beta}+\epsilon p_{\alpha\beta}$ about a four-dimensional Schwarzschild black hole of physical mass $M\gt0$. Write $\mu=GM$, $f=1-2\mu/r$ and $x=r_*$ with $dx/dr=f^{-1}$. The local inverse below uses the open exterior $r\gt2\mu$ and smooth vacuum fields. Radiation amplitudes are first order; their energy flux is second order. We suppress the corresponding overall $\epsilon$ or $\epsilon^2$.

Fix a scalar harmonic $Y=Y_{\ell m}$ with unit sphere norm and $\ell\geq2$, and define

$$L=\ell(\ell+1),\qquad \lambda=\frac{L-2}{2},\qquad D=\lambda r+3\mu,\qquad \Lambda=\frac{2D}{r}.$$

Here $\mu$ is a mass length, whereas $m$ in $Y_{\ell m}$ is the azimuthal integer. On the unit sphere let

$$\begin{aligned} Y_A&=D_A Y, & Y_{AB}&=D_A D_B Y+\frac L2\Omega_{AB}Y,\\ X_A&=-\epsilon_A{}^B D_B Y, & X_{AB}&=D_{(A}X_{B)}, \end{aligned}$$

with $\epsilon_{\theta\phi}=\sin\theta$ and symmetrization of weight one half. The tensors are trace free. Orbit-space indices $a,b$ refer to $(t,r)$, with metric $g_{ab}dx^a dx^b=-fdt^2+f^{-1}dr^2$ and $\epsilon_{tr}=+1$.

For even parity, write $p_{ab}=h_{ab}Y$, $p_{aB}=j_aY_B$ and $p_{AB}=r^2(K\Omega_{AB}Y+\mathcal GY_{AB})$. The angular amplitude $\mathcal G$ is not Newton’s constant. The combinations

$$\begin{aligned} \varepsilon_a&=j_a-\frac{r^2}{2}\nabla_a\mathcal G,\\ \widetilde h_{ab}&=h_{ab}-\nabla_a\varepsilon_b-\nabla_b\varepsilon_a,\\ \widetilde K&=K+\frac L2\mathcal G-\frac2r r^a\varepsilon_a, \qquad r_a=\nabla_a r, \end{aligned}$$

do not change under an infinitesimal coordinate gauge transformation. In odd parity, $p_{aB}=h_aX_B$, $p_{AB}=h_2X_{AB}$ and

$$\widetilde h_a=h_a-\frac{r^2}{2}\nabla_a(h_2/r^2).$$

The complete masters used here are

$$\begin{aligned} \Psi_{\mathrm{even}} &=\frac{2r}{L}\left[ \widetilde K+\frac2\Lambda \left(r^ar^b\widetilde h_{ab}-r r^a\nabla_a\widetilde K\right)\right],\\ \Psi_{\mathrm{odd}} &=\frac{2r}{L-2}\epsilon^{ab} \left(\nabla_a\widetilde h_b-\frac2r r_a\widetilde h_b\right). \end{aligned}$$

Both have dimension length. These are the Zerilli–Moncrief and Cunningham–Price–Moncrief (CPM) normalizations, not arbitrary rescalings of a Schrödinger-like radial function. The definitions and invariant field equations are given in Martel and Poisson 2005, §§ IV–V, Eqs. (4.10)–(4.24) and (5.7)–(5.13), pp. 104003-4–5 and 104003-7, PDF. The even and odd metric pages derive their gauge cancellations.

A master solution means the field with its time dependence, or equivalently its field and velocity on a Cauchy slice with suitable boundary data. A frequency alone, an energy flux, or one radial amplitude without its normalization is not that information.

The complete local vacuum inverse

Even-parity constraints

Choose the Regge–Wheeler representative $j_a=\mathcal G=0$, so the invariant amplitudes equal the metric amplitudes. Write

$$h_{tt}=fH_e,\qquad h_{tr}=H_1,\qquad h_{rr}=H_e/f, \qquad \psi=\Psi_{\mathrm{even}}.$$

Equality of the two diagonal amplitudes follows from the trace-free angular Einstein equation; it is not another gauge choice. Define

$$A=\frac{\lambda(\lambda+1)r^2+3\lambda\mu r+6\mu^2}{r^2D}, \qquad B=\frac{\lambda r^2-3\lambda\mu r-3\mu^2}{rfD}.$$

The full inverse is

$$\begin{aligned} K&=A\psi+f\psi_r,\\ H_e&=rK_r-\frac12\partial_r(\Lambda\psi),\\ H_1&=r\psi_{tr}+B\psi_t =\frac{rK_t-\Lambda\psi_t/2}{f}. \end{aligned}$$

Subscripts denote partial derivatives at fixed other coordinate. The radial derivative acts on $\Lambda$ as well as $\psi$. No time integration or division by frequency appears.

Here is how the constraints select this inverse. Normalize the Einstein projections by

$$\delta G_{ab}=\mathcal Q_{ab}Y,\qquad \delta G_{aB}=\frac12\mathcal Q_aY_B,\qquad \delta G_{AB}=\frac{r^2}{2}\mathcal Q^\flat\Omega_{AB}Y +\frac12\mathcal Q^\sharp Y_{AB}.$$

Raise orbit indices with $g^{ab}$; in particular, $\mathcal Q^{tt}=f^{-2}\mathcal Q_{tt}$, $\mathcal Q^{tr}=-\mathcal Q_{tr}$ and $\mathcal Q^r=f\mathcal Q_r$. The master definition first gives

$$H_e=rK_r+\frac{D}{rf}\left(\frac{\lambda+1}{r}\psi-K\right).$$

Substituting this into the Hamiltonian equation cancels the second derivatives of $K$ and yields

$$\mathcal Q^{tt} =\frac{(\lambda+1)D}{r^3f^2}(A\psi+f\psi_r-K).$$

Thus $\mathcal Q^{tt}=0$ fixes $K$. After that substitution the momentum equation becomes

$$\mathcal Q^{tr} =-\frac{\lambda+1}{r^2} \left(H_1-r\psi_{tr}-B\psi_t\right),$$

which fixes $H_1$. Substitution in the algebraic expression for $H_e$ gives the displayed compact inverse. These are constructive constraints, not merely a count of unknown functions.

The remaining equations impose the master wave equation. Define its covariant residual by

$$\mathcal W_+=-\frac{\psi_{tt}}f+\partial_r(f\psi_r)-U_Z\psi, \qquad U_Z=\frac{ 2\lambda^2(\lambda+1)r^3+6\lambda^2\mu r^2 +18\lambda\mu^2r+18\mu^3}{r^3D^2}.$$

On the full inverse, direct reinsertion into the Einstein projections gives

$$\begin{aligned} \mathcal Q^{tt}=\mathcal Q^{tr}=\mathcal Q^t =\mathcal Q^\sharp&=0,\\ \mathcal Q^r&=rf\,\partial_r\mathcal W_++\beta\mathcal W_+,\\ \mathcal Q^{rr}+\frac fr\mathcal Q^r &=\frac{fD}{r^2}\mathcal W_+,\\ \mathcal Q^\flat&=-\partial_r\mathcal Q^r-\frac2r\mathcal Q^r, \qquad \beta=\frac{\lambda r^2-\lambda\mu r+3\mu^2}{rD}. \end{aligned}$$

Consequently $\mathcal W_+=0$ makes every vacuum projection vanish. Conversely, the third line forces $\mathcal W_+=0$ when the Einstein equations hold, because $fD/r^2\ne0$ in this domain. It does not merely force a derivative of the residual to vanish. Multiplication by $f$ gives $(-\partial_t^2+\partial_x^2-fU_Z)\psi=0$.

This calculation is the reconstruction version of Martel and Poisson 2005, § IV.B–C, Eqs. (4.13)–(4.26), pp. 104003-5–6, PDF. The even metric derivation develops the component equations and Bianchi identities behind the substitution.

Odd parity and stationary information

For the complete CPM field, the local vacuum inverse is

$$\widetilde h_a=\frac12\epsilon_{ab}\nabla^b(r\Psi_{\mathrm{odd}}), \qquad \widetilde h_t=\frac f2\partial_r(r\Psi_{\mathrm{odd}}), \qquad \widetilde h_r=\frac{r}{2f}\partial_t\Psi_{\mathrm{odd}}.$$

The signs use $\epsilon_{tr}=+1$ and hence $\epsilon^{tr}=-1$. Let $\mathcal W_-=(\Box_2-U_{\mathrm{RW}})\Psi_{\mathrm{odd}}$, where $\Box_2=-f^{-1}\partial_t^2+\partial_r(f\partial_r)$ and $U_{\mathrm{RW}}=L/r^2-6\mu/r^3$. Putting this inverse back into the CPM definition returns

$$\Psi_{\mathrm{CPM}}[\widetilde h] =\Psi_{\mathrm{odd}}+\frac{r^2}{L-2}\mathcal W_-.$$

It is an inverse on solutions, not for an arbitrary off-shell function. The remaining odd Einstein equations reduce to derivatives of $r^3\mathcal W_-$, and the full vacuum field/master correspondence is derived on Odd-Parity Metric Perturbations. In particular, the inverse itself retains stationary $\ell\geq2$ solutions.

By contrast, the alternative Regge–Wheeler variable satisfies in vacuum

$$Q_{\mathrm{RW}}=\frac f r\widetilde h_r =\frac12\partial_t\Psi_{\mathrm{odd}}.$$

This different map erases stationary CPM data. The Chandrasekhar differential transformation has its own exceptional kernel. Neither fact is a failure of reconstruction from the original, complete CPM or Zerilli–Moncrief solution.

Restoring a gauge representative

For even parity, arbitrary smooth $j_a,\mathcal G$ specify $\varepsilon_a$ and restore

$$h_{ab}=\widetilde h_{ab}+\nabla_a\varepsilon_b+\nabla_b\varepsilon_a, \qquad K=\widetilde K-\frac L2\mathcal G+\frac2r r^a\varepsilon_a.$$

For odd parity, choose $h_2$ and set $h_a=\widetilde h_a+\frac12r^2\nabla_a(h_2/r^2)$. The vacuum inverse is unique in invariant metric data at fixed $\ell\geq2$; coordinate components are not unique. A locally smooth gauge choice need not preserve the regularity or boundary frame required by a global problem.

A singular source is a separate issue. Differentiating a discontinuous master produces distributions, and matter-dependent reconstruction terms must also be included. One must not insert a sourced master into the vacuum inverse and declare the constraints solved. Use the sourced equations and junction conditions before evaluating one-sided fields or choosing a regularization.

Strain and curvature at future null infinity

Assume an asymptotically flat perturbation with a smooth outgoing expansion, a vacuum neighborhood of future null infinity, areal radius $r$ and retarded time $u=t-x$. Hold the angular coordinates and polarization frame fixed in $u$. Matter confined to a bounded region is compatible with this vacuum exterior; arbitrary finite-radius coordinate components are not already a radiation field.

Denote the limits at fixed $u$ by

$$\psi_{\mathrm e}^{\ell m}(u)=\lim_{r\to\infty}\Psi_{\mathrm{even}}^{\ell m}(u,r), \qquad \psi_{\mathrm o}^{\ell m}(u)=\lim_{r\to\infty}\Psi_{\mathrm{odd}}^{\ell m}(u,r).$$

In an asymptotically radiative representative,

$$p_{AB}^{\mathrm{rad}}=rC_{AB}+O(1),\qquad C_{AB}=\sum_{\ell m} \left(\psi_{\mathrm e}^{\ell m}Y_{AB}^{\ell m} +\psi_{\mathrm o}^{\ell m}X_{AB}^{\ell m}\right).$$

This is the outgoing-limit reconstruction of Martel and Poisson 2005, § VI, Eqs. (6.8)–(6.15), pp. 104003-9–10, PDF.

From an angular metric to strain

Use the unit-sphere dyad $m^A=(1,i\csc\theta)/\sqrt2$. The trace-free metric in an orthonormal angular basis has entries $(h_+,h_\times;h_\times,-h_+)$, whereas $p_{AB}^{\mathrm{rad}}$ uses coordinate components. Contracting the latter with $\bar m^A\bar m^B/r^2$ gives $h_+-ih_\times$ to leading order. Let $N_\ell=L(L-2)=(\ell+2)!/(\ell-2)!$. The tensor-to-spin identities are

$$Y_{AB}^{\ell m}\bar m^A\bar m^B =\frac{\sqrt{N_\ell}}2\,{}_{-2}Y_{\ell m}, \qquad X_{AB}^{\ell m}\bar m^A\bar m^B =-i\frac{\sqrt{N_\ell}}2\,{}_{-2}Y_{\ell m}.$$

They follow by applying the dyad to the trace-free Hessian and its axial rotation; see Martel and Poisson 2005, Appendix A, p. 104003-11, PDF. The spin harmonics have unit norm. Therefore

$$h_+-ih_\times =\frac1{2r}\sum_{\ell m}\sqrt{N_\ell} \left(\psi_{\mathrm e}^{\ell m}-i\psi_{\mathrm o}^{\ell m}\right) {}_{-2}Y_{\ell m}+O(r^{-2}).$$

The factor $1/r$ converts a master of dimension length into dimensionless strain. Changing the axial orientation changes the odd sign in this whole dictionary, not just in one flux formula.

The null tetrad fixes the curvature coefficient

Define the radius-rescaled strain and leading curvature by

$$H:=\lim_{r\to\infty}r(h_+-ih_\times), \qquad \Psi_4^0:=\lim_{r\to\infty}r\Psi_4.$$

Use a future unit time direction $T$, an outward unit radial direction $R$, and an orthonormal polarization pair $e_\theta,e_\phi$ that approach a fixed asymptotic inertial frame. Choose the symmetric null tetrad

$$\ell=\frac{T+R}{\sqrt2},\qquad n=\frac{T-R}{\sqrt2},\qquad m=\frac{e_\theta+i e_\phi}{\sqrt2}, \qquad \ell\!\cdot n=-1,\qquad m\!\cdot\bar m=1,$$

and define the spacetime Weyl contraction

$$\Psi_4=-C_{\alpha\beta\gamma\delta} n^\alpha\bar m^\beta n^\gamma\bar m^\delta.$$

Inner products alone would leave an arbitrary null boost unfixed. To check the coefficient, use a local outgoing transverse-traceless wave $h_{ij}^{\mathrm{TT}}(u)$, with $i,j$ transverse and $\alpha,\beta$ in its time/radial plane. Linearizing the site’s Riemann tensor gives

$$R_{\alpha i\beta j} =-\frac12(\partial_\alpha u)(\partial_\beta u) \,\partial_u^2 h_{ij}^{\mathrm{TT}}.$$

In vacuum it is the leading Weyl tensor. The dyad contraction of $h_{ij}^{\mathrm{TT}}$ is $h_+-ih_\times$, and the chosen ingoing leg has $n(u)\to\sqrt2$. Thus its squared contraction cancels the one half:

$$\Psi_4^0=\partial_u^2H,\qquad \Psi_4=\partial_u^2(h_+-ih_\times)+O(r^{-2}).$$

This is the explicit convention package in Bishop and Rezzolla 2016, §3.3, Eqs. (95)–(103), pp. 22–23, PDF. A constant boost $\ell\mapsto b\ell$, $n\mapsto b^{-1}n$ with $b\gt0$ and a dyad rotation $m\mapsto e^{i\chi}m$ give

$$\Psi_4\mapsto b^{-2}e^{-2i\chi}\Psi_4.$$

For example, $\ell=T+R$, $n=(T-R)/2$ has $b=\sqrt2$ and produces one half of the symmetric-tetrad curvature, despite the same null inner products. A time-dependent angular frame also differentiates the basis and cannot be ignored in the news or curvature.

Writing $\mathcal N=\partial_uH$, curvature data on $[u_0,u_1]$ determine

$$H(u)=H(u_0)+(u-u_0)\mathcal N(u_0) +\int_{u_0}^{u}(u-s)\Psi_4^0(s)\,ds.$$

The initial strain and news must be supplied separately. A constant strain offset leaves both news and curvature unchanged; a linear strain drift leaves curvature unchanged but changes the radiated energy. The Strain and Curvature Mode Decompositions page develops the common frame, Fourier and integration-constant issues.

Positive radiation flux and the real-mode sum

The unit-sphere norms are

$$\int Y_{AB}^{\ell m}\overline{Y}^{AB}_{\ell' m'}d\Omega =\int X_{AB}^{\ell m}\overline{X}^{AB}_{\ell' m'}d\Omega =\frac{N_\ell}{2}\delta_{\ell\ell'}\delta_{mm'}, \qquad \int Y_{AB}^{\ell m}\overline{X}^{AB}_{\ell' m'}d\Omega=0.$$

Indices here are raised with $\Omega^{AB}$, not the inverse metric of a physical sphere. With tensor news $N_{AB}=\partial_u C_{AB}$, the gravitational contribution to Bondi luminosity is

$$\mathcal F_\infty =\frac1{32\pi G}\int N_{AB}N^{AB}d\Omega =\frac1{16\pi G}\int|\mathcal N|^2d\Omega.$$

This is an instantaneous, leading quadratic flux at null infinity, not the averaged finite-radius effective stress tensor. For a real metric the scalar-harmonic coefficients obey $\psi_{\mathrm e,o}^{\ell,-m}=(-1)^m\overline{\psi_{\mathrm e,o}^{\ell m}}$. For $a_m=\partial_u\psi_{\mathrm e}^{\ell m}$ and $b_m=\partial_u\psi_{\mathrm o}^{\ell m}$,

$$|a_m-i b_m|^2=|a_m|^2+|b_m|^2-2\operatorname{Im}(a_m\overline b_m).$$

The interference term need not vanish for one complex $m$ mode. Its sign reverses for $-m$, while the $m=0$ coefficients are real. Summing the complete real perturbation therefore gives

$$\mathcal F_\infty =\sum_{\ell m}C_\ell \left(|\partial_u\psi_{\mathrm e}^{\ell m}|^2 +|\partial_u\psi_{\mathrm o}^{\ell m}|^2\right), \qquad C_\ell=\frac{N_\ell}{64\pi G}.$$

One may instead use a real harmonic basis and obtain the same result. Do not both sum the conjugate pair and add another factor of two. The time average agrees with Martel and Poisson 2005, § VI, Eq. (6.16), p. 104003-10, PDF; the tensor identities and norms are given in their Martel and Poisson 2005, § III, Eqs. (3.6)–(3.11), pp. 104003-3–4, PDF.

The normalization also follows from the reduced action. For a real mode its directed current is $j=-C_\ell\Psi_t\Psi_x$. An outgoing $\Psi=F(t-x)$ has $j=+C_\ell(F')^2$; an ingoing $\Psi=G(t+x)$ has $j=-C_\ell(G')^2$. Consequently, on a finite fixed interval $[x_a,x_b]$ in vacuum, $dE/dt=j(x_a)-j(x_b)$. Here $E$ is the quadratic canonical energy of this mode between the two endpoints on a constant-$t$ slice. Both an ingoing left-end flux and an outgoing right-end flux remove exterior energy. The canonical normalization and finite-cut energy discussion establish what this balance does and does not imply about global decay.

Horizon shear and absorption

Use advanced time $v=t+x$ and an ingoing chart regular at the future Schwarzschild horizon $r_+=2\mu$. Normalize its generator by $k=\partial_v$ relative to the asymptotic time, so $\kappa=1/(4\mu)$. Assume a vacuum neighborhood of the horizon and a small perturbation of the event horizon. In a horizon-adapted gauge with the generators at fixed angular labels, the first-order intrinsic angular metric perturbation is

$$p_{AB}^{\mathcal H} =r_+\sum_{\ell m} \left(\psi_{\mathrm e,H}^{\ell m}Y_{AB}^{\ell m} -\psi_{\mathrm o,H}^{\ell m}X_{AB}^{\ell m}\right),$$

where $\psi_{\mathrm e,o,H}(v)$ are the master limits at fixed $v$. The minus sign in the odd sector is essential for the metric and shear, even though it disappears in the squared flux. It differs from the outgoing infinity dictionary because the horizon limit is ingoing; see Martel and Poisson 2005, § VII, Eqs. (7.2)–(7.4), pp. 104003-10–11, PDF.

Let $q_{AB}^{(0)}=r_+^2\Omega_{AB}$. To first order,

$$\sigma_{AB}^{(1)}=\frac12\partial_v p_{AB}^{\mathcal H}, \qquad q_{(0)}^{AB}=r_+^{-2}\Omega^{AB},\qquad dA_0=r_+^2d\Omega.$$

The shear is trace free at this order. Its positive quadratic energy quantity is

$$\begin{aligned} \mathcal F_H &=\frac1{8\pi G}\int q_{(0)}^{AC}q_{(0)}^{BD} \sigma_{AB}^{(1)}\sigma_{CD}^{(1)}\,dA_0\\ &=\sum_{\ell m}C_\ell \left(|\partial_v\psi_{\mathrm e,H}^{\ell m}|^2 +|\partial_v\psi_{\mathrm o,H}^{\ell m}|^2\right). \end{aligned}$$

Each shear contributes $r_+/2$, the two inverse metrics contribute $r_+^{-4}$, and the area measure contributes $r_+^2$. Together with the tensor norm $N_\ell/2$, this yields $N_\ell/(64\pi G)$, with no leftover power of the horizon radius. It agrees with the positive ingoing canonical flux $-j$.

This does not define an instantaneous horizon mass derivative on arbitrary cuts. Write $\sigma=\epsilon\sigma^{(1)}+\cdots$ and $\theta=\epsilon^2\theta^{(2)}+\cdots$. The vacuum Raychaudhuri equation, with the final stationary boundary condition appropriate to an event horizon, gives at this order

$$(\partial_v-\kappa)\theta^{(2)} =-\sigma_{AB}^{(1)}\sigma^{(1)AB}, \qquad \frac{dA^{(2)}}{dv}=\int\theta^{(2)}dA_0.$$

The background metric raises the shear indices. Integrating between finite cuts $v_0,v_1$ gives

$$\kappa\Delta A^{(2)} =8\pi G\int_{v_0}^{v_1}\mathcal F_H\,dv +\left[\int\theta^{(2)}dA_0\right]_{v_0}^{v_1}.$$

The endpoint expansion term cannot generally be dropped. Between settled initial and final states with vanishing endpoint expansion, the leading physical-process first law identifies $\Delta M^{(2)}=\kappa\Delta A^{(2)}/(8\pi G)$ for the initially nonrotating hole. Then $\Delta M^{(2)}=\int\mathcal F_H\,dv$. An averaged absorption rate gives the corresponding averaged mode sum, under the averaging and slow area-evolution assumptions stated in Poisson 2004, arXiv v2, § IV, Eqs. (4.5)–(4.14) and (4.22)–(4.24), pp. 10–12, PDF. The master-field averaged formula is Martel and Poisson 2005, Eq. (7.5), p. 104003-11, PDF. No matter flux is included in this vacuum horizon calculation.

Schwarzschild-coordinate factors $1/f$ alone do not diagnose a tensor singularity. Transform the reconstructed perturbation to an ingoing representative before testing regularity. Nor does local reconstruction guarantee a smooth event horizon for arbitrary boundary data. The absorption and tidal-heating page develops that response problem further.

Limits of reconstruction

Quantity	Required choice or operation	Information not supplied by that quantity
Metric at $\ell\geq2$	Complete vacuum master, constraints and a gauge representative	Low-multipole or matter completion
Asymptotic strain	Outgoing limit and a fixed angular/polarization frame	The full finite-radius metric
Curvature	Two strain derivatives and an explicit null tetrad	Initial strain and news
Energy flux	Angular integral of squared first derivatives	Waveform phase and polarization

The even monopole includes a mass shift and the odd dipole an angular-momentum shift. The even dipole depends on the center-of-mass frame and the admitted gauge class. These are separate sectors, not extra stationary invariant pieces at a fixed $\ell\geq2$. Consult Low Multipoles, Parameter Shifts, and Gauge Modes before declaring a gauge transformation physically trivial at a boundary.

Likewise, a pointwise singular self-field requires a regularization; a vacuum inverse does not choose one. Spectral solution methods remain in the response chapters, and generic numerical extraction belongs to Computational Relativity and Gravitational Radiation. Within these limits, comparing the reconstructed metric, curvature and energy flux is a powerful consistency check precisely because they test different derivatives and normalizations.

Exercises

Recover a stationary mode before differentiating it

For $\ell=2$, take $\Psi_{\mathrm{odd}}=cr^3$ with constant $c$ of dimension length$^{-2}$. Verify its vacuum wave equation, reconstruct the invariant odd metric, and transform it with $dt=dv-dr/f$. What would be lost if only $Q_{\mathrm{RW}}$ were supplied?

Solution

Since $L=6$,

$$\partial_x^2(cr^3) =f\partial_r(3cfr^2) =6cf(r-\mu) =f\left(\frac6{r^2}-\frac{6\mu}{r^3}\right)cr^3.$$

The inverse gives

$$\widetilde h_t=2cfr^3,\qquad \widetilde h_r=0,\qquad \widetilde h_v=2cfr^3,\qquad \widetilde h_r^{(v,r)}=-2cr^3.$$

The ingoing components are regular at $r=2\mu$, but the field grows at infinity. It is an externally supported tidal-type solution, not isolated asymptotically flat black-hole hair. Its stationary master is nonzero while $Q_{\mathrm{RW}}=\partial_t\Psi_{\mathrm{odd}}/2=0$. Reconstructing from CPM retains the field; inverting its time derivative without stationary data does not.

Compare curvature and power for two polarizations

Consider a real $\ell=2,m=0$ radiation field with

$$\psi_{\mathrm e}=a\cos(\omega u),\qquad \psi_{\mathrm o}=b\sin(\omega u).$$

Here $a,b$ are real lengths and $\omega\gt0$. Find $H_{20}$, $\Psi_{4,20}^0$, instantaneous power and its period average. What changes under a constant null boost by $\sqrt2$?

Solution

Here $N_2=24$ and $C_2=3/(8\pi G)$, so

$$H_{20}=\sqrt6\,[a\cos(\omega u)-ib\sin(\omega u)], \qquad \Psi_{4,20}^0=-\omega^2H_{20}.$$

The mode is already a complex polarization field built from a real metric; do not take another real part or insert an extra one half. Its power is

$$\mathcal F_\infty =\frac{3\omega^2}{8\pi G} \left[a^2\sin^2(\omega u)+b^2\cos^2(\omega u)\right], \qquad \langle\mathcal F_\infty\rangle =\frac{3\omega^2(a^2+b^2)}{16\pi G}.$$

The boost divides $\Psi_{4,20}^0$ by two while leaving the specified strain and Bondi flux unchanged. Applying the unboosted curvature-to-strain formula would wrongly infer half the strain amplitude and one quarter of the power. Even in the correct frame, curvature alone leaves two strain integration constants. A nonzero initial news cannot be discarded merely because it does not appear after a second derivative.

Keep the finite-cut horizon term

On $0\leq v\leq T$, suppose the integrated expansion coefficient is $I(v)=\int\theta^{(2)}dA_0=I_0e^{-\alpha v}$, with $I_0,\alpha,T\gt0$. Let $Q(v)=\int\sigma_{AB}^{(1)}\sigma^{(1)AB}dA_0$. Determine $Q$, $\Delta A^{(2)}$ and the endpoint correction. Can one identify $\int\mathcal F_Hdv$ with a settled-state mass change on these cuts?

Solution

Integrating Raychaudhuri over the background sphere gives $I'-\kappa I=-Q$. Hence

$$Q(v)=(\kappa+\alpha)I_0e^{-\alpha v},\qquad \Delta A^{(2)}=\frac{I_0}{\alpha}(1-e^{-\alpha T}).$$

The two terms on the right side of the finite-cut identity are

$$\int_0^T Q\,dv =\frac{(\kappa+\alpha)I_0}{\alpha}(1-e^{-\alpha T}), \qquad I(T)-I(0)=-I_0(1-e^{-\alpha T}).$$

Their sum is exactly $\kappa\Delta A^{(2)}$. Omitting the endpoint term overestimates that quantity by $I_0(1-e^{-\alpha T})$. These finite cuts have nonzero expansion and are not the settled endpoints needed for the stated mass interpretation. This is an exact finite-interval check of the perturbative expansion equation, not a construction of a complete global black-hole spacetime.

References

• Bishop, Nigel T., and Luciano Rezzolla. “Extraction of Gravitational Waves in Numerical Relativity.” Living Reviews in Relativity 19, 2 (2016). DOI. Open PDF, arXiv v2.
• Martel, Karl, and Eric Poisson. “Gravitational Perturbations of the Schwarzschild Spacetime: A Practical Covariant and Gauge-Invariant Formalism.” Physical Review D 71, 104003 (2005). DOI. Open PDF, arXiv preprint. Claim-adjacent 104003-page locators refer to the journal edition.
• Poisson, Eric. “Absorption of Mass and Angular Momentum by a Black Hole: Time-Domain Formalisms for Gravitational Perturbations, and the Small-Hole/Slow-Motion Approximation.” Physical Review D 70, 084044 (2004). DOI. Open PDF, arXiv v2.