Even-Parity Metric Perturbations

Even-parity, or polar, perturbations of Schwarzschild have seven metric amplitudes, but only one independent dynamical field for each harmonic $(\ell,m)$ with $\ell\ge2$. The odd sector has its own independent field; together the two parity sectors describe gravitational radiation. Here we derive the polar gauge invariants, eliminate the vacuum constraints, and reconstruct the metric locally from the Zerilli–Moncrief field. The reconstruction includes stationary solutions. The monopole and dipole require separate equations, not substitution into a radiative formula.

Required background. PDEs and Green Functions for Gravitation supplies constrained hyperbolic equations and the distinction between an equation and its initial or boundary data.

Helpful background. Black-Hole Geometry and Mechanics fixes the Schwarzschild geometry; Formulations of General Relativity distinguishes gauge conditions from constraints; Killing Fields and Isometry Groups explains the exceptional displacement modes.

Polar metric amplitudes

Consider a first-order vacuum perturbation $g_{\mu\nu}(\eta)=g_{\mu\nu}^{(0)}+\eta p_{\mu\nu}+O(\eta^2)$, with dimensionless $\eta$ and $\delta G_{\mu\nu}[p]=0$. Write $\mu=GM\gt0$. The background is four-dimensional Schwarzschild, with orbit metric $g_{ab}$ and unit-sphere metric $\Omega_{AB}$:

$$ds_0^2=g_{ab}\,dx^a dx^b+r^2\Omega_{AB}\,d\theta^A d\theta^B, \qquad g_{ab}\,dx^a dx^b=-f\,dt^2+f^{-1}dr^2, \qquad f=1-\frac{2\mu}{r}.$$

The coordinate calculation initially uses $r\gt2\mu$. Let $\nabla_a$ and $D_A$ be the orbit and unit-sphere derivatives. Define $r_a=\nabla_a r$, $r^a=g^{ab}r_b$ and $L=\ell(\ell+1)$. For one normalized scalar harmonic $Y=Y^{\ell m}$, use

$$D^A D_A Y=-LY,\qquad Y_A=D_A Y,\qquad Y_{AB}=D_A D_B Y+\frac L2\Omega_{AB}Y.$$

The last harmonic is trace free. These are the conventions of Tensor Harmonics and Parity Decomposition. Complex modes are combined with their conjugates to form a real metric. Suppressing $(\ell,m)$, write

$$p_{ab}=h_{ab}Y,\qquad p_{aB}=j_aY_B,\qquad p_{AB}=r^2\left(K\Omega_{AB}Y+G Y_{AB}\right).$$

Amplitudes	Number of functions on the orbit space	What they describe
$h_{ab}$	3	Perturbation of the orbit metric
$j_a$	2	Mixed orbit–sphere components
$K$	1	Angular trace
$G$	1	Trace-free angular deformation

The count is a list of components, not a count of physical degrees of freedom. Three even gauge functions change their description, and the Einstein equations impose differential constraints. In particular, $K$ differs from the coefficient used in a basis containing $D_A D_B Y$ instead of $Y_{AB}$. The precise decomposition and basis conversion are Martel and Poisson 2005, § IV.A, Eqs. (4.1)–(4.3), p. 104003-4, PDF.

Removing gauge dependence

Use an even generator with covariant components $\Xi_a=\xi_aY$ and $\Xi_A=\xi Y_A$. Here $\Xi=-\xi_{\mathrm{Ch.\,1}}$ is the inverse-identification generator: the following minus signs describe the same transformation as Chapter 1’s active-plus convention with the opposite generator. The amplitude changes are

$$h'_{ab}=h_{ab}-\nabla_a\xi_b-\nabla_b\xi_a,$$ $$j'_a=j_a-\xi_a-\nabla_a\xi+\frac{2r_a}{r}\xi, \qquad G'=G-\frac{2}{r^2}\xi, \qquad K'=K+\frac{L}{r^2}\xi-\frac{2}{r}r^a\xi_a.$$

For $\ell\ge2$, define

$$\varepsilon_a=j_a-\frac{r^2}{2}\nabla_aG, \qquad \varepsilon'_a=\varepsilon_a-\xi_a,$$ $$\widetilde h_{ab} =h_{ab}-\nabla_a\varepsilon_b-\nabla_b\varepsilon_a, \qquad \widetilde K=K+\frac L2G-\frac{2}{r}r^a\varepsilon_a.$$

Both tilded quantities are invariant without using a field equation. Choose $\xi=r^2G/2$ and $\xi_a=\varepsilon_a$ to obtain $j'_a=G'=0$. This is even Regge–Wheeler gauge; its surviving amplitudes equal the invariant ones. The choice is local: whether that generator is allowed in a global problem depends on its regularity and boundary behavior. The transformation and invariants agree with Martel and Poisson 2005, § IV.A, Eqs. (4.6)–(4.12), p. 104003-4, PDF.

For the calculation below, use Regge–Wheeler gauge and omit tildes:

$$h_{tt}=fH_0,\qquad h_{tr}=H_1,\qquad h_{rr}=H_2/f.$$

Thus $H_0,H_1,H_2,K$ name invariant metric data in this representative. They have no independent gauge freedom left at fixed $\ell\ge2$.

Einstein constraints and the master field

Fix the projection normalization before setting any equation to zero:

$$\delta G_{ab}=\mathcal Q_{ab}Y,\qquad \delta G_{aB}=\frac12\mathcal Q_aY_B,$$ $$\delta G_{AB} =\frac{r^2}{2}\mathcal Q^\flat\Omega_{AB}Y +\frac12\mathcal Q^\sharp Y_{AB}, \qquad \mathcal Q^\sharp=H_0-H_2.$$

The trace-free angular equation therefore sets $H_0=H_2\equiv H$. This is an Einstein equation, not a further gauge choice. The projections are related off shell by

$$\nabla^b\mathcal Q_{ab}+\frac{2r^b}{r}\mathcal Q_{ab} -\frac{L}{2r^2}\mathcal Q_a-\frac{r_a}{r}\mathcal Q^\flat=0,$$ $$\nabla^a\mathcal Q_a+\frac{2r^a}{r}\mathcal Q_a -\frac{L-2}{2r^2}\mathcal Q^\sharp+\mathcal Q^\flat=0.$$

These identities explain why checking selected equations can imply the remaining ones; they do not license dropping constraints arbitrarily. The full covariant projections and identities are Martel and Poisson 2005, § IV.B, Eqs. (4.13)–(4.22), p. 104003-5, PDF.

Introduce the constant angular number and two radial functions

$$\lambda=\frac{L-2}{2},\qquad D=\lambda r+3\mu,\qquad \Lambda=\frac{2D}{r}=2\lambda+\frac{6\mu}{r}.$$

With the trace relation imposed, the two projections that contain no second time derivative are

$$\begin{aligned} \mathcal Q^{tt} &=-K_{rr}-\frac{3r-5\mu}{r^2f}K_r+\frac{H_r}{r} +\frac{\lambda+2}{r^2f}H+\frac{\lambda}{r^2f}K,\\ \mathcal Q^{tr} &=K_{tr}+\frac{r-3\mu}{r^2f}K_t -\frac{H_t}{r}-\frac{\lambda+1}{r^2}H_1, \end{aligned}$$

where subscripts $t,r$ denote partial derivatives at fixed other coordinate.

In particular, the indices on $\mathcal Q^{tt}$ and $\mathcal Q^{tr}$ have been raised with the background orbit metric; they are not the covariant $tt$ and $tr$ coefficients. The component equations can be checked directly against Martel and Poisson 2005, arXiv v1, Appendix C, p. 15, PDF. In that appendix their angular $\lambda$ is our $L$, their angular $\mu$ is our $2\lambda$, and their mass $M$ is our length $\mu$.

The Zerilli–Moncrief field is the invariant combination

$$\Psi_{\mathrm{even}} =\frac{2r}{L} \left[\widetilde K+\frac{2}{\Lambda} \left(r^ar^b\widetilde h_{ab} -r\,r^a\nabla_a\widetilde K\right)\right].$$

It has dimension length. This normalization agrees with Martel and Poisson 2005, § IV.C, Eqs. (4.23)–(4.24), p. 104003-5, PDF. Write $\psi=\Psi_{\mathrm{even}}$. In the chosen coordinates it becomes

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

Solve this definition for $H$:

$$H=rK_r+\frac{D}{rf} \left[\frac{\lambda+1}{r}\psi-K\right].$$

Now substitute into the Hamiltonian projection. All $K_{rr}$ terms cancel, leaving the exact identity

$$\mathcal Q^{tt} =\frac{(\lambda+1)D}{r^3f^2} \left(A\psi+f\psi_r-K\right), \qquad A=\frac{\lambda(\lambda+1)r^2+3\lambda\mu r+6\mu^2}{r^2D}.$$

The prefactor is nonzero for $\ell\ge2$ and $r\gt2\mu$. Hence the first constraint gives the local reconstruction

$$K=A\psi+f\psi_r.$$

Substituting this result into the momentum projection similarly gives

$$\mathcal Q^{tr} =-\frac{\lambda+1}{r^2} \left(H_1-r\psi_{tr}-B\psi_t\right), \qquad B=\frac{\lambda r^2-3\lambda\mu r-3\mu^2}{rfD}.$$

Thus $H_1=r\psi_{tr}+B\psi_t$. These cancellations, not a subtraction of component counts, establish how the constraints remove independent metric data.

The wave equation and the complete local inverse

A compact form of the reconstructed amplitudes is

$$\begin{aligned} K&=A\psi+f\psi_r,\\ H_0=H_2=H&=rK_r-\frac{\Lambda}{2}\psi_r -\frac{\Lambda'}{2}\psi,\\ H_1&=r\psi_{tr}+B\psi_t, \qquad \Lambda'=\frac{d\Lambda}{dr}=-\frac{6\mu}{r^2}. \end{aligned}$$

There is no integration in time and no division by frequency. The remaining equation must still be imposed. To see precisely which one, the radial mixed and radial orbit projections are

$$\begin{aligned} \mathcal Q^r &=-(H_1)_t+fH_r+f'H-fK_r,\\ \mathcal Q^{rr} &=-K_{tt}+\frac{(r-\mu)f}{r^2}K_r +\frac{2f}{r}(H_1)_t-\frac{f^2}{r}H_r +\frac{\lambda f}{r^2}(H-K). \end{aligned}$$

Here $f'=df/dr$. Insert the complete inverse and define

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

Direct substitution yields

$$\mathcal Q^{rr}+\frac fr\mathcal Q^r =\frac{fD}{r^2}\mathcal W, \qquad \mathcal Q^r=rf\,\partial_r\mathcal W+\beta\mathcal W, \qquad \beta=\frac{\lambda r^2-\lambda\mu r+3\mu^2}{rD}.$$

The first equality is important: it forces $\mathcal W=0$, not merely a derivative of $\mathcal W$ to vanish. Conversely, a smooth $\psi$ satisfying $\mathcal W=0$ gives $\mathcal Q^r=\mathcal Q^{rr}=0$. The inverse already satisfies $\mathcal Q^{tt}=\mathcal Q^{tr}=\mathcal Q^\sharp=0$. The temporal orbit Bianchi identity then gives $\mathcal Q^t=0$ because $L\ne0$, and the second displayed Bianchi identity gives $\mathcal Q^\flat=0$. Every vacuum projection therefore vanishes.

With $dr_*/dr=f^{-1}$, multiplication by $f$ puts $\mathcal W=0$ into the usual form

$$\left[-\partial_t^2+\partial_{r_*}^2-V_Z(r)\right]\psi=0, \qquad V_Z=fU_Z.$$

The potential agrees with Martel and Poisson 2005, § IV.C, Eqs. (4.25)–(4.26), p. 104003-6, PDF. Both $U_Z$ and $V_Z$ have dimension length$^{-2}$. The barrier $V_Z$ is positive in the open exterior, vanishes with $f$ at the horizon, and approaches $L/r^2$ at infinity. These checks do not specify scattering or quasinormal-mode boundary conditions.

This proves the local vacuum correspondence at fixed $\ell\ge2$. Once $\psi$ is fixed, there is no extra invariant stationary metric piece to add: $\psi=0$ makes the entire inverse zero. Arbitrary smooth $j_a,G$ restore a different metric representative through

$$h_{ab}=\widetilde h_{ab} +\nabla_a\varepsilon_b+\nabla_b\varepsilon_a, \qquad K=\widetilde K-\frac L2G+\frac{2}{r}r^a\varepsilon_a.$$

Global gauge admissibility and the separate low multipoles are additional questions. One dynamical field still requires two Cauchy-data functions, $\psi(t_0,r_*)$ and $\partial_t\psi(t_0,r_*)$.

A regular future-horizon chart

The factors $1/f$ in Schwarzschild components do not establish a tensor singularity. Set $v=t+r_*$ and let $\Phi(v,r)=\psi(t,r)$. The covector relation $dt=dv-dr/f$ gives the tensor transformation

$$h_{vv}=fH,\qquad h_{vr}=H_1-H,\qquad h_{rr}^{(v,r)}=\frac{2(H-H_1)}{f}.$$

For a solution of the wave equation, define

$$S=r\Phi_{rr}+\frac{2\lambda r}{D}\Phi_r, \qquad \gamma=rA-\frac Dr,$$

where every $r$ derivative now holds $v$ fixed. Substitution and use of $2\Phi_{vr}+\partial_r(f\Phi_r)-U_Z\Phi=0$ give

$$\begin{aligned} K&=\Phi_v+f\Phi_r+A\Phi,\\ h_{rr}^{(v,r)}&=S,\qquad h_{vr}=-\frac f2 S,\\ h_{vv}&=r\Phi_{vv}+rf\Phi_{vr}+\gamma\Phi_v+\frac{f^2}{2}S. \end{aligned}$$

All coefficients are finite at $r=2\mu$. Smooth master data through the future horizon therefore reconstruct a regular local representative there. The simplification uses the field equation; it is not an identity for an arbitrary off-shell scalar. Nor does it make an arbitrarily chosen exterior solution smooth at the horizon.

Monopoles, dipoles, and the limits of the reduction

At $\ell=0$, $Y_A=Y_{AB}=0$. The amplitudes $j_a,G$ are absent, while $h_{ab}$ and the angular trace survive. For the vacuum Schwarzschild family, a mass variation is a genuine nonradiative perturbation. In ingoing coordinates its only nonzero metric component can be written

$$p_{vv}=\frac{2G\,\delta M}{r}.$$

This follows by differentiating the exact Schwarzschild metric at fixed $(v,r,\theta,\phi)$. It changes the mass even though it has no radiative master field. If expressed in normalized $Y^{00}=1/\sqrt{4\pi}$ amplitudes, $h_{vv}^{00}=\sqrt{4\pi}\,2G\delta M/r$.

At $\ell=1$, $Y_{AB}=0$ but $Y_A$ survives. There is no amplitude $G$ and no trace-free angular equation imposing $H_0=H_2$ in the manner used above. The remaining gauge transformations and constraint system must be treated directly. A vacuum even dipole has no local radiative degree of freedom; a displacement of the coordinate origin is a useful example. Whether an asymptotic displacement is identified as gauge depends on the allowed changes of frame.

A subtle warning is worth making explicit. At $\ell=1$ the numbers $L=2$ and $\Lambda=6\mu/r$ are nonzero, so the written master expression may look finite. It is nevertheless not a defined gauge-invariant radiative variable in this sector: its construction used a harmonic and an equation that no longer exist. This is a failure of the reduction, not a removable singularity or a field whose value should be assigned zero. See Martel and Poisson 2005, § IV.D, Eqs. (4.28)–(4.36), p. 104003-6, PDF and Low Multipoles, Parameter Shifts, and Gauge Modes.

Matter sources change the constraint and reconstruction relations; the vacuum inverse cannot simply be combined with a sourced wave equation. Conserved smooth sources and distributional jumps are treated in Sourced Master Equations and Junction Conditions. The Zerilli Equation develops the covariant reduction and axial–polar operator relation. There, the geometrized mass is named $m=GM$ and the angular number $\mu=L-2$; on this page $\mu$ denotes the mass length and $2\lambda=L-2$. That notation change does not change the master normalization.

Exercises

A nonzero metric that is pure gauge. Starting from zero perturbation, choose arbitrary smooth $\xi_a,\xi$. Use the amplitude transformations to show that $\widetilde h_{ab}=\widetilde K=\psi=0$. Which step uses the Einstein equation?

Solution

The transformed amplitudes give $\varepsilon_a=-\xi_a$. Hence

$$\widetilde h_{ab} =-\nabla_a\xi_b-\nabla_b\xi_a +\nabla_a\xi_b+\nabla_b\xi_a=0,$$ $$\widetilde K =\frac{L\xi}{r^2}-\frac{2r^a\xi_a}{r} -\frac{L\xi}{r^2}+\frac{2r^a\xi_a}{r}=0.$$

The master is consequently zero. No field equation was used: gauge invariance is an off-shell property. Removing this perturbation in a specified global problem still requires an admissible generator.

A stationary polar quadrupole. For $\ell=2$, take

$$H=c r^2 f,\qquad H_1=0,\qquad K=c(r^2-2\mu^2),$$

where $c$ has dimension length$^{-2}$. Find $\psi$, check the static master equation, and decide whether this is an isolated asymptotically flat black-hole deformation.

Solution

Here $\lambda=2$. The master definition gives

$$\psi=\frac{c\,r(r^3+3\mu r^2-6\mu^3)}{3(2r+3\mu)}.$$

Substitution returns $K=A\psi+f\psi_r$ and $H=rK_r-\Lambda\psi_r/2-\Lambda'\psi/2$. Differentiating the rational expression also gives

$$\partial_r(f\psi_r)=U_Z^{\ell=2}\psi.$$

Thus the complete inverse and all vacuum equations are satisfied, despite $\psi_t=0$. The ingoing metric components are

$$h_{vv}=c r^2 f^2,\qquad h_{vr}=-c r^2 f,\qquad h_{rr}^{(v,r)}=2c r^2.$$

They are regular at the future horizon. But $K\sim cr^2$ and $\psi\sim cr^3/6$ at infinity: this is a growing tidal-type solution, not an isolated asymptotically flat deformation. Perturbation theory also requires $\lvert\eta c\rvert r^2\ll1$ in the region where it is used. Horizon regularity alone is not an asymptotic boundary condition.

A displacement is not a mass change. In the Schwarzschild exterior, let $d$ be a constant length and choose the inverse-identification vector

$$\Xi=d\left(\cos\theta\,\partial_r -\frac{\sin\theta}{r}\,\partial_\theta\right).$$

Compute $p_{\mu\nu}=-\mathcal L_\Xi g_{\mu\nu}^{(0)}$. Why is this an even dipole, and why is its removal different from removing the monopole mass variation?

Solution

The only independent nonzero components are

$$p_{tt}=d f'\cos\theta,\qquad p_{rr}=\frac{d f'}{f^2}\cos\theta,\qquad p_{r\theta}=d(f^{-1}-1)\sin\theta.$$

The angular metric variation cancels. The $\cos\theta$ and its angular derivative are the $\ell=1,m=0$ scalar and vector shapes. This is the Lie derivative of a vacuum metric, so its linearized Einstein tensor vanishes; it changes coordinates rather than the mass. Asymptotically the vector approaches $d\,\partial_z$, a translation of the origin. It can therefore be removed only if the chosen boundary/frame convention allows that translation. No horizon-regularity claim is made for this particular Schwarzschild-coordinate generator.

In contrast, differentiating the mass family changes the asymptotic mass charge. A gauge transformation preserving that charge and the asymptotic time normalization cannot remove $p_{vv}=2G\delta M/r$. Neither example is obtained by extending the $\ell\ge2$ radiative inverse to low multipoles.

References

• Martel, Karl, and Eric Poisson. “Gravitational Perturbations of the Schwarzschild Spacetime: A Practical Covariant and Gauge-Invariant Formalism.” Physical Review D 71 (2005): 104003. DOI. Open journal PDF. Open preprint PDF, v1, including coordinate appendices.