Mode Excitation and Amplitudes

Two mergers that leave the same Kerr mass and spin have the same linear quasinormal frequencies, yet they need not produce the same ringdown waveform. A pole location belongs to the stationary response operator; its residue, its projection against the source, and the chosen field normalization determine the amplitude. This page makes that separation explicit and shows why a coefficient obtained from a finite-window fit need not equal a pole-excitation coefficient.

Helpful background. Einstein Equations supplies the linearized field equation; Waves, Green Functions, and Geometric Optics supplies forced linear evolution and residues; Black-Hole Perturbation Theory supplies QNM poles and excitation factors; Gravitational Radiation and Waveforms supplies the radiative strain; and Black-Hole Dynamics and Collapse supplies the remnant and source history that excite it.

Spectrum, residue, and source projection

Use geometrized units $G=c=1$ for this remnant calculation: $M_f$ is the remnant mass as a time or length. For a physical mass $M_{f,\rm phys}$, restore a time $bM_f$ as $bGM_{f,\rm phys}/c^3$ and an angular frequency $\widehat\omega/M_f$ as $\widehat\omega c^3/(GM_{f,\rm phys})$.

Work in the remnant-aligned inertial Bondi frame of the QNM waveform expansion, with

$$H=r(h_+-ih_\times),\qquad H\propto e^{-i\omega u},$$

unit-normalized spin-$-2$ angular functions, remnant units $M_f\omega$, and $\operatorname{Im}\omega\lt 0$. The retarded radial problem is ingoing at the future horizon and outgoing at future null infinity.

To display the residue without field-dependent powers of radius, first use an asymptotically flat Schwarzschild master equation in Schrödinger form. For one separated channel, let $R^{\rm in}(\omega,r)$ be the homogeneous solution that is ingoing at the horizon. At large radius write

$$R^{\rm in}\sim A_{\rm in}(\omega)e^{-i\omega r_*} +A_{\rm out}(\omega)e^{+i\omega r_*}.$$

A QNM frequency $\omega_n$ is a zero of $A_{\rm in}$: no wave arrives from infinity. Near that zero the Green function has a pole. In the normalization used by Berti, Cardoso, and Starinets, its source-independent excitation factor can be written

$$B_n= \left. \frac{A_{\rm out}} {2\omega\,dA_{\rm in}/d\omega} \right|_{\omega_n},$$

whereas the waveform coefficient has the structure

$$C_n=B_n\,I_n, \qquad I_n=\int_{-\infty}^{+\infty} \frac{I(\omega_n,r)R^{\rm in}(\omega_n,r)} {A_{\rm out}(\omega_n)}\,dr_*.$$

Here $I(\omega,r)$ encodes the initial data or forcing. Thus $\omega_n$ identifies the pole, $B_n$ measures the residue in a declared field and radial convention, and $I_n$ tells how the particular source overlaps that response. This distinction and the corresponding formulas are given in Berti, Cardoso, and Starinets 2009, § 3.2, eqs. (33)–(38), pp. 18–20, PDF.

Changing the progenitor orbit, perturbing pulse, angular content, or source phase changes $I_n$ without moving $\omega_n$. It can even make a permitted mode nearly absent through a small or cancelling projection. Universal frequencies therefore do not imply universal amplitudes.

The integral is literal only when its endpoint behavior permits evaluation at $\omega_n$. Damped-QNM radial functions grow on a constant-time slice, so generic noncompact source data require the analytic continuation or regularization inherited from the retarded Green function; one must not treat a divergent real-axis overlap as an ordinary inner product. For Kerr, the horizon phase uses $\omega-m\Omega_H$, and a Teukolsky variable carries additional radial powers and a different Wronskian normalization. The invariant lesson is the pole/residue/source separation, not a universal value of $B_n$ for every field variable.

Why non-Hermitian projection needs a dual mode

The same logic appears in a finite-dimensional response problem. Let a linear state satisfy

$$\dot x=Lx+s(t),$$

and suppose $Lv_n=-i\omega_n v_n$. Because the QNM problem is non-Hermitian, projection generally uses a dual left covector $w_n$ satisfying

$$w_nL=-i\omega_n w_n, \qquad w_nv_m=\delta_{nm}.$$

After the forcing ends at $u_0$, the modal coefficient is

$$c_n(u_0)=w_nx(u_0) =w_ne^{L(u_0-u_i)}x(u_i) +\int_{u_i}^{u_0}w_ne^{L(u_0-u')}s(u')\,du'.$$

This formula displays both initial-data and source-history excitation. It also supplies a normalization check. If the right mode is rescaled as $v_n'=\lambda_n v_n$, the dual must become $w_n'=\lambda_n^{-1}w_n$. Hence

$$c_n'=\lambda_n^{-1}c_n, \qquad c_n'v_n'=c_nv_n.$$

The coefficient alone is not invariant; the reconstructed field is. A reported amplitude is scientifically meaningful only with its strain variable, angular/radial normalization, tetrad and Bondi frame, reference time, and remnant mass convention.

Pole coefficient versus finite-window amplitude

Suppose the exact waveform on a window is

$$d(u)=Cq(u)+p(u), \qquad q(u)=e^{-i\omega_n(u-u_0)},$$

where $C$ is the pole coefficient and $p$ contains prompt response, tails, other modes, nonlinear terms, background drift, or numerical error. An unweighted least-squares fit with only $q$ returns

$$\widehat C =\frac{\langle q,d\rangle}{\langle q,q\rangle} =C+\frac{\langle q,p\rangle}{\langle q,q\rangle}, \qquad \langle a,b\rangle=\int_{u_0}^{u_1}a^*(u)b(u)\,du.$$

Unless the omitted contribution vanishes or is orthogonal to $q$ in this particular window and norm, $\widehat C\ne C$. Moving either endpoint changes the leakage term. Adding more damped sinusoids can reduce the residual while redistributing coefficients among nearly dependent columns. This is why amplitude stability under start time, end time, model order, and resolution is a stronger physical diagnostic than fit quality alone.

The distinction is especially important near merger, where the stationary-background and pole-dominance assumptions are least secure, and at late times, where rapidly damped terms can fall below the numerical floor. The production evolution and its convergence belong to Computational Relativity; detector-weighted inference belongs to Gravitational-Wave Astronomy.

Exercise: expose omitted-component leakage

On $0\le t\lt \infty$, let a real toy signal be

$$d(t)=3e^{-t}+0.6e^{-2t}.$$

Fit it with the one-column model $\widehat C e^{-t}$ using the unweighted squared-error norm. Find $\widehat C$. Then rescale the model function to $v'(t)=2e^{-t}$ and find its fitted coefficient. What remains invariant?

Solution

With $q=e^{-t}$,

$$\widehat C =3+0.6\frac{\int_0^\infty e^{-3t}dt} {\int_0^\infty e^{-2t}dt} =3+0.6\frac{1/3}{1/2} =3.4.$$

The true coefficient of the first exponential is $3$; the extra $0.4$ is projection of the omitted faster term onto the retained column. For $v'=2q$, reciprocal normalization gives $\widehat C'=\widehat C/2=1.7$. The product is unchanged: $\widehat C'v'=\widehat Cq=3.4e^{-t}$. Thus neither a fitted number without its mode normalization nor a low residual establishes the source’s pole coefficient.

References

• Emanuele Berti, Vitor Cardoso, and Andrei O. Starinets, “Quasinormal Modes of Black Holes and Black Branes,” Classical and Quantum Gravity 26 (2009) 163001. DOI. Open PDF.