Whiting Mode Stability of Kerr

Whiting’s theorem excludes exponentially growing, source-free separated Teukolsky modes on every fixed subextremal Kerr exterior. Its central idea is to transform an assumed unstable mode into an auxiliary radial problem with a positive integral identity, even though Kerr’s stationary energy is not positive in the ergoregion. This page proves that contradiction for the scalar, electromagnetic and gravitational spin sectors. It does not turn mode exclusion into a boundedness or decay theorem for arbitrary initial data.

Required background. PDEs and Green Functions for Gravitation supplies separated boundary problems, integration by parts and integral-transform uniqueness.

Helpful background. Black-Hole Geometry and Mechanics supplies Kerr horizons and the ergoregion; Formulations of General Relativity supplies tetrad reductions; Mathematical Relativity supplies theorem and estimate logic; Black-Hole Perturbation Theory supplies the surrounding mode problem. The local radial equation and angular equation give the detailed preparation for the proof.

The separated Kerr mode problem

For this sustained calculation use geometrized units $G=c=1$, so $M$ and $a$ are lengths; restoring explicit $G$ means replacing $M$ by $GM$. Let $M\gt0$ and $\lvert a\rvert\lt M$, and define

$$\begin{gathered} r_\pm=M\pm\sqrt{M^2-a^2},\qquad d=r_+-r_-\gt0,\qquad \Delta(r)=(r-r_+)(r-r_-),\\ \Omega_H=\frac{a}{r_+^2+a^2},\qquad \kappa=\frac{d}{2(r_+^2+a^2)},\qquad \frac{dr_*}{dr}=\frac{r^2+a^2}{\Delta}. \end{gathered}$$

Consider a nonzero separated massless Teukolsky field

$$\Upsilon_s=e^{-i\omega t+im\phi}S(\theta)R_s(r), \qquad s\in\{0,\pm1,\pm2\},\quad m\in\mathbb Z, \qquad \beta=\operatorname{Im}\omega\gt0.$$

Here $s=0$ is the scalar wave equation, $s=\pm1$ describes extreme Maxwell components, and $s=\pm2$ describes extreme gravitational curvature components. The angular factor must be smooth as a spin-weighted field at both poles. Choose one regular angular branch and normalize it by

$$\int_0^\pi |S|^2\sin\theta\,d\theta=1.$$

This normalization of one eigenfunction does not assert orthogonality of different complex-frequency eigenfunctions. The branch label $\ell$ is suppressed below.

Equations and physical endpoint conditions

Put $c=a\omega$ and relate the angular and radial separation constants by

$$A=\Lambda-s,\qquad \lambda=\Lambda-s+a^2\omega^2-2am\omega,\qquad K(r)=(r^2+a^2)\omega-am.$$

The homogeneous equations are

$$\frac1{\sin\theta}\frac{d}{d\theta} \left(\sin\theta\frac{dS}{d\theta}\right) +\left[ a^2\omega^2\cos^2\theta-2as\omega\cos\theta -\frac{(m+s\cos\theta)^2}{\sin^2\theta}+\Lambda \right]S=0$$

and

$$\Delta^{-s}\frac{d}{dr} \left(\Delta^{s+1}\frac{dR_s}{dr}\right) +\left[ \frac{K^2-2is(r-M)K}{\Delta} +4is\omega r-\lambda \right]R_s=0.$$

The physical branches are purely ingoing at the future horizon and purely outgoing at infinity:

$$\begin{aligned} R_s&=\Delta^{-s}e^{-i(\omega-m\Omega_H)r_*} \bigl(H_0+O(r-r_+)\bigr), &&r\to r_+,\\ R_s&=e^{i\omega r}r^{2iM\omega-2s-1} \bigl(I_0+O(r^{-1})\bigr), &&r\to\infty. \end{aligned}$$

The expansions have the corresponding differentiated asymptotics. A nonzero outgoing homogeneous solution has $I_0\ne0$ for every spin. For $s\le0$, horizon regularity also forces $H_0\ne0$; for positive spin, a smooth weighted horizon solution can have $H_0=0$ at exceptional Frobenius frequencies. Powers of dimensional positive radii mean powers of their ratios to a fixed reference length, absorbed into the arbitrary amplitude. At the horizon the condition means smoothness after removing the displayed singular weight and phase, not smoothness of the raw $R_s$. For $\beta\gt0$, the weighted radial profiles decay on a fixed-$t$ slice. This is compatible with a regular, nonzero spacetime field on the future horizon in ingoing coordinates.

Whiting’s conclusion. No nonzero pair $(S,R_s)$ satisfies all these conditions. The original result is due to Whiting 1989, pp. 1301–1305. A modern full statement, also covering half-integer spin with its corresponding angular regularity and azimuthal labels, is Teixeira da Costa 2020, Theorem 1.1, p. 707. We prove the bosonic sectors used in this volume without assuming that the complex-frequency angular modes form a complete basis.

Why the original energy does not settle the question

For a real-frequency scalar scattering solution, the horizon generator $\chi=\partial_t+\Omega_H\partial_\phi$ measures $k_H=\omega-m\Omega_H$. In a flux-normalized scalar basis, the energy entering the future horizon has sign

$$\mathcal F_H\ \propto\ \omega(\omega-m\Omega_H).$$

It can be negative when $0\lt\omega\lt m\Omega_H$. This is the separated manifestation of the ergoregion: $\partial_t$ is not everywhere timelike. The scalar current identity is given in Teixeira da Costa 2020, Proposition 2.20, p. 729. Raw electromagnetic or gravitational Teukolsky amplitudes are not themselves flux-normalized physical waves.

One can see the obstruction directly, without confusing a real scattering problem with an unstable mode. In the scalar case, multiply the original radial equation by $\overline{R_0}$, integrate, then take the imaginary part after multiplication by $\overline\omega$. The angular identity derived below gives

$$0=\beta\int_{r_+}^\infty \left[ \Delta|R_0'|^2+ \left\{ |\omega|^2\left(\frac{(r^2+a^2)^2}{\Delta} -a^2+a^2C\right) +A_0-\frac{a^2m^2}{\Delta} \right\}|R_0|^2 \right]dr.$$

The negative term proportional to $a^2m^2/\Delta$ prevents the Schwarzschild-style inference that the entire integrand is nonnegative. Here $A_0$ and $C$ are the nonnegative angular quantities defined below. When $a=0$, this particular obstruction disappears. Its presence for Kerr does not prove instability; it shows why this direct identity is insufficient. A driven superradiant scattering event also need not generate repeated feedback or an exponentially growing homogeneous mode.

A convergent radial transformation

First take $s\le0$ and suppress the spin subscript on $R$. Define

$$\xi=\frac{i(am-2Mr_+\omega)}d,\qquad \eta=\frac{i(2Mr_-\omega-am)}d,\qquad \xi+\eta=-2iM\omega.$$

For an independent auxiliary coordinate $x\gt r_+$, set

$$\begin{aligned} F(r)&=(r-r_-)^\eta(r-r_+)^\xi e^{-i\omega r}R(r),\\ \mathcal K(x,r)&=\exp\left[ \frac{2i\omega}{d}(x-r_-)(r-r_-)\right],\\ I(x)&=\int_{r_+}^\infty \mathcal K(x,r)F(r)\,dr,\\ u(x)&=(x-r_-)^{-s}(x-r_+)^{-2iM\omega}e^{-i\omega x}I(x). \end{aligned}$$

The integral changes the whole radial function, not merely its value at $r=x$. All powers use the real logarithm on these positive intervals. The exponential in $\mathcal K$ is dimensionless. This is the radial-transform presentation in Andersson, Ma, Paganini, and Whiting 2017, arXiv v2, Eqs. (44)–(59), pp. 17–19, PDF, translated from their radial unknown $\Delta^{s/2}R$ and their separation constant $\lambda+s$.

Why does the integral exist? The physical conditions give

$$F(r)=(r-r_+)^{2\xi-s}h(r),\qquad \operatorname{Re}(2\xi-s)=\frac{4Mr_+\beta}{d}-s\gt-1$$

near the horizon, with $h$ smooth there. At infinity, $F(r)=O(r^{-2s-1})$: its exponential factors cancel. It need not be integrable, or square-integrable, without an extra weight. The kernel supplies that weight:

$$\lvert\mathcal K(x,r)\rvert =\exp\left[-\frac{2\beta}{d}(x-r_-)(r-r_-)\right].$$

For every $x\ge r_+$ this defeats polynomial growth. It also permits differentiation under the integral. Both the horizon integrability and the infinity damping are needed.

From the kernel to the transformed equation

Here is an explicit way to verify the transformation. Factor the original radial function as

$$R=(r-r_-)^{\eta-s}(r-r_+)^{\xi-s}e^{-i\omega r}g.$$

Then $g$ satisfies $T_rg=0$, where

$$\begin{gathered} T_r=\Delta\partial_r^2+ \left[b(r-r_+)+c_0(r-r_-)+p\Delta\right]\partial_r +\alpha p(r-r_-)+\sigma,\\ b=1+2\eta-s,\quad c_0=1+2\xi-s,\quad p=-2i\omega,\quad \alpha=1-2s,\\ \sigma=-(\lambda+s)-s-2i\omega r_-(1-2s). \end{gathered}$$

Do not confuse $c_0$ with the spheroidal parameter $c=a\omega$. Replace $(b,c_0,\alpha)$ by $(\alpha,b+c_0-\alpha,b)$ to define $\widetilde T_x$, keeping $p,\sigma$ unchanged and replacing $r$ by $x$. Direct differentiation of the exponential shows

$$\widetilde T_x\mathcal K=T_r\mathcal K.$$

The integration-by-parts weight is

$$w(r)=(r-r_-)^{2\eta-s}(r-r_+)^{2\xi-s}e^{-2i\omega r}, \qquad T_r=w^{-1}\partial_r(\Delta w\,\partial_r) +\alpha p(r-r_-)+\sigma.$$

Since $I=\int\mathcal K wg\,dr$, the only boundary term in moving $T_r$ between $\mathcal K$ and $g$ is

$$\left[ \Delta w\left(g\,\partial_r\mathcal K-\mathcal K\,\partial_rg\right) \right]_{r_+}^{\infty}.$$

It vanishes: at the horizon the extra factor from $\Delta$ gives magnitude $O((r-r_+)^{2\operatorname{Re}\xi-s+1})$, and at infinity the kernel damps all remaining powers. Thus $\widetilde T_xI=0$. Applying the displayed prefactor defining $u$ yields the divergence-form equation

$$\frac{d}{dx}\left(\Delta(x)\frac{du}{dx}\right)+V_0(x)u=0,$$

with

$$\begin{aligned} V_0(x)&=\omega^2P(x)-\frac{4am(x-M)}d\,\omega -\Lambda-s^2\frac{x-r_+}{x-r_-},\\ P(x)&=\frac{(x-r_-)(x+r_-)^2}{x-r_+} +\frac{8Mr_-(x-r_-)}d-a^2. \end{aligned}$$

This also follows by expanding Eq. (59) in the cited presentation. It is an auxiliary equation; $V_0$ is generally complex when $\omega$ is complex. Divergence form is not a claim of a self-adjoint physical Kerr operator at that frequency.

What happens at the two endpoints

Inspect how the transform changes the endpoint powers. Write $y=x-r_+$ and $p_\infty=\eta-\xi-1$.

Auxiliary endpoint	Selected transformed behavior	Boundary bracket
$y\to0^+$	$u=y^{-2iM\omega}(u_H+O(y))$	$\Delta\overline u\,u'=O(y^{4M\beta})$
$x\to\infty$	$u=e^{i\omega x}x^{p_\infty}(u_\infty+O(x^{-1}))$	$\Delta\overline u\,u'=O(e^{-2\beta x}x^{-8M^2\beta/d})$

Primes in this table mean $d/dx$. The first row follows because $I(x)$ is smooth at $r_+$. For the second, put $r=r_++z$ in the integral. The horizon part is a Laplace integral of $z^{2\xi-s}h(r_++z)$, whose leading term is

$$I(x)\sim e^{2i\omega(x-r_-)}h(r_+)\Gamma(2\xi-s+1) \left[-\frac{2i\omega(x-r_-)}d\right]^{-(2\xi-s+1)}.$$

The complex power here uses the logarithm in the right half-plane. The portion away from $z=0$ is exponentially smaller. Combining this with the prefactor gives

$$p_\infty=-1+\frac{i(4M^2\omega-2am)}d,\qquad \operatorname{Re}p_\infty=-1-\frac{4M^2\beta}d.$$

Differentiating the Laplace expansion gives the corresponding derivative estimate. At the horizon both quadratic densities $\Delta|u'|^2$ and $P|u|^2$ are $O(y^{4M\beta-1})$, hence integrable; at infinity they decay exponentially. The table therefore justifies both taking limits in the integrated equation and dropping its boundary bracket. It does not identify $u$ with the original physical field.

The positive integral and the contradiction

The remaining obstacle is the complex angular eigenvalue. Define three real moments of the chosen normalized $S$:

$$\begin{aligned} A_0&=\int_0^\pi \left[ |S'|^2+\frac{(m+s\cos\theta)^2}{\sin^2\theta}|S|^2 \right]\sin\theta\,d\theta,\\ B&=\int_0^\pi\cos\theta\,|S|^2\sin\theta\,d\theta,\qquad C=\int_0^\pi\cos^2\theta\,|S|^2\sin\theta\,d\theta. \end{aligned}$$

Here the prime means $d/d\theta$. Pole regularity kills $[\sin\theta\,\overline S S']_0^\pi$. Multiplication of the angular equation by $\overline S$ and integration gives exactly

$$\Lambda=A_0-a^2\omega^2C+2as\omega B, \qquad A_0\ge0,\quad 0\le C\le1.$$

The moments depend on this mode. They are real even though $\Lambda$ need not be. Substitution writes the transformed potential as a quadratic expression with real coefficients:

$$\begin{gathered} V_0=\omega^2D-\omega E-Q,\\ D=P+a^2C,\qquad E=\frac{4am(x-M)}d+2asB,\qquad Q=A_0+s^2\frac{x-r_+}{x-r_-}. \end{gathered}$$

For every $x\gt r_+$, we have $r_-\ge0$, $r_+\gt|a|$, and $(x-r_-)/(x-r_+)\gt1$. Consequently

$$P(x)\gt x^2-a^2\gt0,\qquad D\gt0,\qquad Q\ge0.$$

Now integrate the transformed equation against $\overline u$. The vanishing bracket established above gives

$$0=-\int_{r_+}^\infty\Delta|u'|^2\,dx +\int_{r_+}^\infty(\omega^2D-\omega E-Q)|u|^2\,dx.$$

Multiply by $\overline\omega$ and take the imaginary part. The troublesome term $-|\omega|^2E$ is real and drops out. The result is

$$0=\beta\int_{r_+}^\infty \left[ \Delta|u'|^2+ \left(|\omega|^2D+Q\right)|u|^2 \right]dx.$$

Every term is nonnegative, and $\beta|\omega|^2D$ is strictly positive. Therefore $u=0$. No sign restriction on $\operatorname{Re}\omega$, $m$ or $a$ was introduced. In particular, $A_0$ need not be strictly positive: the argument still works for the scalar monopole. Dimensions also agree: $D$ has units of length squared, $E$ of length, and $Q,V_0$ are dimensionless. This is the explicit upper-half-plane contradiction underlying the modern proof in Teixeira da Costa 2020, § 4.2, pp. 758–759.

Why a vanishing transform implies a vanishing mode

The prefactor is nonzero for $x\gt r_+$, so $u=0$ implies $I=0$. Consider

$$L(z)=\int_{r_+}^\infty e^{-z(r-r_-)}F(r)\,dr, \qquad \operatorname{Re}z\gt0.$$

Horizon integrability and polynomial growth of $F$ make $L$ holomorphic in this half-plane. The identity $I(x)=0$ says that $L$ vanishes on $z=-2i\omega(x-r_-)/d$, a ray inside its domain with interior accumulation points. The identity theorem gives $L=0$ throughout the half-plane.

For any fixed $\epsilon\gt0$, the values $L(\epsilon+i\tau)$ are the Fourier transform in the variable $\rho=r-r_-$ of the integrable function $e^{-\epsilon\rho}F(r_-+\rho)$, extended by zero below $\rho=d$. Fourier uniqueness forces $F=0$, hence $R=0$. This is the injectivity step: the integral transform has not merely erased a possible unstable mode. It uses a damped input, not an unjustified assertion that the original polynomially growing $F$ lies in $L^2$.

Positive spin modes

For $s=1$ or $2$, first apply the Teukolsky–Starobinsky map

$$R_{-s}=\Delta^s(\mathcal D_+)^{2s}(\Delta^sR_{+s}), \qquad \mathcal D_+=\frac{d}{dr}+\frac{iK}{\Delta}.$$

It produces a spin-$-s$ radial solution with $\lambda_{-s}=\lambda_{+s}+2s$. The corresponding angular factor is $S_{-s}(\theta)=S_{+s}(\pi-\theta)$, with the same $\Lambda$ and regularity. The map preserves the two physical endpoint conditions on a hypothetical outgoing homogeneous mode. Importantly, its outgoing infinity coefficient is $(2i\omega)^{2s}I_0$, which is nonzero here. Thus it produces a nontrivial negative-spin mode, already ruled out by the integral proof. The precise boundary-preserving statement and its nondegeneracy are Teixeira da Costa 2020, Proposition 2.14, pp. 723–724, Lemma 2.19, p. 729, and § 4.3, p. 760. This argument does not assume that a Starobinsky constant is nonzero for every arbitrary complex-frequency solution.

We have now excluded all the stated bosonic separated modes, including both gravitational curvature sectors. Reconstructing a metric still requires the constraints, gauge choices and nonradiative mass/angular-momentum completion. Absence of growing curvature modes is not, by itself, the full linearized Einstein stability theorem.

Real frequencies and evolution

The proof above uses $\beta\gt0$ twice: to define an absolutely convergent transform and to obtain a strictly positive integral. At real frequency the integral becomes oscillatory and the final factor $\beta$ vanishes. One cannot replace those steps with an unsupported continuity argument.

For scalar waves, Shlapentokh-Rothman proved nonzero real-frequency mode stability and a quantitative inverse-Wronskian bound. At fixed $(a,M)$, that bound applies on parameter sets for which $|\omega|+|\omega|^{-1}+|m|+|\ell|$ is bounded; see Shlapentokh-Rothman 2015, arXiv v3, Theorems 1.4–1.6, pp. 7–8, PDF. It is not a zero-frequency or uniform extremal-limit estimate. Andersson, Ma, Paganini and Whiting established the general-spin real-axis result, including the nonzero synchronous frequency $\omega=m\Omega_H$, with the appropriate horizon regularity; see Andersson et al. 2017, arXiv v2, Theorem 1 and frequency symmetry, p. 6, PDF. Their theorem is not a new consequence of the positive integral written above.

Damped quasinormal modes have $\operatorname{Im}\omega\lt0$: their selected ingoing and outgoing phase factors grow toward the spatial endpoints on fixed-time slices. They are not excluded. Real-frequency superradiant reflection is also compatible with the theorem: it is a driven scattering problem, not the source-free growing-mode problem.

Neither qualitative mode exclusion nor a bounded-frequency Wronskian estimate alone supplies a complete quasinormal expansion, a global resolvent estimate or bounded arbitrary-data evolution. The latter requires control of trapping, low and high frequencies, and the relevant norms. Continue to Boundedness and Decay of Linear Fields for that distinction, or Canonical Energy and Linearized Gravitational Stability for metric perturbations.

Finally, $d\to0$ is singular in the displayed kernel and coefficients: this proof does not yield extremal mode stability by setting $a=M$. Separate extremal results and horizon-derivative growth require their own hypotheses; see Extremal Horizons and Aretakis Instability. Full nonlinear evolution remains a further problem, treated in Black-Hole Dynamics.

Exercises

Distinguish the raw radial field from its regular weight

Let $\rho=(r-r_+)/d$ and assume $H_0\ne0$. Find the magnitudes of the ingoing $R_s$ and $\Delta^sR_s$ near the horizon when $\beta\gt0$. Must the raw radial function vanish? Test $s=2$ and $\beta=\kappa$.

Solution

Since $r_*=(2\kappa)^{-1}\log\rho+O(1)$ and $\Delta$ is proportional to $\rho$ at leading order,

$$\lvert R_s\rvert\asymp\rho^{-s+\beta/(2\kappa)},\qquad \lvert\Delta^sR_s\rvert\asymp\rho^{\beta/(2\kappa)}.$$

For $s=2$ and $\beta=\kappa$, the first behaves as $\rho^{-3/2}$ and the second as $\rho^{1/2}$. Thus the raw radial function can diverge while satisfying the correct weighted ingoing condition. These are fixed-$t$ radial statements; they do not say that the regular spacetime field vanishes along the future horizon.

Eliminate a complex separation constant

Write $\omega=\omega_R+i\beta$. Derive $\operatorname{Im}\Lambda$ and $\operatorname{Im}\lambda$ from the angular identity. Explain why setting either imaginary part to zero would not be justified.

Solution

The real moments $A_0,B,C$ give

$$\begin{aligned} \operatorname{Im}\Lambda &=2a\beta(sB-a\omega_RC),\\ \operatorname{Im}\lambda &=2a\beta\left[sB+a\omega_R(1-C)-m\right]. \end{aligned}$$

The second formula follows from $\lambda=\Lambda-s+a^2\omega^2-2am\omega$. A unit norm makes these moments real, not the eigenvalue. The proof instead uses the combination

$$\operatorname{Im}(\overline\omega\Lambda) =-\beta\left(A_0+a^2|\omega|^2C\right),$$

whose sign follows from the angular equation and its vanishing pole term.

Test what a stable spectrum does not control

In dimensionless time $\tau$, let

$$A_\epsilon= \begin{pmatrix} -\epsilon&1\\0&-\epsilon \end{pmatrix}, \qquad \epsilon\ge0.$$

Solve $\dot v=A_\epsilon v$ with $v(0)=(0,1)^T$. For $\epsilon\gt0$, can eigenvalues strictly in the left half-plane give a norm bound uniform as $\epsilon\to0^+$? What happens at $\epsilon=0$?

Solution

The nilpotent off-diagonal matrix has square zero, so

$$v(\tau)=e^{-\epsilon\tau} \begin{pmatrix}\tau\\1\end{pmatrix},\qquad \|v(1/\epsilon)\|_2 =e^{-1}\sqrt{1+\epsilon^{-2}} \ge\frac1{e\epsilon}.$$

For each fixed positive $\epsilon$, the solution ultimately decays and is bounded. The bound is not uniform as $\epsilon\to0^+$, despite the two negative eigenvalues. At $\epsilon=0$, there are still no exponentially growing eigenmodes, but $v(\tau)=(\tau,1)^T$ grows polynomially. This finite-dimensional example is not a Kerr solution; it diagnoses the logical gap between qualitative mode exclusion and quantitative evolution control.

References

• Andersson, L., Ma, S., Paganini, C., and Whiting, B. F. (2017). “Mode stability on the real axis.” Journal of Mathematical Physics 58, 072501. DOI. Open PDF, arXiv:1607.02759v2.
• Shlapentokh-Rothman, Y. (2015). “Quantitative mode stability for the wave equation on the Kerr spacetime.” Annales Henri Poincaré 16, 289–345. DOI. Open PDF, arXiv:1302.6902v3.
• Teixeira da Costa, R. (2020). “Mode stability for the Teukolsky equation on extremal and subextremal Kerr spacetimes.” Communications in Mathematical Physics 378, 705–781. DOI. Open PDF.
• Whiting, B. F. (1989). “Mode stability of the Kerr black hole.” Journal of Mathematical Physics 30, 1301–1305. DOI.