Simpler version

Growth rate of Black Hole Instabilities

Kartik Prabhu with Robert M. WaldAPS2015, Baltimore, MD


  • Hollands and Wald [arXiv:1201.0463]
    • for axisymmetric perturbations, \(\mathscr{E} > 0\implies\) mode stability.
    • Negative energy perturbation can not go to stationary solution. (“instability”)
  • show negative energy\(\implies\)unbounded growth. Find the growth rate.


perturbations \(Q = \mathbb{K}P\) define: \[ \omega^2 = \frac{ \langle Q, \mathbb A Q \rangle_{\mathscr H} }{ \langle Q, Q \rangle_{\mathscr H} } = \frac{\mathscr U(Q, Q)}{\mathscr K (P,P)} \]

\(\omega^2 < 0 \) for some \(Q\) \(\implies\) exponential growth;
at least \(\sim e^{|\omega|t}\)


\( (\pi^{ab},h_{ab})\) on \(\Sigma\)


\(( p_{ab},q_{ab})\)

linearised constraints & boundary conditions

\(D \geq 4\), compact bifurcation surface \(B\)

restrict to static or stationary-axisymmetric with \(t\phi\)-reflection isometry.


\(P \equiv p_{ab}\) (\(t\)-odd) \(Q \equiv q_{ab}\) (\(t\)-even)


\(P \equiv (p_{polar}, q_{axial})\) (\(t\phi\)-odd) \(Q \equiv (p_{axial}, q_{polar})\) (\(t\phi\)-even)
canonical transformation

Canonical Energy

\(\mathscr E = \mathscr K(P,P) + \mathscr U(Q,Q) \) (gauge invariant)


For perturbations around a static background or axisymmetric perturbations around stationary-axisymmetric background, the kinetic energy is a symmetric bilinear form on \(P\), gauge-invariant and:

  • \(\mathscr K \geq 0 \)
  • \(\mathscr K = 0 \) iff \(P\) is a pure gauge


Fix a gauge which preserve the reflection symmetry.

\[ \dot Q = \mathbb K P \quad \dot P = -\mathbb U Q \] gauge-fixed ADM evolution equations.

\(\mathscr E = \int P\dot Q - Q\dot P = \langle P, \mathbb KP \rangle + \langle Q, \mathbb UQ \rangle \)

\( \langle -,- \rangle \) is \(L^2\) inner product

Hilbert space

dynamical evolution \(\ddot Q = -\mathbb A Q\) where \(\mathbb A = \mathbb{KU} \)

Define Hilbert space \(\mathscr H\) of \(Q\)s so that: \(\langle Q', Q\rangle_{\mathscr H} = \langle Q', \mathbb K^{-1} Q\rangle \) “inverse KE”

On this we have \(\langle Q, \mathbb AQ\rangle_{\mathscr H} = \langle Q, \mathbb U Q\rangle \)

\(\mathbb A\) is symmetric in \(\mathscr H\)

Spectral Solution

Using the spectral theorem

\[\begin{split} Q_t & = \cos(t{\mathbb A}_+^{1/2})\Pi_+ Q_0 + \sin(t{\mathbb A}_+^{1/2}){\mathbb A}_+^{-1/2} \Pi_+\dot Q_0 \\ & \quad + \Pi_0Q_0 + t \Pi_0 \dot Q_0 \\ & \quad + \cosh(t{\mathbb A}_-^{1/2})\Pi_- Q_0 + \sinh(t{\mathbb A}_-^{1/2}){\mathbb A}_-^{-1/2} \Pi_-\dot Q_0 \end{split}\]

  • \({\mathbb A}_+ \) and \( -{\mathbb A}_- \) are the positive and negative parts of \( {\mathbb A}\)
  • \(\Pi_+, \Pi_0, \Pi_-\) are projections onto the spectrum of \( {\mathbb A} \) respectively.
unique solution due to well-posed IVF for the PDE version.


If the potential energy \(\mathscr U < 0\) on \(\mathscr H\), then the black hole background is unstable in the sense that there exist solutions to the linearised evolution equations which grow without bound (at least exponentially) in time.

Gauge invariant quantities also grow without bound.

\(\mathscr{H}\) caveat

  • Perturbation in \(\mathscr{H}\) if \(\mathbb{K}\) can be inverted
  • Perturbation \(X = £_t X'\) i.e. \(Q = \mathbb K P'\)
  • Similar to Wald’s 1979-result for scalar fields in Schwarzschild \(\psi\vert_B = 0\)

Moral of the Story