# Canonical energy & linear stability of Schwarzschild

Kartik Prabhu with Robert M. Wald APS 2017, Washington D.C.

# Linear stability of Schwarzschild

Dafermos, Holzegel, Rodnianski (arXiv:1601.06467)

Solutions to linearised Einstein equations around Schwarzschild with regular asymptotically flat initial data

1. remain uniformly bounded on the exterior
2. decay inverse polynomially to linearised Kerr

# Teukolsky variable

$$\tilde\gamma_{ab}$$ be linearised perturbation of Schwarzschild

• $$\psi = r^4\psi_4$$ (rescaled Weyl component) satisfies the Teukolsky equation $\left[ (þ' + \rho)(þ + 3\rho ) - ð'ð - 3 \Psi_2 \right]\psi = 0$

$$þ, þ', ð, ð'$$ are GHP derivatives, $$\rho$$ is a GHP spin-coefficient, and $$\Psi_2 = - \frac{M}{r^3}$$ .

# DHR variable

• DHR construct new variable $\Psi := (þ + \rho)(þ + 3\rho)\psi \sim \nabla^4\tilde\gamma$ which satisfies a wave-like Regge-Wheeler equation.

# DHR energy

Conserved and positive energy

$\mathscr E_{\mathrm{DHR}} = \int N \left[ \tfrac{1}{2}\left\vert (þ + þ') \Psi \right\vert^2 + \tfrac{1}{2} \left\vert (D_r - 2\rho)\Psi \right\vert^2 + \left\vert ð\Psi \right\vert^2 + \left\vert ð'\Psi \right\vert^2 + \left( \frac{4}{r^2} - \frac{6M}{r^3} \right) \left\vert \Psi \right\vert^2 \right]$

use $$\mathscr E_{\mathrm{DHR}} \sim (\nabla^3\psi)^2 \sim (\nabla^5\tilde\gamma)^2$$ to get decay estimates on $$\Psi$$, descend down to $$\tilde\gamma_{ab}$$.

# Kerr?

no Regge-Wheeler variable, so DHR method does not easily generalise

Use canonical energy method of Hollands and Wald (arXiv:1201.0463)

• $$\mathscr E_{\mathrm{can}}(\tilde\gamma) \sim (\nabla^1\tilde\gamma)^2$$ (positivity unknown, even in Schwarzschild)
• $$\mathscr E_{\mathrm{DHR}} \sim (\nabla^5\tilde\gamma)^2$$

Can we use canonical energy method to get something $$\sim (\nabla^5\tilde\gamma)^2$$ and positive?

Yes! Let's go back to Schwarzschild…

# Hertz potential

Use $$\psi$$ as a Hertz potential to generate a complex solution to Einstein's equation $\gamma_{ab} = -l_al_b U + l_{(a}m_{b)} V - m_am_b W$ with

$U = ð^2\psi \\ V = \left[ þ ð + ð ( þ + 3\rho ) \right]\psi \\ W = ( þ - \rho ) ( þ + 3\rho )\psi \\$

Note $$\gamma \sim \nabla^4 \tilde\gamma$$

# Canonical energy

Canonical energy of $$\gamma_{ab}$$

$\mathscr E_{\mathrm{can}} = \frac{1}{4}\int N \left[ \tfrac{1}{4} \left\vert (þ + þ' + D_r + 2\rho)V \right\vert^2 + \left\vert (þ + þ')W - ð' V \right\vert^2 + \left\vert D_r W + ð' V \right\vert^2 \right]$

so $$\mathscr E_{\mathrm{can}} \sim (\nabla^3\psi)^2 \sim (\nabla^5\tilde\gamma)^2$$ and positive.

So we have two conserved, positive energies at the same order in $$\tilde\gamma$$. Are they the same?

# Equality!

$\mathscr E_{\mathrm{DHR}} = \int N \left[ \tfrac{1}{2}\left\vert (þ + þ') \Psi \right\vert^2 + \tfrac{1}{2} \left\vert (D_r - 2\rho)\Psi \right\vert^2 + \left\vert ð\Psi \right\vert^2 + \left\vert ð'\Psi \right\vert^2 + \left( \frac{4}{r^2} - \frac{6M}{r^3} \right) \left\vert \Psi \right\vert^2 \right]$
$\mathscr E_{\mathrm{can}} = \frac{1}{4}\int N \left[ \tfrac{1}{4} \left\vert (þ + þ' + D_r + 2\rho)V \right\vert^2 + \left\vert (þ + þ')W - ð' V \right\vert^2 + \left\vert D_r W + ð' V \right\vert^2 \right]$

Repeatedly use Teukolsky equation for $$\psi$$, background Schwarzschild identities, integration-by-parts to get $$\mathscr E_{\mathrm{can}} = 4 \mathscr E_{\mathrm{DHR}}$$.

# Two roads diverged in a wood …

metric perturbation $$\tilde\gamma_{ab}$$ → Teukolsky variable $$\psi$$

1. $$\psi$$ → DHR's Regge-Wheeler-like variable $$\Psi$$ → $$\mathscr E_{\mathrm{DHR}}$$
2. $$\psi$$ as Hertz potential → complex metric $$\gamma_{ab}$$ → canonical energy $$\mathscr E_{\mathrm{can}}$$

In Schwarzschild, both are equivalent.

Also, works for electromagnetism in Schwarzschild using Fackerell-Isper-like equation (see Pasqualotto arXiv:1612.07244)

# … but only one leads to Kerr

In Kerr,

1. no Regge-Wheeler variable, so DHR method does not easily generalise
2. Hertz potential construction still works; $$\mathscr E_{\mathrm{can}} \geq 0$$?

Do electromagnetism in Kerr first.

1. Fackerell-Ipser equation has complex potential so no real, conserved energy.
2. use Hertz potential + canonical energy method: on going

# And both that morning equally lay

neither generalises easily to higher dimensions; curvature components do not decouple so no single Teukolsky variable

What happens? ¯\_(ツ)_/¯