Killing Horizon for Kerr Black Hole

I have some confusion about Killing Horizon of BH.

Since a Killing Horizon (KH) is a null hyper-surface at which killing vector $k^{\mu}$ is null; $k^{\mu}k_{\mu}=0.$ For time translation symmetry $k^{\mu}=\partial_t$ in Kerr BH case.
$k^{\mu}k_{\mu}=(1-\frac{2Mr}{\rho^2})=0 \Rightarrow \rho^2=2Mr$. Which is static limit surface of Kerr BH.

So is SLS a Killing Horizon for Kerr BH?

• Is that a question or an announcement of a discovery? – Luboš Motl Jul 16 '16 at 13:20
• Yes, the Killing horizon corresponds to the ergosphere boundary. – Lawrence B. Crowell Jul 16 '16 at 13:50
• @LawrenceB.Crowell, is there only one KH in any Black Hole. – Rahul kumar walia Jul 16 '16 at 13:53

There are the ${\bf k}_t,~{\bf k}_\phi$ Killing vectors. Another condition is that $${\bf k}_t\cdot{\bf k}_\phi~=~\frac{(2mr~-~Q^2)asin\phi}{\rho^2},$$ for $a~=~J/m$. This is zero for $a~=~0$ or for $Q^2~=~2mr$ or $\phi~=~0,~\pi$. There is also $${\bf k}_\phi\cdot{\bf k}_\phi~=~\frac{(r^2~+~a^2)^2sin^2\phi~-~\Delta a^2sin^4\phi}{\rho^2},$$ for $\Delta~=~r^2~-~2mr~+~a^2~+~Q^2$. The last one is a sort of particle horizon involved with frame dragging.
• Just a small question on notation; $\mathbf{k}^t=(1,0,0,0)$? – Otto Jul 18 '16 at 6:49
• That would be for a local frame. In general $K_t~=~\sqrt{1~-~2mr/\rho}\partial_t$. – Lawrence B. Crowell Jul 18 '16 at 19:49