Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 52 / 3 / b

Write $A=r^2+a^2$, $s=\sin \theta$, $\Sigma =A-a^2{s}^2$, and $\Delta =A-2Mr$. Use ingoing Kerr coordinates, so $dt=dv-Adr/\Delta$ and $d\varphi =d\chi -adr/\Delta$. To see the cancellations without expanding every term, write the Kerr metric in the equivalent formThe combinations becomeThe first square contributes $-\Sigma d{r}^2/\Delta$, cancelling the explicit radial term. Expanding the remaining terms givesThere is no denominator $\Delta$ in this Lorentzian metric. At the outer horizon $r_{+}=M+\sqrt{M^2-a^2}$, $\Sigma >0$ and the components are smooth. The determinant is $-{\Sigma }^2{\sin }^2\theta$, so away from the usual polar-coordinate degeneracy the metric is nondegenerate and extends across $r_{+}$. The axis can be covered by regular angular charts. Thus the Boyer-Lindquist coordinates are singular there, while the ingoing Kerr coordinates are regular at the future horizon.

For physical nonextremality the invariant parameter condition is $M>|a|$, $M>0$. The printed $M>a$ is sufficient when the rotation orientation has been chosen so that $a\ge 0$; without that convention it needs the absolute value.

The change to ingoing Kerr coordinates adds functions of $r$ to $t$ and $\varphi$, and leaves $r,\theta$ unchanged. Differentiating at fixed $r,\theta ,\varphi$ therefore gives ${\partial }_{t}v=1$ and ${\partial }_t\chi =0$. Differentiating at fixed $r,\theta ,t$ gives ${\partial }_{\varphi }\chi =1$ and ${\partial }_{\varphi }v=0$. Hence the two Killing vector fields areThey remain the stationary and axial Killing vector fields; the coordinate change does not mix their generators with ${\partial }_r$.

The inverse Kerr metric in ingoing Kerr coordinates gives the raised normal to a surface of constant $r$:On $r=r_{+}$, $\Delta =0$, so the normal is null and tangent to the horizon. It reduces toA constant linear combination of the two Killing vector fields is again a Killing vector field. Therefore $k+{\Omega }_{H}m$ is normal to this null hypersurface, establishing that it is a Killing horizon, with Kerr horizon angular velocityThe second equality uses $\Delta (r_{+})=0$. This normal calculation proves hypersurface orthogonality on the horizon, rather than merely proving that the proposed vector happens to have zero norm there.

A generator of the Killing horizon is an orbit of $k+{\Omega }_{H}m$. In ingoing Kerr coordinates it has constant $r=r_{+}$ and $\theta$, with $d\chi /dv={\Omega }_H$. Since the coordinate shifts depend only on $r$, this is also $d\varphi /dt={\Omega }_H$ in the limiting Boyer-Lindquist coordinates description.

Thus ${\Omega }_H$ is the angular velocity of the horizon relative to the nonrotating stationary frame at infinity. The stationary Killing vector field $k$ is normalized to unit time translation there, while the axial Killing vector field has $2\pi$-periodic orbits. This normalization makes the Kerr horizon angular velocity physically definite. It describes the rotation of the null generators and the dragging of inertial frames, not a material solid surface rotating through space. In the Schwarzschild black hole limit $a=0$, ${\Omega }_H=0$.

For the scalar wave separation in Kerr spacetime, continue to use $A=r^2+a^2$ and $s=\sin \theta$. First verify the determinant in the hint. Direct multiplication of the covariant $t,\varphi$ components givesHence the block determinant is $-\Delta {\sin }^2\theta$. Inverting this block givesThe other inverse components are $g^{rr}=\Delta /\Sigma$, $g^{\theta \theta }=1/\Sigma$, and $\sqrt{-g}=\Sigma s$. The covariant wave operator on a scalar consequently has the divergence formLet $m_{\varphi }$ denote the azimuthal mode number, to distinguish it from the axial vector $m$. Insert the mode $\Psi =e^{-i\omega t+i{m}_{\varphi }\varphi }R(r)\Theta (\theta )$ in the massless Klein-Gordon equation. Single-valuedness makes $m_{\varphi }$ an integer. The $t,\varphi$ derivatives giveWith $K(r)=A\omega -a{m}_{\varphi }$, division by the mode factor gives, on patches where $R\Theta \ne 0$,The radial and angular expressions must be opposite constants. Defining the separation constant as $\Lambda$, we obtain the two ordinary differential equationsThese equations also hold at zeros of a mode by continuity, without dividing there. Regular angular solutions are scalar spheroidal harmonics, with discrete $\Lambda ={\Lambda }_{\ell m_{\varphi }}(a\omega )$. For $a\omega =0$ the angular equation becomes the associated Legendre function equation, with $\Lambda =\ell (\ell +1)$ and $\ell \ge |m_{\varphi }|$, providing a useful check of the signs and normalization. The radial function here is exactly $R$ in the chosen ansatz, without an additional factor of $1/r$.