Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 311 / 3 / b

Work in the Boyer-Lindquist coordinates exterior $r>r_{+}=M+\sqrt{M^2-a^2}$, away from the axes. There $\Delta >0$, $\Sigma >0$, andInverting the $t,\varphi$ block of the Kerr metric givesThus $W^a=-g^{ab}(dt{)}_b=-{\nabla }^{a}t$ has $g(W,W)=g^{tt}<0$, and the smooth nonvanishing timelike vector field $W$ defines a time orientation. Choose it future-directed, matching increasing $t$ near infinity. For every nonzero future causal vector $Z$, the Lorentzian inner product satisfies $g(W,Z)<0$, henceSo $t$ increases strictly along every regular future causal curve in this exterior. Notice that ${\partial }_t$ itself can be spacelike in the Kerr ergoregion; the timelike object used here is $-\nabla t$. Direct substitution also gives $D=\Delta {\sin }^2\theta$.

For a regular causal curve, orient its tangent $Z$ toward the future; reversing its parametrization does not change $d\varphi /dt$. Since $dt(Z)>0$, use $t$ as parameter and set $\Omega =d\varphi /dt$. The causal inequality givesBoth transverse terms are nonnegative in the exterior, so necessarilyBecause $g_{\varphi \varphi }>0$ and the discriminant is $4D>0$, this is equivalent toEquality requires $dr/dt=d\theta /dt=0$ and a null tangent. The limiting vectors are the two local azimuthal null directions; a curve following one at constant radius need not be a null geodesic. For a curve with radial or polar motion these angular bounds are necessary, while the full causal inequality can impose stricter bounds.

In the open exterior Kerr ergoregion, $g_{tt}>0$ andSince $D=g_{t\varphi }^2-g_{tt}{g}_{\varphi \varphi }<g_{t\varphi }^2$,so even the lower permitted angular velocity is positive. Every regular causal trajectory in the ergoregion has $d\varphi /dt>0$: it must co-rotate with the Kerr black hole. At the outer stationary-limit surface $g_{tt}=0$, the lower bound is zero; strict positivity is asserted in the open ergoregion. The sign reverses if the orientation of $a$ is reversed.