Past exam of the general relativity II course of the University of Oxford / 2017 / 3

A regular hypersurface $f=0$, with $df\ne 0$, is a null hypersurface if its normal covector has zero squared norm:Its normal vector is then also tangent to the hypersurface; the induced metric tensor is degenerate.

For the Kerr metric the normal to a constant-$r$ surface has squared normFor $0<a<M$ the two roots are positive and distinct. To verify that they are genuine regular null hypersurfaces, rather than trusting singular Boyer-Lindquist coordinates, introduce ingoing Kerr coordinatesSubstitution givesThere are no ${\Delta }^{-1}$ terms. At either root $\Sigma >0$, and the determinant is $-{\Sigma }^2{\sin }^2\theta$ away from the angular coordinate poles. Thus the metric tensor extends smoothly across each root. The raised normal on either surface isThis is a nonzero null Killing vector field normal to the surface. Both surfaces are Killing horizons, with horizon angular velocities ${\Omega }_{\pm }$.

The event-horizon terminology in the paper needs qualification. In the usual asymptotically flat Kerr black hole extension, relative to a chosen exterior's future null infinity, the future branch of $r=r_{+}$ is the event horizon, the boundary $\partial J^{-}({\mathcal{I}}^{+})$. In the intervening black-hole region $\Delta <0$, the coordinate $r$ has timelike gradient and future-directed curves move toward smaller $r$; they cannot return across the outer boundary to that exterior. The inner surface $r=r_{-}$ is a Cauchy horizon, not a second event horizon for the same asymptotic end. Nullness is a local property; identifying an event horizon additionally requires this global causal structure.^[ref]

The allowed endpoint $a=0$ provides an especially concrete counterexample to the unrestricted description of both roots. The Kerr metric then becomes Schwarzschild spacetime, $r_{+}=2M$ is its regular event horizon, but $r_{-}=0$ is a curvature singularity:It is not a regular inner null hypersurface. Thus the two-horizon calculation above assumes $0<a<M$.

The metric tensor is independent of $t$ and $\varphi$, soare commuting Killing vector fields. For $0<a<M$ they span the continuous Killing vector field algebra; every constant linear combination is also a Killing vector field. They express stationary spacetime and axial symmetry, respectively. At the allowed endpoint $a=0$ the spherical symmetry of Schwarzschild spacetime enlarges the algebra: in addition to $T$, a basis of rotational Killing vector fields isTheir polar-coordinate expressions are singular at the axis, but the rotational fields are smooth there.

Let $U^a=d{x}^a/ds$ be the tangent to any affinely parametrized geodesic. For a Killing vector field $X$,This is the geodesic conserved quantity from a Killing vector. With $p_a=g_{ab}{U}^b$, choose the energy and axial angular momentum conventionsFor a massive particle with proper time parameter these are per unit rest mass; for a null geodesic their common scale depends on the affine parameter normalization. Expanding in the supplied Boyer-Lindquist coordinates givesThe factor of two between $g_{t\varphi }$ and the coefficient of $dtd\varphi$ is essential for these signs and coefficients.

Reflection across $\theta =\pi /2$ is an isometry; a geodesic initially tangent to its fixed equatorial plane remains in that plane by uniqueness of the geodesic equation. Thus take $\theta =\pi /2$, $\dot{\theta }=0$, and $\Sigma =r^2$. Define the algebraic abbreviationsThe conserved quantities of part (b) becomeFor $\Delta \ne 0$, inversion givesThe null condition can be written without expanding a quadratic square:Substituting the preceding expressions yields the equatorial Kerr null geodesic radial equationAn equivalent useful polynomial form isThese expressions use an affine parameter $s$; a common rescaling of $E$ and $L$ accompanies its rescaling. The inversion was performed in the exterior $r>r_{+}$, where Boyer-Lindquist coordinates are regular. The polynomial radial identity has a smooth continuation to a horizon, but the formulas for $\dot{t}$ and $\dot{\varphi }$ there must be replaced by regular-coordinate expressions. In the limit $a=0$, the result reduces to the Schwarzschild spacetime equation ${\dot{r}}^2=E^2-(1-2M/r)L^2/r^2$.

The assertion about every exterior radius is false for geodesics. A circular orbit here means an equatorial null geodesic at constant $r$, rather than a genuinely radial trajectory. A zero of the radial effective potential alone gives a turning point, not necessarily a circular orbit. The radial geodesic equation must hold as well.

Write $q=L-aE$ andFor an exterior circular null geodesic, the two conditions are $f(r)=f^{\prime }(r)=0$, equivalently $\mathcal{R}(r)={\mathcal{R}}^{\prime }(r)=0$ with $E,L$ fixed. One way to justify the derivative condition without dividing by $\dot{r}=0$ is to use the radial Euler-Lagrange equation. At constant $r$ it givesTogether with $g_{AB}{\dot{x}}^A{\dot{x}}^B=0$, and the fixed conserved $p_A$, this is precisely the double-root condition for $\mathcal{R}$.

NowSetting this to zero and using $f=0$ givesA nontrivial exterior null geodesic cannot have $E=0$ here: the second equation would force $q=L=0$, and then every tangent component would vanish. ThereforeBoth equations, with the same constants $E,L$, are required. They give the requested radius relations in terms of the conserved quantities, while showing why an arbitrary exterior radius is not permitted.

For a more explicit classification let $b=L/E$ be the impact parameter. The second relation gives $b-a=s{r}^{3/2}/\sqrt{M}$ with $s=\pm 1$. Substituting into the first relation yieldsFor $0<a<M$, $s=+1$ gives the prograde equatorial circular Kerr photon orbit and $s=-1$ the retrograde one. The two allowed exterior radii areBoth are strictly greater than $r_{+}$ in the subextremal range. Squaring the signed radius equation gives $r(r-3M{)}^2=4M{a}^2$, but introduces an additional inner root and loses the sign distinguishing the two senses of rotation. At $a=0$ the two exterior radii coincide at the photon sphere $r=3M$, with $b=\pm 3\sqrt{3}M$.

For a direct counterexample to the all-radii claim take $a=0$ and $r_0=4M>r_{+}=2M$. Choose $E=1$ and $L=4\sqrt{2}M$ so that $f(r_0)=0$. The radial geodesic equation nevertheless givesIt is a turning point, not a circular orbit. Alternatively, constant-$r$ null directions exist throughout the exterior because the two angular-velocity roots of $g_{tt}+2\Omega g_{t\varphi }+{\Omega }^2{g}_{\varphi \varphi }=0$ are real; almost all of their integral curves require acceleration and fail the radial geodesic equation. This distinction explains how an all-radii assertion can arise.

Finally, $r=r_{+}$ itself has the null geodesic generators of its Killing horizon, proportional to ${\chi }_{+}={\partial }_v+{\Omega }_{+}{\partial }_{\psi }$. They are a separate horizon case. On the horizon ${\chi }_{+}$ is normal to every tangent, so its contractions with ${\partial }_v$ and ${\partial }_{\psi }$ both vanish: $E=L=0$ for a generator tangent. Its affine parameter can be chosen by rescaling that tangent. The exterior calculation dividing by $E$ does not include it. The presence of these generators does not supply circular null geodesics at the intervening arbitrary exterior radii.