For the aligned magnetic dipole field,Since ${\mathbf{e}}_z=\cos \theta {\mathbf{e}}_r-\sin \theta {\mathbf{e}}_{\theta }$,The corotation velocity is $\mathbf{u}=\Omega r\sin \theta {\mathbf{e}}_{\varphi }$. The current density is therefore the advected charge density,

The horizon area is $A=4\pi r_{+}^2$, whileThus the First law of black-hole mechanicsholds with the Bekenstein-Hawking entropy $S_{BH}=A/4$.

For equilibrium with an infinite fixed-temperature reservoir, local stability requires positive heat capacity. Since $dM/d{r}_{+}>0$, its sign is the sign ofTherefore the Canonical stability of a Schwarzschild--anti-de Sitter black hole occurs precisely forThe equality is the minimum-temperature, marginal case; smaller black holes have negative heat capacity and unstable thermal equilibrium.

LetThe one-form dual to $k={\partial }_v$ is $k^{♭}=-fdv+dr$. For the $r_{+}\to 0$ anti-de Sitter background, ${\bar{k}}^{♭}=-(1+r^2/L^2)dv+dr$, soWith the stated orientation and metric volume form,The Background-subtracted Komar energy is consequently

For a metric $d{s}^2=-f(r)d{v}^2+2dvdr+r^{2}d{\Omega }_2^2$ with a simple Killing horizon, the surface gravity of $k={\partial }_v$ is $\kappa =f^{\prime }(r_{+})/2$. HeresoThe Hawking temperature of the Schwarzschild--anti-de Sitter black hole is therefore

The result is weaker than the pointwise null energy condition used in the elementary proof of the second law of black-hole mechanics: $T_{ab}{U}^a{U}^b$ may be negative over a finite interval, so the expansion and area can have local behavior that the pointwise argument does not control. Nevertheless, the averaged null energy condition for a nonminimally coupled scalar supplies the integrated positivity required by strengthened global focusing and area theorems when their completeness, endpoint, and genericity hypotheses hold. Thus the calculation supports a suitable global second law for $\xi <0$, but the displayed averaged inequality alone does not prove pointwise area monotonicity without those additional assumptions.

Along the affine null geodesic write ${\Phi }^{\prime }=U^a{\nabla }_a\Phi$. Contracting the scalar stress tensor with $U^a{U}^b$ removes every term proportional to $g_{ab}$. Since $G_{ab}=T_{ab}$ andone obtainsPut $D=1-\xi {\Phi }^2$. An integration by parts givesThe stated endpoint condition kills the boundary term. If $\xi <0$, then $D>0$ and $\xi (4\xi -1)>0$, so the remaining integrand is nonnegative. Consequently this nonminimally coupled scalar satisfies the averaged null energy condition:

The Penrose singularity theorem concludes future null-geodesic incompleteness; it does not require a divergent curvature invariant. This spacetime supplies exactly that distinction. For a null geodesic in the static coordinates, the Killing constantsand the null condition giveAs $r\to 0$, $f\to -r_{+}^2/L^2$. A radial geodesic with $J=0$ reaches $r=0$ linearly in finite affine parameter. If $J\ne 0$, then $|\dot{r}|\sim r_{+}|J|/(Lr)$, so it again reaches zero in finite affine parameter. The anti-de Sitter quotient ends these geodesics at $r=0$ even though local curvature remains finite. Thus the trapped circles lead to the BTZ causal singularity, satisfying the theorem through geodesic incompleteness rather than curvature blowup.

When $r_{-}=0$, $\Omega =0$ and $f=(r^2-r_{+}^2)/L^2$. A circle $S$ at fixed $v,r$ has circumference $2\pi r$. Choose its future null normals asFor a one-dimensional transverse surface, each null expansion is the logarithmic derivative of its length:Inside the horizon, $0<r<r_{+}$ implies $f<0$, so both expansions are negative. Every such circle is therefore a trapped surface, specifically a Trapped circle inside a nonrotating BTZ black hole.

For $r\gg R$, the multipole expansion isThe center of mass condition removes the mass dipole. Axisymmetry makes the mixed second moments vanish and gives $\int \rho x^{\prime 2}=\int \rho y^{\prime 2}$. Undoing trace reversal therefore yieldswhereThus $a=J_z/m$ is the specific angular momentum, and the cross term is the dipole part of the slowly rotating weak-field metric.

Stationarity changes $\square$ to the spatial Laplace operator. The Green function of the Poisson equation therefore givesHere the bar denotes the trace-reversed metric perturbation; this is the variable denoted by $h$ in the displayed field equation of the question.

To first order in the angular velocity, $T^{00}=\rho$ and $T^{i0}=\rho v^i$ with $v=(-\Omega y,\Omega x,0)$. The conservation equation ${\partial }_{\mu }{T}^{\mu 0}=0$ and time independence giveThus $\boxed{x{\partial }_y\rho -y{\partial }_x\rho =0}$: the density is invariant under rotations about the z-axis.

Let $K={\partial }_t$ and let the future unit normal to ${\Sigma }_{\tau }$ be $n=f^{-1}{\partial }_t$. The conserved Killing energy is the fluxSubstitution of the Klein-Gordon scalar stress-energy tensor gives exactlyApply the divergence theorem to the slab $[{\tau }_1,{\tau }_2]\times {\mathbb{R}}^3$. The spatial-boundary flux vanishes because $\psi$ decays, and ${\nabla }_a{J}^a=0$ makes the two time-slice fluxes equal. Thus this conserved scalar-field energy in a static spacetime is independent of $\tau$.

The coordinate formula for the Lie derivative isFor $X={\partial }_t$, its components are constant and every metric coefficient is independent of $t$, so ${\mathcal{L}}_{{\partial }_t}g=0$. Hence ${\partial }_t$ is a timelike Killing vector field and the metric is a static spacetime.

Killing equation is ${\nabla }_{(a}{K}_{b)}=0$. For the stress-energy current from a Killing vector $J^a=T^a{}_b{K}^b$,The first term vanishes on shell, while the second contracts the symmetric tensor $T^{ab}$ with the antisymmetric derivative selected by Killing's equation. Thus $\boxed{{\nabla }_a{J}^a=0}$.

Using both metric-variation formulas from part a gives the Klein-Gordon scalar stress-energy tensorIts divergence isbecause the two Hessian terms cancel by symmetry. It vanishes on every solution of the Klein-Gordon equation.

Varying the scalar and integrating by parts givesbecause the boundary term vanishes by the support assumption. The fundamental lemma of the calculus of variations therefore gives the Klein-Gordon equation

The determinant identity $\delta \log |\det g|=g^{ab}\delta g_{ab}$ impliesHence the metric volume form varies as

Vary $g^{ac}{g}_{cb}={\delta }_b^a$. The product rule givesand multiplication by $g^{bd}$ yields the metric variation

In the propagated frame $T=e_0$, and the constant sectional curvature formula reduces geodesic deviation toThe stated temporal initial data give $y^0(t)=0$. Writing $\omega =\sqrt{-K}$ for $K<0$ and $\kappa =\sqrt{K}$ for $K>0$, the spatial displacement isThese formulas are valid to first order in the initial separation and relative velocity.

Parallel propagation means ${\nabla }_T{e}_{\alpha }=0$, where $T=\dot{\lambda }$. By metric compatibility,The inner products therefore retain their initial values ${\eta }_{\alpha \beta }$, so the parallel-propagated orthonormal frame remains orthonormal.