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For a weakly gravitating thin spherical shell of mass $M$, radius $R$ and uniform angular velocity $\Omega$, its interior linearized gravity metric has constant scalar potentials $A=-GM/R$, $C=GM/R$ and $g_{0i}=\omega (\hat{\mathbf{z}}\times \mathbf{x}{)}_i$. The scalar mass shell is a monopole. Its mass current is a dipole: regular interior and decaying exterior radial solutions are proportional to $r$ and $r^{-2}$, and their derivative jump fixes the factor $4/3$. Local inertial axes co-rotate with the shell at $\omega$ relative to asymptotic nonrotating axes. This is the leading interior Lense-Thirring precession, neglecting quadratic rotation and higher gravitational orders.

A gyroscope is a rotating instrument whose rapidly spinning, nearly symmetric rotor provides a direction reference through its angular momentum. In an inertial frame, negligible external torque approximately preserves that direction by conservation of angular momentum. A transverse torque turns the angular momentum instead: the surprising sideways motion of a supported spinning wheel is precession. An ideal freely falling test gyroscope in general relativity has its spin undergo parallel transport along its timelike geodesic. Comparing that direction with distant reference axes measures Lense-Thirring precession; the direction need not remain fixed in arbitrary rotating coordinates.

A gyroscope direction reference is useful precisely because its direction can persist while its housing turns. A spinning-wheel precession experiment deliberately supplies gravitational torque to reveal how that persistence fails. The gyroscope vector geometry explains the direction, and slow steady gyroscope precession gives a numerical prediction with a check on its assumptions.

Use units $c=1$, signature $(+---)$, and the weak-field metric $g_{ab}={\eta }_{ab}+h_{ab}$. Define the trace-reversed metric perturbation by ${\bar{h}}_{ab}=h_{ab}-{\eta }_{ab}h/2$, with $h={\eta }^{ab}{h}_{ab}$. In the Lorenz gauge in linearized gravity, ${\partial }^a{\bar{h}}_{ab}=0$, the sign convention of Q2 givesFor completeness, the linearized Ricci tensor in this convention isSubstituting the Lorenz condition ${\partial }_c{h}^c{}_b={\partial }_{b}h/2$ and trace reversing gives the stated Einstein tensor. For a stationary source, $\square =-{\nabla }^2$, so ${\nabla }^2{\bar{h}}_{ab}=16\pi G{T}_{ab}$.

Keep the monopole terms of order $ϵ=GM/R$ and the leading term linear in the rotation rate. Since $\Omega R=O({ϵ}^{1/2})$, the latter metric term is of order $ϵ\Omega R=O({ϵ}^{3/2})$ on the shell. Thus it would disappear from a strictly order-$ϵ$ truncation; the requested approximation must retain the leading rotation correction as well. Terms $T_{ij}=\rho v_i{v}_j$ generate metric corrections of order $ϵ(\Omega R{)}^2=O({ϵ}^2)$ and are neglected. Corrections to four-velocity normalization and the supporting shell stresses affect the omitted higher orders. To retained order,The minus sign in $T_{0i}$ comes from lowering its spatial index; using $T^{0i}$ in its place would reverse the dragging direction.

The scalar Poisson solution regular at the centre and vanishing at spatial infinity is ${\bar{h}}_{00}=-4GM/r$ outside the shell and ${\bar{h}}_{00}=-4GM/R=-4ϵ$ inside. Continuity and the derivative jump ${\bar{h}}_{00}^{\prime }(R^{+})-{\bar{h}}_{00}^{\prime }(R^{-})=4GM/R^2$ verify the density $M\delta (r-R)/(4\pi R^2)$. Also ${\bar{h}}_{ij}=0$ at this order. Undoing trace reversal givesComparing with the scalar parts of the interior line element yields $A=-ϵ$ and $C=ϵ$.

For the mass-current contribution writeEach nonzero angular component is a degree-one spherical harmonic. The supplied Laplacian eigenvalue therefore givesAway from the shell, the two radial solutions are $r$ and $r^{-2}$. Regularity at $r=0$ and decay at infinity select $F_{\mathrm{in}}=ar$ and $F_{\mathrm{out}}=b/r^2$. Continuity at $R$ requires $b=a{R}^3$, and integrating the differential equation across the shell givesThus $a=4GM\Omega /(3R)=\omega$, with $F_{\mathrm{in}}=\omega r$. Trace reversal leaves mixed components unchanged, so $h_{0i}=\omega (\hat{\mathbf{z}}\times \mathbf{x}{)}_i=\omega (-y,x,0{)}_i$. The paper writes $g_{0i}=-B_i$, giving the frame dragging inside a slowly rotating spherical shellThis dipole matching is valid throughout $r<R$; it does not assume the observation point is very close to the centre. The stationary current is divergence-free, so this solution also satisfies the Lorenz gauge. The exterior result is $h_{0i}=2G(\mathbf{J}\times \mathbf{x}{)}_i/r^3$, where the thin shell has $\mathbf{J}=(2/3)M{R}^2\mathbf{\Omega }$, providing a check on the coefficient.

To identify the freely falling frames inside, let ${\mathcal{R}}_z(\theta )$ be the ordinary spatial rotation by angle $\theta$. Define the inertial coordinates inside a slowly rotating spherical shell byDifferentiating and dropping ${ϵ}^2$, products beyond the retained rotation order, and ${\omega }^2$ giveswhich is precisely the interior metric. Thus the retained interior curvature is zero, despite its nontrivial relation to the nonrotating coordinates anchored at infinity. Timelike geodesics in these local inertial coordinates are straight lines $\mathbf{X}={\mathbf{X}}_0+\mathbf{V}T$, with $|\mathbf{V}|<1$. In the original coordinates this isunderstood to the same perturbative order. Their coordinate equation is $d^2\mathbf{x}/d{t}^2=2\mathbf{\omega }\times d\mathbf{x}/dt$ to first rotation order. There is no interior Newtonian acceleration from the constant potential; the velocity-dependent term is the apparent Coriolis term relative to the locally dragged inertial axes.

A freely transported gyroscope at rest in the interior to the retained order has constant spatial direction in the $\mathbf{X}$ frame, hence its direction in the original axes obeys $d\mathbf{S}/dt=\mathbf{\omega }\times \mathbf{S}$. Local inertial axes precess in the same sense as the shell, at $\omega =(4ϵ/3)\Omega$ relative to distant nonrotating axes. This is the interior Lense-Thirring precession. For a weak shell the dragging is only a small fraction of its angular velocity, not perfect corotation. It is compatible with the equivalence principle: rotation relative to distant axes is a comparison between frames, and local flatness at the retained order does not make that comparison vanish. Quadratic rotation terms, which would include centrifugal effects and additional curvature, are outside the requested leading approximation.

Each annulus has a torque proportional to ${\mathbf{J}}_h\times \delta {\mathbf{J}}_d$, so its scalar product with ${\mathbf{J}}_h$ is zero. Summing over annuli preserves this property even when the proportionality coefficient varies with radius. Therefore $\boxed{\mathbf{T}\cdot {\mathbf{J}}_h=0}$ for the total torque due to Lense-Thirring precession.

The torque specified is the torque on the disk. Assuming a closed disk-plus-hole angular-momentum budget with no external torque, its reaction on the black hole has the opposite sign:Thus ${\dot{\mathbf{J}}}_t=0$. The $K_1$ term describes precession, because it is perpendicular to both spins. Since ${\mathbf{J}}_h\times {\mathbf{J}}_d={\mathbf{J}}_h\times {\mathbf{J}}_t$, the hole precesses about the fixed total vector. The $K_2$ term changes the spin directions dissipatively. Writing ${\mathbf{J}}_{d,\perp }$ for the disk component perpendicular to the hole, the vector triple product identity givesA positive $K_2$ removes this component from the disk and transfers it into the hole's changing direction. This is black-hole and disk alignment under internal torque.

Since $\mathbf{T}$ is perpendicular to ${\mathbf{J}}_h$,Hence $\boxed{J_h\text{ is constant}}$, even if the torque coefficients vary in time.

Now differentiate the scalar product with the conserved total vector. The precession term contributes zero, and the vector triple product identity givesAlso ${\mathbf{J}}_h\cdot {\mathbf{J}}_t=J_h^2+{\mathbf{J}}_h\cdot {\mathbf{J}}_d$, so constancy of $J_h$ givesFinally, $J_t^2=J_h^2+J_d^2+2{\mathbf{J}}_h\cdot {\mathbf{J}}_d$ is constant. Differentiating provesThus the disk's spin magnitude decreases while the hole's magnitude is fixed; their vector sum remains fixed as well.

For the alignment conclusion, assume $J_h,J_t>0$ and let $c=\cos \theta ={\mathbf{J}}_h\cdot {\mathbf{J}}_t/(J_h{J}_t)$. Since ${\mathbf{J}}_h\times {\mathbf{J}}_d={\mathbf{J}}_h\times {\mathbf{J}}_t$,For a noncollinear initial state, $|c(0)|<1$, this integrates exactly toConsequently a fixed positive $K_2$, or more generally a positive coefficient of divergent integrated strength, gives $c(t)\to 1$. In this persistent dissipative evolution, $\boxed{{\mathbf{J}}_h\to J_h{\hat{\mathbf{J}}}_t}$: the hole aligns asymptotically with the total angular momentum. The exactly antiparallel configuration $c=-1$ is a stationary, nongeneric exception; when $J_t=0$ there is no total-vector direction to align with.

If the phrase $K_2>0$ allows arbitrary time dependence, positivity alone does not logically force complete eventual alignment. The persistent-torque criterion for asymptotic spin alignment is the needed qualification. For example, take $J_h=J_t=1$, $K_1=0$, $K_2(t)=e^{-t}$ and $c(0)=0$. The same torque equations give $c(t)=\tanh (1-e^{-t})$, with limit $\tanh 1<1$. The final configurations below use the intended persistent-torque regime, including constant positive $K_2$.

Once the hole has aligned, conservation of angular momentum fixes the disk without any additional torque calculation:The disk and hole therefore finish parallel only if $J_t>J_h$. If $J_t<J_h$, they finish antiparallel: this is counteralignment of a black hole and accretion disk. In terms of the initial disk-hole angle ${\theta }_0$, squaring $J_t$ shows that counteralignment occurs whenIf $J_t=J_h$, the final disk angular momentum vanishes and has no spin direction. Thus alignment of the hole with the total vector does not imply alignment of both component spins.