Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 71 / 3 / Solution

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.