Black-hole and disk alignment under internal torque (${\dot{\mathbf{J}}}_h=-K_1{\mathbf{J}}_h\times {\mathbf{J}}_d-K_2{\mathbf{J}}_h\times ({\mathbf{J}}_h\times {\mathbf{J}}_d)$)

For the opposite internal torques on a black hole and its accretion disk, the total angular momentum ${\mathbf{J}}_t$ and the black-hole spin magnitude $J_h$ are constant. If $A=K_2|{\mathbf{J}}_h\times {\mathbf{J}}_d{|}^2$, then $d({\mathbf{J}}_h\cdot {\mathbf{J}}_t)/dt=A$ and $d{J}_d^2/dt=-2A$. The $K_1$ term produces precession; the positive $K_2$ term increases the alignment of the hole with the total vector while decreasing the disk angular-momentum magnitude. The exact direction equation is $\dot{c}=K_2{J}_h{J}_t(1-c^2)$ for $c=\cos \angle ({\mathbf{J}}_h,{\mathbf{J}}_t)$.

After persistent dissipative alignment of the hole with the conserved total angular momentum, the final vectors are ${\mathbf{J}}_{h,f}=J_h{\hat{\mathbf{J}}}_t$ and ${\mathbf{J}}_{d,f}=(J_t-J_h){\hat{\mathbf{J}}}_t$. The disk is therefore antiparallel to the hole if $J_t<J_h$. Squaring $J_t^2=J_h^2+J_{d,0}^2+2J_h{J}_{d,0}\cos {\theta }_0$ gives the displayed initial-angle criterion. The disk and hole need not finish aligned with each other even though the hole aligns with the total. Equality $J_t=J_h$ leaves zero final disk angular momentum rather than a defined disk spin direction.

For nonzero $J_h,J_t$ and a noncollinear initial configuration, the alignment equation integrates to $\mathrm{artanh}c(t)=\mathrm{artanh}c(0)+J_h{J}_t{\int }_0^t{K}_2(s)ds$. Hence the positive dissipative torque aligns the hole with the total vector if its integrated strength diverges. A fixed positive coefficient is sufficient. Pointwise positivity alone need not suffice for time-dependent coefficients: with $J_h=J_t=1$, $K_1=0$, $K_2=e^{-t}$ and $c(0)=0$, the limiting cosine is $\tanh 1<1$. Exactly antiparallel initial spin and total angular momentum form a stationary, nongeneric exception.