= Solution
Let $\mathcal G$ be the outward <viscous torque in an accretion disk>. For a declining <angular velocity>, the vertically integrated viscous shear stress gives
$$
\mathcal G=-2\pi\nu\Sigma r^3\Omega'(r)>0.
$$
In a steady <accretion disk> without radiated <angular momentum>, the net inward <angular-momentum flux> $Fh-\mathcal G$ is constant and equals the swallowed flux $Fh_0$. Thus <conservation of angular momentum> gives $F(h-h_0)=\mathcal G$. Substituting $F=2\pi r\Sigma u$ cancels the <surface density> and proves
$$
\boxed{u=\frac{\nu r^2[-\Omega'(r)]}{h-h_0}.}
$$
This uses the stipulated phenomenological viscous shear law; it does not replace it by a different fully relativistic stress prescription.
For the orbital calculation let a dot denote a derivative with respect to <proper time>. Substituting the geodesic energy and <specific angular momentum> first integrals into the timelike normalization in <Schwarzschild spacetime> gives
$$
\dot r^2=\frac{\epsilon^2}{c^2}-\left(1-\frac{2m}{r}\right)\left(c^2+\frac{h^2}{r^2}\right).
$$
Differentiate with $h$ and $\epsilon$ constant along the <geodesic>. Where $\dot r\ne0$, division by $2\dot r$ gives
$$
\ddot r=-\frac12\frac d{dr}\left[\left(1-\frac{2m}{r}\right)\left(c^2+\frac{h^2}{r^2}\right)\right]
=\boxed{-\frac{mc^2}{r^2}+\frac{h^2}{r^3}\left(1-\frac{3m}{r}\right).}
$$
This equation also holds at turning points and on <circular orbits>. To justify that without dividing by zero, put $A=1-2m/r$ in the equatorial geodesic <Lagrangian> $L=(Ac^2\dot t^2-A^{-1}\dot r^2-r^2\dot\phi^2)/2$. Its radial <Euler-Lagrange equation> is $\ddot r=(A'/2A)\dot r^2-AA'c^2\dot t^2/2+Ar\dot\phi^2$. Substitution of the timelike normalization cancels the $\dot r^2$ terms and yields $\ddot r=-A'c^2/2+(A/r-A'/2)h^2/r^2$, exactly the displayed radial equation, including when $\dot r=0$.
A <circular orbit> requires $\ddot r=\dot r=0$. Therefore
$$
h^2=\frac{mc^2r^2}{r-3m},\qquad
\boxed{h(r)=rc\sqrt{\frac m{r-3m}}},\qquad r>3m,
$$
choosing the positive sense of rotation. Differentiating the square is simpler than differentiating $h$:
$$
\frac d{dr}h^2=mc^2\frac{r(r-6m)}{(r-3m)^2}.
$$
It is negative for $3m<r<6m$ and positive for $r>6m$, so the global minimum over timelike <circular orbits> is
$$
\boxed{r_0=6m,\qquad h_0=\sqrt{12}\,mc.}
$$
As a stability check, the second radial derivative of $W=(1-2m/r)(c^2+h^2/r^2)$ at a <circular orbit>, holding $h$ fixed for the perturbation, is $W''=2mc^2(r-6m)/[r^3(r-3m)]$. Thus the exterior branch is stable, and $6m$ is the <innermost stable circular orbit>. It is the natural matching radius for the idealized <zero-torque inner boundary condition>.
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