= Solution
For an equatorial <circular orbit> in the <Paczyński-Wiita potential>, radial force balance gives
$$
R\Omega^2=\frac{d\Phi}{dR}=\frac{GM}{(R-R_S)^2},\qquad \Omega^2=\frac{GM}{R(R-R_S)^2}.
$$
Its <specific angular momentum> is $j=R^2\Omega=\sqrt{GM}R^{3/2}/(R-R_S)$. Perturbing the circular radius at fixed $j$ gives the <radial epicyclic frequency> squared $\kappa^2=R^{-3}d(j^2)/dR$. Therefore
$$
\boxed{\kappa^2(R)=\frac{GM(R-3R_S)}{R(R-R_S)^3}=\Omega^2\frac{R-3R_S}{R-R_S}.}
$$
Circular motion is radially stable for $R>3R_S$, marginal at $3R_S$, and unstable between $R_S$ and $3R_S$. Thus \b[the inner edge of the nearly circular thin accretion disk is $R_{\mathrm{in}}=3R_S$], joining a <plunging region of a black-hole accretion disk> inside the <innermost stable circular orbit>. This conclusion concerns the circular thin-disk model, not an assertion that all gas disappears inside this radius.
Take the inward <mass accretion rate> $\dot M$ as positive, and write the positive outward <viscous torque in an accretion disk> as
$$
\mathcal G=-2\pi\nu\Sigma R^3\Omega'(R).
$$
Steady <conservation of angular momentum> makes the net inward flux $C=\dot M j-\mathcal G$ independent of radius. Impose a <zero-torque inner boundary condition> at $3R_S$, appropriate to an approximately stress-free transition to plunging motion. Then
$$
\boxed{C=\dot Mj_{\mathrm{in}}=\frac{3\sqrt3}{2}\dot M\sqrt{GMR_S},\qquad\mathcal G=\dot M(j-j_{\mathrm{in}}).}
$$
This is the <steady accretion disk with arbitrary rotation law> angular-momentum balance.
The local <viscous dissipation> integrated through the full disk thickness is $D=\nu\Sigma(R\Omega')^2=-\mathcal G\Omega'/(2\pi R)$. It is the power per unit planar area summed over both radiating faces. Since
$$
-\Omega'=\frac{\sqrt{GM}(3R-R_S)}{2R^{3/2}(R-R_S)^2},\qquad
\frac{j_{\mathrm{in}}}{j}=\frac{3\sqrt3}{2}\left(\frac{R_S}{R}\right)^{1/2}\frac{R-R_S}{R},
$$
substitution gives the <two-face dissipation of a Paczyński-Wiita disk>:
$$
\boxed{D(R)=\frac{GM\dot M(3R-R_S)}{4\pi R(R-R_S)^3}\left[1-\frac{3\sqrt3}{2}\left(\frac{R_S}{R}\right)^{1/2}\frac{R-R_S}{R}\right].}
$$
In particular $D$ vanishes at the stress-free inner edge and is nonnegative outside it because $j$ increases there. A one-face radiative flux is $D/2$; the total <accretion luminosity> is therefore $L=\int_{3R_S}^\infty2\pi R D\,dR$, with no additional factor of two.
For a direct evaluation put $x=R/R_S$. Then
$$
L=\frac{GM\dot M}{2R_S}\left\{\int_3^\infty\frac{3x-1}{(x-1)^3}dx-\frac{3\sqrt3}{2}\int_3^\infty\frac{3x-1}{x^{3/2}(x-1)^2}dx\right\}.
$$
The first integral is $7/4$, using $(3x-1)/(x-1)^3=3/(x-1)^2+2/(x-1)^3$. For the second,
$$
\frac{d}{dx}\frac1{\sqrt x(x-1)}=-\frac{3x-1}{2x^{3/2}(x-1)^2},
$$
so it equals $2/[\sqrt3(3-1)]=1/\sqrt3$. Hence
$$
\boxed{L=\frac{GM\dot M}{8R_S}=\frac1{16}\dot M c^2,\qquad\epsilon=\frac1{16}.}
$$
This is the <radiative efficiency of black-hole accretion> within this pseudo-Newtonian model.
Independently, the <specific orbital energy> of a circular particle is
$$
E(R)=\frac12R^2\Omega^2+\Phi(R)=\frac{GM(2R_S-R)}{2(R-R_S)^2}.
$$
Thus $E(\infty)=0$ and $E(3R_S)=-GM/(8R_S)=-c^2/16$. The energy released per unit accreted mass is exactly $c^2/16$, confirming the integrated <accretion luminosity>. The <Paczyński-Wiita circular orbit> reproduces the Schwarzschild inner radius but gives an approximate binding energy; the efficiency derived here is for the stated potential.
Back to article page
