= Solution
The stated <standard thin-disk dissipation flux> is summed over the two faces, so the total luminosity is
$$
\begin{aligned}
L_{\rm thin}
&=\int_{R_{\rm ISCO}}^\infty 2\pi R F_{\rm diss}(R)\,dR\\
&=\frac{3GM\dot m}{2}
\int_{R_{\rm ISCO}}^\infty
\left(R^{-2}-R_{\rm ISCO}^{1/2}R^{-5/2}\right)dR\\
&=\boxed{\frac{GM\dot m}{2R_{\rm ISCO}}}.
\end{aligned}
$$
This equals the Newtonian <standard thin-disk luminosity> and the orbital binding energy delivered per unit time at the inner edge. It is half the magnitude $GM\dot m/R_{\rm ISCO}$ of the potential-energy decrease because the other half appears as orbital kinetic energy. In the zero-torque model that remaining mechanical energy passes through the inner edge rather than being dissipated at larger radii. The associated Newtonian <radiative efficiency of black-hole accretion> is $GM/(2R_{\rm ISCO}c^2)$.
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