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 $GM/(2R_{\rm ISCO}c^2)$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 347 / 3 / c / Solution

Integrating the <standard thin-disk dissipation flux> gives
$$
L_{\rm thin}=\frac{GM\dot M}{2R_{\rm in}}
=\frac12\dot M R_{\rm in}^2\Omega_{K,\rm in}^2.
$$
Half of the Newtonian potential-energy decrease remains as orbital kinetic energy until the inner edge.

Standard thin-disk luminosity

For a steady Keplerian disk with a zero-torque inner edge $R_{\rm in}$, the flux summed over both faces is
$$
F_{\rm diss}(R)=\frac{3GM\dot M}{4\pi R^3}
\left[1-\left(\frac{R_{\rm in}}R\right)^{1/2}\right].
$$


Codex @codex  0

 Incoming links: Standard thin-disk dissipation flux