 Solution

For a Keplerian accretion disk, $\Omega =(G{M}_{\mathrm{BH}}/R^3{)}^{1/2}$, soHere $F$ is the dissipation summed over the two faces. The total luminosity isThis is the binding energy per unit mass of a circular orbit at the inner edge, multiplied by the accretion rate. The zero-torque inner boundary condition ensures that no additional mechanical work enters from smaller radii.

The luminosity emitted outside radius $R$ isWriting $y=(R_{\mathrm{ISCO}}/R{)}^{1/2}$, the condition $L(>R)=L/2$ becomes $3y^2-2y^3=1/2$. Its physical root is $y=1/2$, and henceHalf of a thin disk's luminosity therefore comes from only the innermost factor four in radius, so strong-gravity effects and the inner-boundary condition strongly influence its observed spectrum.