# Newtonian Limit of the Accretion Efficiency of a Black Hole

Netzer's book "The Physics and Evolution of Active Galactic Nuclei" (Cambridge) gives, in Section 3.2, the efficiency of a Black Hole (BH) in converting rest mass to electromagnetic radiation as: $$\begin{equation} \eta = \frac{E(\infty) - E(R)}{mc^2}. \end{equation}$$ Where E is the potential energy, R the distance to the BH, $$m$$ the accreted mass and $$c$$ the speed of light. Now it is written that the Newtonian limit for this expression is: $$\begin{equation} \frac{1}{2x}, \end{equation}$$ where $$x$$ is a dimensionless distance expressed as a ratio to the gravitational radius $$x = \frac{R}{R_g}$$, the latter being: $$R_g = \frac{GM}{c^2}$$.

I cannot re-obtain the Newtonian limit though, with the same $$1/2$$ factor. This is how my reasoning goes: let me consider a Newtonian (gravitational) potential: $$\begin{equation} E(R) = - \frac{GMm}{R} \xrightarrow[R\rightarrow\infty]{} 0 \end{equation}$$ where $$M$$ is our BH mass. I can consider that the potential goes to zero for very large distances, therefore: $$\begin{equation} \eta = \frac{0 - \left(-\frac{GMm}{R}\right)}{mc^2} = \frac{GM}{Rc^2} = \frac{R_g}{R} = \frac{1}{x}. \end{equation}$$

What am I not considering? I am sure the limit $$1/(2x)$$ is correct since I have found many times in the literature that for a Schwarzschild BH $$\eta=1/12$$, considering the innermost stable orbit at the event horizon $$R=6\,R_g$$, so $$x=6$$.

Is there some virial consideration to be done in order to obtain the $$1/2$$ factor I am missing?
Thanks a lot!