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!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.