# Shadows of Rotating Black Holes

In Shadows of Rotating Black Holes it was mentioned that for Kerr-Newman with mass M angular momentum a and electrical charge $Q$, the apparent position of a photon moving on a closed orbit with radius $r$ in the $(x, y)$ reference frame of an asymptotic observer located in the angle of latitude theta is given by

$$x=\frac{r D+rQ^2-M(r^2-a^2)}{r(r-M)\sin \theta}$$ $$y=\frac{4r^2D}{(r-M)^2}-(x-a\sin\theta)^2.$$

I am confused about something. If $r$ is the radius of closed orbit of photon then doesn't that mean that the photon never reaches us because its moving around the black hole at radius $r$. But if the photon doesn't reach us then how do we see this apparent shadow of black hole if no photons enter our eyes?

• Key phrase from the linked document: "Such orbits describe the limit of the innermost photon orbits coming from infinity and escaping back to infinity."
– user12029
Jan 30 '17 at 21:44
• @NeuroFuzzy, but it also says that the orbits are closed.
– MrDi
Jan 30 '17 at 21:46
• @NeuroFuzzy, also why doesn't the shape of the shadow depend on the position between the observer and the black hole?
– MrDi
Jan 30 '17 at 21:50

You're right that photons on closed orbits will never reach us, but these orbits are unstable, so a photon on an orbit even slightly deviant from the formal closed orbit will either escape to infinity or plunge into the black hole. So photons on the closed orbit at $r$ will not escape, but those on orbits at $r+\delta r$ where $\delta r$ is very small will. The closed orbits are useful in describing the shadow as a limiting case.
• Thanks for answer. What does $\theta$ represent? Is it the angle that the photon escapes to?