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?

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

  • $\begingroup$ Thanks for answer. What does $\theta$ represent? Is it the angle that the photon escapes to? $\endgroup$
    – MrDi
    Jan 30 '17 at 22:03
  • $\begingroup$ It looks like it's just the latitude of the observer, measured relative to the spin axis of the hole, but I'd suggest double checking rather than just trusting me... If that's the case, then it would make sense that the photon escapes along this direction, since that is where you'd be interested in making the measurement of the shadow shape. $\endgroup$
    – Kyle Oman
    Jan 30 '17 at 22:14

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