complete version. One can check immediately that $F(\chi,k)$ only reduces to $K(k)$ at the starting point, where $r=r_{\rm eq}$ and so $\chi=\pi/2$, which correctly yields $\theta(r\!=\!r_{\rm eq})=\pi/2$. The second expression (\[theta(r)’\]) contains a different amplitude $\chi'$ which is related to $\chi$ by $$\begin{aligned}
\lefteqn{
  \sin^2\chi'
   = \frac{1-\sin^2\chi}{1-k^2\sin^2\chi}} \nonumber \\
\lefteqn{\quad
   = \frac{1}{k^2}\,
      \frac{4M\left(1-\frac{r_{\rm eq}}{r}\right)}
           {\sqrt{(r_{\rm eq}-2M)(r_{\rm eq}+6M)}+r_{\rm eq}-2M-4M\frac{r_{\rm eq}}{r}}
  \;.}\end{aligned}$$ The complementary modulus $k'$ which is related to $k$ by $k'^2=1-k^2$ is given by the same expression (\[2k2\]) as $k$, just with a [*plus*]{} sign after the 1; their product is therefore quite short, $$k^2 k'^2=\frac{4M\,(r_{\rm eq}-3M)}{(r_{\rm eq}-2M)(r_{\rm eq}+6M)} \;.$$

The latitude $\theta$ of all our photons decreases from $\pi/2$ until they cross the symmetry axis $\theta=0$. From there $\theta$ increases back, which is ensured by the sign $\epsilon^\theta$ on the left hand side of (\[theta(r)\]). In describing the photon trajectories, this sign should only appear in front of $\sin\theta$ terms: actually, one can effectively treat $\sin\theta$ (as well as $\theta$ itself) as positive for photons which have not yet crossed the axis while as negative for those which have already crossed it. Such a distinction will be important in evaluation of the photon effect on the particle, because at 