Consider an observer located at radius $r_o$ from a Schwarzschild black hole of radius $r_s$. The observer may be inside the event horizon ($r_o < r_s$).
Suppose the observer receives a light ray from a direction which is at angle $\alpha$ with respect to the radial direction, which points outward from the black hole.
What is the original, incident angle of the light ray $\varphi$ from infinity? That is, if we were to trace the light ray back through time, which point on the celestial sphere did it come from?
See the diagram below:
According to this source, we have
$$ \varphi = \int_{r_o}^\infty \frac{\mathrm{d}r}{r^2 \sqrt{ \frac{1}{b^2} - \frac{1}{r^2} \left(1-\frac{r_s}{r}\right) }} $$
where $b$ is the impact parameter, which is the distance between the ray at infinity and a ray parallel to it that plunges directly into the black hole. The impact parameter is given here as
$$ b = m c \frac{h}{E} = m c r^2 \frac{\mathrm{d}\varphi}{\mathrm{d}\tau} \frac{1}{mc^2} \frac{\mathrm{d}\tau}{\mathrm{d}t} \left(1 - \frac{r_s}{r}\right)^{-1} = \frac{r^2}{c} \left( 1 - \frac{r_s}{r} \right)^{-1} \frac{\mathrm{d}\varphi}{\mathrm{d}t} $$
taking $r = r_o$. Therefore
$$ \frac{1}{b^2} = \frac{c^2}{r^4} \left( 1 - \frac{r_s}{r} \right)^2 \left( \frac{\mathrm{d}t}{\mathrm{d}\varphi} \right)^2 $$
To find the last term, recall that the metric is given by
$$ 0 = \left( 1- \frac{r_s}{r} \right) c^2 \mathrm{d}t^2 - \left( 1- \frac{r_s}{r} \right)^{-1} \mathrm{d}r^2 - r^2 \mathrm{d}\varphi^2 $$
for a light ray. Dividing by $\mathrm{d}\varphi^2$ yields
\begin{align} 0 &= \left( 1- \frac{r_s}{r} \right) c^2 \left(\frac{\mathrm{d}t}{\mathrm{d}\varphi}\right)^2 - \left( 1- \frac{r_s}{r} \right)^{-1} \left(\frac{\mathrm{d}r}{\mathrm{d}\varphi}\right)^2 - r^2 \\ \left( 1- \frac{r_s}{r} \right) c^2 \left(\frac{\mathrm{d}t}{\mathrm{d}\varphi}\right)^2 &= \left( 1- \frac{r_s}{r} \right)^{-1} \left(\frac{\mathrm{d}r}{\mathrm{d}\varphi}\right)^2 + r^2 \\ c^2 \left(\frac{\mathrm{d}t}{\mathrm{d}\varphi}\right)^2 &= \left( 1- \frac{r_s}{r} \right)^{-2} \left(\frac{\mathrm{d}r}{\mathrm{d}\varphi}\right)^2 + r^2 \left( 1- \frac{r_s}{r} \right)^{-1} \\ \left(\frac{\mathrm{d}t}{\mathrm{d}\varphi}\right)^2 &= \frac{1}{c^2} \left( 1- \frac{r_s}{r} \right)^{-2} \left(\frac{\mathrm{d}r}{\mathrm{d}\varphi}\right)^2 + \frac{r^2}{c^2} \left( 1- \frac{r_s}{r} \right)^{-1} \end{align}
Hence
\begin{align} \frac{1}{b^2} &= \frac{c^2}{r^4} \left( 1 - \frac{r_s}{r} \right)^2 \left( \frac{1}{c^2} \left( 1- \frac{r_s}{r} \right)^{-2} \left(\frac{\mathrm{d}r}{\mathrm{d}\varphi}\right)^2 + \frac{r^2}{c^2} \left( 1- \frac{r_s}{r} \right)^{-1} \right) \\ &= \frac{1}{r^4} \left( \left(\frac{\mathrm{d}r}{\mathrm{d}\varphi}\right)^2 + r^2 \left( 1- \frac{r_s}{r} \right) \right) \\ &= \frac{1}{r^4} \left(\frac{\mathrm{d}r}{\mathrm{d}\varphi}\right)^2 + \frac{1}{r^2} \left( 1- \frac{r_s}{r} \right) \end{align}
The relationship between $\mathrm{d}\varphi$ and $\mathrm{d}r$ can be derived as follows:
$$ \tan \alpha = r \frac{\mathrm{d}\varphi}{\mathrm{d}r} $$
$$ \frac{\mathrm{d}\varphi}{\mathrm{d}r} = \frac{\tan \alpha}{r} $$
$$ \left(\frac{\mathrm{d}r}{\mathrm{d}\varphi}\right)^2 = r^2 \cot^2 \alpha $$
Therefore
\begin{align} \frac{1}{b^2} &= \frac{1}{r^4} r^2 \cot^2 \alpha + \frac{1}{r^2} \left( 1- \frac{r_s}{r} \right) \\ &= \frac{1}{r^2} \cot^2 \alpha + \frac{1}{r^2} \left( 1 - \frac{r_s}{r} \right) \\ &= \frac{1}{r^2} \left( \cot^2 \alpha + 1 - \frac{r_s}{r} \right) \\ &= \frac{1}{r^2} \left( \csc^2 \alpha - \frac{r_s}{r} \right) \end{align}
and thus
$$ \varphi = \int_{r_o}^\infty \frac{\mathrm{d}r}{r^2 \sqrt{ \frac{1}{r_o^2} \left( \csc^2 \alpha - \frac{r_s}{r_o} \right) - \frac{1}{r^2} \left(1-\frac{r_s}{r}\right) }} $$
Is this expression correct? Is there a closed form in terms of elliptic functions or other special functions?
