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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:

enter image description here

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:

enter image description here

$$ \tan \alpha = r \frac{\mathrm{d}\varphi}{\mathrm{d}r} $$

$$ \frac{\mathrm{d}\varphi}{\mathrm{d}r} = \frac{\tan \alpha}{r} $$

$$ \left(\frac{\mathrm{d}r}{\mathrm{d}\varphi}\right)^2 = r^2 \cot^2 \alpha $$

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?

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  • $\begingroup$ Why do you ask? Your reference on PG coordinates has the graphic. Not that I believe it, but why do you want more, and since you seem to know where to find how to do it, why ask here? Or are you asking if there is anything wrong with the description you referenced? $\endgroup$ – Bob Bee Jul 3 '16 at 2:15
  • $\begingroup$ @BobBee The reference gives neither the value of $I$ nor the final expression (if there is a closed form). $\endgroup$ – user76284 Jul 3 '16 at 2:31
  • $\begingroup$ Well, somebody has to work it through. Have you given it a try? $\endgroup$ – Bob Bee Jul 3 '16 at 2:42
  • $\begingroup$ This seems like a homework problem What have you tried to solve it? $\endgroup$ – Rob Jeffries Jul 3 '16 at 7:58
  • 1
    $\begingroup$ @RobJeffries This is not a homework problem. I have updated the question with more details. $\endgroup$ – user76284 Jul 3 '16 at 15:10

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