Wikipedia states:
Here, $b$ can be interpreted as the distance of closest approach.
Yes, but that interpretation on Wikipedia is made for massless particles with closest approach much bigger than the Schwarzschild radius $r_s$. In general that interpretation is not true.
On the other hand, OP's above equations (v1) are for massive particle, so let us focus on the massive case in what follows.
As OP writes
$$
ac~=~h~=~\frac{L}{m_0}
$$
is the specific angular momentum $h$. In general,
$$b~=~c\frac{L}{E}~=~\frac{m_0c}{E}h~=~a\frac{m_0c^2}{E}$$
is the ratio between the (specific) angular momentum and the (specific) energy times the speed of light. The quantities $L$ and $E$ are constants of motion, which in turn reflect (some of) the Killing symmetries of the Schwarzschild metric.
The equation for the closest distance to the black hole can be deduced from the radial geodesic equation
$$\frac{1}{c^2}\left( \frac{dr}{d\tau} \right)^{2} = \left(\frac{E}{m_0c^2}\right)^{2} - \left( 1 - \frac{r_{s}}{r} \right) \left( \left(\frac{a}{r}\right)^2 + 1 \right) $$
for a massive particle in the equatorial plane by putting
$$\frac{dr}{d\tau}~=~0$$
and solve this third-order equation for $r$.
References:
- S. Carroll, Lecture Notes on General Relativity, Chapter 7, p.172-179. The pdf file is available from his website.
The light ray in general relativity travels along the null geodesic, which is determined by the simple equation
$$ g_{\mu \nu} (x) \frac{dx^{\mu}}{d\tau} \frac{dx^{\nu}}{d\tau} = 0, $$
where $g_{\mu \nu}$ in your case is the Schwarzschild metric. Using this equation and the initial condition (angle $\alpha$) you should be able to trace your light ray back in time and restore it's initial position at any $t$.
There is a small subtlety: the evolution parameter ($\tau$) is arbitrary, which means that the equation above has multiple solutions. This is due to the reparametrization invariance, which is the gauge symmetry of the relativistic particle. In order to receive definite answers, you can use the $\tau = t = x^0$ gauge (by simply saying that your $\tau$ is the physical time of an observer at infinity).
Update:
OK, I will answer your questions.
The $sn$ function is the solution of the differential equation above
Why? $sn$ is defined on the complex plane, as far as I know.
You don't. This equation tells you how $u=1/r$ depends on $\phi$.
$a$ and $b$ are the parameters which you should fix with your initial conditions.
But I don't see why you need those equations. The shape of an orbit of a planet? I thought you needed to trace the light ray back in time? Why didn't you do what I suggested in this answer?
Best Answer
The relationship between $b$ and $r_0$ for the Schwarzschild metric is:
$$ b = \frac{r_0}{\sqrt{1 - \tfrac{r_s}{r0}}} $$
where $r_s$ is the radius of the event horizon. See this paper for the gory details.
The equations given in the two articles are derived in the weak field limit i.e. $r \gg r_s$ so $b \approx r_0$ anyway and it doesn't make any real difference which you use. If I were writing the article I would describe $b$ as the impact parameter and not the distance of closest approach, but I don't feel strongly enough about it to want to edit the offending Wikipedia article.