When does light reach a shell observer in Schwarzschild metric? I am trying to simulate the trajectory of light in the Schwarzschild metric (as seen by a far away observer) with fixed $\theta = \pi/2$. According to my source (Chapter 18, section 18.5) the trajectory is then governed by:
$$\frac{dr}{dt} = \dot{r}$$
$$\frac{d\phi}{dt} = \dot{\phi}$$
$$\frac{d\dot{r}}{dt} = \frac{-4M^2+2Mr+(r-5M)r^3\dot{\phi}^2}{r^3}$$
$$\frac{d\dot{\phi}}{dt} = \frac{2(-3M+r)\dot{r}\dot{\phi}}{(2M-r)r}$$
I have a situtation where shell observer sits at $(r_T, \phi_T)$ and I know that $r(0) = r_0$, $\phi(0) = \phi_0$, $r(T) = r_T$, $\phi(T) = \phi_T$ where $r_0$, $r_T$,$\phi_0$ and $\phi_T$ are known, but $T$ is unknown. It seems to me that I need an additional constraint to figure out $T$ since I have 4 equations (the ones above), but 5 unknowns ($r(t), \dot{r}(t), \phi(t), \dot{\phi}(t), T$).
Do I need an additional constraint to figure out $T$ and what would that constraint be?
 A: The way the equations are presented seems unnecessarily obscure, as there are only two equations that matter:
$$ \frac{d^2r}{dt^2} = \frac{-4M^2+2Mr+(r-5M)r^3}{r^3}\,\dot{\phi}^2 $$
$$ \frac{d^2\phi}{dt^2} = \frac{2(-3M+r)}{(2M-r)r} \, \dot{r}\dot{\phi} $$
These come from the geodesic equation expressed using coordinate time.
So you start at some convenient $(r, \phi)$ with initial coordinate velocity $(\dot{r}, \dot{\phi})$ and integrate forward in time to calculate $r$, $\phi$, $\dot{r}$ and $\dot{\phi}$ as a function of the coordinate time $t$.
You get to pick whatever initial values for $r$, $\phi$, $\dot{r}$ and $\dot{\phi}$ you want, but obviously $\dot{r}$ and $\dot{\phi}$ are related because you're describing a light beam. The relationship comes from the Schwarzschild metric. For a light ray $ds = 0$, and we can take $\theta = \pi/2$ and $d\theta = 0$, so we get:
$$ 0 = -\left(1-\frac{2M}{r}\right)dt^2 + \frac{dr^2}{\left(1-\frac{2M}{r}\right)} + r^2d\phi^2 $$
or:
$$ \left(1-\frac{2M}{r}\right) = \frac{\dot{r}^2}{\left(1-\frac{2M}{r}\right)} + r^2\dot{\phi}^2 $$
