Sign of $dr$ in Schwarzschild geodesics There is an equation that relates energy $E$, angular momentum $L$ and other constants and variables to find $\left(\frac{dr}{d\tau}\right)^2$ in a plane.
$$\left(\frac{dr}{d\tau}\right)^2=\frac{E^2}{m^2c^2}-\left(1-\frac{2GM}{c^2r}\right)\left(c^2+\frac{L^2}{m^2r^2}\right)$$
So, $\frac{dr}{d\tau}$ is:
$$\frac{dr}{d\tau}=\pm\sqrt{\frac{E^2}{m^2c^2}-\left(1-\frac{2GM}{c^2r}\right)\left(c^2+\frac{L^2}{m^2r^2}\right)}$$
So, which sign should I use? Is there a method to find which sign is to be used? Perhaps, if $|d\phi|$ is increasing, so $r$ is decreasing and then $dr$ is negative, and vice versa. But I don't know how much this is true, especially near the event horizon.
 A: To determine a geodesic you have to fix its initial point and its initial tangent vector everything at $t=0$. We can always assume $\theta_0 = \pi/2$, whereas  $\phi_0$, and $r_0$ are arbitrary. Surely $d\theta/dt|_0=0$, $dt/dt|_0=1$, while $d\phi/dt|_0= \dot{\phi}_0$ and $dr/dt|_{0}= \dot{r}_0$ are arbitrary.
Using these values you determine the values of $L^2$ and $E^2$ that, as they being constant, they can determined with data at $t=0$. So $E^2$ and $L^2$ are known.
Therefore, up to the sign, the right-hand side of 
$$\frac{dr}{d\tau}=\pm\sqrt{\frac{E^2}{m^2c^2}-\left(1-\frac{2GM}{c^2r}\right)\left(c^2+\frac{L^2}{m^2r^2}\right)}\qquad (1)$$
is already known once you fix the initial conditions.
Unless $\dot{r}_0=0$, that number has a sign. This is the sign you have to choose in (1), since, by continuity (and the considered functions are $C^2$ at least) the sign does not change around $t=0$.
If you assume $\dot{r}_0=0$, the sign is determined by the other initial conditions with a more careful analysis. 
