# 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.

• I don't think there is a global answer. It's like ask, for the harmonic oscillator, what is the sign of $\dfrac{dx}{dt}$, if $(\dfrac{dx}{dt})^2 = \dfrac{2}{m}(E - \frac{1}{2}m \omega^2 x^2)$. It depends only on the initial conditions. – Trimok Jan 16 '14 at 18:48
• How does it depend on the initial conditions? If I know them, it is possible to answer the question, isn't it? – Ale Jan 16 '14 at 19:03
• If you know the initial position and the initial (vector) speed (compatible with $E$ and $L$), and because you know the equation of the movement, you should be able to answer the question. You may consider (if necessary) different "phases" of the movement which beginning/end correspond to $\frac{dr}{dt}=0$ (so you have a change of sign from one "phase" to the other). – Trimok Jan 16 '14 at 19:14

## 1 Answer

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.

• Does the formula say that a particle can't change its direction from outwards to inwards (with respect to a black hole) or vice versa? should it be $\left|\frac{dr}{dt}\right|$? What about e.g. stable orbits around black holes? – Otto Nov 3 '15 at 13:50