How to Determine the Sign of the Square Root in ODEs for Photon Orbits?

[nocks]

hi there,
I'm trying to plot r against $$\phi$$ by solving the following ODEs using runge-kutta. The problem I'm having is with the square root. How do I know when it will be positive and when it will be negative? If this is a simple question I apologise I'm not that great with the maths :).

E and L and M are constants.
r>0.
$$V^2(r) = \left(1 - \frac{2M}{r} \right)\frac{L^2}{r^2}$$
$$\frac{d \phi}{d \lambda} &= \frac{L}{r^2}$$
$$\frac{dr}{d\lambda} &= \pm\sqrt{E^2 - V^2(r)}$$

Thanks

 

[matematikawan]

$\pm$ sign shouldn't be there in the first place. You have to decide whether it is positive or negative sign. Just look back at your model again.

 

[nocks]

$\pm$ sign shouldn't be there in the first place. You have to decide whether it is positive or negative sign. Just look back at your model again.

The model is of a photon's path around a black hole. The $$\pm$$ is due to the fact that the photons distance from the black hole can increase and decrease in the same orbit.

I've been advised to differentiate dr/d$$\lambda$$ to get around the $$\pm$$ but I am unsure as to how this will solve my problem.

I have the $$d^2r/d \lambda ^2$$ as $$\frac{d^2r}{d\lambda^2} &= \frac{L^2(r - 3M)}{r^4}$$