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.
<br /> V^2(r) = \left(1 - \frac{2M}{r} \right)\frac{L^2}{r^2}<br />
<br /> \frac{d \phi}{d \lambda} &amp;= \frac{L}{r^2} <br />
<br /> \frac{dr}{d\lambda} &amp;= \pm\sqrt{E^2 - V^2(r)}<br />

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} &amp;= \frac{L^2(r - 3M)}{r^4}