# Schwarzchild Metric ODE for θ solution

Consider the schwarzchild metric and one of its geodesic differential equation $$r^2\sin(\theta)\cos(\theta)\phi'^2-r^2 \theta''=0$$ where $$f'\equiv \frac{d}{d \tau}:\forall f \in \mathbb{R}$$. How would I go about solving this ODE for $$\theta''$$since it includes both $$\phi'$$ and $$\theta''$$? Will one part of the equation always go to zero? Thats the case for the geodesic on a sphere. Note that is 1 equation for a set of 4 ODE's

• For one, it seems to simplify to: $\sin(\theta)\cos(\theta)\phi'^2-\theta''=0$
– Gert
Nov 4 '21 at 1:22
• yes, but how would I start to solve will the $\phi'$go to zero?
– aygx
Nov 4 '21 at 2:01
• I see. There's not enough information to solve this. You need a second ODE, at least. But I think my simplification holds, I think, for $r\neq 0$
– Gert
Nov 4 '21 at 6:53
• They are four simultaneous equations, and can be solved using RK4. You need to turn them into eight first-order equations first though. This is a standard procedure for second order ODEs. Nov 4 '21 at 11:06