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

  • $\begingroup$ For one, it seems to simplify to: $\sin(\theta)\cos(\theta)\phi'^2-\theta''=0$ $\endgroup$
    – Gert
    Nov 4 '21 at 1:22
  • $\begingroup$ yes, but how would I start to solve will the $\phi'$go to zero? $\endgroup$
    – aygx
    Nov 4 '21 at 2:01
  • $\begingroup$ 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 $r\neq 0$ $\endgroup$
    – Gert
    Nov 4 '21 at 6:53
  • $\begingroup$ 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. $\endgroup$
    – m4r35n357
    Nov 4 '21 at 11:06

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