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Suppose that you want to solve the geodesic equation for the Schwarzschild metric. The geodesic equation is $$\frac{d^2 x^{\mu}}{d \tau^2}+\Gamma^{\mu}_{\rho\sigma}\frac{d x^{\rho}}{d \tau}\frac{d x^{\sigma}}{d \tau}=0.$$ What would be the quickest wat to solve this differential equation?

All the videos and research I have done just shown the derivation of the geodesic equation using the Euler-Lagrange equation from calculus of variations.

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  • $\begingroup$ Quickest in what sense? $\endgroup$
    – TimRias
    Oct 30, 2021 at 20:41
  • $\begingroup$ Quickest in the sense of least tedious and quickest in the overall work I have to do. $\endgroup$
    – aygx
    Oct 30, 2021 at 21:08
  • $\begingroup$ The Euler method is as easy and quick as it gets, bit it isn't very good ;) en.wikipedia.org/wiki/Euler_method (code included) $\endgroup$
    – m4r35n357
    Oct 30, 2021 at 21:12

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I usually find the Lagrangian approach faster because it is easier to spot conserved quantities with it and that makes solving things easier. For the Schwarzschild spacetime if you calculate the geodesic equation directly from the Christoffel symbols then you get: $$ \begin{array}{c} 0=\frac{2 M \dot r \dot t}{r (r-2 M)}+\ddot t \\ 0=\frac{M \dot r^2}{2 M r-r^2}+(2 M-r) \left(\dot \theta^2+\sin ^2(\theta ) \dot \phi^2\right)+\frac{M (r-2 M) \dot t^2}{r^3}+\ddot r \\ 0=\ddot \theta+\sin (\theta ) (-\cos (\theta )) \dot \phi^2+\frac{2 \ \dot \theta \dot r}{r} \\ 0=\ddot \phi+\frac{2 \dot \phi \left(\dot r+r \dot \theta \cot (\theta \ )\right)}{r} \\ \end{array} $$

If you calculate using the Lagrangian you get $$\begin{array}{c} 0= -\frac{\partial }{\partial \lambda }\left(\left(\frac{2 M}{r}-1 \right) \dot t \right) \\ 0= -\frac{\partial }{\partial \lambda }\frac{\dot r}{1-\frac{2 \ M}{r}}-\frac{M \dot r^2}{(r-2 M)^2}-\frac{M \dot t^2}{r^2}+r \ \left(\dot \theta^2+\sin ^2(\theta ) \dot \phi^2\right) \\ 0= r^2 \sin (\theta ) \cos (\theta ) \dot \phi^2-\frac{\partial \ \left(r^2 \dot \theta\right)}{\partial \lambda } \\ 0= -\frac{\partial \left(r^2 \sin ^2(\theta ) \dot \phi\right)}{\partial \lambda } \\ \end{array}$$

For me, it is better to immediately see the conserved energy and angular momentum in the Lagrangian equations than the ones based on the Christoffel symbols.

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  • $\begingroup$ Thanks, isn't there a method using using killing vectors? $\endgroup$
    – aygx
    Nov 1, 2021 at 19:39

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