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P: 665
 Quote by Anthony I think the OP might be making life difficult for himself. With any form of variational problem, you should try to use Noether's theorem to integrate your equations up, rather than try to go from the Euler-Lagrange equations themselves. This will invariably make your life easier. Your Lagrangian density reads: $$\mathcal{L} = \left( 1- \frac{1}{r}\right) \left( \frac{\mathrm{d} t}{\mathrm{d} \lambda}\right)^2 - \left( 1- \frac{1}{r}\right)^{-1} \left( \frac{\mathrm{d} r}{\mathrm{d} \lambda}\right)^2 - r^2 \left( \frac{\mathrm{d} \theta}{\mathrm{d} \lambda}\right)^2 - r^2 \sin^2\theta \left( \frac{\mathrm{d} \phi}{\mathrm{d} \lambda}\right)^2$$ using natural units. It is clear that the vector fields: $$\frac{\partial}{\partial t} \quad \textrm{and}\quad \frac{\partial}{\partial \phi}$$ are Killing, so Noether's theorem integrates up two of the Euler-Lagrange equations for you and gives you two constants of motion: $$\left(1-\frac{1}{r}\right) \frac{\mathrm{d} t}{\mathrm{d}\lambda} = \mathrm{const} \,(=E) \quad \textrm{and} \quad r^2 \sin^2\theta \frac{\mathrm{d} \phi}{\mathrm{d} \lambda} = \mathrm{const} \,(=h) \qquad (*)$$ i.e. on a given geodesic, these quantities remain unchanged. Similarly, since $$\partial_\lambda \mathcal{L}=0$$, we know $$\mathcal{L}$$ remains constant, and we set it to {+1,-1,0} depending on whether you're interested in timelike, spacelike or nulll geodesics. Call this constant k. Note that all our ODEs are now 1st order. Setting $$\theta = \pi/2$$ (validity can be deduced from the $$\theta$$ E-L equation) and using (*) in $$\mathcal{L}=k$$ gives the ODE: $$\left( \frac{\mathrm{d} r}{\mathrm{d}\lambda}\right)^2 = E^2 - \left( 1-\frac{1}{r}\right) \left(k + \frac{h^2}{r^2}\right)$$ If you'd prefer to parameterise your geodesics using $$\phi$$, use the second of the constraints in (*) again and you get: $$\frac{h^2}{r^4} \left( \frac{\mathrm{d} r}{\mathrm{d} \phi}\right)^2 = E^2 - \left( 1-\frac{1}{r}\right) \left( k + \frac{h^2}{r^2}\right)$$ If you'd prefer to do all this using tensors, just apply Noether's theorem in the form: if $$L_V g=0$$ (i.e. $$V$$ is a Killing vector) then $$V^\mu \dot{x}^\nu g_{\mu\nu} = \mathrm{const}$$.
Exactly! This simply is based on the proposition given in post #389 and by this perfect explanation we are done here.

AB