The problem is to calculate the precession of Mercury using a naive approach:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]{{d(\gamma m \vec v)} \over {dt}} = {-GmM \over {r^2}}[/tex]

So apparently a Lagrangian of the form

[tex] - mc^2 \sqrt {c^2 - \dot r^2 - r^2 \dot \theta ^2 } - {{Gmm} \over r}[/tex]

gives us a Lagrangian that obeys the previous line when using a force that can be made as the -gradient of a potential. But nevermind, the real problem im facing is finding r(t) and [tex]\theta(t)[/tex].

Since energy is conserved, we find the following relations

[tex]{{\partial L} \over {\partial \dot \theta}} = {{mc^2r^2\dot \theta} \over {\sqrt{c^2 - \dot r^2 - r^2 \dot \theta^2}}} = \alpha[/tex]

[tex]{{\partial L} \over {\partial \dot r}} = {{mc^2 \dot r} \over {\sqrt{c^2 - \dot r^2 - r^2 \dot \theta^2}}} = {{\alpha \dot r} \over {r^2 \dot \theta}}[/tex]

At this point, using the fact that [tex]\sum {\dot q_i {{\partial L} \over {\partial \dot q_i }}} - L = \beta[/tex] where [tex]\beta[/tex] is a constant in this case (the energy), we arrive at this simplified equation

[tex]\alpha \dot \theta + {{\alpha \dot r^2} \over {r^2 \dot \theta}} + {{m^2 c^4 r^2 \dot \theta} \over {\alpha}} + {GmM \over r} = \beta[/tex]

From here I have no idea where to go. I don't see any obvious way to decouple the time derivatives of the thetas and r's. Any guesses on where to go from here?

*EDIT* And yes, I do realize I messed up the Lagrangian slight at the start. It should read

[tex] L = -mc {\sqrt {c^2 - \dot r^2 - r^2 \dot \theta ^2}}[/tex]

but it shouldn't effect the actual problem im having here (Set c = 1 and i win :D)

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# Homework Help: Mercury precession with naive GR

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