# Homework Help: Mercury precession with naive GR

1. Sep 8, 2010

### Pengwuino

The problem is to calculate the precession of Mercury using a naive approach:

$${{d(\gamma m \vec v)} \over {dt}} = {-GmM \over {r^2}}$$

So apparently a Lagrangian of the form

$$- mc^2 \sqrt {c^2 - \dot r^2 - r^2 \dot \theta ^2 } - {{Gmm} \over r}$$

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 $$\theta(t)$$.

Since energy is conserved, we find the following relations

$${{\partial L} \over {\partial \dot \theta}} = {{mc^2r^2\dot \theta} \over {\sqrt{c^2 - \dot r^2 - r^2 \dot \theta^2}}} = \alpha$$
$${{\partial L} \over {\partial \dot r}} = {{mc^2 \dot r} \over {\sqrt{c^2 - \dot r^2 - r^2 \dot \theta^2}}} = {{\alpha \dot r} \over {r^2 \dot \theta}}$$

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

$$\alpha \dot \theta + {{\alpha \dot r^2} \over {r^2 \dot \theta}} + {{m^2 c^4 r^2 \dot \theta} \over {\alpha}} + {GmM \over r} = \beta$$

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

$$L = -mc {\sqrt {c^2 - \dot r^2 - r^2 \dot \theta ^2}}$$

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

Last edited: Sep 8, 2010
2. Sep 8, 2010

### diazona

My guess would be to use the fact that the Lagrangian is independent of θ. That means the conjugate momentum (α) is conserved, so you can solve for $\dot\theta$ in terms of α and use that to eliminate $\dot\theta$ from your equations. At least, that's the way it's normally done in the nonrelativistic case (so, no guarantees it'll work here).