Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Mercury precession with naive GR

  1. Sep 8, 2010 #1


    User Avatar
    Gold Member

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

    [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)
    Last edited: Sep 8, 2010
  2. jcsd
  3. Sep 8, 2010 #2


    User Avatar
    Homework Helper

    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 [itex]\dot\theta[/itex] in terms of α and use that to eliminate [itex]\dot\theta[/itex] from your equations. At least, that's the way it's normally done in the nonrelativistic case (so, no guarantees it'll work here).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook