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Equations of motion from Lagrangian and metric

  1. Mar 29, 2012 #1
    Disregard. I done figured it out.


    1. The problem statement, all variables and given/known data
    Find equations of motion for the metric:

    [tex]ds^2 = dr^2 + r^2 d\phi^2[/tex]


    2. Relevant equations

    [tex]L = g_{ab} \dot{x}^a \dot{x}^b[/tex]

    [tex]\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x}^a} \right) = \frac{\partial L}{\partial x^a}[/tex]

    3. The attempt at a solution

    So I know to just cycle the free index in the lagrangian from r to phi to get the corresponding equations of motion, but my phi solution isn't coming out right. For r, I get

    [tex]L = \dot{r}^2 + r^2 \dot{\phi}^2[/tex]

    which gives

    [tex]\ddot{r} - r\dot{\phi}^2 = 0.[/tex]

    For phi, I get:

    [tex]\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\phi}} \right) = \frac{\partial L}{\partial \phi}[/tex]


    which yields:

    [tex] \frac{d}{dt}\left( 2r^2 \dot{\phi} \right) = 0[/tex]

    The solution lists the other equation of motion as:

    [tex]\ddot{\phi} + 2 \frac{\dot{r}}{r}\dot{\phi} = 0[/tex]

    I'm having trouble bridging the last bit. If I apply the time differential operator on the dotted phi, I can see where the double dotted phi comes from, but does that mean I have to do some kind of wacky product rule business on the r as well?


    edit: got it, nevermind. For those who google this years from now and hate how people always say they got it without showing how they got it, here's the step I was too tired to see:


    [tex] \frac{d}{dt}\left( 2r^2 \dot{\phi} \right) = 0[/tex]

    apply product rule

    [tex] 2r^2 \ddot{\phi} + 4 r \dot{r} \dot{\phi} = 0[/tex]

    divide by [tex]2r^2[/tex]

    gives

    [tex]\ddot{\phi} + 2 \frac{\dot{r}}{r} \dot{\phi} = 0 [/tex]

    Just goes to show if you stare at it long enough it can come to you. Plus making lengthy tex-based posts on PP can help flesh out whatever misunderstandings you may have!
     
    Last edited: Mar 29, 2012
  2. jcsd
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