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Radial Equation for Two-Body Central Forces

  1. Oct 30, 2009 #1
    I'm getting two different radial equations depending on when I plug in the angular momentum piece. Here's the Lagrangian:

    [tex]L = \frac{1}{2} \mu (\dot{r}^2 + r^2 \dot{\phi}^2) - U(r)[/tex]

    The Euler-Lagrange equation for phi gives angular momentum (conserved), which can be solved for [tex]\dot{\phi}[/tex]:

    [tex] \dot{\phi} = \frac{l}{\mu r^2} [/tex]

    Now, let's find the radial equation (that is, the Euler-Lagrange equation for r):

    Method 1: Substitute angular momentum piece into Lagrangian, then find the Euler-Lagrange equation for r.

    [tex]L = \frac{1}{2} \mu \dot{r}^2 + \frac{l^2}{2 \mu r^2} - U(r)[/tex]

    [tex] \mu \ddot{r} = \frac{-l^2}{\mu r^3} - \frac{dU}{dr} [/tex]

    Method 2: Find the Euler-Lagrange equation for r, then substitute angular momentum piece into the radial equation.

    [tex] \mu \ddot{r} = \mu r \dot{\phi}^2 - \frac{dU}{dr} [/tex]

    [tex] \mu \ddot{r} = \frac{l^2}{\mu r^3} - \frac{dU}{dr} [/tex]

    These two radial equations have opposite signs for the "centrifugal term." Which is correct, and why is the other wrong?
     
  2. jcsd
  3. Nov 1, 2009 #2
    Remember that when you solved for [tex] \dot{\phi} [/tex], you held r constant. Similarly, [tex] \phi [/tex] should be held constant when you solve for the radial equation.
     
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